on the nature of undecidability within computing
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/ d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit?usp=sharing
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computing
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit?usp=sharing
The section 1 "the halting problem" should clearly define what the
expression "halting problem" means. It might be useful to first
define "halting question" as that would simplify the definitions of
"halting problem" and "halting decider". In each of these it is
also inportant to be clear what is the scope of the problem and
what kind of solutions are allowed. For example, the halting
problem of finite state machines is Turing decidable and the halting
problem of Turing machines is decidable with a simple oracle.
One should also clearly state that the halting of the program "und"
is not undecidable: there are partial halt deciders that can tell
whether "und" halts. In order to make this clear as soon as possible
it might be better to define "partial halt decider" before the
introduction of "und".
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computing
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit?usp=sharing
The section 1 "the halting problem" should clearly define what the
expression "halting problem" means. It might be useful to first
define "halting question" as that would simplify the definitions of
"halting problem" and "halting decider". In each of these it is
also inportant to be clear what is the scope of the problem and
what kind of solutions are allowed. For example, the halting
problem of finite state machines is Turing decidable and the halting
problem of Turing machines is decidable with a simple oracle.
One should also clearly state that the halting of the program "und"
is not undecidable: there are partial halt deciders that can tell
whether "und" halts. In order to make this clear as soon as possible
it might be better to define "partial halt decider" before the
introduction of "und".
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computing
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit?usp=sharing
The section 1 "the halting problem" should clearly define what the
expression "halting problem" means. It might be useful to first
define "halting question" as that would simplify the definitions of
"halting problem" and "halting decider". In each of these it is
also inportant to be clear what is the scope of the problem and
what kind of solutions are allowed. For example, the halting
problem of finite state machines is Turing decidable and the halting
problem of Turing machines is decidable with a simple oracle.
One should also clearly state that the halting of the program "und"
is not undecidable: there are partial halt deciders that can tell
whether "und" halts. In order to make this clear as soon as possible
it might be better to define "partial halt decider" before the
introduction of "und".
could u read the whole paper please?
On 9/12/2026 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:You came here for 5th grade math?
on the nature of undecidability within computing
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit?usp=sharing
The section 1 "the halting problem" should clearly define what the
expression "halting problem" means. It might be useful to first
define "halting question" as that would simplify the definitions of
"halting problem" and "halting decider". In each of these it is
also inportant to be clear what is the scope of the problem and
what kind of solutions are allowed. For example, the halting
problem of finite state machines is Turing decidable and the halting
problem of Turing machines is decidable with a simple oracle.
One should also clearly state that the halting of the program "und"
is not undecidable: there are partial halt deciders that can tell
whether "und" halts. In order to make this clear as soon as possible
it might be better to define "partial halt decider" before the
introduction of "und".
On 12/09/2026 19:02, dart200 wrote:
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computing
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit?usp=sharing
The section 1 "the halting problem" should clearly define what the
expression "halting problem" means. It might be useful to first
define "halting question" as that would simplify the definitions of
"halting problem" and "halting decider". In each of these it is
also inportant to be clear what is the scope of the problem and
what kind of solutions are allowed. For example, the halting
problem of finite state machines is Turing decidable and the halting
problem of Turing machines is decidable with a simple oracle.
One should also clearly state that the halting of the program "und"
is not undecidable: there are partial halt deciders that can tell
whether "und" halts. In order to make this clear as soon as possible
it might be better to define "partial halt decider" before the
introduction of "und".
could u read the whole paper please?
Irrelevant. The section 1 should be readable before reading the rest
of the paper.
The sections 7 and later are even more irrelevant.
On 9/13/26 2:41 AM, Mikko wrote:
On 12/09/2026 19:02, dart200 wrote:
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computing
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit?usp=sharing
The section 1 "the halting problem" should clearly define what the
expression "halting problem" means. It might be useful to first
define "halting question" as that would simplify the definitions of
"halting problem" and "halting decider". In each of these it is
also inportant to be clear what is the scope of the problem and
what kind of solutions are allowed. For example, the halting
problem of finite state machines is Turing decidable and the halting
problem of Turing machines is decidable with a simple oracle.
"simple" is a weird term for something we provably can't build, and that doesn't make the haltinging problem "decidable" since we can't build, compute, or actually decide with the oracle
recursive undecidability is discussed in -o4
One should also clearly state that the halting of the program "und"
is not undecidable: there are partial halt deciders that can tell
whether "und" halts. In order to make this clear as soon as possible
it might be better to define "partial halt decider" before the
introduction of "und".
other classifiers are discussed in -o5.2
could u read the whole paper please?
Irrelevant. The section 1 should be readable before reading the rest
of the paper.
the core of the halting problem is the fact that no total decider is possible with turing machines,
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:
On 12/09/2026 19:02, dart200 wrote:
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computing
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit?usp=sharing
The section 1 "the halting problem" should clearly define what the
expression "halting problem" means. It might be useful to first
define "halting question" as that would simplify the definitions of
"halting problem" and "halting decider". In each of these it is
also inportant to be clear what is the scope of the problem and
what kind of solutions are allowed. For example, the halting
problem of finite state machines is Turing decidable and the halting >>>>> problem of Turing machines is decidable with a simple oracle.
"simple" is a weird term for something we provably can't build, and
that doesn't make the haltinging problem "decidable" since we can't
build, compute, or actually decide with the oracle
A "simple oracle" is an oracle that is simpler than most other oracles,
just like a "big mouse" is a mouse that is bigger than most other mice although not as big as a little elephant.
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the program "und"
is not undecidable: there are partial halt deciders that can tell
whether "und" halts. In order to make this clear as soon as possible >>>>> it might be better to define "partial halt decider" before the
introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
could u read the whole paper please?
Irrelevant. The section 1 should be readable before reading the rest
of the paper.
the core of the halting problem is the fact that no total decider is
possible with turing machines,
No, it is not even a peripheral part of the halting problem. The usual halting problem is about Turing machines but a similar problem can be
posed about other classes of machines. Although usually not explicitly stated, by convention a proof that there is no decider that decides
halting is usually said to solve the problem.
On 9/14/26 1:26 AM, Mikko wrote:
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:
On 12/09/2026 19:02, dart200 wrote:
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computing
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit?usp=sharing
The section 1 "the halting problem" should clearly define what the >>>>>> expression "halting problem" means. It might be useful to first
define "halting question" as that would simplify the definitions of >>>>>> "halting problem" and "halting decider". In each of these it is
also inportant to be clear what is the scope of the problem and
what kind of solutions are allowed. For example, the halting
problem of finite state machines is Turing decidable and the halting >>>>>> problem of Turing machines is decidable with a simple oracle.
"simple" is a weird term for something we provably can't build, and
that doesn't make the haltinging problem "decidable" since we can't
build, compute, or actually decide with the oracle
A "simple oracle" is an oracle that is simpler than most other oracles,
just like a "big mouse" is a mouse that is bigger than most other mice
although not as big as a little elephant.
like i said: recursive undecidability leading to the concept of "simple"
vs "complex" oracles is discussing in -o4, i will not be discussing them
in -o1
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the program "und" >>>>>> is not undecidable: there are partial halt deciders that can tell
whether "und" halts. In order to make this clear as soon as possible >>>>>> it might be better to define "partial halt decider" before the
introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
that's why partial deciders are discussed in -o5.2 not -o1, you brought those up not me
could u read the whole paper please?>>>>Irrelevant. The section 1 should be readable before reading the rest
of the paper.
the core of the halting problem is the fact that no total decider is
possible with turing machines,
No, it is not even a peripheral part of the halting problem. The usual
halting problem is about Turing machines but a similar problem can be
posed about other classes of machines. Although usually not explicitly
stated, by convention a proof that there is no decider that decides
halting is usually said to solve the problem.
i love you start the paragraph by disagreeing with me and then end with restating my claim.
am i supposed to take this level of critique serious? what do u expect
me to learn from this???
On 14/09/2026 11:45, dart200 wrote:
On 9/14/26 1:26 AM, Mikko wrote:
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:
On 12/09/2026 19:02, dart200 wrote:
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computingThe section 1 "the halting problem" should clearly define what the >>>>>>> expression "halting problem" means. It might be useful to first
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit?usp=sharing >>>>>>>
define "halting question" as that would simplify the definitions of >>>>>>> "halting problem" and "halting decider". In each of these it is
also inportant to be clear what is the scope of the problem and
what kind of solutions are allowed. For example, the halting
problem of finite state machines is Turing decidable and the halting >>>>>>> problem of Turing machines is decidable with a simple oracle.
"simple" is a weird term for something we provably can't build, and
that doesn't make the haltinging problem "decidable" since we can't
build, compute, or actually decide with the oracle
A "simple oracle" is an oracle that is simpler than most other oracles,
just like a "big mouse" is a mouse that is bigger than most other mice
although not as big as a little elephant.
like i said: recursive undecidability leading to the concept of
"simple" vs "complex" oracles is discussing in -o4, i will not be
discussing them in -o1
That's OK, although there should be a short description of each section
at the end of the introduction.
But the section 1 gives the impression that the author does not know
what the halting problem is.
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the program "und" >>>>>>> is not undecidable: there are partial halt deciders that can tell >>>>>>> whether "und" halts. In order to make this clear as soon as possible >>>>>>> it might be better to define "partial halt decider" before the
introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
that's why partial deciders are discussed in -o5.2 not -o1, you brought
those up not me
The section 1 is about the halting problem. You may discuss partial and
total solutions of the problem in the same section or elsewhere. But
it is important to point out about "und" that its halting is
determinable. Otherwise a reader could be confused or get the impression
that you are don't know.
could u read the whole paper please?>>>>Irrelevant. The section 1 should be readable before reading the rest >>>>> of the paper.
the core of the halting problem is the fact that no total decider is
possible with turing machines,
No, it is not even a peripheral part of the halting problem. The usual
halting problem is about Turing machines but a similar problem can be
posed about other classes of machines. Although usually not explicitly
stated, by convention a proof that there is no decider that decides
halting is usually said to solve the problem.
i love you start the paragraph by disagreeing with me and then end
with restating my claim.
You must keep apart the commoents about your error and comments
about your presentation.
am i supposed to take this level of critique serious? what do u expect
me to learn from this???
That depends on whether you intend to write a serious article.
On 9/15/26 1:18 AM, Mikko wrote:
On 14/09/2026 11:45, dart200 wrote:
On 9/14/26 1:26 AM, Mikko wrote:
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:
On 12/09/2026 19:02, dart200 wrote:
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computingThe section 1 "the halting problem" should clearly define what the >>>>>>>> expression "halting problem" means. It might be useful to first >>>>>>>> define "halting question" as that would simplify the definitions of >>>>>>>> "halting problem" and "halting decider". In each of these it is >>>>>>>> also inportant to be clear what is the scope of the problem and >>>>>>>> what kind of solutions are allowed. For example, the halting
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit?usp=sharing >>>>>>>>
problem of finite state machines is Turing decidable and the
halting
problem of Turing machines is decidable with a simple oracle.
"simple" is a weird term for something we provably can't build, and >>>>> that doesn't make the haltinging problem "decidable" since we can't >>>>> build, compute, or actually decide with the oracle
A "simple oracle" is an oracle that is simpler than most other oracles, >>>> just like a "big mouse" is a mouse that is bigger than most other mice >>>> although not as big as a little elephant.
like i said: recursive undecidability leading to the concept of
"simple" vs "complex" oracles is discussing in -o4, i will not be
discussing them in -o1
That's OK, although there should be a short description of each section
at the end of the introduction.
But the section 1 gives the impression that the author does not know
what the halting problem is.
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the program "und" >>>>>>>> is not undecidable: there are partial halt deciders that can tell >>>>>>>> whether "und" halts. In order to make this clear as soon as
possible
it might be better to define "partial halt decider" before the >>>>>>>> introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
that's why partial deciders are discussed in -o5.2 not -o1, you brought >>> those up not me
The section 1 is about the halting problem. You may discuss partial and
total solutions of the problem in the same section or elsewhere. But
it is important to point out about "und" that its halting is
determinable. Otherwise a reader could be confused or get the impression
that you are don't know.
und as specified in -o1 does not even exist, an argument which takes up
to the end of -o4 to explain
On 15/09/2026 19:34, dart200 wrote:
On 9/15/26 1:18 AM, Mikko wrote:
On 14/09/2026 11:45, dart200 wrote:
On 9/14/26 1:26 AM, Mikko wrote:
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:
On 12/09/2026 19:02, dart200 wrote:
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computingThe section 1 "the halting problem" should clearly define what the >>>>>>>>> expression "halting problem" means. It might be useful to first >>>>>>>>> define "halting question" as that would simplify the
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit?usp=sharing >>>>>>>>>
definitions of
"halting problem" and "halting decider". In each of these it is >>>>>>>>> also inportant to be clear what is the scope of the problem and >>>>>>>>> what kind of solutions are allowed. For example, the halting >>>>>>>>> problem of finite state machines is Turing decidable and the >>>>>>>>> halting
problem of Turing machines is decidable with a simple oracle.
"simple" is a weird term for something we provably can't build,
and that doesn't make the haltinging problem "decidable" since we >>>>>> can't build, compute, or actually decide with the oracle
A "simple oracle" is an oracle that is simpler than most other
oracles,
just like a "big mouse" is a mouse that is bigger than most other mice >>>>> although not as big as a little elephant.
like i said: recursive undecidability leading to the concept of
"simple" vs "complex" oracles is discussing in -o4, i will not be
discussing them in -o1
That's OK, although there should be a short description of each section
at the end of the introduction.
But the section 1 gives the impression that the author does not know
what the halting problem is.
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the program >>>>>>>>> "und"
is not undecidable: there are partial halt deciders that can tell >>>>>>>>> whether "und" halts. In order to make this clear as soon as >>>>>>>>> possible
it might be better to define "partial halt decider" before the >>>>>>>>> introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
that's why partial deciders are discussed in -o5.2 not -o1, you
brought those up not me
The section 1 is about the halting problem. You may discuss partial and
total solutions of the problem in the same section or elsewhere. But
it is important to point out about "und" that its halting is
determinable. Otherwise a reader could be confused or get the impression >>> that you are don't know.
und as specified in -o1 does not even exist, an argument which takes up
to the end of -o4 to explain
The section 1 says otherwise: the words "can be constructed" mean that
it does exist.
On 15/09/2026 19:34, dart200 wrote:
On 9/15/26 1:18 AM, Mikko wrote:
On 14/09/2026 11:45, dart200 wrote:
On 9/14/26 1:26 AM, Mikko wrote:
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:
On 12/09/2026 19:02, dart200 wrote:
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computingThe section 1 "the halting problem" should clearly define what the >>>>>>>>> expression "halting problem" means. It might be useful to first >>>>>>>>> define "halting question" as that would simplify the
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit?usp=sharing >>>>>>>>>
definitions of
"halting problem" and "halting decider". In each of these it is >>>>>>>>> also inportant to be clear what is the scope of the problem and >>>>>>>>> what kind of solutions are allowed. For example, the halting >>>>>>>>> problem of finite state machines is Turing decidable and the >>>>>>>>> halting
problem of Turing machines is decidable with a simple oracle.
"simple" is a weird term for something we provably can't build,
and that doesn't make the haltinging problem "decidable" since we >>>>>> can't build, compute, or actually decide with the oracle
A "simple oracle" is an oracle that is simpler than most other
oracles,
just like a "big mouse" is a mouse that is bigger than most other mice >>>>> although not as big as a little elephant.
like i said: recursive undecidability leading to the concept of
"simple" vs "complex" oracles is discussing in -o4, i will not be
discussing them in -o1
That's OK, although there should be a short description of each section
at the end of the introduction.
But the section 1 gives the impression that the author does not know
what the halting problem is.
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the program >>>>>>>>> "und"
is not undecidable: there are partial halt deciders that can tell >>>>>>>>> whether "und" halts. In order to make this clear as soon as >>>>>>>>> possible
it might be better to define "partial halt decider" before the >>>>>>>>> introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
that's why partial deciders are discussed in -o5.2 not -o1, you
brought those up not me
The section 1 is about the halting problem. You may discuss partial and
total solutions of the problem in the same section or elsewhere. But
it is important to point out about "und" that its halting is
determinable. Otherwise a reader could be confused or get the impression >>> that you are don't know.
und as specified in -o1 does not even exist, an argument which takes up
to the end of -o4 to explain
The section 1 says otherwise: the words "can be constructed" mean that
it does exist.
On 9/16/26 12:40 AM, Mikko wrote:
On 15/09/2026 19:34, dart200 wrote:
On 9/15/26 1:18 AM, Mikko wrote:
On 14/09/2026 11:45, dart200 wrote:
On 9/14/26 1:26 AM, Mikko wrote:
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:
On 12/09/2026 19:02, dart200 wrote:"simple" is a weird term for something we provably can't build, >>>>>>> and that doesn't make the haltinging problem "decidable" since we >>>>>>> can't build, compute, or actually decide with the oracle
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computingThe section 1 "the halting problem" should clearly define what >>>>>>>>>> the
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit?usp=sharing >>>>>>>>>>
expression "halting problem" means. It might be useful to first >>>>>>>>>> define "halting question" as that would simplify the
definitions of
"halting problem" and "halting decider". In each of these it is >>>>>>>>>> also inportant to be clear what is the scope of the problem and >>>>>>>>>> what kind of solutions are allowed. For example, the halting >>>>>>>>>> problem of finite state machines is Turing decidable and the >>>>>>>>>> halting
problem of Turing machines is decidable with a simple oracle. >>>>>>>
A "simple oracle" is an oracle that is simpler than most other
oracles,
just like a "big mouse" is a mouse that is bigger than most other >>>>>> mice
although not as big as a little elephant.
like i said: recursive undecidability leading to the concept of
"simple" vs "complex" oracles is discussing in -o4, i will not be
discussing them in -o1
That's OK, although there should be a short description of each section >>>> at the end of the introduction.
But the section 1 gives the impression that the author does not know
what the halting problem is.
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the program >>>>>>>>>> "und"
is not undecidable: there are partial halt deciders that can tell >>>>>>>>>> whether "und" halts. In order to make this clear as soon as >>>>>>>>>> possible
it might be better to define "partial halt decider" before the >>>>>>>>>> introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
that's why partial deciders are discussed in -o5.2 not -o1, you
brought those up not me
The section 1 is about the halting problem. You may discuss partial and >>>> total solutions of the problem in the same section or elsewhere. But
it is important to point out about "und" that its halting is
determinable. Otherwise a reader could be confused or get the
impression
that you are don't know.
und as specified in -o1 does not even exist, an argument which takes
up to the end of -o4 to explain
The section 1 says otherwise: the words "can be constructed" mean that
it does exist.
it exist as a hypothetical problem, constructed from a hypothetical
decider, that presents a real limit to turing machine computability, but does not exist in the actual enumeration of turing machines, and does
not limit our ability to decide on any given machine, as we are not an addressable computing machine.
that is the point of the entire paper, and it takes 26 pages to explain
why. i'm not changing that specific wording there based on confusion
that will not resolve until you actually read the paper dud
On 12/09/2026 20:32, Dude wrote:
On 9/12/2026 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:You came here for 5th grade math?
on the nature of undecidability within computing
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit?usp=sharing
The section 1 "the halting problem" should clearly define what the
expression "halting problem" means. It might be useful to first
define "halting question" as that would simplify the definitions of
"halting problem" and "halting decider". In each of these it is
also inportant to be clear what is the scope of the problem and
what kind of solutions are allowed. For example, the halting
problem of finite state machines is Turing decidable and the halting
problem of Turing machines is decidable with a simple oracle.
One should also clearly state that the halting of the program "und"
is not undecidable: there are partial halt deciders that can tell
whether "und" halts. In order to make this clear as soon as possible
it might be better to define "partial halt decider" before the
introduction of "und".
No, thank you.
it in sci.math.
On 9/16/2026 9:26 AM, dart200 wrote:
On 9/16/26 12:40 AM, Mikko wrote:
On 15/09/2026 19:34, dart200 wrote:
On 9/15/26 1:18 AM, Mikko wrote:
On 14/09/2026 11:45, dart200 wrote:
On 9/14/26 1:26 AM, Mikko wrote:
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:
On 12/09/2026 19:02, dart200 wrote:"simple" is a weird term for something we provably can't build, >>>>>>>> and that doesn't make the haltinging problem "decidable" since >>>>>>>> we can't build, compute, or actually decide with the oracle
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computingThe section 1 "the halting problem" should clearly define >>>>>>>>>>> what the
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit?usp=sharing >>>>>>>>>>>
expression "halting problem" means. It might be useful to first >>>>>>>>>>> define "halting question" as that would simplify the
definitions of
"halting problem" and "halting decider". In each of these it is >>>>>>>>>>> also inportant to be clear what is the scope of the problem and >>>>>>>>>>> what kind of solutions are allowed. For example, the halting >>>>>>>>>>> problem of finite state machines is Turing decidable and the >>>>>>>>>>> halting
problem of Turing machines is decidable with a simple oracle. >>>>>>>>
A "simple oracle" is an oracle that is simpler than most other
oracles,
just like a "big mouse" is a mouse that is bigger than most other >>>>>>> mice
although not as big as a little elephant.
like i said: recursive undecidability leading to the concept of
"simple" vs "complex" oracles is discussing in -o4, i will not be >>>>>> discussing them in -o1
That's OK, although there should be a short description of each
section
at the end of the introduction.
But the section 1 gives the impression that the author does not know >>>>> what the halting problem is.
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the program >>>>>>>>>>> "und"
is not undecidable: there are partial halt deciders that can >>>>>>>>>>> tell
whether "und" halts. In order to make this clear as soon as >>>>>>>>>>> possible
it might be better to define "partial halt decider" before the >>>>>>>>>>> introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
that's why partial deciders are discussed in -o5.2 not -o1, you
brought those up not me
The section 1 is about the halting problem. You may discuss partial >>>>> and
total solutions of the problem in the same section or elsewhere. But >>>>> it is important to point out about "und" that its halting is
determinable. Otherwise a reader could be confused or get the
impression
that you are don't know.
und as specified in -o1 does not even exist, an argument which takes
up to the end of -o4 to explain
The section 1 says otherwise: the words "can be constructed" mean that
it does exist.
it exist as a hypothetical problem, constructed from a hypothetical
decider, that presents a real limit to turing machine computability,
but does not exist in the actual enumeration of turing machines, and
does not limit our ability to decide on any given machine, as we are
not an addressable computing machine.
that is the point of the entire paper, and it takes 26 pages to
explain why. i'm not changing that specific wording there based on
confusion that will not resolve until you actually read the paper dud
Can you decide random numbers? Say a black box in isolation. You ask it
to run a process, when you wait for a signal that the process is done.
Well, do you get a signal or not? Say the black box takes some results
from a TRNG. Sometimes it halts after some random time. Sometimes it
does not halt. How does your system account for the black box?
On 9/16/26 1:54 PM, Chris M. Thomasson wrote:
On 9/16/2026 9:26 AM, dart200 wrote:
On 9/16/26 12:40 AM, Mikko wrote:
On 15/09/2026 19:34, dart200 wrote:
On 9/15/26 1:18 AM, Mikko wrote:
On 14/09/2026 11:45, dart200 wrote:
On 9/14/26 1:26 AM, Mikko wrote:
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:
On 12/09/2026 19:02, dart200 wrote:"simple" is a weird term for something we provably can't build, >>>>>>>>> and that doesn't make the haltinging problem "decidable" since >>>>>>>>> we can't build, compute, or actually decide with the oracle
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computing
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit? >>>>>>>>>>>>> usp=sharing
The section 1 "the halting problem" should clearly define >>>>>>>>>>>> what the
expression "halting problem" means. It might be useful to first >>>>>>>>>>>> define "halting question" as that would simplify the
definitions of
"halting problem" and "halting decider". In each of these it is >>>>>>>>>>>> also inportant to be clear what is the scope of the problem and >>>>>>>>>>>> what kind of solutions are allowed. For example, the halting >>>>>>>>>>>> problem of finite state machines is Turing decidable and the >>>>>>>>>>>> halting
problem of Turing machines is decidable with a simple oracle. >>>>>>>>>
A "simple oracle" is an oracle that is simpler than most other >>>>>>>> oracles,
just like a "big mouse" is a mouse that is bigger than most
other mice
although not as big as a little elephant.
like i said: recursive undecidability leading to the concept of >>>>>>> "simple" vs "complex" oracles is discussing in -o4, i will not be >>>>>>> discussing them in -o1
That's OK, although there should be a short description of each
section
at the end of the introduction.
But the section 1 gives the impression that the author does not know >>>>>> what the halting problem is.
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the >>>>>>>>>>>> program "und"
is not undecidable: there are partial halt deciders that can >>>>>>>>>>>> tell
whether "und" halts. In order to make this clear as soon as >>>>>>>>>>>> possible
it might be better to define "partial halt decider" before the >>>>>>>>>>>> introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
that's why partial deciders are discussed in -o5.2 not -o1, you >>>>>>> brought those up not me
The section 1 is about the halting problem. You may discuss
partial and
total solutions of the problem in the same section or elsewhere. But >>>>>> it is important to point out about "und" that its halting is
determinable. Otherwise a reader could be confused or get the
impression
that you are don't know.
und as specified in -o1 does not even exist, an argument which takes >>>>> up to the end of -o4 to explain
The section 1 says otherwise: the words "can be constructed" mean that >>>> it does exist.
it exist as a hypothetical problem, constructed from a hypothetical
decider, that presents a real limit to turing machine computability,
but does not exist in the actual enumeration of turing machines, and
does not limit our ability to decide on any given machine, as we are
not an addressable computing machine.
that is the point of the entire paper, and it takes 26 pages to
explain why. i'm not changing that specific wording there based on
confusion that will not resolve until you actually read the paper dud
Can you decide random numbers? Say a black box in isolation. You ask
it to run a process, when you wait for a signal that the process is
done. Well, do you get a signal or not? Say the black box takes some
results from a TRNG. Sometimes it halts after some random time.
Sometimes it does not halt. How does your system account for the black
box?
the turing machine model does not allow for randomness and is _strictly_ deterministic
it's kinda nuts u don't know a rather basic fact of computing theory,
but judging from the quality of ur comments over the year i've been
posting it doesn't surprise me
On 9/16/2026 6:12 PM, dart200 wrote:
On 9/16/26 1:54 PM, Chris M. Thomasson wrote:
On 9/16/2026 9:26 AM, dart200 wrote:
On 9/16/26 12:40 AM, Mikko wrote:
On 15/09/2026 19:34, dart200 wrote:
On 9/15/26 1:18 AM, Mikko wrote:
On 14/09/2026 11:45, dart200 wrote:
On 9/14/26 1:26 AM, Mikko wrote:
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:A "simple oracle" is an oracle that is simpler than most other >>>>>>>> oracles,
On 12/09/2026 19:02, dart200 wrote:"simple" is a weird term for something we provably can't build, >>>>>>>>> and that doesn't make the haltinging problem "decidable" since >>>>>>>>> we can't build, compute, or actually decide with the oracle >>>>>>>>
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computing
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit? >>>>>>>>>>>>> usp=sharing
The section 1 "the halting problem" should clearly define >>>>>>>>>>>> what the
expression "halting problem" means. It might be useful to first >>>>>>>>>>>> define "halting question" as that would simplify the >>>>>>>>>>>> definitions of
"halting problem" and "halting decider". In each of these it is >>>>>>>>>>>> also inportant to be clear what is the scope of the problem and >>>>>>>>>>>> what kind of solutions are allowed. For example, the halting >>>>>>>>>>>> problem of finite state machines is Turing decidable and the >>>>>>>>>>>> halting
problem of Turing machines is decidable with a simple oracle. >>>>>>>>>
just like a "big mouse" is a mouse that is bigger than most >>>>>>>> other mice
although not as big as a little elephant.
like i said: recursive undecidability leading to the concept of >>>>>>> "simple" vs "complex" oracles is discussing in -o4, i will not be >>>>>>> discussing them in -o1
That's OK, although there should be a short description of each >>>>>> section
at the end of the introduction.
But the section 1 gives the impression that the author does not know >>>>>> what the halting problem is.
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the >>>>>>>>>>>> program "und"
is not undecidable: there are partial halt deciders that can >>>>>>>>>>>> tell
whether "und" halts. In order to make this clear as soon as >>>>>>>>>>>> possible
it might be better to define "partial halt decider" before the >>>>>>>>>>>> introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
that's why partial deciders are discussed in -o5.2 not -o1, you >>>>>>> brought those up not me
The section 1 is about the halting problem. You may discuss
partial and
total solutions of the problem in the same section or elsewhere. But >>>>>> it is important to point out about "und" that its halting is
determinable. Otherwise a reader could be confused or get the
impression
that you are don't know.
und as specified in -o1 does not even exist, an argument which takes >>>>> up to the end of -o4 to explain
The section 1 says otherwise: the words "can be constructed" mean that >>>> it does exist.
it exist as a hypothetical problem, constructed from a hypothetical
decider, that presents a real limit to turing machine computability,
but does not exist in the actual enumeration of turing machines, and
does not limit our ability to decide on any given machine, as we are
not an addressable computing machine.
that is the point of the entire paper, and it takes 26 pages to
explain why. i'm not changing that specific wording there based on
confusion that will not resolve until you actually read the paper dud
Can you decide random numbers? Say a black box in isolation. You ask
it to run a process, when you wait for a signal that the process is
done. Well, do you get a signal or not? Say the black box takes some
results from a TRNG. Sometimes it halts after some random time.
Sometimes it does not halt. How does your system account for the black
box?
the turing machine model does not allow for randomness and is _strictly_ deterministic
it's kinda nuts u don't know a rather basic fact of computing theory,
but judging from the quality of ur comments over the year i've been posting it doesn't surprise me
You must be new around here. Nobody in Particular has posted here since
at least 2014.
Dude <punditster@gmail.com> posted:
On 9/16/2026 6:12 PM, dart200 wrote:
On 9/16/26 1:54 PM, Chris M. Thomasson wrote:You must be new around here. Nobody in Particular has posted here since
On 9/16/2026 9:26 AM, dart200 wrote:
On 9/16/26 12:40 AM, Mikko wrote:
On 15/09/2026 19:34, dart200 wrote:
On 9/15/26 1:18 AM, Mikko wrote:
On 14/09/2026 11:45, dart200 wrote:
On 9/14/26 1:26 AM, Mikko wrote:
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:A "simple oracle" is an oracle that is simpler than most other >>>>>>>>>> oracles,
On 12/09/2026 19:02, dart200 wrote:"simple" is a weird term for something we provably can't build, >>>>>>>>>>> and that doesn't make the haltinging problem "decidable" since >>>>>>>>>>> we can't build, compute, or actually decide with the oracle >>>>>>>>>>
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computing >>>>>>>>>>>>>>> and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit? >>>>>>>>>>>>>>> usp=sharing
The section 1 "the halting problem" should clearly define >>>>>>>>>>>>>> what the
expression "halting problem" means. It might be useful to first >>>>>>>>>>>>>> define "halting question" as that would simplify the >>>>>>>>>>>>>> definitions of
"halting problem" and "halting decider". In each of these it is >>>>>>>>>>>>>> also inportant to be clear what is the scope of the problem and >>>>>>>>>>>>>> what kind of solutions are allowed. For example, the halting >>>>>>>>>>>>>> problem of finite state machines is Turing decidable and the >>>>>>>>>>>>>> halting
problem of Turing machines is decidable with a simple oracle. >>>>>>>>>>>
just like a "big mouse" is a mouse that is bigger than most >>>>>>>>>> other mice
although not as big as a little elephant.
like i said: recursive undecidability leading to the concept of >>>>>>>>> "simple" vs "complex" oracles is discussing in -o4, i will not be >>>>>>>>> discussing them in -o1
That's OK, although there should be a short description of each >>>>>>>> section
at the end of the introduction.
But the section 1 gives the impression that the author does not know >>>>>>>> what the halting problem is.
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the >>>>>>>>>>>>>> program "und"
is not undecidable: there are partial halt deciders that can >>>>>>>>>>>>>> tell
whether "und" halts. In order to make this clear as soon as >>>>>>>>>>>>>> possible
it might be better to define "partial halt decider" before the >>>>>>>>>>>>>> introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
that's why partial deciders are discussed in -o5.2 not -o1, you >>>>>>>>> brought those up not me
The section 1 is about the halting problem. You may discuss
partial and
total solutions of the problem in the same section or elsewhere. But >>>>>>>> it is important to point out about "und" that its halting is
determinable. Otherwise a reader could be confused or get the
impression
that you are don't know.
und as specified in -o1 does not even exist, an argument which takes >>>>>>> up to the end of -o4 to explain
The section 1 says otherwise: the words "can be constructed" mean that >>>>>> it does exist.
it exist as a hypothetical problem, constructed from a hypothetical
decider, that presents a real limit to turing machine computability, >>>>> but does not exist in the actual enumeration of turing machines, and >>>>> does not limit our ability to decide on any given machine, as we are >>>>> not an addressable computing machine.
that is the point of the entire paper, and it takes 26 pages to
explain why. i'm not changing that specific wording there based on
confusion that will not resolve until you actually read the paper dud >>>>>
Can you decide random numbers? Say a black box in isolation. You ask
it to run a process, when you wait for a signal that the process is
done. Well, do you get a signal or not? Say the black box takes some
results from a TRNG. Sometimes it halts after some random time.
Sometimes it does not halt. How does your system account for the black >>>> box?
the turing machine model does not allow for randomness and is _strictly_ >>> deterministic
it's kinda nuts u don't know a rather basic fact of computing theory,
but judging from the quality of ur comments over the year i've been
posting it doesn't surprise me
at least 2014.
He wasn't replying to me.
On 9/16/26 12:40 AM, Mikko wrote:
On 15/09/2026 19:34, dart200 wrote:
On 9/15/26 1:18 AM, Mikko wrote:
On 14/09/2026 11:45, dart200 wrote:
On 9/14/26 1:26 AM, Mikko wrote:
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:
On 12/09/2026 19:02, dart200 wrote:"simple" is a weird term for something we provably can't build, >>>>>>> and that doesn't make the haltinging problem "decidable" since we >>>>>>> can't build, compute, or actually decide with the oracle
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computingThe section 1 "the halting problem" should clearly define what >>>>>>>>>> the
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit?usp=sharing >>>>>>>>>>
expression "halting problem" means. It might be useful to first >>>>>>>>>> define "halting question" as that would simplify the
definitions of
"halting problem" and "halting decider". In each of these it is >>>>>>>>>> also inportant to be clear what is the scope of the problem and >>>>>>>>>> what kind of solutions are allowed. For example, the halting >>>>>>>>>> problem of finite state machines is Turing decidable and the >>>>>>>>>> halting
problem of Turing machines is decidable with a simple oracle. >>>>>>>
A "simple oracle" is an oracle that is simpler than most other
oracles,
just like a "big mouse" is a mouse that is bigger than most other >>>>>> mice
although not as big as a little elephant.
like i said: recursive undecidability leading to the concept of
"simple" vs "complex" oracles is discussing in -o4, i will not be
discussing them in -o1
That's OK, although there should be a short description of each section >>>> at the end of the introduction.
But the section 1 gives the impression that the author does not know
what the halting problem is.
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the program >>>>>>>>>> "und"
is not undecidable: there are partial halt deciders that can tell >>>>>>>>>> whether "und" halts. In order to make this clear as soon as >>>>>>>>>> possible
it might be better to define "partial halt decider" before the >>>>>>>>>> introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
that's why partial deciders are discussed in -o5.2 not -o1, you
brought those up not me
The section 1 is about the halting problem. You may discuss partial and >>>> total solutions of the problem in the same section or elsewhere. But
it is important to point out about "und" that its halting is
determinable. Otherwise a reader could be confused or get the
impression
that you are don't know.
und as specified in -o1 does not even exist, an argument which takes
up to the end of -o4 to explain
The section 1 says otherwise: the words "can be constructed" mean that
it does exist.
it exist as a hypothetical problem, constructed from a hypothetical
decider, that presents a real limit to turing machine computability, but does not exist in the actual enumeration of turing machines,
and does not limit our ability to decide on any given machine, as we
are not an addressable computing machine.
On 9/13/2026 2:43 AM, Mikko wrote:
On 12/09/2026 20:32, Dude wrote:
On 9/12/2026 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:You came here for 5th grade math?
on the nature of undecidability within computing
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit?usp=sharing
The section 1 "the halting problem" should clearly define what the
expression "halting problem" means. It might be useful to first
define "halting question" as that would simplify the definitions of
"halting problem" and "halting decider". In each of these it is
also inportant to be clear what is the scope of the problem and
what kind of solutions are allowed. For example, the halting
problem of finite state machines is Turing decidable and the halting
problem of Turing machines is decidable with a simple oracle.
One should also clearly state that the halting of the program "und"
is not undecidable: there are partial halt deciders that can tell
whether "und" halts. In order to make this clear as soon as possible
it might be better to define "partial halt decider" before the
introduction of "und".
No, thank you.
"You are soaking in it." - Madge
-aIf you wnat to offer some 5th grade math you can do
it in sci.math.You came here to refute the church-turing thesis?
On 16/09/2026 19:26, dart200 wrote:
On 9/16/26 12:40 AM, Mikko wrote:
On 15/09/2026 19:34, dart200 wrote:
On 9/15/26 1:18 AM, Mikko wrote:
On 14/09/2026 11:45, dart200 wrote:
On 9/14/26 1:26 AM, Mikko wrote:
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:
On 12/09/2026 19:02, dart200 wrote:"simple" is a weird term for something we provably can't build, >>>>>>>> and that doesn't make the haltinging problem "decidable" since >>>>>>>> we can't build, compute, or actually decide with the oracle
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computingThe section 1 "the halting problem" should clearly define >>>>>>>>>>> what the
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit?usp=sharing >>>>>>>>>>>
expression "halting problem" means. It might be useful to first >>>>>>>>>>> define "halting question" as that would simplify the
definitions of
"halting problem" and "halting decider". In each of these it is >>>>>>>>>>> also inportant to be clear what is the scope of the problem and >>>>>>>>>>> what kind of solutions are allowed. For example, the halting >>>>>>>>>>> problem of finite state machines is Turing decidable and the >>>>>>>>>>> halting
problem of Turing machines is decidable with a simple oracle. >>>>>>>>
A "simple oracle" is an oracle that is simpler than most other
oracles,
just like a "big mouse" is a mouse that is bigger than most other >>>>>>> mice
although not as big as a little elephant.
like i said: recursive undecidability leading to the concept of
"simple" vs "complex" oracles is discussing in -o4, i will not be >>>>>> discussing them in -o1
That's OK, although there should be a short description of each
section
at the end of the introduction.
But the section 1 gives the impression that the author does not know >>>>> what the halting problem is.
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the program >>>>>>>>>>> "und"
is not undecidable: there are partial halt deciders that can >>>>>>>>>>> tell
whether "und" halts. In order to make this clear as soon as >>>>>>>>>>> possible
it might be better to define "partial halt decider" before the >>>>>>>>>>> introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
that's why partial deciders are discussed in -o5.2 not -o1, you
brought those up not me
The section 1 is about the halting problem. You may discuss partial >>>>> and
total solutions of the problem in the same section or elsewhere. But >>>>> it is important to point out about "und" that its halting is
determinable. Otherwise a reader could be confused or get the
impression
that you are don't know.
und as specified in -o1 does not even exist, an argument which takes
up to the end of -o4 to explain
The section 1 says otherwise: the words "can be constructed" mean that
it does exist.
it exist as a hypothetical problem, constructed from a hypothetical
decider, that presents a real limit to turing machine computability,
but does not exist in the actual enumeration of turing machines,
In the world of computing theory everything is hypothetical. The section
1 does not call anything hypothetical. It merely constructs "und" from "halts" already specified and "loop" that is not specified.
Although "und" is called "the halting problem" it obviously isn't any problem, just a specification of a computation.
and does not limit our ability to decide on any given machine, as we
are not an addressable computing machine.
That is not relevant to this discussion about the section 1. Even if
the section 1 is only a prelude to the main topic it should not give
the impression that the author cannot write anyting worth of reading.
On 9/18/26 1:54 AM, Mikko wrote:
On 16/09/2026 19:26, dart200 wrote:
On 9/16/26 12:40 AM, Mikko wrote:
On 15/09/2026 19:34, dart200 wrote:
On 9/15/26 1:18 AM, Mikko wrote:
On 14/09/2026 11:45, dart200 wrote:
On 9/14/26 1:26 AM, Mikko wrote:
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:
On 12/09/2026 19:02, dart200 wrote:"simple" is a weird term for something we provably can't build, >>>>>>>>> and that doesn't make the haltinging problem "decidable" since >>>>>>>>> we can't build, compute, or actually decide with the oracle
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computingThe section 1 "the halting problem" should clearly define >>>>>>>>>>>> what the
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit?usp=sharing >>>>>>>>>>>>
expression "halting problem" means. It might be useful to first >>>>>>>>>>>> define "halting question" as that would simplify the
definitions of
"halting problem" and "halting decider". In each of these it is >>>>>>>>>>>> also inportant to be clear what is the scope of the problem and >>>>>>>>>>>> what kind of solutions are allowed. For example, the halting >>>>>>>>>>>> problem of finite state machines is Turing decidable and the >>>>>>>>>>>> halting
problem of Turing machines is decidable with a simple oracle. >>>>>>>>>
A "simple oracle" is an oracle that is simpler than most other >>>>>>>> oracles,
just like a "big mouse" is a mouse that is bigger than most
other mice
although not as big as a little elephant.
like i said: recursive undecidability leading to the concept of >>>>>>> "simple" vs "complex" oracles is discussing in -o4, i will not be >>>>>>> discussing them in -o1
That's OK, although there should be a short description of each
section
at the end of the introduction.
But the section 1 gives the impression that the author does not know >>>>>> what the halting problem is.
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the >>>>>>>>>>>> program "und"
is not undecidable: there are partial halt deciders that can >>>>>>>>>>>> tell
whether "und" halts. In order to make this clear as soon as >>>>>>>>>>>> possible
it might be better to define "partial halt decider" before the >>>>>>>>>>>> introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
that's why partial deciders are discussed in -o5.2 not -o1, you >>>>>>> brought those up not me
The section 1 is about the halting problem. You may discuss
partial and
total solutions of the problem in the same section or elsewhere. But >>>>>> it is important to point out about "und" that its halting is
determinable. Otherwise a reader could be confused or get the
impression
that you are don't know.
und as specified in -o1 does not even exist, an argument which takes >>>>> up to the end of -o4 to explain
The section 1 says otherwise: the words "can be constructed" mean that >>>> it does exist.
it exist as a hypothetical problem, constructed from a hypothetical
decider, that presents a real limit to turing machine computability,
but does not exist in the actual enumeration of turing machines,
In the world of computing theory everything is hypothetical. The section
1 does not call anything hypothetical. It merely constructs "und" from
"halts" already specified and "loop" that is not specified.
if u can't figure out what halts does contrasted from loop in that pseudo-code, when we are discussing the halting problem in a group
called comp.theory, this paper is just not for you, at this time
i have to be selective who i spend my time on at this point, i don't
have personal time to waste on someone who spends more time responding
to me, than it would take to just read the 3rd page
Although "und" is called "the halting problem" it obviously isn't any
problem, just a specification of a computation.
and does not limit our ability to decide on any given machine, as we
are not an addressable computing machine.
That is not relevant to this discussion about the section 1. Even if
the undecidability of und is literally the core justification for why we can't build a total halting decider in turing machine computing, while
the church-turing thesis asserts we can only compute something that
turing machines can compute ... so therefor undecidability of und the literally the core justification for why we aren't building total
halting deciders.
the section 1 is only a prelude to the main topic it should not give
the impression that the author cannot write anyting worth of reading.
prove ur not a retard dud, tell me:
does und() halt or loop forever?
On 2026-09-18 03:33, dart200 wrote:
On 9/18/26 1:54 AM, Mikko wrote:
On 16/09/2026 19:26, dart200 wrote:
On 9/16/26 12:40 AM, Mikko wrote:
On 15/09/2026 19:34, dart200 wrote:
On 9/15/26 1:18 AM, Mikko wrote:
On 14/09/2026 11:45, dart200 wrote:
On 9/14/26 1:26 AM, Mikko wrote:
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:
On 12/09/2026 19:02, dart200 wrote:"simple" is a weird term for something we provably can't
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computing
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit? >>>>>>>>>>>>>> usp=sharing
The section 1 "the halting problem" should clearly define >>>>>>>>>>>>> what the
expression "halting problem" means. It might be useful to >>>>>>>>>>>>> first
define "halting question" as that would simplify the >>>>>>>>>>>>> definitions of
"halting problem" and "halting decider". In each of these >>>>>>>>>>>>> it is
also inportant to be clear what is the scope of the problem >>>>>>>>>>>>> and
what kind of solutions are allowed. For example, the halting >>>>>>>>>>>>> problem of finite state machines is Turing decidable and >>>>>>>>>>>>> the halting
problem of Turing machines is decidable with a simple oracle. >>>>>>>>>>
build, and that doesn't make the haltinging problem
"decidable" since we can't build, compute, or actually decide >>>>>>>>>> with the oracle
A "simple oracle" is an oracle that is simpler than most other >>>>>>>>> oracles,
just like a "big mouse" is a mouse that is bigger than most >>>>>>>>> other mice
although not as big as a little elephant.
like i said: recursive undecidability leading to the concept of >>>>>>>> "simple" vs "complex" oracles is discussing in -o4, i will not be >>>>>>>> discussing them in -o1
That's OK, although there should be a short description of each >>>>>>> section
at the end of the introduction.
But the section 1 gives the impression that the author does not know >>>>>>> what the halting problem is.
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the >>>>>>>>>>>>> program "und"
is not undecidable: there are partial halt deciders that >>>>>>>>>>>>> can tell
whether "und" halts. In order to make this clear as soon as >>>>>>>>>>>>> possible
it might be better to define "partial halt decider" before the >>>>>>>>>>>>> introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
that's why partial deciders are discussed in -o5.2 not -o1, you >>>>>>>> brought those up not me
The section 1 is about the halting problem. You may discuss
partial and
total solutions of the problem in the same section or elsewhere. But >>>>>>> it is important to point out about "und" that its halting is
determinable. Otherwise a reader could be confused or get the
impression
that you are don't know.
und as specified in -o1 does not even exist, an argument which
takes up to the end of -o4 to explain
The section 1 says otherwise: the words "can be constructed" mean that >>>>> it does exist.
it exist as a hypothetical problem, constructed from a hypothetical
decider, that presents a real limit to turing machine computability,
but does not exist in the actual enumeration of turing machines,
In the world of computing theory everything is hypothetical. The section >>> 1 does not call anything hypothetical. It merely constructs "und" from
"halts" already specified and "loop" that is not specified.
if u can't figure out what halts does contrasted from loop in that
pseudo-code, when we are discussing the halting problem in a group
called comp.theory, this paper is just not for you, at this time
i have to be selective who i spend my time on at this point, i don't
have personal time to waste on someone who spends more time responding
to me, than it would take to just read the 3rd page
Although "und" is called "the halting problem" it obviously isn't any
problem, just a specification of a computation.
and does not limit our ability to decide on any given machine, as we >>> -a> are not an addressable computing machine.
That is not relevant to this discussion about the section 1. Even if
the undecidability of und is literally the core justification for why
we can't build a total halting decider in turing machine computing,
while the church-turing thesis asserts we can only compute something
that turing machines can compute ... so therefor undecidability of und
the literally the core justification for why we aren't building total
halting deciders.
und() isn't undecidable. It just cannot be correctly decided by halts(). When we talk about things being (un)decidable we are normally talking
about sets, and und() isn't a set so it doesn't really make sense to
claim that it is undecidable.
the section 1 is only a prelude to the main topic it should not give
the impression that the author cannot write anyting worth of reading.
prove ur not a retard dud, tell me:
does und() halt or loop forever?
Obviously he cannot answer that because whether und() halts or loops is entirely dependent on how halts() is implemented. You don't actually
provide an implementation of halts(), just a specification of what
it's /supposed/ to do. The entire point of the halting problem proofs is that it isn't actually possible to implement something which does what
you claim halts() does.
Andr|-
On 9/18/26 2:54 PM, Andr|- G. Isaak wrote:
On 2026-09-18 03:33, dart200 wrote:
On 9/18/26 1:54 AM, Mikko wrote:
On 16/09/2026 19:26, dart200 wrote:
On 9/16/26 12:40 AM, Mikko wrote:
On 15/09/2026 19:34, dart200 wrote:
On 9/15/26 1:18 AM, Mikko wrote:
On 14/09/2026 11:45, dart200 wrote:
On 9/14/26 1:26 AM, Mikko wrote:
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:
On 12/09/2026 19:02, dart200 wrote:"simple" is a weird term for something we provably can't >>>>>>>>>>> build, and that doesn't make the haltinging problem
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computing >>>>>>>>>>>>>>> and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit? >>>>>>>>>>>>>>> usp=sharing
The section 1 "the halting problem" should clearly define >>>>>>>>>>>>>> what the
expression "halting problem" means. It might be useful to >>>>>>>>>>>>>> first
define "halting question" as that would simplify the >>>>>>>>>>>>>> definitions of
"halting problem" and "halting decider". In each of these >>>>>>>>>>>>>> it is
also inportant to be clear what is the scope of the >>>>>>>>>>>>>> problem and
what kind of solutions are allowed. For example, the halting >>>>>>>>>>>>>> problem of finite state machines is Turing decidable and >>>>>>>>>>>>>> the halting
problem of Turing machines is decidable with a simple oracle. >>>>>>>>>>>
"decidable" since we can't build, compute, or actually decide >>>>>>>>>>> with the oracle
A "simple oracle" is an oracle that is simpler than most other >>>>>>>>>> oracles,
just like a "big mouse" is a mouse that is bigger than most >>>>>>>>>> other mice
although not as big as a little elephant.
like i said: recursive undecidability leading to the concept of >>>>>>>>> "simple" vs "complex" oracles is discussing in -o4, i will not >>>>>>>>> be discussing them in -o1
That's OK, although there should be a short description of each >>>>>>>> section
at the end of the introduction.
But the section 1 gives the impression that the author does not >>>>>>>> know
what the halting problem is.
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the >>>>>>>>>>>>>> program "und"
is not undecidable: there are partial halt deciders that >>>>>>>>>>>>>> can tell
whether "und" halts. In order to make this clear as soon >>>>>>>>>>>>>> as possible
it might be better to define "partial halt decider" before >>>>>>>>>>>>>> the
introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
that's why partial deciders are discussed in -o5.2 not -o1, you >>>>>>>>> brought those up not me
The section 1 is about the halting problem. You may discuss
partial and
total solutions of the problem in the same section or elsewhere. >>>>>>>> But
it is important to point out about "und" that its halting is
determinable. Otherwise a reader could be confused or get the >>>>>>>> impression
that you are don't know.
und as specified in -o1 does not even exist, an argument which
takes up to the end of -o4 to explain
The section 1 says otherwise: the words "can be constructed" mean >>>>>> that
it does exist.
it exist as a hypothetical problem, constructed from a hypothetical >>>>> decider, that presents a real limit to turing machine
computability, but does not exist in the actual enumeration of
turing machines,
In the world of computing theory everything is hypothetical. The
section
1 does not call anything hypothetical. It merely constructs "und" from >>>> "halts" already specified and "loop" that is not specified.
if u can't figure out what halts does contrasted from loop in that
pseudo-code, when we are discussing the halting problem in a group
called comp.theory, this paper is just not for you, at this time
i have to be selective who i spend my time on at this point, i don't
have personal time to waste on someone who spends more time
responding to me, than it would take to just read the 3rd page
Although "und" is called "the halting problem" it obviously isn't any
problem, just a specification of a computation.
and does not limit our ability to decide on any given machine, as we >>>> -a> are not an addressable computing machine.
That is not relevant to this discussion about the section 1. Even if
the undecidability of und is literally the core justification for why
we can't build a total halting decider in turing machine computing,
while the church-turing thesis asserts we can only compute something
that turing machines can compute ... so therefor undecidability of
und the literally the core justification for why we aren't building
total halting deciders.
und() isn't undecidable. It just cannot be correctly decided by
halts(). When we talk about things being (un)decidable we are normally
talking about sets, and und() isn't a set so it doesn't really make
sense to claim that it is undecidable.
und() as hypothetical object not being in a set is what makes it "undecidable", because we cannot "decide" it into a set ... that's what
a "decider" ought to do ...
the section 1 is only a prelude to the main topic it should not give
the impression that the author cannot write anyting worth of reading.
prove ur not a retard dud, tell me:
does und() halt or loop forever?
Obviously he cannot answer that because whether und() halts or loops
is entirely dependent on how halts() is implemented. You don't
actually provide an implementation of halts(), just a specification of
what it's /supposed/ to do. The entire point of the halting problem
proofs is that it isn't actually possible to implement something which
does what you claim halts() does.
great, glad we gotten past the 2nd page. could maybe someone could
actually read how i've gotten past that sticking point.
cause und() doesn't exist in the proper enumeration of turing machines, which isn't a hypothetical, and instead a very real countable infinite
set of objects that you can physically compute and produce. in that set there is no und() machine, as und() does not have a runtime which can be decided upon, while any real turing machine definition _must_ have a decidable runtime. it _cannot_ get stuck in a state of uncertain
future... either there is a next state, or the damn thing halts!
the part that we got stuck on is the fact that within turing machines,
only subsets of machines are decidable by any given set classifier. this
is _not_ a limitation of computation entirely, but a limitation caused
by the addressable nature of the turing machine model.
the part that no one realized about turing machine computation is that _despite_ this limitation, we *can still* use a partial recognizer
(-o5.2) to enumerate across all output sequences (which is not the same
as all machines which produce all output sequences). furthermore that enumeration *cannot* be used to construct either a total diagonal, or a total anti-diagonal, for reasons ur gunna have to read in an update i
will post in day or so (-o6.1-2 is getting overhauled a bit), because i
just discerned that in last few days. those are computable sequences,
but _not_ turing computable sequences
until then ur homework is up until the end of -o5, tho we all know no one here is reading past -o1 cause ur all just looking for a fucking brainrot nitpick vs actually trying to understand what i've discovered
Andr|-
i'm redefining undecidability with computing to be a property of a particular machine in respect to a particular set-classifier machine (or
set of them if a machine is an input paradox to multiple classifiers)
you're going to try to forget what you learned for a bit, cause i'm rewriting the textbook on this particular matter
or just continue to be a useless twat wasting all our quite limited time
on this earth, my god do u have any sense of mortality???
On 2026-09-18 21:36, dart200 wrote:
On 9/18/26 2:54 PM, Andr|- G. Isaak wrote:
On 2026-09-18 03:33, dart200 wrote:
On 9/18/26 1:54 AM, Mikko wrote:
On 16/09/2026 19:26, dart200 wrote:
On 9/16/26 12:40 AM, Mikko wrote:
On 15/09/2026 19:34, dart200 wrote:
On 9/15/26 1:18 AM, Mikko wrote:
On 14/09/2026 11:45, dart200 wrote:
On 9/14/26 1:26 AM, Mikko wrote:
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:
On 12/09/2026 19:02, dart200 wrote:
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computing >>>>>>>>>>>>>>>> and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit? >>>>>>>>>>>>>>>> usp=sharing
The section 1 "the halting problem" should clearly define >>>>>>>>>>>>>>> what the
expression "halting problem" means. It might be useful to >>>>>>>>>>>>>>> first
define "halting question" as that would simplify the >>>>>>>>>>>>>>> definitions of
"halting problem" and "halting decider". In each of these >>>>>>>>>>>>>>> it is
also inportant to be clear what is the scope of the >>>>>>>>>>>>>>> problem and
what kind of solutions are allowed. For example, the halting >>>>>>>>>>>>>>> problem of finite state machines is Turing decidable and >>>>>>>>>>>>>>> the halting
problem of Turing machines is decidable with a simple >>>>>>>>>>>>>>> oracle.
"simple" is a weird term for something we provably can't >>>>>>>>>>>> build, and that doesn't make the haltinging problem
"decidable" since we can't build, compute, or actually >>>>>>>>>>>> decide with the oracle
A "simple oracle" is an oracle that is simpler than most >>>>>>>>>>> other oracles,
just like a "big mouse" is a mouse that is bigger than most >>>>>>>>>>> other mice
although not as big as a little elephant.
like i said: recursive undecidability leading to the concept >>>>>>>>>> of "simple" vs "complex" oracles is discussing in -o4, i will >>>>>>>>>> not be discussing them in -o1
That's OK, although there should be a short description of each >>>>>>>>> section
at the end of the introduction.
But the section 1 gives the impression that the author does not >>>>>>>>> know
what the halting problem is.
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the >>>>>>>>>>>>>>> program "und"
is not undecidable: there are partial halt deciders that >>>>>>>>>>>>>>> can tell
whether "und" halts. In order to make this clear as soon >>>>>>>>>>>>>>> as possible
it might be better to define "partial halt decider" >>>>>>>>>>>>>>> before the
introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
that's why partial deciders are discussed in -o5.2 not -o1, you >>>>>>>>>> brought those up not me
The section 1 is about the halting problem. You may discuss >>>>>>>>> partial and
total solutions of the problem in the same section or
elsewhere. But
it is important to point out about "und" that its halting is >>>>>>>>> determinable. Otherwise a reader could be confused or get the >>>>>>>>> impression
that you are don't know.
und as specified in -o1 does not even exist, an argument which >>>>>>>> takes up to the end of -o4 to explain
The section 1 says otherwise: the words "can be constructed" mean >>>>>>> that
it does exist.
it exist as a hypothetical problem, constructed from a
hypothetical decider, that presents a real limit to turing machine >>>>>> computability, but does not exist in the actual enumeration of
turing machines,
In the world of computing theory everything is hypothetical. The
section
1 does not call anything hypothetical. It merely constructs "und" from >>>>> "halts" already specified and "loop" that is not specified.
if u can't figure out what halts does contrasted from loop in that
pseudo-code, when we are discussing the halting problem in a group
called comp.theory, this paper is just not for you, at this time
i have to be selective who i spend my time on at this point, i don't
have personal time to waste on someone who spends more time
responding to me, than it would take to just read the 3rd page
Although "und" is called "the halting problem" it obviously isn't any >>>>> problem, just a specification of a computation.
and does not limit our ability to decide on any given machine,as we
are not an addressable computing machine.
That is not relevant to this discussion about the section 1. Even if
the undecidability of und is literally the core justification for
why we can't build a total halting decider in turing machine
computing, while the church-turing thesis asserts we can only
compute something that turing machines can compute ... so therefor
undecidability of und the literally the core justification for why
we aren't building total halting deciders.
und() isn't undecidable. It just cannot be correctly decided by
halts(). When we talk about things being (un)decidable we are
normally talking about sets, and und() isn't a set so it doesn't
really make sense to claim that it is undecidable.
und() as hypothetical object not being in a set is what makes it
"undecidable", because we cannot "decide" it into a set ... that's
what a "decider" ought to do ...
What do you mean by 'not being in a set'? It belongs to any number of
sets. And you don't 'decide things into a set'. Things are elements of a
set or they are not. Your above passage is essentially gibberish, suggesting, that you don't understand what '(un)decidable' means.
the section 1 is only a prelude to the main topic it should not give >>>>> the impression that the author cannot write anyting worth of reading. >>>>>
prove ur not a retard dud, tell me:
does und() halt or loop forever?
Obviously he cannot answer that because whether und() halts or loops
is entirely dependent on how halts() is implemented. You don't
actually provide an implementation of halts(), just a specification
of what it's /supposed/ to do. The entire point of the halting
problem proofs is that it isn't actually possible to implement
something which does what you claim halts() does.
great, glad we gotten past the 2nd page. could maybe someone could
actually read how i've gotten past that sticking point.
Between the first and second pages you've demonstrated that you have no
idea what you're talking about. Why should anyone feel motivated to go
past the 2nd page? If you really have something to contribute you should make sure that your initial presentation is both coherent and accurate.
You have failed to do that in your introductory section.
cause und() doesn't exist in the proper enumeration of turing
machines, which isn't a hypothetical, and instead a very real
countable infinite set of objects that you can physically compute and
produce. in that set there is no und() machine, as und() does not have
a runtime which can be decided upon, while any real turing machine
definition _must_ have a decidable runtime. it _cannot_ get stuck in a
state of uncertain future... either there is a next state, or the damn
thing halts!
The notion of 'runtimes' is entirely missing from computational theory.
und() does not exist in the enumeration of turing machines for the
simple reason that halts(), upon which it relies, does not exist in the enumeration of turing machines.
the part that we got stuck on is the fact that within turing machines,
only subsets of machines are decidable by any given set classifier.
this is _not_ a limitation of computation entirely, but a limitation
caused by the addressable nature of the turing machine model.
I have no idea what any of the above means. Sets or problems are
decidable. Individual machines are not decidable. And I have absolutely
no idea what you mean by 'the addressable nature of the turing machine model'.
the part that no one realized about turing machine computation is that
_despite_ this limitation, we *can still* use a partial recognizer
(-o5.2) to enumerate across all output sequences (which is not the same
as all machines which produce all output sequences). furthermore that
enumeration *cannot* be used to construct either a total diagonal, or
a total anti-diagonal, for reasons ur gunna have to read in an update
i will post in day or so (-o6.1-2 is getting overhauled a bit), because
i just discerned that in last few days. those are computable
sequences, but _not_ turing computable sequences
until then ur homework is up until the end of -o5, tho we all know no
one here is reading past -o1 cause ur all just looking for a fucking
brainrot nitpick vs actually trying to understand what i've discovered
Andr|-
i'm redefining undecidability with computing to be a property of a
particular machine in respect to a particular set-classifier machine
(or set of them if a machine is an input paradox to multiple classifiers)
Decidablity is already precisely defined. If you want to "redefine" it, you're no longer talking about anything which of interest to anyone else.
you're going to try to forget what you learned for a bit, cause i'm
rewriting the textbook on this particular matter
No one is interested in your 'rewritten' textbook. Decidability is a
property of sets or problems, not a property of machines. If you're
interested in some specific property of machines you need to (a) define
that property, and (b) give it a name that isn't already in use for something else.
or just continue to be a useless twat wasting all our quite limited
time on this earth, my god do u have any sense of mortality???
Seriously? I don't see what debates about decidability have to do with morality.
Andr|-
On 9/18/26 1:54 AM, Mikko wrote:
On 16/09/2026 19:26, dart200 wrote:
On 9/16/26 12:40 AM, Mikko wrote:
On 15/09/2026 19:34, dart200 wrote:
On 9/15/26 1:18 AM, Mikko wrote:
On 14/09/2026 11:45, dart200 wrote:
On 9/14/26 1:26 AM, Mikko wrote:
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:
On 12/09/2026 19:02, dart200 wrote:"simple" is a weird term for something we provably can't build, >>>>>>>>> and that doesn't make the haltinging problem "decidable" since >>>>>>>>> we can't build, compute, or actually decide with the oracle
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computing
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit? >>>>>>>>>>>>> usp=sharing
The section 1 "the halting problem" should clearly define >>>>>>>>>>>> what the
expression "halting problem" means. It might be useful to first >>>>>>>>>>>> define "halting question" as that would simplify the
definitions of
"halting problem" and "halting decider". In each of these it is >>>>>>>>>>>> also inportant to be clear what is the scope of the problem and >>>>>>>>>>>> what kind of solutions are allowed. For example, the halting >>>>>>>>>>>> problem of finite state machines is Turing decidable and the >>>>>>>>>>>> halting
problem of Turing machines is decidable with a simple oracle. >>>>>>>>>
A "simple oracle" is an oracle that is simpler than most other >>>>>>>> oracles,
just like a "big mouse" is a mouse that is bigger than most
other mice
although not as big as a little elephant.
like i said: recursive undecidability leading to the concept of >>>>>>> "simple" vs "complex" oracles is discussing in -o4, i will not be >>>>>>> discussing them in -o1
That's OK, although there should be a short description of each
section
at the end of the introduction.
But the section 1 gives the impression that the author does not know >>>>>> what the halting problem is.
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the >>>>>>>>>>>> program "und"
is not undecidable: there are partial halt deciders that can >>>>>>>>>>>> tell
whether "und" halts. In order to make this clear as soon as >>>>>>>>>>>> possible
it might be better to define "partial halt decider" before the >>>>>>>>>>>> introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
that's why partial deciders are discussed in -o5.2 not -o1, you >>>>>>> brought those up not me
The section 1 is about the halting problem. You may discuss
partial and
total solutions of the problem in the same section or elsewhere. But >>>>>> it is important to point out about "und" that its halting is
determinable. Otherwise a reader could be confused or get the
impression
that you are don't know.
und as specified in -o1 does not even exist, an argument which takes >>>>> up to the end of -o4 to explain
The section 1 says otherwise: the words "can be constructed" mean that >>>> it does exist.
it exist as a hypothetical problem, constructed from a hypothetical
decider, that presents a real limit to turing machine computability,
but does not exist in the actual enumeration of turing machines,
In the world of computing theory everything is hypothetical. The section
1 does not call anything hypothetical. It merely constructs "und" from
"halts" already specified and "loop" that is not specified.
if u can't figure out what halts does contrasted from loop in that pseudo-code, when we are discussing the halting problem in a group
called comp.theory, this paper is just not for you, at this time
i have to be selective who i spend my time on at this point, i don't
have personal time to waste on someone who spends more time responding
to me, than it would take to just read the 3rd page
Although "und" is called "the halting problem" it obviously isn't any
problem, just a specification of a computation.
and does not limit our ability to decide on any given machine, as we
are not an addressable computing machine.
That is not relevant to this discussion about the section 1. Even if
the undecidability of und is literally the core justification for why we can't build a total halting decider in turing machine computing, while
the church-turing thesis asserts we can only compute something that
turing machines can compute ... so therefor undecidability of und the literally the core justification for why we aren't building total
halting deciders.
the section 1 is only a prelude to the main topic it should not give
the impression that the author cannot write anyting worth of reading.
prove ur not a retard dud, tell me:
does und() halt or loop forever?
On 18/09/2026 12:33, dart200 wrote:
On 9/18/26 1:54 AM, Mikko wrote:
On 16/09/2026 19:26, dart200 wrote:
On 9/16/26 12:40 AM, Mikko wrote:
On 15/09/2026 19:34, dart200 wrote:
On 9/15/26 1:18 AM, Mikko wrote:
On 14/09/2026 11:45, dart200 wrote:
On 9/14/26 1:26 AM, Mikko wrote:
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:
On 12/09/2026 19:02, dart200 wrote:"simple" is a weird term for something we provably can't
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computing
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit? >>>>>>>>>>>>>> usp=sharing
The section 1 "the halting problem" should clearly define >>>>>>>>>>>>> what the
expression "halting problem" means. It might be useful to >>>>>>>>>>>>> first
define "halting question" as that would simplify the >>>>>>>>>>>>> definitions of
"halting problem" and "halting decider". In each of these >>>>>>>>>>>>> it is
also inportant to be clear what is the scope of the problem >>>>>>>>>>>>> and
what kind of solutions are allowed. For example, the halting >>>>>>>>>>>>> problem of finite state machines is Turing decidable and >>>>>>>>>>>>> the halting
problem of Turing machines is decidable with a simple oracle. >>>>>>>>>>
build, and that doesn't make the haltinging problem
"decidable" since we can't build, compute, or actually decide >>>>>>>>>> with the oracle
A "simple oracle" is an oracle that is simpler than most other >>>>>>>>> oracles,
just like a "big mouse" is a mouse that is bigger than most >>>>>>>>> other mice
although not as big as a little elephant.
like i said: recursive undecidability leading to the concept of >>>>>>>> "simple" vs "complex" oracles is discussing in -o4, i will not be >>>>>>>> discussing them in -o1
That's OK, although there should be a short description of each >>>>>>> section
at the end of the introduction.
But the section 1 gives the impression that the author does not know >>>>>>> what the halting problem is.
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the >>>>>>>>>>>>> program "und"
is not undecidable: there are partial halt deciders that >>>>>>>>>>>>> can tell
whether "und" halts. In order to make this clear as soon as >>>>>>>>>>>>> possible
it might be better to define "partial halt decider" before the >>>>>>>>>>>>> introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
that's why partial deciders are discussed in -o5.2 not -o1, you >>>>>>>> brought those up not me
The section 1 is about the halting problem. You may discuss
partial and
total solutions of the problem in the same section or elsewhere. But >>>>>>> it is important to point out about "und" that its halting is
determinable. Otherwise a reader could be confused or get the
impression
that you are don't know.
und as specified in -o1 does not even exist, an argument which
takes up to the end of -o4 to explain
The section 1 says otherwise: the words "can be constructed" mean that >>>>> it does exist.
it exist as a hypothetical problem, constructed from a hypothetical
decider, that presents a real limit to turing machine computability,
but does not exist in the actual enumeration of turing machines,
In the world of computing theory everything is hypothetical. The section >>> 1 does not call anything hypothetical. It merely constructs "und" from
"halts" already specified and "loop" that is not specified.
if u can't figure out what halts does contrasted from loop in that
pseudo-code, when we are discussing the halting problem in a group
called comp.theory, this paper is just not for you, at this time
An attempt to change the topic indicates that you don't want to
discuss honestly but instead want to deceive.
The topic of my comment clearly was the article at https://www.academia.edu/175392427 , not informal--
discussions here.
i have to be selective who i spend my time on at this point, i don't
have personal time to waste on someone who spends more time responding
to me, than it would take to just read the 3rd page
Although "und" is called "the halting problem" it obviously isn't any
problem, just a specification of a computation.
and does not limit our ability to decide on any given machine, as we >>> -a> are not an addressable computing machine.
That is not relevant to this discussion about the section 1. Even if
the undecidability of und is literally the core justification for why
we can't build a total halting decider in turing machine computing,
while the church-turing thesis asserts we can only compute something
that turing machines can compute ... so therefor undecidability of und
the literally the core justification for why we aren't building total
halting deciders.
the section 1 is only a prelude to the main topic it should not give
the impression that the author cannot write anyting worth of reading.
prove ur not a retard dud, tell me:
does und() halt or loop forever?
as constructed, the halting paradox und() can neither be decided into
the set, nor it's complement, by the decider
i don't think u even know what turing machines are, to be frank
and therefore u have lost the proof of the existence for a machine that cannot be decided into a set
in order to prove a set "undecidable" you need an example of an object
that is cannot be decidable into that set or it's complement.
object would "undecidable". holy fuck. you can't just continually assert undecidability of a set, YOU NEED FUCKING PROOF OF AN OBJECT THAT CANNOT
BE DECIDED UPON
you don't actually have that with a hypothetical machine that DOES NOT FUCKING EXIST...
On 9/19/2026 2:19 AM, dart200 wrote:
as constructed, the halting paradox und() can neither be decided into
the set, nor it's complement, by the decider
False.-a Given algorithm halts1 and algorithm und1 which uses algorithm halts1, halts1(und1)==1 and und1 does not halt. Also, Ggven algorithm
halts0 and algorithm und0 which uses algorithm halts0, halts0(und0)==0
and und0 halts.-a So halts1 decides und1 wrong and halts0 decides und0 wrong.
i don't think u even know what turing machines are, to be frank
You're demonstrating quite clearly that it is you who don't understand
what turing machines are.
A Turing machine is essentially an algorithm, i.e. a fixed immutable
sequence of instructions, and is therefore defined by the instructions themselves, not where the instructions physically reside (i.e. the body
of the function halts()).
and therefore u have lost the proof of the existence for a machine
that cannot be decided into a set
That is not what the proof is.
The proof is that there is no single turing machine that can decide
whether any arbitrary turing machine with a given input will halt.
in order to prove a set "undecidable" you need an example of an object
that is cannot be decidable into that set or it's complement.
False.-a You need to show that there exists no algorithm that can
correctly place all objects in the correct set.
_that_
object would "undecidable". holy fuck. you can't just continually
assert undecidability of a set, YOU NEED FUCKING PROOF OF AN OBJECT
THAT CANNOT BE DECIDED UPON
you don't actually have that with a hypothetical machine that DOES NOT
FUCKING EXIST...
It seem you're failing to understand proof by contradiction, otherwise
you wouldn't be asserting this.
On 9/19/26 7:45 AM, dbush wrote:
On 9/19/2026 2:19 AM, dart200 wrote:
as constructed, the halting paradox und() can neither be decided into
the set, nor it's complement, by the decider
False.-a Given algorithm halts1 and algorithm und1 which uses algorithm
halts1, halts1(und1)==1 and und1 does not halt. Also, Ggven algorithm
halts0 and algorithm und0 which uses algorithm halts0, halts0(und0)==0
and und0 halts.-a So halts1 decides und1 wrong and halts0 decides und0
wrong.
i think u just supported my point by presenting two cases where a
machine failed to be decided correctly
i don't think u even know what turing machines are, to be frank
You're demonstrating quite clearly that it is you who don't understand
what turing machines are.
A Turing machine is essentially an algorithm, i.e. a fixed immutable
equating the turing machine model with the concept of an algorithm _is_
the church-turing thesis, which has not been proven, that my paper refutes
algorithms are more fundamental that turing machines, and not all
algorithms can be constructed into a turing machine
sequence of instructions, and is therefore defined by the instructions
themselves, not where the instructions physically reside (i.e. the
body of the function halts()).
and therefore u have lost the proof of the existence for a machine
that cannot be decided into a set
That is not what the proof is.
The proof is that there is no single turing machine that can decide
whether any arbitrary turing machine with a given input will halt.
in order to prove a set "undecidable" you need an example of an
object that is cannot be decidable into that set or it's complement.
False.-a You need to show that there exists no algorithm that can
correctly place all objects in the correct set.
which u do by presenting objects that cannot be decided, like you
literally just did with und1 and und0...
ofc then u also did correctly decide them regardless of what halts1 and halts0 returned, so what aglorithm did u use to do that dud?
...cause that's exactly what my paper is on...
_that_
object would "undecidable". holy fuck. you can't just continually
assert undecidability of a set, YOU NEED FUCKING PROOF OF AN OBJECT
THAT CANNOT BE DECIDED UPON
you don't actually have that with a hypothetical machine that DOES
NOT FUCKING EXIST...
It seem you're failing to understand proof by contradiction, otherwise
you wouldn't be asserting this.
ur failing to understand that refuted ur own claims with this very post
On 9/19/2026 1:57 PM, dart200 wrote:
On 9/19/26 7:45 AM, dbush wrote:
On 9/19/2026 2:19 AM, dart200 wrote:
as constructed, the halting paradox und() can neither be decided
into the set, nor it's complement, by the decider
False.-a Given algorithm halts1 and algorithm und1 which uses
algorithm halts1, halts1(und1)==1 and und1 does not halt. Also, Ggven
algorithm halts0 and algorithm und0 which uses algorithm halts0,
halts0(und0)==0 and und0 halts.-a So halts1 decides und1 wrong and
halts0 decides und0 wrong.
i think u just supported my point by presenting two cases where a
machine failed to be decided correctly
Nope.-a Two different machine decided differently by two different deciders.
i don't think u even know what turing machines are, to be frank
You're demonstrating quite clearly that it is you who don't
understand what turing machines are.
A Turing machine is essentially an algorithm, i.e. a fixed immutable
equating the turing machine model with the concept of an algorithm
_is_ the church-turing thesis, which has not been proven, that my
paper refutes
algorithms are more fundamental that turing machines, and not all
algorithms can be constructed into a turing machine
sequence of instructions, and is therefore defined by the
instructions themselves, not where the instructions physically reside
(i.e. the body of the function halts()).
and therefore u have lost the proof of the existence for a machine
that cannot be decided into a set
That is not what the proof is.
The proof is that there is no single turing machine that can decide
whether any arbitrary turing machine with a given input will halt.
in order to prove a set "undecidable" you need an example of an
object that is cannot be decidable into that set or it's complement.
False.-a You need to show that there exists no algorithm that can
correctly place all objects in the correct set.
which u do by presenting objects that cannot be decided, like you
literally just did with und1 and und0...
No, they are two separate objects which can be decided.-a Specifically,
und1 is correctly decided by a decider that just returns 0, and und0 is correctly decider by a decider that just returns 1.
ofc then u also did correctly decide them regardless of what halts1
and halts0 returned, so what aglorithm did u use to do that dud?
halts1 is one algorithm, halts0 is another algorithm.-a Just to keep
things simple, halts1 returns 1 in all cases and halts0 returns 0 in all cases.
i'm not sure you even know what "correctness" is my dud...
...cause that's exactly what my paper is on...
_that_
object would "undecidable". holy fuck. you can't just continually
assert undecidability of a set, YOU NEED FUCKING PROOF OF AN OBJECT
THAT CANNOT BE DECIDED UPON
you don't actually have that with a hypothetical machine that DOES
NOT FUCKING EXIST...
It seem you're failing to understand proof by contradiction,
otherwise you wouldn't be asserting this.
ur failing to understand that refuted ur own claims with this very post
No, you simply don't understand what a turing machine is.
On 9/19/26 11:17 AM, dbush wrote:
On 9/19/2026 1:57 PM, dart200 wrote:
On 9/19/26 7:45 AM, dbush wrote:
On 9/19/2026 2:19 AM, dart200 wrote:
as constructed, the halting paradox und() can neither be decided
into the set, nor it's complement, by the decider
False.-a Given algorithm halts1 and algorithm und1 which uses
algorithm halts1, halts1(und1)==1 and und1 does not halt. Also,
Ggven algorithm halts0 and algorithm und0 which uses algorithm
halts0, halts0(und0)==0 and und0 halts.-a So halts1 decides und1
wrong and halts0 decides und0 wrong.
i think u just supported my point by presenting two cases where a
machine failed to be decided correctly
Nope.-a Two different machine decided differently by two different
deciders.
i'm not sure how you think you can claim they are "deciders" if there's
no way to extract _truthful_ information from _any_ output,
are we just randomly labeling things now?
i don't think u even know what turing machines are, to be frank
You're demonstrating quite clearly that it is you who don't
understand what turing machines are.
A Turing machine is essentially an algorithm, i.e. a fixed immutable
equating the turing machine model with the concept of an algorithm
_is_ the church-turing thesis, which has not been proven, that my
paper refutes
algorithms are more fundamental that turing machines, and not all
algorithms can be constructed into a turing machine
sequence of instructions, and is therefore defined by the
instructions themselves, not where the instructions physically
reside (i.e. the body of the function halts()).
and therefore u have lost the proof of the existence for a machine
that cannot be decided into a set
That is not what the proof is.
The proof is that there is no single turing machine that can decide
whether any arbitrary turing machine with a given input will halt.
in order to prove a set "undecidable" you need an example of anFalse.-a You need to show that there exists no algorithm that can
object that is cannot be decidable into that set or it's complement. >>>>
correctly place all objects in the correct set.
which u do by presenting objects that cannot be decided, like you
literally just did with und1 and und0...
No, they are two separate objects which can be decided.-a Specifically,
und1 is correctly decided by a decider that just returns 0, and und0
is correctly decider by a decider that just returns 1.
i'm sorry, your definition of "correct" is an aglo that says a
terminating machine is non-terminating, and another which says a non- terminating machine is terminating ...
i honestly can't believe u just wrote that in all seriousness, might as
well start claiming 1=0 is "correct" at this point
ofc then u also did correctly decide them regardless of what halts1
and halts0 returned, so what aglorithm did u use to do that dud?
halts1 is one algorithm, halts0 is another algorithm.-a Just to keep
things simple, halts1 returns 1 in all cases and halts0 returns 0 in
all cases.
those are not genuine decision algorithms as _no_ information can be extracted from _any_ of the results, such that their output _cannot_ be further used in _any_ computation possible
they do not genuinely classify the input to any degree, ur just
mislabeling things to cover up for century old fallacy...
...two fallacies actually, one is covered in -o5, the next is covered in -o6
i'm not sure you even know what "correctness" is my dud...
...cause that's exactly what my paper is on...
_that_
object would "undecidable". holy fuck. you can't just continually
assert undecidability of a set, YOU NEED FUCKING PROOF OF AN OBJECT >>>>> THAT CANNOT BE DECIDED UPON
you don't actually have that with a hypothetical machine that DOES
NOT FUCKING EXIST...
It seem you're failing to understand proof by contradiction,
otherwise you wouldn't be asserting this.
ur failing to understand that refuted ur own claims with this very post
No, you simply don't understand what a turing machine is.
and u didn't respond to the fact that despite halt1 and halt0 being incorrect, u still classified und1 and und0 into sets truthfully.
what aglo did u use to do that
and why doesn't turing's proof prevent
you from doing that if all algorithms can be encapsulated by the turing machine model?
On 9/16/26 1:54 PM, Chris M. Thomasson wrote:
On 9/16/2026 9:26 AM, dart200 wrote:
On 9/16/26 12:40 AM, Mikko wrote:
On 15/09/2026 19:34, dart200 wrote:
On 9/15/26 1:18 AM, Mikko wrote:
On 14/09/2026 11:45, dart200 wrote:
On 9/14/26 1:26 AM, Mikko wrote:
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:
On 12/09/2026 19:02, dart200 wrote:"simple" is a weird term for something we provably can't build, >>>>>>>>> and that doesn't make the haltinging problem "decidable" since >>>>>>>>> we can't build, compute, or actually decide with the oracle
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computing
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit? >>>>>>>>>>>>> usp=sharing
The section 1 "the halting problem" should clearly define >>>>>>>>>>>> what the
expression "halting problem" means. It might be useful to first >>>>>>>>>>>> define "halting question" as that would simplify the
definitions of
"halting problem" and "halting decider". In each of these it is >>>>>>>>>>>> also inportant to be clear what is the scope of the problem and >>>>>>>>>>>> what kind of solutions are allowed. For example, the halting >>>>>>>>>>>> problem of finite state machines is Turing decidable and the >>>>>>>>>>>> halting
problem of Turing machines is decidable with a simple oracle. >>>>>>>>>
A "simple oracle" is an oracle that is simpler than most other >>>>>>>> oracles,
just like a "big mouse" is a mouse that is bigger than most
other mice
although not as big as a little elephant.
like i said: recursive undecidability leading to the concept of >>>>>>> "simple" vs "complex" oracles is discussing in -o4, i will not be >>>>>>> discussing them in -o1
That's OK, although there should be a short description of each
section
at the end of the introduction.
But the section 1 gives the impression that the author does not know >>>>>> what the halting problem is.
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the >>>>>>>>>>>> program "und"
is not undecidable: there are partial halt deciders that can >>>>>>>>>>>> tell
whether "und" halts. In order to make this clear as soon as >>>>>>>>>>>> possible
it might be better to define "partial halt decider" before the >>>>>>>>>>>> introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
that's why partial deciders are discussed in -o5.2 not -o1, you >>>>>>> brought those up not me
The section 1 is about the halting problem. You may discuss
partial and
total solutions of the problem in the same section or elsewhere. But >>>>>> it is important to point out about "und" that its halting is
determinable. Otherwise a reader could be confused or get the
impression
that you are don't know.
und as specified in -o1 does not even exist, an argument which takes >>>>> up to the end of -o4 to explain
The section 1 says otherwise: the words "can be constructed" mean that >>>> it does exist.
it exist as a hypothetical problem, constructed from a hypothetical
decider, that presents a real limit to turing machine computability,
but does not exist in the actual enumeration of turing machines, and
does not limit our ability to decide on any given machine, as we are
not an addressable computing machine.
that is the point of the entire paper, and it takes 26 pages to
explain why. i'm not changing that specific wording there based on
confusion that will not resolve until you actually read the paper dud
Can you decide random numbers? Say a black box in isolation. You ask
it to run a process, when you wait for a signal that the process is
done. Well, do you get a signal or not? Say the black box takes some
results from a TRNG. Sometimes it halts after some random time.
Sometimes it does not halt. How does your system account for the black
box?
the turing machine model does not allow for randomness and is _strictly_ deterministic
it's kinda nuts u don't know a rather basic fact of computing theory,
but judging from the quality of ur comments over the year i've been
posting it doesn't surprise me
the turing machine model does not allow for randomness and is _strictly_ deterministic
it's kinda nuts u don't know a rather basic fact of computing theory,
but judging from the quality of ur comments over the year i've been
posting it doesn't surprise me
On 9/16/2026 6:12 PM, dart200 wrote:
On 9/16/26 1:54 PM, Chris M. Thomasson wrote:
On 9/16/2026 9:26 AM, dart200 wrote:
On 9/16/26 12:40 AM, Mikko wrote:
On 15/09/2026 19:34, dart200 wrote:
On 9/15/26 1:18 AM, Mikko wrote:
On 14/09/2026 11:45, dart200 wrote:
On 9/14/26 1:26 AM, Mikko wrote:
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:
On 12/09/2026 19:02, dart200 wrote:"simple" is a weird term for something we provably can't
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computing
and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit? >>>>>>>>>>>>>> usp=sharing
The section 1 "the halting problem" should clearly define >>>>>>>>>>>>> what the
expression "halting problem" means. It might be useful to >>>>>>>>>>>>> first
define "halting question" as that would simplify the >>>>>>>>>>>>> definitions of
"halting problem" and "halting decider". In each of these >>>>>>>>>>>>> it is
also inportant to be clear what is the scope of the problem >>>>>>>>>>>>> and
what kind of solutions are allowed. For example, the halting >>>>>>>>>>>>> problem of finite state machines is Turing decidable and >>>>>>>>>>>>> the halting
problem of Turing machines is decidable with a simple oracle. >>>>>>>>>>
build, and that doesn't make the haltinging problem
"decidable" since we can't build, compute, or actually decide >>>>>>>>>> with the oracle
A "simple oracle" is an oracle that is simpler than most other >>>>>>>>> oracles,
just like a "big mouse" is a mouse that is bigger than most >>>>>>>>> other mice
although not as big as a little elephant.
like i said: recursive undecidability leading to the concept of >>>>>>>> "simple" vs "complex" oracles is discussing in -o4, i will not be >>>>>>>> discussing them in -o1
That's OK, although there should be a short description of each >>>>>>> section
at the end of the introduction.
But the section 1 gives the impression that the author does not know >>>>>>> what the halting problem is.
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the >>>>>>>>>>>>> program "und"
is not undecidable: there are partial halt deciders that >>>>>>>>>>>>> can tell
whether "und" halts. In order to make this clear as soon as >>>>>>>>>>>>> possible
it might be better to define "partial halt decider" before the >>>>>>>>>>>>> introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
that's why partial deciders are discussed in -o5.2 not -o1, you >>>>>>>> brought those up not me
The section 1 is about the halting problem. You may discuss
partial and
total solutions of the problem in the same section or elsewhere. But >>>>>>> it is important to point out about "und" that its halting is
determinable. Otherwise a reader could be confused or get the
impression
that you are don't know.
und as specified in -o1 does not even exist, an argument which
takes up to the end of -o4 to explain
The section 1 says otherwise: the words "can be constructed" mean that >>>>> it does exist.
it exist as a hypothetical problem, constructed from a hypothetical
decider, that presents a real limit to turing machine computability,
but does not exist in the actual enumeration of turing machines, and
does not limit our ability to decide on any given machine, as we are
not an addressable computing machine.
that is the point of the entire paper, and it takes 26 pages to
explain why. i'm not changing that specific wording there based on
confusion that will not resolve until you actually read the paper dud
Can you decide random numbers? Say a black box in isolation. You ask
it to run a process, when you wait for a signal that the process is
done. Well, do you get a signal or not? Say the black box takes some
results from a TRNG. Sometimes it halts after some random time.
Sometimes it does not halt. How does your system account for the
black box?
the turing machine model does not allow for randomness and is
_strictly_ deterministic
it's kinda nuts u don't know a rather basic fact of computing theory,
but judging from the quality of ur comments over the year i've been
posting it doesn't surprise me
Say the guy who can solve the halting problem? ;^o Actually, are you
trying to do that?
If not, wtf! You remind me of PO. Sorry for that cut.
On 9/19/2026 2:40 PM, dart200 wrote:
On 9/19/26 11:17 AM, dbush wrote:
On 9/19/2026 1:57 PM, dart200 wrote:
On 9/19/26 7:45 AM, dbush wrote:
On 9/19/2026 2:19 AM, dart200 wrote:
as constructed, the halting paradox und() can neither be decided
into the set, nor it's complement, by the decider
False.-a Given algorithm halts1 and algorithm und1 which uses
algorithm halts1, halts1(und1)==1 and und1 does not halt. Also,
Ggven algorithm halts0 and algorithm und0 which uses algorithm
halts0, halts0(und0)==0 and und0 halts.-a So halts1 decides und1
wrong and halts0 decides und0 wrong.
i think u just supported my point by presenting two cases where a
machine failed to be decided correctly
Nope.-a Two different machine decided differently by two different
deciders.
i'm not sure how you think you can claim they are "deciders" if
there's no way to extract _truthful_ information from _any_ output,
are we just randomly labeling things now?
If an algorithm returns 0 or 1 for all inputs, they are a decider.
Whether they are correct for that input depends on whether the result matches the mathematical mapping they're trying to reproduce.
equating the turing machine model with the concept of an algorithm
i don't think u even know what turing machines are, to be frank
You're demonstrating quite clearly that it is you who don't
understand what turing machines are.
A Turing machine is essentially an algorithm, i.e. a fixed immutable >>>>
_is_ the church-turing thesis, which has not been proven, that my
paper refutes
algorithms are more fundamental that turing machines, and not all
algorithms can be constructed into a turing machine
sequence of instructions, and is therefore defined by the
instructions themselves, not where the instructions physically
reside (i.e. the body of the function halts()).
and therefore u have lost the proof of the existence for a machine >>>>>> that cannot be decided into a set
That is not what the proof is.
The proof is that there is no single turing machine that can decide >>>>> whether any arbitrary turing machine with a given input will halt.
in order to prove a set "undecidable" you need an example of anFalse.-a You need to show that there exists no algorithm that can
object that is cannot be decidable into that set or it's complement. >>>>>
correctly place all objects in the correct set.
which u do by presenting objects that cannot be decided, like you
literally just did with und1 and und0...
No, they are two separate objects which can be decided.
Specifically, und1 is correctly decided by a decider that just
returns 0, and und0 is correctly decider by a decider that just
returns 1.
i'm sorry, your definition of "correct" is an aglo that says a
terminating machine is non-terminating, and another which says a non-
terminating machine is terminating ...
Read it again.-a Algorithm und1 does not halt, so a decider that returns
0 when given und1 as input is correct for that input.-a Similarly,
algorithm und0 does halt, so a decider that returns 1 when given und0 as input is correct for that input.
i honestly can't believe u just wrote that in all seriousness, might
as well start claiming 1=0 is "correct" at this point
ofc then u also did correctly decide them regardless of what halts1
and halts0 returned, so what aglorithm did u use to do that dud?
halts1 is one algorithm, halts0 is another algorithm.-a Just to keep
things simple, halts1 returns 1 in all cases and halts0 returns 0 in
all cases.
those are not genuine decision algorithms as _no_ information can be
extracted from _any_ of the results, such that their output _cannot_
be further used in _any_ computation possible
Sure they are.-a They return either 0 or 1 for all inputs.-a Whether they are correct for a given input depends on whether the return matches the mathematical mapping they're trying to reproduce.
they do not genuinely classify the input to any degree, ur just
mislabeling things to cover up for century old fallacy...
They do classify.-a They just don't correct map the halting function for
all inputs (and in fact no algorithm does)
...two fallacies actually, one is covered in -o5, the next is covered
in -o6
i'm not sure you even know what "correctness" is my dud...
...cause that's exactly what my paper is on...
_that_
object would "undecidable". holy fuck. you can't just continually >>>>>> assert undecidability of a set, YOU NEED FUCKING PROOF OF AN
OBJECT THAT CANNOT BE DECIDED UPON
you don't actually have that with a hypothetical machine that DOES >>>>>> NOT FUCKING EXIST...
It seem you're failing to understand proof by contradiction,
otherwise you wouldn't be asserting this.
ur failing to understand that refuted ur own claims with this very post >>>>
No, you simply don't understand what a turing machine is.
and u didn't respond to the fact that despite halt1 and halt0 being
incorrect, u still classified und1 and und0 into sets truthfully.
what aglo did u use to do that
Not one but multiple.-a und1 is correctly decided by algorithm halt0 and und0 is correctly decider by algorithm halt1.
and why doesn't turing's proof prevent you from doing that if all
algorithms can be encapsulated by the turing machine model?
Turing's proof shows that no single algorithm can map the halting function.-a It doesn't say anything about multiple algorithms.
On 9/19/26 12:13 PM, dbush wrote:
On 9/19/2026 2:40 PM, dart200 wrote:
On 9/19/26 11:17 AM, dbush wrote:
On 9/19/2026 1:57 PM, dart200 wrote:
On 9/19/26 7:45 AM, dbush wrote:
On 9/19/2026 2:19 AM, dart200 wrote:
as constructed, the halting paradox und() can neither be decided >>>>>>> into the set, nor it's complement, by the decider
False.-a Given algorithm halts1 and algorithm und1 which uses
algorithm halts1, halts1(und1)==1 and und1 does not halt. Also,
Ggven algorithm halts0 and algorithm und0 which uses algorithm
halts0, halts0(und0)==0 and und0 halts.-a So halts1 decides und1
wrong and halts0 decides und0 wrong.
i think u just supported my point by presenting two cases where a
machine failed to be decided correctly
Nope.-a Two different machine decided differently by two different
deciders.
i'm not sure how you think you can claim they are "deciders" if
there's no way to extract _truthful_ information from _any_ output,
are we just randomly labeling things now?
If an algorithm returns 0 or 1 for all inputs, they are a decider.
Whether they are correct for that input depends on whether the result
matches the mathematical mapping they're trying to reproduce.
equating the turing machine model with the concept of an algorithm
i don't think u even know what turing machines are, to be frank
You're demonstrating quite clearly that it is you who don't
understand what turing machines are.
A Turing machine is essentially an algorithm, i.e. a fixed immutable >>>>>
_is_ the church-turing thesis, which has not been proven, that my
paper refutes
algorithms are more fundamental that turing machines, and not all
algorithms can be constructed into a turing machine
sequence of instructions, and is therefore defined by the
instructions themselves, not where the instructions physically
reside (i.e. the body of the function halts()).
and therefore u have lost the proof of the existence for a
machine that cannot be decided into a set
That is not what the proof is.
The proof is that there is no single turing machine that can
decide whether any arbitrary turing machine with a given input
will halt.
in order to prove a set "undecidable" you need an example of an >>>>>>> object that is cannot be decidable into that set or it's complement. >>>>>>False.-a You need to show that there exists no algorithm that can >>>>>> correctly place all objects in the correct set.
which u do by presenting objects that cannot be decided, like you
literally just did with und1 and und0...
No, they are two separate objects which can be decided.
Specifically, und1 is correctly decided by a decider that just
returns 0, and und0 is correctly decider by a decider that just
returns 1.
i'm sorry, your definition of "correct" is an aglo that says a
terminating machine is non-terminating, and another which says a non-
terminating machine is terminating ...
Read it again.-a Algorithm und1 does not halt, so a decider that
returns 0 when given und1 as input is correct for that input.
Similarly, algorithm und0 does halt, so a decider that returns 1 when
given und0 as input is correct for that input.
i honestly can't believe u just wrote that in all seriousness, might
as well start claiming 1=0 is "correct" at this point
ofc then u also did correctly decide them regardless of what halts1 >>>>> and halts0 returned, so what aglorithm did u use to do that dud?
halts1 is one algorithm, halts0 is another algorithm.-a Just to keep
things simple, halts1 returns 1 in all cases and halts0 returns 0 in
all cases.
those are not genuine decision algorithms as _no_ information can be
extracted from _any_ of the results, such that their output _cannot_
be further used in _any_ computation possible
Sure they are.-a They return either 0 or 1 for all inputs.-a Whether
they are correct for a given input depends on whether the return
matches the mathematical mapping they're trying to reproduce.
they do not genuinely classify the input to any degree, ur just
mislabeling things to cover up for century old fallacy...
They do classify.-a They just don't correct map the halting function for
so you are actually claiming that broken clock "correctly" tells time because it's correct twice a day???
i don't know how to respond to that level of brainrot dud, it's clearly
not mathematically sound
all inputs (and in fact no algorithm does)
...two fallacies actually, one is covered in -o5, the next is covered
in -o6
i'm not sure you even know what "correctness" is my dud...
...cause that's exactly what my paper is on...
_that_
object would "undecidable". holy fuck. you can't just continually >>>>>>> assert undecidability of a set, YOU NEED FUCKING PROOF OF AN
OBJECT THAT CANNOT BE DECIDED UPON
you don't actually have that with a hypothetical machine that
DOES NOT FUCKING EXIST...
It seem you're failing to understand proof by contradiction,
otherwise you wouldn't be asserting this.
ur failing to understand that refuted ur own claims with this very
post
No, you simply don't understand what a turing machine is.
and u didn't respond to the fact that despite halt1 and halt0 being
incorrect, u still classified und1 and und0 into sets truthfully.
what aglo did u use to do that
Not one but multiple.-a und1 is correctly decided by algorithm halt0
and und0 is correctly decider by algorithm halt1.
yes dud, by which algorithm did you use to combine those in order to
provide actually useful knowledge on what those machines objectively do ...
and why doesn't turing's proof prevent you from doing that if all
algorithms can be encapsulated by the turing machine model?
Turing's proof shows that no single algorithm can map the halting
function.-a It doesn't say anything about multiple algorithms.
adjacent classifiers are covered in -o7.1 of the revised version
On 9/19/2026 6:40 PM, dart200 wrote:
On 9/19/26 12:13 PM, dbush wrote:
On 9/19/2026 2:40 PM, dart200 wrote:so you are actually claiming that broken clock "correctly" tells time
On 9/19/26 11:17 AM, dbush wrote:
On 9/19/2026 1:57 PM, dart200 wrote:
On 9/19/26 7:45 AM, dbush wrote:
On 9/19/2026 2:19 AM, dart200 wrote:
as constructed, the halting paradox und() can neither be decided >>>>>>>> into the set, nor it's complement, by the decider
False.-a Given algorithm halts1 and algorithm und1 which uses
algorithm halts1, halts1(und1)==1 and und1 does not halt. Also, >>>>>>> Ggven algorithm halts0 and algorithm und0 which uses algorithm
halts0, halts0(und0)==0 and und0 halts.-a So halts1 decides und1 >>>>>>> wrong and halts0 decides und0 wrong.
i think u just supported my point by presenting two cases where a >>>>>> machine failed to be decided correctly
Nope.-a Two different machine decided differently by two different
deciders.
i'm not sure how you think you can claim they are "deciders" if
there's no way to extract _truthful_ information from _any_ output,
are we just randomly labeling things now?
If an algorithm returns 0 or 1 for all inputs, they are a decider.
Whether they are correct for that input depends on whether the result
matches the mathematical mapping they're trying to reproduce.
equating the turing machine model with the concept of an algorithm >>>>>> _is_ the church-turing thesis, which has not been proven, that my >>>>>> paper refutes
i don't think u even know what turing machines are, to be frank >>>>>>>You're demonstrating quite clearly that it is you who don't
understand what turing machines are.
A Turing machine is essentially an algorithm, i.e. a fixed immutable >>>>>>
algorithms are more fundamental that turing machines, and not all >>>>>> algorithms can be constructed into a turing machine
sequence of instructions, and is therefore defined by the
instructions themselves, not where the instructions physically
reside (i.e. the body of the function halts()).
and therefore u have lost the proof of the existence for a
machine that cannot be decided into a set
That is not what the proof is.
The proof is that there is no single turing machine that can
decide whether any arbitrary turing machine with a given input
will halt.
in order to prove a set "undecidable" you need an example of an >>>>>>>> object that is cannot be decidable into that set or it's
complement.
False.-a You need to show that there exists no algorithm that can >>>>>>> correctly place all objects in the correct set.
which u do by presenting objects that cannot be decided, like you >>>>>> literally just did with und1 and und0...
No, they are two separate objects which can be decided.
Specifically, und1 is correctly decided by a decider that just
returns 0, and und0 is correctly decider by a decider that just
returns 1.
i'm sorry, your definition of "correct" is an aglo that says a
terminating machine is non-terminating, and another which says a
non- terminating machine is terminating ...
Read it again.-a Algorithm und1 does not halt, so a decider that
returns 0 when given und1 as input is correct for that input.
Similarly, algorithm und0 does halt, so a decider that returns 1 when
given und0 as input is correct for that input.
i honestly can't believe u just wrote that in all seriousness, might
as well start claiming 1=0 is "correct" at this point
ofc then u also did correctly decide them regardless of what
halts1 and halts0 returned, so what aglorithm did u use to do that >>>>>> dud?
halts1 is one algorithm, halts0 is another algorithm.-a Just to keep >>>>> things simple, halts1 returns 1 in all cases and halts0 returns 0
in all cases.
those are not genuine decision algorithms as _no_ information can be
extracted from _any_ of the results, such that their output _cannot_
be further used in _any_ computation possible
Sure they are.-a They return either 0 or 1 for all inputs.-a Whether
they are correct for a given input depends on whether the return
matches the mathematical mapping they're trying to reproduce.
they do not genuinely classify the input to any degree, ur just
mislabeling things to cover up for century old fallacy...
They do classify.-a They just don't correct map the halting function for >>
because it's correct twice a day???
It correctly tells time for some subset of times.
If a broken analog clock is stuck at 4:45, and the current time is 4:45, that clock is correct at that time.
It doesn't matter "why" it is correct, just that it is.
In the same way, a prospective halt decider that always returns 0
correctly reports the halt status of all non-halting computations. "How"
it got that result is irrelevant.
i don't know how to respond to that level of brainrot dud, it's
clearly not mathematically sound
If an algorithm maps a specific input to an output that matches a mathematical function, it correctly maps that value for that function.
"How" is irrelevant.
all inputs (and in fact no algorithm does)
...two fallacies actually, one is covered in -o5, the next is covered >>>> in -o6
i'm not sure you even know what "correctness" is my dud...
...cause that's exactly what my paper is on...
_that_
object would "undecidable". holy fuck. you can't just
continually assert undecidability of a set, YOU NEED FUCKING
PROOF OF AN OBJECT THAT CANNOT BE DECIDED UPON
you don't actually have that with a hypothetical machine that >>>>>>>> DOES NOT FUCKING EXIST...
It seem you're failing to understand proof by contradiction,
otherwise you wouldn't be asserting this.
ur failing to understand that refuted ur own claims with this very >>>>>> post
No, you simply don't understand what a turing machine is.
and u didn't respond to the fact that despite halt1 and halt0 being
incorrect, u still classified und1 and und0 into sets truthfully.
what aglo did u use to do that
Not one but multiple.-a und1 is correctly decided by algorithm halt0
and und0 is correctly decider by algorithm halt1.
yes dud, by which algorithm did you use to combine those in order to
provide actually useful knowledge on what those machines objectively
do ...
It's not about "useful knowledge".-a It's about whether those algorithms
map the halting function for specific inputs.
But for the sake of argument let's say I created some algorithm, i.e. a fixed immutable sequence of instructions, that we'll call halts2 that
uses *only* a machine description and that machine's input to determine whether that machine with the given input will halt, such that
halts2(und0) report 1 and halts2(und1) returns 0.-a We can then construct
an algorithm und2 based on algorithm halts2 that either halts or does
not halt and halts2(und2) returns the incorrect answer, and either halts0(und2) or halts1(und2) will be correct.
What turing's proof shows is that any algorithm that attempts to map the halting function will have at least one case it gets wrong.
and why doesn't turing's proof prevent you from doing that if all
algorithms can be encapsulated by the turing machine model?
Turing's proof shows that no single algorithm can map the halting
function.-a It doesn't say anything about multiple algorithms.
adjacent classifiers are covered in -o7.1 of the revised version
On 9/19/26 4:27 PM, dbush wrote:
On 9/19/2026 6:40 PM, dart200 wrote:
On 9/19/26 12:13 PM, dbush wrote:
On 9/19/2026 2:40 PM, dart200 wrote:so you are actually claiming that broken clock "correctly" tells time
On 9/19/26 11:17 AM, dbush wrote:
On 9/19/2026 1:57 PM, dart200 wrote:
On 9/19/26 7:45 AM, dbush wrote:
On 9/19/2026 2:19 AM, dart200 wrote:
as constructed, the halting paradox und() can neither be
decided into the set, nor it's complement, by the decider
False.-a Given algorithm halts1 and algorithm und1 which uses >>>>>>>> algorithm halts1, halts1(und1)==1 and und1 does not halt. Also, >>>>>>>> Ggven algorithm halts0 and algorithm und0 which uses algorithm >>>>>>>> halts0, halts0(und0)==0 and und0 halts.-a So halts1 decides und1 >>>>>>>> wrong and halts0 decides und0 wrong.
i think u just supported my point by presenting two cases where a >>>>>>> machine failed to be decided correctly
Nope.-a Two different machine decided differently by two different >>>>>> deciders.
i'm not sure how you think you can claim they are "deciders" if
there's no way to extract _truthful_ information from _any_ output,
are we just randomly labeling things now?
If an algorithm returns 0 or 1 for all inputs, they are a decider.
Whether they are correct for that input depends on whether the
result matches the mathematical mapping they're trying to reproduce.
i don't think u even know what turing machines are, to be frank >>>>>>>>You're demonstrating quite clearly that it is you who don't
understand what turing machines are.
A Turing machine is essentially an algorithm, i.e. a fixed
immutable
equating the turing machine model with the concept of an
algorithm _is_ the church-turing thesis, which has not been
proven, that my paper refutes
algorithms are more fundamental that turing machines, and not all >>>>>>> algorithms can be constructed into a turing machine
sequence of instructions, and is therefore defined by the
instructions themselves, not where the instructions physically >>>>>>>> reside (i.e. the body of the function halts()).
and therefore u have lost the proof of the existence for a
machine that cannot be decided into a set
That is not what the proof is.
The proof is that there is no single turing machine that can
decide whether any arbitrary turing machine with a given input >>>>>>>> will halt.
in order to prove a set "undecidable" you need an example of an >>>>>>>>> object that is cannot be decidable into that set or it's
complement.
False.-a You need to show that there exists no algorithm that can >>>>>>>> correctly place all objects in the correct set.
which u do by presenting objects that cannot be decided, like you >>>>>>> literally just did with und1 and und0...
No, they are two separate objects which can be decided.
Specifically, und1 is correctly decided by a decider that just
returns 0, and und0 is correctly decider by a decider that just
returns 1.
i'm sorry, your definition of "correct" is an aglo that says a
terminating machine is non-terminating, and another which says a
non- terminating machine is terminating ...
Read it again.-a Algorithm und1 does not halt, so a decider that
returns 0 when given und1 as input is correct for that input.
Similarly, algorithm und0 does halt, so a decider that returns 1
when given und0 as input is correct for that input.
i honestly can't believe u just wrote that in all seriousness,
might as well start claiming 1=0 is "correct" at this point
ofc then u also did correctly decide them regardless of what
halts1 and halts0 returned, so what aglorithm did u use to do
that dud?
halts1 is one algorithm, halts0 is another algorithm.-a Just to
keep things simple, halts1 returns 1 in all cases and halts0
returns 0 in all cases.
those are not genuine decision algorithms as _no_ information can
be extracted from _any_ of the results, such that their output
_cannot_ be further used in _any_ computation possible
Sure they are.-a They return either 0 or 1 for all inputs.-a Whether
they are correct for a given input depends on whether the return
matches the mathematical mapping they're trying to reproduce.
they do not genuinely classify the input to any degree, ur just
mislabeling things to cover up for century old fallacy...
They do classify.-a They just don't correct map the halting function for >>>
because it's correct twice a day???
It correctly tells time for some subset of times.
wow u really are gunna double down on that nonsense, eh? that much time
on ur hands??
If a broken analog clock is stuck at 4:45, and the current time is
4:45, that clock is correct at that time.
It doesn't matter "why" it is correct, just that it is.
bro an algorithm that randomly produces a correct response at times
without a specific reason why is just not a "correct" algorithm by any meaningful notion of "correct"
maybe ur a set theorist dud instead of computing guy, and don't care
about trivial concerns like the ability to extract useful meaning from
the response... but in computing we actually care about practical
matters, and you can't get that from constant return functions, lol.
a constant return function is _not_ a decider, you calling it a decider
is just useless stupidity, else a constant return function is apparently
a decider for _all_ semantic decision problems ever ... which is just crazy
i will not debate this further because it's not worth my time
In the same way, a prospective halt decider that always returns 0
correctly reports the halt status of all non-halting computations.
"How" it got that result is irrelevant.
i don't know how to respond to that level of brainrot dud, it's
clearly not mathematically sound
If an algorithm maps a specific input to an output that matches a
mathematical function, it correctly maps that value for that function.
"How" is irrelevant.
all inputs (and in fact no algorithm does)
...two fallacies actually, one is covered in -o5, the next is
covered in -o6
i'm not sure you even know what "correctness" is my dud...
...cause that's exactly what my paper is on...
_that_
object would "undecidable". holy fuck. you can't just
continually assert undecidability of a set, YOU NEED FUCKING >>>>>>>>> PROOF OF AN OBJECT THAT CANNOT BE DECIDED UPON
you don't actually have that with a hypothetical machine that >>>>>>>>> DOES NOT FUCKING EXIST...
It seem you're failing to understand proof by contradiction,
otherwise you wouldn't be asserting this.
ur failing to understand that refuted ur own claims with this
very post
No, you simply don't understand what a turing machine is.
and u didn't respond to the fact that despite halt1 and halt0 being >>>>> incorrect, u still classified und1 and und0 into sets truthfully.
what aglo did u use to do that
Not one but multiple.-a und1 is correctly decided by algorithm halt0
and und0 is correctly decider by algorithm halt1.
yes dud, by which algorithm did you use to combine those in order to
provide actually useful knowledge on what those machines objectively
do ...
It's not about "useful knowledge".-a It's about whether those
algorithms map the halting function for specific inputs.
But for the sake of argument let's say I created some algorithm, i.e.
a fixed immutable sequence of instructions, that we'll call halts2
that uses *only* a machine description and that machine's input to
determine whether that machine with the given input will halt, such
that halts2(und0) report 1 and halts2(und1) returns 0.-a We can then
construct an algorithm und2 based on algorithm halts2 that either
halts or does not halt and halts2(und2) returns the incorrect answer,
and either halts0(und2) or halts1(und2) will be correct.
What turing's proof shows is that any algorithm that attempts to map
the halting function will have at least one case it gets wrong.
good glad we've gotten to what was described on the 2nd page of my 26
page paper
heck if u even read the 1st page of it, u'd know i'm responding to a published paper (from jan of this year) claiming very specifically it's incorrect to attribute the halting problem to turing
but we both know u have no intention of reading it, so whatever dud
and why doesn't turing's proof prevent you from doing that if all
algorithms can be encapsulated by the turing machine model?
Turing's proof shows that no single algorithm can map the halting
function.-a It doesn't say anything about multiple algorithms.
adjacent classifiers are covered in -o7.1 of the revised version
On 9/19/2026 7:02 PM, Dude wrote:
[...]
Dart reminds me of PO's old dribble. That is a horrible thing to say,
but well... Shit.
On 9/19/26 4:27 PM, dbush wrote:
On 9/19/2026 6:40 PM, dart200 wrote:
On 9/19/26 12:13 PM, dbush wrote:
On 9/19/2026 2:40 PM, dart200 wrote:so you are actually claiming that broken clock "correctly" tells time
On 9/19/26 11:17 AM, dbush wrote:
On 9/19/2026 1:57 PM, dart200 wrote:
On 9/19/26 7:45 AM, dbush wrote:
On 9/19/2026 2:19 AM, dart200 wrote:
as constructed, the halting paradox und() can neither be
decided into the set, nor it's complement, by the decider
False.-a Given algorithm halts1 and algorithm und1 which uses >>>>>>>> algorithm halts1, halts1(und1)==1 and und1 does not halt. Also, >>>>>>>> Ggven algorithm halts0 and algorithm und0 which uses algorithm >>>>>>>> halts0, halts0(und0)==0 and und0 halts.-a So halts1 decides und1 >>>>>>>> wrong and halts0 decides und0 wrong.
i think u just supported my point by presenting two cases where a >>>>>>> machine failed to be decided correctly
Nope.-a Two different machine decided differently by two different >>>>>> deciders.
i'm not sure how you think you can claim they are "deciders" if
there's no way to extract _truthful_ information from _any_ output,
are we just randomly labeling things now?
If an algorithm returns 0 or 1 for all inputs, they are a decider.
Whether they are correct for that input depends on whether the
result matches the mathematical mapping they're trying to reproduce.
i don't think u even know what turing machines are, to be frank >>>>>>>>You're demonstrating quite clearly that it is you who don't
understand what turing machines are.
A Turing machine is essentially an algorithm, i.e. a fixed
immutable
equating the turing machine model with the concept of an
algorithm _is_ the church-turing thesis, which has not been
proven, that my paper refutes
algorithms are more fundamental that turing machines, and not all >>>>>>> algorithms can be constructed into a turing machine
sequence of instructions, and is therefore defined by the
instructions themselves, not where the instructions physically >>>>>>>> reside (i.e. the body of the function halts()).
and therefore u have lost the proof of the existence for a
machine that cannot be decided into a set
That is not what the proof is.
The proof is that there is no single turing machine that can
decide whether any arbitrary turing machine with a given input >>>>>>>> will halt.
in order to prove a set "undecidable" you need an example of an >>>>>>>>> object that is cannot be decidable into that set or it's
complement.
False.-a You need to show that there exists no algorithm that can >>>>>>>> correctly place all objects in the correct set.
which u do by presenting objects that cannot be decided, like you >>>>>>> literally just did with und1 and und0...
No, they are two separate objects which can be decided.
Specifically, und1 is correctly decided by a decider that just
returns 0, and und0 is correctly decider by a decider that just
returns 1.
i'm sorry, your definition of "correct" is an aglo that says a
terminating machine is non-terminating, and another which says a
non- terminating machine is terminating ...
Read it again.-a Algorithm und1 does not halt, so a decider that
returns 0 when given und1 as input is correct for that input.
Similarly, algorithm und0 does halt, so a decider that returns 1
when given und0 as input is correct for that input.
i honestly can't believe u just wrote that in all seriousness,
might as well start claiming 1=0 is "correct" at this point
ofc then u also did correctly decide them regardless of what
halts1 and halts0 returned, so what aglorithm did u use to do
that dud?
halts1 is one algorithm, halts0 is another algorithm.-a Just to
keep things simple, halts1 returns 1 in all cases and halts0
returns 0 in all cases.
those are not genuine decision algorithms as _no_ information can
be extracted from _any_ of the results, such that their output
_cannot_ be further used in _any_ computation possible
Sure they are.-a They return either 0 or 1 for all inputs.-a Whether
they are correct for a given input depends on whether the return
matches the mathematical mapping they're trying to reproduce.
they do not genuinely classify the input to any degree, ur just
mislabeling things to cover up for century old fallacy...
They do classify.-a They just don't correct map the halting function for >>>
because it's correct twice a day???
It correctly tells time for some subset of times.
wow u really are gunna double down on that nonsense, eh? that much time
on ur hands??
If a broken analog clock is stuck at 4:45, and the current time is
4:45, that clock is correct at that time.
It doesn't matter "why" it is correct, just that it is.
bro an algorithm that randomly produces a correct response at times
without a specific reason why is just not a "correct" algorithm by any meaningful notion of "correct"
maybe ur a set theorist dud instead of computing guy, and don't care
about trivial concerns like the ability to extract useful meaning from
the response... but in computing we actually care about practical
matters, and you can't get that from constant return functions, lol.
a constant return function is _not_ a decider, you calling it a decider
is just useless stupidity, else a constant return function is apparently
a decider for _all_ semantic decision problems ever ... which is just crazy
i will not debate this further because it's not worth my time
In the same way, a prospective halt decider that always returns 0
correctly reports the halt status of all non-halting computations.
"How" it got that result is irrelevant.
i don't know how to respond to that level of brainrot dud, it's
clearly not mathematically sound
If an algorithm maps a specific input to an output that matches a
mathematical function, it correctly maps that value for that function.
"How" is irrelevant.
all inputs (and in fact no algorithm does)
...two fallacies actually, one is covered in -o5, the next is
covered in -o6
i'm not sure you even know what "correctness" is my dud...
...cause that's exactly what my paper is on...
_that_
object would "undecidable". holy fuck. you can't just
continually assert undecidability of a set, YOU NEED FUCKING >>>>>>>>> PROOF OF AN OBJECT THAT CANNOT BE DECIDED UPON
you don't actually have that with a hypothetical machine that >>>>>>>>> DOES NOT FUCKING EXIST...
It seem you're failing to understand proof by contradiction,
otherwise you wouldn't be asserting this.
ur failing to understand that refuted ur own claims with this
very post
No, you simply don't understand what a turing machine is.
and u didn't respond to the fact that despite halt1 and halt0 being >>>>> incorrect, u still classified und1 and und0 into sets truthfully.
what aglo did u use to do that
Not one but multiple.-a und1 is correctly decided by algorithm halt0
and und0 is correctly decider by algorithm halt1.
yes dud, by which algorithm did you use to combine those in order to
provide actually useful knowledge on what those machines objectively
do ...
It's not about "useful knowledge".-a It's about whether those
algorithms map the halting function for specific inputs.
But for the sake of argument let's say I created some algorithm, i.e.
a fixed immutable sequence of instructions, that we'll call halts2
that uses *only* a machine description and that machine's input to
determine whether that machine with the given input will halt, such
that halts2(und0) report 1 and halts2(und1) returns 0.-a We can then
construct an algorithm und2 based on algorithm halts2 that either
halts or does not halt and halts2(und2) returns the incorrect answer,
and either halts0(und2) or halts1(und2) will be correct.
What turing's proof shows is that any algorithm that attempts to map
the halting function will have at least one case it gets wrong.
good glad we've gotten to what was described on the 2nd page of my 26
page paper
heck if u even read the 1st page of it, u'd know i'm responding to a published paper (from jan of this year) claiming very specifically it's incorrect to attribute the halting problem to turing
but we both know u have no intention of reading it, so whatever dud
and why doesn't turing's proof prevent you from doing that if all
algorithms can be encapsulated by the turing machine model?
Turing's proof shows that no single algorithm can map the halting
function.-a It doesn't say anything about multiple algorithms.
adjacent classifiers are covered in -o7.1 of the revised version
On 9/19/2026 9:20 PM, dart200 wrote:
On 9/19/26 4:27 PM, dbush wrote:
On 9/19/2026 6:40 PM, dart200 wrote:
On 9/19/26 12:13 PM, dbush wrote:
On 9/19/2026 2:40 PM, dart200 wrote:
On 9/19/26 11:17 AM, dbush wrote:
On 9/19/2026 1:57 PM, dart200 wrote:
On 9/19/26 7:45 AM, dbush wrote:
On 9/19/2026 2:19 AM, dart200 wrote:
as constructed, the halting paradox und() can neither be
decided into the set, nor it's complement, by the decider
False.-a Given algorithm halts1 and algorithm und1 which uses >>>>>>>>> algorithm halts1, halts1(und1)==1 and und1 does not halt. Also, >>>>>>>>> Ggven algorithm halts0 and algorithm und0 which uses algorithm >>>>>>>>> halts0, halts0(und0)==0 and und0 halts.-a So halts1 decides und1 >>>>>>>>> wrong and halts0 decides und0 wrong.
i think u just supported my point by presenting two cases where >>>>>>>> a machine failed to be decided correctly
Nope.-a Two different machine decided differently by two different >>>>>>> deciders.
i'm not sure how you think you can claim they are "deciders" if
there's no way to extract _truthful_ information from _any_ output, >>>>>>
are we just randomly labeling things now?
If an algorithm returns 0 or 1 for all inputs, they are a decider.
Whether they are correct for that input depends on whether the
result matches the mathematical mapping they're trying to reproduce. >>>>>
i don't think u even know what turing machines are, to be frank >>>>>>>>>You're demonstrating quite clearly that it is you who don't >>>>>>>>> understand what turing machines are.
A Turing machine is essentially an algorithm, i.e. a fixed
immutable
equating the turing machine model with the concept of an
algorithm _is_ the church-turing thesis, which has not been
proven, that my paper refutes
algorithms are more fundamental that turing machines, and not >>>>>>>> all algorithms can be constructed into a turing machine
sequence of instructions, and is therefore defined by the
instructions themselves, not where the instructions physically >>>>>>>>> reside (i.e. the body of the function halts()).
and therefore u have lost the proof of the existence for a >>>>>>>>>> machine that cannot be decided into a set
That is not what the proof is.
The proof is that there is no single turing machine that can >>>>>>>>> decide whether any arbitrary turing machine with a given input >>>>>>>>> will halt.
in order to prove a set "undecidable" you need an example of >>>>>>>>>> an object that is cannot be decidable into that set or it's >>>>>>>>>> complement.
False.-a You need to show that there exists no algorithm that >>>>>>>>> can correctly place all objects in the correct set.
which u do by presenting objects that cannot be decided, like >>>>>>>> you literally just did with und1 and und0...
No, they are two separate objects which can be decided.
Specifically, und1 is correctly decided by a decider that just
returns 0, and und0 is correctly decider by a decider that just >>>>>>> returns 1.
i'm sorry, your definition of "correct" is an aglo that says a
terminating machine is non-terminating, and another which says a
non- terminating machine is terminating ...
Read it again.-a Algorithm und1 does not halt, so a decider that
returns 0 when given und1 as input is correct for that input.
Similarly, algorithm und0 does halt, so a decider that returns 1
when given und0 as input is correct for that input.
i honestly can't believe u just wrote that in all seriousness,
might as well start claiming 1=0 is "correct" at this point
ofc then u also did correctly decide them regardless of what
halts1 and halts0 returned, so what aglorithm did u use to do >>>>>>>> that dud?
halts1 is one algorithm, halts0 is another algorithm.-a Just to >>>>>>> keep things simple, halts1 returns 1 in all cases and halts0
returns 0 in all cases.
those are not genuine decision algorithms as _no_ information can >>>>>> be extracted from _any_ of the results, such that their output
_cannot_ be further used in _any_ computation possible
Sure they are.-a They return either 0 or 1 for all inputs.-a Whether >>>>> they are correct for a given input depends on whether the return
matches the mathematical mapping they're trying to reproduce.
they do not genuinely classify the input to any degree, ur just
mislabeling things to cover up for century old fallacy...
They do classify.-a They just don't correct map the halting function >>>>> for
so you are actually claiming that broken clock "correctly" tells
time because it's correct twice a day???
It correctly tells time for some subset of times.
wow u really are gunna double down on that nonsense, eh? that much
time on ur hands??
If a broken analog clock is stuck at 4:45, and the current time is
4:45, that clock is correct at that time.
It doesn't matter "why" it is correct, just that it is.
bro an algorithm that randomly produces a correct response at times
without a specific reason why is just not a "correct" algorithm by any
meaningful notion of "correct"
The only requirement for an algorithm to be correct is to reproduce a mathematical mapping.-a How it does so is irrelevant.-a I'll illustrate
with a simple example.
Suppose there's a mathematical function foo whose domain is the integers from 1 to 5 which maps its domain as follows:
1 -> 457
2 -> 26
3 -> 72
4 -> 983
5 -> 235
The following algorithm correctly maps this function:
int bar(int x) {
-a-a-a if (x==1) return 457;
-a-a-a if (x==2) return 26;
-a-a-a if (x==3) return 72;
-a-a-a if (x==4) return 983;
-a-a-a if (x==5) return 235;
-a-a-a return 0;
}
maybe ur a set theorist dud instead of computing guy, and don't care
about trivial concerns like the ability to extract useful meaning from
the response... but in computing we actually care about practical
matters, and you can't get that from constant return functions, lol.
a constant return function is _not_ a decider, you calling it a
decider is just useless stupidity, else a constant return function is
apparently a decider for _all_ semantic decision problems ever ...
which is just crazy
It absolutely is a decider.-a It maps all inputs to either 0 or 1, which
is what defines a decider.
i will not debate this further because it's not worth my time
In the same way, a prospective halt decider that always returns 0
correctly reports the halt status of all non-halting computations.
"How" it got that result is irrelevant.
i don't know how to respond to that level of brainrot dud, it's
clearly not mathematically sound
If an algorithm maps a specific input to an output that matches a
mathematical function, it correctly maps that value for that
function. "How" is irrelevant.
all inputs (and in fact no algorithm does)
...two fallacies actually, one is covered in -o5, the next is
covered in -o6
i'm not sure you even know what "correctness" is my dud...
...cause that's exactly what my paper is on...
_that_
object would "undecidable". holy fuck. you can't just
continually assert undecidability of a set, YOU NEED FUCKING >>>>>>>>>> PROOF OF AN OBJECT THAT CANNOT BE DECIDED UPON
you don't actually have that with a hypothetical machine that >>>>>>>>>> DOES NOT FUCKING EXIST...
It seem you're failing to understand proof by contradiction, >>>>>>>>> otherwise you wouldn't be asserting this.
ur failing to understand that refuted ur own claims with this >>>>>>>> very post
No, you simply don't understand what a turing machine is.
and u didn't respond to the fact that despite halt1 and halt0
being incorrect, u still classified und1 and und0 into sets
truthfully.
what aglo did u use to do that
Not one but multiple.-a und1 is correctly decided by algorithm halt0 >>>>> and und0 is correctly decider by algorithm halt1.
yes dud, by which algorithm did you use to combine those in order to
provide actually useful knowledge on what those machines objectively
do ...
It's not about "useful knowledge".-a It's about whether those
algorithms map the halting function for specific inputs.
But for the sake of argument let's say I created some algorithm, i.e.
a fixed immutable sequence of instructions, that we'll call halts2
that uses *only* a machine description and that machine's input to
determine whether that machine with the given input will halt, such
that halts2(und0) report 1 and halts2(und1) returns 0.-a We can then
construct an algorithm und2 based on algorithm halts2 that either
halts or does not halt and halts2(und2) returns the incorrect answer,
and either halts0(und2) or halts1(und2) will be correct.
What turing's proof shows is that any algorithm that attempts to map
the halting function will have at least one case it gets wrong.
good glad we've gotten to what was described on the 2nd page of my 26
page paper
heck if u even read the 1st page of it, u'd know i'm responding to a
published paper (from jan of this year) claiming very specifically
it's incorrect to attribute the halting problem to turing
but we both know u have no intention of reading it, so whatever dud
and why doesn't turing's proof prevent you from doing that if all >>>>>> algorithms can be encapsulated by the turing machine model?
Turing's proof shows that no single algorithm can map the halting
function.-a It doesn't say anything about multiple algorithms.
adjacent classifiers are covered in -o7.1 of the revised version
On 9/19/26 1:10 AM, Mikko wrote:
On 18/09/2026 12:33, dart200 wrote:
On 9/18/26 1:54 AM, Mikko wrote:
On 16/09/2026 19:26, dart200 wrote:
On 9/16/26 12:40 AM, Mikko wrote:
On 15/09/2026 19:34, dart200 wrote:
On 9/15/26 1:18 AM, Mikko wrote:
On 14/09/2026 11:45, dart200 wrote:
On 9/14/26 1:26 AM, Mikko wrote:
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:
On 12/09/2026 19:02, dart200 wrote:"simple" is a weird term for something we provably can't >>>>>>>>>>> build, and that doesn't make the haltinging problem
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computing >>>>>>>>>>>>>>> and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit? >>>>>>>>>>>>>>> usp=sharing
The section 1 "the halting problem" should clearly define >>>>>>>>>>>>>> what the
expression "halting problem" means. It might be useful to >>>>>>>>>>>>>> first
define "halting question" as that would simplify the >>>>>>>>>>>>>> definitions of
"halting problem" and "halting decider". In each of these >>>>>>>>>>>>>> it is
also inportant to be clear what is the scope of the >>>>>>>>>>>>>> problem and
what kind of solutions are allowed. For example, the halting >>>>>>>>>>>>>> problem of finite state machines is Turing decidable and >>>>>>>>>>>>>> the halting
problem of Turing machines is decidable with a simple oracle. >>>>>>>>>>>
"decidable" since we can't build, compute, or actually decide >>>>>>>>>>> with the oracle
A "simple oracle" is an oracle that is simpler than most other >>>>>>>>>> oracles,
just like a "big mouse" is a mouse that is bigger than most >>>>>>>>>> other mice
although not as big as a little elephant.
like i said: recursive undecidability leading to the concept of >>>>>>>>> "simple" vs "complex" oracles is discussing in -o4, i will not >>>>>>>>> be discussing them in -o1
That's OK, although there should be a short description of each >>>>>>>> section
at the end of the introduction.
But the section 1 gives the impression that the author does not >>>>>>>> know
what the halting problem is.
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the >>>>>>>>>>>>>> program "und"
is not undecidable: there are partial halt deciders that >>>>>>>>>>>>>> can tell
whether "und" halts. In order to make this clear as soon >>>>>>>>>>>>>> as possible
it might be better to define "partial halt decider" before >>>>>>>>>>>>>> the
introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
that's why partial deciders are discussed in -o5.2 not -o1, you >>>>>>>>> brought those up not me
The section 1 is about the halting problem. You may discuss
partial and
total solutions of the problem in the same section or elsewhere. >>>>>>>> But
it is important to point out about "und" that its halting is
determinable. Otherwise a reader could be confused or get the >>>>>>>> impression
that you are don't know.
und as specified in -o1 does not even exist, an argument which
takes up to the end of -o4 to explain
The section 1 says otherwise: the words "can be constructed" mean >>>>>> that
it does exist.
it exist as a hypothetical problem, constructed from a hypothetical >>>>> decider, that presents a real limit to turing machine
computability, but does not exist in the actual enumeration of
turing machines,
In the world of computing theory everything is hypothetical. The
section
1 does not call anything hypothetical. It merely constructs "und" from >>>> "halts" already specified and "loop" that is not specified.
if u can't figure out what halts does contrasted from loop in that
pseudo-code, when we are discussing the halting problem in a group
called comp.theory, this paper is just not for you, at this time
An attempt to change the topic indicates that you don't want to
discuss honestly but instead want to deceive.
while u literally just changed the topic u dishonest fuck
On 9/19/26 8:17 PM, dbush wrote:
On 9/19/2026 9:20 PM, dart200 wrote:
On 9/19/26 4:27 PM, dbush wrote:
On 9/19/2026 6:40 PM, dart200 wrote:
On 9/19/26 12:13 PM, dbush wrote:
On 9/19/2026 2:40 PM, dart200 wrote:
On 9/19/26 11:17 AM, dbush wrote:
On 9/19/2026 1:57 PM, dart200 wrote:
On 9/19/26 7:45 AM, dbush wrote:
On 9/19/2026 2:19 AM, dart200 wrote:
as constructed, the halting paradox und() can neither be >>>>>>>>>>> decided into the set, nor it's complement, by the decider >>>>>>>>>>False.-a Given algorithm halts1 and algorithm und1 which uses >>>>>>>>>> algorithm halts1, halts1(und1)==1 and und1 does not halt. >>>>>>>>>> Also, Ggven algorithm halts0 and algorithm und0 which uses >>>>>>>>>> algorithm halts0, halts0(und0)==0 and und0 halts.-a So halts1 >>>>>>>>>> decides und1 wrong and halts0 decides und0 wrong.
i think u just supported my point by presenting two cases where >>>>>>>>> a machine failed to be decided correctly
Nope.-a Two different machine decided differently by two
different deciders.
i'm not sure how you think you can claim they are "deciders" if >>>>>>> there's no way to extract _truthful_ information from _any_ output, >>>>>>>
are we just randomly labeling things now?
If an algorithm returns 0 or 1 for all inputs, they are a decider. >>>>>> Whether they are correct for that input depends on whether the
result matches the mathematical mapping they're trying to reproduce. >>>>>>
i don't think u even know what turing machines are, to be frank >>>>>>>>>>You're demonstrating quite clearly that it is you who don't >>>>>>>>>> understand what turing machines are.
A Turing machine is essentially an algorithm, i.e. a fixed >>>>>>>>>> immutable
equating the turing machine model with the concept of an
algorithm _is_ the church-turing thesis, which has not been >>>>>>>>> proven, that my paper refutes
algorithms are more fundamental that turing machines, and not >>>>>>>>> all algorithms can be constructed into a turing machine
sequence of instructions, and is therefore defined by the >>>>>>>>>> instructions themselves, not where the instructions physically >>>>>>>>>> reside (i.e. the body of the function halts()).
and therefore u have lost the proof of the existence for a >>>>>>>>>>> machine that cannot be decided into a set
That is not what the proof is.
The proof is that there is no single turing machine that can >>>>>>>>>> decide whether any arbitrary turing machine with a given input >>>>>>>>>> will halt.
in order to prove a set "undecidable" you need an example of >>>>>>>>>>> an object that is cannot be decidable into that set or it's >>>>>>>>>>> complement.
False.-a You need to show that there exists no algorithm that >>>>>>>>>> can correctly place all objects in the correct set.
which u do by presenting objects that cannot be decided, like >>>>>>>>> you literally just did with und1 and und0...
No, they are two separate objects which can be decided.
Specifically, und1 is correctly decided by a decider that just >>>>>>>> returns 0, and und0 is correctly decider by a decider that just >>>>>>>> returns 1.
i'm sorry, your definition of "correct" is an aglo that says a
terminating machine is non-terminating, and another which says a >>>>>>> non- terminating machine is terminating ...
Read it again.-a Algorithm und1 does not halt, so a decider that
returns 0 when given und1 as input is correct for that input.
Similarly, algorithm und0 does halt, so a decider that returns 1
when given und0 as input is correct for that input.
i honestly can't believe u just wrote that in all seriousness,
might as well start claiming 1=0 is "correct" at this point
ofc then u also did correctly decide them regardless of what >>>>>>>>> halts1 and halts0 returned, so what aglorithm did u use to do >>>>>>>>> that dud?
halts1 is one algorithm, halts0 is another algorithm.-a Just to >>>>>>>> keep things simple, halts1 returns 1 in all cases and halts0
returns 0 in all cases.
those are not genuine decision algorithms as _no_ information can >>>>>>> be extracted from _any_ of the results, such that their output
_cannot_ be further used in _any_ computation possible
Sure they are.-a They return either 0 or 1 for all inputs.-a Whether >>>>>> they are correct for a given input depends on whether the return
matches the mathematical mapping they're trying to reproduce.
they do not genuinely classify the input to any degree, ur just >>>>>>> mislabeling things to cover up for century old fallacy...
They do classify.-a They just don't correct map the halting
function for
so you are actually claiming that broken clock "correctly" tells
time because it's correct twice a day???
It correctly tells time for some subset of times.
wow u really are gunna double down on that nonsense, eh? that much
time on ur hands??
If a broken analog clock is stuck at 4:45, and the current time is
4:45, that clock is correct at that time.
It doesn't matter "why" it is correct, just that it is.
bro an algorithm that randomly produces a correct response at times
without a specific reason why is just not a "correct" algorithm by
any meaningful notion of "correct"
The only requirement for an algorithm to be correct is to reproduce a
mathematical mapping.-a How it does so is irrelevant.-a I'll illustrate
with a simple example.
Suppose there's a mathematical function foo whose domain is the
integers from 1 to 5 which maps its domain as follows:
1 -> 457
2 -> 26
3 -> 72
4 -> 983
5 -> 235
The following algorithm correctly maps this function:
int bar(int x) {
-a-a-a-a if (x==1) return 457;
-a-a-a-a if (x==2) return 26;
-a-a-a-a if (x==3) return 72;
-a-a-a-a if (x==4) return 983;
-a-a-a-a if (x==5) return 235;
-a-a-a-a return 0;
}
while ur arguing that actually
int foo(int x) {
-a if (x==1) return 0;
-a if (x==2) return 0;
-a if (x==3) return 0;
-a if (x==4) return 0;
-a if (x==5) return 0;
-a return 0;
}
correctly maps the function
maybe ur a set theorist dud instead of computing guy, and don't care
about trivial concerns like the ability to extract useful meaning
from the response... but in computing we actually care about
practical matters, and you can't get that from constant return
functions, lol.
a constant return function is _not_ a decider, you calling it a
decider is just useless stupidity, else a constant return function is
apparently a decider for _all_ semantic decision problems ever ...
which is just crazy
It absolutely is a decider.-a It maps all inputs to either 0 or 1,
which is what defines a decider.
this is fking incredible
i honestly can't believe ur really trying to argue that a constant
return function can be labeled as a correct halting decider machine.
i'm not gunna waste my time debating that, it's clearly not so
and if that's kinda of argument ur going to use as an excuse to not read
my paper, ur clearly not worth my time
i cannot wait to use people like u when i give talks in the future on
much a fucking shitshow this experience has been
i will not debate this further because it's not worth my time
In the same way, a prospective halt decider that always returns 0
correctly reports the halt status of all non-halting computations.
"How" it got that result is irrelevant.
i don't know how to respond to that level of brainrot dud, it's
clearly not mathematically sound
If an algorithm maps a specific input to an output that matches a
mathematical function, it correctly maps that value for that
function. "How" is irrelevant.
all inputs (and in fact no algorithm does)
...two fallacies actually, one is covered in -o5, the next is
covered in -o6
i'm not sure you even know what "correctness" is my dud...
...cause that's exactly what my paper is on...
_that_
object would "undecidable". holy fuck. you can't just
continually assert undecidability of a set, YOU NEED FUCKING >>>>>>>>>>> PROOF OF AN OBJECT THAT CANNOT BE DECIDED UPON
you don't actually have that with a hypothetical machine that >>>>>>>>>>> DOES NOT FUCKING EXIST...
It seem you're failing to understand proof by contradiction, >>>>>>>>>> otherwise you wouldn't be asserting this.
ur failing to understand that refuted ur own claims with this >>>>>>>>> very post
No, you simply don't understand what a turing machine is.
and u didn't respond to the fact that despite halt1 and halt0
being incorrect, u still classified und1 and und0 into sets
truthfully.
what aglo did u use to do that
Not one but multiple.-a und1 is correctly decided by algorithm
halt0 and und0 is correctly decider by algorithm halt1.
yes dud, by which algorithm did you use to combine those in order
to provide actually useful knowledge on what those machines
objectively do ...
It's not about "useful knowledge".-a It's about whether those
algorithms map the halting function for specific inputs.
But for the sake of argument let's say I created some algorithm,
i.e. a fixed immutable sequence of instructions, that we'll call
halts2 that uses *only* a machine description and that machine's
input to determine whether that machine with the given input will
halt, such that halts2(und0) report 1 and halts2(und1) returns 0.
We can then construct an algorithm und2 based on algorithm halts2
that either halts or does not halt and halts2(und2) returns the
incorrect answer, and either halts0(und2) or halts1(und2) will be
correct.
What turing's proof shows is that any algorithm that attempts to map
the halting function will have at least one case it gets wrong.
good glad we've gotten to what was described on the 2nd page of my 26
page paper
heck if u even read the 1st page of it, u'd know i'm responding to a
published paper (from jan of this year) claiming very specifically
it's incorrect to attribute the halting problem to turing
but we both know u have no intention of reading it, so whatever dud
and why doesn't turing's proof prevent you from doing that if all >>>>>>> algorithms can be encapsulated by the turing machine model?
Turing's proof shows that no single algorithm can map the halting >>>>>> function.-a It doesn't say anything about multiple algorithms.
adjacent classifiers are covered in -o7.1 of the revised version
On 9/20/2026 1:43 AM, dart200 wrote:
On 9/19/26 8:17 PM, dbush wrote:
On 9/19/2026 9:20 PM, dart200 wrote:
On 9/19/26 4:27 PM, dbush wrote:
On 9/19/2026 6:40 PM, dart200 wrote:
On 9/19/26 12:13 PM, dbush wrote:
On 9/19/2026 2:40 PM, dart200 wrote:
On 9/19/26 11:17 AM, dbush wrote:
On 9/19/2026 1:57 PM, dart200 wrote:
On 9/19/26 7:45 AM, dbush wrote:
On 9/19/2026 2:19 AM, dart200 wrote:
as constructed, the halting paradox und() can neither be >>>>>>>>>>>> decided into the set, nor it's complement, by the decider >>>>>>>>>>>False.-a Given algorithm halts1 and algorithm und1 which uses >>>>>>>>>>> algorithm halts1, halts1(und1)==1 and und1 does not halt. >>>>>>>>>>> Also, Ggven algorithm halts0 and algorithm und0 which uses >>>>>>>>>>> algorithm halts0, halts0(und0)==0 and und0 halts.-a So halts1 >>>>>>>>>>> decides und1 wrong and halts0 decides und0 wrong.
i think u just supported my point by presenting two cases >>>>>>>>>> where a machine failed to be decided correctly
Nope.-a Two different machine decided differently by two
different deciders.
i'm not sure how you think you can claim they are "deciders" if >>>>>>>> there's no way to extract _truthful_ information from _any_ output, >>>>>>>>
are we just randomly labeling things now?
If an algorithm returns 0 or 1 for all inputs, they are a
decider. Whether they are correct for that input depends on
whether the result matches the mathematical mapping they're
trying to reproduce.
i don't think u even know what turing machines are, to be frank >>>>>>>>>>>You're demonstrating quite clearly that it is you who don't >>>>>>>>>>> understand what turing machines are.
A Turing machine is essentially an algorithm, i.e. a fixed >>>>>>>>>>> immutable
equating the turing machine model with the concept of an
algorithm _is_ the church-turing thesis, which has not been >>>>>>>>>> proven, that my paper refutes
algorithms are more fundamental that turing machines, and not >>>>>>>>>> all algorithms can be constructed into a turing machine
sequence of instructions, and is therefore defined by the >>>>>>>>>>> instructions themselves, not where the instructions
physically reside (i.e. the body of the function halts()). >>>>>>>>>>>
and therefore u have lost the proof of the existence for a >>>>>>>>>>>> machine that cannot be decided into a set
That is not what the proof is.
The proof is that there is no single turing machine that can >>>>>>>>>>> decide whether any arbitrary turing machine with a given >>>>>>>>>>> input will halt.
in order to prove a set "undecidable" you need an example of >>>>>>>>>>>> an object that is cannot be decidable into that set or it's >>>>>>>>>>>> complement.
False.-a You need to show that there exists no algorithm that >>>>>>>>>>> can correctly place all objects in the correct set.
which u do by presenting objects that cannot be decided, like >>>>>>>>>> you literally just did with und1 and und0...
No, they are two separate objects which can be decided.
Specifically, und1 is correctly decided by a decider that just >>>>>>>>> returns 0, and und0 is correctly decider by a decider that just >>>>>>>>> returns 1.
i'm sorry, your definition of "correct" is an aglo that says a >>>>>>>> terminating machine is non-terminating, and another which says a >>>>>>>> non- terminating machine is terminating ...
Read it again.-a Algorithm und1 does not halt, so a decider that >>>>>>> returns 0 when given und1 as input is correct for that input.
Similarly, algorithm und0 does halt, so a decider that returns 1 >>>>>>> when given und0 as input is correct for that input.
i honestly can't believe u just wrote that in all seriousness, >>>>>>>> might as well start claiming 1=0 is "correct" at this point
ofc then u also did correctly decide them regardless of what >>>>>>>>>> halts1 and halts0 returned, so what aglorithm did u use to do >>>>>>>>>> that dud?
halts1 is one algorithm, halts0 is another algorithm.-a Just to >>>>>>>>> keep things simple, halts1 returns 1 in all cases and halts0 >>>>>>>>> returns 0 in all cases.
those are not genuine decision algorithms as _no_ information >>>>>>>> can be extracted from _any_ of the results, such that their
output _cannot_ be further used in _any_ computation possible
Sure they are.-a They return either 0 or 1 for all inputs.
Whether they are correct for a given input depends on whether the >>>>>>> return matches the mathematical mapping they're trying to reproduce. >>>>>>>
they do not genuinely classify the input to any degree, ur just >>>>>>>> mislabeling things to cover up for century old fallacy...
They do classify.-a They just don't correct map the halting
function for
so you are actually claiming that broken clock "correctly" tells
time because it's correct twice a day???
It correctly tells time for some subset of times.
wow u really are gunna double down on that nonsense, eh? that much
time on ur hands??
If a broken analog clock is stuck at 4:45, and the current time is
4:45, that clock is correct at that time.
It doesn't matter "why" it is correct, just that it is.
bro an algorithm that randomly produces a correct response at times
without a specific reason why is just not a "correct" algorithm by
any meaningful notion of "correct"
The only requirement for an algorithm to be correct is to reproduce a
mathematical mapping.-a How it does so is irrelevant.-a I'll illustrate >>> with a simple example.
Suppose there's a mathematical function foo whose domain is the
integers from 1 to 5 which maps its domain as follows:
1 -> 457
2 -> 26
3 -> 72
4 -> 983
5 -> 235
The following algorithm correctly maps this function:
int bar(int x) {
-a-a-a-a if (x==1) return 457;
-a-a-a-a if (x==2) return 26;
-a-a-a-a if (x==3) return 72;
-a-a-a-a if (x==4) return 983;
-a-a-a-a if (x==5) return 235;
-a-a-a-a return 0;
}
while ur arguing that actually
int foo(int x) {
-a-a if (x==1) return 0;
-a-a if (x==2) return 0;
-a-a if (x==3) return 0;
-a-a if (x==4) return 0;
-a-a if (x==5) return 0;
-a-a return 0;
}
correctly maps the function
No, I'm arguing that:
int foo(int x) {
-a-a if (x==1) return 0;
-a-a if (x==2) return 26;
-a-a if (x==3) return 0;
-a-a if (x==4) return 0;
-a-a if (x==5) return 0;
-a-a return 0;
}
Correctly maps the function for the input 2.
maybe ur a set theorist dud instead of computing guy, and don't care
about trivial concerns like the ability to extract useful meaning
from the response... but in computing we actually care about
practical matters, and you can't get that from constant return
functions, lol.
a constant return function is _not_ a decider, you calling it a
decider is just useless stupidity, else a constant return function
is apparently a decider for _all_ semantic decision problems
ever ... which is just crazy
It absolutely is a decider.-a It maps all inputs to either 0 or 1,
which is what defines a decider.
this is fking incredible
i honestly can't believe ur really trying to argue that a constant
return function can be labeled as a correct halting decider machine.
I never said that.
I said it correctly decides halting for one class of machines, namely
either all that halt or all that do not halt, not all machines.
i'm not gunna waste my time debating that, it's clearly not so
and if that's kinda of argument ur going to use as an excuse to not
read my paper, ur clearly not worth my time
i cannot wait to use people like u when i give talks in the future on
much a fucking shitshow this experience has been
You mean people where you don't bother to read the one or two sentences they've written and conclude the opposite of what they said?
You won't be taken seriously if you display such a low level of reading comprehension.
i will not debate this further because it's not worth my time
In the same way, a prospective halt decider that always returns 0
correctly reports the halt status of all non-halting computations.
"How" it got that result is irrelevant.
i don't know how to respond to that level of brainrot dud, it's
clearly not mathematically sound
If an algorithm maps a specific input to an output that matches a
mathematical function, it correctly maps that value for that
function. "How" is irrelevant.
all inputs (and in fact no algorithm does)
...two fallacies actually, one is covered in -o5, the next is >>>>>>>> covered in -o6
i'm not sure you even know what "correctness" is my dud...
...cause that's exactly what my paper is on...
_that_
object would "undecidable". holy fuck. you can't just >>>>>>>>>>>> continually assert undecidability of a set, YOU NEED FUCKING >>>>>>>>>>>> PROOF OF AN OBJECT THAT CANNOT BE DECIDED UPON
you don't actually have that with a hypothetical machine >>>>>>>>>>>> that DOES NOT FUCKING EXIST...
It seem you're failing to understand proof by contradiction, >>>>>>>>>>> otherwise you wouldn't be asserting this.
ur failing to understand that refuted ur own claims with this >>>>>>>>>> very post
No, you simply don't understand what a turing machine is.
and u didn't respond to the fact that despite halt1 and halt0 >>>>>>>> being incorrect, u still classified und1 and und0 into sets
truthfully.
what aglo did u use to do that
Not one but multiple.-a und1 is correctly decided by algorithm
halt0 and und0 is correctly decider by algorithm halt1.
yes dud, by which algorithm did you use to combine those in order >>>>>> to provide actually useful knowledge on what those machines
objectively do ...
It's not about "useful knowledge".-a It's about whether those
algorithms map the halting function for specific inputs.
But for the sake of argument let's say I created some algorithm,
i.e. a fixed immutable sequence of instructions, that we'll call
halts2 that uses *only* a machine description and that machine's
input to determine whether that machine with the given input will
halt, such that halts2(und0) report 1 and halts2(und1) returns 0.
We can then construct an algorithm und2 based on algorithm halts2
that either halts or does not halt and halts2(und2) returns the
incorrect answer, and either halts0(und2) or halts1(und2) will be
correct.
What turing's proof shows is that any algorithm that attempts to
map the halting function will have at least one case it gets wrong.
good glad we've gotten to what was described on the 2nd page of my
26 page paper
heck if u even read the 1st page of it, u'd know i'm responding to a
published paper (from jan of this year) claiming very specifically
it's incorrect to attribute the halting problem to turing
but we both know u have no intention of reading it, so whatever dud
and why doesn't turing's proof prevent you from doing that if >>>>>>>> all algorithms can be encapsulated by the turing machine model? >>>>>>>>
Turing's proof shows that no single algorithm can map the halting >>>>>>> function.-a It doesn't say anything about multiple algorithms.
adjacent classifiers are covered in -o7.1 of the revised version
On 19/09/2026 11:26, dart200 wrote:
On 9/19/26 1:10 AM, Mikko wrote:
On 18/09/2026 12:33, dart200 wrote:
On 9/18/26 1:54 AM, Mikko wrote:
On 16/09/2026 19:26, dart200 wrote:
On 9/16/26 12:40 AM, Mikko wrote:
On 15/09/2026 19:34, dart200 wrote:
On 9/15/26 1:18 AM, Mikko wrote:
On 14/09/2026 11:45, dart200 wrote:
On 9/14/26 1:26 AM, Mikko wrote:
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:
On 12/09/2026 19:02, dart200 wrote:
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computing >>>>>>>>>>>>>>>> and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit? >>>>>>>>>>>>>>>> usp=sharing
The section 1 "the halting problem" should clearly define >>>>>>>>>>>>>>> what the
expression "halting problem" means. It might be useful to >>>>>>>>>>>>>>> first
define "halting question" as that would simplify the >>>>>>>>>>>>>>> definitions of
"halting problem" and "halting decider". In each of these >>>>>>>>>>>>>>> it is
also inportant to be clear what is the scope of the >>>>>>>>>>>>>>> problem and
what kind of solutions are allowed. For example, the halting >>>>>>>>>>>>>>> problem of finite state machines is Turing decidable and >>>>>>>>>>>>>>> the halting
problem of Turing machines is decidable with a simple >>>>>>>>>>>>>>> oracle.
"simple" is a weird term for something we provably can't >>>>>>>>>>>> build, and that doesn't make the haltinging problem
"decidable" since we can't build, compute, or actually >>>>>>>>>>>> decide with the oracle
A "simple oracle" is an oracle that is simpler than most >>>>>>>>>>> other oracles,
just like a "big mouse" is a mouse that is bigger than most >>>>>>>>>>> other mice
although not as big as a little elephant.
like i said: recursive undecidability leading to the concept >>>>>>>>>> of "simple" vs "complex" oracles is discussing in -o4, i will >>>>>>>>>> not be discussing them in -o1
That's OK, although there should be a short description of each >>>>>>>>> section
at the end of the introduction.
But the section 1 gives the impression that the author does not >>>>>>>>> know
what the halting problem is.
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the >>>>>>>>>>>>>>> program "und"
is not undecidable: there are partial halt deciders that >>>>>>>>>>>>>>> can tell
whether "und" halts. In order to make this clear as soon >>>>>>>>>>>>>>> as possible
it might be better to define "partial halt decider" >>>>>>>>>>>>>>> before the
introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
that's why partial deciders are discussed in -o5.2 not -o1, you >>>>>>>>>> brought those up not me
The section 1 is about the halting problem. You may discuss >>>>>>>>> partial and
total solutions of the problem in the same section or
elsewhere. But
it is important to point out about "und" that its halting is >>>>>>>>> determinable. Otherwise a reader could be confused or get the >>>>>>>>> impression
that you are don't know.
und as specified in -o1 does not even exist, an argument which >>>>>>>> takes up to the end of -o4 to explain
The section 1 says otherwise: the words "can be constructed" mean >>>>>>> that
it does exist.
it exist as a hypothetical problem, constructed from a
hypothetical decider, that presents a real limit to turing machine >>>>>> computability, but does not exist in the actual enumeration of
turing machines,
In the world of computing theory everything is hypothetical. The
section
1 does not call anything hypothetical. It merely constructs "und" from >>>>> "halts" already specified and "loop" that is not specified.
if u can't figure out what halts does contrasted from loop in that
pseudo-code, when we are discussing the halting problem in a group
called comp.theory, this paper is just not for you, at this time
An attempt to change the topic indicates that you don't want to
discuss honestly but instead want to deceive.
while u literally just changed the topic u dishonest fuck
A meta comment about the discussion is not a change of topic. I already
said all that needs be said about the topic until someone asks or tries
to support false claims. Until then, the best to forget everything what
you have said.
On 9/20/26 6:23 AM, dbush wrote:
On 9/20/2026 1:43 AM, dart200 wrote:
On 9/19/26 8:17 PM, dbush wrote:
On 9/19/2026 9:20 PM, dart200 wrote:
On 9/19/26 4:27 PM, dbush wrote:
On 9/19/2026 6:40 PM, dart200 wrote:
On 9/19/26 12:13 PM, dbush wrote:
On 9/19/2026 2:40 PM, dart200 wrote:
On 9/19/26 11:17 AM, dbush wrote:
On 9/19/2026 1:57 PM, dart200 wrote:
On 9/19/26 7:45 AM, dbush wrote:
On 9/19/2026 2:19 AM, dart200 wrote:
as constructed, the halting paradox und() can neither be >>>>>>>>>>>>> decided into the set, nor it's complement, by the decider >>>>>>>>>>>>False.-a Given algorithm halts1 and algorithm und1 which uses >>>>>>>>>>>> algorithm halts1, halts1(und1)==1 and und1 does not halt. >>>>>>>>>>>> Also, Ggven algorithm halts0 and algorithm und0 which uses >>>>>>>>>>>> algorithm halts0, halts0(und0)==0 and und0 halts.-a So halts1 >>>>>>>>>>>> decides und1 wrong and halts0 decides und0 wrong.
i think u just supported my point by presenting two cases >>>>>>>>>>> where a machine failed to be decided correctly
Nope.-a Two different machine decided differently by two
different deciders.
i'm not sure how you think you can claim they are "deciders" if >>>>>>>>> there's no way to extract _truthful_ information from _any_ >>>>>>>>> output,
are we just randomly labeling things now?
If an algorithm returns 0 or 1 for all inputs, they are a
decider. Whether they are correct for that input depends on
whether the result matches the mathematical mapping they're
trying to reproduce.
i don't think u even know what turing machines are, to be >>>>>>>>>>>>> frank
You're demonstrating quite clearly that it is you who don't >>>>>>>>>>>> understand what turing machines are.
A Turing machine is essentially an algorithm, i.e. a fixed >>>>>>>>>>>> immutable
equating the turing machine model with the concept of an >>>>>>>>>>> algorithm _is_ the church-turing thesis, which has not been >>>>>>>>>>> proven, that my paper refutes
algorithms are more fundamental that turing machines, and not >>>>>>>>>>> all algorithms can be constructed into a turing machine >>>>>>>>>>>> sequence of instructions, and is therefore defined by the >>>>>>>>>>>> instructions themselves, not where the instructions
physically reside (i.e. the body of the function halts()). >>>>>>>>>>>>
and therefore u have lost the proof of the existence for a >>>>>>>>>>>>> machine that cannot be decided into a set
That is not what the proof is.
The proof is that there is no single turing machine that can >>>>>>>>>>>> decide whether any arbitrary turing machine with a given >>>>>>>>>>>> input will halt.
in order to prove a set "undecidable" you need an example >>>>>>>>>>>>> of an object that is cannot be decidable into that set or >>>>>>>>>>>>> it's complement.
False.-a You need to show that there exists no algorithm that >>>>>>>>>>>> can correctly place all objects in the correct set.
which u do by presenting objects that cannot be decided, like >>>>>>>>>>> you literally just did with und1 and und0...
No, they are two separate objects which can be decided.
Specifically, und1 is correctly decided by a decider that just >>>>>>>>>> returns 0, and und0 is correctly decider by a decider that >>>>>>>>>> just returns 1.
i'm sorry, your definition of "correct" is an aglo that says a >>>>>>>>> terminating machine is non-terminating, and another which says >>>>>>>>> a non- terminating machine is terminating ...
Read it again.-a Algorithm und1 does not halt, so a decider that >>>>>>>> returns 0 when given und1 as input is correct for that input. >>>>>>>> Similarly, algorithm und0 does halt, so a decider that returns 1 >>>>>>>> when given und0 as input is correct for that input.
Sure they are.-a They return either 0 or 1 for all inputs.
i honestly can't believe u just wrote that in all seriousness, >>>>>>>>> might as well start claiming 1=0 is "correct" at this point
ofc then u also did correctly decide them regardless of what >>>>>>>>>>> halts1 and halts0 returned, so what aglorithm did u use to do >>>>>>>>>>> that dud?
halts1 is one algorithm, halts0 is another algorithm.-a Just to >>>>>>>>>> keep things simple, halts1 returns 1 in all cases and halts0 >>>>>>>>>> returns 0 in all cases.
those are not genuine decision algorithms as _no_ information >>>>>>>>> can be extracted from _any_ of the results, such that their >>>>>>>>> output _cannot_ be further used in _any_ computation possible >>>>>>>>
Whether they are correct for a given input depends on whether >>>>>>>> the return matches the mathematical mapping they're trying to >>>>>>>> reproduce.
they do not genuinely classify the input to any degree, ur just >>>>>>>>> mislabeling things to cover up for century old fallacy...
They do classify.-a They just don't correct map the halting
function for
so you are actually claiming that broken clock "correctly" tells >>>>>>> time because it's correct twice a day???
It correctly tells time for some subset of times.
wow u really are gunna double down on that nonsense, eh? that much
time on ur hands??
If a broken analog clock is stuck at 4:45, and the current time is >>>>>> 4:45, that clock is correct at that time.
It doesn't matter "why" it is correct, just that it is.
bro an algorithm that randomly produces a correct response at times >>>>> without a specific reason why is just not a "correct" algorithm by
any meaningful notion of "correct"
The only requirement for an algorithm to be correct is to reproduce
a mathematical mapping.-a How it does so is irrelevant.-a I'll
illustrate with a simple example.
Suppose there's a mathematical function foo whose domain is the
integers from 1 to 5 which maps its domain as follows:
1 -> 457
2 -> 26
3 -> 72
4 -> 983
5 -> 235
The following algorithm correctly maps this function:
int bar(int x) {
-a-a-a-a if (x==1) return 457;
-a-a-a-a if (x==2) return 26;
-a-a-a-a if (x==3) return 72;
-a-a-a-a if (x==4) return 983;
-a-a-a-a if (x==5) return 235;
-a-a-a-a return 0;
}
while ur arguing that actually
int foo(int x) {
-a-a if (x==1) return 0;
-a-a if (x==2) return 0;
-a-a if (x==3) return 0;
-a-a if (x==4) return 0;
-a-a if (x==5) return 0;
-a-a return 0;
}
correctly maps the function
No, I'm arguing that:
int foo(int x) {
-a-a-a if (x==1) return 0;
-a-a-a if (x==2) return 26;
-a-a-a if (x==3) return 0;
-a-a-a if (x==4) return 0;
-a-a-a if (x==5) return 0;
-a-a-a return 0;
}
Correctly maps the function for the input 2.
maybe ur a set theorist dud instead of computing guy, and don't
care about trivial concerns like the ability to extract useful
meaning from the response... but in computing we actually care
about practical matters, and you can't get that from constant
return functions, lol.
a constant return function is _not_ a decider, you calling it a
decider is just useless stupidity, else a constant return function
is apparently a decider for _all_ semantic decision problems
ever ... which is just crazy
It absolutely is a decider.-a It maps all inputs to either 0 or 1,
which is what defines a decider.
this is fking incredible
i honestly can't believe ur really trying to argue that a constant
return function can be labeled as a correct halting decider machine.
I never said that.
I said it correctly decides halting for one class of machines, namely
either all that halt or all that do not halt, not all machines.
the correctness of an algorithm is judged across all input, not just a subset.
i will not debate this further, ur attempt at a point is a
useless bastardization of computing theory
i'm not gunna waste my time debating that, it's clearly not so
and if that's kinda of argument ur going to use as an excuse to not
read my paper, ur clearly not worth my time
i cannot wait to use people like u when i give talks in the future on
much a fucking shitshow this experience has been
You mean people where you don't bother to read the one or two
sentences they've written and conclude the opposite of what they said?
You won't be taken seriously if you display such a low level of
reading comprehension.
you don't even read what u write:
-a| It absolutely is a decider.-a It maps all inputs to
-a| either 0 or 1, which is what defines a decider.
u don't even mention the any notion of correctness in ur definition there
i will not debate this further because it's not worth my time
good glad we've gotten to what was described on the 2nd page of my
In the same way, a prospective halt decider that always returns 0 >>>>>> correctly reports the halt status of all non-halting computations. >>>>>> "How" it got that result is irrelevant.
i don't know how to respond to that level of brainrot dud, it's >>>>>>> clearly not mathematically sound
If an algorithm maps a specific input to an output that matches a >>>>>> mathematical function, it correctly maps that value for that
function. "How" is irrelevant.
all inputs (and in fact no algorithm does)
...two fallacies actually, one is covered in -o5, the next is >>>>>>>>> covered in -o6
i'm not sure you even know what "correctness" is my dud...
...cause that's exactly what my paper is on...
_that_
object would "undecidable". holy fuck. you can't just >>>>>>>>>>>>> continually assert undecidability of a set, YOU NEED >>>>>>>>>>>>> FUCKING PROOF OF AN OBJECT THAT CANNOT BE DECIDED UPON >>>>>>>>>>>>>
you don't actually have that with a hypothetical machine >>>>>>>>>>>>> that DOES NOT FUCKING EXIST...
It seem you're failing to understand proof by contradiction, >>>>>>>>>>>> otherwise you wouldn't be asserting this.
ur failing to understand that refuted ur own claims with this >>>>>>>>>>> very post
No, you simply don't understand what a turing machine is.
and u didn't respond to the fact that despite halt1 and halt0 >>>>>>>>> being incorrect, u still classified und1 and und0 into sets >>>>>>>>> truthfully.
what aglo did u use to do that
Not one but multiple.-a und1 is correctly decided by algorithm >>>>>>>> halt0 and und0 is correctly decider by algorithm halt1.
yes dud, by which algorithm did you use to combine those in order >>>>>>> to provide actually useful knowledge on what those machines
objectively do ...
It's not about "useful knowledge".-a It's about whether those
algorithms map the halting function for specific inputs.
But for the sake of argument let's say I created some algorithm,
i.e. a fixed immutable sequence of instructions, that we'll call
halts2 that uses *only* a machine description and that machine's
input to determine whether that machine with the given input will >>>>>> halt, such that halts2(und0) report 1 and halts2(und1) returns 0. >>>>>> We can then construct an algorithm und2 based on algorithm halts2 >>>>>> that either halts or does not halt and halts2(und2) returns the
incorrect answer, and either halts0(und2) or halts1(und2) will be >>>>>> correct.
What turing's proof shows is that any algorithm that attempts to
map the halting function will have at least one case it gets wrong. >>>>>
26 page paper
heck if u even read the 1st page of it, u'd know i'm responding to
a published paper (from jan of this year) claiming very
specifically it's incorrect to attribute the halting problem to turing >>>>>
but we both know u have no intention of reading it, so whatever dud
and why doesn't turing's proof prevent you from doing that if >>>>>>>>> all algorithms can be encapsulated by the turing machine model? >>>>>>>>>
Turing's proof shows that no single algorithm can map the
halting function.-a It doesn't say anything about multiple
algorithms.
adjacent classifiers are covered in -o7.1 of the revised version >>>>>>>
On 9/20/2026 11:58 AM, dart200 wrote:
On 9/20/26 6:23 AM, dbush wrote:
On 9/20/2026 1:43 AM, dart200 wrote:
On 9/19/26 8:17 PM, dbush wrote:
On 9/19/2026 9:20 PM, dart200 wrote:
On 9/19/26 4:27 PM, dbush wrote:
On 9/19/2026 6:40 PM, dart200 wrote:
On 9/19/26 12:13 PM, dbush wrote:
On 9/19/2026 2:40 PM, dart200 wrote:
On 9/19/26 11:17 AM, dbush wrote:
On 9/19/2026 1:57 PM, dart200 wrote:
On 9/19/26 7:45 AM, dbush wrote:
On 9/19/2026 2:19 AM, dart200 wrote:i think u just supported my point by presenting two cases >>>>>>>>>>>> where a machine failed to be decided correctly
as constructed, the halting paradox und() can neither be >>>>>>>>>>>>>> decided into the set, nor it's complement, by the decider >>>>>>>>>>>>>False.-a Given algorithm halts1 and algorithm und1 which >>>>>>>>>>>>> uses algorithm halts1, halts1(und1)==1 and und1 does not >>>>>>>>>>>>> halt. Also, Ggven algorithm halts0 and algorithm und0 which >>>>>>>>>>>>> uses algorithm halts0, halts0(und0)==0 and und0 halts.-a So >>>>>>>>>>>>> halts1 decides und1 wrong and halts0 decides und0 wrong. >>>>>>>>>>>>
Nope.-a Two different machine decided differently by two >>>>>>>>>>> different deciders.
i'm not sure how you think you can claim they are "deciders" >>>>>>>>>> if there's no way to extract _truthful_ information from _any_ >>>>>>>>>> output,
are we just randomly labeling things now?
If an algorithm returns 0 or 1 for all inputs, they are a
decider. Whether they are correct for that input depends on >>>>>>>>> whether the result matches the mathematical mapping they're >>>>>>>>> trying to reproduce.
i don't think u even know what turing machines are, to be >>>>>>>>>>>>>> frank
You're demonstrating quite clearly that it is you who don't >>>>>>>>>>>>> understand what turing machines are.
A Turing machine is essentially an algorithm, i.e. a fixed >>>>>>>>>>>>> immutable
equating the turing machine model with the concept of an >>>>>>>>>>>> algorithm _is_ the church-turing thesis, which has not been >>>>>>>>>>>> proven, that my paper refutes
algorithms are more fundamental that turing machines, and >>>>>>>>>>>> not all algorithms can be constructed into a turing machine >>>>>>>>>>>>> sequence of instructions, and is therefore defined by the >>>>>>>>>>>>> instructions themselves, not where the instructions >>>>>>>>>>>>> physically reside (i.e. the body of the function halts()). >>>>>>>>>>>>>
which u do by presenting objects that cannot be decided, >>>>>>>>>>>> like you literally just did with und1 and und0...and therefore u have lost the proof of the existence for a >>>>>>>>>>>>>> machine that cannot be decided into a set
That is not what the proof is.
The proof is that there is no single turing machine that >>>>>>>>>>>>> can decide whether any arbitrary turing machine with a >>>>>>>>>>>>> given input will halt.
in order to prove a set "undecidable" you need an example >>>>>>>>>>>>>> of an object that is cannot be decidable into that set or >>>>>>>>>>>>>> it's complement.
False.-a You need to show that there exists no algorithm >>>>>>>>>>>>> that can correctly place all objects in the correct set. >>>>>>>>>>>>
No, they are two separate objects which can be decided. >>>>>>>>>>> Specifically, und1 is correctly decided by a decider that >>>>>>>>>>> just returns 0, and und0 is correctly decider by a decider >>>>>>>>>>> that just returns 1.
i'm sorry, your definition of "correct" is an aglo that says a >>>>>>>>>> terminating machine is non-terminating, and another which says >>>>>>>>>> a non- terminating machine is terminating ...
Read it again.-a Algorithm und1 does not halt, so a decider that >>>>>>>>> returns 0 when given und1 as input is correct for that input. >>>>>>>>> Similarly, algorithm und0 does halt, so a decider that returns >>>>>>>>> 1 when given und0 as input is correct for that input.
Sure they are.-a They return either 0 or 1 for all inputs.
i honestly can't believe u just wrote that in all seriousness, >>>>>>>>>> might as well start claiming 1=0 is "correct" at this point >>>>>>>>>>
ofc then u also did correctly decide them regardless of what >>>>>>>>>>>> halts1 and halts0 returned, so what aglorithm did u use to >>>>>>>>>>>> do that dud?
halts1 is one algorithm, halts0 is another algorithm.-a Just >>>>>>>>>>> to keep things simple, halts1 returns 1 in all cases and >>>>>>>>>>> halts0 returns 0 in all cases.
those are not genuine decision algorithms as _no_ information >>>>>>>>>> can be extracted from _any_ of the results, such that their >>>>>>>>>> output _cannot_ be further used in _any_ computation possible >>>>>>>>>
Whether they are correct for a given input depends on whether >>>>>>>>> the return matches the mathematical mapping they're trying to >>>>>>>>> reproduce.
They do classify.-a They just don't correct map the halting >>>>>>>>> function for
they do not genuinely classify the input to any degree, ur >>>>>>>>>> just mislabeling things to cover up for century old fallacy... >>>>>>>>>
so you are actually claiming that broken clock "correctly" tells >>>>>>>> time because it's correct twice a day???
It correctly tells time for some subset of times.
wow u really are gunna double down on that nonsense, eh? that much >>>>>> time on ur hands??
If a broken analog clock is stuck at 4:45, and the current time >>>>>>> is 4:45, that clock is correct at that time.
It doesn't matter "why" it is correct, just that it is.
bro an algorithm that randomly produces a correct response at
times without a specific reason why is just not a "correct"
algorithm by any meaningful notion of "correct"
The only requirement for an algorithm to be correct is to reproduce >>>>> a mathematical mapping.-a How it does so is irrelevant.-a I'll
illustrate with a simple example.
Suppose there's a mathematical function foo whose domain is the
integers from 1 to 5 which maps its domain as follows:
1 -> 457
2 -> 26
3 -> 72
4 -> 983
5 -> 235
The following algorithm correctly maps this function:
int bar(int x) {
-a-a-a-a if (x==1) return 457;
-a-a-a-a if (x==2) return 26;
-a-a-a-a if (x==3) return 72;
-a-a-a-a if (x==4) return 983;
-a-a-a-a if (x==5) return 235;
-a-a-a-a return 0;
}
while ur arguing that actually
int foo(int x) {
-a-a if (x==1) return 0;
-a-a if (x==2) return 0;
-a-a if (x==3) return 0;
-a-a if (x==4) return 0;
-a-a if (x==5) return 0;
-a-a return 0;
}
correctly maps the function
No, I'm arguing that:
int foo(int x) {
-a-a-a if (x==1) return 0;
-a-a-a if (x==2) return 26;
-a-a-a if (x==3) return 0;
-a-a-a if (x==4) return 0;
-a-a-a if (x==5) return 0;
-a-a-a return 0;
}
Correctly maps the function for the input 2.
maybe ur a set theorist dud instead of computing guy, and don't
care about trivial concerns like the ability to extract useful
meaning from the response... but in computing we actually care
about practical matters, and you can't get that from constant
return functions, lol.
a constant return function is _not_ a decider, you calling it a
decider is just useless stupidity, else a constant return function >>>>>> is apparently a decider for _all_ semantic decision problems
ever ... which is just crazy
It absolutely is a decider.-a It maps all inputs to either 0 or 1,
which is what defines a decider.
this is fking incredible
i honestly can't believe ur really trying to argue that a constant
return function can be labeled as a correct halting decider machine.
I never said that.
I said it correctly decides halting for one class of machines, namely
either all that halt or all that do not halt, not all machines.
the correctness of an algorithm is judged across all input, not just a
subset.
And it has been proved that any attempt to produce an algorithm that can tell if any arbitrary algorithm with a given input will halt will fail
to produce such an algorithm.
i will not debate this further, ur attempt at a point is a useless
bastardization of computing theory
i'm not gunna waste my time debating that, it's clearly not so
and if that's kinda of argument ur going to use as an excuse to not
read my paper, ur clearly not worth my time
i cannot wait to use people like u when i give talks in the future
on much a fucking shitshow this experience has been
You mean people where you don't bother to read the one or two
sentences they've written and conclude the opposite of what they said?
You won't be taken seriously if you display such a low level of
reading comprehension.
you don't even read what u write:
-a-a| It absolutely is a decider.-a It maps all inputs to
-a-a| either 0 or 1, which is what defines a decider.
u don't even mention the any notion of correctness in ur definition there
Correctness depends on what mathematical function it's mapping.-a A
decider that takes an integer as input always returns 1 is a correct
decider for a mathematical mapping that maps all integers to 1.
i will not debate this further because it's not worth my time
good glad we've gotten to what was described on the 2nd page of my >>>>>> 26 page paper
In the same way, a prospective halt decider that always returns 0 >>>>>>> correctly reports the halt status of all non-halting
computations. "How" it got that result is irrelevant.
i don't know how to respond to that level of brainrot dud, it's >>>>>>>> clearly not mathematically sound
If an algorithm maps a specific input to an output that matches a >>>>>>> mathematical function, it correctly maps that value for that
function. "How" is irrelevant.
all inputs (and in fact no algorithm does)
...two fallacies actually, one is covered in -o5, the next is >>>>>>>>>> covered in -o6
and u didn't respond to the fact that despite halt1 and halt0 >>>>>>>>>> being incorrect, u still classified und1 and und0 into sets >>>>>>>>>> truthfully.
...cause that's exactly what my paper is on...
ur failing to understand that refuted ur own claims with >>>>>>>>>>>> this very post
_that_
object would "undecidable". holy fuck. you can't just >>>>>>>>>>>>>> continually assert undecidability of a set, YOU NEED >>>>>>>>>>>>>> FUCKING PROOF OF AN OBJECT THAT CANNOT BE DECIDED UPON >>>>>>>>>>>>>>
you don't actually have that with a hypothetical machine >>>>>>>>>>>>>> that DOES NOT FUCKING EXIST...
It seem you're failing to understand proof by
contradiction, otherwise you wouldn't be asserting this. >>>>>>>>>>>>
No, you simply don't understand what a turing machine is. >>>>>>>>>> i'm not sure you even know what "correctness" is my dud... >>>>>>>>>>
what aglo did u use to do that
Not one but multiple.-a und1 is correctly decided by algorithm >>>>>>>>> halt0 and und0 is correctly decider by algorithm halt1.
yes dud, by which algorithm did you use to combine those in
order to provide actually useful knowledge on what those
machines objectively do ...
It's not about "useful knowledge".-a It's about whether those
algorithms map the halting function for specific inputs.
But for the sake of argument let's say I created some algorithm, >>>>>>> i.e. a fixed immutable sequence of instructions, that we'll call >>>>>>> halts2 that uses *only* a machine description and that machine's >>>>>>> input to determine whether that machine with the given input will >>>>>>> halt, such that halts2(und0) report 1 and halts2(und1) returns 0. >>>>>>> We can then construct an algorithm und2 based on algorithm halts2 >>>>>>> that either halts or does not halt and halts2(und2) returns the >>>>>>> incorrect answer, and either halts0(und2) or halts1(und2) will be >>>>>>> correct.
What turing's proof shows is that any algorithm that attempts to >>>>>>> map the halting function will have at least one case it gets wrong. >>>>>>
heck if u even read the 1st page of it, u'd know i'm responding to >>>>>> a published paper (from jan of this year) claiming very
specifically it's incorrect to attribute the halting problem to
turing
but we both know u have no intention of reading it, so whatever dud >>>>>>
and why doesn't turing's proof prevent you from doing that if >>>>>>>>>> all algorithms can be encapsulated by the turing machine model? >>>>>>>>>>
Turing's proof shows that no single algorithm can map the
halting function.-a It doesn't say anything about multiple
algorithms.
adjacent classifiers are covered in -o7.1 of the revised version >>>>>>>>
If not, wtf! You remind me of PO. Sorry for that cut.
i don't care for ur fallacy by association brainrot
On 9/20/26 1:27 PM, dbush wrote:
On 9/20/2026 11:58 AM, dart200 wrote:
On 9/20/26 6:23 AM, dbush wrote:
On 9/20/2026 1:43 AM, dart200 wrote:
On 9/19/26 8:17 PM, dbush wrote:
On 9/19/2026 9:20 PM, dart200 wrote:
On 9/19/26 4:27 PM, dbush wrote:
On 9/19/2026 6:40 PM, dart200 wrote:
On 9/19/26 12:13 PM, dbush wrote:
On 9/19/2026 2:40 PM, dart200 wrote:
On 9/19/26 11:17 AM, dbush wrote:
On 9/19/2026 1:57 PM, dart200 wrote:
On 9/19/26 7:45 AM, dbush wrote:
On 9/19/2026 2:19 AM, dart200 wrote:
as constructed, the halting paradox und() can neither be >>>>>>>>>>>>>>> decided into the set, nor it's complement, by the decider >>>>>>>>>>>>>>False.-a Given algorithm halts1 and algorithm und1 which >>>>>>>>>>>>>> uses algorithm halts1, halts1(und1)==1 and und1 does not >>>>>>>>>>>>>> halt. Also, Ggven algorithm halts0 and algorithm und0 >>>>>>>>>>>>>> which uses algorithm halts0, halts0(und0)==0 and und0 >>>>>>>>>>>>>> halts.-a So halts1 decides und1 wrong and halts0 decides >>>>>>>>>>>>>> und0 wrong.
i think u just supported my point by presenting two cases >>>>>>>>>>>>> where a machine failed to be decided correctly
Nope.-a Two different machine decided differently by two >>>>>>>>>>>> different deciders.
i'm not sure how you think you can claim they are "deciders" >>>>>>>>>>> if there's no way to extract _truthful_ information from >>>>>>>>>>> _any_ output,
are we just randomly labeling things now?
If an algorithm returns 0 or 1 for all inputs, they are a >>>>>>>>>> decider. Whether they are correct for that input depends on >>>>>>>>>> whether the result matches the mathematical mapping they're >>>>>>>>>> trying to reproduce.
i don't think u even know what turing machines are, to be >>>>>>>>>>>>>>> frank
You're demonstrating quite clearly that it is you who >>>>>>>>>>>>>> don't understand what turing machines are.
A Turing machine is essentially an algorithm, i.e. a fixed >>>>>>>>>>>>>> immutable
equating the turing machine model with the concept of an >>>>>>>>>>>>> algorithm _is_ the church-turing thesis, which has not been >>>>>>>>>>>>> proven, that my paper refutes
algorithms are more fundamental that turing machines, and >>>>>>>>>>>>> not all algorithms can be constructed into a turing machine >>>>>>>>>>>>>> sequence of instructions, and is therefore defined by the >>>>>>>>>>>>>> instructions themselves, not where the instructions >>>>>>>>>>>>>> physically reside (i.e. the body of the function halts()). >>>>>>>>>>>>>>
which u do by presenting objects that cannot be decided, >>>>>>>>>>>>> like you literally just did with und1 and und0...and therefore u have lost the proof of the existence for >>>>>>>>>>>>>>> a machine that cannot be decided into a set
That is not what the proof is.
The proof is that there is no single turing machine that >>>>>>>>>>>>>> can decide whether any arbitrary turing machine with a >>>>>>>>>>>>>> given input will halt.
in order to prove a set "undecidable" you need an example >>>>>>>>>>>>>>> of an object that is cannot be decidable into that set or >>>>>>>>>>>>>>> it's complement.
False.-a You need to show that there exists no algorithm >>>>>>>>>>>>>> that can correctly place all objects in the correct set. >>>>>>>>>>>>>
No, they are two separate objects which can be decided. >>>>>>>>>>>> Specifically, und1 is correctly decided by a decider that >>>>>>>>>>>> just returns 0, and und0 is correctly decider by a decider >>>>>>>>>>>> that just returns 1.
i'm sorry, your definition of "correct" is an aglo that says >>>>>>>>>>> a terminating machine is non-terminating, and another which >>>>>>>>>>> says a non- terminating machine is terminating ...
Read it again.-a Algorithm und1 does not halt, so a decider >>>>>>>>>> that returns 0 when given und1 as input is correct for that >>>>>>>>>> input. Similarly, algorithm und0 does halt, so a decider that >>>>>>>>>> returns 1 when given und0 as input is correct for that input. >>>>>>>>>>
Sure they are.-a They return either 0 or 1 for all inputs. >>>>>>>>>> Whether they are correct for a given input depends on whether >>>>>>>>>> the return matches the mathematical mapping they're trying to >>>>>>>>>> reproduce.
i honestly can't believe u just wrote that in all
seriousness, might as well start claiming 1=0 is "correct" at >>>>>>>>>>> this point
ofc then u also did correctly decide them regardless of >>>>>>>>>>>>> what halts1 and halts0 returned, so what aglorithm did u >>>>>>>>>>>>> use to do that dud?
halts1 is one algorithm, halts0 is another algorithm.-a Just >>>>>>>>>>>> to keep things simple, halts1 returns 1 in all cases and >>>>>>>>>>>> halts0 returns 0 in all cases.
those are not genuine decision algorithms as _no_ information >>>>>>>>>>> can be extracted from _any_ of the results, such that their >>>>>>>>>>> output _cannot_ be further used in _any_ computation possible >>>>>>>>>>
They do classify.-a They just don't correct map the halting >>>>>>>>>> function for
they do not genuinely classify the input to any degree, ur >>>>>>>>>>> just mislabeling things to cover up for century old fallacy... >>>>>>>>>>
so you are actually claiming that broken clock "correctly"
tells time because it's correct twice a day???
It correctly tells time for some subset of times.
wow u really are gunna double down on that nonsense, eh? that
much time on ur hands??
If a broken analog clock is stuck at 4:45, and the current time >>>>>>>> is 4:45, that clock is correct at that time.
It doesn't matter "why" it is correct, just that it is.
bro an algorithm that randomly produces a correct response at
times without a specific reason why is just not a "correct"
algorithm by any meaningful notion of "correct"
The only requirement for an algorithm to be correct is to
reproduce a mathematical mapping.-a How it does so is irrelevant. >>>>>> I'll illustrate with a simple example.
Suppose there's a mathematical function foo whose domain is the
integers from 1 to 5 which maps its domain as follows:
1 -> 457
2 -> 26
3 -> 72
4 -> 983
5 -> 235
The following algorithm correctly maps this function:
int bar(int x) {
-a-a-a-a if (x==1) return 457;
-a-a-a-a if (x==2) return 26;
-a-a-a-a if (x==3) return 72;
-a-a-a-a if (x==4) return 983;
-a-a-a-a if (x==5) return 235;
-a-a-a-a return 0;
}
while ur arguing that actually
int foo(int x) {
-a-a if (x==1) return 0;
-a-a if (x==2) return 0;
-a-a if (x==3) return 0;
-a-a if (x==4) return 0;
-a-a if (x==5) return 0;
-a-a return 0;
}
correctly maps the function
No, I'm arguing that:
int foo(int x) {
-a-a-a if (x==1) return 0;
-a-a-a if (x==2) return 26;
-a-a-a if (x==3) return 0;
-a-a-a if (x==4) return 0;
-a-a-a if (x==5) return 0;
-a-a-a return 0;
}
Correctly maps the function for the input 2.
maybe ur a set theorist dud instead of computing guy, and don't >>>>>>> care about trivial concerns like the ability to extract useful
meaning from the response... but in computing we actually care
about practical matters, and you can't get that from constant
return functions, lol.
a constant return function is _not_ a decider, you calling it a >>>>>>> decider is just useless stupidity, else a constant return
function is apparently a decider for _all_ semantic decision
problems ever ... which is just crazy
It absolutely is a decider.-a It maps all inputs to either 0 or 1, >>>>>> which is what defines a decider.
this is fking incredible
i honestly can't believe ur really trying to argue that a constant
return function can be labeled as a correct halting decider machine.
I never said that.
I said it correctly decides halting for one class of machines,
namely either all that halt or all that do not halt, not all machines.
the correctness of an algorithm is judged across all input, not just
a subset.
And it has been proved that any attempt to produce an algorithm that
can tell if any arbitrary algorithm with a given input will halt will
fail to produce such an algorithm.
great, glad we got to the 2nd page of my paper. if u have any further comments about the rest of it, that would be great rather than
repeatedly wasting time on this
i will not debate this further, ur attempt at a point is a useless
bastardization of computing theory
i'm not gunna waste my time debating that, it's clearly not so
and if that's kinda of argument ur going to use as an excuse to not >>>>> read my paper, ur clearly not worth my time
i cannot wait to use people like u when i give talks in the future
on much a fucking shitshow this experience has been
You mean people where you don't bother to read the one or two
sentences they've written and conclude the opposite of what they said? >>>>
You won't be taken seriously if you display such a low level of
reading comprehension.
you don't even read what u write:
-a-a| It absolutely is a decider.-a It maps all inputs to
-a-a| either 0 or 1, which is what defines a decider.
u don't even mention the any notion of correctness in ur definition
there
Correctness depends on what mathematical function it's mapping.-a A
right and we're talking about a correct _decider_, not some other function
decider that takes an integer as input always returns 1 is a correct
decider for a mathematical mapping that maps all integers to 1.
that's not a decider by any notion of what that term means in computing theory
a decider is _not_ just a synonym for turing machine that outputs some random-ass function, it is a specific type of machine that outputs
answers to some decision problem via some meaningful specification of correctness, which a constant return machine does not do
i will not debate this further because it's not worth my time
good glad we've gotten to what was described on the 2nd page of >>>>>>> my 26 page paper
In the same way, a prospective halt decider that always returns >>>>>>>> 0 correctly reports the halt status of all non-halting
computations. "How" it got that result is irrelevant.
i don't know how to respond to that level of brainrot dud, it's >>>>>>>>> clearly not mathematically sound
If an algorithm maps a specific input to an output that matches >>>>>>>> a mathematical function, it correctly maps that value for that >>>>>>>> function. "How" is irrelevant.
all inputs (and in fact no algorithm does)
...two fallacies actually, one is covered in -o5, the next is >>>>>>>>>>> covered in -o6
and u didn't respond to the fact that despite halt1 and halt0 >>>>>>>>>>> being incorrect, u still classified und1 and und0 into sets >>>>>>>>>>> truthfully.
...cause that's exactly what my paper is on...
ur failing to understand that refuted ur own claims with >>>>>>>>>>>>> this very post
_that_
object would "undecidable". holy fuck. you can't just >>>>>>>>>>>>>>> continually assert undecidability of a set, YOU NEED >>>>>>>>>>>>>>> FUCKING PROOF OF AN OBJECT THAT CANNOT BE DECIDED UPON >>>>>>>>>>>>>>>
you don't actually have that with a hypothetical machine >>>>>>>>>>>>>>> that DOES NOT FUCKING EXIST...
It seem you're failing to understand proof by
contradiction, otherwise you wouldn't be asserting this. >>>>>>>>>>>>>
No, you simply don't understand what a turing machine is. >>>>>>>>>>> i'm not sure you even know what "correctness" is my dud... >>>>>>>>>>>
what aglo did u use to do that
Not one but multiple.-a und1 is correctly decided by algorithm >>>>>>>>>> halt0 and und0 is correctly decider by algorithm halt1.
yes dud, by which algorithm did you use to combine those in >>>>>>>>> order to provide actually useful knowledge on what those
machines objectively do ...
It's not about "useful knowledge".-a It's about whether those >>>>>>>> algorithms map the halting function for specific inputs.
But for the sake of argument let's say I created some algorithm, >>>>>>>> i.e. a fixed immutable sequence of instructions, that we'll call >>>>>>>> halts2 that uses *only* a machine description and that machine's >>>>>>>> input to determine whether that machine with the given input
will halt, such that halts2(und0) report 1 and halts2(und1)
returns 0. We can then construct an algorithm und2 based on
algorithm halts2 that either halts or does not halt and
halts2(und2) returns the incorrect answer, and either
halts0(und2) or halts1(und2) will be correct.
What turing's proof shows is that any algorithm that attempts to >>>>>>>> map the halting function will have at least one case it gets wrong. >>>>>>>
heck if u even read the 1st page of it, u'd know i'm responding >>>>>>> to a published paper (from jan of this year) claiming very
specifically it's incorrect to attribute the halting problem to >>>>>>> turing
but we both know u have no intention of reading it, so whatever dud >>>>>>>
and why doesn't turing's proof prevent you from doing that if >>>>>>>>>>> all algorithms can be encapsulated by the turing machine model? >>>>>>>>>>>
Turing's proof shows that no single algorithm can map the >>>>>>>>>> halting function.-a It doesn't say anything about multiple >>>>>>>>>> algorithms.
adjacent classifiers are covered in -o7.1 of the revised version >>>>>>>>>
On 9/19/2026 2:38 PM, dart200 wrote:
[...]
If not, wtf! You remind me of PO. Sorry for that cut.
i don't care for ur fallacy by association brainrot
Sigh. I only said you kind of do remind me of the way PO dealt with the halting program.
On 9/18/26 1:54 AM, Mikko wrote:<SNIP>
On 16/09/2026 19:26, dart200 wrote:
In the world of computing theory everything is hypothetical. The section
1 does not call anything hypothetical. It merely constructs "und" from
"halts" already specified and "loop" that is not specified.
if u can't figure out what halts does contrasted from loop in that pseudo-code, when we are discussing the halting problem in a group
called comp.theory, this paper is just not for you, at this time
i have to be selective who i spend my time on at this point, i don't
have personal time to waste on someone who spends more time responding
to me, than it would take to just read the 3rd page
Although "und" is called "the halting problem" it obviously isn't any
problem, just a specification of a computation.
and does not limit our ability to decide on any given machine, as we
are not an addressable computing machine.
That is not relevant to this discussion about the section 1. Even if
the undecidability of und is literally the core justification for why we can't build a total halting decider in turing machine computing, while
the church-turing thesis asserts we can only compute something that
turing machines can compute ... so therefor undecidability of und the literally the core justification for why we aren't building total
halting deciders.
the section 1 is only a prelude to the main topic it should not give
the impression that the author cannot write anyting worth of reading.
prove ur not a retard dud, tell me:
does und() halt or loop forever?
On 9/19/2026 2:38 PM, dart200 wrote:
[...]
If not, wtf! You remind me of PO. Sorry for that cut.
i don't care for ur fallacy by association brainrot
Sigh. I only said you kind of do remind me of the way PO dealt with the halting program.
On 9/20/2026 11:58 AM, dart200 wrote:
On 9/20/26 6:23 AM, dbush wrote:
On 9/20/2026 1:43 AM, dart200 wrote:
On 9/19/26 8:17 PM, dbush wrote:
On 9/19/2026 9:20 PM, dart200 wrote:
On 9/19/26 4:27 PM, dbush wrote:
On 9/19/2026 6:40 PM, dart200 wrote:
On 9/19/26 12:13 PM, dbush wrote:
On 9/19/2026 2:40 PM, dart200 wrote:
On 9/19/26 11:17 AM, dbush wrote:
On 9/19/2026 1:57 PM, dart200 wrote:
On 9/19/26 7:45 AM, dbush wrote:
On 9/19/2026 2:19 AM, dart200 wrote:i think u just supported my point by presenting two cases >>>>>>>>>>>> where a machine failed to be decided correctly
as constructed, the halting paradox und() can neither be >>>>>>>>>>>>>> decided into the set, nor it's complement, by the decider >>>>>>>>>>>>>False.-a Given algorithm halts1 and algorithm und1 which >>>>>>>>>>>>> uses algorithm halts1, halts1(und1)==1 and und1 does not >>>>>>>>>>>>> halt. Also, Ggven algorithm halts0 and algorithm und0 which >>>>>>>>>>>>> uses algorithm halts0, halts0(und0)==0 and und0 halts.-a So >>>>>>>>>>>>> halts1 decides und1 wrong and halts0 decides und0 wrong. >>>>>>>>>>>>
Nope.-a Two different machine decided differently by two >>>>>>>>>>> different deciders.
i'm not sure how you think you can claim they are "deciders" >>>>>>>>>> if there's no way to extract _truthful_ information from _any_ >>>>>>>>>> output,
are we just randomly labeling things now?
If an algorithm returns 0 or 1 for all inputs, they are a
decider. Whether they are correct for that input depends on >>>>>>>>> whether the result matches the mathematical mapping they're >>>>>>>>> trying to reproduce.
i don't think u even know what turing machines are, to be >>>>>>>>>>>>>> frank
You're demonstrating quite clearly that it is you who don't >>>>>>>>>>>>> understand what turing machines are.
A Turing machine is essentially an algorithm, i.e. a fixed >>>>>>>>>>>>> immutable
equating the turing machine model with the concept of an >>>>>>>>>>>> algorithm _is_ the church-turing thesis, which has not been >>>>>>>>>>>> proven, that my paper refutes
algorithms are more fundamental that turing machines, and >>>>>>>>>>>> not all algorithms can be constructed into a turing machine >>>>>>>>>>>>> sequence of instructions, and is therefore defined by the >>>>>>>>>>>>> instructions themselves, not where the instructions >>>>>>>>>>>>> physically reside (i.e. the body of the function halts()). >>>>>>>>>>>>>
which u do by presenting objects that cannot be decided, >>>>>>>>>>>> like you literally just did with und1 and und0...and therefore u have lost the proof of the existence for a >>>>>>>>>>>>>> machine that cannot be decided into a set
That is not what the proof is.
The proof is that there is no single turing machine that >>>>>>>>>>>>> can decide whether any arbitrary turing machine with a >>>>>>>>>>>>> given input will halt.
in order to prove a set "undecidable" you need an example >>>>>>>>>>>>>> of an object that is cannot be decidable into that set or >>>>>>>>>>>>>> it's complement.
False.-a You need to show that there exists no algorithm >>>>>>>>>>>>> that can correctly place all objects in the correct set. >>>>>>>>>>>>
No, they are two separate objects which can be decided. >>>>>>>>>>> Specifically, und1 is correctly decided by a decider that >>>>>>>>>>> just returns 0, and und0 is correctly decider by a decider >>>>>>>>>>> that just returns 1.
i'm sorry, your definition of "correct" is an aglo that says a >>>>>>>>>> terminating machine is non-terminating, and another which says >>>>>>>>>> a non- terminating machine is terminating ...
Read it again.-a Algorithm und1 does not halt, so a decider that >>>>>>>>> returns 0 when given und1 as input is correct for that input. >>>>>>>>> Similarly, algorithm und0 does halt, so a decider that returns >>>>>>>>> 1 when given und0 as input is correct for that input.
Sure they are.-a They return either 0 or 1 for all inputs.
i honestly can't believe u just wrote that in all seriousness, >>>>>>>>>> might as well start claiming 1=0 is "correct" at this point >>>>>>>>>>
ofc then u also did correctly decide them regardless of what >>>>>>>>>>>> halts1 and halts0 returned, so what aglorithm did u use to >>>>>>>>>>>> do that dud?
halts1 is one algorithm, halts0 is another algorithm.-a Just >>>>>>>>>>> to keep things simple, halts1 returns 1 in all cases and >>>>>>>>>>> halts0 returns 0 in all cases.
those are not genuine decision algorithms as _no_ information >>>>>>>>>> can be extracted from _any_ of the results, such that their >>>>>>>>>> output _cannot_ be further used in _any_ computation possible >>>>>>>>>
Whether they are correct for a given input depends on whether >>>>>>>>> the return matches the mathematical mapping they're trying to >>>>>>>>> reproduce.
They do classify.-a They just don't correct map the halting >>>>>>>>> function for
they do not genuinely classify the input to any degree, ur >>>>>>>>>> just mislabeling things to cover up for century old fallacy... >>>>>>>>>
so you are actually claiming that broken clock "correctly" tells >>>>>>>> time because it's correct twice a day???
It correctly tells time for some subset of times.
wow u really are gunna double down on that nonsense, eh? that much >>>>>> time on ur hands??
If a broken analog clock is stuck at 4:45, and the current time >>>>>>> is 4:45, that clock is correct at that time.
It doesn't matter "why" it is correct, just that it is.
bro an algorithm that randomly produces a correct response at
times without a specific reason why is just not a "correct"
algorithm by any meaningful notion of "correct"
The only requirement for an algorithm to be correct is to reproduce >>>>> a mathematical mapping.-a How it does so is irrelevant.-a I'll
illustrate with a simple example.
Suppose there's a mathematical function foo whose domain is the
integers from 1 to 5 which maps its domain as follows:
1 -> 457
2 -> 26
3 -> 72
4 -> 983
5 -> 235
The following algorithm correctly maps this function:
int bar(int x) {
-a-a-a-a if (x==1) return 457;
-a-a-a-a if (x==2) return 26;
-a-a-a-a if (x==3) return 72;
-a-a-a-a if (x==4) return 983;
-a-a-a-a if (x==5) return 235;
-a-a-a-a return 0;
}
while ur arguing that actually
int foo(int x) {
-a-a if (x==1) return 0;
-a-a if (x==2) return 0;
-a-a if (x==3) return 0;
-a-a if (x==4) return 0;
-a-a if (x==5) return 0;
-a-a return 0;
}
correctly maps the function
No, I'm arguing that:
int foo(int x) {
-a-a-a if (x==1) return 0;
-a-a-a if (x==2) return 26;
-a-a-a if (x==3) return 0;
-a-a-a if (x==4) return 0;
-a-a-a if (x==5) return 0;
-a-a-a return 0;
}
Correctly maps the function for the input 2.
maybe ur a set theorist dud instead of computing guy, and don't
care about trivial concerns like the ability to extract useful
meaning from the response... but in computing we actually care
about practical matters, and you can't get that from constant
return functions, lol.
a constant return function is _not_ a decider, you calling it a
decider is just useless stupidity, else a constant return function >>>>>> is apparently a decider for _all_ semantic decision problems
ever ... which is just crazy
It absolutely is a decider.-a It maps all inputs to either 0 or 1,
which is what defines a decider.
this is fking incredible
i honestly can't believe ur really trying to argue that a constant
return function can be labeled as a correct halting decider machine.
I never said that.
I said it correctly decides halting for one class of machines, namely
either all that halt or all that do not halt, not all machines.
the correctness of an algorithm is judged across all input, not just a
subset.
And it has been proved that any attempt to produce an algorithm that can tell if any arbitrary algorithm with a given input will halt will fail
to produce such an algorithm.
i will not debate this further, ur attempt at a point is a useless
bastardization of computing theory
i'm not gunna waste my time debating that, it's clearly not so
and if that's kinda of argument ur going to use as an excuse to not
read my paper, ur clearly not worth my time
i cannot wait to use people like u when i give talks in the future
on much a fucking shitshow this experience has been
You mean people where you don't bother to read the one or two
sentences they've written and conclude the opposite of what they said?
You won't be taken seriously if you display such a low level of
reading comprehension.
you don't even read what u write:
-a-a| It absolutely is a decider.-a It maps all inputs to
-a-a| either 0 or 1, which is what defines a decider.
u don't even mention the any notion of correctness in ur definition there
Correctness depends on what mathematical function it's mapping.-a A
decider that takes an integer as input always returns 1 is a correct
decider for a mathematical mapping that maps all integers to 1.
i will not debate this further because it's not worth my time
good glad we've gotten to what was described on the 2nd page of my >>>>>> 26 page paper
In the same way, a prospective halt decider that always returns 0 >>>>>>> correctly reports the halt status of all non-halting
computations. "How" it got that result is irrelevant.
i don't know how to respond to that level of brainrot dud, it's >>>>>>>> clearly not mathematically sound
If an algorithm maps a specific input to an output that matches a >>>>>>> mathematical function, it correctly maps that value for that
function. "How" is irrelevant.
all inputs (and in fact no algorithm does)
...two fallacies actually, one is covered in -o5, the next is >>>>>>>>>> covered in -o6
and u didn't respond to the fact that despite halt1 and halt0 >>>>>>>>>> being incorrect, u still classified und1 and und0 into sets >>>>>>>>>> truthfully.
...cause that's exactly what my paper is on...
ur failing to understand that refuted ur own claims with >>>>>>>>>>>> this very post
_that_
object would "undecidable". holy fuck. you can't just >>>>>>>>>>>>>> continually assert undecidability of a set, YOU NEED >>>>>>>>>>>>>> FUCKING PROOF OF AN OBJECT THAT CANNOT BE DECIDED UPON >>>>>>>>>>>>>>
you don't actually have that with a hypothetical machine >>>>>>>>>>>>>> that DOES NOT FUCKING EXIST...
It seem you're failing to understand proof by
contradiction, otherwise you wouldn't be asserting this. >>>>>>>>>>>>
No, you simply don't understand what a turing machine is. >>>>>>>>>> i'm not sure you even know what "correctness" is my dud... >>>>>>>>>>
what aglo did u use to do that
Not one but multiple.-a und1 is correctly decided by algorithm >>>>>>>>> halt0 and und0 is correctly decider by algorithm halt1.
yes dud, by which algorithm did you use to combine those in
order to provide actually useful knowledge on what those
machines objectively do ...
It's not about "useful knowledge".-a It's about whether those
algorithms map the halting function for specific inputs.
But for the sake of argument let's say I created some algorithm, >>>>>>> i.e. a fixed immutable sequence of instructions, that we'll call >>>>>>> halts2 that uses *only* a machine description and that machine's >>>>>>> input to determine whether that machine with the given input will >>>>>>> halt, such that halts2(und0) report 1 and halts2(und1) returns 0. >>>>>>> We can then construct an algorithm und2 based on algorithm halts2 >>>>>>> that either halts or does not halt and halts2(und2) returns the >>>>>>> incorrect answer, and either halts0(und2) or halts1(und2) will be >>>>>>> correct.
What turing's proof shows is that any algorithm that attempts to >>>>>>> map the halting function will have at least one case it gets wrong. >>>>>>
heck if u even read the 1st page of it, u'd know i'm responding to >>>>>> a published paper (from jan of this year) claiming very
specifically it's incorrect to attribute the halting problem to
turing
but we both know u have no intention of reading it, so whatever dud >>>>>>
and why doesn't turing's proof prevent you from doing that if >>>>>>>>>> all algorithms can be encapsulated by the turing machine model? >>>>>>>>>>
Turing's proof shows that no single algorithm can map the
halting function.-a It doesn't say anything about multiple
algorithms.
adjacent classifiers are covered in -o7.1 of the revised version >>>>>>>>
On 9/20/26 2:53 PM, Chris M. Thomasson wrote:
On 9/19/2026 2:38 PM, dart200 wrote:
[...]
If not, wtf! You remind me of PO. Sorry for that cut.
i don't care for ur fallacy by association brainrot
Sigh. I only said you kind of do remind me of the way PO dealt with
the halting program.
and that statement has no bearing on the correctness of my arguments, u brain-rotted boomer
On 9/20/26 1:27 PM, dbush wrote:
On 9/20/2026 11:58 AM, dart200 wrote:
On 9/20/26 6:23 AM, dbush wrote:
On 9/20/2026 1:43 AM, dart200 wrote:
On 9/19/26 8:17 PM, dbush wrote:
On 9/19/2026 9:20 PM, dart200 wrote:
On 9/19/26 4:27 PM, dbush wrote:
On 9/19/2026 6:40 PM, dart200 wrote:
On 9/19/26 12:13 PM, dbush wrote:
On 9/19/2026 2:40 PM, dart200 wrote:
On 9/19/26 11:17 AM, dbush wrote:
On 9/19/2026 1:57 PM, dart200 wrote:
On 9/19/26 7:45 AM, dbush wrote:
On 9/19/2026 2:19 AM, dart200 wrote:
as constructed, the halting paradox und() can neither be >>>>>>>>>>>>>>> decided into the set, nor it's complement, by the decider >>>>>>>>>>>>>>False.-a Given algorithm halts1 and algorithm und1 which >>>>>>>>>>>>>> uses algorithm halts1, halts1(und1)==1 and und1 does not >>>>>>>>>>>>>> halt. Also, Ggven algorithm halts0 and algorithm und0 >>>>>>>>>>>>>> which uses algorithm halts0, halts0(und0)==0 and und0 >>>>>>>>>>>>>> halts.-a So halts1 decides und1 wrong and halts0 decides >>>>>>>>>>>>>> und0 wrong.
i think u just supported my point by presenting two cases >>>>>>>>>>>>> where a machine failed to be decided correctly
Nope.-a Two different machine decided differently by two >>>>>>>>>>>> different deciders.
i'm not sure how you think you can claim they are "deciders" >>>>>>>>>>> if there's no way to extract _truthful_ information from >>>>>>>>>>> _any_ output,
are we just randomly labeling things now?
If an algorithm returns 0 or 1 for all inputs, they are a >>>>>>>>>> decider. Whether they are correct for that input depends on >>>>>>>>>> whether the result matches the mathematical mapping they're >>>>>>>>>> trying to reproduce.
i don't think u even know what turing machines are, to be >>>>>>>>>>>>>>> frank
You're demonstrating quite clearly that it is you who >>>>>>>>>>>>>> don't understand what turing machines are.
A Turing machine is essentially an algorithm, i.e. a fixed >>>>>>>>>>>>>> immutable
equating the turing machine model with the concept of an >>>>>>>>>>>>> algorithm _is_ the church-turing thesis, which has not been >>>>>>>>>>>>> proven, that my paper refutes
algorithms are more fundamental that turing machines, and >>>>>>>>>>>>> not all algorithms can be constructed into a turing machine >>>>>>>>>>>>>> sequence of instructions, and is therefore defined by the >>>>>>>>>>>>>> instructions themselves, not where the instructions >>>>>>>>>>>>>> physically reside (i.e. the body of the function halts()). >>>>>>>>>>>>>>
which u do by presenting objects that cannot be decided, >>>>>>>>>>>>> like you literally just did with und1 and und0...and therefore u have lost the proof of the existence for >>>>>>>>>>>>>>> a machine that cannot be decided into a set
That is not what the proof is.
The proof is that there is no single turing machine that >>>>>>>>>>>>>> can decide whether any arbitrary turing machine with a >>>>>>>>>>>>>> given input will halt.
in order to prove a set "undecidable" you need an example >>>>>>>>>>>>>>> of an object that is cannot be decidable into that set or >>>>>>>>>>>>>>> it's complement.
False.-a You need to show that there exists no algorithm >>>>>>>>>>>>>> that can correctly place all objects in the correct set. >>>>>>>>>>>>>
No, they are two separate objects which can be decided. >>>>>>>>>>>> Specifically, und1 is correctly decided by a decider that >>>>>>>>>>>> just returns 0, and und0 is correctly decider by a decider >>>>>>>>>>>> that just returns 1.
i'm sorry, your definition of "correct" is an aglo that says >>>>>>>>>>> a terminating machine is non-terminating, and another which >>>>>>>>>>> says a non- terminating machine is terminating ...
Read it again.-a Algorithm und1 does not halt, so a decider >>>>>>>>>> that returns 0 when given und1 as input is correct for that >>>>>>>>>> input. Similarly, algorithm und0 does halt, so a decider that >>>>>>>>>> returns 1 when given und0 as input is correct for that input. >>>>>>>>>>
Sure they are.-a They return either 0 or 1 for all inputs. >>>>>>>>>> Whether they are correct for a given input depends on whether >>>>>>>>>> the return matches the mathematical mapping they're trying to >>>>>>>>>> reproduce.
i honestly can't believe u just wrote that in all
seriousness, might as well start claiming 1=0 is "correct" at >>>>>>>>>>> this point
ofc then u also did correctly decide them regardless of >>>>>>>>>>>>> what halts1 and halts0 returned, so what aglorithm did u >>>>>>>>>>>>> use to do that dud?
halts1 is one algorithm, halts0 is another algorithm.-a Just >>>>>>>>>>>> to keep things simple, halts1 returns 1 in all cases and >>>>>>>>>>>> halts0 returns 0 in all cases.
those are not genuine decision algorithms as _no_ information >>>>>>>>>>> can be extracted from _any_ of the results, such that their >>>>>>>>>>> output _cannot_ be further used in _any_ computation possible >>>>>>>>>>
They do classify.-a They just don't correct map the halting >>>>>>>>>> function for
they do not genuinely classify the input to any degree, ur >>>>>>>>>>> just mislabeling things to cover up for century old fallacy... >>>>>>>>>>
so you are actually claiming that broken clock "correctly"
tells time because it's correct twice a day???
It correctly tells time for some subset of times.
wow u really are gunna double down on that nonsense, eh? that
much time on ur hands??
If a broken analog clock is stuck at 4:45, and the current time >>>>>>>> is 4:45, that clock is correct at that time.
It doesn't matter "why" it is correct, just that it is.
bro an algorithm that randomly produces a correct response at
times without a specific reason why is just not a "correct"
algorithm by any meaningful notion of "correct"
The only requirement for an algorithm to be correct is to
reproduce a mathematical mapping.-a How it does so is irrelevant. >>>>>> I'll illustrate with a simple example.
Suppose there's a mathematical function foo whose domain is the
integers from 1 to 5 which maps its domain as follows:
1 -> 457
2 -> 26
3 -> 72
4 -> 983
5 -> 235
The following algorithm correctly maps this function:
int bar(int x) {
-a-a-a-a if (x==1) return 457;
-a-a-a-a if (x==2) return 26;
-a-a-a-a if (x==3) return 72;
-a-a-a-a if (x==4) return 983;
-a-a-a-a if (x==5) return 235;
-a-a-a-a return 0;
}
while ur arguing that actually
int foo(int x) {
-a-a if (x==1) return 0;
-a-a if (x==2) return 0;
-a-a if (x==3) return 0;
-a-a if (x==4) return 0;
-a-a if (x==5) return 0;
-a-a return 0;
}
correctly maps the function
No, I'm arguing that:
int foo(int x) {
-a-a-a if (x==1) return 0;
-a-a-a if (x==2) return 26;
-a-a-a if (x==3) return 0;
-a-a-a if (x==4) return 0;
-a-a-a if (x==5) return 0;
-a-a-a return 0;
}
Correctly maps the function for the input 2.
maybe ur a set theorist dud instead of computing guy, and don't >>>>>>> care about trivial concerns like the ability to extract useful
meaning from the response... but in computing we actually care
about practical matters, and you can't get that from constant
return functions, lol.
a constant return function is _not_ a decider, you calling it a >>>>>>> decider is just useless stupidity, else a constant return
function is apparently a decider for _all_ semantic decision
problems ever ... which is just crazy
It absolutely is a decider.-a It maps all inputs to either 0 or 1, >>>>>> which is what defines a decider.
this is fking incredible
i honestly can't believe ur really trying to argue that a constant
return function can be labeled as a correct halting decider machine.
I never said that.
I said it correctly decides halting for one class of machines,
namely either all that halt or all that do not halt, not all machines.
the correctness of an algorithm is judged across all input, not just
a subset.
And it has been proved that any attempt to produce an algorithm that
can tell if any arbitrary algorithm with a given input will halt will
fail to produce such an algorithm.
idk why u keep repeating -o1 understanding at me, i understand turing's proof better than he did
and the fault there is that's only proven for any given machine, not any given algorithm. those are not actually the same thing, as infinite
machines compute any give algorithm, and we only need to correctly
decide on one of those machines to decide it for the algorithm. this is discussed in -o5
i will not debate this further, ur attempt at a point is a useless
bastardization of computing theory
i'm not gunna waste my time debating that, it's clearly not so
and if that's kinda of argument ur going to use as an excuse to not >>>>> read my paper, ur clearly not worth my time
i cannot wait to use people like u when i give talks in the future
on much a fucking shitshow this experience has been
You mean people where you don't bother to read the one or two
sentences they've written and conclude the opposite of what they said? >>>>
You won't be taken seriously if you display such a low level of
reading comprehension.
you don't even read what u write:
-a-a| It absolutely is a decider.-a It maps all inputs to
-a-a| either 0 or 1, which is what defines a decider.
u don't even mention the any notion of correctness in ur definition
there
Correctness depends on what mathematical function it's mapping.-a A
decider that takes an integer as input always returns 1 is a correct
decider for a mathematical mapping that maps all integers to 1.
ur asserting all machines that halt with binary output from some input
are deciders, and i see with some googling that such is consensus theory
on the matter. well i reject that claim as banal and useless
and it certainly isn't a halting decider, partial or otherwise.
no
computably useful information about the semantics of the input can be extracted from the output of a constant function, so therefore it does
not function as a halting decider
i will not debate this further because it's not worth my time
good glad we've gotten to what was described on the 2nd page of >>>>>>> my 26 page paper
In the same way, a prospective halt decider that always returns >>>>>>>> 0 correctly reports the halt status of all non-halting
computations. "How" it got that result is irrelevant.
i don't know how to respond to that level of brainrot dud, it's >>>>>>>>> clearly not mathematically sound
If an algorithm maps a specific input to an output that matches >>>>>>>> a mathematical function, it correctly maps that value for that >>>>>>>> function. "How" is irrelevant.
all inputs (and in fact no algorithm does)
...two fallacies actually, one is covered in -o5, the next is >>>>>>>>>>> covered in -o6
and u didn't respond to the fact that despite halt1 and halt0 >>>>>>>>>>> being incorrect, u still classified und1 and und0 into sets >>>>>>>>>>> truthfully.
...cause that's exactly what my paper is on...
ur failing to understand that refuted ur own claims with >>>>>>>>>>>>> this very post
_that_
object would "undecidable". holy fuck. you can't just >>>>>>>>>>>>>>> continually assert undecidability of a set, YOU NEED >>>>>>>>>>>>>>> FUCKING PROOF OF AN OBJECT THAT CANNOT BE DECIDED UPON >>>>>>>>>>>>>>>
you don't actually have that with a hypothetical machine >>>>>>>>>>>>>>> that DOES NOT FUCKING EXIST...
It seem you're failing to understand proof by
contradiction, otherwise you wouldn't be asserting this. >>>>>>>>>>>>>
No, you simply don't understand what a turing machine is. >>>>>>>>>>> i'm not sure you even know what "correctness" is my dud... >>>>>>>>>>>
what aglo did u use to do that
Not one but multiple.-a und1 is correctly decided by algorithm >>>>>>>>>> halt0 and und0 is correctly decider by algorithm halt1.
yes dud, by which algorithm did you use to combine those in >>>>>>>>> order to provide actually useful knowledge on what those
machines objectively do ...
It's not about "useful knowledge".-a It's about whether those >>>>>>>> algorithms map the halting function for specific inputs.
But for the sake of argument let's say I created some algorithm, >>>>>>>> i.e. a fixed immutable sequence of instructions, that we'll call >>>>>>>> halts2 that uses *only* a machine description and that machine's >>>>>>>> input to determine whether that machine with the given input
will halt, such that halts2(und0) report 1 and halts2(und1)
returns 0. We can then construct an algorithm und2 based on
algorithm halts2 that either halts or does not halt and
halts2(und2) returns the incorrect answer, and either
halts0(und2) or halts1(und2) will be correct.
What turing's proof shows is that any algorithm that attempts to >>>>>>>> map the halting function will have at least one case it gets wrong. >>>>>>>
heck if u even read the 1st page of it, u'd know i'm responding >>>>>>> to a published paper (from jan of this year) claiming very
specifically it's incorrect to attribute the halting problem to >>>>>>> turing
but we both know u have no intention of reading it, so whatever dud >>>>>>>
and why doesn't turing's proof prevent you from doing that if >>>>>>>>>>> all algorithms can be encapsulated by the turing machine model? >>>>>>>>>>>
Turing's proof shows that no single algorithm can map the >>>>>>>>>> halting function.-a It doesn't say anything about multiple >>>>>>>>>> algorithms.
adjacent classifiers are covered in -o7.1 of the revised version >>>>>>>>>
On 9/20/2026 7:11 PM, dart200 wrote:
On 9/20/26 2:53 PM, Chris M. Thomasson wrote:
On 9/19/2026 2:38 PM, dart200 wrote:
[...]
If not, wtf! You remind me of PO. Sorry for that cut.
i don't care for ur fallacy by association brainrot
Sigh. I only said you kind of do remind me of the way PO dealt with
the halting program.
and that statement has no bearing on the correctness of my arguments,
u brain-rotted boomer
You cannot predict a random number and you cannot solve the halting
problem. Sigh.
On 9/20/2026 10:35 PM, dart200 wrote:
On 9/20/26 1:27 PM, dbush wrote:
On 9/20/2026 11:58 AM, dart200 wrote:
On 9/20/26 6:23 AM, dbush wrote:
On 9/20/2026 1:43 AM, dart200 wrote:the correctness of an algorithm is judged across all input, not just
On 9/19/26 8:17 PM, dbush wrote:
On 9/19/2026 9:20 PM, dart200 wrote:
On 9/19/26 4:27 PM, dbush wrote:
On 9/19/2026 6:40 PM, dart200 wrote:
On 9/19/26 12:13 PM, dbush wrote:
On 9/19/2026 2:40 PM, dart200 wrote:
On 9/19/26 11:17 AM, dbush wrote:
On 9/19/2026 1:57 PM, dart200 wrote:
On 9/19/26 7:45 AM, dbush wrote:
On 9/19/2026 2:19 AM, dart200 wrote:
as constructed, the halting paradox und() can neither be >>>>>>>>>>>>>>>> decided into the set, nor it's complement, by the decider >>>>>>>>>>>>>>>False.-a Given algorithm halts1 and algorithm und1 which >>>>>>>>>>>>>>> uses algorithm halts1, halts1(und1)==1 and und1 does not >>>>>>>>>>>>>>> halt. Also, Ggven algorithm halts0 and algorithm und0 >>>>>>>>>>>>>>> which uses algorithm halts0, halts0(und0)==0 and und0 >>>>>>>>>>>>>>> halts.-a So halts1 decides und1 wrong and halts0 decides >>>>>>>>>>>>>>> und0 wrong.
i think u just supported my point by presenting two cases >>>>>>>>>>>>>> where a machine failed to be decided correctly
Nope.-a Two different machine decided differently by two >>>>>>>>>>>>> different deciders.
i'm not sure how you think you can claim they are "deciders" >>>>>>>>>>>> if there's no way to extract _truthful_ information from >>>>>>>>>>>> _any_ output,
are we just randomly labeling things now?
If an algorithm returns 0 or 1 for all inputs, they are a >>>>>>>>>>> decider. Whether they are correct for that input depends on >>>>>>>>>>> whether the result matches the mathematical mapping they're >>>>>>>>>>> trying to reproduce.
i don't think u even know what turing machines are, to >>>>>>>>>>>>>>>> be frank
You're demonstrating quite clearly that it is you who >>>>>>>>>>>>>>> don't understand what turing machines are.
A Turing machine is essentially an algorithm, i.e. a >>>>>>>>>>>>>>> fixed immutable
equating the turing machine model with the concept of an >>>>>>>>>>>>>> algorithm _is_ the church-turing thesis, which has not >>>>>>>>>>>>>> been proven, that my paper refutes
algorithms are more fundamental that turing machines, and >>>>>>>>>>>>>> not all algorithms can be constructed into a turing machine >>>>>>>>>>>>>>> sequence of instructions, and is therefore defined by the >>>>>>>>>>>>>>> instructions themselves, not where the instructions >>>>>>>>>>>>>>> physically reside (i.e. the body of the function halts()). >>>>>>>>>>>>>>>
which u do by presenting objects that cannot be decided, >>>>>>>>>>>>>> like you literally just did with und1 and und0...and therefore u have lost the proof of the existence for >>>>>>>>>>>>>>>> a machine that cannot be decided into a set
That is not what the proof is.
The proof is that there is no single turing machine that >>>>>>>>>>>>>>> can decide whether any arbitrary turing machine with a >>>>>>>>>>>>>>> given input will halt.
in order to prove a set "undecidable" you need an >>>>>>>>>>>>>>>> example of an object that is cannot be decidable into >>>>>>>>>>>>>>>> that set or it's complement.
False.-a You need to show that there exists no algorithm >>>>>>>>>>>>>>> that can correctly place all objects in the correct set. >>>>>>>>>>>>>>
No, they are two separate objects which can be decided. >>>>>>>>>>>>> Specifically, und1 is correctly decided by a decider that >>>>>>>>>>>>> just returns 0, and und0 is correctly decider by a decider >>>>>>>>>>>>> that just returns 1.
i'm sorry, your definition of "correct" is an aglo that says >>>>>>>>>>>> a terminating machine is non-terminating, and another which >>>>>>>>>>>> says a non- terminating machine is terminating ...
Read it again.-a Algorithm und1 does not halt, so a decider >>>>>>>>>>> that returns 0 when given und1 as input is correct for that >>>>>>>>>>> input. Similarly, algorithm und0 does halt, so a decider that >>>>>>>>>>> returns 1 when given und0 as input is correct for that input. >>>>>>>>>>>
i honestly can't believe u just wrote that in all
seriousness, might as well start claiming 1=0 is "correct" >>>>>>>>>>>> at this point
ofc then u also did correctly decide them regardless of >>>>>>>>>>>>>> what halts1 and halts0 returned, so what aglorithm did u >>>>>>>>>>>>>> use to do that dud?
halts1 is one algorithm, halts0 is another algorithm.-a Just >>>>>>>>>>>>> to keep things simple, halts1 returns 1 in all cases and >>>>>>>>>>>>> halts0 returns 0 in all cases.
those are not genuine decision algorithms as _no_
information can be extracted from _any_ of the results, such >>>>>>>>>>>> that their output _cannot_ be further used in _any_
computation possible
Sure they are.-a They return either 0 or 1 for all inputs. >>>>>>>>>>> Whether they are correct for a given input depends on whether >>>>>>>>>>> the return matches the mathematical mapping they're trying to >>>>>>>>>>> reproduce.
They do classify.-a They just don't correct map the halting >>>>>>>>>>> function for
they do not genuinely classify the input to any degree, ur >>>>>>>>>>>> just mislabeling things to cover up for century old fallacy... >>>>>>>>>>>
so you are actually claiming that broken clock "correctly" >>>>>>>>>> tells time because it's correct twice a day???
It correctly tells time for some subset of times.
wow u really are gunna double down on that nonsense, eh? that >>>>>>>> much time on ur hands??
If a broken analog clock is stuck at 4:45, and the current time >>>>>>>>> is 4:45, that clock is correct at that time.
It doesn't matter "why" it is correct, just that it is.
bro an algorithm that randomly produces a correct response at >>>>>>>> times without a specific reason why is just not a "correct"
algorithm by any meaningful notion of "correct"
The only requirement for an algorithm to be correct is to
reproduce a mathematical mapping.-a How it does so is irrelevant. >>>>>>> I'll illustrate with a simple example.
Suppose there's a mathematical function foo whose domain is the >>>>>>> integers from 1 to 5 which maps its domain as follows:
1 -> 457
2 -> 26
3 -> 72
4 -> 983
5 -> 235
The following algorithm correctly maps this function:
int bar(int x) {
-a-a-a-a if (x==1) return 457;
-a-a-a-a if (x==2) return 26;
-a-a-a-a if (x==3) return 72;
-a-a-a-a if (x==4) return 983;
-a-a-a-a if (x==5) return 235;
-a-a-a-a return 0;
}
while ur arguing that actually
int foo(int x) {
-a-a if (x==1) return 0;
-a-a if (x==2) return 0;
-a-a if (x==3) return 0;
-a-a if (x==4) return 0;
-a-a if (x==5) return 0;
-a-a return 0;
}
correctly maps the function
No, I'm arguing that:
int foo(int x) {
-a-a-a if (x==1) return 0;
-a-a-a if (x==2) return 26;
-a-a-a if (x==3) return 0;
-a-a-a if (x==4) return 0;
-a-a-a if (x==5) return 0;
-a-a-a return 0;
}
Correctly maps the function for the input 2.
I never said that.
maybe ur a set theorist dud instead of computing guy, and don't >>>>>>>> care about trivial concerns like the ability to extract useful >>>>>>>> meaning from the response... but in computing we actually care >>>>>>>> about practical matters, and you can't get that from constant >>>>>>>> return functions, lol.
a constant return function is _not_ a decider, you calling it a >>>>>>>> decider is just useless stupidity, else a constant return
function is apparently a decider for _all_ semantic decision
problems ever ... which is just crazy
It absolutely is a decider.-a It maps all inputs to either 0 or 1, >>>>>>> which is what defines a decider.
this is fking incredible
i honestly can't believe ur really trying to argue that a constant >>>>>> return function can be labeled as a correct halting decider machine. >>>>>
I said it correctly decides halting for one class of machines,
namely either all that halt or all that do not halt, not all machines. >>>>
a subset.
And it has been proved that any attempt to produce an algorithm that
can tell if any arbitrary algorithm with a given input will halt will
fail to produce such an algorithm.
idk why u keep repeating -o1 understanding at me, i understand turing's
proof better than he did
and the fault there is that's only proven for any given machine, not
any given algorithm. those are not actually the same thing, as
infinite machines compute any give algorithm, and we only need to
correctly decide on one of those machines to decide it for the
algorithm. this is discussed in -o5
They are one in the same.-a A turing machine is essentially a
manifestation of an algorithm.-a Until you can show an algorithm that is
not a turing machine, or vice versa, Church-Turning stands.
i will not debate this further, ur attempt at a point is a useless
bastardization of computing theory
i'm not gunna waste my time debating that, it's clearly not so
and if that's kinda of argument ur going to use as an excuse to
not read my paper, ur clearly not worth my time
i cannot wait to use people like u when i give talks in the future >>>>>> on much a fucking shitshow this experience has been
You mean people where you don't bother to read the one or two
sentences they've written and conclude the opposite of what they said? >>>>>
You won't be taken seriously if you display such a low level of
reading comprehension.
you don't even read what u write:
-a-a| It absolutely is a decider.-a It maps all inputs to
-a-a| either 0 or 1, which is what defines a decider.
u don't even mention the any notion of correctness in ur definition
there
Correctness depends on what mathematical function it's mapping.-a A
decider that takes an integer as input always returns 1 is a correct
decider for a mathematical mapping that maps all integers to 1.
ur asserting all machines that halt with binary output from some input
are deciders, and i see with some googling that such is consensus
theory on the matter. well i reject that claim as banal and useless
It's not a claim.-a It's a definition.-a And if you reject it, you are essentially lying by redefining terms.
and it certainly isn't a halting decider, partial or otherwise.
It is a partial halt decider that correctly decides all halting machines.
no computably useful information about the semantics of the input can
be extracted from the output of a constant function, so therefore it
does not function as a halting decider
i will not debate this further because it's not worth my time
In the same way, a prospective halt decider that always returns >>>>>>>>> 0 correctly reports the halt status of all non-halting
computations. "How" it got that result is irrelevant.
i don't know how to respond to that level of brainrot dud, >>>>>>>>>> it's clearly not mathematically sound
If an algorithm maps a specific input to an output that matches >>>>>>>>> a mathematical function, it correctly maps that value for that >>>>>>>>> function. "How" is irrelevant.
all inputs (and in fact no algorithm does)
...two fallacies actually, one is covered in -o5, the next is >>>>>>>>>>>> covered in -o6
and u didn't respond to the fact that despite halt1 and >>>>>>>>>>>> halt0 being incorrect, u still classified und1 and und0 into >>>>>>>>>>>> sets truthfully.
...cause that's exactly what my paper is on...
ur failing to understand that refuted ur own claims with >>>>>>>>>>>>>> this very post
_that_
object would "undecidable". holy fuck. you can't just >>>>>>>>>>>>>>>> continually assert undecidability of a set, YOU NEED >>>>>>>>>>>>>>>> FUCKING PROOF OF AN OBJECT THAT CANNOT BE DECIDED UPON >>>>>>>>>>>>>>>>
you don't actually have that with a hypothetical machine >>>>>>>>>>>>>>>> that DOES NOT FUCKING EXIST...
It seem you're failing to understand proof by
contradiction, otherwise you wouldn't be asserting this. >>>>>>>>>>>>>>
No, you simply don't understand what a turing machine is. >>>>>>>>>>>> i'm not sure you even know what "correctness" is my dud... >>>>>>>>>>>>
what aglo did u use to do that
Not one but multiple.-a und1 is correctly decided by algorithm >>>>>>>>>>> halt0 and und0 is correctly decider by algorithm halt1.
yes dud, by which algorithm did you use to combine those in >>>>>>>>>> order to provide actually useful knowledge on what those
machines objectively do ...
It's not about "useful knowledge".-a It's about whether those >>>>>>>>> algorithms map the halting function for specific inputs.
But for the sake of argument let's say I created some
algorithm, i.e. a fixed immutable sequence of instructions, >>>>>>>>> that we'll call halts2 that uses *only* a machine description >>>>>>>>> and that machine's input to determine whether that machine with >>>>>>>>> the given input will halt, such that halts2(und0) report 1 and >>>>>>>>> halts2(und1) returns 0. We can then construct an algorithm und2 >>>>>>>>> based on algorithm halts2 that either halts or does not halt >>>>>>>>> and halts2(und2) returns the incorrect answer, and either
halts0(und2) or halts1(und2) will be correct.
What turing's proof shows is that any algorithm that attempts >>>>>>>>> to map the halting function will have at least one case it gets >>>>>>>>> wrong.
good glad we've gotten to what was described on the 2nd page of >>>>>>>> my 26 page paper
heck if u even read the 1st page of it, u'd know i'm responding >>>>>>>> to a published paper (from jan of this year) claiming very
specifically it's incorrect to attribute the halting problem to >>>>>>>> turing
but we both know u have no intention of reading it, so whatever dud >>>>>>>>
and why doesn't turing's proof prevent you from doing that >>>>>>>>>>>> if all algorithms can be encapsulated by the turing machine >>>>>>>>>>>> model?
Turing's proof shows that no single algorithm can map the >>>>>>>>>>> halting function.-a It doesn't say anything about multiple >>>>>>>>>>> algorithms.
adjacent classifiers are covered in -o7.1 of the revised version >>>>>>>>>>
On 9/20/26 2:25 AM, Mikko wrote:
On 19/09/2026 11:26, dart200 wrote:
On 9/19/26 1:10 AM, Mikko wrote:
On 18/09/2026 12:33, dart200 wrote:
On 9/18/26 1:54 AM, Mikko wrote:
On 16/09/2026 19:26, dart200 wrote:
On 9/16/26 12:40 AM, Mikko wrote:
On 15/09/2026 19:34, dart200 wrote:
On 9/15/26 1:18 AM, Mikko wrote:
On 14/09/2026 11:45, dart200 wrote:
On 9/14/26 1:26 AM, Mikko wrote:
On 13/09/2026 20:00, dart200 wrote:
On 9/13/26 2:41 AM, Mikko wrote:
On 12/09/2026 19:02, dart200 wrote:
On 9/12/26 2:42 AM, Mikko wrote:
On 12/09/2026 02:20, dart200 wrote:
on the nature of undecidability within computing >>>>>>>>>>>>>>>>> and refuting the church-turing thesis
preprint:
https://doi.org/10.5281/zenodo.22715823
https://www.academia.edu/175392427
comment on the live draft:
https://docs.google.com/document/
d/1BfuvPBT0RYnGvvOaiSQcpjLnG5FT3xwjBMpZMlmEeZg/edit? >>>>>>>>>>>>>>>>> usp=sharing
The section 1 "the halting problem" should clearly >>>>>>>>>>>>>>>> define what the
expression "halting problem" means. It might be useful >>>>>>>>>>>>>>>> to first
define "halting question" as that would simplify the >>>>>>>>>>>>>>>> definitions of
"halting problem" and "halting decider". In each of >>>>>>>>>>>>>>>> these it is
also inportant to be clear what is the scope of the >>>>>>>>>>>>>>>> problem and
what kind of solutions are allowed. For example, the >>>>>>>>>>>>>>>> halting
problem of finite state machines is Turing decidable and >>>>>>>>>>>>>>>> the halting
problem of Turing machines is decidable with a simple >>>>>>>>>>>>>>>> oracle.
"simple" is a weird term for something we provably can't >>>>>>>>>>>>> build, and that doesn't make the haltinging problem >>>>>>>>>>>>> "decidable" since we can't build, compute, or actually >>>>>>>>>>>>> decide with the oracle
A "simple oracle" is an oracle that is simpler than most >>>>>>>>>>>> other oracles,
just like a "big mouse" is a mouse that is bigger than most >>>>>>>>>>>> other mice
although not as big as a little elephant.
like i said: recursive undecidability leading to the concept >>>>>>>>>>> of "simple" vs "complex" oracles is discussing in -o4, i will >>>>>>>>>>> not be discussing them in -o1
That's OK, although there should be a short description of >>>>>>>>>> each section
at the end of the introduction.
But the section 1 gives the impression that the author does >>>>>>>>>> not know
what the halting problem is.
recursive undecidability is discussed in -o4
So not relevant to a discussion of section 1.
One should also clearly state that the halting of the >>>>>>>>>>>>>>>> program "und"
is not undecidable: there are partial halt deciders that >>>>>>>>>>>>>>>> can tell
whether "und" halts. In order to make this clear as soon >>>>>>>>>>>>>>>> as possible
it might be better to define "partial halt decider" >>>>>>>>>>>>>>>> before the
introduction of "und".
other classifiers are discussed in -o5.2
So not relevant to a discuddion of section 1.
that's why partial deciders are discussed in -o5.2 not -o1, you >>>>>>>>>>> brought those up not me
The section 1 is about the halting problem. You may discuss >>>>>>>>>> partial and
total solutions of the problem in the same section or
elsewhere. But
it is important to point out about "und" that its halting is >>>>>>>>>> determinable. Otherwise a reader could be confused or get the >>>>>>>>>> impression
that you are don't know.
und as specified in -o1 does not even exist, an argument which >>>>>>>>> takes up to the end of -o4 to explain
The section 1 says otherwise: the words "can be constructed"
mean that
it does exist.
it exist as a hypothetical problem, constructed from a
hypothetical decider, that presents a real limit to turing
machine computability, but does not exist in the actual
enumeration of turing machines,
In the world of computing theory everything is hypothetical. The
section
1 does not call anything hypothetical. It merely constructs "und" >>>>>> from
"halts" already specified and "loop" that is not specified.
if u can't figure out what halts does contrasted from loop in that
pseudo-code, when we are discussing the halting problem in a group
called comp.theory, this paper is just not for you, at this time
An attempt to change the topic indicates that you don't want to
discuss honestly but instead want to deceive.
while u literally just changed the topic u dishonest fuck
A meta comment about the discussion is not a change of topic. I already
god ur a fucking twat bro, can't even admit u changed the topic
On 9/20/2026 7:11 PM, dart200 wrote:
On 9/20/26 2:53 PM, Chris M. Thomasson wrote:
On 9/19/2026 2:38 PM, dart200 wrote:
[...]
If not, wtf! You remind me of PO. Sorry for that cut.
i don't care for ur fallacy by association brainrot
Sigh. I only said you kind of do remind me of the way PO dealt with
the halting program.
and that statement has no bearing on the correctness of my arguments,
u brain-rotted boomer
You cannot predict a random number and you cannot solve the halting
problem. Sigh.
On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote:
They are one in the same.-a A turing machine is essentially a
manifestation of an algorithm.-a Until you can show an algorithm that
is not a turing machine, or vice versa, Church-Turning stands.
that's what -o7 is on dud! maybe u should read the fking paper instead of dicking about in -o1!
it ultimately discusses how to compute both a total diagonal and total anti-diagonal across the enumeration of turing computable sequences, a computation which obviously cannot be done with a turing machine, as per turing's original proof on the matter. and demonstrated that is a 26
page paper which shall not be able to condense into a damn usenet post
ur asserting all machines that halt with binary output from some
input are deciders, and i see with some googling that such is
consensus theory on the matter. well i reject that claim as banal and
useless
It's not a claim.-a It's a definition.-a And if you reject it, you are
essentially lying by redefining terms.
says the dud trying to declare a constant return as actually a "partial halting decider". u can't make this shit up!
and it certainly isn't a halting decider, partial or otherwise.
It is a partial halt decider that correctly decides all halting machines.
i reject that definition
because it has lead you to the confusion of you
trying to claim that a constant return function is a partial halting decider, when no meaningful information on the semantics can be
extracted from the return, as it will mix in non-halting machine in with that return, so no meaningful decision was produced
saying something is a "correct decision" must be judged across the
entirety of the output, not just cherry-picked examples
why? because i cannot computably use a constant return function to
decide whether a machine halts or not, across as partial subset or
otherwise
-a(m) -> {
-a-a if (halts(m))
-a-a-a-a m()
-a-a else
-a-a-a-a print "m does not halt!"
-a}
if halts() is just the constant return function () -> 1 then all input machines will be run, and the if statement trying to function as a guard against non-halting input is functionally useless
the requirement for a specific decider is there is a way to extract information about what it is supposed to be decided upon. a constant
return function decides all possible input strings, which is trivially decidable with a constant return function,
it is _not_ a halting decider, partial or otherwise.
anyways, i cover types of classifiers, machine that classify an input machine into a semantic set via a binary output, in -o5.2--- Synchronet 3.22a-Linux NewsLink 1.2
read the paper cause this discussion is over until u do regardless of
how much more misinformed bs u try to post
On 9/21/2026 2:00 AM, dart200 wrote:
On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote:
They are one in the same.-a A turing machine is essentially a
manifestation of an algorithm.-a Until you can show an algorithm that
is not a turing machine, or vice versa, Church-Turning stands.
that's what -o7 is on dud! maybe u should read the fking paper instead
of dicking about in -o1!
it ultimately discusses how to compute both a total diagonal and total
anti-diagonal across the enumeration of turing computable sequences, a
computation which obviously cannot be done with a turing machine, as
per turing's original proof on the matter. and demonstrated that is a
26 page paper which shall not be able to condense into a damn usenet post
Which you do by getting unspecified input from a human, which means it's
not an algorithm.-a This also means it takes input other than a machine description and the input to that machine, which are the only allowed
inputs to a halt decider.
Anything you're doing regarding the address of a function is simply
adding another input which means you're changing the question.
ur asserting all machines that halt with binary output from some
input are deciders, and i see with some googling that such is
consensus theory on the matter. well i reject that claim as banal
and useless
It's not a claim.-a It's a definition.-a And if you reject it, you are
essentially lying by redefining terms.
says the dud trying to declare a constant return as actually a
"partial halting decider". u can't make this shit up!
It fits the definition.
and it certainly isn't a halting decider, partial or otherwise.
It is a partial halt decider that correctly decides all halting
machines.
i reject that definition
In other words you intend to lie by redefining terms.
because it has lead you to the confusion of you trying to claim that a
constant return function is a partial halting decider, when no
meaningful information on the semantics can be extracted from the
return, as it will mix in non-halting machine in with that return, so
no meaningful decision was produced
Which is no different from any other partial halt decider.
saying something is a "correct decision" must be judged across the
entirety of the output, not just cherry-picked examples
In other words, any partial halt decider is simply not correct.
why? because i cannot computably use a constant return function to
decide whether a machine halts or not, across as partial subset or
otherwise
-a-a(m) -> {
-a-a-a if (halts(m))
-a-a-a-a-a m()
-a-a-a else
-a-a-a-a-a print "m does not halt!"
-a-a}
if halts() is just the constant return function () -> 1 then all input
machines will be run, and the if statement trying to function as a
guard against non-halting input is functionally useless
So such a function wouldn't work in that specific instance.-a And in fact
no function will.
the requirement for a specific decider is there is a way to extract
information about what it is supposed to be decided upon. a constant
return function decides all possible input strings, which is trivially
decidable with a constant return function,
it is _not_ a halting decider, partial or otherwise.
Any algorithm that takes a description of an algorithm and its input and returns either 0 or 1 for all inputs is by definition a partial halt decider.
--
anyways, i cover types of classifiers, machine that classify an input
machine into a semantic set via a binary output, in -o5.2
read the paper cause this discussion is over until u do regardless of
how much more misinformed bs u try to post
On 9/21/26 6:34 AM, dbush wrote:
On 9/21/2026 2:00 AM, dart200 wrote:
On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote:
They are one in the same.-a A turing machine is essentially a
manifestation of an algorithm.-a Until you can show an algorithm that >>>> is not a turing machine, or vice versa, Church-Turning stands.
that's what -o7 is on dud! maybe u should read the fking paper instead
of dicking about in -o1!
it ultimately discusses how to compute both a total diagonal and
total anti-diagonal across the enumeration of turing computable
sequences, a computation which obviously cannot be done with a turing
machine, as per turing's original proof on the matter. and
demonstrated that is a 26 page paper which shall not be able to
condense into a damn usenet post
Which you do by getting unspecified input from a human, which means
it's not an algorithm.-a This also means it takes input other than a
machine description and the input to that machine, which are the only
allowed inputs to a halt decider.
Anything you're doing regarding the address of a function is simply
adding another input which means you're changing the question.
-o7.6 - truly begging a question
ur asserting all machines that halt with binary output from some
input are deciders, and i see with some googling that such is
consensus theory on the matter. well i reject that claim as banal
and useless
It's not a claim.-a It's a definition.-a And if you reject it, you are >>>> essentially lying by redefining terms.
says the dud trying to declare a constant return as actually a
"partial halting decider". u can't make this shit up!
It fits the definition.
and it certainly isn't a halting decider, partial or otherwise.
It is a partial halt decider that correctly decides all halting
machines.
i reject that definition
In other words you intend to lie by redefining terms.
because it has lead you to the confusion of you trying to claim that
a constant return function is a partial halting decider, when no
meaningful information on the semantics can be extracted from the
return, as it will mix in non-halting machine in with that return, so
no meaningful decision was produced
Which is no different from any other partial halt decider.
saying something is a "correct decision" must be judged across the
entirety of the output, not just cherry-picked examples
In other words, any partial halt decider is simply not correct.
why? because i cannot computably use a constant return function to
decide whether a machine halts or not, across as partial subset or
otherwise
-a-a(m) -> {
-a-a-a if (halts(m))
-a-a-a-a-a m()
-a-a-a else
-a-a-a-a-a print "m does not halt!"
-a-a}
if halts() is just the constant return function () -> 1 then all
input machines will be run, and the if statement trying to function
as a guard against non-halting input is functionally useless
So such a function wouldn't work in that specific instance.-a And in
fact no function will.
the a halting partial recognizer (from -o5.2) will prevent any non- terminating function from being run and therefore functionally act as a guard
the requirement for a specific decider is there is a way to extract
information about what it is supposed to be decided upon. a constant
return function decides all possible input strings, which is
trivially decidable with a constant return function,
it is _not_ a halting decider, partial or otherwise.
Any algorithm that takes a description of an algorithm and its input
and returns either 0 or 1 for all inputs is by definition a partial
halt decider.
i'm not going to argue against something so stupid
anyways, i cover types of classifiers, machine that classify an input
machine into a semantic set via a binary output, in -o5.2
read the paper cause this discussion is over until u do regardless of
how much more misinformed bs u try to post
On 9/21/2026 1:18 PM, dart200 wrote:
On 9/21/26 6:34 AM, dbush wrote:
On 9/21/2026 2:00 AM, dart200 wrote:
On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote:
They are one in the same.-a A turing machine is essentially a
manifestation of an algorithm.-a Until you can show an algorithm
that is not a turing machine, or vice versa, Church-Turning stands.
that's what -o7 is on dud! maybe u should read the fking paper
instead of dicking about in -o1!
it ultimately discusses how to compute both a total diagonal and
total anti-diagonal across the enumeration of turing computable
sequences, a computation which obviously cannot be done with a
turing machine, as per turing's original proof on the matter. and
demonstrated that is a 26 page paper which shall not be able to
condense into a damn usenet post
Which you do by getting unspecified input from a human, which means
it's not an algorithm.-a This also means it takes input other than a
machine description and the input to that machine, which are the only
allowed inputs to a halt decider.
Anything you're doing regarding the address of a function is simply
adding another input which means you're changing the question.
-o7.6 - truly begging a question
So what the human agent is recording becomes an input.-a And therefore whatever uses it is disqualified from being a halt decider, partial or otherwise.
Also:
"We will start by amending the basic turing machine into a terminal
machine"
Which means you no longer have a Turing machine so nothing that follows applies to Turing machines.
ur asserting all machines that halt with binary output from some
input are deciders, and i see with some googling that such is
consensus theory on the matter. well i reject that claim as banal >>>>>> and useless
It's not a claim.-a It's a definition.-a And if you reject it, you
are essentially lying by redefining terms.
says the dud trying to declare a constant return as actually a
"partial halting decider". u can't make this shit up!
It fits the definition.
No response to this, so I and others reading this will have to assume
you agree
and it certainly isn't a halting decider, partial or otherwise.
It is a partial halt decider that correctly decides all halting
machines.
i reject that definition
In other words you intend to lie by redefining terms.
The lack of response to this is telling.
because it has lead you to the confusion of you trying to claim that
a constant return function is a partial halting decider, when no
meaningful information on the semantics can be extracted from the
return, as it will mix in non-halting machine in with that return,
so no meaningful decision was produced
Which is no different from any other partial halt decider.
saying something is a "correct decision" must be judged across the
entirety of the output, not just cherry-picked examples
In other words, any partial halt decider is simply not correct.
why? because i cannot computably use a constant return function to
decide whether a machine halts or not, across as partial subset or
otherwise
-a-a(m) -> {
-a-a-a if (halts(m))
-a-a-a-a-a m()
-a-a-a else
-a-a-a-a-a print "m does not halt!"
-a-a}
if halts() is just the constant return function () -> 1 then all
input machines will be run, and the if statement trying to function
as a guard against non-halting input is functionally useless
So such a function wouldn't work in that specific instance.-a And in
fact no function will.
the a halting partial recognizer (from -o5.2) will prevent any non-
terminating function from being run and therefore functionally act as
a guard
So will a constant function that returns 0.-a Or (for example) a function that doesn't find a "return" instruction among the first N instructions.
So it's not particularly interesting.
the requirement for a specific decider is there is a way to extract
information about what it is supposed to be decided upon. a constant
return function decides all possible input strings, which is
trivially decidable with a constant return function,
it is _not_ a halting decider, partial or otherwise.
Any algorithm that takes a description of an algorithm and its input
and returns either 0 or 1 for all inputs is by definition a partial
halt decider.
i'm not going to argue against something so stupid
Arguing against definitions would be stupid.
anyways, i cover types of classifiers, machine that classify an
input machine into a semantic set via a binary output, in -o5.2
read the paper cause this discussion is over until u do regardless
of how much more misinformed bs u try to post
On 9/21/26 11:53 AM, dbush wrote:
On 9/21/2026 1:18 PM, dart200 wrote:
On 9/21/26 6:34 AM, dbush wrote:
On 9/21/2026 2:00 AM, dart200 wrote:
On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote:that's what -o7 is on dud! maybe u should read the fking paper
They are one in the same.-a A turing machine is essentially a
manifestation of an algorithm.-a Until you can show an algorithm
that is not a turing machine, or vice versa, Church-Turning stands. >>>>>
instead of dicking about in -o1!
it ultimately discusses how to compute both a total diagonal and
total anti-diagonal across the enumeration of turing computable
sequences, a computation which obviously cannot be done with a
turing machine, as per turing's original proof on the matter. and
demonstrated that is a 26 page paper which shall not be able to
condense into a damn usenet post
Which you do by getting unspecified input from a human, which means
it's not an algorithm.-a This also means it takes input other than a
machine description and the input to that machine, which are the
only allowed inputs to a halt decider.
Anything you're doing regarding the address of a function is simply
adding another input which means you're changing the question.
-o7.6 - truly begging a question
So what the human agent is recording becomes an input.-a And therefore
whatever uses it is disqualified from being a halt decider, partial or
otherwise.
the human agent produces a response back to the terminal machine, yes
but he also computes things on the side that persist between terminal interactions, and it's in this side record where he can compute things outside the turing machine model
Also:
"We will start by amending the basic turing machine into a terminal
machine"
Which means you no longer have a Turing machine so nothing that
follows applies to Turing machines.
_duh_ ... the total solution to turing machine halting can only be
computed from _outside_ the turing machine model
this does not it make not computation, as it still a method that can
produce a sequence, including both a true diagonal across turing
computable numbers and the true anti-diagonal across turing computable sequences, both of which are outside the scope of turing computation
ur asserting all machines that halt with binary output from some >>>>>>> input are deciders, and i see with some googling that such is
consensus theory on the matter. well i reject that claim as banal >>>>>>> and useless
It's not a claim.-a It's a definition.-a And if you reject it, you >>>>>> are essentially lying by redefining terms.
says the dud trying to declare a constant return as actually a
"partial halting decider". u can't make this shit up!
It fits the definition.
No response to this, so I and others reading this will have to assume
you agree
only if ur ignorant to what a argumentum ex silentio fallacy is
and it certainly isn't a halting decider, partial or otherwise.
It is a partial halt decider that correctly decides all halting
machines.
i reject that definition
In other words you intend to lie by redefining terms.
The lack of response to this is telling.
because it has lead you to the confusion of you trying to claim
that a constant return function is a partial halting decider, when
no meaningful information on the semantics can be extracted from
the return, as it will mix in non-halting machine in with that
return, so no meaningful decision was produced
Which is no different from any other partial halt decider.
saying something is a "correct decision" must be judged across the
entirety of the output, not just cherry-picked examples
In other words, any partial halt decider is simply not correct.
why? because i cannot computably use a constant return function to
decide whether a machine halts or not, across as partial subset or
otherwise
-a-a(m) -> {
-a-a-a if (halts(m))
-a-a-a-a-a m()
-a-a-a else
-a-a-a-a-a print "m does not halt!"
-a-a}
if halts() is just the constant return function () -> 1 then all
input machines will be run, and the if statement trying to function >>>>> as a guard against non-halting input is functionally useless
So such a function wouldn't work in that specific instance.-a And in
fact no function will.
the a halting partial recognizer (from -o5.2) will prevent any non-
terminating function from being run and therefore functionally act as
a guard
So will a constant function that returns 0.-a Or (for example) a
function that doesn't find a "return" instruction among the first N
instructions.
So it's not particularly interesting.
as demonstrated by -o6, for any given possible output/turing-computable sequence there _exists_ a machine that both halts, and is within the set decidable by partial recognizer halts()
which makes it incredibly useful, and beyond anything previously acknowledged within the theory of computing
the requirement for a specific decider is there is a way to extract >>>>> information about what it is supposed to be decided upon. a
constant return function decides all possible input strings, which
is trivially decidable with a constant return function,
it is _not_ a halting decider, partial or otherwise.
Any algorithm that takes a description of an algorithm and its input
and returns either 0 or 1 for all inputs is by definition a partial
halt decider.
i'm not going to argue against something so stupid
Arguing against definitions would be stupid.
twisting my words is also stupid
anyways, i cover types of classifiers, machine that classify an
input machine into a semantic set via a binary output, in -o5.2
read the paper cause this discussion is over until u do regardless
of how much more misinformed bs u try to post
Chris M. Thomasson wrote:
On 9/20/2026 7:11 PM, dart200 wrote:
On 9/20/26 2:53 PM, Chris M. Thomasson wrote:
On 9/19/2026 2:38 PM, dart200 wrote:
[...]
If not, wtf! You remind me of PO. Sorry for that cut.
i don't care for ur fallacy by association brainrot
Sigh. I only said you kind of do remind me of the way PO dealt with
the halting program.
and that statement has no bearing on the correctness of my arguments,
u brain-rotted boomer
You cannot predict a random number and you cannot solve the halting
problem. Sigh.
You're quite the disappointment, dart200. We'll have to ask some of the other guys to get this done.
On 9/21/2026 4:06 PM, dart200 wrote:
On 9/21/26 11:53 AM, dbush wrote:
On 9/21/2026 1:18 PM, dart200 wrote:
On 9/21/26 6:34 AM, dbush wrote:
On 9/21/2026 2:00 AM, dart200 wrote:
On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote:that's what -o7 is on dud! maybe u should read the fking paper
They are one in the same.-a A turing machine is essentially a
manifestation of an algorithm.-a Until you can show an algorithm >>>>>>> that is not a turing machine, or vice versa, Church-Turning stands. >>>>>>
instead of dicking about in -o1!
it ultimately discusses how to compute both a total diagonal and
total anti-diagonal across the enumeration of turing computable
sequences, a computation which obviously cannot be done with a
turing machine, as per turing's original proof on the matter. and >>>>>> demonstrated that is a 26 page paper which shall not be able to
condense into a damn usenet post
Which you do by getting unspecified input from a human, which means >>>>> it's not an algorithm.-a This also means it takes input other than a >>>>> machine description and the input to that machine, which are the
only allowed inputs to a halt decider.
Anything you're doing regarding the address of a function is simply >>>>> adding another input which means you're changing the question.
-o7.6 - truly begging a question
So what the human agent is recording becomes an input.-a And therefore
whatever uses it is disqualified from being a halt decider, partial
or otherwise.
the human agent produces a response back to the terminal machine, yes
but he also computes things on the side that persist between terminal
interactions, and it's in this side record where he can compute things
outside the turing machine model
And that side record is what disqualifies it from being a halt decider.
Also:
"We will start by amending the basic turing machine into a terminal
machine"
Which means you no longer have a Turing machine so nothing that
follows applies to Turing machines.
_duh_ ... the total solution to turing machine halting can only be
computed from _outside_ the turing machine model
Which is already well-known and therefore uninteresting.
this does not it make not computation, as it still a method that can
produce a sequence, including both a true diagonal across turing
computable numbers and the true anti-diagonal across turing computable
sequences, both of which are outside the scope of turing computation
It makes it a computation that has input in addition to a machine description and that machine's input, and therefore disqualified from
being a halt decider.
Which is the same mistake PO made.
ur asserting all machines that halt with binary output from some >>>>>>>> input are deciders, and i see with some googling that such is >>>>>>>> consensus theory on the matter. well i reject that claim as
banal and useless
It's not a claim.-a It's a definition.-a And if you reject it, you >>>>>>> are essentially lying by redefining terms.
says the dud trying to declare a constant return as actually a
"partial halting decider". u can't make this shit up!
It fits the definition.
No response to this, so I and others reading this will have to assume
you agree
only if ur ignorant to what a argumentum ex silentio fallacy is
Which only applies to historical accounts, not a running conversation.
It is a partial halt decider that correctly decides all halting >>>>>>> machines.
and it certainly isn't a halting decider, partial or otherwise. >>>>>>>
i reject that definition
In other words you intend to lie by redefining terms.
The lack of response to this is telling.
because it has lead you to the confusion of you trying to claim
that a constant return function is a partial halting decider, when >>>>>> no meaningful information on the semantics can be extracted from
the return, as it will mix in non-halting machine in with that
return, so no meaningful decision was produced
Which is no different from any other partial halt decider.
saying something is a "correct decision" must be judged across the >>>>>> entirety of the output, not just cherry-picked examples
In other words, any partial halt decider is simply not correct.
So such a function wouldn't work in that specific instance.-a And in >>>>> fact no function will.
why? because i cannot computably use a constant return function to >>>>>> decide whether a machine halts or not, across as partial subset or >>>>>> otherwise
-a-a(m) -> {
-a-a-a if (halts(m))
-a-a-a-a-a m()
-a-a-a else
-a-a-a-a-a print "m does not halt!"
-a-a}
if halts() is just the constant return function () -> 1 then all
input machines will be run, and the if statement trying to
function as a guard against non-halting input is functionally useless >>>>>
the a halting partial recognizer (from -o5.2) will prevent any non-
terminating function from being run and therefore functionally act
as a guard
So will a constant function that returns 0.-a Or (for example) a
function that doesn't find a "return" instruction among the first N
instructions.
So it's not particularly interesting.
as demonstrated by -o6, for any given possible output/turing-computable
sequence there _exists_ a machine that both halts, and is within the
set decidable by partial recognizer halts()
which makes it incredibly useful, and beyond anything previously
acknowledged within the theory of computing
the requirement for a specific decider is there is a way to
extract information about what it is supposed to be decided upon. >>>>>> a constant return function decides all possible input strings,
which is trivially decidable with a constant return function,
it is _not_ a halting decider, partial or otherwise.
Any algorithm that takes a description of an algorithm and its
input and returns either 0 or 1 for all inputs is by definition a
partial halt decider.
i'm not going to argue against something so stupid
Arguing against definitions would be stupid.
twisting my words is also stupid
anyways, i cover types of classifiers, machine that classify an
input machine into a semantic set via a binary output, in -o5.2
read the paper cause this discussion is over until u do regardless >>>>>> of how much more misinformed bs u try to post
On 9/21/26 1:41 PM, dbush wrote:
On 9/21/2026 4:06 PM, dart200 wrote:
On 9/21/26 11:53 AM, dbush wrote:
On 9/21/2026 1:18 PM, dart200 wrote:
On 9/21/26 6:34 AM, dbush wrote:
On 9/21/2026 2:00 AM, dart200 wrote:-o7.6 - truly begging a question
On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote:that's what -o7 is on dud! maybe u should read the fking paper
They are one in the same.-a A turing machine is essentially a >>>>>>>> manifestation of an algorithm.-a Until you can show an algorithm >>>>>>>> that is not a turing machine, or vice versa, Church-Turning stands. >>>>>>>
instead of dicking about in -o1!
it ultimately discusses how to compute both a total diagonal and >>>>>>> total anti-diagonal across the enumeration of turing computable >>>>>>> sequences, a computation which obviously cannot be done with a
turing machine, as per turing's original proof on the matter. and >>>>>>> demonstrated that is a 26 page paper which shall not be able to >>>>>>> condense into a damn usenet post
Which you do by getting unspecified input from a human, which
means it's not an algorithm.-a This also means it takes input other >>>>>> than a machine description and the input to that machine, which
are the only allowed inputs to a halt decider.
Anything you're doing regarding the address of a function is
simply adding another input which means you're changing the question. >>>>>
So what the human agent is recording becomes an input.-a And
therefore whatever uses it is disqualified from being a halt
decider, partial or otherwise.
the human agent produces a response back to the terminal machine, yes
but he also computes things on the side that persist between terminal
interactions, and it's in this side record where he can compute
things outside the turing machine model
And that side record is what disqualifies it from being a halt decider.
the knowledge is still computable, even if it can't be used in total
within the turing machine model
Also:
"We will start by amending the basic turing machine into a terminal
machine"
Which means you no longer have a Turing machine so nothing that
follows applies to Turing machines.
_duh_ ... the total solution to turing machine halting can only be
computed from _outside_ the turing machine model
Which is already well-known and therefore uninteresting.
not by a mechanically realizable form it isn't, oracle machines can be
made real, and are largely useless beyond describing something that realizable computing is surely not.
this does not it make not computation, as it still a method that can
produce a sequence, including both a true diagonal across turing
computable numbers and the true anti-diagonal across turing
computable sequences, both of which are outside the scope of turing
computation
It makes it a computation that has input in addition to a machine
description and that machine's input, and therefore disqualified from
being a halt decider.
like i already said, this is addressed in -o7.6
Which is the same mistake PO made.
u do know that begging the question is a fallacy, eh?
ur asserting all machines that halt with binary output from >>>>>>>>> some input are deciders, and i see with some googling that such >>>>>>>>> is consensus theory on the matter. well i reject that claim as >>>>>>>>> banal and useless
It's not a claim.-a It's a definition.-a And if you reject it, you >>>>>>>> are essentially lying by redefining terms.
says the dud trying to declare a constant return as actually a
"partial halting decider". u can't make this shit up!
It fits the definition.
No response to this, so I and others reading this will have to
assume you agree
only if ur ignorant to what a argumentum ex silentio fallacy is
Which only applies to historical accounts, not a running conversation.
wow, ur actually gunna try to double down on a fallacy i literally just named? jesus fucking christ...
It is a partial halt decider that correctly decides all halting >>>>>>>> machines.
and it certainly isn't a halting decider, partial or otherwise. >>>>>>>>
i reject that definition
In other words you intend to lie by redefining terms.
The lack of response to this is telling.
because it has lead you to the confusion of you trying to claim >>>>>>> that a constant return function is a partial halting decider,
when no meaningful information on the semantics can be extracted >>>>>>> from the return, as it will mix in non-halting machine in with
that return, so no meaningful decision was produced
Which is no different from any other partial halt decider.
saying something is a "correct decision" must be judged across
the entirety of the output, not just cherry-picked examples
In other words, any partial halt decider is simply not correct.
why? because i cannot computably use a constant return function >>>>>>> to decide whether a machine halts or not, across as partial
subset or otherwise
-a-a(m) -> {
-a-a-a if (halts(m))
-a-a-a-a-a m()
-a-a-a else
-a-a-a-a-a print "m does not halt!"
-a-a}
if halts() is just the constant return function () -> 1 then all >>>>>>> input machines will be run, and the if statement trying to
function as a guard against non-halting input is functionally
useless
So such a function wouldn't work in that specific instance.-a And >>>>>> in fact no function will.
the a halting partial recognizer (from -o5.2) will prevent any non- >>>>> terminating function from being run and therefore functionally act
as a guard
So will a constant function that returns 0.-a Or (for example) a
function that doesn't find a "return" instruction among the first N
instructions.
So it's not particularly interesting.
as demonstrated by -o6, for any given possible output/turing-
computable sequence there _exists_ a machine that both halts, and is
within the set decidable by partial recognizer halts()
which makes it incredibly useful, and beyond anything previously
acknowledged within the theory of computing
the requirement for a specific decider is there is a way to
extract information about what it is supposed to be decided upon. >>>>>>> a constant return function decides all possible input strings,
which is trivially decidable with a constant return function,
it is _not_ a halting decider, partial or otherwise.
Any algorithm that takes a description of an algorithm and its
input and returns either 0 or 1 for all inputs is by definition a >>>>>> partial halt decider.
i'm not going to argue against something so stupid
Arguing against definitions would be stupid.
twisting my words is also stupid
anyways, i cover types of classifiers, machine that classify an >>>>>>> input machine into a semantic set via a binary output, in -o5.2
read the paper cause this discussion is over until u do
regardless of how much more misinformed bs u try to post
On 9/21/2026 11:10 PM, dart200 wrote:
On 9/21/26 1:41 PM, dbush wrote:
On 9/21/2026 4:06 PM, dart200 wrote:
On 9/21/26 11:53 AM, dbush wrote:
On 9/21/2026 1:18 PM, dart200 wrote:
On 9/21/26 6:34 AM, dbush wrote:
On 9/21/2026 2:00 AM, dart200 wrote:
On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote:
They are one in the same.-a A turing machine is essentially a >>>>>>>>> manifestation of an algorithm.-a Until you can show an algorithm >>>>>>>>> that is not a turing machine, or vice versa, Church-Turning >>>>>>>>> stands.
that's what -o7 is on dud! maybe u should read the fking paper >>>>>>>> instead of dicking about in -o1!
it ultimately discusses how to compute both a total diagonal and >>>>>>>> total anti-diagonal across the enumeration of turing computable >>>>>>>> sequences, a computation which obviously cannot be done with a >>>>>>>> turing machine, as per turing's original proof on the matter. >>>>>>>> and demonstrated that is a 26 page paper which shall not be able >>>>>>>> to condense into a damn usenet post
Which you do by getting unspecified input from a human, which
means it's not an algorithm.-a This also means it takes input
other than a machine description and the input to that machine, >>>>>>> which are the only allowed inputs to a halt decider.
Anything you're doing regarding the address of a function is
simply adding another input which means you're changing the
question.
-o7.6 - truly begging a question
So what the human agent is recording becomes an input.-a And
therefore whatever uses it is disqualified from being a halt
decider, partial or otherwise.
the human agent produces a response back to the terminal machine, yes
but he also computes things on the side that persist between
terminal interactions, and it's in this side record where he can
compute things outside the turing machine model
And that side record is what disqualifies it from being a halt decider.
the knowledge is still computable, even if it can't be used in total
within the turing machine model
It is computable in the turing machine model from a machine description
and "extra" data, not just a machine description, so not a halt decider, partial or otherwise.
Also:
"We will start by amending the basic turing machine into a terminal >>>>> machine"
Which means you no longer have a Turing machine so nothing that
follows applies to Turing machines.
_duh_ ... the total solution to turing machine halting can only be
computed from _outside_ the turing machine model
Which is already well-known and therefore uninteresting.
not by a mechanically realizable form it isn't, oracle machines can be
made real, and are largely useless beyond describing something that
realizable computing is surely not.
this does not it make not computation, as it still a method that can
produce a sequence, including both a true diagonal across turing
computable numbers and the true anti-diagonal across turing
computable sequences, both of which are outside the scope of turing
computation
It makes it a computation that has input in addition to a machine
description and that machine's input, and therefore disqualified from
being a halt decider.
like i already said, this is addressed in -o7.6
And you addressed it by using extra input disqualifying it from being a
halt decider.
Which is the same mistake PO made.
u do know that begging the question is a fallacy, eh?
Identifying a fundamental mistake is not a fallacy.
ur asserting all machines that halt with binary output from >>>>>>>>>> some input are deciders, and i see with some googling that >>>>>>>>>> such is consensus theory on the matter. well i reject that >>>>>>>>>> claim as banal and useless
It's not a claim.-a It's a definition.-a And if you reject it, >>>>>>>>> you are essentially lying by redefining terms.
says the dud trying to declare a constant return as actually a >>>>>>>> "partial halting decider". u can't make this shit up!
It fits the definition.
No response to this, so I and others reading this will have to
assume you agree
only if ur ignorant to what a argumentum ex silentio fallacy is
Which only applies to historical accounts, not a running conversation.
wow, ur actually gunna try to double down on a fallacy i literally
just named? jesus fucking christ...
You mean a fallacy that doesn't apply in this case?
It is a partial halt decider that correctly decides all halting >>>>>>>>> machines.
and it certainly isn't a halting decider, partial or otherwise. >>>>>>>>>
i reject that definition
In other words you intend to lie by redefining terms.
The lack of response to this is telling.
because it has lead you to the confusion of you trying to claim >>>>>>>> that a constant return function is a partial halting decider, >>>>>>>> when no meaningful information on the semantics can be extracted >>>>>>>> from the return, as it will mix in non-halting machine in with >>>>>>>> that return, so no meaningful decision was produced
Which is no different from any other partial halt decider.
saying something is a "correct decision" must be judged across >>>>>>>> the entirety of the output, not just cherry-picked examples
In other words, any partial halt decider is simply not correct.
why? because i cannot computably use a constant return function >>>>>>>> to decide whether a machine halts or not, across as partial
subset or otherwise
-a-a(m) -> {
-a-a-a if (halts(m))
-a-a-a-a-a m()
-a-a-a else
-a-a-a-a-a print "m does not halt!"
-a-a}
if halts() is just the constant return function () -> 1 then all >>>>>>>> input machines will be run, and the if statement trying to
function as a guard against non-halting input is functionally >>>>>>>> useless
So such a function wouldn't work in that specific instance.-a And >>>>>>> in fact no function will.
the a halting partial recognizer (from -o5.2) will prevent any non- >>>>>> terminating function from being run and therefore functionally act >>>>>> as a guard
So will a constant function that returns 0.-a Or (for example) a
function that doesn't find a "return" instruction among the first N >>>>> instructions.
So it's not particularly interesting.
as demonstrated by -o6, for any given possible output/turing-
computable sequence there _exists_ a machine that both halts, and is
within the set decidable by partial recognizer halts()
which makes it incredibly useful, and beyond anything previously
acknowledged within the theory of computing
the requirement for a specific decider is there is a way to
extract information about what it is supposed to be decided
upon. a constant return function decides all possible input
strings, which is trivially decidable with a constant return
function,
it is _not_ a halting decider, partial or otherwise.
Any algorithm that takes a description of an algorithm and its
input and returns either 0 or 1 for all inputs is by definition a >>>>>>> partial halt decider.
i'm not going to argue against something so stupid
Arguing against definitions would be stupid.
twisting my words is also stupid
anyways, i cover types of classifiers, machine that classify an >>>>>>>> input machine into a semantic set via a binary output, in -o5.2 >>>>>>>>
read the paper cause this discussion is over until u do
regardless of how much more misinformed bs u try to post
On 9/21/26 8:18 PM, dbush wrote:
On 9/21/2026 11:10 PM, dart200 wrote:
On 9/21/26 1:41 PM, dbush wrote:
On 9/21/2026 4:06 PM, dart200 wrote:the knowledge is still computable, even if it can't be used in total
On 9/21/26 11:53 AM, dbush wrote:
On 9/21/2026 1:18 PM, dart200 wrote:
On 9/21/26 6:34 AM, dbush wrote:
On 9/21/2026 2:00 AM, dart200 wrote:
On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote:
They are one in the same.-a A turing machine is essentially a >>>>>>>>>> manifestation of an algorithm.-a Until you can show an
algorithm that is not a turing machine, or vice versa, Church- >>>>>>>>>> Turning stands.
that's what -o7 is on dud! maybe u should read the fking paper >>>>>>>>> instead of dicking about in -o1!
it ultimately discusses how to compute both a total diagonal >>>>>>>>> and total anti-diagonal across the enumeration of turing
computable sequences, a computation which obviously cannot be >>>>>>>>> done with a turing machine, as per turing's original proof on >>>>>>>>> the matter. and demonstrated that is a 26 page paper which
shall not be able to condense into a damn usenet post
Which you do by getting unspecified input from a human, which >>>>>>>> means it's not an algorithm.-a This also means it takes input >>>>>>>> other than a machine description and the input to that machine, >>>>>>>> which are the only allowed inputs to a halt decider.
Anything you're doing regarding the address of a function is
simply adding another input which means you're changing the
question.
-o7.6 - truly begging a question
So what the human agent is recording becomes an input.-a And
therefore whatever uses it is disqualified from being a halt
decider, partial or otherwise.
the human agent produces a response back to the terminal machine, yes >>>>>
but he also computes things on the side that persist between
terminal interactions, and it's in this side record where he can
compute things outside the turing machine model
And that side record is what disqualifies it from being a halt decider. >>>
within the turing machine model
It is computable in the turing machine model from a machine
description and "extra" data, not just a machine description, so not a
halt decider, partial or otherwise.
not that ur going to read it, but this argument is responded to in -o7.5 "addressable simulation vs objective mechanics"
Also:
"We will start by amending the basic turing machine into a
terminal machine"
Which means you no longer have a Turing machine so nothing that
follows applies to Turing machines.
_duh_ ... the total solution to turing machine halting can only be
computed from _outside_ the turing machine model
Which is already well-known and therefore uninteresting.
not by a mechanically realizable form it isn't, oracle machines can
be made real, and are largely useless beyond describing something
that realizable computing is surely not.
this does not it make not computation, as it still a method that
can produce a sequence, including both a true diagonal across
turing computable numbers and the true anti-diagonal across turing
computable sequences, both of which are outside the scope of turing >>>>> computation
It makes it a computation that has input in addition to a machine
description and that machine's input, and therefore disqualified
from being a halt decider.
like i already said, this is addressed in -o7.6
And you addressed it by using extra input disqualifying it from being
a halt decider.
idk why ur lying about reading the section. or any of the paper at all.
but i guess that aligns with all the blatant fallacies u keep committing
Which is the same mistake PO made.
u do know that begging the question is a fallacy, eh?
Identifying a fundamental mistake is not a fallacy.
ur not identifying a mistake, ur just continually begging the question
that the unproven church-turing thesis is in fact true,
when u do not in fact have a proof showing that it is true,
while i'm presenting a proof that is it _not_ true
On 9/21/2026 11:32 PM, dart200 wrote:
On 9/21/26 8:18 PM, dbush wrote:
On 9/21/2026 11:10 PM, dart200 wrote:
On 9/21/26 1:41 PM, dbush wrote:
On 9/21/2026 4:06 PM, dart200 wrote:
On 9/21/26 11:53 AM, dbush wrote:
On 9/21/2026 1:18 PM, dart200 wrote:
On 9/21/26 6:34 AM, dbush wrote:
On 9/21/2026 2:00 AM, dart200 wrote:
On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote:
They are one in the same.-a A turing machine is essentially a >>>>>>>>>>> manifestation of an algorithm.-a Until you can show an
algorithm that is not a turing machine, or vice versa,
Church- Turning stands.
that's what -o7 is on dud! maybe u should read the fking paper >>>>>>>>>> instead of dicking about in -o1!
it ultimately discusses how to compute both a total diagonal >>>>>>>>>> and total anti-diagonal across the enumeration of turing
computable sequences, a computation which obviously cannot be >>>>>>>>>> done with a turing machine, as per turing's original proof on >>>>>>>>>> the matter. and demonstrated that is a 26 page paper which >>>>>>>>>> shall not be able to condense into a damn usenet post
Which you do by getting unspecified input from a human, which >>>>>>>>> means it's not an algorithm.-a This also means it takes input >>>>>>>>> other than a machine description and the input to that machine, >>>>>>>>> which are the only allowed inputs to a halt decider.
Anything you're doing regarding the address of a function is >>>>>>>>> simply adding another input which means you're changing the >>>>>>>>> question.
-o7.6 - truly begging a question
So what the human agent is recording becomes an input.-a And
therefore whatever uses it is disqualified from being a halt
decider, partial or otherwise.
the human agent produces a response back to the terminal machine, yes >>>>>>
but he also computes things on the side that persist between
terminal interactions, and it's in this side record where he can
compute things outside the turing machine model
And that side record is what disqualifies it from being a halt
decider.
the knowledge is still computable, even if it can't be used in total
within the turing machine model
It is computable in the turing machine model from a machine
description and "extra" data, not just a machine description, so not
a halt decider, partial or otherwise.
not that ur going to read it, but this argument is responded to in
-o7.5 "addressable simulation vs objective mechanics"
You think you found something turing computable but you haven't.-a You
just have the equivalent of a turing machine / algorithm that takes an additional input that a halt decider doesn't have.
Also:
"We will start by amending the basic turing machine into a
terminal machine"
Which means you no longer have a Turing machine so nothing that >>>>>>> follows applies to Turing machines.
_duh_ ... the total solution to turing machine halting can only be >>>>>> computed from _outside_ the turing machine model
Which is already well-known and therefore uninteresting.
not by a mechanically realizable form it isn't, oracle machines can
be made real, and are largely useless beyond describing something
that realizable computing is surely not.
this does not it make not computation, as it still a method that
can produce a sequence, including both a true diagonal across
turing computable numbers and the true anti-diagonal across turing >>>>>> computable sequences, both of which are outside the scope of
turing computation
It makes it a computation that has input in addition to a machine
description and that machine's input, and therefore disqualified
from being a halt decider.
like i already said, this is addressed in -o7.6
And you addressed it by using extra input disqualifying it from being
a halt decider.
idk why ur lying about reading the section. or any of the paper at
all. but i guess that aligns with all the blatant fallacies u keep
committing
Which is the same mistake PO made.
u do know that begging the question is a fallacy, eh?
Identifying a fundamental mistake is not a fallacy.
ur not identifying a mistake, ur just continually begging the question
that the unproven church-turing thesis is in fact true,
when u do not in fact have a proof showing that it is true,
while i'm presenting a proof that is it _not_ true
You've proven no such thing.-a The only thing you've proved is that you don't understand that your "model" is simply an algorithm / turning
machine with an extra inputs (i.e. the what the human records and the addresses of functions) that isn't allowed for a halt decider.
Which, again, is the same mistake PO made.
On 9/21/26 8:48 PM, dbush wrote:
On 9/21/2026 11:32 PM, dart200 wrote:
On 9/21/26 8:18 PM, dbush wrote:
On 9/21/2026 11:10 PM, dart200 wrote:
On 9/21/26 1:41 PM, dbush wrote:
On 9/21/2026 4:06 PM, dart200 wrote:
On 9/21/26 11:53 AM, dbush wrote:
On 9/21/2026 1:18 PM, dart200 wrote:
On 9/21/26 6:34 AM, dbush wrote:
On 9/21/2026 2:00 AM, dart200 wrote:
On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote:
They are one in the same.-a A turing machine is essentially a >>>>>>>>>>>> manifestation of an algorithm.-a Until you can show an >>>>>>>>>>>> algorithm that is not a turing machine, or vice versa, >>>>>>>>>>>> Church- Turning stands.
that's what -o7 is on dud! maybe u should read the fking paper >>>>>>>>>>> instead of dicking about in -o1!
it ultimately discusses how to compute both a total diagonal >>>>>>>>>>> and total anti-diagonal across the enumeration of turing >>>>>>>>>>> computable sequences, a computation which obviously cannot be >>>>>>>>>>> done with a turing machine, as per turing's original proof on >>>>>>>>>>> the matter. and demonstrated that is a 26 page paper which >>>>>>>>>>> shall not be able to condense into a damn usenet post
Which you do by getting unspecified input from a human, which >>>>>>>>>> means it's not an algorithm.-a This also means it takes input >>>>>>>>>> other than a machine description and the input to that
machine, which are the only allowed inputs to a halt decider. >>>>>>>>>>
Anything you're doing regarding the address of a function is >>>>>>>>>> simply adding another input which means you're changing the >>>>>>>>>> question.
-o7.6 - truly begging a question
So what the human agent is recording becomes an input.-a And
therefore whatever uses it is disqualified from being a halt
decider, partial or otherwise.
the human agent produces a response back to the terminal machine, >>>>>>> yes
but he also computes things on the side that persist between
terminal interactions, and it's in this side record where he can >>>>>>> compute things outside the turing machine model
And that side record is what disqualifies it from being a halt
decider.
the knowledge is still computable, even if it can't be used in
total within the turing machine model
It is computable in the turing machine model from a machine
description and "extra" data, not just a machine description, so not
a halt decider, partial or otherwise.
not that ur going to read it, but this argument is responded to in
-o7.5 "addressable simulation vs objective mechanics"
You think you found something turing computable but you haven't.-a You
u haven't read my paper so u don't know what i've argued
just have the equivalent of a turing machine / algorithm that takes an
additional input that a halt decider doesn't have.
i won't be convinced until you tell me what the example from -o7.5 my
paper, sim_cir_sr(und_cir_sr), is supposed to output...
and if u just brush it off like i suspect u will, i won't reply further because it's clearly ur not interested in genuine engagement,
nvm all the name fallacies committed in this discussion. i honestly
can't quite believe you tried to double down on an argumentum ex
silentio, i never imagined encountering someone so ungodly retarded
Also:
"We will start by amending the basic turing machine into a
terminal machine"
Which means you no longer have a Turing machine so nothing that >>>>>>>> follows applies to Turing machines.
_duh_ ... the total solution to turing machine halting can only >>>>>>> be computed from _outside_ the turing machine model
Which is already well-known and therefore uninteresting.
not by a mechanically realizable form it isn't, oracle machines can >>>>> be made real, and are largely useless beyond describing something
that realizable computing is surely not.
this does not it make not computation, as it still a method that >>>>>>> can produce a sequence, including both a true diagonal across
turing computable numbers and the true anti-diagonal across
turing computable sequences, both of which are outside the scope >>>>>>> of turing computation
It makes it a computation that has input in addition to a machine >>>>>> description and that machine's input, and therefore disqualified
from being a halt decider.
like i already said, this is addressed in -o7.6
And you addressed it by using extra input disqualifying it from
being a halt decider.
idk why ur lying about reading the section. or any of the paper at
all. but i guess that aligns with all the blatant fallacies u keep
committing
Which is the same mistake PO made.
u do know that begging the question is a fallacy, eh?
Identifying a fundamental mistake is not a fallacy.
ur not identifying a mistake, ur just continually begging the
question that the unproven church-turing thesis is in fact true,
when u do not in fact have a proof showing that it is true,
while i'm presenting a proof that is it _not_ true
You've proven no such thing.-a The only thing you've proved is that you
don't understand that your "model" is simply an algorithm / turning
machine with an extra inputs (i.e. the what the human records and the
addresses of functions) that isn't allowed for a halt decider.
ur still just begging the question
Which, again, is the same mistake PO made.
and ur trying to reinforce it with a fallacy by association
On 9/22/2026 1:32 AM, dart200 wrote:
On 9/21/26 8:48 PM, dbush wrote:
On 9/21/2026 11:32 PM, dart200 wrote:
On 9/21/26 8:18 PM, dbush wrote:
On 9/21/2026 11:10 PM, dart200 wrote:
On 9/21/26 1:41 PM, dbush wrote:
On 9/21/2026 4:06 PM, dart200 wrote:
On 9/21/26 11:53 AM, dbush wrote:
On 9/21/2026 1:18 PM, dart200 wrote:
On 9/21/26 6:34 AM, dbush wrote:
On 9/21/2026 2:00 AM, dart200 wrote:
On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote:
They are one in the same.-a A turing machine is essentially >>>>>>>>>>>>> a manifestation of an algorithm.-a Until you can show an >>>>>>>>>>>>> algorithm that is not a turing machine, or vice versa, >>>>>>>>>>>>> Church- Turning stands.
that's what -o7 is on dud! maybe u should read the fking >>>>>>>>>>>> paper instead of dicking about in -o1!
it ultimately discusses how to compute both a total diagonal >>>>>>>>>>>> and total anti-diagonal across the enumeration of turing >>>>>>>>>>>> computable sequences, a computation which obviously cannot >>>>>>>>>>>> be done with a turing machine, as per turing's original >>>>>>>>>>>> proof on the matter. and demonstrated that is a 26 page >>>>>>>>>>>> paper which shall not be able to condense into a damn usenet >>>>>>>>>>>> post
Which you do by getting unspecified input from a human, which >>>>>>>>>>> means it's not an algorithm.-a This also means it takes input >>>>>>>>>>> other than a machine description and the input to that
machine, which are the only allowed inputs to a halt decider. >>>>>>>>>>>
Anything you're doing regarding the address of a function is >>>>>>>>>>> simply adding another input which means you're changing the >>>>>>>>>>> question.
-o7.6 - truly begging a question
So what the human agent is recording becomes an input.-a And >>>>>>>>> therefore whatever uses it is disqualified from being a halt >>>>>>>>> decider, partial or otherwise.
the human agent produces a response back to the terminal
machine, yes
but he also computes things on the side that persist between
terminal interactions, and it's in this side record where he can >>>>>>>> compute things outside the turing machine model
And that side record is what disqualifies it from being a halt
decider.
the knowledge is still computable, even if it can't be used in
total within the turing machine model
It is computable in the turing machine model from a machine
description and "extra" data, not just a machine description, so
not a halt decider, partial or otherwise.
not that ur going to read it, but this argument is responded to in
-o7.5 "addressable simulation vs objective mechanics"
You think you found something turing computable but you haven't.-a You
u haven't read my paper so u don't know what i've argued
just have the equivalent of a turing machine / algorithm that takes
an additional input that a halt decider doesn't have.
i won't be convinced until you tell me what the example from -o7.5 my
paper, sim_cir_sr(und_cir_sr), is supposed to output...
You still don't understand.
The agent's side record is not outside Turing computability.-a In the
Turing model it is *part of the input*, i.e. it would be somewhere on tape.
This side record is effectively a global variable, and the contents of
any global variable are an input to the algorithm.
Just because this side record isn't passed as a parameter to sim_cir_sr
or und_cir_sr doesn't mean it's not an input.
What you're doing is no different from putting the fixed set of steps
that a human is doing in a separate function and putting the human's
side record in a global variable.
The fact that the human's side record can't be accessed by the machine
being analyzed doesn't make it non-Turing computable.
and if u just brush it off like i suspect u will, i won't reply
further because it's clearly ur not interested in genuine engagement,
nvm all the name fallacies committed in this discussion. i honestly
can't quite believe you tried to double down on an argumentum ex
silentio, i never imagined encountering someone so ungodly retarded
Also:
"We will start by amending the basic turing machine into a
terminal machine"
Which means you no longer have a Turing machine so nothing that >>>>>>>>> follows applies to Turing machines.
_duh_ ... the total solution to turing machine halting can only >>>>>>>> be computed from _outside_ the turing machine model
Which is already well-known and therefore uninteresting.
not by a mechanically realizable form it isn't, oracle machines
can be made real, and are largely useless beyond describing
something that realizable computing is surely not.
this does not it make not computation, as it still a method that >>>>>>>> can produce a sequence, including both a true diagonal across >>>>>>>> turing computable numbers and the true anti-diagonal across
turing computable sequences, both of which are outside the scope >>>>>>>> of turing computation
It makes it a computation that has input in addition to a machine >>>>>>> description and that machine's input, and therefore disqualified >>>>>>> from being a halt decider.
like i already said, this is addressed in -o7.6
And you addressed it by using extra input disqualifying it from
being a halt decider.
idk why ur lying about reading the section. or any of the paper at
all. but i guess that aligns with all the blatant fallacies u keep
committing
Which is the same mistake PO made.
u do know that begging the question is a fallacy, eh?
Identifying a fundamental mistake is not a fallacy.
ur not identifying a mistake, ur just continually begging the
question that the unproven church-turing thesis is in fact true,
when u do not in fact have a proof showing that it is true,
while i'm presenting a proof that is it _not_ true
You've proven no such thing.-a The only thing you've proved is that
you don't understand that your "model" is simply an algorithm /
turning machine with an extra inputs (i.e. the what the human records
and the addresses of functions) that isn't allowed for a halt decider.
ur still just begging the question
Which, again, is the same mistake PO made.
and ur trying to reinforce it with a fallacy by association
Nope, just pointing out that you don't understand that you made the same mistake that King Crank made.
On 9/22/26 5:26 AM, dbush wrote:
On 9/22/2026 1:32 AM, dart200 wrote:
On 9/21/26 8:48 PM, dbush wrote:
On 9/21/2026 11:32 PM, dart200 wrote:u haven't read my paper so u don't know what i've argued
On 9/21/26 8:18 PM, dbush wrote:
On 9/21/2026 11:10 PM, dart200 wrote:
On 9/21/26 1:41 PM, dbush wrote:
On 9/21/2026 4:06 PM, dart200 wrote:
On 9/21/26 11:53 AM, dbush wrote:
On 9/21/2026 1:18 PM, dart200 wrote:
On 9/21/26 6:34 AM, dbush wrote:
On 9/21/2026 2:00 AM, dart200 wrote:
On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote:
They are one in the same.-a A turing machine is essentially >>>>>>>>>>>>>> a manifestation of an algorithm.-a Until you can show an >>>>>>>>>>>>>> algorithm that is not a turing machine, or vice versa, >>>>>>>>>>>>>> Church- Turning stands.
that's what -o7 is on dud! maybe u should read the fking >>>>>>>>>>>>> paper instead of dicking about in -o1!
it ultimately discusses how to compute both a total >>>>>>>>>>>>> diagonal and total anti-diagonal across the enumeration of >>>>>>>>>>>>> turing computable sequences, a computation which obviously >>>>>>>>>>>>> cannot be done with a turing machine, as per turing's >>>>>>>>>>>>> original proof on the matter. and demonstrated that is a 26 >>>>>>>>>>>>> page paper which shall not be able to condense into a damn >>>>>>>>>>>>> usenet post
Which you do by getting unspecified input from a human, >>>>>>>>>>>> which means it's not an algorithm.-a This also means it takes >>>>>>>>>>>> input other than a machine description and the input to that >>>>>>>>>>>> machine, which are the only allowed inputs to a halt decider. >>>>>>>>>>>>
Anything you're doing regarding the address of a function is >>>>>>>>>>>> simply adding another input which means you're changing the >>>>>>>>>>>> question.
-o7.6 - truly begging a question
So what the human agent is recording becomes an input.-a And >>>>>>>>>> therefore whatever uses it is disqualified from being a halt >>>>>>>>>> decider, partial or otherwise.
the human agent produces a response back to the terminal
machine, yes
but he also computes things on the side that persist between >>>>>>>>> terminal interactions, and it's in this side record where he >>>>>>>>> can compute things outside the turing machine model
And that side record is what disqualifies it from being a halt >>>>>>>> decider.
the knowledge is still computable, even if it can't be used in
total within the turing machine model
It is computable in the turing machine model from a machine
description and "extra" data, not just a machine description, so
not a halt decider, partial or otherwise.
not that ur going to read it, but this argument is responded to in
-o7.5 "addressable simulation vs objective mechanics"
You think you found something turing computable but you haven't.-a You >>>
just have the equivalent of a turing machine / algorithm that takes
an additional input that a halt decider doesn't have.
i won't be convinced until you tell me what the example from -o7.5 my
paper, sim_cir_sr(und_cir_sr), is supposed to output...
You still don't understand.
The agent's side record is not outside Turing computability.-a In the
Turing model it is *part of the input*, i.e. it would be somewhere on
tape.
This side record is effectively a global variable, and the contents of
any global variable are an input to the algorithm.
Just because this side record isn't passed as a parameter to
sim_cir_sr or und_cir_sr doesn't mean it's not an input.
What you're doing is no different from putting the fixed set of steps
that a human is doing in a separate function and putting the human's
side record in a global variable.
The fact that the human's side record can't be accessed by the machine
being analyzed doesn't make it non-Turing computable.
again, u ramble on without giving me what sim_cir_sr(und_cir_sr) is
supposed to output, and worse u think u can explain what's going on
without even reading the passage explaining what that is. nuts.
and if u just brush it off like i suspect u will, i won't reply
further because it's clearly ur not interested in genuine engagement,
nvm all the name fallacies committed in this discussion. i honestly
can't quite believe you tried to double down on an argumentum ex
silentio, i never imagined encountering someone so ungodly retarded
Also:
"We will start by amending the basic turing machine into a >>>>>>>>>> terminal machine"
Which means you no longer have a Turing machine so nothing >>>>>>>>>> that follows applies to Turing machines.
_duh_ ... the total solution to turing machine halting can only >>>>>>>>> be computed from _outside_ the turing machine model
Which is already well-known and therefore uninteresting.
not by a mechanically realizable form it isn't, oracle machines >>>>>>> can be made real, and are largely useless beyond describing
something that realizable computing is surely not.
this does not it make not computation, as it still a method >>>>>>>>> that can produce a sequence, including both a true diagonal >>>>>>>>> across turing computable numbers and the true anti-diagonal >>>>>>>>> across turing computable sequences, both of which are outside >>>>>>>>> the scope of turing computation
It makes it a computation that has input in addition to a
machine description and that machine's input, and therefore
disqualified from being a halt decider.
like i already said, this is addressed in -o7.6
And you addressed it by using extra input disqualifying it from
being a halt decider.
idk why ur lying about reading the section. or any of the paper at
all. but i guess that aligns with all the blatant fallacies u keep
committing
Which is the same mistake PO made.
u do know that begging the question is a fallacy, eh?
Identifying a fundamental mistake is not a fallacy.
ur not identifying a mistake, ur just continually begging the
question that the unproven church-turing thesis is in fact true,
when u do not in fact have a proof showing that it is true,
while i'm presenting a proof that is it _not_ true
You've proven no such thing.-a The only thing you've proved is that
you don't understand that your "model" is simply an algorithm /
turning machine with an extra inputs (i.e. the what the human
records and the addresses of functions) that isn't allowed for a
halt decider.
ur still just begging the question
Which, again, is the same mistake PO made.
and ur trying to reinforce it with a fallacy by association
Nope, just pointing out that you don't understand that you made the
same mistake that King Crank made.
and if that mistake is you continually begging that the ct-thesis is
correct ...
On 9/22/2026 11:39 AM, dart200 wrote:
On 9/22/26 5:26 AM, dbush wrote:
On 9/22/2026 1:32 AM, dart200 wrote:
On 9/21/26 8:48 PM, dbush wrote:
On 9/21/2026 11:32 PM, dart200 wrote:u haven't read my paper so u don't know what i've argued
On 9/21/26 8:18 PM, dbush wrote:
On 9/21/2026 11:10 PM, dart200 wrote:
On 9/21/26 1:41 PM, dbush wrote:
On 9/21/2026 4:06 PM, dart200 wrote:
On 9/21/26 11:53 AM, dbush wrote:
On 9/21/2026 1:18 PM, dart200 wrote:
On 9/21/26 6:34 AM, dbush wrote:
On 9/21/2026 2:00 AM, dart200 wrote:
On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote:
They are one in the same.-a A turing machine is >>>>>>>>>>>>>>> essentially a manifestation of an algorithm.-a Until you >>>>>>>>>>>>>>> can show an algorithm that is not a turing machine, or >>>>>>>>>>>>>>> vice versa, Church- Turning stands.
that's what -o7 is on dud! maybe u should read the fking >>>>>>>>>>>>>> paper instead of dicking about in -o1!
it ultimately discusses how to compute both a total >>>>>>>>>>>>>> diagonal and total anti-diagonal across the enumeration of >>>>>>>>>>>>>> turing computable sequences, a computation which obviously >>>>>>>>>>>>>> cannot be done with a turing machine, as per turing's >>>>>>>>>>>>>> original proof on the matter. and demonstrated that is a >>>>>>>>>>>>>> 26 page paper which shall not be able to condense into a >>>>>>>>>>>>>> damn usenet post
Which you do by getting unspecified input from a human, >>>>>>>>>>>>> which means it's not an algorithm.-a This also means it >>>>>>>>>>>>> takes input other than a machine description and the input >>>>>>>>>>>>> to that machine, which are the only allowed inputs to a >>>>>>>>>>>>> halt decider.
Anything you're doing regarding the address of a function >>>>>>>>>>>>> is simply adding another input which means you're changing >>>>>>>>>>>>> the question.
-o7.6 - truly begging a question
So what the human agent is recording becomes an input.-a And >>>>>>>>>>> therefore whatever uses it is disqualified from being a halt >>>>>>>>>>> decider, partial or otherwise.
the human agent produces a response back to the terminal
machine, yes
but he also computes things on the side that persist between >>>>>>>>>> terminal interactions, and it's in this side record where he >>>>>>>>>> can compute things outside the turing machine model
And that side record is what disqualifies it from being a halt >>>>>>>>> decider.
the knowledge is still computable, even if it can't be used in >>>>>>>> total within the turing machine model
It is computable in the turing machine model from a machine
description and "extra" data, not just a machine description, so >>>>>>> not a halt decider, partial or otherwise.
not that ur going to read it, but this argument is responded to in >>>>>> -o7.5 "addressable simulation vs objective mechanics"
You think you found something turing computable but you haven't.-a You >>>>
just have the equivalent of a turing machine / algorithm that takes >>>>> an additional input that a halt decider doesn't have.
i won't be convinced until you tell me what the example from -o7.5 my >>>> paper, sim_cir_sr(und_cir_sr), is supposed to output...
You still don't understand.
The agent's side record is not outside Turing computability.-a In the
Turing model it is *part of the input*, i.e. it would be somewhere on
tape.
This side record is effectively a global variable, and the contents
of any global variable are an input to the algorithm.
Just because this side record isn't passed as a parameter to
sim_cir_sr or und_cir_sr doesn't mean it's not an input.
What you're doing is no different from putting the fixed set of steps
that a human is doing in a separate function and putting the human's
side record in a global variable.
The fact that the human's side record can't be accessed by the
machine being analyzed doesn't make it non-Turing computable.
again, u ramble on without giving me what sim_cir_sr(und_cir_sr) is
supposed to output, and worse u think u can explain what's going on
without even reading the passage explaining what that is. nuts.
As derived from above, if the steps a human performs is replaced with a function with the the reads/writes to the side record going into a
global variable, and that global variable is set to the contents of the human's side record prior to the given call, it will behave the same as
if und_cir_sr is called.
On 9/22/2026 11:39 AM, dart200 wrote:
On 9/22/26 5:26 AM, dbush wrote:
On 9/22/2026 1:32 AM, dart200 wrote:
On 9/21/26 8:48 PM, dbush wrote:
On 9/21/2026 11:32 PM, dart200 wrote:u haven't read my paper so u don't know what i've argued
On 9/21/26 8:18 PM, dbush wrote:
On 9/21/2026 11:10 PM, dart200 wrote:
On 9/21/26 1:41 PM, dbush wrote:
On 9/21/2026 4:06 PM, dart200 wrote:
On 9/21/26 11:53 AM, dbush wrote:
On 9/21/2026 1:18 PM, dart200 wrote:
On 9/21/26 6:34 AM, dbush wrote:
On 9/21/2026 2:00 AM, dart200 wrote:
On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote:
They are one in the same.-a A turing machine is >>>>>>>>>>>>>>> essentially a manifestation of an algorithm.-a Until you >>>>>>>>>>>>>>> can show an algorithm that is not a turing machine, or >>>>>>>>>>>>>>> vice versa, Church- Turning stands.
that's what -o7 is on dud! maybe u should read the fking >>>>>>>>>>>>>> paper instead of dicking about in -o1!
it ultimately discusses how to compute both a total >>>>>>>>>>>>>> diagonal and total anti-diagonal across the enumeration of >>>>>>>>>>>>>> turing computable sequences, a computation which obviously >>>>>>>>>>>>>> cannot be done with a turing machine, as per turing's >>>>>>>>>>>>>> original proof on the matter. and demonstrated that is a >>>>>>>>>>>>>> 26 page paper which shall not be able to condense into a >>>>>>>>>>>>>> damn usenet post
Which you do by getting unspecified input from a human, >>>>>>>>>>>>> which means it's not an algorithm.-a This also means it >>>>>>>>>>>>> takes input other than a machine description and the input >>>>>>>>>>>>> to that machine, which are the only allowed inputs to a >>>>>>>>>>>>> halt decider.
Anything you're doing regarding the address of a function >>>>>>>>>>>>> is simply adding another input which means you're changing >>>>>>>>>>>>> the question.
-o7.6 - truly begging a question
So what the human agent is recording becomes an input.-a And >>>>>>>>>>> therefore whatever uses it is disqualified from being a halt >>>>>>>>>>> decider, partial or otherwise.
the human agent produces a response back to the terminal
machine, yes
but he also computes things on the side that persist between >>>>>>>>>> terminal interactions, and it's in this side record where he >>>>>>>>>> can compute things outside the turing machine model
And that side record is what disqualifies it from being a halt >>>>>>>>> decider.
the knowledge is still computable, even if it can't be used in >>>>>>>> total within the turing machine model
It is computable in the turing machine model from a machine
description and "extra" data, not just a machine description, so >>>>>>> not a halt decider, partial or otherwise.
not that ur going to read it, but this argument is responded to in >>>>>> -o7.5 "addressable simulation vs objective mechanics"
You think you found something turing computable but you haven't.-a You >>>>
just have the equivalent of a turing machine / algorithm that takes >>>>> an additional input that a halt decider doesn't have.
i won't be convinced until you tell me what the example from -o7.5 my >>>> paper, sim_cir_sr(und_cir_sr), is supposed to output...
You still don't understand.
The agent's side record is not outside Turing computability.-a In the
Turing model it is *part of the input*, i.e. it would be somewhere on
tape.
This side record is effectively a global variable, and the contents
of any global variable are an input to the algorithm.
Just because this side record isn't passed as a parameter to
sim_cir_sr or und_cir_sr doesn't mean it's not an input.
What you're doing is no different from putting the fixed set of steps
that a human is doing in a separate function and putting the human's
side record in a global variable.
The fact that the human's side record can't be accessed by the
machine being analyzed doesn't make it non-Turing computable.
again, u ramble on without giving me what sim_cir_sr(und_cir_sr) is
supposed to output, and worse u think u can explain what's going on
without even reading the passage explaining what that is. nuts.
As derived from above, if the steps a human performs is replaced with a function with the the reads/writes to the side record going into a
global variable, and that global variable is set to the contents of the human's side record prior to the given call, it will behave the same as
if und_cir_sr is called.-a Because that's what algorithms do: give the
same results for the same input.
You don't seem to understand how you're modeling the algorithm abstraction.
and if u just brush it off like i suspect u will, i won't reply
further because it's clearly ur not interested in genuine engagement,
nvm all the name fallacies committed in this discussion. i honestly
can't quite believe you tried to double down on an argumentum ex
silentio, i never imagined encountering someone so ungodly retarded
Also:
"We will start by amending the basic turing machine into a >>>>>>>>>>> terminal machine"
Which means you no longer have a Turing machine so nothing >>>>>>>>>>> that follows applies to Turing machines.
_duh_ ... the total solution to turing machine halting can >>>>>>>>>> only be computed from _outside_ the turing machine model
Which is already well-known and therefore uninteresting.
not by a mechanically realizable form it isn't, oracle machines >>>>>>>> can be made real, and are largely useless beyond describing
something that realizable computing is surely not.
this does not it make not computation, as it still a method >>>>>>>>>> that can produce a sequence, including both a true diagonal >>>>>>>>>> across turing computable numbers and the true anti-diagonal >>>>>>>>>> across turing computable sequences, both of which are outside >>>>>>>>>> the scope of turing computation
It makes it a computation that has input in addition to a
machine description and that machine's input, and therefore >>>>>>>>> disqualified from being a halt decider.
like i already said, this is addressed in -o7.6
And you addressed it by using extra input disqualifying it from >>>>>>> being a halt decider.
idk why ur lying about reading the section. or any of the paper at >>>>>> all. but i guess that aligns with all the blatant fallacies u keep >>>>>> committing
Which is the same mistake PO made.
u do know that begging the question is a fallacy, eh?
Identifying a fundamental mistake is not a fallacy.
ur not identifying a mistake, ur just continually begging the
question that the unproven church-turing thesis is in fact true,
when u do not in fact have a proof showing that it is true,
while i'm presenting a proof that is it _not_ true
You've proven no such thing.-a The only thing you've proved is that >>>>> you don't understand that your "model" is simply an algorithm /
turning machine with an extra inputs (i.e. the what the human
records and the addresses of functions) that isn't allowed for a
halt decider.
ur still just begging the question
Which, again, is the same mistake PO made.
and ur trying to reinforce it with a fallacy by association
Nope, just pointing out that you don't understand that you made the
same mistake that King Crank made.
and if that mistake is you continually begging that the ct-thesis is
correct ...
No, I'm pointing out that you didn't refute it because you didn't do
what you think you did.
On 9/22/26 9:52 AM, dbush wrote:
On 9/22/2026 11:39 AM, dart200 wrote:
On 9/22/26 5:26 AM, dbush wrote:
On 9/22/2026 1:32 AM, dart200 wrote:
On 9/21/26 8:48 PM, dbush wrote:
On 9/21/2026 11:32 PM, dart200 wrote:u haven't read my paper so u don't know what i've argued
On 9/21/26 8:18 PM, dbush wrote:
On 9/21/2026 11:10 PM, dart200 wrote:
On 9/21/26 1:41 PM, dbush wrote:
On 9/21/2026 4:06 PM, dart200 wrote:
On 9/21/26 11:53 AM, dbush wrote:
On 9/21/2026 1:18 PM, dart200 wrote:
On 9/21/26 6:34 AM, dbush wrote:
On 9/21/2026 2:00 AM, dart200 wrote:
On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote:
They are one in the same.-a A turing machine is >>>>>>>>>>>>>>>> essentially a manifestation of an algorithm.-a Until you >>>>>>>>>>>>>>>> can show an algorithm that is not a turing machine, or >>>>>>>>>>>>>>>> vice versa, Church- Turning stands.
that's what -o7 is on dud! maybe u should read the fking >>>>>>>>>>>>>>> paper instead of dicking about in -o1!
it ultimately discusses how to compute both a total >>>>>>>>>>>>>>> diagonal and total anti-diagonal across the enumeration >>>>>>>>>>>>>>> of turing computable sequences, a computation which >>>>>>>>>>>>>>> obviously cannot be done with a turing machine, as per >>>>>>>>>>>>>>> turing's original proof on the matter. and demonstrated >>>>>>>>>>>>>>> that is a 26 page paper which shall not be able to >>>>>>>>>>>>>>> condense into a damn usenet post
Which you do by getting unspecified input from a human, >>>>>>>>>>>>>> which means it's not an algorithm.-a This also means it >>>>>>>>>>>>>> takes input other than a machine description and the input >>>>>>>>>>>>>> to that machine, which are the only allowed inputs to a >>>>>>>>>>>>>> halt decider.
Anything you're doing regarding the address of a function >>>>>>>>>>>>>> is simply adding another input which means you're changing >>>>>>>>>>>>>> the question.
-o7.6 - truly begging a question
So what the human agent is recording becomes an input.-a And >>>>>>>>>>>> therefore whatever uses it is disqualified from being a halt >>>>>>>>>>>> decider, partial or otherwise.
the human agent produces a response back to the terminal >>>>>>>>>>> machine, yes
but he also computes things on the side that persist between >>>>>>>>>>> terminal interactions, and it's in this side record where he >>>>>>>>>>> can compute things outside the turing machine model
And that side record is what disqualifies it from being a halt >>>>>>>>>> decider.
the knowledge is still computable, even if it can't be used in >>>>>>>>> total within the turing machine model
It is computable in the turing machine model from a machine
description and "extra" data, not just a machine description, so >>>>>>>> not a halt decider, partial or otherwise.
not that ur going to read it, but this argument is responded to >>>>>>> in -o7.5 "addressable simulation vs objective mechanics"
You think you found something turing computable but you haven't.-a You >>>>>
just have the equivalent of a turing machine / algorithm that
takes an additional input that a halt decider doesn't have.
i won't be convinced until you tell me what the example from -o7.5
my paper, sim_cir_sr(und_cir_sr), is supposed to output...
You still don't understand.
The agent's side record is not outside Turing computability.-a In the >>>> Turing model it is *part of the input*, i.e. it would be somewhere
on tape.
This side record is effectively a global variable, and the contents
of any global variable are an input to the algorithm.
Just because this side record isn't passed as a parameter to
sim_cir_sr or und_cir_sr doesn't mean it's not an input.
What you're doing is no different from putting the fixed set of
steps that a human is doing in a separate function and putting the
human's side record in a global variable.
The fact that the human's side record can't be accessed by the
machine being analyzed doesn't make it non-Turing computable.
again, u ramble on without giving me what sim_cir_sr(und_cir_sr) is
supposed to output, and worse u think u can explain what's going on
without even reading the passage explaining what that is. nuts.
As derived from above, if the steps a human performs is replaced with
a function with the the reads/writes to the side record going into a
global variable, and that global variable is set to the contents of
the human's side record prior to the given call, it will behave the
same as if und_cir_sr is called.-a Because that's what algorithms do:
give the same results for the same input.
You don't seem to understand how you're modeling the algorithm
abstraction.
u don't seem to understand how ur begging the question of the ct-thesis being true... eh?
and if u just brush it off like i suspect u will, i won't reply
further because it's clearly ur not interested in genuine engagement, >>>>>
nvm all the name fallacies committed in this discussion. i honestly >>>>> can't quite believe you tried to double down on an argumentum ex
silentio, i never imagined encountering someone so ungodly retarded
Also:
"We will start by amending the basic turing machine into a >>>>>>>>>>>> terminal machine"
Which means you no longer have a Turing machine so nothing >>>>>>>>>>>> that follows applies to Turing machines.
_duh_ ... the total solution to turing machine halting can >>>>>>>>>>> only be computed from _outside_ the turing machine model
Which is already well-known and therefore uninteresting.
not by a mechanically realizable form it isn't, oracle machines >>>>>>>>> can be made real, and are largely useless beyond describing >>>>>>>>> something that realizable computing is surely not.
this does not it make not computation, as it still a method >>>>>>>>>>> that can produce a sequence, including both a true diagonal >>>>>>>>>>> across turing computable numbers and the true anti-diagonal >>>>>>>>>>> across turing computable sequences, both of which are outside >>>>>>>>>>> the scope of turing computation
It makes it a computation that has input in addition to a >>>>>>>>>> machine description and that machine's input, and therefore >>>>>>>>>> disqualified from being a halt decider.
like i already said, this is addressed in -o7.6
And you addressed it by using extra input disqualifying it from >>>>>>>> being a halt decider.
idk why ur lying about reading the section. or any of the paper >>>>>>> at all. but i guess that aligns with all the blatant fallacies u >>>>>>> keep committing
Which is the same mistake PO made.
u do know that begging the question is a fallacy, eh?
Identifying a fundamental mistake is not a fallacy.
ur not identifying a mistake, ur just continually begging the
question that the unproven church-turing thesis is in fact true, >>>>>>>
when u do not in fact have a proof showing that it is true,
while i'm presenting a proof that is it _not_ true
You've proven no such thing.-a The only thing you've proved is that >>>>>> you don't understand that your "model" is simply an algorithm /
turning machine with an extra inputs (i.e. the what the human
records and the addresses of functions) that isn't allowed for a
halt decider.
ur still just begging the question
Which, again, is the same mistake PO made.
and ur trying to reinforce it with a fallacy by association
Nope, just pointing out that you don't understand that you made the
same mistake that King Crank made.
and if that mistake is you continually begging that the ct-thesis is
correct ...
No, I'm pointing out that you didn't refute it because you didn't do
what you think you did.
suppose sim_cir_sr() successfully simulates and outputs the agent's
decision process for the side record...
so how does the simulated agent within sim_cir_sr respond to the input
of und_cir_sr?
0) does the simulated agent simulate sim_cir_sr(und_cir_sr) recursively without ever reaching a decision, such that sim_cir_sr(und_cir_sr) never returns, and thereby und_cir_sr will recurse infinitely without output (making it circular)
1) does the simulated agent decide that und_cir_sr is circular, so sim_cir_sr(und_cir_sr) => true, causing und_cir_sr to output an infinite sequence (making it circle-free)
2) does the simulated agent decide that und_cir_sr is circle-free, so sim_cir_sr(und_cir_sr) => false, causing und_cir_sr to halt (making it circular)
these all exist as possible sim_cir_sr machines in the total machines enumeration, which one are you asserting correctly simulates the
idealized agent acting externally to the turing machine model?
On 9/22/2026 2:54 PM, dart200 wrote:
On 9/22/26 9:52 AM, dbush wrote:
On 9/22/2026 11:39 AM, dart200 wrote:
On 9/22/26 5:26 AM, dbush wrote:
On 9/22/2026 1:32 AM, dart200 wrote:
On 9/21/26 8:48 PM, dbush wrote:
On 9/21/2026 11:32 PM, dart200 wrote:
On 9/21/26 8:18 PM, dbush wrote:
On 9/21/2026 11:10 PM, dart200 wrote:
On 9/21/26 1:41 PM, dbush wrote:
On 9/21/2026 4:06 PM, dart200 wrote:
On 9/21/26 11:53 AM, dbush wrote:
On 9/21/2026 1:18 PM, dart200 wrote:
On 9/21/26 6:34 AM, dbush wrote:
On 9/21/2026 2:00 AM, dart200 wrote:
On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote:
They are one in the same.-a A turing machine is >>>>>>>>>>>>>>>>> essentially a manifestation of an algorithm.-a Until you >>>>>>>>>>>>>>>>> can show an algorithm that is not a turing machine, or >>>>>>>>>>>>>>>>> vice versa, Church- Turning stands.
that's what -o7 is on dud! maybe u should read the fking >>>>>>>>>>>>>>>> paper instead of dicking about in -o1!
it ultimately discusses how to compute both a total >>>>>>>>>>>>>>>> diagonal and total anti-diagonal across the enumeration >>>>>>>>>>>>>>>> of turing computable sequences, a computation which >>>>>>>>>>>>>>>> obviously cannot be done with a turing machine, as per >>>>>>>>>>>>>>>> turing's original proof on the matter. and demonstrated >>>>>>>>>>>>>>>> that is a 26 page paper which shall not be able to >>>>>>>>>>>>>>>> condense into a damn usenet post
Which you do by getting unspecified input from a human, >>>>>>>>>>>>>>> which means it's not an algorithm.-a This also means it >>>>>>>>>>>>>>> takes input other than a machine description and the >>>>>>>>>>>>>>> input to that machine, which are the only allowed inputs >>>>>>>>>>>>>>> to a halt decider.
Anything you're doing regarding the address of a function >>>>>>>>>>>>>>> is simply adding another input which means you're >>>>>>>>>>>>>>> changing the question.
-o7.6 - truly begging a question
So what the human agent is recording becomes an input.-a And >>>>>>>>>>>>> therefore whatever uses it is disqualified from being a >>>>>>>>>>>>> halt decider, partial or otherwise.
the human agent produces a response back to the terminal >>>>>>>>>>>> machine, yes
but he also computes things on the side that persist between >>>>>>>>>>>> terminal interactions, and it's in this side record where he >>>>>>>>>>>> can compute things outside the turing machine model
And that side record is what disqualifies it from being a >>>>>>>>>>> halt decider.
the knowledge is still computable, even if it can't be used in >>>>>>>>>> total within the turing machine model
It is computable in the turing machine model from a machine >>>>>>>>> description and "extra" data, not just a machine description, >>>>>>>>> so not a halt decider, partial or otherwise.
not that ur going to read it, but this argument is responded to >>>>>>>> in -o7.5 "addressable simulation vs objective mechanics"
You think you found something turing computable but you haven't. >>>>>>> You
u haven't read my paper so u don't know what i've argued
just have the equivalent of a turing machine / algorithm that
takes an additional input that a halt decider doesn't have.
i won't be convinced until you tell me what the example from -o7.5 >>>>>> my paper, sim_cir_sr(und_cir_sr), is supposed to output...
You still don't understand.
The agent's side record is not outside Turing computability.-a In
the Turing model it is *part of the input*, i.e. it would be
somewhere on tape.
This side record is effectively a global variable, and the contents >>>>> of any global variable are an input to the algorithm.
Just because this side record isn't passed as a parameter to
sim_cir_sr or und_cir_sr doesn't mean it's not an input.
What you're doing is no different from putting the fixed set of
steps that a human is doing in a separate function and putting the
human's side record in a global variable.
The fact that the human's side record can't be accessed by the
machine being analyzed doesn't make it non-Turing computable.
again, u ramble on without giving me what sim_cir_sr(und_cir_sr) is
supposed to output, and worse u think u can explain what's going on
without even reading the passage explaining what that is. nuts.
As derived from above, if the steps a human performs is replaced with
a function with the the reads/writes to the side record going into a
global variable, and that global variable is set to the contents of
the human's side record prior to the given call, it will behave the
same as if und_cir_sr is called.-a Because that's what algorithms do:
give the same results for the same input.
You don't seem to understand how you're modeling the algorithm
abstraction.
u don't seem to understand how ur begging the question of the ct-
thesis being true... eh?
No, you don't seem to understand that you haven't show that you're
refuted it.
and if u just brush it off like i suspect u will, i won't reply
further because it's clearly ur not interested in genuine engagement, >>>>>>
nvm all the name fallacies committed in this discussion. i
honestly can't quite believe you tried to double down on an
argumentum ex silentio, i never imagined encountering someone so
ungodly retarded
Which is already well-known and therefore uninteresting.
Also:
"We will start by amending the basic turing machine into a >>>>>>>>>>>>> terminal machine"
Which means you no longer have a Turing machine so nothing >>>>>>>>>>>>> that follows applies to Turing machines.
_duh_ ... the total solution to turing machine halting can >>>>>>>>>>>> only be computed from _outside_ the turing machine model >>>>>>>>>>>
not by a mechanically realizable form it isn't, oracle
machines can be made real, and are largely useless beyond >>>>>>>>>> describing something that realizable computing is surely not. >>>>>>>>>>
this does not it make not computation, as it still a method >>>>>>>>>>>> that can produce a sequence, including both a true diagonal >>>>>>>>>>>> across turing computable numbers and the true anti-diagonal >>>>>>>>>>>> across turing computable sequences, both of which are >>>>>>>>>>>> outside the scope of turing computation
It makes it a computation that has input in addition to a >>>>>>>>>>> machine description and that machine's input, and therefore >>>>>>>>>>> disqualified from being a halt decider.
like i already said, this is addressed in -o7.6
And you addressed it by using extra input disqualifying it from >>>>>>>>> being a halt decider.
idk why ur lying about reading the section. or any of the paper >>>>>>>> at all. but i guess that aligns with all the blatant fallacies u >>>>>>>> keep committing
Which is the same mistake PO made.
u do know that begging the question is a fallacy, eh?
Identifying a fundamental mistake is not a fallacy.
ur not identifying a mistake, ur just continually begging the >>>>>>>> question that the unproven church-turing thesis is in fact true, >>>>>>>>
when u do not in fact have a proof showing that it is true,
while i'm presenting a proof that is it _not_ true
You've proven no such thing.-a The only thing you've proved is
that you don't understand that your "model" is simply an
algorithm / turning machine with an extra inputs (i.e. the what >>>>>>> the human records and the addresses of functions) that isn't
allowed for a halt decider.
ur still just begging the question
Which, again, is the same mistake PO made.
and ur trying to reinforce it with a fallacy by association
Nope, just pointing out that you don't understand that you made the >>>>> same mistake that King Crank made.
and if that mistake is you continually begging that the ct-thesis is
correct ...
No, I'm pointing out that you didn't refute it because you didn't do
what you think you did.
suppose sim_cir_sr() successfully simulates and outputs the agent's
decision process for the side record...
so how does the simulated agent within sim_cir_sr respond to the input
of und_cir_sr?
0) does the simulated agent simulate sim_cir_sr(und_cir_sr)
recursively without ever reaching a decision, such that
sim_cir_sr(und_cir_sr) never returns, and thereby und_cir_sr will
recurse infinitely without output (making it circular)
1) does the simulated agent decide that und_cir_sr is circular, so
sim_cir_sr(und_cir_sr) => true, causing und_cir_sr to output an
infinite sequence (making it circle-free)
2) does the simulated agent decide that und_cir_sr is circle-free, so
sim_cir_sr(und_cir_sr) => false, causing und_cir_sr to halt (making it
circular)
these all exist as possible sim_cir_sr machines in the total machines
enumeration, which one are you asserting correctly simulates the
idealized agent acting externally to the turing machine model?
None of those simulate und_cir_sr correctly because they aren't setting
the side record to the same value that it has when und_cir_sr is
originally called, meaning it doesn't actually simulate "itself".
That means your abstraction is broken and you don't actually have an algorithm.
To fix it, the fixed steps the human agent runs need to be converted to
a function which read/writes a global variable, and that global variable need to be set to the same value it had when und_cir_sr was first called.
On 9/22/26 12:21 PM, dbush wrote:
On 9/22/2026 2:54 PM, dart200 wrote:
On 9/22/26 9:52 AM, dbush wrote:
On 9/22/2026 11:39 AM, dart200 wrote:
On 9/22/26 5:26 AM, dbush wrote:
On 9/22/2026 1:32 AM, dart200 wrote:
On 9/21/26 8:48 PM, dbush wrote:
On 9/21/2026 11:32 PM, dart200 wrote:
On 9/21/26 8:18 PM, dbush wrote:
On 9/21/2026 11:10 PM, dart200 wrote:
On 9/21/26 1:41 PM, dbush wrote:
On 9/21/2026 4:06 PM, dart200 wrote:
On 9/21/26 11:53 AM, dbush wrote:And that side record is what disqualifies it from being a >>>>>>>>>>>> halt decider.
On 9/21/2026 1:18 PM, dart200 wrote:
On 9/21/26 6:34 AM, dbush wrote:
On 9/21/2026 2:00 AM, dart200 wrote:
On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote:that's what -o7 is on dud! maybe u should read the fking >>>>>>>>>>>>>>>>> paper instead of dicking about in -o1!
They are one in the same.-a A turing machine is >>>>>>>>>>>>>>>>>> essentially a manifestation of an algorithm.-a Until >>>>>>>>>>>>>>>>>> you can show an algorithm that is not a turing >>>>>>>>>>>>>>>>>> machine, or vice versa, Church- Turning stands. >>>>>>>>>>>>>>>>>
it ultimately discusses how to compute both a total >>>>>>>>>>>>>>>>> diagonal and total anti-diagonal across the enumeration >>>>>>>>>>>>>>>>> of turing computable sequences, a computation which >>>>>>>>>>>>>>>>> obviously cannot be done with a turing machine, as per >>>>>>>>>>>>>>>>> turing's original proof on the matter. and demonstrated >>>>>>>>>>>>>>>>> that is a 26 page paper which shall not be able to >>>>>>>>>>>>>>>>> condense into a damn usenet post
Which you do by getting unspecified input from a human, >>>>>>>>>>>>>>>> which means it's not an algorithm.-a This also means it >>>>>>>>>>>>>>>> takes input other than a machine description and the >>>>>>>>>>>>>>>> input to that machine, which are the only allowed inputs >>>>>>>>>>>>>>>> to a halt decider.
Anything you're doing regarding the address of a >>>>>>>>>>>>>>>> function is simply adding another input which means >>>>>>>>>>>>>>>> you're changing the question.
-o7.6 - truly begging a question
So what the human agent is recording becomes an input. >>>>>>>>>>>>>> And therefore whatever uses it is disqualified from being >>>>>>>>>>>>>> a halt decider, partial or otherwise.
the human agent produces a response back to the terminal >>>>>>>>>>>>> machine, yes
but he also computes things on the side that persist >>>>>>>>>>>>> between terminal interactions, and it's in this side record >>>>>>>>>>>>> where he can compute things outside the turing machine model >>>>>>>>>>>>
the knowledge is still computable, even if it can't be used >>>>>>>>>>> in total within the turing machine model
It is computable in the turing machine model from a machine >>>>>>>>>> description and "extra" data, not just a machine description, >>>>>>>>>> so not a halt decider, partial or otherwise.
not that ur going to read it, but this argument is responded to >>>>>>>>> in -o7.5 "addressable simulation vs objective mechanics"
You think you found something turing computable but you haven't. >>>>>>>> You
u haven't read my paper so u don't know what i've argued
just have the equivalent of a turing machine / algorithm that >>>>>>>> takes an additional input that a halt decider doesn't have.
i won't be convinced until you tell me what the example from -o7.5 >>>>>>> my paper, sim_cir_sr(und_cir_sr), is supposed to output...
You still don't understand.
The agent's side record is not outside Turing computability.-a In >>>>>> the Turing model it is *part of the input*, i.e. it would be
somewhere on tape.
This side record is effectively a global variable, and the
contents of any global variable are an input to the algorithm.
Just because this side record isn't passed as a parameter to
sim_cir_sr or und_cir_sr doesn't mean it's not an input.
What you're doing is no different from putting the fixed set of
steps that a human is doing in a separate function and putting the >>>>>> human's side record in a global variable.
The fact that the human's side record can't be accessed by the
machine being analyzed doesn't make it non-Turing computable.
again, u ramble on without giving me what sim_cir_sr(und_cir_sr) is >>>>> supposed to output, and worse u think u can explain what's going on >>>>> without even reading the passage explaining what that is. nuts.
As derived from above, if the steps a human performs is replaced
with a function with the the reads/writes to the side record going
into a global variable, and that global variable is set to the
contents of the human's side record prior to the given call, it will
behave the same as if und_cir_sr is called.-a Because that's what
algorithms do: give the same results for the same input.
You don't seem to understand how you're modeling the algorithm
abstraction.
u don't seem to understand how ur begging the question of the ct-
thesis being true... eh?
No, you don't seem to understand that you haven't show that you're
refuted it.
again dud, if u haven't my paper, so u can't actually know the logic i presented! this mind reading bs ur keep trying to assert is a _classic_ cognitive distortion, so add that to the list of fallacies u've shat out
in this thread
and if u just brush it off like i suspect u will, i won't reply >>>>>>> further because it's clearly ur not interested in genuine
engagement,
nvm all the name fallacies committed in this discussion. i
honestly can't quite believe you tried to double down on an
argumentum ex silentio, i never imagined encountering someone so >>>>>>> ungodly retarded
not by a mechanically realizable form it isn't, oracle
Which is already well-known and therefore uninteresting. >>>>>>>>>>>
Also:
"We will start by amending the basic turing machine into a >>>>>>>>>>>>>> terminal machine"
Which means you no longer have a Turing machine so nothing >>>>>>>>>>>>>> that follows applies to Turing machines.
_duh_ ... the total solution to turing machine halting can >>>>>>>>>>>>> only be computed from _outside_ the turing machine model >>>>>>>>>>>>
machines can be made real, and are largely useless beyond >>>>>>>>>>> describing something that realizable computing is surely not. >>>>>>>>>>>
this does not it make not computation, as it still a method >>>>>>>>>>>>> that can produce a sequence, including both a true diagonal >>>>>>>>>>>>> across turing computable numbers and the true anti-diagonal >>>>>>>>>>>>> across turing computable sequences, both of which are >>>>>>>>>>>>> outside the scope of turing computation
It makes it a computation that has input in addition to a >>>>>>>>>>>> machine description and that machine's input, and therefore >>>>>>>>>>>> disqualified from being a halt decider.
like i already said, this is addressed in -o7.6
And you addressed it by using extra input disqualifying it >>>>>>>>>> from being a halt decider.
idk why ur lying about reading the section. or any of the paper >>>>>>>>> at all. but i guess that aligns with all the blatant fallacies >>>>>>>>> u keep committing
Which is the same mistake PO made.
u do know that begging the question is a fallacy, eh?
Identifying a fundamental mistake is not a fallacy.
ur not identifying a mistake, ur just continually begging the >>>>>>>>> question that the unproven church-turing thesis is in fact true, >>>>>>>>>
when u do not in fact have a proof showing that it is true,
while i'm presenting a proof that is it _not_ true
You've proven no such thing.-a The only thing you've proved is >>>>>>>> that you don't understand that your "model" is simply an
algorithm / turning machine with an extra inputs (i.e. the what >>>>>>>> the human records and the addresses of functions) that isn't
allowed for a halt decider.
ur still just begging the question
Which, again, is the same mistake PO made.
and ur trying to reinforce it with a fallacy by association
Nope, just pointing out that you don't understand that you made
the same mistake that King Crank made.
and if that mistake is you continually begging that the ct-thesis
is correct ...
No, I'm pointing out that you didn't refute it because you didn't do
what you think you did.
suppose sim_cir_sr() successfully simulates and outputs the agent's
decision process for the side record...
so how does the simulated agent within sim_cir_sr respond to the
input of und_cir_sr?
0) does the simulated agent simulate sim_cir_sr(und_cir_sr)
recursively without ever reaching a decision, such that
sim_cir_sr(und_cir_sr) never returns, and thereby und_cir_sr will
recurse infinitely without output (making it circular)
1) does the simulated agent decide that und_cir_sr is circular, so
sim_cir_sr(und_cir_sr) => true, causing und_cir_sr to output an
infinite sequence (making it circle-free)
2) does the simulated agent decide that und_cir_sr is circle-free, so
sim_cir_sr(und_cir_sr) => false, causing und_cir_sr to halt (making
it circular)
these all exist as possible sim_cir_sr machines in the total machines
enumeration, which one are you asserting correctly simulates the
idealized agent acting externally to the turing machine model?
None of those simulate und_cir_sr correctly because they aren't
setting the side record to the same value that it has when und_cir_sr
is originally called, meaning it doesn't actually simulate "itself".
That means your abstraction is broken and you don't actually have an
algorithm.
in the case of 0) the idealized agent will record und_cir_sr as circular
in the case of 1) the idealized agent will record und_cir_sr as circle-free
in the case of 2) the idealized agent will record und_cir_sr as circular
how so i know that's possible ... because we just did it u moron! u
can't pin and contradict the general ability to determine what a
particular machine does, you can only pin addressable models like turing machine. yes we _are_ doing a form of computing because it _can_ be used
to produce sequences with confidence, including some outside that of
turing computability
To fix it, the fixed steps the human agent runs need to be converted
to a function which read/writes a global variable, and that global
variable need to be set to the same value it had when und_cir_sr was
first called.
lol, turing machines don't have "global variables" dud, they just have a tape, and if the value is output to the tape anywhere, it can be picked
out by a simulation and contradicted by a paradox
there is _no_ way to simulate the general ability to decide on
paradoxical turing machines within turing machines.
On 9/23/2026 4:12 AM, dart200 wrote:
On 9/22/26 12:21 PM, dbush wrote:
On 9/22/2026 2:54 PM, dart200 wrote:
On 9/22/26 9:52 AM, dbush wrote:
On 9/22/2026 11:39 AM, dart200 wrote:
On 9/22/26 5:26 AM, dbush wrote:
On 9/22/2026 1:32 AM, dart200 wrote:
On 9/21/26 8:48 PM, dbush wrote:You still don't understand.
On 9/21/2026 11:32 PM, dart200 wrote:
On 9/21/26 8:18 PM, dbush wrote:You think you found something turing computable but you
On 9/21/2026 11:10 PM, dart200 wrote:
On 9/21/26 1:41 PM, dbush wrote:
On 9/21/2026 4:06 PM, dart200 wrote:
On 9/21/26 11:53 AM, dbush wrote:
On 9/21/2026 1:18 PM, dart200 wrote:
On 9/21/26 6:34 AM, dbush wrote:
On 9/21/2026 2:00 AM, dart200 wrote:
On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote:that's what -o7 is on dud! maybe u should read the >>>>>>>>>>>>>>>>>> fking paper instead of dicking about in -o1! >>>>>>>>>>>>>>>>>>
They are one in the same.-a A turing machine is >>>>>>>>>>>>>>>>>>> essentially a manifestation of an algorithm.-a Until >>>>>>>>>>>>>>>>>>> you can show an algorithm that is not a turing >>>>>>>>>>>>>>>>>>> machine, or vice versa, Church- Turning stands. >>>>>>>>>>>>>>>>>>
it ultimately discusses how to compute both a total >>>>>>>>>>>>>>>>>> diagonal and total anti-diagonal across the >>>>>>>>>>>>>>>>>> enumeration of turing computable sequences, a >>>>>>>>>>>>>>>>>> computation which obviously cannot be done with a >>>>>>>>>>>>>>>>>> turing machine, as per turing's original proof on the >>>>>>>>>>>>>>>>>> matter. and demonstrated that is a 26 page paper which >>>>>>>>>>>>>>>>>> shall not be able to condense into a damn usenet post >>>>>>>>>>>>>>>>>>
Which you do by getting unspecified input from a human, >>>>>>>>>>>>>>>>> which means it's not an algorithm.-a This also means it >>>>>>>>>>>>>>>>> takes input other than a machine description and the >>>>>>>>>>>>>>>>> input to that machine, which are the only allowed >>>>>>>>>>>>>>>>> inputs to a halt decider.
Anything you're doing regarding the address of a >>>>>>>>>>>>>>>>> function is simply adding another input which means >>>>>>>>>>>>>>>>> you're changing the question.
-o7.6 - truly begging a question
So what the human agent is recording becomes an input. >>>>>>>>>>>>>>> And therefore whatever uses it is disqualified from being >>>>>>>>>>>>>>> a halt decider, partial or otherwise.
the human agent produces a response back to the terminal >>>>>>>>>>>>>> machine, yes
but he also computes things on the side that persist >>>>>>>>>>>>>> between terminal interactions, and it's in this side >>>>>>>>>>>>>> record where he can compute things outside the turing >>>>>>>>>>>>>> machine model
And that side record is what disqualifies it from being a >>>>>>>>>>>>> halt decider.
the knowledge is still computable, even if it can't be used >>>>>>>>>>>> in total within the turing machine model
It is computable in the turing machine model from a machine >>>>>>>>>>> description and "extra" data, not just a machine description, >>>>>>>>>>> so not a halt decider, partial or otherwise.
not that ur going to read it, but this argument is responded >>>>>>>>>> to in -o7.5 "addressable simulation vs objective mechanics" >>>>>>>>>
haven't. You
u haven't read my paper so u don't know what i've argued
just have the equivalent of a turing machine / algorithm that >>>>>>>>> takes an additional input that a halt decider doesn't have.
i won't be convinced until you tell me what the example from
-o7.5 my paper, sim_cir_sr(und_cir_sr), is supposed to output... >>>>>>>
The agent's side record is not outside Turing computability.-a In >>>>>>> the Turing model it is *part of the input*, i.e. it would be
somewhere on tape.
This side record is effectively a global variable, and the
contents of any global variable are an input to the algorithm.
Just because this side record isn't passed as a parameter to
sim_cir_sr or und_cir_sr doesn't mean it's not an input.
What you're doing is no different from putting the fixed set of >>>>>>> steps that a human is doing in a separate function and putting
the human's side record in a global variable.
The fact that the human's side record can't be accessed by the
machine being analyzed doesn't make it non-Turing computable.
again, u ramble on without giving me what sim_cir_sr(und_cir_sr)
is supposed to output, and worse u think u can explain what's
going on without even reading the passage explaining what that is. >>>>>> nuts.
As derived from above, if the steps a human performs is replaced
with a function with the the reads/writes to the side record going
into a global variable, and that global variable is set to the
contents of the human's side record prior to the given call, it
will behave the same as if und_cir_sr is called.-a Because that's
what algorithms do: give the same results for the same input.
You don't seem to understand how you're modeling the algorithm
abstraction.
u don't seem to understand how ur begging the question of the ct-
thesis being true... eh?
No, you don't seem to understand that you haven't show that you're
refuted it.
again dud, if u haven't my paper, so u can't actually know the logic i
presented! this mind reading bs ur keep trying to assert is a
_classic_ cognitive distortion, so add that to the list of fallacies
u've shat out in this thread
Surprised you left this in considering 1) I showed that both above and below, and 2) you *responded* to me doing so further down *in this same message*.
It seems I'm not the one that's not reading.-a Maybe, like Olcott, you've gotten too far in to be able to admit your mistakes.
and if u just brush it off like i suspect u will, i won't reply >>>>>>>> further because it's clearly ur not interested in genuine
engagement,
nvm all the name fallacies committed in this discussion. i
honestly can't quite believe you tried to double down on an
argumentum ex silentio, i never imagined encountering someone so >>>>>>>> ungodly retarded
not by a mechanically realizable form it isn't, oracle >>>>>>>>>>>> machines can be made real, and are largely useless beyond >>>>>>>>>>>> describing something that realizable computing is surely not. >>>>>>>>>>>>
Which is already well-known and therefore uninteresting. >>>>>>>>>>>>
Also:
"We will start by amending the basic turing machine into >>>>>>>>>>>>>>> a terminal machine"
Which means you no longer have a Turing machine so >>>>>>>>>>>>>>> nothing that follows applies to Turing machines.
_duh_ ... the total solution to turing machine halting can >>>>>>>>>>>>>> only be computed from _outside_ the turing machine model >>>>>>>>>>>>>
this does not it make not computation, as it still a >>>>>>>>>>>>>> method that can produce a sequence, including both a true >>>>>>>>>>>>>> diagonal across turing computable numbers and the true >>>>>>>>>>>>>> anti-diagonal across turing computable sequences, both of >>>>>>>>>>>>>> which are outside the scope of turing computation
It makes it a computation that has input in addition to a >>>>>>>>>>>>> machine description and that machine's input, and therefore >>>>>>>>>>>>> disqualified from being a halt decider.
like i already said, this is addressed in -o7.6
And you addressed it by using extra input disqualifying it >>>>>>>>>>> from being a halt decider.
idk why ur lying about reading the section. or any of the >>>>>>>>>> paper at all. but i guess that aligns with all the blatant >>>>>>>>>> fallacies u keep committing
Which is the same mistake PO made.
u do know that begging the question is a fallacy, eh?
Identifying a fundamental mistake is not a fallacy.
ur not identifying a mistake, ur just continually begging the >>>>>>>>>> question that the unproven church-turing thesis is in fact true, >>>>>>>>>>
when u do not in fact have a proof showing that it is true, >>>>>>>>>>
while i'm presenting a proof that is it _not_ true
You've proven no such thing.-a The only thing you've proved is >>>>>>>>> that you don't understand that your "model" is simply an
algorithm / turning machine with an extra inputs (i.e. the what >>>>>>>>> the human records and the addresses of functions) that isn't >>>>>>>>> allowed for a halt decider.
ur still just begging the question
Which, again, is the same mistake PO made.
and ur trying to reinforce it with a fallacy by association
Nope, just pointing out that you don't understand that you made >>>>>>> the same mistake that King Crank made.
and if that mistake is you continually begging that the ct-thesis >>>>>> is correct ...
No, I'm pointing out that you didn't refute it because you didn't
do what you think you did.
suppose sim_cir_sr() successfully simulates and outputs the agent's
decision process for the side record...
so how does the simulated agent within sim_cir_sr respond to the
input of und_cir_sr?
0) does the simulated agent simulate sim_cir_sr(und_cir_sr)
recursively without ever reaching a decision, such that
sim_cir_sr(und_cir_sr) never returns, and thereby und_cir_sr will
recurse infinitely without output (making it circular)
1) does the simulated agent decide that und_cir_sr is circular, so
sim_cir_sr(und_cir_sr) => true, causing und_cir_sr to output an
infinite sequence (making it circle-free)
2) does the simulated agent decide that und_cir_sr is circle-free,
so sim_cir_sr(und_cir_sr) => false, causing und_cir_sr to halt
(making it circular)
these all exist as possible sim_cir_sr machines in the total
machines enumeration, which one are you asserting correctly
simulates the idealized agent acting externally to the turing
machine model?
None of those simulate und_cir_sr correctly because they aren't
setting the side record to the same value that it has when und_cir_sr
is originally called, meaning it doesn't actually simulate "itself".
That means your abstraction is broken and you don't actually have an
algorithm.
in the case of 0) the idealized agent will record und_cir_sr as circular
No, it records und_cir_sr(side_record_value_0a) when und_cir_sr(side_record_value_0) is called.
in the case of 1) the idealized agent will record und_cir_sr as
circle-free
No, it records und_cir_sr(side_record_value_1a) when und_cir_sr(side_record_value_1) is called.
in the case of 2) the idealized agent will record und_cir_sr as circular
No, it records und_cir_sr(side_record_value_2a) when und_cir_sr(side_record_value_2) is called.
how so i know that's possible ... because we just did it u moron! u
can't pin and contradict the general ability to determine what a
particular machine does, you can only pin addressable models like
turing machine. yes we _are_ doing a form of computing because it
_can_ be used to produce sequences with confidence, including some
outside that of turing computability
So what you really did is break the rules of a deterministic algorithm
by using the side record and by not simulating what you though you were simulating.
No deterministic algorithm, no refutation of church-turing
To fix it, the fixed steps the human agent runs need to be converted
to a function which read/writes a global variable, and that global
variable need to be set to the same value it had when und_cir_sr was
first called.
lol, turing machines don't have "global variables" dud, they just have
a tape, and if the value is output to the tape anywhere, it can be
picked out by a simulation and contradicted by a paradox
Strawman.-a "it can be picked out" means you're talking about changing
the code which means you're no longer talking about the same machine.
there is _no_ way to simulate the general ability to decide on
paradoxical turing machines within turing machines.
There are no "paradoxical" turing machine in the way you're thinking of them.-a Once you change the instructions, you no longer have the same
turing machine.
On 9/23/26 5:17 AM, dbush wrote:
On 9/23/2026 4:12 AM, dart200 wrote:
On 9/22/26 12:21 PM, dbush wrote:
On 9/22/2026 2:54 PM, dart200 wrote:
On 9/22/26 9:52 AM, dbush wrote:
On 9/22/2026 11:39 AM, dart200 wrote:
On 9/22/26 5:26 AM, dbush wrote:
On 9/22/2026 1:32 AM, dart200 wrote:
On 9/21/26 8:48 PM, dbush wrote:You still don't understand.
On 9/21/2026 11:32 PM, dart200 wrote:
On 9/21/26 8:18 PM, dbush wrote:You think you found something turing computable but you
On 9/21/2026 11:10 PM, dart200 wrote:not that ur going to read it, but this argument is responded >>>>>>>>>>> to in -o7.5 "addressable simulation vs objective mechanics" >>>>>>>>>>
On 9/21/26 1:41 PM, dbush wrote:
On 9/21/2026 4:06 PM, dart200 wrote:
On 9/21/26 11:53 AM, dbush wrote:
On 9/21/2026 1:18 PM, dart200 wrote:
On 9/21/26 6:34 AM, dbush wrote:
On 9/21/2026 2:00 AM, dart200 wrote:
On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote:that's what -o7 is on dud! maybe u should read the >>>>>>>>>>>>>>>>>>> fking paper instead of dicking about in -o1! >>>>>>>>>>>>>>>>>>>
They are one in the same.-a A turing machine is >>>>>>>>>>>>>>>>>>>> essentially a manifestation of an algorithm.-a Until >>>>>>>>>>>>>>>>>>>> you can show an algorithm that is not a turing >>>>>>>>>>>>>>>>>>>> machine, or vice versa, Church- Turning stands. >>>>>>>>>>>>>>>>>>>
it ultimately discusses how to compute both a total >>>>>>>>>>>>>>>>>>> diagonal and total anti-diagonal across the >>>>>>>>>>>>>>>>>>> enumeration of turing computable sequences, a >>>>>>>>>>>>>>>>>>> computation which obviously cannot be done with a >>>>>>>>>>>>>>>>>>> turing machine, as per turing's original proof on the >>>>>>>>>>>>>>>>>>> matter. and demonstrated that is a 26 page paper >>>>>>>>>>>>>>>>>>> which shall not be able to condense into a damn >>>>>>>>>>>>>>>>>>> usenet post
Which you do by getting unspecified input from a >>>>>>>>>>>>>>>>>> human, which means it's not an algorithm.-a This also >>>>>>>>>>>>>>>>>> means it takes input other than a machine description >>>>>>>>>>>>>>>>>> and the input to that machine, which are the only >>>>>>>>>>>>>>>>>> allowed inputs to a halt decider.
Anything you're doing regarding the address of a >>>>>>>>>>>>>>>>>> function is simply adding another input which means >>>>>>>>>>>>>>>>>> you're changing the question.
-o7.6 - truly begging a question
So what the human agent is recording becomes an input. >>>>>>>>>>>>>>>> And therefore whatever uses it is disqualified from >>>>>>>>>>>>>>>> being a halt decider, partial or otherwise.
the human agent produces a response back to the terminal >>>>>>>>>>>>>>> machine, yes
but he also computes things on the side that persist >>>>>>>>>>>>>>> between terminal interactions, and it's in this side >>>>>>>>>>>>>>> record where he can compute things outside the turing >>>>>>>>>>>>>>> machine model
And that side record is what disqualifies it from being a >>>>>>>>>>>>>> halt decider.
the knowledge is still computable, even if it can't be used >>>>>>>>>>>>> in total within the turing machine model
It is computable in the turing machine model from a machine >>>>>>>>>>>> description and "extra" data, not just a machine
description, so not a halt decider, partial or otherwise. >>>>>>>>>>>
haven't. You
u haven't read my paper so u don't know what i've argued
just have the equivalent of a turing machine / algorithm that >>>>>>>>>> takes an additional input that a halt decider doesn't have. >>>>>>>>>i won't be convinced until you tell me what the example from >>>>>>>>> -o7.5 my paper, sim_cir_sr(und_cir_sr), is supposed to output... >>>>>>>>
The agent's side record is not outside Turing computability.-a In >>>>>>>> the Turing model it is *part of the input*, i.e. it would be
somewhere on tape.
This side record is effectively a global variable, and the
contents of any global variable are an input to the algorithm. >>>>>>>>
Just because this side record isn't passed as a parameter to
sim_cir_sr or und_cir_sr doesn't mean it's not an input.
What you're doing is no different from putting the fixed set of >>>>>>>> steps that a human is doing in a separate function and putting >>>>>>>> the human's side record in a global variable.
The fact that the human's side record can't be accessed by the >>>>>>>> machine being analyzed doesn't make it non-Turing computable.
again, u ramble on without giving me what sim_cir_sr(und_cir_sr) >>>>>>> is supposed to output, and worse u think u can explain what's
going on without even reading the passage explaining what that
is. nuts.
As derived from above, if the steps a human performs is replaced
with a function with the the reads/writes to the side record going >>>>>> into a global variable, and that global variable is set to the
contents of the human's side record prior to the given call, it
will behave the same as if und_cir_sr is called.-a Because that's >>>>>> what algorithms do: give the same results for the same input.
You don't seem to understand how you're modeling the algorithm
abstraction.
u don't seem to understand how ur begging the question of the ct-
thesis being true... eh?
No, you don't seem to understand that you haven't show that you're
refuted it.
again dud, if u haven't my paper, so u can't actually know the logic
i presented! this mind reading bs ur keep trying to assert is a
_classic_ cognitive distortion, so add that to the list of fallacies
u've shat out in this thread
Surprised you left this in considering 1) I showed that both above and
below, and 2) you *responded* to me doing so further down *in this
same message*.
It seems I'm not the one that's not reading.-a Maybe, like Olcott,
you've gotten too far in to be able to admit your mistakes.
dud i've named 3 fallacies and one cognitive distortion in ur arguments
thus far,
i'm confident you have nothing to offer me in terms of any meaningful refutations because u still haven't grasped what i'm even arguing, and
have been trying to cover over that lack of understanding with blatant errors in reasoning
and if u just brush it off like i suspect u will, i won't reply >>>>>>>>> further because it's clearly ur not interested in genuine
engagement,
nvm all the name fallacies committed in this discussion. i
honestly can't quite believe you tried to double down on an >>>>>>>>> argumentum ex silentio, i never imagined encountering someone >>>>>>>>> so ungodly retarded
not by a mechanically realizable form it isn't, oracle >>>>>>>>>>>>> machines can be made real, and are largely useless beyond >>>>>>>>>>>>> describing something that realizable computing is surely not. >>>>>>>>>>>>>
Which is already well-known and therefore uninteresting. >>>>>>>>>>>>>
_duh_ ... the total solution to turing machine halting >>>>>>>>>>>>>>> can only be computed from _outside_ the turing machine model >>>>>>>>>>>>>>
Also:
"We will start by amending the basic turing machine into >>>>>>>>>>>>>>>> a terminal machine"
Which means you no longer have a Turing machine so >>>>>>>>>>>>>>>> nothing that follows applies to Turing machines. >>>>>>>>>>>>>>>
It makes it a computation that has input in addition to a >>>>>>>>>>>>>> machine description and that machine's input, and >>>>>>>>>>>>>> therefore disqualified from being a halt decider.
this does not it make not computation, as it still a >>>>>>>>>>>>>>> method that can produce a sequence, including both a true >>>>>>>>>>>>>>> diagonal across turing computable numbers and the true >>>>>>>>>>>>>>> anti-diagonal across turing computable sequences, both of >>>>>>>>>>>>>>> which are outside the scope of turing computation >>>>>>>>>>>>>>
like i already said, this is addressed in -o7.6
And you addressed it by using extra input disqualifying it >>>>>>>>>>>> from being a halt decider.
idk why ur lying about reading the section. or any of the >>>>>>>>>>> paper at all. but i guess that aligns with all the blatant >>>>>>>>>>> fallacies u keep committing
Identifying a fundamental mistake is not a fallacy.
Which is the same mistake PO made.
u do know that begging the question is a fallacy, eh? >>>>>>>>>>>>
ur not identifying a mistake, ur just continually begging the >>>>>>>>>>> question that the unproven church-turing thesis is in fact true, >>>>>>>>>>>
when u do not in fact have a proof showing that it is true, >>>>>>>>>>>
while i'm presenting a proof that is it _not_ true
You've proven no such thing.-a The only thing you've proved is >>>>>>>>>> that you don't understand that your "model" is simply an
algorithm / turning machine with an extra inputs (i.e. the >>>>>>>>>> what the human records and the addresses of functions) that >>>>>>>>>> isn't allowed for a halt decider.
ur still just begging the question
Which, again, is the same mistake PO made.
and ur trying to reinforce it with a fallacy by association
Nope, just pointing out that you don't understand that you made >>>>>>>> the same mistake that King Crank made.
and if that mistake is you continually begging that the ct-thesis >>>>>>> is correct ...
No, I'm pointing out that you didn't refute it because you didn't >>>>>> do what you think you did.
suppose sim_cir_sr() successfully simulates and outputs the agent's >>>>> decision process for the side record...
so how does the simulated agent within sim_cir_sr respond to the
input of und_cir_sr?
0) does the simulated agent simulate sim_cir_sr(und_cir_sr)
recursively without ever reaching a decision, such that
sim_cir_sr(und_cir_sr) never returns, and thereby und_cir_sr will
recurse infinitely without output (making it circular)
1) does the simulated agent decide that und_cir_sr is circular, so
sim_cir_sr(und_cir_sr) => true, causing und_cir_sr to output an
infinite sequence (making it circle-free)
2) does the simulated agent decide that und_cir_sr is circle-free,
so sim_cir_sr(und_cir_sr) => false, causing und_cir_sr to halt
(making it circular)
these all exist as possible sim_cir_sr machines in the total
machines enumeration, which one are you asserting correctly
simulates the idealized agent acting externally to the turing
machine model?
None of those simulate und_cir_sr correctly because they aren't
setting the side record to the same value that it has when
und_cir_sr is originally called, meaning it doesn't actually
simulate "itself".
That means your abstraction is broken and you don't actually have an
algorithm.
in the case of 0) the idealized agent will record und_cir_sr as circular
No, it records und_cir_sr(side_record_value_0a) when
und_cir_sr(side_record_value_0) is called.
in the case of 1) the idealized agent will record und_cir_sr as
circle-free
No, it records und_cir_sr(side_record_value_1a) when
und_cir_sr(side_record_value_1) is called.
in the case of 2) the idealized agent will record und_cir_sr as circular >>>
No, it records und_cir_sr(side_record_value_2a) when
und_cir_sr(side_record_value_2) is called.
how so i know that's possible ... because we just did it u moron! u
can't pin and contradict the general ability to determine what a
particular machine does, you can only pin addressable models like
turing machine. yes we _are_ doing a form of computing because it
_can_ be used to produce sequences with confidence, including some
outside that of turing computability
So what you really did is break the rules of a deterministic algorithm
by using the side record and by not simulating what you though you
were simulating.
No deterministic algorithm, no refutation of church-turing
the algorithm is detailed in -o7.6 ... which u still haven't even opened
To fix it, the fixed steps the human agent runs need to be converted
to a function which read/writes a global variable, and that global
variable need to be set to the same value it had when und_cir_sr was
first called.
lol, turing machines don't have "global variables" dud, they just
have a tape, and if the value is output to the tape anywhere, it can
be picked out by a simulation and contradicted by a paradox
Strawman.-a "it can be picked out" means you're talking about changing
the code which means you're no longer talking about the same machine.
that's incorrect. reading values from the tape during a step of the computation does not change the machine which is being simulated ...
and if a machine is simulating itself, like in the case an
undecidability paradox within computing, then this can be used to pick
out values behind any form of data encapsulation that you might suggest
to hide data away from a paradox. there's no where to hide output dud. there's no "global variable" that might prevent a paradox from being formed
there is _no_ way to simulate a general ability to decide on paradoxical turing machines, from within turing machines
unless like what ... u lost track what turing's proof was even about
cause ur so confused in asserting that i'm wrong???
there is _no_ way to simulate the general ability to decide on
paradoxical turing machines within turing machines.
There are no "paradoxical" turing machine in the way you're thinking
of them.-a Once you change the instructions, you no longer have the
same turing machine.
i never suggested changing any instructions dud, idk where u even pulled that from
On 9/23/26 5:17 AM, dbush wrote:
On 9/23/2026 4:12 AM, dart200 wrote:
On 9/22/26 12:21 PM, dbush wrote:
On 9/22/2026 2:54 PM, dart200 wrote:
On 9/22/26 9:52 AM, dbush wrote:
On 9/22/2026 11:39 AM, dart200 wrote:
On 9/22/26 5:26 AM, dbush wrote:
On 9/22/2026 1:32 AM, dart200 wrote:
On 9/21/26 8:48 PM, dbush wrote:You still don't understand.
On 9/21/2026 11:32 PM, dart200 wrote:
On 9/21/26 8:18 PM, dbush wrote:You think you found something turing computable but you
On 9/21/2026 11:10 PM, dart200 wrote:not that ur going to read it, but this argument is responded >>>>>>>>>>> to in -o7.5 "addressable simulation vs objective mechanics" >>>>>>>>>>
On 9/21/26 1:41 PM, dbush wrote:
On 9/21/2026 4:06 PM, dart200 wrote:
On 9/21/26 11:53 AM, dbush wrote:
On 9/21/2026 1:18 PM, dart200 wrote:
On 9/21/26 6:34 AM, dbush wrote:
On 9/21/2026 2:00 AM, dart200 wrote:
On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote:that's what -o7 is on dud! maybe u should read the >>>>>>>>>>>>>>>>>>> fking paper instead of dicking about in -o1! >>>>>>>>>>>>>>>>>>>
They are one in the same.-a A turing machine is >>>>>>>>>>>>>>>>>>>> essentially a manifestation of an algorithm.-a Until >>>>>>>>>>>>>>>>>>>> you can show an algorithm that is not a turing >>>>>>>>>>>>>>>>>>>> machine, or vice versa, Church- Turning stands. >>>>>>>>>>>>>>>>>>>
it ultimately discusses how to compute both a total >>>>>>>>>>>>>>>>>>> diagonal and total anti-diagonal across the >>>>>>>>>>>>>>>>>>> enumeration of turing computable sequences, a >>>>>>>>>>>>>>>>>>> computation which obviously cannot be done with a >>>>>>>>>>>>>>>>>>> turing machine, as per turing's original proof on the >>>>>>>>>>>>>>>>>>> matter. and demonstrated that is a 26 page paper >>>>>>>>>>>>>>>>>>> which shall not be able to condense into a damn >>>>>>>>>>>>>>>>>>> usenet post
Which you do by getting unspecified input from a >>>>>>>>>>>>>>>>>> human, which means it's not an algorithm.-a This also >>>>>>>>>>>>>>>>>> means it takes input other than a machine description >>>>>>>>>>>>>>>>>> and the input to that machine, which are the only >>>>>>>>>>>>>>>>>> allowed inputs to a halt decider.
Anything you're doing regarding the address of a >>>>>>>>>>>>>>>>>> function is simply adding another input which means >>>>>>>>>>>>>>>>>> you're changing the question.
-o7.6 - truly begging a question
So what the human agent is recording becomes an input. >>>>>>>>>>>>>>>> And therefore whatever uses it is disqualified from >>>>>>>>>>>>>>>> being a halt decider, partial or otherwise.
the human agent produces a response back to the terminal >>>>>>>>>>>>>>> machine, yes
but he also computes things on the side that persist >>>>>>>>>>>>>>> between terminal interactions, and it's in this side >>>>>>>>>>>>>>> record where he can compute things outside the turing >>>>>>>>>>>>>>> machine model
And that side record is what disqualifies it from being a >>>>>>>>>>>>>> halt decider.
the knowledge is still computable, even if it can't be used >>>>>>>>>>>>> in total within the turing machine model
It is computable in the turing machine model from a machine >>>>>>>>>>>> description and "extra" data, not just a machine
description, so not a halt decider, partial or otherwise. >>>>>>>>>>>
haven't. You
u haven't read my paper so u don't know what i've argued
just have the equivalent of a turing machine / algorithm that >>>>>>>>>> takes an additional input that a halt decider doesn't have. >>>>>>>>>i won't be convinced until you tell me what the example from >>>>>>>>> -o7.5 my paper, sim_cir_sr(und_cir_sr), is supposed to output... >>>>>>>>
The agent's side record is not outside Turing computability.-a In >>>>>>>> the Turing model it is *part of the input*, i.e. it would be
somewhere on tape.
This side record is effectively a global variable, and the
contents of any global variable are an input to the algorithm. >>>>>>>>
Just because this side record isn't passed as a parameter to
sim_cir_sr or und_cir_sr doesn't mean it's not an input.
What you're doing is no different from putting the fixed set of >>>>>>>> steps that a human is doing in a separate function and putting >>>>>>>> the human's side record in a global variable.
The fact that the human's side record can't be accessed by the >>>>>>>> machine being analyzed doesn't make it non-Turing computable.
again, u ramble on without giving me what sim_cir_sr(und_cir_sr) >>>>>>> is supposed to output, and worse u think u can explain what's
going on without even reading the passage explaining what that
is. nuts.
As derived from above, if the steps a human performs is replaced
with a function with the the reads/writes to the side record going >>>>>> into a global variable, and that global variable is set to the
contents of the human's side record prior to the given call, it
will behave the same as if und_cir_sr is called.-a Because that's >>>>>> what algorithms do: give the same results for the same input.
You don't seem to understand how you're modeling the algorithm
abstraction.
u don't seem to understand how ur begging the question of the ct-
thesis being true... eh?
No, you don't seem to understand that you haven't show that you're
refuted it.
again dud, if u haven't my paper, so u can't actually know the logic
i presented! this mind reading bs ur keep trying to assert is a
_classic_ cognitive distortion, so add that to the list of fallacies
u've shat out in this thread
Surprised you left this in considering 1) I showed that both above and
below, and 2) you *responded* to me doing so further down *in this
same message*.
It seems I'm not the one that's not reading.-a Maybe, like Olcott,
you've gotten too far in to be able to admit your mistakes.
dud i've named 3 fallacies and one cognitive distortion in ur arguments
thus far,
i'm confident you have nothing to offer me in terms of any meaningful refutations because u still haven't grasped what i'm even arguing, and
have been trying to cover over that lack of understanding with blatant errors in reasoning
and if u just brush it off like i suspect u will, i won't reply >>>>>>>>> further because it's clearly ur not interested in genuine
engagement,
nvm all the name fallacies committed in this discussion. i
honestly can't quite believe you tried to double down on an >>>>>>>>> argumentum ex silentio, i never imagined encountering someone >>>>>>>>> so ungodly retarded
not by a mechanically realizable form it isn't, oracle >>>>>>>>>>>>> machines can be made real, and are largely useless beyond >>>>>>>>>>>>> describing something that realizable computing is surely not. >>>>>>>>>>>>>
Which is already well-known and therefore uninteresting. >>>>>>>>>>>>>
_duh_ ... the total solution to turing machine halting >>>>>>>>>>>>>>> can only be computed from _outside_ the turing machine model >>>>>>>>>>>>>>
Also:
"We will start by amending the basic turing machine into >>>>>>>>>>>>>>>> a terminal machine"
Which means you no longer have a Turing machine so >>>>>>>>>>>>>>>> nothing that follows applies to Turing machines. >>>>>>>>>>>>>>>
It makes it a computation that has input in addition to a >>>>>>>>>>>>>> machine description and that machine's input, and >>>>>>>>>>>>>> therefore disqualified from being a halt decider.
this does not it make not computation, as it still a >>>>>>>>>>>>>>> method that can produce a sequence, including both a true >>>>>>>>>>>>>>> diagonal across turing computable numbers and the true >>>>>>>>>>>>>>> anti-diagonal across turing computable sequences, both of >>>>>>>>>>>>>>> which are outside the scope of turing computation >>>>>>>>>>>>>>
like i already said, this is addressed in -o7.6
And you addressed it by using extra input disqualifying it >>>>>>>>>>>> from being a halt decider.
idk why ur lying about reading the section. or any of the >>>>>>>>>>> paper at all. but i guess that aligns with all the blatant >>>>>>>>>>> fallacies u keep committing
Identifying a fundamental mistake is not a fallacy.
Which is the same mistake PO made.
u do know that begging the question is a fallacy, eh? >>>>>>>>>>>>
ur not identifying a mistake, ur just continually begging the >>>>>>>>>>> question that the unproven church-turing thesis is in fact true, >>>>>>>>>>>
when u do not in fact have a proof showing that it is true, >>>>>>>>>>>
while i'm presenting a proof that is it _not_ true
You've proven no such thing.-a The only thing you've proved is >>>>>>>>>> that you don't understand that your "model" is simply an
algorithm / turning machine with an extra inputs (i.e. the >>>>>>>>>> what the human records and the addresses of functions) that >>>>>>>>>> isn't allowed for a halt decider.
ur still just begging the question
Which, again, is the same mistake PO made.
and ur trying to reinforce it with a fallacy by association
Nope, just pointing out that you don't understand that you made >>>>>>>> the same mistake that King Crank made.
and if that mistake is you continually begging that the ct-thesis >>>>>>> is correct ...
No, I'm pointing out that you didn't refute it because you didn't >>>>>> do what you think you did.
suppose sim_cir_sr() successfully simulates and outputs the agent's >>>>> decision process for the side record...
so how does the simulated agent within sim_cir_sr respond to the
input of und_cir_sr?
0) does the simulated agent simulate sim_cir_sr(und_cir_sr)
recursively without ever reaching a decision, such that
sim_cir_sr(und_cir_sr) never returns, and thereby und_cir_sr will
recurse infinitely without output (making it circular)
1) does the simulated agent decide that und_cir_sr is circular, so
sim_cir_sr(und_cir_sr) => true, causing und_cir_sr to output an
infinite sequence (making it circle-free)
2) does the simulated agent decide that und_cir_sr is circle-free,
so sim_cir_sr(und_cir_sr) => false, causing und_cir_sr to halt
(making it circular)
these all exist as possible sim_cir_sr machines in the total
machines enumeration, which one are you asserting correctly
simulates the idealized agent acting externally to the turing
machine model?
None of those simulate und_cir_sr correctly because they aren't
setting the side record to the same value that it has when
und_cir_sr is originally called, meaning it doesn't actually
simulate "itself".
That means your abstraction is broken and you don't actually have an
algorithm.
in the case of 0) the idealized agent will record und_cir_sr as circular
No, it records und_cir_sr(side_record_value_0a) when
und_cir_sr(side_record_value_0) is called.
in the case of 1) the idealized agent will record und_cir_sr as
circle-free
No, it records und_cir_sr(side_record_value_1a) when
und_cir_sr(side_record_value_1) is called.
in the case of 2) the idealized agent will record und_cir_sr as circular >>>
No, it records und_cir_sr(side_record_value_2a) when
und_cir_sr(side_record_value_2) is called.
how so i know that's possible ... because we just did it u moron! u
can't pin and contradict the general ability to determine what a
particular machine does, you can only pin addressable models like
turing machine. yes we _are_ doing a form of computing because it
_can_ be used to produce sequences with confidence, including some
outside that of turing computability
So what you really did is break the rules of a deterministic algorithm
by using the side record and by not simulating what you though you
were simulating.
No deterministic algorithm, no refutation of church-turing
the algorithm is detailed in -o7.6 ... which u still haven't even opened
To fix it, the fixed steps the human agent runs need to be converted
to a function which read/writes a global variable, and that global
variable need to be set to the same value it had when und_cir_sr was
first called.
lol, turing machines don't have "global variables" dud, they just
have a tape, and if the value is output to the tape anywhere, it can
be picked out by a simulation and contradicted by a paradox
Strawman.-a "it can be picked out" means you're talking about changing
the code which means you're no longer talking about the same machine.
that's incorrect. reading values from the tape during a step of the computation does not change the machine which is being simulated ...
and if a machine is simulating itself,
like in the case an
undecidability paradox within computing, then this can be used to pick
out values behind any form of data encapsulation that you might suggest
to hide data away from a paradox. there's no where to hide output dud. there's no "global variable" that might prevent a paradox from being formed
there is _no_ way to simulate a general ability to decide on paradoxical turing machines, from within turing machines
unless like what ... u lost track what turing's proof was even about
cause ur so confused in asserting that i'm wrong???
there is _no_ way to simulate the general ability to decide on
paradoxical turing machines within turing machines.
There are no "paradoxical" turing machine in the way you're thinking
of them.-a Once you change the instructions, you no longer have the
same turing machine.
i never suggested changing any instructions dud, idk where u even pulled that from
On 9/23/2026 3:17 PM, dart200 wrote:
On 9/23/26 5:17 AM, dbush wrote:
On 9/23/2026 4:12 AM, dart200 wrote:
On 9/22/26 12:21 PM, dbush wrote:
On 9/22/2026 2:54 PM, dart200 wrote:
On 9/22/26 9:52 AM, dbush wrote:
On 9/22/2026 11:39 AM, dart200 wrote:
On 9/22/26 5:26 AM, dbush wrote:
On 9/22/2026 1:32 AM, dart200 wrote:again, u ramble on without giving me what sim_cir_sr(und_cir_sr) >>>>>>>> is supposed to output, and worse u think u can explain what's >>>>>>>> going on without even reading the passage explaining what that >>>>>>>> is. nuts.
On 9/21/26 8:48 PM, dbush wrote:You still don't understand.
On 9/21/2026 11:32 PM, dart200 wrote:
On 9/21/26 8:18 PM, dbush wrote:You think you found something turing computable but you >>>>>>>>>>> haven't. You
On 9/21/2026 11:10 PM, dart200 wrote:not that ur going to read it, but this argument is responded >>>>>>>>>>>> to in -o7.5 "addressable simulation vs objective mechanics" >>>>>>>>>>>
On 9/21/26 1:41 PM, dbush wrote:
On 9/21/2026 4:06 PM, dart200 wrote:
On 9/21/26 11:53 AM, dbush wrote:
On 9/21/2026 1:18 PM, dart200 wrote:
On 9/21/26 6:34 AM, dbush wrote:
On 9/21/2026 2:00 AM, dart200 wrote:
On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>that's what -o7 is on dud! maybe u should read the >>>>>>>>>>>>>>>>>>>> fking paper instead of dicking about in -o1! >>>>>>>>>>>>>>>>>>>>
They are one in the same.-a A turing machine is >>>>>>>>>>>>>>>>>>>>> essentially a manifestation of an algorithm.-a Until >>>>>>>>>>>>>>>>>>>>> you can show an algorithm that is not a turing >>>>>>>>>>>>>>>>>>>>> machine, or vice versa, Church- Turning stands. >>>>>>>>>>>>>>>>>>>>
it ultimately discusses how to compute both a total >>>>>>>>>>>>>>>>>>>> diagonal and total anti-diagonal across the >>>>>>>>>>>>>>>>>>>> enumeration of turing computable sequences, a >>>>>>>>>>>>>>>>>>>> computation which obviously cannot be done with a >>>>>>>>>>>>>>>>>>>> turing machine, as per turing's original proof on >>>>>>>>>>>>>>>>>>>> the matter. and demonstrated that is a 26 page paper >>>>>>>>>>>>>>>>>>>> which shall not be able to condense into a damn >>>>>>>>>>>>>>>>>>>> usenet post
Which you do by getting unspecified input from a >>>>>>>>>>>>>>>>>>> human, which means it's not an algorithm.-a This also >>>>>>>>>>>>>>>>>>> means it takes input other than a machine description >>>>>>>>>>>>>>>>>>> and the input to that machine, which are the only >>>>>>>>>>>>>>>>>>> allowed inputs to a halt decider.
Anything you're doing regarding the address of a >>>>>>>>>>>>>>>>>>> function is simply adding another input which means >>>>>>>>>>>>>>>>>>> you're changing the question.
-o7.6 - truly begging a question
So what the human agent is recording becomes an input. >>>>>>>>>>>>>>>>> And therefore whatever uses it is disqualified from >>>>>>>>>>>>>>>>> being a halt decider, partial or otherwise.
the human agent produces a response back to the terminal >>>>>>>>>>>>>>>> machine, yes
but he also computes things on the side that persist >>>>>>>>>>>>>>>> between terminal interactions, and it's in this side >>>>>>>>>>>>>>>> record where he can compute things outside the turing >>>>>>>>>>>>>>>> machine model
And that side record is what disqualifies it from being a >>>>>>>>>>>>>>> halt decider.
the knowledge is still computable, even if it can't be >>>>>>>>>>>>>> used in total within the turing machine model
It is computable in the turing machine model from a machine >>>>>>>>>>>>> description and "extra" data, not just a machine
description, so not a halt decider, partial or otherwise. >>>>>>>>>>>>
u haven't read my paper so u don't know what i've argued
just have the equivalent of a turing machine / algorithm that >>>>>>>>>>> takes an additional input that a halt decider doesn't have. >>>>>>>>>>i won't be convinced until you tell me what the example from >>>>>>>>>> -o7.5 my paper, sim_cir_sr(und_cir_sr), is supposed to output... >>>>>>>>>
The agent's side record is not outside Turing computability. >>>>>>>>> In the Turing model it is *part of the input*, i.e. it would be >>>>>>>>> somewhere on tape.
This side record is effectively a global variable, and the
contents of any global variable are an input to the algorithm. >>>>>>>>>
Just because this side record isn't passed as a parameter to >>>>>>>>> sim_cir_sr or und_cir_sr doesn't mean it's not an input.
What you're doing is no different from putting the fixed set of >>>>>>>>> steps that a human is doing in a separate function and putting >>>>>>>>> the human's side record in a global variable.
The fact that the human's side record can't be accessed by the >>>>>>>>> machine being analyzed doesn't make it non-Turing computable. >>>>>>>>
As derived from above, if the steps a human performs is replaced >>>>>>> with a function with the the reads/writes to the side record
going into a global variable, and that global variable is set to >>>>>>> the contents of the human's side record prior to the given call, >>>>>>> it will behave the same as if und_cir_sr is called.-a Because
that's what algorithms do: give the same results for the same input. >>>>>>>
You don't seem to understand how you're modeling the algorithm
abstraction.
u don't seem to understand how ur begging the question of the ct- >>>>>> thesis being true... eh?
No, you don't seem to understand that you haven't show that you're
refuted it.
again dud, if u haven't my paper, so u can't actually know the logic
i presented! this mind reading bs ur keep trying to assert is a
_classic_ cognitive distortion, so add that to the list of fallacies
u've shat out in this thread
Surprised you left this in considering 1) I showed that both above
and below, and 2) you *responded* to me doing so further down *in
this same message*.
It seems I'm not the one that's not reading.-a Maybe, like Olcott,
you've gotten too far in to be able to admit your mistakes.
dud i've named 3 fallacies and one cognitive distortion in ur
arguments thus far,
i'm confident you have nothing to offer me in terms of any meaningful
refutations because u still haven't grasped what i'm even arguing, and
have been trying to cover over that lack of understanding with blatant
errors in reasoning
I've grasped that you have a fundamental misunderstanding of what you're attempting to do, and that you're just too far gone to see or admit.
and if u just brush it off like i suspect u will, i won't >>>>>>>>>> reply further because it's clearly ur not interested in
genuine engagement,
nvm all the name fallacies committed in this discussion. i >>>>>>>>>> honestly can't quite believe you tried to double down on an >>>>>>>>>> argumentum ex silentio, i never imagined encountering someone >>>>>>>>>> so ungodly retarded
not by a mechanically realizable form it isn't, oracle >>>>>>>>>>>>>> machines can be made real, and are largely useless beyond >>>>>>>>>>>>>> describing something that realizable computing is surely not. >>>>>>>>>>>>>>
_duh_ ... the total solution to turing machine halting >>>>>>>>>>>>>>>> can only be computed from _outside_ the turing machine >>>>>>>>>>>>>>>> model
Also:
"We will start by amending the basic turing machine >>>>>>>>>>>>>>>>> into a terminal machine"
Which means you no longer have a Turing machine so >>>>>>>>>>>>>>>>> nothing that follows applies to Turing machines. >>>>>>>>>>>>>>>>
Which is already well-known and therefore uninteresting. >>>>>>>>>>>>>>
like i already said, this is addressed in -o7.6
It makes it a computation that has input in addition to a >>>>>>>>>>>>>>> machine description and that machine's input, and >>>>>>>>>>>>>>> therefore disqualified from being a halt decider. >>>>>>>>>>>>>>
this does not it make not computation, as it still a >>>>>>>>>>>>>>>> method that can produce a sequence, including both a >>>>>>>>>>>>>>>> true diagonal across turing computable numbers and the >>>>>>>>>>>>>>>> true anti-diagonal across turing computable sequences, >>>>>>>>>>>>>>>> both of which are outside the scope of turing computation >>>>>>>>>>>>>>>
And you addressed it by using extra input disqualifying it >>>>>>>>>>>>> from being a halt decider.
idk why ur lying about reading the section. or any of the >>>>>>>>>>>> paper at all. but i guess that aligns with all the blatant >>>>>>>>>>>> fallacies u keep committing
Identifying a fundamental mistake is not a fallacy.
Which is the same mistake PO made.
u do know that begging the question is a fallacy, eh? >>>>>>>>>>>>>
ur not identifying a mistake, ur just continually begging >>>>>>>>>>>> the question that the unproven church-turing thesis is in >>>>>>>>>>>> fact true,
when u do not in fact have a proof showing that it is true, >>>>>>>>>>>>
while i'm presenting a proof that is it _not_ true
You've proven no such thing.-a The only thing you've proved is >>>>>>>>>>> that you don't understand that your "model" is simply an >>>>>>>>>>> algorithm / turning machine with an extra inputs (i.e. the >>>>>>>>>>> what the human records and the addresses of functions) that >>>>>>>>>>> isn't allowed for a halt decider.
ur still just begging the question
Which, again, is the same mistake PO made.
and ur trying to reinforce it with a fallacy by association >>>>>>>>>>
Nope, just pointing out that you don't understand that you made >>>>>>>>> the same mistake that King Crank made.
and if that mistake is you continually begging that the ct-
thesis is correct ...
No, I'm pointing out that you didn't refute it because you didn't >>>>>>> do what you think you did.
suppose sim_cir_sr() successfully simulates and outputs the
agent's decision process for the side record...
so how does the simulated agent within sim_cir_sr respond to the
input of und_cir_sr?
0) does the simulated agent simulate sim_cir_sr(und_cir_sr)
recursively without ever reaching a decision, such that
sim_cir_sr(und_cir_sr) never returns, and thereby und_cir_sr will >>>>>> recurse infinitely without output (making it circular)
1) does the simulated agent decide that und_cir_sr is circular, so >>>>>> sim_cir_sr(und_cir_sr) => true, causing und_cir_sr to output an
infinite sequence (making it circle-free)
2) does the simulated agent decide that und_cir_sr is circle-free, >>>>>> so sim_cir_sr(und_cir_sr) => false, causing und_cir_sr to halt
(making it circular)
these all exist as possible sim_cir_sr machines in the total
machines enumeration, which one are you asserting correctly
simulates the idealized agent acting externally to the turing
machine model?
None of those simulate und_cir_sr correctly because they aren't
setting the side record to the same value that it has when
und_cir_sr is originally called, meaning it doesn't actually
simulate "itself".
That means your abstraction is broken and you don't actually have
an algorithm.
in the case of 0) the idealized agent will record und_cir_sr as
circular
No, it records und_cir_sr(side_record_value_0a) when
und_cir_sr(side_record_value_0) is called.
in the case of 1) the idealized agent will record und_cir_sr as
circle-free
No, it records und_cir_sr(side_record_value_1a) when
und_cir_sr(side_record_value_1) is called.
in the case of 2) the idealized agent will record und_cir_sr as
circular
No, it records und_cir_sr(side_record_value_2a) when
und_cir_sr(side_record_value_2) is called.
how so i know that's possible ... because we just did it u moron! u
can't pin and contradict the general ability to determine what a
particular machine does, you can only pin addressable models like
turing machine. yes we _are_ doing a form of computing because it
_can_ be used to produce sequences with confidence, including some
outside that of turing computability
So what you really did is break the rules of a deterministic
algorithm by using the side record and by not simulating what you
though you were simulating.
No deterministic algorithm, no refutation of church-turing
the algorithm is detailed in -o7.6 ... which u still haven't even opened
Not a deterministic algorithm, as I previously stated and which you made
no attempt to refute, and therefore irrelevant to church-turning
To fix it, the fixed steps the human agent runs need to be
converted to a function which read/writes a global variable, and
that global variable need to be set to the same value it had when
und_cir_sr was first called.
lol, turing machines don't have "global variables" dud, they just
have a tape, and if the value is output to the tape anywhere, it can
be picked out by a simulation and contradicted by a paradox
Strawman.-a "it can be picked out" means you're talking about changing
the code which means you're no longer talking about the same machine.
that's incorrect. reading values from the tape during a step of the
computation does not change the machine which is being simulated ...
No, but the code you've shown doesn't do that.-a So if you change the
code to do that, it's no longer the same machine.
and if a machine is simulating itself,
Which your code isn't doing because the side data isn't the same when starting the simulation as it was when the machine was invoked.
like in the case an undecidability paradox within computing, then this
can be used to pick out values behind any form of data encapsulation
that you might suggest to hide data away from a paradox. there's no
where to hide output dud. there's no "global variable" that might
prevent a paradox from being formed
"can be used" meaning it's not being used now, and therefore irrelevant.
there is _no_ way to simulate a general ability to decide on
paradoxical turing machines, from within turing machines
So a paradoxical machine is one that a decider gets wrong?-a So this machine:
void foo() { return; }
Is a paridoxical machine to this one:
int bar(void *p) { return 0; }
Because machine bar can't successfully report the status of machine foo.
-aYes?
unless like what ... u lost track what turing's proof was even about
cause ur so confused in asserting that i'm wrong???
there is _no_ way to simulate the general ability to decide on
paradoxical turing machines within turing machines.
There are no "paradoxical" turing machine in the way you're thinking
of them.-a Once you change the instructions, you no longer have the
same turing machine.
i never suggested changing any instructions dud, idk where u even
pulled that from
Phrases like "can be used" and "paradoxical machine" imply changing code which you can't do.
Turing machines are defined by their actual instructions, not where instructions can physically reside.--
On 9/23/26 12:43 PM, dbush wrote:
On 9/23/2026 3:17 PM, dart200 wrote:
On 9/23/26 5:17 AM, dbush wrote:
On 9/23/2026 4:12 AM, dart200 wrote:
On 9/22/26 12:21 PM, dbush wrote:
On 9/22/2026 2:54 PM, dart200 wrote:
On 9/22/26 9:52 AM, dbush wrote:
On 9/22/2026 11:39 AM, dart200 wrote:
On 9/22/26 5:26 AM, dbush wrote:
On 9/22/2026 1:32 AM, dart200 wrote:again, u ramble on without giving me what
On 9/21/26 8:48 PM, dbush wrote:You still don't understand.
On 9/21/2026 11:32 PM, dart200 wrote:
On 9/21/26 8:18 PM, dbush wrote:
On 9/21/2026 11:10 PM, dart200 wrote:not that ur going to read it, but this argument is
On 9/21/26 1:41 PM, dbush wrote:
On 9/21/2026 4:06 PM, dart200 wrote:
On 9/21/26 11:53 AM, dbush wrote:
On 9/21/2026 1:18 PM, dart200 wrote:the human agent produces a response back to the >>>>>>>>>>>>>>>>> terminal machine, yes
On 9/21/26 6:34 AM, dbush wrote:
On 9/21/2026 2:00 AM, dart200 wrote:
On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>
They are one in the same.-a A turing machine is >>>>>>>>>>>>>>>>>>>>>> essentially a manifestation of an algorithm. >>>>>>>>>>>>>>>>>>>>>> Until you can show an algorithm that is not a >>>>>>>>>>>>>>>>>>>>>> turing machine, or vice versa, Church- Turning >>>>>>>>>>>>>>>>>>>>>> stands.
that's what -o7 is on dud! maybe u should read the >>>>>>>>>>>>>>>>>>>>> fking paper instead of dicking about in -o1! >>>>>>>>>>>>>>>>>>>>>
it ultimately discusses how to compute both a total >>>>>>>>>>>>>>>>>>>>> diagonal and total anti-diagonal across the >>>>>>>>>>>>>>>>>>>>> enumeration of turing computable sequences, a >>>>>>>>>>>>>>>>>>>>> computation which obviously cannot be done with a >>>>>>>>>>>>>>>>>>>>> turing machine, as per turing's original proof on >>>>>>>>>>>>>>>>>>>>> the matter. and demonstrated that is a 26 page >>>>>>>>>>>>>>>>>>>>> paper which shall not be able to condense into a >>>>>>>>>>>>>>>>>>>>> damn usenet post
Which you do by getting unspecified input from a >>>>>>>>>>>>>>>>>>>> human, which means it's not an algorithm.-a This also >>>>>>>>>>>>>>>>>>>> means it takes input other than a machine >>>>>>>>>>>>>>>>>>>> description and the input to that machine, which are >>>>>>>>>>>>>>>>>>>> the only allowed inputs to a halt decider. >>>>>>>>>>>>>>>>>>>>
Anything you're doing regarding the address of a >>>>>>>>>>>>>>>>>>>> function is simply adding another input which means >>>>>>>>>>>>>>>>>>>> you're changing the question.
-o7.6 - truly begging a question
So what the human agent is recording becomes an input. >>>>>>>>>>>>>>>>>> And therefore whatever uses it is disqualified from >>>>>>>>>>>>>>>>>> being a halt decider, partial or otherwise. >>>>>>>>>>>>>>>>>
but he also computes things on the side that persist >>>>>>>>>>>>>>>>> between terminal interactions, and it's in this side >>>>>>>>>>>>>>>>> record where he can compute things outside the turing >>>>>>>>>>>>>>>>> machine model
And that side record is what disqualifies it from being >>>>>>>>>>>>>>>> a halt decider.
the knowledge is still computable, even if it can't be >>>>>>>>>>>>>>> used in total within the turing machine model
It is computable in the turing machine model from a >>>>>>>>>>>>>> machine description and "extra" data, not just a machine >>>>>>>>>>>>>> description, so not a halt decider, partial or otherwise. >>>>>>>>>>>>>
responded to in -o7.5 "addressable simulation vs objective >>>>>>>>>>>>> mechanics"
You think you found something turing computable but you >>>>>>>>>>>> haven't. You
u haven't read my paper so u don't know what i've argued >>>>>>>>>>>
just have the equivalent of a turing machine / algorithm >>>>>>>>>>>> that takes an additional input that a halt decider doesn't >>>>>>>>>>>> have.
i won't be convinced until you tell me what the example from >>>>>>>>>>> -o7.5 my paper, sim_cir_sr(und_cir_sr), is supposed to output... >>>>>>>>>>
The agent's side record is not outside Turing computability. >>>>>>>>>> In the Turing model it is *part of the input*, i.e. it would >>>>>>>>>> be somewhere on tape.
This side record is effectively a global variable, and the >>>>>>>>>> contents of any global variable are an input to the algorithm. >>>>>>>>>>
Just because this side record isn't passed as a parameter to >>>>>>>>>> sim_cir_sr or und_cir_sr doesn't mean it's not an input.
What you're doing is no different from putting the fixed set >>>>>>>>>> of steps that a human is doing in a separate function and >>>>>>>>>> putting the human's side record in a global variable.
The fact that the human's side record can't be accessed by the >>>>>>>>>> machine being analyzed doesn't make it non-Turing computable. >>>>>>>>>
sim_cir_sr(und_cir_sr) is supposed to output, and worse u think >>>>>>>>> u can explain what's going on without even reading the passage >>>>>>>>> explaining what that is. nuts.
As derived from above, if the steps a human performs is replaced >>>>>>>> with a function with the the reads/writes to the side record
going into a global variable, and that global variable is set to >>>>>>>> the contents of the human's side record prior to the given call, >>>>>>>> it will behave the same as if und_cir_sr is called.-a Because >>>>>>>> that's what algorithms do: give the same results for the same >>>>>>>> input.
You don't seem to understand how you're modeling the algorithm >>>>>>>> abstraction.
u don't seem to understand how ur begging the question of the ct- >>>>>>> thesis being true... eh?
No, you don't seem to understand that you haven't show that you're >>>>>> refuted it.
again dud, if u haven't my paper, so u can't actually know the
logic i presented! this mind reading bs ur keep trying to assert is >>>>> a _classic_ cognitive distortion, so add that to the list of
fallacies u've shat out in this thread
Surprised you left this in considering 1) I showed that both above
and below, and 2) you *responded* to me doing so further down *in
this same message*.
It seems I'm not the one that's not reading.-a Maybe, like Olcott,
you've gotten too far in to be able to admit your mistakes.
dud i've named 3 fallacies and one cognitive distortion in ur
arguments thus far,
i'm confident you have nothing to offer me in terms of any meaningful
refutations because u still haven't grasped what i'm even arguing,
and have been trying to cover over that lack of understanding with
blatant errors in reasoning
I've grasped that you have a fundamental misunderstanding of what
you're attempting to do, and that you're just too far gone to see or
admit.
this is called gaslighting, will u please stop? it's abusive and
completely irrelevant to a genuine discussion
and if u just brush it off like i suspect u will, i won't >>>>>>>>>>> reply further because it's clearly ur not interested in >>>>>>>>>>> genuine engagement,
nvm all the name fallacies committed in this discussion. i >>>>>>>>>>> honestly can't quite believe you tried to double down on an >>>>>>>>>>> argumentum ex silentio, i never imagined encountering someone >>>>>>>>>>> so ungodly retarded
not by a mechanically realizable form it isn't, oracle >>>>>>>>>>>>>>> machines can be made real, and are largely useless beyond >>>>>>>>>>>>>>> describing something that realizable computing is surely >>>>>>>>>>>>>>> not.
_duh_ ... the total solution to turing machine halting >>>>>>>>>>>>>>>>> can only be computed from _outside_ the turing machine >>>>>>>>>>>>>>>>> model
Also:
"We will start by amending the basic turing machine >>>>>>>>>>>>>>>>>> into a terminal machine"
Which means you no longer have a Turing machine so >>>>>>>>>>>>>>>>>> nothing that follows applies to Turing machines. >>>>>>>>>>>>>>>>>
Which is already well-known and therefore uninteresting. >>>>>>>>>>>>>>>
like i already said, this is addressed in -o7.6
It makes it a computation that has input in addition to >>>>>>>>>>>>>>>> a machine description and that machine's input, and >>>>>>>>>>>>>>>> therefore disqualified from being a halt decider. >>>>>>>>>>>>>>>
this does not it make not computation, as it still a >>>>>>>>>>>>>>>>> method that can produce a sequence, including both a >>>>>>>>>>>>>>>>> true diagonal across turing computable numbers and the >>>>>>>>>>>>>>>>> true anti-diagonal across turing computable sequences, >>>>>>>>>>>>>>>>> both of which are outside the scope of turing computation >>>>>>>>>>>>>>>>
And you addressed it by using extra input disqualifying it >>>>>>>>>>>>>> from being a halt decider.
idk why ur lying about reading the section. or any of the >>>>>>>>>>>>> paper at all. but i guess that aligns with all the blatant >>>>>>>>>>>>> fallacies u keep committing
ur not identifying a mistake, ur just continually begging >>>>>>>>>>>>> the question that the unproven church-turing thesis is in >>>>>>>>>>>>> fact true,
Identifying a fundamental mistake is not a fallacy. >>>>>>>>>>>>>
Which is the same mistake PO made.
u do know that begging the question is a fallacy, eh? >>>>>>>>>>>>>>
when u do not in fact have a proof showing that it is true, >>>>>>>>>>>>>
while i'm presenting a proof that is it _not_ true
You've proven no such thing.-a The only thing you've proved >>>>>>>>>>>> is that you don't understand that your "model" is simply an >>>>>>>>>>>> algorithm / turning machine with an extra inputs (i.e. the >>>>>>>>>>>> what the human records and the addresses of functions) that >>>>>>>>>>>> isn't allowed for a halt decider.
ur still just begging the question
Which, again, is the same mistake PO made.
and ur trying to reinforce it with a fallacy by association >>>>>>>>>>>
Nope, just pointing out that you don't understand that you >>>>>>>>>> made the same mistake that King Crank made.
and if that mistake is you continually begging that the ct- >>>>>>>>> thesis is correct ...
No, I'm pointing out that you didn't refute it because you
didn't do what you think you did.
suppose sim_cir_sr() successfully simulates and outputs the
agent's decision process for the side record...
so how does the simulated agent within sim_cir_sr respond to the >>>>>>> input of und_cir_sr?
0) does the simulated agent simulate sim_cir_sr(und_cir_sr)
recursively without ever reaching a decision, such that
sim_cir_sr(und_cir_sr) never returns, and thereby und_cir_sr will >>>>>>> recurse infinitely without output (making it circular)
1) does the simulated agent decide that und_cir_sr is circular, >>>>>>> so sim_cir_sr(und_cir_sr) => true, causing und_cir_sr to output >>>>>>> an infinite sequence (making it circle-free)
2) does the simulated agent decide that und_cir_sr is circle-
free, so sim_cir_sr(und_cir_sr) => false, causing und_cir_sr to >>>>>>> halt (making it circular)
these all exist as possible sim_cir_sr machines in the total
machines enumeration, which one are you asserting correctly
simulates the idealized agent acting externally to the turing
machine model?
None of those simulate und_cir_sr correctly because they aren't
setting the side record to the same value that it has when
und_cir_sr is originally called, meaning it doesn't actually
simulate "itself".
That means your abstraction is broken and you don't actually have >>>>>> an algorithm.
in the case of 0) the idealized agent will record und_cir_sr as
circular
No, it records und_cir_sr(side_record_value_0a) when
und_cir_sr(side_record_value_0) is called.
in the case of 1) the idealized agent will record und_cir_sr as
circle-free
No, it records und_cir_sr(side_record_value_1a) when
und_cir_sr(side_record_value_1) is called.
in the case of 2) the idealized agent will record und_cir_sr as
circular
No, it records und_cir_sr(side_record_value_2a) when
und_cir_sr(side_record_value_2) is called.
how so i know that's possible ... because we just did it u moron! u >>>>> can't pin and contradict the general ability to determine what a
particular machine does, you can only pin addressable models like
turing machine. yes we _are_ doing a form of computing because it
_can_ be used to produce sequences with confidence, including some
outside that of turing computability
So what you really did is break the rules of a deterministic
algorithm by using the side record and by not simulating what you
though you were simulating.
what rule??? lol deterministic only means running it always has the same result given the same input ... which a total decision algo run by the idealized agent certainly does
No deterministic algorithm, no refutation of church-turing
the algorithm is detailed in -o7.6 ... which u still haven't even opened
Not a deterministic algorithm, as I previously stated and which you
made no attempt to refute, and therefore irrelevant to church-turning
u haven't even read the algo, so how are you going to explain why specifically it's non-deterministic ... ???
To fix it, the fixed steps the human agent runs need to be
converted to a function which read/writes a global variable, and
that global variable need to be set to the same value it had when >>>>>> und_cir_sr was first called.
lol, turing machines don't have "global variables" dud, they just
have a tape, and if the value is output to the tape anywhere, it
can be picked out by a simulation and contradicted by a paradox
Strawman.-a "it can be picked out" means you're talking about
changing the code which means you're no longer talking about the
same machine.
that's incorrect. reading values from the tape during a step of the
computation does not change the machine which is being simulated ...
No, but the code you've shown doesn't do that.-a So if you change the
code to do that, it's no longer the same machine.
ofc the code does that
... getting "output" from a simulation requires
reading the output from tape of the simulated machine, it can do that
for any value, even ones that are specified to be "output"
and if a machine is simulating itself,
Which your code isn't doing because the side data isn't the same when
starting the simulation as it was when the machine was invoked.
idk what u mean by "side data" there only data on the tape when it comes
to turing machines, there is nothing on the side of data on the tape.
like in the case an undecidability paradox within computing, then
this can be used to pick out values behind any form of data
encapsulation that you might suggest to hide data away from a
paradox. there's no where to hide output dud. there's no "global
variable" that might prevent a paradox from being formed
"can be used" meaning it's not being used now, and therefore irrelevant.
there is _no_ way to simulate a general ability to decide on
paradoxical turing machines, from within turing machines
So a paradoxical machine is one that a decider gets wrong?-a So this
machine:
void foo() { return; }
Is a paridoxical machine to this one:
int bar(void *p) { return 0; }
Because machine bar can't successfully report the status of machine
foo. -a-aYes?
no. first, we went over this dud: constant return function are not and
will never qualify as a halting decider as they do not output useful information in regards to the halting status Efn+. the fact a machine happens to output the correct answer does not qualify them as a halting decider, a decider must output trustworthy information which is judged across the entirety of the output space, not just single instances
second, an actual paradox is necessarily undecidable due to the
construction of the machine itself, the form is generalized in -o3 of my paper. if it doesn't fit that form, it's not a paradox, and is not undecidable in respect to any classifier let alone in totality
unless like what ... u lost track what turing's proof was even about
cause ur so confused in asserting that i'm wrong???
there is _no_ way to simulate the general ability to decide on
paradoxical turing machines within turing machines.
There are no "paradoxical" turing machine in the way you're thinking
of them.-a Once you change the instructions, you no longer have the
same turing machine.
i never suggested changing any instructions dud, idk where u even
pulled that from
Phrases like "can be used" and "paradoxical machine" imply changing
code which you can't do.
this is theoretical computing dud, all possibilities must be accounted
for and robustly handled
Turing machines are defined by their actual instructions, not where
instructions can physically reside.
On 9/23/2026 9:35 PM, dart200 wrote:
On 9/23/26 12:43 PM, dbush wrote:
On 9/23/2026 3:17 PM, dart200 wrote:
On 9/23/26 5:17 AM, dbush wrote:
On 9/23/2026 4:12 AM, dart200 wrote:
On 9/22/26 12:21 PM, dbush wrote:
On 9/22/2026 2:54 PM, dart200 wrote:
On 9/22/26 9:52 AM, dbush wrote:
On 9/22/2026 11:39 AM, dart200 wrote:
On 9/22/26 5:26 AM, dbush wrote:
On 9/22/2026 1:32 AM, dart200 wrote:
On 9/21/26 8:48 PM, dbush wrote:You still don't understand.
On 9/21/2026 11:32 PM, dart200 wrote:
On 9/21/26 8:18 PM, dbush wrote:
On 9/21/2026 11:10 PM, dart200 wrote:not that ur going to read it, but this argument is >>>>>>>>>>>>>> responded to in -o7.5 "addressable simulation vs objective >>>>>>>>>>>>>> mechanics"
On 9/21/26 1:41 PM, dbush wrote:
On 9/21/2026 4:06 PM, dart200 wrote:
On 9/21/26 11:53 AM, dbush wrote:
On 9/21/2026 1:18 PM, dart200 wrote:the human agent produces a response back to the >>>>>>>>>>>>>>>>>> terminal machine, yes
On 9/21/26 6:34 AM, dbush wrote:
On 9/21/2026 2:00 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>> On 9/20/26 8:32 PM, dbush wrote:
On 9/20/2026 10:35 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>
They are one in the same.-a A turing machine is >>>>>>>>>>>>>>>>>>>>>>> essentially a manifestation of an algorithm. >>>>>>>>>>>>>>>>>>>>>>> Until you can show an algorithm that is not a >>>>>>>>>>>>>>>>>>>>>>> turing machine, or vice versa, Church- Turning >>>>>>>>>>>>>>>>>>>>>>> stands.
that's what -o7 is on dud! maybe u should read the >>>>>>>>>>>>>>>>>>>>>> fking paper instead of dicking about in -o1! >>>>>>>>>>>>>>>>>>>>>>
it ultimately discusses how to compute both a >>>>>>>>>>>>>>>>>>>>>> total diagonal and total anti-diagonal across the >>>>>>>>>>>>>>>>>>>>>> enumeration of turing computable sequences, a >>>>>>>>>>>>>>>>>>>>>> computation which obviously cannot be done with a >>>>>>>>>>>>>>>>>>>>>> turing machine, as per turing's original proof on >>>>>>>>>>>>>>>>>>>>>> the matter. and demonstrated that is a 26 page >>>>>>>>>>>>>>>>>>>>>> paper which shall not be able to condense into a >>>>>>>>>>>>>>>>>>>>>> damn usenet post
Which you do by getting unspecified input from a >>>>>>>>>>>>>>>>>>>>> human, which means it's not an algorithm.-a This >>>>>>>>>>>>>>>>>>>>> also means it takes input other than a machine >>>>>>>>>>>>>>>>>>>>> description and the input to that machine, which >>>>>>>>>>>>>>>>>>>>> are the only allowed inputs to a halt decider. >>>>>>>>>>>>>>>>>>>>>
Anything you're doing regarding the address of a >>>>>>>>>>>>>>>>>>>>> function is simply adding another input which means >>>>>>>>>>>>>>>>>>>>> you're changing the question.
-o7.6 - truly begging a question
So what the human agent is recording becomes an >>>>>>>>>>>>>>>>>>> input. And therefore whatever uses it is disqualified >>>>>>>>>>>>>>>>>>> from being a halt decider, partial or otherwise. >>>>>>>>>>>>>>>>>>
but he also computes things on the side that persist >>>>>>>>>>>>>>>>>> between terminal interactions, and it's in this side >>>>>>>>>>>>>>>>>> record where he can compute things outside the turing >>>>>>>>>>>>>>>>>> machine model
And that side record is what disqualifies it from being >>>>>>>>>>>>>>>>> a halt decider.
the knowledge is still computable, even if it can't be >>>>>>>>>>>>>>>> used in total within the turing machine model
It is computable in the turing machine model from a >>>>>>>>>>>>>>> machine description and "extra" data, not just a machine >>>>>>>>>>>>>>> description, so not a halt decider, partial or otherwise. >>>>>>>>>>>>>>
You think you found something turing computable but you >>>>>>>>>>>>> haven't. You
u haven't read my paper so u don't know what i've argued >>>>>>>>>>>>
just have the equivalent of a turing machine / algorithm >>>>>>>>>>>>> that takes an additional input that a halt decider doesn't >>>>>>>>>>>>> have.
i won't be convinced until you tell me what the example from >>>>>>>>>>>> -o7.5 my paper, sim_cir_sr(und_cir_sr), is supposed to output... >>>>>>>>>>>
The agent's side record is not outside Turing computability. >>>>>>>>>>> In the Turing model it is *part of the input*, i.e. it would >>>>>>>>>>> be somewhere on tape.
This side record is effectively a global variable, and the >>>>>>>>>>> contents of any global variable are an input to the algorithm. >>>>>>>>>>>
Just because this side record isn't passed as a parameter to >>>>>>>>>>> sim_cir_sr or und_cir_sr doesn't mean it's not an input. >>>>>>>>>>>
What you're doing is no different from putting the fixed set >>>>>>>>>>> of steps that a human is doing in a separate function and >>>>>>>>>>> putting the human's side record in a global variable.
The fact that the human's side record can't be accessed by >>>>>>>>>>> the machine being analyzed doesn't make it non-Turing
computable.
again, u ramble on without giving me what
sim_cir_sr(und_cir_sr) is supposed to output, and worse u >>>>>>>>>> think u can explain what's going on without even reading the >>>>>>>>>> passage explaining what that is. nuts.
As derived from above, if the steps a human performs is
replaced with a function with the the reads/writes to the side >>>>>>>>> record going into a global variable, and that global variable >>>>>>>>> is set to the contents of the human's side record prior to the >>>>>>>>> given call, it will behave the same as if und_cir_sr is
called.-a Because that's what algorithms do: give the same
results for the same input.
You don't seem to understand how you're modeling the algorithm >>>>>>>>> abstraction.
u don't seem to understand how ur begging the question of the >>>>>>>> ct- thesis being true... eh?
No, you don't seem to understand that you haven't show that
you're refuted it.
again dud, if u haven't my paper, so u can't actually know the
logic i presented! this mind reading bs ur keep trying to assert
is a _classic_ cognitive distortion, so add that to the list of
fallacies u've shat out in this thread
Surprised you left this in considering 1) I showed that both above
and below, and 2) you *responded* to me doing so further down *in
this same message*.
It seems I'm not the one that's not reading.-a Maybe, like Olcott,
you've gotten too far in to be able to admit your mistakes.
dud i've named 3 fallacies and one cognitive distortion in ur
arguments thus far,
i'm confident you have nothing to offer me in terms of any
meaningful refutations because u still haven't grasped what i'm even
arguing, and have been trying to cover over that lack of
understanding with blatant errors in reasoning
I've grasped that you have a fundamental misunderstanding of what
you're attempting to do, and that you're just too far gone to see or
admit.
this is called gaslighting, will u please stop? it's abusive and
completely irrelevant to a genuine discussion
Pointing out your misconceptions is not gaslighting.
and if u just brush it off like i suspect u will, i won't >>>>>>>>>>>> reply further because it's clearly ur not interested in >>>>>>>>>>>> genuine engagement,
nvm all the name fallacies committed in this discussion. i >>>>>>>>>>>> honestly can't quite believe you tried to double down on an >>>>>>>>>>>> argumentum ex silentio, i never imagined encountering >>>>>>>>>>>> someone so ungodly retarded
And you addressed it by using extra input disqualifying >>>>>>>>>>>>>>> it from being a halt decider.
not by a mechanically realizable form it isn't, oracle >>>>>>>>>>>>>>>> machines can be made real, and are largely useless >>>>>>>>>>>>>>>> beyond describing something that realizable computing is >>>>>>>>>>>>>>>> surely not.
_duh_ ... the total solution to turing machine halting >>>>>>>>>>>>>>>>>> can only be computed from _outside_ the turing machine >>>>>>>>>>>>>>>>>> model
Also:
"We will start by amending the basic turing machine >>>>>>>>>>>>>>>>>>> into a terminal machine"
Which means you no longer have a Turing machine so >>>>>>>>>>>>>>>>>>> nothing that follows applies to Turing machines. >>>>>>>>>>>>>>>>>>
Which is already well-known and therefore uninteresting. >>>>>>>>>>>>>>>>
like i already said, this is addressed in -o7.6 >>>>>>>>>>>>>>>
It makes it a computation that has input in addition to >>>>>>>>>>>>>>>>> a machine description and that machine's input, and >>>>>>>>>>>>>>>>> therefore disqualified from being a halt decider. >>>>>>>>>>>>>>>>
this does not it make not computation, as it still a >>>>>>>>>>>>>>>>>> method that can produce a sequence, including both a >>>>>>>>>>>>>>>>>> true diagonal across turing computable numbers and the >>>>>>>>>>>>>>>>>> true anti-diagonal across turing computable sequences, >>>>>>>>>>>>>>>>>> both of which are outside the scope of turing computation >>>>>>>>>>>>>>>>>
idk why ur lying about reading the section. or any of the >>>>>>>>>>>>>> paper at all. but i guess that aligns with all the blatant >>>>>>>>>>>>>> fallacies u keep committing
ur not identifying a mistake, ur just continually begging >>>>>>>>>>>>>> the question that the unproven church-turing thesis is in >>>>>>>>>>>>>> fact true,
Identifying a fundamental mistake is not a fallacy. >>>>>>>>>>>>>>
Which is the same mistake PO made.
u do know that begging the question is a fallacy, eh? >>>>>>>>>>>>>>>
when u do not in fact have a proof showing that it is true, >>>>>>>>>>>>>>
while i'm presenting a proof that is it _not_ true
You've proven no such thing.-a The only thing you've proved >>>>>>>>>>>>> is that you don't understand that your "model" is simply an >>>>>>>>>>>>> algorithm / turning machine with an extra inputs (i.e. the >>>>>>>>>>>>> what the human records and the addresses of functions) that >>>>>>>>>>>>> isn't allowed for a halt decider.
ur still just begging the question
Which, again, is the same mistake PO made.
and ur trying to reinforce it with a fallacy by association >>>>>>>>>>>>
Nope, just pointing out that you don't understand that you >>>>>>>>>>> made the same mistake that King Crank made.
and if that mistake is you continually begging that the ct- >>>>>>>>>> thesis is correct ...
No, I'm pointing out that you didn't refute it because you
didn't do what you think you did.
suppose sim_cir_sr() successfully simulates and outputs the
agent's decision process for the side record...
so how does the simulated agent within sim_cir_sr respond to the >>>>>>>> input of und_cir_sr?
0) does the simulated agent simulate sim_cir_sr(und_cir_sr)
recursively without ever reaching a decision, such that
sim_cir_sr(und_cir_sr) never returns, and thereby und_cir_sr
will recurse infinitely without output (making it circular)
1) does the simulated agent decide that und_cir_sr is circular, >>>>>>>> so sim_cir_sr(und_cir_sr) => true, causing und_cir_sr to output >>>>>>>> an infinite sequence (making it circle-free)
2) does the simulated agent decide that und_cir_sr is circle- >>>>>>>> free, so sim_cir_sr(und_cir_sr) => false, causing und_cir_sr to >>>>>>>> halt (making it circular)
these all exist as possible sim_cir_sr machines in the total
machines enumeration, which one are you asserting correctly
simulates the idealized agent acting externally to the turing >>>>>>>> machine model?
None of those simulate und_cir_sr correctly because they aren't >>>>>>> setting the side record to the same value that it has when
und_cir_sr is originally called, meaning it doesn't actually
simulate "itself".
That means your abstraction is broken and you don't actually have >>>>>>> an algorithm.
in the case of 0) the idealized agent will record und_cir_sr as
circular
No, it records und_cir_sr(side_record_value_0a) when
und_cir_sr(side_record_value_0) is called.
in the case of 1) the idealized agent will record und_cir_sr as
circle-free
No, it records und_cir_sr(side_record_value_1a) when
und_cir_sr(side_record_value_1) is called.
in the case of 2) the idealized agent will record und_cir_sr as
circular
No, it records und_cir_sr(side_record_value_2a) when
und_cir_sr(side_record_value_2) is called.
how so i know that's possible ... because we just did it u moron! >>>>>> u can't pin and contradict the general ability to determine what a >>>>>> particular machine does, you can only pin addressable models like >>>>>> turing machine. yes we _are_ doing a form of computing because it >>>>>> _can_ be used to produce sequences with confidence, including some >>>>>> outside that of turing computability
So what you really did is break the rules of a deterministic
algorithm by using the side record and by not simulating what you
though you were simulating.
what rule??? lol deterministic only means running it always has the
same result given the same input ... which a total decision algo run
by the idealized agent certainly does
So what exactly are the inputs to the human agent?
Not a deterministic algorithm, as I previously stated and which you
No deterministic algorithm, no refutation of church-turing
the algorithm is detailed in -o7.6 ... which u still haven't even opened >>>
made no attempt to refute, and therefore irrelevant to church-turning
u haven't even read the algo, so how are you going to explain why
specifically it's non-deterministic ... ???
You seem to think that the human's side record is not part of the input.
-aIf so, that makes it non-deterministic.
To fix it, the fixed steps the human agent runs need to be
converted to a function which read/writes a global variable, and >>>>>>> that global variable need to be set to the same value it had when >>>>>>> und_cir_sr was first called.
lol, turing machines don't have "global variables" dud, they just >>>>>> have a tape, and if the value is output to the tape anywhere, it
can be picked out by a simulation and contradicted by a paradox
Strawman.-a "it can be picked out" means you're talking about
changing the code which means you're no longer talking about the
same machine.
that's incorrect. reading values from the tape during a step of the
computation does not change the machine which is being simulated ...
No, but the code you've shown doesn't do that.-a So if you change the
code to do that, it's no longer the same machine.
ofc the code does that
Your code is reading the part of the tape where the mechanized human
agent is writing its side record?
No?
Then your code doesn't do that, and talking about what it "can" do necessarily means changing the code to something not being decided on.
... getting "output" from a simulation requires reading the output
from tape of the simulated machine, it can do that for any value, even
ones that are specified to be "output"
and if a machine is simulating itself,
Which your code isn't doing because the side data isn't the same when
starting the simulation as it was when the machine was invoked.
idk what u mean by "side data" there only data on the tape when it
comes to turing machines, there is nothing on the side of data on the
tape.
I'm referring specifically to what the mechanized version of the human
agent writes to the tape.
like in the case an undecidability paradox within computing, then
this can be used to pick out values behind any form of data
encapsulation that you might suggest to hide data away from a
paradox. there's no where to hide output dud. there's no "global
variable" that might prevent a paradox from being formed
"can be used" meaning it's not being used now, and therefore irrelevant. >>>
there is _no_ way to simulate a general ability to decide on
paradoxical turing machines, from within turing machines
So a paradoxical machine is one that a decider gets wrong?-a So this
machine:
void foo() { return; }
Is a paridoxical machine to this one:
int bar(void *p) { return 0; }
Because machine bar can't successfully report the status of machine
foo. -a-aYes?
no. first, we went over this dud: constant return function are not and
will never qualify as a halting decider as they do not output useful
information in regards to the halting status Efn+. the fact a machine
happens to output the correct answer does not qualify them as a
halting decider, a decider must output trustworthy information which
is judged across the entirety of the output space, not just single
instances
In other words, it's no different from *any* partial halt decider.
That's not even mentioning ill-defined weasel words like "useful",
"happens to", "trustworthy", and "judged".
second, an actual paradox is necessarily undecidable due to the
construction of the machine itself, the form is generalized in -o3 of
my paper. if it doesn't fit that form, it's not a paradox, and is not
undecidable in respect to any classifier let alone in totality
Alright, let's satisfy your arbitrary requirement of a non-constant
function and "useful":
void foo() {
-a-a-a puts("hello");
}
int bar(void *p) {
-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a return 1;
-a-a-a } else {
-a-a-a-a-a-a-a return 0;
-a-a-a }
}
Given that the first instruction of "foo" is not a "return" instruction, bar(foo) will return 0 even though foo() will halt.
Is the machine foo "undecidable" by machine bar?
unless like what ... u lost track what turing's proof was even about
cause ur so confused in asserting that i'm wrong???
there is _no_ way to simulate the general ability to decide on
paradoxical turing machines within turing machines.
There are no "paradoxical" turing machine in the way you're
thinking of them.-a Once you change the instructions, you no longer >>>>> have the same turing machine.
i never suggested changing any instructions dud, idk where u even
pulled that from
Phrases like "can be used" and "paradoxical machine" imply changing
code which you can't do.
this is theoretical computing dud, all possibilities must be accounted
for and robustly handled
Without conflating one for another, which you seem to be doing.
Turing machines are defined by their actual instructions, not where
instructions can physically reside.
On 9/23/26 7:08 PM, dbush wrote:
On 9/23/2026 9:35 PM, dart200 wrote:
On 9/23/26 12:43 PM, dbush wrote:
On 9/23/2026 3:17 PM, dart200 wrote:
On 9/23/26 5:17 AM, dbush wrote:
So what you really did is break the rules of a deterministic
algorithm by using the side record and by not simulating what you >>>>>> though you were simulating.
what rule??? lol deterministic only means running it always has the
same result given the same input ... which a total decision algo run
by the idealized agent certainly does
So what exactly are the inputs to the human agent?
the agent gets the machine description like any decision algo, that
would be clear if u just read the paper
No deterministic algorithm, no refutation of church-turing
the algorithm is detailed in -o7.6 ... which u still haven't even
opened
Not a deterministic algorithm, as I previously stated and which you
made no attempt to refute, and therefore irrelevant to church-turning
u haven't even read the algo, so how are you going to explain why
specifically it's non-deterministic ... ???
You seem to think that the human's side record is not part of the
input. -a-aIf so, that makes it non-deterministic.
the side-record is nothing more than the aggregated output of the agent decision algo run across all turing machines. any particular n-th record does not influence any future records beyond n. or before n for that matter
To fix it, the fixed steps the human agent runs need to be
converted to a function which read/writes a global variable, and >>>>>>>> that global variable need to be set to the same value it had
when und_cir_sr was first called.
lol, turing machines don't have "global variables" dud, they just >>>>>>> have a tape, and if the value is output to the tape anywhere, it >>>>>>> can be picked out by a simulation and contradicted by a paradox
Strawman.-a "it can be picked out" means you're talking about
changing the code which means you're no longer talking about the
same machine.
that's incorrect. reading values from the tape during a step of the >>>>> computation does not change the machine which is being simulated ...
No, but the code you've shown doesn't do that.-a So if you change the >>>> code to do that, it's no longer the same machine.
ofc the code does that
Your code is reading the part of the tape where the mechanized human
agent is writing its side record?
No?
Then your code doesn't do that, and talking about what it "can" do
necessarily means changing the code to something not being decided on.
u really don't get it: claiming a machine to be a total decider is
subject to the full enumerations of machines, meaning if an input can
exist, it will exist
there exists a paradox that contradicts any possible location you could write a decision bit on a tape, and therefore there will exist a machine that defies the decision, no matter where u try to write it
it is _not_ possible to simulate the side result created by the agent running a total decision algorithm.
it is _not_ even possible to create
a total decision algorithm with turing machines, like seriously have u forgotten the point of turing's proof???
... getting "output" from a simulation requires reading the output
from tape of the simulated machine, it can do that for any value,
even ones that are specified to be "output"
and if a machine is simulating itself,
Which your code isn't doing because the side data isn't the same
when starting the simulation as it was when the machine was invoked.
idk what u mean by "side data" there only data on the tape when it
comes to turing machines, there is nothing on the side of data on the
tape.
I'm referring specifically to what the mechanized version of the human
agent writes to the tape.
like in the case an undecidability paradox within computing, then
this can be used to pick out values behind any form of data
encapsulation that you might suggest to hide data away from a
paradox. there's no where to hide output dud. there's no "global
variable" that might prevent a paradox from being formed
"can be used" meaning it's not being used now, and therefore
irrelevant.
there is _no_ way to simulate a general ability to decide on
paradoxical turing machines, from within turing machines
So a paradoxical machine is one that a decider gets wrong?-a So this
machine:
void foo() { return; }
Is a paridoxical machine to this one:
int bar(void *p) { return 0; }
Because machine bar can't successfully report the status of machine
foo. -a-aYes?
no. first, we went over this dud: constant return function are not
and will never qualify as a halting decider as they do not output
useful information in regards to the halting status Efn+. the fact a
machine happens to output the correct answer does not qualify them as
a halting decider, a decider must output trustworthy information
which is judged across the entirety of the output space, not just
single instances
In other words, it's no different from *any* partial halt decider.
That's not even mentioning ill-defined weasel words like "useful",
"happens to", "trustworthy", and "judged".
dud if ur not here to effectively compute useful knowledge, idk what ur doing discussing the theory of computing.
let me not mince words. the existing theory on the matter of deciders is thus:
- a decider is not allowed a false positive or false negative, and never allowed to diverge. we cannot build any non-trivial decider (within
turing machines)
a constant return 1 function, if use as a halting decider is abound in
false positives and therefore is completely disqualified as being a
halting decider.
- a recognizer is also not allowed false positives or false negatives,
it is however allowed to diverge on some negatives, but never a
positive. a halting recognizer is allowed to diverge on some non-halting input, but never halting input. this is sometimes called a partial
decider, but i'm going to label as it recognizer to align the rather
well known sipser on this, and because i have further definitions to propose. we can build a halting recognizer, but not a non-halting recognizer.
a constant return 1 function does not satisfy this for halting either,
as again it's abound in false positives.
in my paper i propose two more types:
- a partial decider is still _not_ allowed false positive or negatives.
but it is allowed to diverge on some positives and some negatives. we
can be build withing turing machines both halting and non-halting
partial deciders
a constant return 1 function still does not satisfy this for halting either
- a partial recognizer is still not allowed false positives, but _does_ allow false negative ... but _only_ in the case where returning positive would be a false positive. this exception is allowed because a partial recognizer therefore does not need to diverge, ever, and always halts.
this can also be built within turing machines for the halting and non- halting problems.
a constant return 1 function again still does not satisfy these
requirements for halting
anyways, that was all covered in section -o5.2 which u never read
i really don't know why i'm humoring you dud
second, an actual paradox is necessarily undecidable due to the
construction of the machine itself, the form is generalized in -o3 of
my paper. if it doesn't fit that form, it's not a paradox, and is not
undecidable in respect to any classifier let alone in totality
Alright, let's satisfy your arbitrary requirement of a non-constant
function and "useful":
void foo() {
-a-a-a-a puts("hello");
}
int bar(void *p) {
-a-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a-a return 1;
-a-a-a-a } else {
-a-a-a-a-a-a-a-a return 0;
-a-a-a-a }
}
Given that the first instruction of "foo" is not a "return"
instruction, bar(foo) will return 0 even though foo() will halt.
Is the machine foo "undecidable" by machine bar?
On 9/24/2026 5:38 AM, dart200 wrote:
On 9/23/26 7:08 PM, dbush wrote:
On 9/23/2026 9:35 PM, dart200 wrote:
On 9/23/26 12:43 PM, dbush wrote:
On 9/23/2026 3:17 PM, dart200 wrote:
On 9/23/26 5:17 AM, dbush wrote:
So what you really did is break the rules of a deterministic
algorithm by using the side record and by not simulating what you >>>>>>> though you were simulating.
what rule??? lol deterministic only means running it always has the
same result given the same input ... which a total decision algo run
by the idealized agent certainly does
So what exactly are the inputs to the human agent?
the agent gets the machine description like any decision algo, that
would be clear if u just read the paper
So you don't consider the side record as part of the human's input. That makes it a non-deterministic algorithm, and therefore irrelevant to church-turning.
As I said, you have a fundamental misunderstanding of the problem, and
you just demonstrated it quite clearly.
u haven't even read the algo, so how are you going to explain why
No deterministic algorithm, no refutation of church-turing
the algorithm is detailed in -o7.6 ... which u still haven't even >>>>>> opened
Not a deterministic algorithm, as I previously stated and which you >>>>> made no attempt to refute, and therefore irrelevant to church-turning >>>>
specifically it's non-deterministic ... ???
You seem to think that the human's side record is not part of the
input. -a-aIf so, that makes it non-deterministic.
the side-record is nothing more than the aggregated output of the
agent decision algo run across all turing machines. any particular n-
th record does not influence any future records beyond n. or before n
for that matter
And if the human later reads it, that makes it an input.
So choose how you're wrong:
- The side record is not an input, so you have a non-determinstic
algorithm that is irrelevant to church-turning
- The side record is an input, and your abstraction is broken with your simulator not actually simulating itself.
To fix it, the fixed steps the human agent runs need to be
converted to a function which read/writes a global variable, >>>>>>>>> and that global variable need to be set to the same value it >>>>>>>>> had when und_cir_sr was first called.
lol, turing machines don't have "global variables" dud, they
just have a tape, and if the value is output to the tape
anywhere, it can be picked out by a simulation and contradicted >>>>>>>> by a paradox
Strawman.-a "it can be picked out" means you're talking about
changing the code which means you're no longer talking about the >>>>>>> same machine.
that's incorrect. reading values from the tape during a step of
the computation does not change the machine which is being
simulated ...
No, but the code you've shown doesn't do that.-a So if you change
the code to do that, it's no longer the same machine.
ofc the code does that
Your code is reading the part of the tape where the mechanized human
agent is writing its side record?
No?
Then your code doesn't do that, and talking about what it "can" do
necessarily means changing the code to something not being decided on.
u really don't get it: claiming a machine to be a total decider is
subject to the full enumerations of machines, meaning if an input can
exist, it will exist
That has nothing to do with the fact that you can't talk about changing
code and still claim it's the same machine.
there exists a paradox that contradicts any possible location you
could write a decision bit on a tape, and therefore there will exist a
machine that defies the decision, no matter where u try to write it
it is _not_ possible to simulate the side result created by the agent
running a total decision algorithm.
1) you can't read the side result because your abstraction is broken
2) the human agent isn't a halt decider, partial or otherwise, because
it takes an input that disqualifies it
it is _not_ even possible to create a total decision algorithm with
turing machines, like seriously have u forgotten the point of turing's
proof???
... getting "output" from a simulation requires reading the output
from tape of the simulated machine, it can do that for any value,
even ones that are specified to be "output"
and if a machine is simulating itself,
Which your code isn't doing because the side data isn't the same
when starting the simulation as it was when the machine was invoked.
idk what u mean by "side data" there only data on the tape when it
comes to turing machines, there is nothing on the side of data on
the tape.
I'm referring specifically to what the mechanized version of the
human agent writes to the tape.
like in the case an undecidability paradox within computing, then >>>>>> this can be used to pick out values behind any form of data
encapsulation that you might suggest to hide data away from a
paradox. there's no where to hide output dud. there's no "global
variable" that might prevent a paradox from being formed
"can be used" meaning it's not being used now, and therefore
irrelevant.
there is _no_ way to simulate a general ability to decide on
paradoxical turing machines, from within turing machines
So a paradoxical machine is one that a decider gets wrong?-a So this >>>>> machine:
void foo() { return; }
Is a paridoxical machine to this one:
int bar(void *p) { return 0; }
Because machine bar can't successfully report the status of machine >>>>> foo. -a-aYes?
no. first, we went over this dud: constant return function are not
and will never qualify as a halting decider as they do not output
useful information in regards to the halting status Efn+. the fact a
machine happens to output the correct answer does not qualify them
as a halting decider, a decider must output trustworthy information
which is judged across the entirety of the output space, not just
single instances
In other words, it's no different from *any* partial halt decider.
That's not even mentioning ill-defined weasel words like "useful",
"happens to", "trustworthy", and "judged".
dud if ur not here to effectively compute useful knowledge, idk what
ur doing discussing the theory of computing.
let me not mince words. the existing theory on the matter of deciders
is thus:
- a decider is not allowed a false positive or false negative, and
never allowed to diverge. we cannot build any non-trivial decider
(within turing machines)
a constant return 1 function, if use as a halting decider is abound in
false positives and therefore is completely disqualified as being a
halting decider.
- a recognizer is also not allowed false positives or false negatives,
it is however allowed to diverge on some negatives, but never a
positive. a halting recognizer is allowed to diverge on some non-
halting input, but never halting input. this is sometimes called a
partial decider, but i'm going to label as it recognizer to align the
rather well known sipser on this, and because i have further
definitions to propose. we can build a halting recognizer, but not a
non-halting recognizer.
a constant return 1 function does not satisfy this for halting either,
as again it's abound in false positives.
in my paper i propose two more types:
- a partial decider is still _not_ allowed false positive or
negatives. but it is allowed to diverge on some positives and some
negatives. we can be build withing turing machines both halting and
non-halting partial deciders
a constant return 1 function still does not satisfy this for halting
either
- a partial recognizer is still not allowed false positives, but
_does_ allow false negative ... but _only_ in the case where returning
positive would be a false positive. this exception is allowed because
a partial recognizer therefore does not need to diverge, ever, and
always halts. this can also be built within turing machines for the
halting and non- halting problems.
a constant return 1 function again still does not satisfy these
requirements for halting
anyways, that was all covered in section -o5.2 which u never read
i really don't know why i'm humoring you dud
second, an actual paradox is necessarily undecidable due to the
construction of the machine itself, the form is generalized in -o3 of >>>> my paper. if it doesn't fit that form, it's not a paradox, and is
not undecidable in respect to any classifier let alone in totality
Alright, let's satisfy your arbitrary requirement of a non-constant
function and "useful":
void foo() {
-a-a-a-a puts("hello");
}
int bar(void *p) {
-a-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a-a return 1;
-a-a-a-a } else {
-a-a-a-a-a-a-a-a return 0;
-a-a-a-a }
}
Given that the first instruction of "foo" is not a "return"
instruction, bar(foo) will return 0 even though foo() will halt.
Is the machine foo "undecidable" by machine bar?
No response to this?-a It goes directly to your definition of
"undecidable" which seems core to your argument.
Without an answer, this deems your use of the term unclear, and--
therefore makes your entire argument null and void.
On 9/24/26 5:37 AM, dbush wrote:
On 9/24/2026 5:38 AM, dart200 wrote:
On 9/23/26 7:08 PM, dbush wrote:
On 9/23/2026 9:35 PM, dart200 wrote:
On 9/23/26 12:43 PM, dbush wrote:
On 9/23/2026 3:17 PM, dart200 wrote:
On 9/23/26 5:17 AM, dbush wrote:
So what you really did is break the rules of a deterministic
algorithm by using the side record and by not simulating what >>>>>>>> you though you were simulating.
what rule??? lol deterministic only means running it always has the >>>>> same result given the same input ... which a total decision algo
run by the idealized agent certainly does
So what exactly are the inputs to the human agent?
the agent gets the machine description like any decision algo, that
would be clear if u just read the paper
So you don't consider the side record as part of the human's input.
That makes it a non-deterministic algorithm, and therefore irrelevant
to church-turning.
As I said, you have a fundamental misunderstanding of the problem, and
you just demonstrated it quite clearly.
that ur quite sure of, even if the reason keep changing
No deterministic algorithm, no refutation of church-turing
the algorithm is detailed in -o7.6 ... which u still haven't even >>>>>>> opened
Not a deterministic algorithm, as I previously stated and which
you made no attempt to refute, and therefore irrelevant to church- >>>>>> turning
u haven't even read the algo, so how are you going to explain why
specifically it's non-deterministic ... ???
You seem to think that the human's side record is not part of the
input. -a-aIf so, that makes it non-deterministic.
the side-record is nothing more than the aggregated output of the
agent decision algo run across all turing machines. any particular n-
th record does not influence any future records beyond n. or before n
for that matter
And if the human later reads it, that makes it an input.
i'm sorry, please explain to me why you think the agent needs to read
from the side record after writing to it???
So choose how you're wrong:
- The side record is not an input, so you have a non-determinstic
algorithm that is irrelevant to church-turning
- The side record is an input, and your abstraction is broken with
your simulator not actually simulating itself.
To fix it, the fixed steps the human agent runs need to be >>>>>>>>>> converted to a function which read/writes a global variable, >>>>>>>>>> and that global variable need to be set to the same value it >>>>>>>>>> had when und_cir_sr was first called.
lol, turing machines don't have "global variables" dud, they >>>>>>>>> just have a tape, and if the value is output to the tape
anywhere, it can be picked out by a simulation and contradicted >>>>>>>>> by a paradox
Strawman.-a "it can be picked out" means you're talking about >>>>>>>> changing the code which means you're no longer talking about the >>>>>>>> same machine.
that's incorrect. reading values from the tape during a step of >>>>>>> the computation does not change the machine which is being
simulated ...
No, but the code you've shown doesn't do that.-a So if you change >>>>>> the code to do that, it's no longer the same machine.
ofc the code does that
Your code is reading the part of the tape where the mechanized human
agent is writing its side record?
No?
Then your code doesn't do that, and talking about what it "can" do
necessarily means changing the code to something not being decided on.
u really don't get it: claiming a machine to be a total decider is
subject to the full enumerations of machines, meaning if an input can
exist, it will exist
That has nothing to do with the fact that you can't talk about
changing code and still claim it's the same machine.
no one is changing the goal post as the goal post of a decider is
handling _all_ possible input
there exists a paradox that contradicts any possible location you
could write a decision bit on a tape, and therefore there will exist
a machine that defies the decision, no matter where u try to write it
it is _not_ possible to simulate the side result created by the agent
running a total decision algorithm.
1) you can't read the side result because your abstraction is broken
2) the human agent isn't a halt decider, partial or otherwise, because
it takes an input that disqualifies it
it is _not_ even possible to create a total decision algorithm with
turing machines, like seriously have u forgotten the point of
turing's proof???
... getting "output" from a simulation requires reading the output
from tape of the simulated machine, it can do that for any value,
even ones that are specified to be "output"
idk what u mean by "side data" there only data on the tape when it
and if a machine is simulating itself,
Which your code isn't doing because the side data isn't the same
when starting the simulation as it was when the machine was invoked. >>>>>
comes to turing machines, there is nothing on the side of data on
the tape.
I'm referring specifically to what the mechanized version of the
human agent writes to the tape.
like in the case an undecidability paradox within computing, then >>>>>>> this can be used to pick out values behind any form of data
encapsulation that you might suggest to hide data away from a
paradox. there's no where to hide output dud. there's no "global >>>>>>> variable" that might prevent a paradox from being formed
"can be used" meaning it's not being used now, and therefore
irrelevant.
there is _no_ way to simulate a general ability to decide on
paradoxical turing machines, from within turing machines
So a paradoxical machine is one that a decider gets wrong?-a So
this machine:
void foo() { return; }
Is a paridoxical machine to this one:
int bar(void *p) { return 0; }
Because machine bar can't successfully report the status of
machine foo. -a-aYes?
no. first, we went over this dud: constant return function are not
and will never qualify as a halting decider as they do not output
useful information in regards to the halting status Efn+. the fact a >>>>> machine happens to output the correct answer does not qualify them
as a halting decider, a decider must output trustworthy information >>>>> which is judged across the entirety of the output space, not just
single instances
In other words, it's no different from *any* partial halt decider.
That's not even mentioning ill-defined weasel words like "useful",
"happens to", "trustworthy", and "judged".
dud if ur not here to effectively compute useful knowledge, idk what
ur doing discussing the theory of computing.
let me not mince words. the existing theory on the matter of deciders
is thus:
- a decider is not allowed a false positive or false negative, and
never allowed to diverge. we cannot build any non-trivial decider
(within turing machines)
a constant return 1 function, if use as a halting decider is abound
in false positives and therefore is completely disqualified as being
a halting decider.
- a recognizer is also not allowed false positives or false
negatives, it is however allowed to diverge on some negatives, but
never a positive. a halting recognizer is allowed to diverge on some
non- halting input, but never halting input. this is sometimes called
a partial decider, but i'm going to label as it recognizer to align
the rather well known sipser on this, and because i have further
definitions to propose. we can build a halting recognizer, but not a
non-halting recognizer.
a constant return 1 function does not satisfy this for halting
either, as again it's abound in false positives.
in my paper i propose two more types:
- a partial decider is still _not_ allowed false positive or
negatives. but it is allowed to diverge on some positives and some
negatives. we can be build withing turing machines both halting and
non-halting partial deciders
a constant return 1 function still does not satisfy this for halting
either
- a partial recognizer is still not allowed false positives, but
_does_ allow false negative ... but _only_ in the case where
returning positive would be a false positive. this exception is
allowed because a partial recognizer therefore does not need to
diverge, ever, and always halts. this can also be built within turing
machines for the halting and non- halting problems.
a constant return 1 function again still does not satisfy these
requirements for halting
anyways, that was all covered in section -o5.2 which u never read
i really don't know why i'm humoring you dud
Alright, let's satisfy your arbitrary requirement of a non-constant
second, an actual paradox is necessarily undecidable due to the
construction of the machine itself, the form is generalized in -o3
of my paper. if it doesn't fit that form, it's not a paradox, and
is not undecidable in respect to any classifier let alone in totality >>>>
function and "useful":
void foo() {
-a-a-a-a puts("hello");
}
int bar(void *p) {
-a-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a-a return 1;
-a-a-a-a } else {
-a-a-a-a-a-a-a-a return 0;
-a-a-a-a }
}
Given that the first instruction of "foo" is not a "return"
instruction, bar(foo) will return 0 even though foo() will halt.
Is the machine foo "undecidable" by machine bar?
No response to this?-a It goes directly to your definition of
"undecidable" which seems core to your argument.
bar isn't a halting decider in the first place, no idea what it's
supposed to be
Without an answer, this deems your use of the term unclear, and
therefore makes your entire argument null and void.
On 9/24/2026 2:03 PM, dart200 wrote:
On 9/24/26 5:37 AM, dbush wrote:
On 9/24/2026 5:38 AM, dart200 wrote:
On 9/23/26 7:08 PM, dbush wrote:
On 9/23/2026 9:35 PM, dart200 wrote:
On 9/23/26 12:43 PM, dbush wrote:
On 9/23/2026 3:17 PM, dart200 wrote:
On 9/23/26 5:17 AM, dbush wrote:
So what you really did is break the rules of a deterministic >>>>>>>>> algorithm by using the side record and by not simulating what >>>>>>>>> you though you were simulating.
what rule??? lol deterministic only means running it always has
the same result given the same input ... which a total decision
algo run by the idealized agent certainly does
So what exactly are the inputs to the human agent?
the agent gets the machine description like any decision algo, that
would be clear if u just read the paper
So you don't consider the side record as part of the human's input.
That makes it a non-deterministic algorithm, and therefore irrelevant
to church-turning.
As I said, you have a fundamental misunderstanding of the problem,
and you just demonstrated it quite clearly.
that ur quite sure of, even if the reason keep changing
I see you made no attempt to refute that you have a non-deterministic algorithm that is therefore not applicable to church-turning.-a Do you agree?
No deterministic algorithm, no refutation of church-turing
the algorithm is detailed in -o7.6 ... which u still haven't even >>>>>>>> opened
Not a deterministic algorithm, as I previously stated and which >>>>>>> you made no attempt to refute, and therefore irrelevant to
church- turning
u haven't even read the algo, so how are you going to explain why >>>>>> specifically it's non-deterministic ... ???
You seem to think that the human's side record is not part of the
input. -a-aIf so, that makes it non-deterministic.
the side-record is nothing more than the aggregated output of the
agent decision algo run across all turing machines. any particular
n- th record does not influence any future records beyond n. or
before n for that matter
And if the human later reads it, that makes it an input.
i'm sorry, please explain to me why you think the agent needs to read
from the side record after writing to it???
If the agent never reads the record, why does it need to write it?
So choose how you're wrong:
- The side record is not an input, so you have a non-determinstic
algorithm that is irrelevant to church-turning
- The side record is an input, and your abstraction is broken with
your simulator not actually simulating itself.
No response to this?
u really don't get it: claiming a machine to be a total decider is
To fix it, the fixed steps the human agent runs need to be >>>>>>>>>>> converted to a function which read/writes a global variable, >>>>>>>>>>> and that global variable need to be set to the same value it >>>>>>>>>>> had when und_cir_sr was first called.
lol, turing machines don't have "global variables" dud, they >>>>>>>>>> just have a tape, and if the value is output to the tape
anywhere, it can be picked out by a simulation and
contradicted by a paradox
Strawman.-a "it can be picked out" means you're talking about >>>>>>>>> changing the code which means you're no longer talking about >>>>>>>>> the same machine.
that's incorrect. reading values from the tape during a step of >>>>>>>> the computation does not change the machine which is being
simulated ...
No, but the code you've shown doesn't do that.-a So if you change >>>>>>> the code to do that, it's no longer the same machine.
ofc the code does that
Your code is reading the part of the tape where the mechanized
human agent is writing its side record?
No?
Then your code doesn't do that, and talking about what it "can" do
necessarily means changing the code to something not being decided on. >>>>
subject to the full enumerations of machines, meaning if an input
can exist, it will exist
That has nothing to do with the fact that you can't talk about
changing code and still claim it's the same machine.
no one is changing the goal post as the goal post of a decider is
handling _all_ possible input
*A* decider.-a There is no "can" or "can not", only "does" or "does not".
there exists a paradox that contradicts any possible location you
could write a decision bit on a tape, and therefore there will exist
a machine that defies the decision, no matter where u try to write it
it is _not_ possible to simulate the side result created by the
agent running a total decision algorithm.
1) you can't read the side result because your abstraction is broken
2) the human agent isn't a halt decider, partial or otherwise,
because it takes an input that disqualifies it
No response to this?
it is _not_ even possible to create a total decision algorithm with
turing machines, like seriously have u forgotten the point of
turing's proof???
... getting "output" from a simulation requires reading the output >>>>>> from tape of the simulated machine, it can do that for any value, >>>>>> even ones that are specified to be "output"
idk what u mean by "side data" there only data on the tape when it >>>>>> comes to turing machines, there is nothing on the side of data on >>>>>> the tape.
and if a machine is simulating itself,
Which your code isn't doing because the side data isn't the same >>>>>>> when starting the simulation as it was when the machine was invoked. >>>>>>
I'm referring specifically to what the mechanized version of the
human agent writes to the tape.
like in the case an undecidability paradox within computing,
then this can be used to pick out values behind any form of data >>>>>>>> encapsulation that you might suggest to hide data away from a >>>>>>>> paradox. there's no where to hide output dud. there's no "global >>>>>>>> variable" that might prevent a paradox from being formed
"can be used" meaning it's not being used now, and therefore
irrelevant.
there is _no_ way to simulate a general ability to decide on
paradoxical turing machines, from within turing machines
So a paradoxical machine is one that a decider gets wrong?-a So >>>>>>> this machine:
void foo() { return; }
Is a paridoxical machine to this one:
int bar(void *p) { return 0; }
Because machine bar can't successfully report the status of
machine foo. -a-aYes?
no. first, we went over this dud: constant return function are not >>>>>> and will never qualify as a halting decider as they do not output >>>>>> useful information in regards to the halting status Efn+. the fact a >>>>>> machine happens to output the correct answer does not qualify them >>>>>> as a halting decider, a decider must output trustworthy
information which is judged across the entirety of the output
space, not just single instances
In other words, it's no different from *any* partial halt decider.
That's not even mentioning ill-defined weasel words like "useful",
"happens to", "trustworthy", and "judged".
dud if ur not here to effectively compute useful knowledge, idk what
ur doing discussing the theory of computing.
let me not mince words. the existing theory on the matter of
deciders is thus:
- a decider is not allowed a false positive or false negative, and
never allowed to diverge. we cannot build any non-trivial decider
(within turing machines)
a constant return 1 function, if use as a halting decider is abound
in false positives and therefore is completely disqualified as being
a halting decider.
- a recognizer is also not allowed false positives or false
negatives, it is however allowed to diverge on some negatives, but
never a positive. a halting recognizer is allowed to diverge on some
non- halting input, but never halting input. this is sometimes
called a partial decider, but i'm going to label as it recognizer to
align the rather well known sipser on this, and because i have
further definitions to propose. we can build a halting recognizer,
but not a non-halting recognizer.
a constant return 1 function does not satisfy this for halting
either, as again it's abound in false positives.
in my paper i propose two more types:
- a partial decider is still _not_ allowed false positive or
negatives. but it is allowed to diverge on some positives and some
negatives. we can be build withing turing machines both halting and
non-halting partial deciders
a constant return 1 function still does not satisfy this for halting
either
- a partial recognizer is still not allowed false positives, but
_does_ allow false negative ... but _only_ in the case where
returning positive would be a false positive. this exception is
allowed because a partial recognizer therefore does not need to
diverge, ever, and always halts. this can also be built within
turing machines for the halting and non- halting problems.
a constant return 1 function again still does not satisfy these
requirements for halting
anyways, that was all covered in section -o5.2 which u never read
i really don't know why i'm humoring you dud
Alright, let's satisfy your arbitrary requirement of a non-constant >>>>> function and "useful":
second, an actual paradox is necessarily undecidable due to the
construction of the machine itself, the form is generalized in -o3 >>>>>> of my paper. if it doesn't fit that form, it's not a paradox, and >>>>>> is not undecidable in respect to any classifier let alone in totality >>>>>
void foo() {
-a-a-a-a puts("hello");
}
int bar(void *p) {
-a-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a-a return 1;
-a-a-a-a } else {
-a-a-a-a-a-a-a-a return 0;
-a-a-a-a }
}
Given that the first instruction of "foo" is not a "return"
instruction, bar(foo) will return 0 even though foo() will halt.
Is the machine foo "undecidable" by machine bar?
No response to this?-a It goes directly to your definition of
"undecidable" which seems core to your argument.
bar isn't a halting decider in the first place, no idea what it's
supposed to be
As noted above, it's a partial halt decider that reports halting if the first instruction is a "return" (0xc3 is the x86 instruction "ret") and reports non-halting otherwise.
Again, is the machine foo "undecidable" by machine bar?
It would be a shame to discard your whole argument just because your
terms aren't well defined.
Without an answer, this deems your use of the term unclear, and
therefore makes your entire argument null and void.
On 9/24/26 11:23 AM, dbush wrote:
On 9/24/2026 2:03 PM, dart200 wrote:
On 9/24/26 5:37 AM, dbush wrote:
On 9/24/2026 5:38 AM, dart200 wrote:
On 9/23/26 7:08 PM, dbush wrote:
On 9/23/2026 9:35 PM, dart200 wrote:
On 9/23/26 12:43 PM, dbush wrote:
On 9/23/2026 3:17 PM, dart200 wrote:
On 9/23/26 5:17 AM, dbush wrote:
So what you really did is break the rules of a deterministic >>>>>>>>>> algorithm by using the side record and by not simulating what >>>>>>>>>> you though you were simulating.
what rule??? lol deterministic only means running it always has >>>>>>> the same result given the same input ... which a total decision >>>>>>> algo run by the idealized agent certainly does
So what exactly are the inputs to the human agent?
the agent gets the machine description like any decision algo, that >>>>> would be clear if u just read the paper
So you don't consider the side record as part of the human's input.
That makes it a non-deterministic algorithm, and therefore
irrelevant to church-turning.
As I said, you have a fundamental misunderstanding of the problem,
and you just demonstrated it quite clearly.
that ur quite sure of, even if the reason keep changing
I see you made no attempt to refute that you have a non-deterministic
algorithm that is therefore not applicable to church-turning.-a Do you
agree?
No deterministic algorithm, no refutation of church-turing
the algorithm is detailed in -o7.6 ... which u still haven't >>>>>>>>> even opened
Not a deterministic algorithm, as I previously stated and which >>>>>>>> you made no attempt to refute, and therefore irrelevant to
church- turning
u haven't even read the algo, so how are you going to explain why >>>>>>> specifically it's non-deterministic ... ???
You seem to think that the human's side record is not part of the >>>>>> input. -a-aIf so, that makes it non-deterministic.
the side-record is nothing more than the aggregated output of the
agent decision algo run across all turing machines. any particular
n- th record does not influence any future records beyond n. or
before n for that matter
And if the human later reads it, that makes it an input.
i'm sorry, please explain to me why you think the agent needs to read
from the side record after writing to it???
If the agent never reads the record, why does it need to write it?
to clearly demonstrate that the sequence is produced, to demonstrate
that the ability to objectively decide on any given turing machine is
not confounded by the fact no turing machine itself can be a total
machine decider
So choose how you're wrong:
- The side record is not an input, so you have a non-determinstic
algorithm that is irrelevant to church-turning
- The side record is an input, and your abstraction is broken with
your simulator not actually simulating itself.
No response to this?
nothing written to the side record is used further in the computation,
so the agent never needs to read from it again
To fix it, the fixed steps the human agent runs need to be >>>>>>>>>>>> converted to a function which read/writes a global variable, >>>>>>>>>>>> and that global variable need to be set to the same value it >>>>>>>>>>>> had when und_cir_sr was first called.
lol, turing machines don't have "global variables" dud, they >>>>>>>>>>> just have a tape, and if the value is output to the tape >>>>>>>>>>> anywhere, it can be picked out by a simulation and
contradicted by a paradox
Strawman.-a "it can be picked out" means you're talking about >>>>>>>>>> changing the code which means you're no longer talking about >>>>>>>>>> the same machine.
that's incorrect. reading values from the tape during a step of >>>>>>>>> the computation does not change the machine which is being
simulated ...
No, but the code you've shown doesn't do that.-a So if you change >>>>>>>> the code to do that, it's no longer the same machine.
ofc the code does that
Your code is reading the part of the tape where the mechanized
human agent is writing its side record?
No?
Then your code doesn't do that, and talking about what it "can" do >>>>>> necessarily means changing the code to something not being decided >>>>>> on.
u really don't get it: claiming a machine to be a total decider is
subject to the full enumerations of machines, meaning if an input
can exist, it will exist
That has nothing to do with the fact that you can't talk about
changing code and still claim it's the same machine.
no one is changing the goal post as the goal post of a decider is
handling _all_ possible input
*A* decider.-a There is no "can" or "can not", only "does" or "does not".
there exists a machine that does a paradox for any location you can
write a decision to the tape
there exists a paradox that contradicts any possible location you
could write a decision bit on a tape, and therefore there will
exist a machine that defies the decision, no matter where u try to
write it
it is _not_ possible to simulate the side result created by the
agent running a total decision algorithm.
1) you can't read the side result because your abstraction is broken
2) the human agent isn't a halt decider, partial or otherwise,
because it takes an input that disqualifies it
No response to this?
neither of those are correct: false dichotomy
it is _not_ even possible to create a total decision algorithm with >>>>> turing machines, like seriously have u forgotten the point of
turing's proof???
... getting "output" from a simulation requires reading the
output from tape of the simulated machine, it can do that for any >>>>>>> value, even ones that are specified to be "output"
and if a machine is simulating itself,
Which your code isn't doing because the side data isn't the same >>>>>>>> when starting the simulation as it was when the machine was
invoked.
idk what u mean by "side data" there only data on the tape when >>>>>>> it comes to turing machines, there is nothing on the side of data >>>>>>> on the tape.
I'm referring specifically to what the mechanized version of the
human agent writes to the tape.
like in the case an undecidability paradox within computing, >>>>>>>>> then this can be used to pick out values behind any form of >>>>>>>>> data encapsulation that you might suggest to hide data away >>>>>>>>> from a paradox. there's no where to hide output dud. there's no >>>>>>>>> "global variable" that might prevent a paradox from being formed >>>>>>>>"can be used" meaning it's not being used now, and therefore
irrelevant.
there is _no_ way to simulate a general ability to decide on >>>>>>>>> paradoxical turing machines, from within turing machines
So a paradoxical machine is one that a decider gets wrong?-a So >>>>>>>> this machine:
void foo() { return; }
Is a paridoxical machine to this one:
int bar(void *p) { return 0; }
Because machine bar can't successfully report the status of
machine foo. -a-aYes?
no. first, we went over this dud: constant return function are
not and will never qualify as a halting decider as they do not
output useful information in regards to the halting status Efn+. >>>>>>> the fact a machine happens to output the correct answer does not >>>>>>> qualify them as a halting decider, a decider must output
trustworthy information which is judged across the entirety of
the output space, not just single instances
In other words, it's no different from *any* partial halt decider. >>>>>> That's not even mentioning ill-defined weasel words like "useful", >>>>>> "happens to", "trustworthy", and "judged".
dud if ur not here to effectively compute useful knowledge, idk
what ur doing discussing the theory of computing.
let me not mince words. the existing theory on the matter of
deciders is thus:
- a decider is not allowed a false positive or false negative, and
never allowed to diverge. we cannot build any non-trivial decider
(within turing machines)
a constant return 1 function, if use as a halting decider is abound >>>>> in false positives and therefore is completely disqualified as
being a halting decider.
- a recognizer is also not allowed false positives or false
negatives, it is however allowed to diverge on some negatives, but
never a positive. a halting recognizer is allowed to diverge on
some non- halting input, but never halting input. this is sometimes >>>>> called a partial decider, but i'm going to label as it recognizer
to align the rather well known sipser on this, and because i have
further definitions to propose. we can build a halting recognizer,
but not a non-halting recognizer.
a constant return 1 function does not satisfy this for halting
either, as again it's abound in false positives.
in my paper i propose two more types:
- a partial decider is still _not_ allowed false positive or
negatives. but it is allowed to diverge on some positives and some
negatives. we can be build withing turing machines both halting and >>>>> non-halting partial deciders
a constant return 1 function still does not satisfy this for
halting either
- a partial recognizer is still not allowed false positives, but
_does_ allow false negative ... but _only_ in the case where
returning positive would be a false positive. this exception is
allowed because a partial recognizer therefore does not need to
diverge, ever, and always halts. this can also be built within
turing machines for the halting and non- halting problems.
a constant return 1 function again still does not satisfy these
requirements for halting
i just defined my terms here quite clearly. i suppose i'll thank you for encouraging me to develop that, i'll may add it to my paper
anyways, that was all covered in section -o5.2 which u never read
i really don't know why i'm humoring you dud
second, an actual paradox is necessarily undecidable due to the >>>>>>> construction of the machine itself, the form is generalized in -o3 >>>>>>> of my paper. if it doesn't fit that form, it's not a paradox, and >>>>>>> is not undecidable in respect to any classifier let alone in
totality
Alright, let's satisfy your arbitrary requirement of a non-
constant function and "useful":
void foo() {
-a-a-a-a puts("hello");
}
int bar(void *p) {
-a-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a-a return 1;
-a-a-a-a } else {
-a-a-a-a-a-a-a-a return 0;
-a-a-a-a }
}
Given that the first instruction of "foo" is not a "return"
instruction, bar(foo) will return 0 even though foo() will halt.
Is the machine foo "undecidable" by machine bar?
No response to this?-a It goes directly to your definition of
"undecidable" which seems core to your argument.
bar isn't a halting decider in the first place, no idea what it's
supposed to be
As noted above, it's a partial halt decider that reports halting if
the first instruction is a "return" (0xc3 is the x86 instruction
"ret") and reports non-halting otherwise.
Again, is the machine foo "undecidable" by machine bar?
It would be a shame to discard your whole argument just because your
terms aren't well defined.
a partial halting decider is not allowed _any_ false positives or false negatives, it must handle the entire input space of machines without a _single_ false positive or false negative. that's not even my assertion, that's the established consensus on the matter
clearly bar would have a ton of false negatives and thereby be
disqualified as a partial halting decider
Without an answer, this deems your use of the term unclear, and
therefore makes your entire argument null and void.
On 9/20/26 7:51 PM, Chris M. Thomasson wrote:
On 9/20/2026 7:11 PM, dart200 wrote:
On 9/20/26 2:53 PM, Chris M. Thomasson wrote:
On 9/19/2026 2:38 PM, dart200 wrote:
[...]
If not, wtf! You remind me of PO. Sorry for that cut.
i don't care for ur fallacy by association brainrot
Sigh. I only said you kind of do remind me of the way PO dealt with
the halting program.
and that statement has no bearing on the correctness of my arguments,
u brain-rotted boomer
You cannot predict a random number and you cannot solve the halting
problem. Sigh.
turing machines do not involve random numbers dud. if u knew literally anything about basic computing theory u'd know that.
but u don't, so go back to ur fractals, eh?
On 9/20/2026 10:39 PM, dart200 wrote:
On 9/20/26 7:51 PM, Chris M. Thomasson wrote:
On 9/20/2026 7:11 PM, dart200 wrote:
On 9/20/26 2:53 PM, Chris M. Thomasson wrote:
On 9/19/2026 2:38 PM, dart200 wrote:
[...]
If not, wtf! You remind me of PO. Sorry for that cut.
i don't care for ur fallacy by association brainrot
Sigh. I only said you kind of do remind me of the way PO dealt with >>>>> the halting program.
and that statement has no bearing on the correctness of my
arguments, u brain-rotted boomer
You cannot predict a random number and you cannot solve the halting
problem. Sigh.
turing machines do not involve random numbers dud. if u knew literally
anything about basic computing theory u'd know that.
but u don't, so go back to ur fractals, eh?
Well, ponder on CSPRNG?
On 9/24/2026 11:03 AM, dart200 wrote:
[...]
You really are PO! ;^o Woha!
On 9/24/2026 4:09 PM, dart200 wrote:
On 9/24/26 11:23 AM, dbush wrote:
On 9/24/2026 2:03 PM, dart200 wrote:
On 9/24/26 5:37 AM, dbush wrote:
On 9/24/2026 5:38 AM, dart200 wrote:
On 9/23/26 7:08 PM, dbush wrote:
On 9/23/2026 9:35 PM, dart200 wrote:
On 9/23/26 12:43 PM, dbush wrote:
On 9/23/2026 3:17 PM, dart200 wrote:
On 9/23/26 5:17 AM, dbush wrote:
So what you really did is break the rules of a deterministic >>>>>>>>>>> algorithm by using the side record and by not simulating what >>>>>>>>>>> you though you were simulating.
what rule??? lol deterministic only means running it always has >>>>>>>> the same result given the same input ... which a total decision >>>>>>>> algo run by the idealized agent certainly does
So what exactly are the inputs to the human agent?
the agent gets the machine description like any decision algo,
that would be clear if u just read the paper
So you don't consider the side record as part of the human's input. >>>>> That makes it a non-deterministic algorithm, and therefore
irrelevant to church-turning.
As I said, you have a fundamental misunderstanding of the problem,
and you just demonstrated it quite clearly.
that ur quite sure of, even if the reason keep changing
I see you made no attempt to refute that you have a non-deterministic
algorithm that is therefore not applicable to church-turning.-a Do you
agree?
Still no response?
the algorithm is detailed in -o7.6 ... which u still haven't >>>>>>>>>> even opened
No deterministic algorithm, no refutation of church-turing >>>>>>>>>>
Not a deterministic algorithm, as I previously stated and which >>>>>>>>> you made no attempt to refute, and therefore irrelevant to
church- turning
u haven't even read the algo, so how are you going to explain >>>>>>>> why specifically it's non-deterministic ... ???
You seem to think that the human's side record is not part of the >>>>>>> input. -a-aIf so, that makes it non-deterministic.
the side-record is nothing more than the aggregated output of the >>>>>> agent decision algo run across all turing machines. any particular >>>>>> n- th record does not influence any future records beyond n. or
before n for that matter
And if the human later reads it, that makes it an input.
i'm sorry, please explain to me why you think the agent needs to
read from the side record after writing to it???
If the agent never reads the record, why does it need to write it?
to clearly demonstrate that the sequence is produced, to demonstrate
that the ability to objectively decide on any given turing machine is
not confounded by the fact no turing machine itself can be a total
machine decider
So it's just a log?
Meaning that if the side record doesn't exist the
agent will behave exactly the same?
So choose how you're wrong:
- The side record is not an input, so you have a non-determinstic
algorithm that is irrelevant to church-turning
- The side record is an input, and your abstraction is broken with
your simulator not actually simulating itself.
No response to this?
nothing written to the side record is used further in the computation,
so the agent never needs to read from it again
"Again".-a So it does read from the side record.
That makes it an input.
Which means that when und_cir_sr calls sim_cir_sr(und_cir_sr) after
calling Tdp, it's not simulating itself.
That also means the agent is disqualified from being a halt decider/ recognizer, partial or otherwise.
So your abstraction is broken.
there exists a machine that does a paradox for any location you can
To fix it, the fixed steps the human agent runs need to be >>>>>>>>>>>>> converted to a function which read/writes a global
variable, and that global variable need to be set to the >>>>>>>>>>>>> same value it had when und_cir_sr was first called.
lol, turing machines don't have "global variables" dud, they >>>>>>>>>>>> just have a tape, and if the value is output to the tape >>>>>>>>>>>> anywhere, it can be picked out by a simulation and
contradicted by a paradox
Strawman.-a "it can be picked out" means you're talking about >>>>>>>>>>> changing the code which means you're no longer talking about >>>>>>>>>>> the same machine.
that's incorrect. reading values from the tape during a step >>>>>>>>>> of the computation does not change the machine which is being >>>>>>>>>> simulated ...
No, but the code you've shown doesn't do that.-a So if you
change the code to do that, it's no longer the same machine.
ofc the code does that
Your code is reading the part of the tape where the mechanized
human agent is writing its side record?
No?
Then your code doesn't do that, and talking about what it "can" >>>>>>> do necessarily means changing the code to something not being
decided on.
u really don't get it: claiming a machine to be a total decider is >>>>>> subject to the full enumerations of machines, meaning if an input >>>>>> can exist, it will exist
That has nothing to do with the fact that you can't talk about
changing code and still claim it's the same machine.
no one is changing the goal post as the goal post of a decider is
handling _all_ possible input
*A* decider.-a There is no "can" or "can not", only "does" or "does not". >>
write a decision to the tape
You "can" write means changing the code of the decider, which is not allowed.-a Otherwise you're not talking about the same decider.
Turing machines are defined by their instructions, not where they1) the agent does not read from the side record
physically reside.
there exists a paradox that contradicts any possible location you >>>>>> could write a decision bit on a tape, and therefore there will
exist a machine that defies the decision, no matter where u try to >>>>>> write it
it is _not_ possible to simulate the side result created by the
agent running a total decision algorithm.
1) you can't read the side result because your abstraction is broken >>>>> 2) the human agent isn't a halt decider, partial or otherwise,
because it takes an input that disqualifies it
No response to this?
neither of those are correct: false dichotomy
That wasn't an either/or.-a Both apply.
it is _not_ even possible to create a total decision algorithm
with turing machines, like seriously have u forgotten the point of >>>>>> turing's proof???
... getting "output" from a simulation requires reading the
output from tape of the simulated machine, it can do that for >>>>>>>> any value, even ones that are specified to be "output"
and if a machine is simulating itself,
Which your code isn't doing because the side data isn't the >>>>>>>>> same when starting the simulation as it was when the machine >>>>>>>>> was invoked.
idk what u mean by "side data" there only data on the tape when >>>>>>>> it comes to turing machines, there is nothing on the side of
data on the tape.
I'm referring specifically to what the mechanized version of the >>>>>>> human agent writes to the tape.
like in the case an undecidability paradox within computing, >>>>>>>>>> then this can be used to pick out values behind any form of >>>>>>>>>> data encapsulation that you might suggest to hide data away >>>>>>>>>> from a paradox. there's no where to hide output dud. there's >>>>>>>>>> no "global variable" that might prevent a paradox from being >>>>>>>>>> formed
"can be used" meaning it's not being used now, and therefore >>>>>>>>> irrelevant.
there is _no_ way to simulate a general ability to decide on >>>>>>>>>> paradoxical turing machines, from within turing machines
So a paradoxical machine is one that a decider gets wrong?-a So >>>>>>>>> this machine:
void foo() { return; }
Is a paridoxical machine to this one:
int bar(void *p) { return 0; }
Because machine bar can't successfully report the status of >>>>>>>>> machine foo. -a-aYes?
no. first, we went over this dud: constant return function are >>>>>>>> not and will never qualify as a halting decider as they do not >>>>>>>> output useful information in regards to the halting status Efn+. >>>>>>>> the fact a machine happens to output the correct answer does not >>>>>>>> qualify them as a halting decider, a decider must output
trustworthy information which is judged across the entirety of >>>>>>>> the output space, not just single instances
In other words, it's no different from *any* partial halt
decider. That's not even mentioning ill-defined weasel words like >>>>>>> "useful", "happens to", "trustworthy", and "judged".
dud if ur not here to effectively compute useful knowledge, idk
what ur doing discussing the theory of computing.
let me not mince words. the existing theory on the matter of
deciders is thus:
- a decider is not allowed a false positive or false negative, and >>>>>> never allowed to diverge. we cannot build any non-trivial decider >>>>>> (within turing machines)
a constant return 1 function, if use as a halting decider is
abound in false positives and therefore is completely disqualified >>>>>> as being a halting decider.
- a recognizer is also not allowed false positives or false
negatives, it is however allowed to diverge on some negatives, but >>>>>> never a positive. a halting recognizer is allowed to diverge on
some non- halting input, but never halting input. this is
sometimes called a partial decider, but i'm going to label as it
recognizer to align the rather well known sipser on this, and
because i have further definitions to propose. we can build a
halting recognizer, but not a non-halting recognizer.
a constant return 1 function does not satisfy this for halting
either, as again it's abound in false positives.
in my paper i propose two more types:
- a partial decider is still _not_ allowed false positive or
negatives. but it is allowed to diverge on some positives and some >>>>>> negatives. we can be build withing turing machines both halting
and non-halting partial deciders
a constant return 1 function still does not satisfy this for
halting either
- a partial recognizer is still not allowed false positives, but
_does_ allow false negative ... but _only_ in the case where
returning positive would be a false positive. this exception is
allowed because a partial recognizer therefore does not need to
diverge, ever, and always halts. this can also be built within
turing machines for the halting and non- halting problems.
a constant return 1 function again still does not satisfy these
requirements for halting
i just defined my terms here quite clearly. i suppose i'll thank you
for encouraging me to develop that, i'll may add it to my paper
anyways, that was all covered in section -o5.2 which u never read
i really don't know why i'm humoring you dud
second, an actual paradox is necessarily undecidable due to the >>>>>>>> construction of the machine itself, the form is generalized in >>>>>>>> -o3 of my paper. if it doesn't fit that form, it's not a paradox, >>>>>>>> and is not undecidable in respect to any classifier let alone in >>>>>>>> totality
Alright, let's satisfy your arbitrary requirement of a non-
constant function and "useful":
void foo() {
-a-a-a-a puts("hello");
}
int bar(void *p) {
-a-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a-a return 1;
-a-a-a-a } else {
-a-a-a-a-a-a-a-a return 0;
-a-a-a-a }
}
Given that the first instruction of "foo" is not a "return"
instruction, bar(foo) will return 0 even though foo() will halt. >>>>>>>
Is the machine foo "undecidable" by machine bar?
No response to this?-a It goes directly to your definition of
"undecidable" which seems core to your argument.
bar isn't a halting decider in the first place, no idea what it's
supposed to be
As noted above, it's a partial halt decider that reports halting if
the first instruction is a "return" (0xc3 is the x86 instruction
"ret") and reports non-halting otherwise.
Again, is the machine foo "undecidable" by machine bar?
It would be a shame to discard your whole argument just because your
terms aren't well defined.
a partial halting decider is not allowed _any_ false positives or
false negatives, it must handle the entire input space of machines
without a _single_ false positive or false negative. that's not even
my assertion, that's the established consensus on the matter
clearly bar would have a ton of false negatives and thereby be
disqualified as a partial halting decider
Alright, let's satisfy one more arbitrary requirement to see if we can
get you to define your terms:
void foo() {
-a-a-a puts("hello");
}
int bar(void *p) {
-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a return 1;
-a-a-a } else {
-a-a-a-a-a-a-a while (1);
-a-a-a }
}
That takes care of the false negatives.-a So no more dodging the point:
Is the machine foo "undecidable" by machine bar?
Failure to answer again will deem your term "undecidable" ill-defined,
and consequently the rest of your paper.
Without an answer, this deems your use of the term unclear, and
therefore makes your entire argument null and void.
a partial halting decider is not allowed _any_ false positives or false negatives, it must handle the entire input space of machines without a _single_ false positive or false negative.
that's not even my assertion, that's the established consensus on the matter
clearly bar would have a ton of false negatives and thereby be
disqualified as a partial halting decider
On 9/24/26 3:10 PM, Chris M. Thomasson wrote:
On 9/20/2026 10:39 PM, dart200 wrote:
On 9/20/26 7:51 PM, Chris M. Thomasson wrote:
On 9/20/2026 7:11 PM, dart200 wrote:
On 9/20/26 2:53 PM, Chris M. Thomasson wrote:
On 9/19/2026 2:38 PM, dart200 wrote:
[...]
If not, wtf! You remind me of PO. Sorry for that cut.
i don't care for ur fallacy by association brainrot
Sigh. I only said you kind of do remind me of the way PO dealt
with the halting program.
and that statement has no bearing on the correctness of my
arguments, u brain-rotted boomer
You cannot predict a random number and you cannot solve the halting
problem. Sigh.
turing machines do not involve random numbers dud. if u knew
literally anything about basic computing theory u'd know that.
but u don't, so go back to ur fractals, eh?
Well, ponder on CSPRNG?
pseudo-random number generators are fully deterministic and therefore entirely decidable in terms of whether it the machine which produces it
is circle-free or not (they are obviously). for any decision ofc one
needs a description of the machine including any input (like a seed),
and that combination can be used to decide on the semantics of the
described computation
when u say "random" that does not describe what can be produced by a
turing machine, as they can only compute pseudo-random sequences that
are inherently predictable, not truly random ones. if u meant pseudo-
random you should have said that...
seriously, bro: back to ur fractals, eh?
On 2026-09-24 14:09, dart200 wrote:
a partial halting decider is not allowed _any_ false positives or false negatives, it must handle
the entire input space of machines without a _single_ false positive or false negative.
You're very confused here. If it is not allowed any false negatives or false positives, then you're
talking about a /total/ halt decider. A /partial/ halt decider is something which correctly decides
/some/ instances of the halting problem but which fails to correctly decide /all/ instances. That's
why that pesky word 'partial' is added.
On 9/24/26 1:54 PM, dbush wrote:
On 9/24/2026 4:09 PM, dart200 wrote:
On 9/24/26 11:23 AM, dbush wrote:
On 9/24/2026 2:03 PM, dart200 wrote:
On 9/24/26 5:37 AM, dbush wrote:
On 9/24/2026 5:38 AM, dart200 wrote:
On 9/23/26 7:08 PM, dbush wrote:
On 9/23/2026 9:35 PM, dart200 wrote:
On 9/23/26 12:43 PM, dbush wrote:
On 9/23/2026 3:17 PM, dart200 wrote:
On 9/23/26 5:17 AM, dbush wrote:
So what you really did is break the rules of a deterministic >>>>>>>>>>>> algorithm by using the side record and by not simulating >>>>>>>>>>>> what you though you were simulating.
what rule??? lol deterministic only means running it always has >>>>>>>>> the same result given the same input ... which a total decision >>>>>>>>> algo run by the idealized agent certainly does
So what exactly are the inputs to the human agent?
the agent gets the machine description like any decision algo,
that would be clear if u just read the paper
So you don't consider the side record as part of the human's
input. That makes it a non-deterministic algorithm, and therefore >>>>>> irrelevant to church-turning.
As I said, you have a fundamental misunderstanding of the problem, >>>>>> and you just demonstrated it quite clearly.
that ur quite sure of, even if the reason keep changing
I see you made no attempt to refute that you have a non-
deterministic algorithm that is therefore not applicable to church-
turning.-a Do you agree?
Still no response?
the algorithm is detailed in -o7.6 ... which u still haven't >>>>>>>>>>> even opened
No deterministic algorithm, no refutation of church-turing >>>>>>>>>>>
Not a deterministic algorithm, as I previously stated and >>>>>>>>>> which you made no attempt to refute, and therefore irrelevant >>>>>>>>>> to church- turning
u haven't even read the algo, so how are you going to explain >>>>>>>>> why specifically it's non-deterministic ... ???
You seem to think that the human's side record is not part of >>>>>>>> the input. -a-aIf so, that makes it non-deterministic.
the side-record is nothing more than the aggregated output of the >>>>>>> agent decision algo run across all turing machines. any
particular n- th record does not influence any future records
beyond n. or before n for that matter
And if the human later reads it, that makes it an input.
i'm sorry, please explain to me why you think the agent needs to
read from the side record after writing to it???
If the agent never reads the record, why does it need to write it?
to clearly demonstrate that the sequence is produced, to demonstrate
that the ability to objectively decide on any given turing machine is
not confounded by the fact no turing machine itself can be a total
machine decider
So it's just a log?
it is a computed sequence,
Meaning that if the side record doesn't exist the agent will behave
exactly the same?
yes beside the process of writing computed results to the side record
So choose how you're wrong:
- The side record is not an input, so you have a non-determinstic >>>>>> algorithm that is irrelevant to church-turning
- The side record is an input, and your abstraction is broken with >>>>>> your simulator not actually simulating itself.
No response to this?
nothing written to the side record is used further in the
computation, so the agent never needs to read from it again
"Again".-a So it does read from the side record.
word-focus fallacy
now we're at 4 named fallacies, 1 named cognitive distortion, and 1
named abuse tactic
talk about brain rot
That makes it an input.
Which means that when und_cir_sr calls sim_cir_sr(und_cir_sr) after
calling Tdp, it's not simulating itself.
That also means the agent is disqualified from being a halt decider/
recognizer, partial or otherwise.
So your abstraction is broken.
ofc the code does that
lol, turing machines don't have "global variables" dud, >>>>>>>>>>>>> they just have a tape, and if the value is output to the >>>>>>>>>>>>> tape anywhere, it can be picked out by a simulation and >>>>>>>>>>>>> contradicted by a paradox
To fix it, the fixed steps the human agent runs need to be >>>>>>>>>>>>>> converted to a function which read/writes a global >>>>>>>>>>>>>> variable, and that global variable need to be set to the >>>>>>>>>>>>>> same value it had when und_cir_sr was first called. >>>>>>>>>>>>>
Strawman.-a "it can be picked out" means you're talking about >>>>>>>>>>>> changing the code which means you're no longer talking about >>>>>>>>>>>> the same machine.
that's incorrect. reading values from the tape during a step >>>>>>>>>>> of the computation does not change the machine which is being >>>>>>>>>>> simulated ...
No, but the code you've shown doesn't do that.-a So if you >>>>>>>>>> change the code to do that, it's no longer the same machine. >>>>>>>>>
Your code is reading the part of the tape where the mechanized >>>>>>>> human agent is writing its side record?
No?
Then your code doesn't do that, and talking about what it "can" >>>>>>>> do necessarily means changing the code to something not being >>>>>>>> decided on.
u really don't get it: claiming a machine to be a total decider >>>>>>> is subject to the full enumerations of machines, meaning if an
input can exist, it will exist
That has nothing to do with the fact that you can't talk about
changing code and still claim it's the same machine.
no one is changing the goal post as the goal post of a decider is
handling _all_ possible input
*A* decider.-a There is no "can" or "can not", only "does" or "does
not".
there exists a machine that does a paradox for any location you can
write a decision to the tape
You "can" write means changing the code of the decider, which is not
allowed.-a Otherwise you're not talking about the same decider.
for any location on the tape that a decider does attempt to write a decision, of which there are infinite locations on the tape, and
infinite deciders writing to any location on the tape,
there does exist infinite paradoxical machines which will read from that specific bit and produce a paradox for it in regards to the written
decision
there is _no_ way to produce a total halting decider for turing
machines, within turing machines
1) the agent does not read from the side record
Turing machines are defined by their instructions, not where they
physically reside.
there exists a paradox that contradicts any possible location you >>>>>>> could write a decision bit on a tape, and therefore there will
exist a machine that defies the decision, no matter where u try >>>>>>> to write it
it is _not_ possible to simulate the side result created by the >>>>>>> agent running a total decision algorithm.
1) you can't read the side result because your abstraction is broken >>>>>> 2) the human agent isn't a halt decider, partial or otherwise,
because it takes an input that disqualifies it
No response to this?
neither of those are correct: false dichotomy
That wasn't an either/or.-a Both apply.
2) the agent does not utilize information from the side record while computing the decision for any given input machine
neither apply
it is _not_ even possible to create a total decision algorithm
with turing machines, like seriously have u forgotten the point >>>>>>> of turing's proof???
... getting "output" from a simulation requires reading the >>>>>>>>> output from tape of the simulated machine, it can do that for >>>>>>>>> any value, even ones that are specified to be "output"
and if a machine is simulating itself,
Which your code isn't doing because the side data isn't the >>>>>>>>>> same when starting the simulation as it was when the machine >>>>>>>>>> was invoked.
idk what u mean by "side data" there only data on the tape when >>>>>>>>> it comes to turing machines, there is nothing on the side of >>>>>>>>> data on the tape.
I'm referring specifically to what the mechanized version of the >>>>>>>> human agent writes to the tape.
like in the case an undecidability paradox within computing, >>>>>>>>>>> then this can be used to pick out values behind any form of >>>>>>>>>>> data encapsulation that you might suggest to hide data away >>>>>>>>>>> from a paradox. there's no where to hide output dud. there's >>>>>>>>>>> no "global variable" that might prevent a paradox from being >>>>>>>>>>> formed
"can be used" meaning it's not being used now, and therefore >>>>>>>>>> irrelevant.
there is _no_ way to simulate a general ability to decide on >>>>>>>>>>> paradoxical turing machines, from within turing machines
So a paradoxical machine is one that a decider gets wrong?-a So >>>>>>>>>> this machine:
void foo() { return; }
Is a paridoxical machine to this one:
int bar(void *p) { return 0; }
Because machine bar can't successfully report the status of >>>>>>>>>> machine foo. -a-aYes?
no. first, we went over this dud: constant return function are >>>>>>>>> not and will never qualify as a halting decider as they do not >>>>>>>>> output useful information in regards to the halting status Efn+. >>>>>>>>> the fact a machine happens to output the correct answer does >>>>>>>>> not qualify them as a halting decider, a decider must output >>>>>>>>> trustworthy information which is judged across the entirety of >>>>>>>>> the output space, not just single instances
In other words, it's no different from *any* partial halt
decider. That's not even mentioning ill-defined weasel words
like "useful", "happens to", "trustworthy", and "judged".
dud if ur not here to effectively compute useful knowledge, idk >>>>>>> what ur doing discussing the theory of computing.
let me not mince words. the existing theory on the matter of
deciders is thus:
- a decider is not allowed a false positive or false negative,
and never allowed to diverge. we cannot build any non-trivial
decider (within turing machines)
a constant return 1 function, if use as a halting decider is
abound in false positives and therefore is completely
disqualified as being a halting decider.
- a recognizer is also not allowed false positives or false
negatives, it is however allowed to diverge on some negatives,
but never a positive. a halting recognizer is allowed to diverge >>>>>>> on some non- halting input, but never halting input. this is
sometimes called a partial decider, but i'm going to label as it >>>>>>> recognizer to align the rather well known sipser on this, and
because i have further definitions to propose. we can build a
halting recognizer, but not a non-halting recognizer.
a constant return 1 function does not satisfy this for halting
either, as again it's abound in false positives.
in my paper i propose two more types:
- a partial decider is still _not_ allowed false positive or
negatives. but it is allowed to diverge on some positives and
some negatives. we can be build withing turing machines both
halting and non-halting partial deciders
a constant return 1 function still does not satisfy this for
halting either
- a partial recognizer is still not allowed false positives, but >>>>>>> _does_ allow false negative ... but _only_ in the case where
returning positive would be a false positive. this exception is >>>>>>> allowed because a partial recognizer therefore does not need to >>>>>>> diverge, ever, and always halts. this can also be built within
turing machines for the halting and non- halting problems.
a constant return 1 function again still does not satisfy these >>>>>>> requirements for halting
i just defined my terms here quite clearly. i suppose i'll thank you
for encouraging me to develop that, i'll may add it to my paper
anyways, that was all covered in section -o5.2 which u never read >>>>>>>
i really don't know why i'm humoring you dud
second, an actual paradox is necessarily undecidable due to the >>>>>>>>> construction of the machine itself, the form is generalized in >>>>>>>>> -o3 of my paper. if it doesn't fit that form, it's not a
paradox, and is not undecidable in respect to any classifier >>>>>>>>> let alone in totality
Alright, let's satisfy your arbitrary requirement of a non-
constant function and "useful":
void foo() {
-a-a-a-a puts("hello");
}
int bar(void *p) {
-a-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a-a return 1;
-a-a-a-a } else {
-a-a-a-a-a-a-a-a return 0;
-a-a-a-a }
}
Given that the first instruction of "foo" is not a "return"
instruction, bar(foo) will return 0 even though foo() will halt. >>>>>>>>
Is the machine foo "undecidable" by machine bar?
No response to this?-a It goes directly to your definition of
"undecidable" which seems core to your argument.
bar isn't a halting decider in the first place, no idea what it's
supposed to be
As noted above, it's a partial halt decider that reports halting if
the first instruction is a "return" (0xc3 is the x86 instruction
"ret") and reports non-halting otherwise.
Again, is the machine foo "undecidable" by machine bar?
It would be a shame to discard your whole argument just because your
terms aren't well defined.
a partial halting decider is not allowed _any_ false positives or
false negatives, it must handle the entire input space of machines
without a _single_ false positive or false negative. that's not even
my assertion, that's the established consensus on the matter
clearly bar would have a ton of false negatives and thereby be
disqualified as a partial halting decider
Alright, let's satisfy one more arbitrary requirement to see if we can
get you to define your terms:
void foo() {
-a-a-a-a puts("hello");
}
int bar(void *p) {
-a-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a-a return 1;
-a-a-a-a } else {
-a-a-a-a-a-a-a-a while (1);
-a-a-a-a }
}
That takes care of the false negatives.-a So no more dodging the point:
Is the machine foo "undecidable" by machine bar?
Failure to answer again will deem your term "undecidable" ill-defined,
and consequently the rest of your paper.
if i'm reading this correctly, bar now diverges (does not return) on the halting input foo
i will clarify the definition for "partial decider": divergence is only allowed when both positive and negative returns would be false positive
or false negative respectively (aka undecidable input to the partial decider), so this would not quality as a partial halting decider
Without an answer, this deems your use of the term unclear, and
therefore makes your entire argument null and void.
On 9/24/2026 3:44 PM, dart200 wrote:
On 9/24/26 3:10 PM, Chris M. Thomasson wrote:
On 9/20/2026 10:39 PM, dart200 wrote:
On 9/20/26 7:51 PM, Chris M. Thomasson wrote:
On 9/20/2026 7:11 PM, dart200 wrote:
On 9/20/26 2:53 PM, Chris M. Thomasson wrote:
On 9/19/2026 2:38 PM, dart200 wrote:
[...]
If not, wtf! You remind me of PO. Sorry for that cut.
i don't care for ur fallacy by association brainrot
Sigh. I only said you kind of do remind me of the way PO dealt
with the halting program.
and that statement has no bearing on the correctness of my
arguments, u brain-rotted boomer
You cannot predict a random number and you cannot solve the halting >>>>> problem. Sigh.
turing machines do not involve random numbers dud. if u knew
literally anything about basic computing theory u'd know that.
but u don't, so go back to ur fractals, eh?
Well, ponder on CSPRNG?
pseudo-random number generators are fully deterministic and therefore
entirely decidable in terms of whether it the machine which produces
it is circle-free or not (they are obviously). for any decision ofc
one needs a description of the machine including any input (like a
seed), and that combination can be used to decide on the semantics of
the described computation
when u say "random" that does not describe what can be produced by a
turing machine, as they can only compute pseudo-random sequences that
are inherently predictable, not truly random ones. if u meant pseudo-
random you should have said that...
seriously, bro: back to ur fractals, eh?
You have no idea how to get around the halting problem.
On 9/24/26 4:35 PM, Chris M. Thomasson wrote:
On 9/24/2026 3:44 PM, dart200 wrote:
On 9/24/26 3:10 PM, Chris M. Thomasson wrote:
On 9/20/2026 10:39 PM, dart200 wrote:
On 9/20/26 7:51 PM, Chris M. Thomasson wrote:
On 9/20/2026 7:11 PM, dart200 wrote:
On 9/20/26 2:53 PM, Chris M. Thomasson wrote:
On 9/19/2026 2:38 PM, dart200 wrote:
[...]
If not, wtf! You remind me of PO. Sorry for that cut.
i don't care for ur fallacy by association brainrot
Sigh. I only said you kind of do remind me of the way PO dealt >>>>>>>> with the halting program.
and that statement has no bearing on the correctness of my
arguments, u brain-rotted boomer
You cannot predict a random number and you cannot solve the
halting problem. Sigh.
turing machines do not involve random numbers dud. if u knew
literally anything about basic computing theory u'd know that.
but u don't, so go back to ur fractals, eh?
Well, ponder on CSPRNG?
pseudo-random number generators are fully deterministic and therefore
entirely decidable in terms of whether it the machine which produces
it is circle-free or not (they are obviously). for any decision ofc
one needs a description of the machine including any input (like a
seed), and that combination can be used to decide on the semantics of
the described computation
when u say "random" that does not describe what can be produced by a
turing machine, as they can only compute pseudo-random sequences that
are inherently predictable, not truly random ones. if u meant pseudo-
random you should have said that...
seriously, bro: back to ur fractals, eh?
You have no idea how to get around the halting problem.
u have no idea what the halting problem even is, or the paradox that
makes it undecidable
and why anyone else isn't correcting u is just beyond me
On 9/24/2026 3:44 PM, dart200 wrote:
[...]
Oh my. Please refrain from calling yourself God, PO.......
On 9/24/2026 6:16 PM, dart200 wrote:
On 9/24/26 4:35 PM, Chris M. Thomasson wrote:You cannot solve the halting problem for every possible program.
On 9/24/2026 3:44 PM, dart200 wrote:
On 9/24/26 3:10 PM, Chris M. Thomasson wrote:
On 9/20/2026 10:39 PM, dart200 wrote:
On 9/20/26 7:51 PM, Chris M. Thomasson wrote:
On 9/20/2026 7:11 PM, dart200 wrote:
On 9/20/26 2:53 PM, Chris M. Thomasson wrote:
On 9/19/2026 2:38 PM, dart200 wrote:
[...]
If not, wtf! You remind me of PO. Sorry for that cut.
i don't care for ur fallacy by association brainrot
Sigh. I only said you kind of do remind me of the way PO dealt >>>>>>>>> with the halting program.
and that statement has no bearing on the correctness of my
arguments, u brain-rotted boomer
You cannot predict a random number and you cannot solve the
halting problem. Sigh.
turing machines do not involve random numbers dud. if u knew
literally anything about basic computing theory u'd know that.
but u don't, so go back to ur fractals, eh?
Well, ponder on CSPRNG?
pseudo-random number generators are fully deterministic and
therefore entirely decidable in terms of whether it the machine
which produces it is circle-free or not (they are obviously). for
any decision ofc one needs a description of the machine including
any input (like a seed), and that combination can be used to decide
on the semantics of the described computation
when u say "random" that does not describe what can be produced by a
turing machine, as they can only compute pseudo-random sequences
that are inherently predictable, not truly random ones. if u meant
pseudo- random you should have said that...
seriously, bro: back to ur fractals, eh?
You have no idea how to get around the halting problem.
u have no idea what the halting problem even is, or the paradox that
makes it undecidable
and why anyone else isn't correcting u is just beyond me
But, you can get around it in practice by restricting your languages, setting resource limits, or using automated parameters.
Alan Turing proved that no single program can correctly predict whether
any arbitrary code will finish running or loop forever.
However, real-world software engineering bypasses this theoretical limit every day.
Where's Noah?
On 9/24/2026 4:37 PM, Chris M. Thomasson wrote:
On 9/24/2026 3:44 PM, dart200 wrote:
[...]
Oh my. Please refrain from calling yourself God, PO.......
It's a 5th grader response.
You can use restricted programming languages or subsets (such as MISRA
C, SPARK, or Rocq) that are not fully Turing-complete.
By banning unbounded loops and wild recursion, you ensure that every
valid program in the language is guaranteed to finish
On 25/09/2026 00:19, Andr|- G. Isaak wrote:
On 2026-09-24 14:09, dart200 wrote:I don't know - what you're describing seems to already have a term: it's
a partial halting decider is not allowed _any_ false positives or
false negatives, it must handle the entire input space of machines
without a _single_ false positive or false negative.
You're very confused here. If it is not allowed any false negatives or
false positives, then you're talking about a /total/ halt decider. A /
partial/ halt decider is something which correctly decides /some/
instances of the halting problem but which fails to correctly decide /
all/ instances. That's why that pesky word 'partial' is added.
a "decider".-a (But just not a "halt decider" since it gets some inputs wrong by the halting criterion.)
The way "partial halt decider" [PHD] has typically been used /on this newsgroup/ is to indicate a TM with three final "outcomes":
--a HALTS-a-a-a-a-a-a-a [input halts]
--a NEVERHALTS-a-a-a-a-a-a-a [input never halts]
--a PASS-a-a-a-a-a-a-a-a-a-a-a ["pass" - the PHD makes no halting decision]
So the PHD may always PASS for a given input, but when it does indicate HALTS/NEVERHALTS that must /correctly/ match the input's halting behaviour.
Quite a few posters over the years have casually suggested that "PHD" is
a well known term meaning [whatever - generally something like my definition, rather than yours...] but I can't say I've seen any use of
the term in books I've used!-a Is it /really/ a standard term???-a If
"PHD" is more of a comp.theory common term, then posters should define
their usage for the term when they use it.
There are variations in how it might reasonably be defined, e.g. I
suspect requiring PASS to be an actual "decision" by the PHD [e.g. transition to a 3rd TM halt state] is substantially different from
allowing a PHD to implicitly PASS by virtue of never halting.-a Since
this NG has often discussed concrete programs that run until they detect
a definite HALTS or NEVERHALTS pattern, my suggestion is to allow PHDs
to include programs that may never halt...-a [so PO's code would then be classed a PHD if its bugs/logic-errors were fixed.]
Mike.
On 9/24/2026 7:13 PM, dart200 wrote:
On 9/24/26 1:54 PM, dbush wrote:
On 9/24/2026 4:09 PM, dart200 wrote:
On 9/24/26 11:23 AM, dbush wrote:
On 9/24/2026 2:03 PM, dart200 wrote:
On 9/24/26 5:37 AM, dbush wrote:
On 9/24/2026 5:38 AM, dart200 wrote:
On 9/23/26 7:08 PM, dbush wrote:
On 9/23/2026 9:35 PM, dart200 wrote:
On 9/23/26 12:43 PM, dbush wrote:
On 9/23/2026 3:17 PM, dart200 wrote:
On 9/23/26 5:17 AM, dbush wrote:
So what you really did is break the rules of a
deterministic algorithm by using the side record and by not >>>>>>>>>>>>> simulating what you though you were simulating.
what rule??? lol deterministic only means running it always >>>>>>>>>> has the same result given the same input ... which a total >>>>>>>>>> decision algo run by the idealized agent certainly does
So what exactly are the inputs to the human agent?
the agent gets the machine description like any decision algo, >>>>>>>> that would be clear if u just read the paper
So you don't consider the side record as part of the human's
input. That makes it a non-deterministic algorithm, and therefore >>>>>>> irrelevant to church-turning.
As I said, you have a fundamental misunderstanding of the
problem, and you just demonstrated it quite clearly.
that ur quite sure of, even if the reason keep changing
I see you made no attempt to refute that you have a non-
deterministic algorithm that is therefore not applicable to church- >>>>> turning.-a Do you agree?
Still no response?
the algorithm is detailed in -o7.6 ... which u still haven't >>>>>>>>>>>> even opened
No deterministic algorithm, no refutation of church-turing >>>>>>>>>>>>
Not a deterministic algorithm, as I previously stated and >>>>>>>>>>> which you made no attempt to refute, and therefore irrelevant >>>>>>>>>>> to church- turning
u haven't even read the algo, so how are you going to explain >>>>>>>>>> why specifically it's non-deterministic ... ???
You seem to think that the human's side record is not part of >>>>>>>>> the input. -a-aIf so, that makes it non-deterministic.
the side-record is nothing more than the aggregated output of >>>>>>>> the agent decision algo run across all turing machines. any
particular n- th record does not influence any future records >>>>>>>> beyond n. or before n for that matter
And if the human later reads it, that makes it an input.
i'm sorry, please explain to me why you think the agent needs to
read from the side record after writing to it???
If the agent never reads the record, why does it need to write it?
to clearly demonstrate that the sequence is produced, to demonstrate
that the ability to objectively decide on any given turing machine
is not confounded by the fact no turing machine itself can be a
total machine decider
So it's just a log?
it is a computed sequence,
Meaning that if the side record doesn't exist the agent will behave
exactly the same?
yes beside the process of writing computed results to the side record
Are you sure?
7.3 "It is these recorded side effects that the idealized human agent
can exploit"
That implies the agent does in fact read the side record.
So choose how you're wrong:
- The side record is not an input, so you have a non-determinstic >>>>>>> algorithm that is irrelevant to church-turning
- The side record is an input, and your abstraction is broken
with your simulator not actually simulating itself.
No response to this?
nothing written to the side record is used further in the
computation, so the agent never needs to read from it again
"Again".-a So it does read from the side record.
word-focus fallacy
Now you're just making things up.
You wouldn't have said "again" if the human agent *never* read from the
side record.-a In fact, you wouldn't have a whole section on it in your paper, otherwise anything in it could simply be derived from the agent's fixed algorithm.
You've basically just confirmed that you're lying about what the agent
does because you know you made a major fundamental mistake and won't
admit it.
now we're at 4 named fallacies, 1 named cognitive distortion, and 1
named abuse tactic
Just because you gave the name of a fallacy doesn't mean it applies.
talk about brain rot
That makes it an input.
Which means that when und_cir_sr calls sim_cir_sr(und_cir_sr) after
calling Tdp, it's not simulating itself.
That also means the agent is disqualified from being a halt decider/
recognizer, partial or otherwise.
So your abstraction is broken.
ofc the code does that
lol, turing machines don't have "global variables" dud, >>>>>>>>>>>>>> they just have a tape, and if the value is output to the >>>>>>>>>>>>>> tape anywhere, it can be picked out by a simulation and >>>>>>>>>>>>>> contradicted by a paradox
To fix it, the fixed steps the human agent runs need to >>>>>>>>>>>>>>> be converted to a function which read/writes a global >>>>>>>>>>>>>>> variable, and that global variable need to be set to the >>>>>>>>>>>>>>> same value it had when und_cir_sr was first called. >>>>>>>>>>>>>>
Strawman.-a "it can be picked out" means you're talking >>>>>>>>>>>>> about changing the code which means you're no longer >>>>>>>>>>>>> talking about the same machine.
that's incorrect. reading values from the tape during a step >>>>>>>>>>>> of the computation does not change the machine which is >>>>>>>>>>>> being simulated ...
No, but the code you've shown doesn't do that.-a So if you >>>>>>>>>>> change the code to do that, it's no longer the same machine. >>>>>>>>>>
Your code is reading the part of the tape where the mechanized >>>>>>>>> human agent is writing its side record?
No?
Then your code doesn't do that, and talking about what it "can" >>>>>>>>> do necessarily means changing the code to something not being >>>>>>>>> decided on.
u really don't get it: claiming a machine to be a total decider >>>>>>>> is subject to the full enumerations of machines, meaning if an >>>>>>>> input can exist, it will exist
That has nothing to do with the fact that you can't talk about
changing code and still claim it's the same machine.
no one is changing the goal post as the goal post of a decider is >>>>>> handling _all_ possible input
*A* decider.-a There is no "can" or "can not", only "does" or "does >>>>> not".
there exists a machine that does a paradox for any location you can
write a decision to the tape
You "can" write means changing the code of the decider, which is not
allowed.-a Otherwise you're not talking about the same decider.
for any location on the tape that a decider does attempt to write a
decision, of which there are infinite locations on the tape, and
infinite deciders writing to any location on the tape,
there does exist infinite paradoxical machines which will read from
that specific bit and produce a paradox for it in regards to the
written decision
there is _no_ way to produce a total halting decider for turing
machines, within turing machines
Which is basically Turing's proof.
1) the agent does not read from the side record
Turing machines are defined by their instructions, not where they
physically reside.
there exists a paradox that contradicts any possible location >>>>>>>> you could write a decision bit on a tape, and therefore there >>>>>>>> will exist a machine that defies the decision, no matter where u >>>>>>>> try to write it
it is _not_ possible to simulate the side result created by the >>>>>>>> agent running a total decision algorithm.
1) you can't read the side result because your abstraction is broken >>>>>>> 2) the human agent isn't a halt decider, partial or otherwise,
because it takes an input that disqualifies it
No response to this?
neither of those are correct: false dichotomy
That wasn't an either/or.-a Both apply.
2) the agent does not utilize information from the side record while
computing the decision for any given input machine
neither apply
I don't believe you.
If the agent *never* uses the side record, that makes the side record entirely irrelevant, and it means that any machine can use the agent's
algorithm to exactly reproduce what the agent does if it doesn't use the
side data.-a You wouldn't have added a whole section to your paper about
the side record and made it the focus of your "refutation" if the agent didn't use it.
You're backtracking because you've been caught in a fundamental mistake
and won't admit it.
it is _not_ even possible to create a total decision algorithm >>>>>>>> with turing machines, like seriously have u forgotten the point >>>>>>>> of turing's proof???
... getting "output" from a simulation requires reading the >>>>>>>>>> output from tape of the simulated machine, it can do that for >>>>>>>>>> any value, even ones that are specified to be "output"
and if a machine is simulating itself,
Which your code isn't doing because the side data isn't the >>>>>>>>>>> same when starting the simulation as it was when the machine >>>>>>>>>>> was invoked.
idk what u mean by "side data" there only data on the tape >>>>>>>>>> when it comes to turing machines, there is nothing on the side >>>>>>>>>> of data on the tape.
I'm referring specifically to what the mechanized version of >>>>>>>>> the human agent writes to the tape.
like in the case an undecidability paradox within computing, >>>>>>>>>>>> then this can be used to pick out values behind any form of >>>>>>>>>>>> data encapsulation that you might suggest to hide data away >>>>>>>>>>>> from a paradox. there's no where to hide output dud. there's >>>>>>>>>>>> no "global variable" that might prevent a paradox from being >>>>>>>>>>>> formed
"can be used" meaning it's not being used now, and therefore >>>>>>>>>>> irrelevant.
So a paradoxical machine is one that a decider gets wrong? >>>>>>>>>>> So this machine:
there is _no_ way to simulate a general ability to decide on >>>>>>>>>>>> paradoxical turing machines, from within turing machines >>>>>>>>>>>
void foo() { return; }
Is a paridoxical machine to this one:
int bar(void *p) { return 0; }
Because machine bar can't successfully report the status of >>>>>>>>>>> machine foo. -a-aYes?
no. first, we went over this dud: constant return function are >>>>>>>>>> not and will never qualify as a halting decider as they do not >>>>>>>>>> output useful information in regards to the halting status Efn+. >>>>>>>>>> the fact a machine happens to output the correct answer does >>>>>>>>>> not qualify them as a halting decider, a decider must output >>>>>>>>>> trustworthy information which is judged across the entirety of >>>>>>>>>> the output space, not just single instances
In other words, it's no different from *any* partial halt
decider. That's not even mentioning ill-defined weasel words >>>>>>>>> like "useful", "happens to", "trustworthy", and "judged".
dud if ur not here to effectively compute useful knowledge, idk >>>>>>>> what ur doing discussing the theory of computing.
let me not mince words. the existing theory on the matter of
deciders is thus:
- a decider is not allowed a false positive or false negative, >>>>>>>> and never allowed to diverge. we cannot build any non-trivial >>>>>>>> decider (within turing machines)
a constant return 1 function, if use as a halting decider is
abound in false positives and therefore is completely
disqualified as being a halting decider.
- a recognizer is also not allowed false positives or false
negatives, it is however allowed to diverge on some negatives, >>>>>>>> but never a positive. a halting recognizer is allowed to diverge >>>>>>>> on some non- halting input, but never halting input. this is
sometimes called a partial decider, but i'm going to label as it >>>>>>>> recognizer to align the rather well known sipser on this, and >>>>>>>> because i have further definitions to propose. we can build a >>>>>>>> halting recognizer, but not a non-halting recognizer.
a constant return 1 function does not satisfy this for halting >>>>>>>> either, as again it's abound in false positives.
in my paper i propose two more types:
- a partial decider is still _not_ allowed false positive or
negatives. but it is allowed to diverge on some positives and >>>>>>>> some negatives. we can be build withing turing machines both
halting and non-halting partial deciders
a constant return 1 function still does not satisfy this for
halting either
- a partial recognizer is still not allowed false positives, but >>>>>>>> _does_ allow false negative ... but _only_ in the case where
returning positive would be a false positive. this exception is >>>>>>>> allowed because a partial recognizer therefore does not need to >>>>>>>> diverge, ever, and always halts. this can also be built within >>>>>>>> turing machines for the halting and non- halting problems.
a constant return 1 function again still does not satisfy these >>>>>>>> requirements for halting
i just defined my terms here quite clearly. i suppose i'll thank you
for encouraging me to develop that, i'll may add it to my paper
anyways, that was all covered in section -o5.2 which u never read >>>>>>>>
i really don't know why i'm humoring you dud
second, an actual paradox is necessarily undecidable due to >>>>>>>>>> the construction of the machine itself, the form is
generalized in -o3 of my paper. if it doesn't fit that form, >>>>>>>>>> it's not a paradox, and is not undecidable in respect to any >>>>>>>>>> classifier let alone in totality
Alright, let's satisfy your arbitrary requirement of a non- >>>>>>>>> constant function and "useful":
void foo() {
-a-a-a-a puts("hello");
}
int bar(void *p) {
-a-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a-a return 1;
-a-a-a-a } else {
-a-a-a-a-a-a-a-a return 0;
-a-a-a-a }
}
Given that the first instruction of "foo" is not a "return" >>>>>>>>> instruction, bar(foo) will return 0 even though foo() will halt. >>>>>>>>>
Is the machine foo "undecidable" by machine bar?
No response to this?-a It goes directly to your definition of
"undecidable" which seems core to your argument.
bar isn't a halting decider in the first place, no idea what it's >>>>>> supposed to be
As noted above, it's a partial halt decider that reports halting if >>>>> the first instruction is a "return" (0xc3 is the x86 instruction
"ret") and reports non-halting otherwise.
Again, is the machine foo "undecidable" by machine bar?
It would be a shame to discard your whole argument just because
your terms aren't well defined.
a partial halting decider is not allowed _any_ false positives or
false negatives, it must handle the entire input space of machines
without a _single_ false positive or false negative. that's not even
my assertion, that's the established consensus on the matter
clearly bar would have a ton of false negatives and thereby be
disqualified as a partial halting decider
Alright, let's satisfy one more arbitrary requirement to see if we
can get you to define your terms:
void foo() {
-a-a-a-a puts("hello");
}
int bar(void *p) {
-a-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a-a return 1;
-a-a-a-a } else {
-a-a-a-a-a-a-a-a while (1);
-a-a-a-a }
}
That takes care of the false negatives.-a So no more dodging the point:
Is the machine foo "undecidable" by machine bar?
Failure to answer again will deem your term "undecidable" ill-
defined, and consequently the rest of your paper.
if i'm reading this correctly, bar now diverges (does not return) on
the halting input foo
i will clarify the definition for "partial decider": divergence is
only allowed when both positive and negative returns would be false
positive or false negative respectively (aka undecidable input to the
partial decider), so this would not quality as a partial halting decider
So yet again you dodge the question on whether foo is "undecidable" by machine bar.
That make that term ill-defined.
And because you whole paper is based on the use of that word, it is
reduced to meaninglessness.
Without an answer, this deems your use of the term unclear, and >>>>>>> therefore makes your entire argument null and void.
On 9/24/26 5:44 PM, Mike Terry wrote:
On 25/09/2026 00:19, Andr|- G. Isaak wrote:
On 2026-09-24 14:09, dart200 wrote:I don't know - what you're describing seems to already have a term:
a partial halting decider is not allowed _any_ false positives or
false negatives, it must handle the entire input space of machines
without a _single_ false positive or false negative.
You're very confused here. If it is not allowed any false negatives
or false positives, then you're talking about a /total/ halt decider.
A / partial/ halt decider is something which correctly decides /some/
instances of the halting problem but which fails to correctly
decide / all/ instances. That's why that pesky word 'partial' is added.
it's a "decider".-a (But just not a "halt decider" since it gets some
inputs wrong by the halting criterion.)
The way "partial halt decider" [PHD] has typically been used /on this
newsgroup/ is to indicate a TM with three final "outcomes":
--a HALTS-a-a-a-a-a-a-a [input halts]
--a NEVERHALTS-a-a-a-a-a-a-a [input never halts]
--a PASS-a-a-a-a-a-a-a-a-a-a-a ["pass" - the PHD makes no halting decision] >>
So the PHD may always PASS for a given input, but when it does
indicate HALTS/NEVERHALTS that must /correctly/ match the input's
halting behaviour.
Quite a few posters over the years have casually suggested that "PHD"
is a well known term meaning [whatever - generally something like my
definition, rather than yours...] but I can't say I've seen any use of
the term in books I've used!-a Is it /really/ a standard term???-a If
"PHD" is more of a comp.theory common term, then posters should define
their usage for the term when they use it.
There are variations in how it might reasonably be defined, e.g. I
suspect requiring PASS to be an actual "decision" by the PHD [e.g.
transition to a 3rd TM halt state] is substantially different from
allowing a PHD to implicitly PASS by virtue of never halting.-a Since
this NG has often discussed concrete programs that run until they
detect a definite HALTS or NEVERHALTS pattern, my suggestion is to
allow PHDs to include programs that may never halt...-a [so PO's code
would then be classed a PHD if its bugs/logic-errors were fixed.]
Mike.
well ... a major problem here is every other decade since the 50s the
terms have changed a bit, and this has caused a lot of confusion
as for the current convention, as far as i can tell the gold standard
for terminology in the theory of computing is from michael sipser of
MIT. to understand what the current conventional theory is, one needs to consider types of sets are considered relevant to decidability: turing- recognizable and turing-decidable
https://broman.dev/download/ Introduction%20to%20the%20Theory%20of%20Computation%203rd%20Edition.pdf#page=194
the textbook calls these "languages" but by language its referring to a
set of strings where the strings are machine definitions, so those "languages" are actually just a particular set of machines
a decidable set of machines means we can enumerate both the set and it's complement, allowing us to build a decider for that set to produce a accept/reject decision for each and every possible input. this cannot be done for halting machines due to the halting problem paradoxes
a recognizable or semi-decidable set of machines is one we can
enumerate, but not it's complement. this we can build a recognizer for,
but not a decider. a recognizer will accept all within the set, but may diverge/loop for some in the complement. this can be built for halting machines, as we can enumerate halting machines using a dovetailing algo,
but we cannot enumerate non-halting machines as such, and therefore a non-halting recognizer cannot be built.
to reiterate a recognizer recognizes turing-recognizable sets of
machines, while a decider deciders turing-decidable sets of machines,
and conventional theory does not recognize other forms of classifiers at
the moment. i think some books refer to a recognizer as a "partial
decider" for semi-decidable sets like halting, but i'm going to call it
a recognizer.
and to be clear: conventional theory does _not_ consider a form of
partial decider where only a subset of the primary set is recognized.
the only form is a recognizer where only a subset of the complement is recognized, but the entirety of primary set is still recognized.
anything less is not considered decidable by convention theory and is entirely ignored in potential,
which is what my paper takes of advantage to poke holes
On 9/24/2026 4:37 PM, Chris M. Thomasson wrote:
On 9/24/2026 3:44 PM, dart200 wrote:
[...]
Oh my. Please refrain from calling yourself God, PO.......
It's a 5th grader response.
You can use restricted programming languages or subsets (such as MISRA
C, SPARK, or Rocq) that are not fully Turing-complete.
By banning unbounded loops and wild recursion, you ensure that every
valid program in the language is guaranteed to finish
On 9/24/26 6:00 PM, dbush wrote:
On 9/24/2026 7:13 PM, dart200 wrote:
On 9/24/26 1:54 PM, dbush wrote:
On 9/24/2026 4:09 PM, dart200 wrote:
On 9/24/26 11:23 AM, dbush wrote:
On 9/24/2026 2:03 PM, dart200 wrote:
On 9/24/26 5:37 AM, dbush wrote:
On 9/24/2026 5:38 AM, dart200 wrote:
On 9/23/26 7:08 PM, dbush wrote:
On 9/23/2026 9:35 PM, dart200 wrote:
On 9/23/26 12:43 PM, dbush wrote:
On 9/23/2026 3:17 PM, dart200 wrote:
On 9/23/26 5:17 AM, dbush wrote:
So what you really did is break the rules of a
deterministic algorithm by using the side record and by >>>>>>>>>>>>>> not simulating what you though you were simulating.
what rule??? lol deterministic only means running it always >>>>>>>>>>> has the same result given the same input ... which a total >>>>>>>>>>> decision algo run by the idealized agent certainly does
So what exactly are the inputs to the human agent?
the agent gets the machine description like any decision algo, >>>>>>>>> that would be clear if u just read the paper
So you don't consider the side record as part of the human's
input. That makes it a non-deterministic algorithm, and
therefore irrelevant to church-turning.
As I said, you have a fundamental misunderstanding of the
problem, and you just demonstrated it quite clearly.
that ur quite sure of, even if the reason keep changing
I see you made no attempt to refute that you have a non-
deterministic algorithm that is therefore not applicable to
church- turning.-a Do you agree?
Still no response?
the algorithm is detailed in -o7.6 ... which u still haven't >>>>>>>>>>>>> even opened
No deterministic algorithm, no refutation of church-turing >>>>>>>>>>>>>
Not a deterministic algorithm, as I previously stated and >>>>>>>>>>>> which you made no attempt to refute, and therefore
irrelevant to church- turning
u haven't even read the algo, so how are you going to explain >>>>>>>>>>> why specifically it's non-deterministic ... ???
You seem to think that the human's side record is not part of >>>>>>>>>> the input. -a-aIf so, that makes it non-deterministic.
the side-record is nothing more than the aggregated output of >>>>>>>>> the agent decision algo run across all turing machines. any >>>>>>>>> particular n- th record does not influence any future records >>>>>>>>> beyond n. or before n for that matter
And if the human later reads it, that makes it an input.
i'm sorry, please explain to me why you think the agent needs to >>>>>>> read from the side record after writing to it???
If the agent never reads the record, why does it need to write it?
to clearly demonstrate that the sequence is produced, to
demonstrate that the ability to objectively decide on any given
turing machine is not confounded by the fact no turing machine
itself can be a total machine decider
So it's just a log?
it is a computed sequence,
Meaning that if the side record doesn't exist the agent will behave
exactly the same?
yes beside the process of writing computed results to the side record
Are you sure?
7.3 "It is these recorded side effects that the idealized human agent
can exploit"
That implies the agent does in fact read the side record.
lol are you trying to educate me about the paper i wrote??? yes i'm
sure, as there is literally no need for the agent to read from the side record, as no halting decision for any turing machine requires doing
so ... SO WHY WOULD HE???
the record is there to formalize the notion of computing a sequence, not because it's used for later computations. and not because we would
actually want to compute the total halting set, that would be as bad as
busy beaver, so far beyond practical. it's to prove that such
computation exist in theory, so we know that proving so is possible for
any given machine
So choose how you're wrong:
- The side record is not an input, so you have a non-
determinstic algorithm that is irrelevant to church-turning
- The side record is an input, and your abstraction is broken >>>>>>>> with your simulator not actually simulating itself.
No response to this?
nothing written to the side record is used further in the
computation, so the agent never needs to read from it again
"Again".-a So it does read from the side record.
word-focus fallacy
Now you're just making things up.
You wouldn't have said "again" if the human agent *never* read from the
do u honestly think this is a genuine line of inquiry?
side record.-a In fact, you wouldn't have a whole section on it in your
paper, otherwise anything in it could simply be derived from the
agent's fixed algorithm.
You've basically just confirmed that you're lying about what the agent
does because you know you made a major fundamental mistake and won't
admit it.
bro unless u can tell me why the agent reads from the side record in the future, then he does not, cause he doesn't need to, and this line of
trying to refute my paper is a failure
now we're at 4 named fallacies, 1 named cognitive distortion, and 1
named abuse tactic
Just because you gave the name of a fallacy doesn't mean it applies.
just because u can't identify forms of fallacies does not negate their application. let's add #5 and #6 for ur latest post: bad-faith interpretation + strawman
talk about brain rot
That makes it an input.
Which means that when und_cir_sr calls sim_cir_sr(und_cir_sr) after
calling Tdp, it's not simulating itself.
That also means the agent is disqualified from being a halt decider/
recognizer, partial or otherwise.
So your abstraction is broken.
ofc the code does that
lol, turing machines don't have "global variables" dud, >>>>>>>>>>>>>>> they just have a tape, and if the value is output to the >>>>>>>>>>>>>>> tape anywhere, it can be picked out by a simulation and >>>>>>>>>>>>>>> contradicted by a paradox
To fix it, the fixed steps the human agent runs need to >>>>>>>>>>>>>>>> be converted to a function which read/writes a global >>>>>>>>>>>>>>>> variable, and that global variable need to be set to the >>>>>>>>>>>>>>>> same value it had when und_cir_sr was first called. >>>>>>>>>>>>>>>
Strawman.-a "it can be picked out" means you're talking >>>>>>>>>>>>>> about changing the code which means you're no longer >>>>>>>>>>>>>> talking about the same machine.
that's incorrect. reading values from the tape during a >>>>>>>>>>>>> step of the computation does not change the machine which >>>>>>>>>>>>> is being simulated ...
No, but the code you've shown doesn't do that.-a So if you >>>>>>>>>>>> change the code to do that, it's no longer the same machine. >>>>>>>>>>>
Your code is reading the part of the tape where the mechanized >>>>>>>>>> human agent is writing its side record?
No?
Then your code doesn't do that, and talking about what it >>>>>>>>>> "can" do necessarily means changing the code to something not >>>>>>>>>> being decided on.
u really don't get it: claiming a machine to be a total decider >>>>>>>>> is subject to the full enumerations of machines, meaning if an >>>>>>>>> input can exist, it will exist
That has nothing to do with the fact that you can't talk about >>>>>>>> changing code and still claim it's the same machine.
no one is changing the goal post as the goal post of a decider is >>>>>>> handling _all_ possible input
*A* decider.-a There is no "can" or "can not", only "does" or "does >>>>>> not".
there exists a machine that does a paradox for any location you can >>>>> write a decision to the tape
You "can" write means changing the code of the decider, which is not
allowed.-a Otherwise you're not talking about the same decider.
for any location on the tape that a decider does attempt to write a
decision, of which there are infinite locations on the tape, and
infinite deciders writing to any location on the tape,
there does exist infinite paradoxical machines which will read from
that specific bit and produce a paradox for it in regards to the
written decision
there is _no_ way to produce a total halting decider for turing
machines, within turing machines
Which is basically Turing's proof.
so why are you trying to argue u can simulate the agent's
computation ... u can't do so within turing machines
1) the agent does not read from the side record
Turing machines are defined by their instructions, not where they
physically reside.
there exists a paradox that contradicts any possible location >>>>>>>>> you could write a decision bit on a tape, and therefore there >>>>>>>>> will exist a machine that defies the decision, no matter where >>>>>>>>> u try to write it
it is _not_ possible to simulate the side result created by the >>>>>>>>> agent running a total decision algorithm.
1) you can't read the side result because your abstraction is >>>>>>>> broken
2) the human agent isn't a halt decider, partial or otherwise, >>>>>>>> because it takes an input that disqualifies it
No response to this?
neither of those are correct: false dichotomy
That wasn't an either/or.-a Both apply.
2) the agent does not utilize information from the side record while
computing the decision for any given input machine
neither apply
I don't believe you.
If the agent *never* uses the side record, that makes the side record
entirely irrelevant, and it means that any machine can use the agent's
it's purpose is to demonstrate the existence of the computation, if u
don't care for the demonstration, ur free to remain in error
algorithm to exactly reproduce what the agent does if it doesn't use the
u keep asserting that, but you have not yet once told me how such a simulation would handle it's own undecidability paradox: und_cir_sr
side data.-a You wouldn't have added a whole section to your paper
about the side record and made it the focus of your "refutation" if
the agent didn't use it.
turing machines don't utilize their own output to justify their
existence... they just output it. why would an agent need to?
i wonder if ur about to commit fallacy #7: special pleading
You're backtracking because you've been caught in a fundamental
mistake and won't admit it.
it is _not_ even possible to create a total decision algorithm >>>>>>>>> with turing machines, like seriously have u forgotten the point >>>>>>>>> of turing's proof???
... getting "output" from a simulation requires reading the >>>>>>>>>>> output from tape of the simulated machine, it can do that for >>>>>>>>>>> any value, even ones that are specified to be "output"
and if a machine is simulating itself,
Which your code isn't doing because the side data isn't the >>>>>>>>>>>> same when starting the simulation as it was when the machine >>>>>>>>>>>> was invoked.
idk what u mean by "side data" there only data on the tape >>>>>>>>>>> when it comes to turing machines, there is nothing on the >>>>>>>>>>> side of data on the tape.
I'm referring specifically to what the mechanized version of >>>>>>>>>> the human agent writes to the tape.
In other words, it's no different from *any* partial halt >>>>>>>>>> decider. That's not even mentioning ill-defined weasel words >>>>>>>>>> like "useful", "happens to", "trustworthy", and "judged".
like in the case an undecidability paradox within
computing, then this can be used to pick out values behind >>>>>>>>>>>>> any form of data encapsulation that you might suggest to >>>>>>>>>>>>> hide data away from a paradox. there's no where to hide >>>>>>>>>>>>> output dud. there's no "global variable" that might prevent >>>>>>>>>>>>> a paradox from being formed
"can be used" meaning it's not being used now, and therefore >>>>>>>>>>>> irrelevant.
So a paradoxical machine is one that a decider gets wrong? >>>>>>>>>>>> So this machine:
there is _no_ way to simulate a general ability to decide >>>>>>>>>>>>> on paradoxical turing machines, from within turing machines >>>>>>>>>>>>
void foo() { return; }
Is a paridoxical machine to this one:
int bar(void *p) { return 0; }
Because machine bar can't successfully report the status of >>>>>>>>>>>> machine foo. -a-aYes?
no. first, we went over this dud: constant return function >>>>>>>>>>> are not and will never qualify as a halting decider as they >>>>>>>>>>> do not output useful information in regards to the halting >>>>>>>>>>> status Efn+. the fact a machine happens to output the correct >>>>>>>>>>> answer does not qualify them as a halting decider, a decider >>>>>>>>>>> must output trustworthy information which is judged across >>>>>>>>>>> the entirety of the output space, not just single instances >>>>>>>>>>
dud if ur not here to effectively compute useful knowledge, idk >>>>>>>>> what ur doing discussing the theory of computing.
let me not mince words. the existing theory on the matter of >>>>>>>>> deciders is thus:
- a decider is not allowed a false positive or false negative, >>>>>>>>> and never allowed to diverge. we cannot build any non-trivial >>>>>>>>> decider (within turing machines)
a constant return 1 function, if use as a halting decider is >>>>>>>>> abound in false positives and therefore is completely
disqualified as being a halting decider.
- a recognizer is also not allowed false positives or false >>>>>>>>> negatives, it is however allowed to diverge on some negatives, >>>>>>>>> but never a positive. a halting recognizer is allowed to
diverge on some non- halting input, but never halting input. >>>>>>>>> this is sometimes called a partial decider, but i'm going to >>>>>>>>> label as it recognizer to align the rather well known sipser on >>>>>>>>> this, and because i have further definitions to propose. we can >>>>>>>>> build a halting recognizer, but not a non-halting recognizer. >>>>>>>>>
a constant return 1 function does not satisfy this for halting >>>>>>>>> either, as again it's abound in false positives.
in my paper i propose two more types:
- a partial decider is still _not_ allowed false positive or >>>>>>>>> negatives. but it is allowed to diverge on some positives and >>>>>>>>> some negatives. we can be build withing turing machines both >>>>>>>>> halting and non-halting partial deciders
a constant return 1 function still does not satisfy this for >>>>>>>>> halting either
- a partial recognizer is still not allowed false positives, >>>>>>>>> but _does_ allow false negative ... but _only_ in the case
where returning positive would be a false positive. this
exception is allowed because a partial recognizer therefore >>>>>>>>> does not need to diverge, ever, and always halts. this can also >>>>>>>>> be built within turing machines for the halting and non-
halting problems.
a constant return 1 function again still does not satisfy these >>>>>>>>> requirements for halting
i just defined my terms here quite clearly. i suppose i'll thank
you for encouraging me to develop that, i'll may add it to my paper
anyways, that was all covered in section -o5.2 which u never read >>>>>>>>>
i really don't know why i'm humoring you dud
second, an actual paradox is necessarily undecidable due to >>>>>>>>>>> the construction of the machine itself, the form is
generalized in -o3 of my paper. if it doesn't fit that form, >>>>>>>>>>> it's not a paradox, and is not undecidable in respect to any >>>>>>>>>>> classifier let alone in totality
Alright, let's satisfy your arbitrary requirement of a non- >>>>>>>>>> constant function and "useful":
void foo() {
-a-a-a-a puts("hello");
}
int bar(void *p) {
-a-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a-a return 1;
-a-a-a-a } else {
-a-a-a-a-a-a-a-a return 0;
-a-a-a-a }
}
Given that the first instruction of "foo" is not a "return" >>>>>>>>>> instruction, bar(foo) will return 0 even though foo() will halt. >>>>>>>>>>
Is the machine foo "undecidable" by machine bar?
No response to this?-a It goes directly to your definition of >>>>>>>> "undecidable" which seems core to your argument.
bar isn't a halting decider in the first place, no idea what it's >>>>>>> supposed to be
As noted above, it's a partial halt decider that reports halting
if the first instruction is a "return" (0xc3 is the x86
instruction "ret") and reports non-halting otherwise.
Again, is the machine foo "undecidable" by machine bar?
It would be a shame to discard your whole argument just because
your terms aren't well defined.
a partial halting decider is not allowed _any_ false positives or
false negatives, it must handle the entire input space of machines
without a _single_ false positive or false negative. that's not
even my assertion, that's the established consensus on the matter
clearly bar would have a ton of false negatives and thereby be
disqualified as a partial halting decider
Alright, let's satisfy one more arbitrary requirement to see if we
can get you to define your terms:
void foo() {
-a-a-a-a puts("hello");
}
int bar(void *p) {
-a-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a-a return 1;
-a-a-a-a } else {
-a-a-a-a-a-a-a-a while (1);
-a-a-a-a }
}
That takes care of the false negatives.-a So no more dodging the point: >>>>
Is the machine foo "undecidable" by machine bar?
Failure to answer again will deem your term "undecidable" ill-
defined, and consequently the rest of your paper.
if i'm reading this correctly, bar now diverges (does not return) on
the halting input foo
i will clarify the definition for "partial decider": divergence is
only allowed when both positive and negative returns would be false
positive or false negative respectively (aka undecidable input to the
partial decider), so this would not quality as a partial halting decider >>>
So yet again you dodge the question on whether foo is "undecidable" by
machine bar.
bar is _not_ halting decider, partial or otherwise, so asking if the
halting semantics of foo is "undecidable" to bar is a category fallacy
wowee fallacy #8
That make that term ill-defined.
the term is quite well defined: if a classifier/decider is stuck between returning a false positive and a false negative for a particular input,
then that input is undecidable to the classifier/decider
And because you whole paper is based on the use of that word, it is
reduced to meaninglessness.
a fallacy laden critique full of cognitive distortion and abuse is never going to convince me dud
Without an answer, this deems your use of the term unclear, and >>>>>>>> therefore makes your entire argument null and void.
On 25/09/2026 00:19, Andr|- G. Isaak wrote:
On 2026-09-24 14:09, dart200 wrote:I don't know - what you're describing seems to already have a term: it's
a partial halting decider is not allowed _any_ false positives or
false negatives, it must handle the entire input space of machines
without a _single_ false positive or false negative.
You're very confused here. If it is not allowed any false negatives or
false positives, then you're talking about a /total/ halt decider. A
/partial/ halt decider is something which correctly decides /some/
instances of the halting problem but which fails to correctly decide
/all/ instances. That's why that pesky word 'partial' is added.
a "decider".-a (But just not a "halt decider" since it gets some inputs wrong by the halting criterion.)
The way "partial halt decider" [PHD] has typically been used /on this newsgroup/ is to indicate a TM with three final "outcomes":
--a HALTS-a-a-a-a-a-a-a [input halts]
--a NEVERHALTS-a-a-a-a-a-a-a [input never halts]
--a PASS-a-a-a-a-a-a-a-a-a-a-a ["pass" - the PHD makes no halting decision]
So the PHD may always PASS for a given input, but when it does indicate HALTS/NEVERHALTS that must /correctly/ match the input's halting behaviour.
Quite a few posters over the years have casually suggested that "PHD" is
a well known term meaning [whatever - generally something like my definition, rather than yours...] but I can't say I've seen any use of
the term in books I've used!-a Is it /really/ a standard term???-a If
"PHD" is more of a comp.theory common term, then posters should define
their usage for the term when they use it.
There are variations in how it might reasonably be defined, e.g. I
suspect requiring PASS to be an actual "decision" by the PHD [e.g. transition to a 3rd TM halt state] is substantially different from
allowing a PHD to implicitly PASS by virtue of never halting.-a Since
this NG has often discussed concrete programs that run until they detect
a definite HALTS or NEVERHALTS pattern, my suggestion is to allow PHDs
to include programs that may never halt...-a [so PO's code would then be classed a PHD if its bugs/logic-errors were fixed.]
On 9/25/2026 12:13 AM, dart200 wrote:
On 9/24/26 6:00 PM, dbush wrote:
On 9/24/2026 7:13 PM, dart200 wrote:
On 9/24/26 1:54 PM, dbush wrote:
On 9/24/2026 4:09 PM, dart200 wrote:
On 9/24/26 11:23 AM, dbush wrote:
On 9/24/2026 2:03 PM, dart200 wrote:
On 9/24/26 5:37 AM, dbush wrote:
On 9/24/2026 5:38 AM, dart200 wrote:
On 9/23/26 7:08 PM, dbush wrote:
On 9/23/2026 9:35 PM, dart200 wrote:
On 9/23/26 12:43 PM, dbush wrote:So what exactly are the inputs to the human agent?
On 9/23/2026 3:17 PM, dart200 wrote:what rule??? lol deterministic only means running it always >>>>>>>>>>>> has the same result given the same input ... which a total >>>>>>>>>>>> decision algo run by the idealized agent certainly does >>>>>>>>>>>
On 9/23/26 5:17 AM, dbush wrote:
So what you really did is break the rules of a
deterministic algorithm by using the side record and by >>>>>>>>>>>>>>> not simulating what you though you were simulating. >>>>>>>>>>>>
the agent gets the machine description like any decision algo, >>>>>>>>>> that would be clear if u just read the paper
So you don't consider the side record as part of the human's >>>>>>>>> input. That makes it a non-deterministic algorithm, and
therefore irrelevant to church-turning.
As I said, you have a fundamental misunderstanding of the
problem, and you just demonstrated it quite clearly.
that ur quite sure of, even if the reason keep changing
I see you made no attempt to refute that you have a non-
deterministic algorithm that is therefore not applicable to
church- turning.-a Do you agree?
Still no response?
to clearly demonstrate that the sequence is produced, to
You seem to think that the human's side record is not part of >>>>>>>>>>> the input. -a-aIf so, that makes it non-deterministic.
the algorithm is detailed in -o7.6 ... which u still >>>>>>>>>>>>>> haven't even opened
No deterministic algorithm, no refutation of church-turing >>>>>>>>>>>>>>
Not a deterministic algorithm, as I previously stated and >>>>>>>>>>>>> which you made no attempt to refute, and therefore
irrelevant to church- turning
u haven't even read the algo, so how are you going to >>>>>>>>>>>> explain why specifically it's non-deterministic ... ??? >>>>>>>>>>>
the side-record is nothing more than the aggregated output of >>>>>>>>>> the agent decision algo run across all turing machines. any >>>>>>>>>> particular n- th record does not influence any future records >>>>>>>>>> beyond n. or before n for that matter
And if the human later reads it, that makes it an input.
i'm sorry, please explain to me why you think the agent needs to >>>>>>>> read from the side record after writing to it???
If the agent never reads the record, why does it need to write it? >>>>>>
demonstrate that the ability to objectively decide on any given
turing machine is not confounded by the fact no turing machine
itself can be a total machine decider
So it's just a log?
it is a computed sequence,
Meaning that if the side record doesn't exist the agent will behave >>>>> exactly the same?
yes beside the process of writing computed results to the side record
Are you sure?
7.3 "It is these recorded side effects that the idealized human agent
can exploit"
That implies the agent does in fact read the side record.
lol are you trying to educate me about the paper i wrote??? yes i'm
sure, as there is literally no need for the agent to read from the
side record, as no halting decision for any turing machine requires
doing so ... SO WHY WOULD HE???
Then anything the agent does can be accurately simulated by a turing machine.
the record is there to formalize the notion of computing a sequence,
not because it's used for later computations. and not because we would
actually want to compute the total halting set, that would be as bad
as busy beaver, so far beyond practical. it's to prove that such
computation exist in theory, so we know that proving so is possible
for any given machine
So nothing more than a log, and therefore irrelevant.
So choose how you're wrong:
- The side record is not an input, so you have a non-
determinstic algorithm that is irrelevant to church-turning
- The side record is an input, and your abstraction is broken >>>>>>>>> with your simulator not actually simulating itself.
No response to this?
nothing written to the side record is used further in the
computation, so the agent never needs to read from it again
"Again".-a So it does read from the side record.
word-focus fallacy
Now you're just making things up.
You wouldn't have said "again" if the human agent *never* read from the
do u honestly think this is a genuine line of inquiry?
Yes, as it means that either 1) you're trying to hide something, or 2)
you have a very poor grasp of English.-a Both of which bring everything
you say into question.
side record.-a In fact, you wouldn't have a whole section on it in
your paper, otherwise anything in it could simply be derived from the
agent's fixed algorithm.
You've basically just confirmed that you're lying about what the
agent does because you know you made a major fundamental mistake and
won't admit it.
bro unless u can tell me why the agent reads from the side record in
the future, then he does not, cause he doesn't need to, and this line
of trying to refute my paper is a failure
now we're at 4 named fallacies, 1 named cognitive distortion, and 1
named abuse tactic
Just because you gave the name of a fallacy doesn't mean it applies.
just because u can't identify forms of fallacies does not negate their
application. let's add #5 and #6 for ur latest post: bad-faith
interpretation + strawman
I explained why some of them don't apply and your response was "nope, I
gave the name of a fallacy, therefore it applies" without explaining why
I was wrong.
I guess that means you'd be OK with me just naming a fallacy when you
say something I don't like and you'll just accept it regardless of what
you have to say about it.
talk about brain rot
That makes it an input.
Which means that when und_cir_sr calls sim_cir_sr(und_cir_sr) after >>>>> calling Tdp, it's not simulating itself.
That also means the agent is disqualified from being a halt
decider/ recognizer, partial or otherwise.
So your abstraction is broken.
ofc the code does that
lol, turing machines don't have "global variables" dud, >>>>>>>>>>>>>>>> they just have a tape, and if the value is output to the >>>>>>>>>>>>>>>> tape anywhere, it can be picked out by a simulation and >>>>>>>>>>>>>>>> contradicted by a paradox
To fix it, the fixed steps the human agent runs need to >>>>>>>>>>>>>>>>> be converted to a function which read/writes a global >>>>>>>>>>>>>>>>> variable, and that global variable need to be set to >>>>>>>>>>>>>>>>> the same value it had when und_cir_sr was first called. >>>>>>>>>>>>>>>>
Strawman.-a "it can be picked out" means you're talking >>>>>>>>>>>>>>> about changing the code which means you're no longer >>>>>>>>>>>>>>> talking about the same machine.
that's incorrect. reading values from the tape during a >>>>>>>>>>>>>> step of the computation does not change the machine which >>>>>>>>>>>>>> is being simulated ...
No, but the code you've shown doesn't do that.-a So if you >>>>>>>>>>>>> change the code to do that, it's no longer the same machine. >>>>>>>>>>>>
Your code is reading the part of the tape where the
mechanized human agent is writing its side record?
No?
Then your code doesn't do that, and talking about what it >>>>>>>>>>> "can" do necessarily means changing the code to something not >>>>>>>>>>> being decided on.
u really don't get it: claiming a machine to be a total
decider is subject to the full enumerations of machines,
meaning if an input can exist, it will exist
That has nothing to do with the fact that you can't talk about >>>>>>>>> changing code and still claim it's the same machine.
no one is changing the goal post as the goal post of a decider >>>>>>>> is handling _all_ possible input
*A* decider.-a There is no "can" or "can not", only "does" or
"does not".
there exists a machine that does a paradox for any location you
can write a decision to the tape
You "can" write means changing the code of the decider, which is
not allowed.-a Otherwise you're not talking about the same decider.
for any location on the tape that a decider does attempt to write a
decision, of which there are infinite locations on the tape, and
infinite deciders writing to any location on the tape,
there does exist infinite paradoxical machines which will read from
that specific bit and produce a paradox for it in regards to the
written decision
there is _no_ way to produce a total halting decider for turing
machines, within turing machines
Which is basically Turing's proof.
so why are you trying to argue u can simulate the agent's
computation ... u can't do so within turing machines
If the agent is using *only* its fixed sequence of steps and a
description of the algorithm to decide on to make a decision, which is
what you now seem to be claiming, it can be exactly replicated.
1) the agent does not read from the side record
Turing machines are defined by their instructions, not where they
physically reside.
there exists a paradox that contradicts any possible location >>>>>>>>>> you could write a decision bit on a tape, and therefore there >>>>>>>>>> will exist a machine that defies the decision, no matter where >>>>>>>>>> u try to write it
it is _not_ possible to simulate the side result created by >>>>>>>>>> the agent running a total decision algorithm.
1) you can't read the side result because your abstraction is >>>>>>>>> broken
2) the human agent isn't a halt decider, partial or otherwise, >>>>>>>>> because it takes an input that disqualifies it
No response to this?
neither of those are correct: false dichotomy
That wasn't an either/or.-a Both apply.
2) the agent does not utilize information from the side record while
computing the decision for any given input machine
neither apply
I don't believe you.
If the agent *never* uses the side record, that makes the side record
entirely irrelevant, and it means that any machine can use the agent's
it's purpose is to demonstrate the existence of the computation, if u
don't care for the demonstration, ur free to remain in error
The existence of the computation is in the agent's algorithm which is
fixed.
algorithm to exactly reproduce what the agent does if it doesn't use the >>u keep asserting that, but you have not yet once told me how such a
simulation would handle it's own undecidability paradox: und_cir_sr
side data.-a You wouldn't have added a whole section to your paper
about the side record and made it the focus of your "refutation" if
the agent didn't use it.
turing machines don't utilize their own output to justify their
existence... they just output it. why would an agent need to?
i wonder if ur about to commit fallacy #7: special pleading
You're backtracking because you've been caught in a fundamental
mistake and won't admit it.
it is _not_ even possible to create a total decision algorithm >>>>>>>>>> with turing machines, like seriously have u forgotten the >>>>>>>>>> point of turing's proof???
dud if ur not here to effectively compute useful knowledge, >>>>>>>>>> idk what ur doing discussing the theory of computing.
... getting "output" from a simulation requires reading the >>>>>>>>>>>> output from tape of the simulated machine, it can do that >>>>>>>>>>>> for any value, even ones that are specified to be "output" >>>>>>>>>>>>
and if a machine is simulating itself,
Which your code isn't doing because the side data isn't the >>>>>>>>>>>>> same when starting the simulation as it was when the >>>>>>>>>>>>> machine was invoked.
idk what u mean by "side data" there only data on the tape >>>>>>>>>>>> when it comes to turing machines, there is nothing on the >>>>>>>>>>>> side of data on the tape.
I'm referring specifically to what the mechanized version of >>>>>>>>>>> the human agent writes to the tape.
In other words, it's no different from *any* partial halt >>>>>>>>>>> decider. That's not even mentioning ill-defined weasel words >>>>>>>>>>> like "useful", "happens to", "trustworthy", and "judged". >>>>>>>>>>
like in the case an undecidability paradox within >>>>>>>>>>>>>> computing, then this can be used to pick out values behind >>>>>>>>>>>>>> any form of data encapsulation that you might suggest to >>>>>>>>>>>>>> hide data away from a paradox. there's no where to hide >>>>>>>>>>>>>> output dud. there's no "global variable" that might >>>>>>>>>>>>>> prevent a paradox from being formed
"can be used" meaning it's not being used now, and
therefore irrelevant.
So a paradoxical machine is one that a decider gets wrong? >>>>>>>>>>>>> So this machine:
there is _no_ way to simulate a general ability to decide >>>>>>>>>>>>>> on paradoxical turing machines, from within turing machines >>>>>>>>>>>>>
void foo() { return; }
Is a paridoxical machine to this one:
int bar(void *p) { return 0; }
Because machine bar can't successfully report the status of >>>>>>>>>>>>> machine foo. -a-aYes?
no. first, we went over this dud: constant return function >>>>>>>>>>>> are not and will never qualify as a halting decider as they >>>>>>>>>>>> do not output useful information in regards to the halting >>>>>>>>>>>> status Efn+. the fact a machine happens to output the correct >>>>>>>>>>>> answer does not qualify them as a halting decider, a decider >>>>>>>>>>>> must output trustworthy information which is judged across >>>>>>>>>>>> the entirety of the output space, not just single instances >>>>>>>>>>>
let me not mince words. the existing theory on the matter of >>>>>>>>>> deciders is thus:
- a decider is not allowed a false positive or false negative, >>>>>>>>>> and never allowed to diverge. we cannot build any non-trivial >>>>>>>>>> decider (within turing machines)
a constant return 1 function, if use as a halting decider is >>>>>>>>>> abound in false positives and therefore is completely
disqualified as being a halting decider.
- a recognizer is also not allowed false positives or false >>>>>>>>>> negatives, it is however allowed to diverge on some negatives, >>>>>>>>>> but never a positive. a halting recognizer is allowed to
diverge on some non- halting input, but never halting input. >>>>>>>>>> this is sometimes called a partial decider, but i'm going to >>>>>>>>>> label as it recognizer to align the rather well known sipser >>>>>>>>>> on this, and because i have further definitions to propose. we >>>>>>>>>> can build a halting recognizer, but not a non-halting recognizer. >>>>>>>>>>
a constant return 1 function does not satisfy this for halting >>>>>>>>>> either, as again it's abound in false positives.
in my paper i propose two more types:
- a partial decider is still _not_ allowed false positive or >>>>>>>>>> negatives. but it is allowed to diverge on some positives and >>>>>>>>>> some negatives. we can be build withing turing machines both >>>>>>>>>> halting and non-halting partial deciders
a constant return 1 function still does not satisfy this for >>>>>>>>>> halting either
- a partial recognizer is still not allowed false positives, >>>>>>>>>> but _does_ allow false negative ... but _only_ in the case >>>>>>>>>> where returning positive would be a false positive. this
exception is allowed because a partial recognizer therefore >>>>>>>>>> does not need to diverge, ever, and always halts. this can >>>>>>>>>> also be built within turing machines for the halting and non- >>>>>>>>>> halting problems.
a constant return 1 function again still does not satisfy >>>>>>>>>> these requirements for halting
i just defined my terms here quite clearly. i suppose i'll thank
you for encouraging me to develop that, i'll may add it to my paper >>>>>>
anyways, that was all covered in section -o5.2 which u never read >>>>>>>>>>
i really don't know why i'm humoring you dud
second, an actual paradox is necessarily undecidable due to >>>>>>>>>>>> the construction of the machine itself, the form is
generalized in -o3 of my paper. if it doesn't fit that form, >>>>>>>>>>>> it's not a paradox, and is not undecidable in respect to any >>>>>>>>>>>> classifier let alone in totality
Alright, let's satisfy your arbitrary requirement of a non- >>>>>>>>>>> constant function and "useful":
void foo() {
-a-a-a-a puts("hello");
}
int bar(void *p) {
-a-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a-a return 1;
-a-a-a-a } else {
-a-a-a-a-a-a-a-a return 0;
-a-a-a-a }
}
Given that the first instruction of "foo" is not a "return" >>>>>>>>>>> instruction, bar(foo) will return 0 even though foo() will halt. >>>>>>>>>>>
Is the machine foo "undecidable" by machine bar?
No response to this?-a It goes directly to your definition of >>>>>>>>> "undecidable" which seems core to your argument.
bar isn't a halting decider in the first place, no idea what
it's supposed to be
As noted above, it's a partial halt decider that reports halting >>>>>>> if the first instruction is a "return" (0xc3 is the x86
instruction "ret") and reports non-halting otherwise.
Again, is the machine foo "undecidable" by machine bar?
It would be a shame to discard your whole argument just because >>>>>>> your terms aren't well defined.
a partial halting decider is not allowed _any_ false positives or >>>>>> false negatives, it must handle the entire input space of machines >>>>>> without a _single_ false positive or false negative. that's not
even my assertion, that's the established consensus on the matter
clearly bar would have a ton of false negatives and thereby be
disqualified as a partial halting decider
Alright, let's satisfy one more arbitrary requirement to see if we
can get you to define your terms:
void foo() {
-a-a-a-a puts("hello");
}
int bar(void *p) {
-a-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a-a return 1;
-a-a-a-a } else {
-a-a-a-a-a-a-a-a while (1);
-a-a-a-a }
}
That takes care of the false negatives.-a So no more dodging the point: >>>>>
Is the machine foo "undecidable" by machine bar?
Failure to answer again will deem your term "undecidable" ill-
defined, and consequently the rest of your paper.
if i'm reading this correctly, bar now diverges (does not return) on
the halting input foo
i will clarify the definition for "partial decider": divergence is
only allowed when both positive and negative returns would be false
positive or false negative respectively (aka undecidable input to
the partial decider), so this would not quality as a partial halting
decider
So yet again you dodge the question on whether foo is "undecidable"
by machine bar.
bar is _not_ halting decider, partial or otherwise, so asking if the
halting semantics of foo is "undecidable" to bar is a category fallacy
wowee fallacy #8
Nope, you committed the fallacy of moving the goalposts after I provided something that satisfied your arbitrary requirements in order to dodge
the question.
The exact code placed in function bar is irrelevant to the question.
That make that term ill-defined.
the term is quite well defined: if a classifier/decider is stuck
between returning a false positive and a false negative for a
particular input, then that input is undecidable to the classifier/
decider
Category error.-a There is no "stuck between" doing two things.-a A classifier/decider does one thing and one thing only: what its fixed instructions tell it to do.
Turing machines are defined by their instruction, not by where
instructions physically reside.
And not understanding that is your core mistake.-aThe same mistake made
by PO.
And because you whole paper is based on the use of that word, it is
reduced to meaninglessness.
a fallacy laden critique full of cognitive distortion and abuse is
never going to convince me dud
Without an answer, this deems your use of the term unclear, and >>>>>>>>> therefore makes your entire argument null and void.
On 2026-09-24 18:44, Mike Terry wrote:
On 25/09/2026 00:19, Andr|- G. Isaak wrote:
On 2026-09-24 14:09, dart200 wrote:I don't know - what you're describing seems to already have a term:
a partial halting decider is not allowed _any_ false positives or
false negatives, it must handle the entire input space of machines
without a _single_ false positive or false negative.
You're very confused here. If it is not allowed any false negatives
or false positives, then you're talking about a /total/ halt decider.
A /partial/ halt decider is something which correctly decides /some/
instances of the halting problem but which fails to correctly
decide /all/ instances. That's why that pesky word 'partial' is added.
it's a "decider".-a (But just not a "halt decider" since it gets some
inputs wrong by the halting criterion.)
The way "partial halt decider" [PHD] has typically been used /on this
newsgroup/ is to indicate a TM with three final "outcomes":
--a HALTS-a-a-a-a-a-a-a [input halts]
--a NEVERHALTS-a-a-a-a-a-a-a [input never halts]
--a PASS-a-a-a-a-a-a-a-a-a-a-a ["pass" - the PHD makes no halting decision] >>
So the PHD may always PASS for a given input, but when it does
indicate HALTS/NEVERHALTS that must /correctly/ match the input's
halting behaviour.
Quite a few posters over the years have casually suggested that "PHD"
is a well known term meaning [whatever - generally something like my
definition, rather than yours...] but I can't say I've seen any use of
the term in books I've used!-a Is it /really/ a standard term???-a If
"PHD" is more of a comp.theory common term, then posters should define
their usage for the term when they use it.
There are variations in how it might reasonably be defined, e.g. I
suspect requiring PASS to be an actual "decision" by the PHD [e.g.
transition to a 3rd TM halt state] is substantially different from
allowing a PHD to implicitly PASS by virtue of never halting.-a Since
this NG has often discussed concrete programs that run until they
detect a definite HALTS or NEVERHALTS pattern, my suggestion is to
allow PHDs to include programs that may never halt...-a [so PO's code
would then be classed a PHD if its bugs/logic-errors were fixed.]
I think you may be correct that this term really is one specific to this newsgroup rather than a term in wide usage.
However, the usage I gave seems consistent with what dbush intends by
the term since he has used both the universal acceptor and the universal rejector as examples of partial halt deciders.
The usage you suggest would certainly describe something more useful
than my own, but it's not clear to me that that is the usage which
others have been using on this group.
Andr|-
On 9/25/26 5:16 AM, dbush wrote:
On 9/25/2026 12:13 AM, dart200 wrote:
On 9/24/26 6:00 PM, dbush wrote:
On 9/24/2026 7:13 PM, dart200 wrote:
On 9/24/26 1:54 PM, dbush wrote:Are you sure?
On 9/24/2026 4:09 PM, dart200 wrote:
On 9/24/26 11:23 AM, dbush wrote:
On 9/24/2026 2:03 PM, dart200 wrote:
On 9/24/26 5:37 AM, dbush wrote:
On 9/24/2026 5:38 AM, dart200 wrote:
On 9/23/26 7:08 PM, dbush wrote:
On 9/23/2026 9:35 PM, dart200 wrote:
On 9/23/26 12:43 PM, dbush wrote:So what exactly are the inputs to the human agent?
On 9/23/2026 3:17 PM, dart200 wrote:what rule??? lol deterministic only means running it always >>>>>>>>>>>>> has the same result given the same input ... which a total >>>>>>>>>>>>> decision algo run by the idealized agent certainly does >>>>>>>>>>>>
On 9/23/26 5:17 AM, dbush wrote:
So what you really did is break the rules of a >>>>>>>>>>>>>>>> deterministic algorithm by using the side record and by >>>>>>>>>>>>>>>> not simulating what you though you were simulating. >>>>>>>>>>>>>
the agent gets the machine description like any decision >>>>>>>>>>> algo, that would be clear if u just read the paper
So you don't consider the side record as part of the human's >>>>>>>>>> input. That makes it a non-deterministic algorithm, and
therefore irrelevant to church-turning.
As I said, you have a fundamental misunderstanding of the >>>>>>>>>> problem, and you just demonstrated it quite clearly.
that ur quite sure of, even if the reason keep changing
I see you made no attempt to refute that you have a non-
deterministic algorithm that is therefore not applicable to
church- turning.-a Do you agree?
Still no response?
to clearly demonstrate that the sequence is produced, to
the side-record is nothing more than the aggregated output of >>>>>>>>>>> the agent decision algo run across all turing machines. any >>>>>>>>>>> particular n- th record does not influence any future records >>>>>>>>>>> beyond n. or before n for that matter
You seem to think that the human's side record is not part >>>>>>>>>>>> of the input. -a-aIf so, that makes it non-deterministic. >>>>>>>>>>>
the algorithm is detailed in -o7.6 ... which u still >>>>>>>>>>>>>>> haven't even opened
No deterministic algorithm, no refutation of church-turing >>>>>>>>>>>>>>>
Not a deterministic algorithm, as I previously stated and >>>>>>>>>>>>>> which you made no attempt to refute, and therefore >>>>>>>>>>>>>> irrelevant to church- turning
u haven't even read the algo, so how are you going to >>>>>>>>>>>>> explain why specifically it's non-deterministic ... ??? >>>>>>>>>>>>
And if the human later reads it, that makes it an input.
i'm sorry, please explain to me why you think the agent needs >>>>>>>>> to read from the side record after writing to it???
If the agent never reads the record, why does it need to write it? >>>>>>>
demonstrate that the ability to objectively decide on any given >>>>>>> turing machine is not confounded by the fact no turing machine
itself can be a total machine decider
So it's just a log?
it is a computed sequence,
Meaning that if the side record doesn't exist the agent will
behave exactly the same?
yes beside the process of writing computed results to the side record >>>>
7.3 "It is these recorded side effects that the idealized human
agent can exploit"
That implies the agent does in fact read the side record.
lol are you trying to educate me about the paper i wrote??? yes i'm
sure, as there is literally no need for the agent to read from the
side record, as no halting decision for any turing machine requires
doing so ... SO WHY WOULD HE???
Then anything the agent does can be accurately simulated by a turing
machine.
do you understand that repeating the unproven ct-thesis at me is not a proof?
the record is there to formalize the notion of computing a sequence,
not because it's used for later computations. and not because we
would actually want to compute the total halting set, that would be
as bad as busy beaver, so far beyond practical. it's to prove that
such computation exist in theory, so we know that proving so is
possible for any given machine
So nothing more than a log, and therefore irrelevant.
it's information that is written down and therefore definitively
computed.
if that isn't meaningful to you, then perhaps ur not really
that interested in the theory of computing
do u honestly think this is a genuine line of inquiry?
So choose how you're wrong:
- The side record is not an input, so you have a non-
determinstic algorithm that is irrelevant to church-turning >>>>>>>>>> - The side record is an input, and your abstraction is broken >>>>>>>>>> with your simulator not actually simulating itself.
No response to this?
nothing written to the side record is used further in the
computation, so the agent never needs to read from it again
"Again".-a So it does read from the side record.
word-focus fallacy
Now you're just making things up.
You wouldn't have said "again" if the human agent *never* read from the >>>
Yes, as it means that either 1) you're trying to hide something, or 2)
you have a very poor grasp of English.-a Both of which bring everything
you say into question.
i won't respond to abusive belittling
side record.-a In fact, you wouldn't have a whole section on it in
your paper, otherwise anything in it could simply be derived from
the agent's fixed algorithm.
You've basically just confirmed that you're lying about what the
agent does because you know you made a major fundamental mistake and
won't admit it.
bro unless u can tell me why the agent reads from the side record in
the future, then he does not, cause he doesn't need to, and this line
of trying to refute my paper is a failure
now we're at 4 named fallacies, 1 named cognitive distortion, and 1 >>>>> named abuse tactic
Just because you gave the name of a fallacy doesn't mean it applies.
just because u can't identify forms of fallacies does not negate
their application. let's add #5 and #6 for ur latest post: bad-faith
interpretation + strawman
I explained why some of them don't apply and your response was "nope,
I gave the name of a fallacy, therefore it applies" without explaining
why I was wrong.
I guess that means you'd be OK with me just naming a fallacy when you
say something I don't like and you'll just accept it regardless of
what you have to say about it.
it would be incredible if u named even just one
talk about brain rot
for any location on the tape that a decider does attempt to write a >>>>> decision, of which there are infinite locations on the tape, and
That makes it an input.
Which means that when und_cir_sr calls sim_cir_sr(und_cir_sr)
after calling Tdp, it's not simulating itself.
That also means the agent is disqualified from being a halt
decider/ recognizer, partial or otherwise.
So your abstraction is broken.
ofc the code does that
To fix it, the fixed steps the human agent runs need >>>>>>>>>>>>>>>>>> to be converted to a function which read/writes a >>>>>>>>>>>>>>>>>> global variable, and that global variable need to be >>>>>>>>>>>>>>>>>> set to the same value it had when und_cir_sr was first >>>>>>>>>>>>>>>>>> called.
lol, turing machines don't have "global variables" dud, >>>>>>>>>>>>>>>>> they just have a tape, and if the value is output to >>>>>>>>>>>>>>>>> the tape anywhere, it can be picked out by a simulation >>>>>>>>>>>>>>>>> and contradicted by a paradox
Strawman.-a "it can be picked out" means you're talking >>>>>>>>>>>>>>>> about changing the code which means you're no longer >>>>>>>>>>>>>>>> talking about the same machine.
that's incorrect. reading values from the tape during a >>>>>>>>>>>>>>> step of the computation does not change the machine which >>>>>>>>>>>>>>> is being simulated ...
No, but the code you've shown doesn't do that.-a So if you >>>>>>>>>>>>>> change the code to do that, it's no longer the same machine. >>>>>>>>>>>>>
Your code is reading the part of the tape where the
mechanized human agent is writing its side record?
No?
Then your code doesn't do that, and talking about what it >>>>>>>>>>>> "can" do necessarily means changing the code to something >>>>>>>>>>>> not being decided on.
u really don't get it: claiming a machine to be a total >>>>>>>>>>> decider is subject to the full enumerations of machines, >>>>>>>>>>> meaning if an input can exist, it will exist
That has nothing to do with the fact that you can't talk about >>>>>>>>>> changing code and still claim it's the same machine.
no one is changing the goal post as the goal post of a decider >>>>>>>>> is handling _all_ possible input
*A* decider.-a There is no "can" or "can not", only "does" or >>>>>>>> "does not".
there exists a machine that does a paradox for any location you >>>>>>> can write a decision to the tape
You "can" write means changing the code of the decider, which is
not allowed.-a Otherwise you're not talking about the same decider. >>>>>
infinite deciders writing to any location on the tape,
there does exist infinite paradoxical machines which will read from >>>>> that specific bit and produce a paradox for it in regards to the
written decision
there is _no_ way to produce a total halting decider for turing
machines, within turing machines
Which is basically Turing's proof.
so why are you trying to argue u can simulate the agent's
computation ... u can't do so within turing machines
If the agent is using *only* its fixed sequence of steps and a
description of the algorithm to decide on to make a decision, which is
what you now seem to be claiming, it can be exactly replicated.
again just continually begging the question that the ct-thesis is true.
u don't know have to proof of the assertion that all of realizable
computing is encapsulated within turing machines,
and u do not seem to be able to understand that u do not
1) the agent does not read from the side record
Turing machines are defined by their instructions, not where they >>>>>> physically reside.
there exists a paradox that contradicts any possible location >>>>>>>>>>> you could write a decision bit on a tape, and therefore there >>>>>>>>>>> will exist a machine that defies the decision, no matter >>>>>>>>>>> where u try to write it
it is _not_ possible to simulate the side result created by >>>>>>>>>>> the agent running a total decision algorithm.
1) you can't read the side result because your abstraction is >>>>>>>>>> broken
2) the human agent isn't a halt decider, partial or otherwise, >>>>>>>>>> because it takes an input that disqualifies it
No response to this?
neither of those are correct: false dichotomy
That wasn't an either/or.-a Both apply.
2) the agent does not utilize information from the side record
while computing the decision for any given input machine
neither apply
I don't believe you.
If the agent *never* uses the side record, that makes the side
record entirely irrelevant, and it means that any machine can use
the agent's
it's purpose is to demonstrate the existence of the computation, if u
don't care for the demonstration, ur free to remain in error
The existence of the computation is in the agent's algorithm which is
fixed.
algorithm to exactly reproduce what the agent does if it doesn't use
the
u keep asserting that, but you have not yet once told me how such a
simulation would handle it's own undecidability paradox: und_cir_sr
side data.-a You wouldn't have added a whole section to your paper
about the side record and made it the focus of your "refutation" if
the agent didn't use it.
turing machines don't utilize their own output to justify their
existence... they just output it. why would an agent need to?
i wonder if ur about to commit fallacy #7: special pleading
You're backtracking because you've been caught in a fundamental
mistake and won't admit it.
it is _not_ even possible to create a total decision
algorithm with turing machines, like seriously have u
forgotten the point of turing's proof???
dud if ur not here to effectively compute useful knowledge, >>>>>>>>>>> idk what ur doing discussing the theory of computing.
... getting "output" from a simulation requires reading the >>>>>>>>>>>>> output from tape of the simulated machine, it can do that >>>>>>>>>>>>> for any value, even ones that are specified to be "output" >>>>>>>>>>>>>
and if a machine is simulating itself,
Which your code isn't doing because the side data isn't >>>>>>>>>>>>>> the same when starting the simulation as it was when the >>>>>>>>>>>>>> machine was invoked.
idk what u mean by "side data" there only data on the tape >>>>>>>>>>>>> when it comes to turing machines, there is nothing on the >>>>>>>>>>>>> side of data on the tape.
I'm referring specifically to what the mechanized version of >>>>>>>>>>>> the human agent writes to the tape.
like in the case an undecidability paradox within >>>>>>>>>>>>>>> computing, then this can be used to pick out values >>>>>>>>>>>>>>> behind any form of data encapsulation that you might >>>>>>>>>>>>>>> suggest to hide data away from a paradox. there's no >>>>>>>>>>>>>>> where to hide output dud. there's no "global variable" >>>>>>>>>>>>>>> that might prevent a paradox from being formed
"can be used" meaning it's not being used now, and >>>>>>>>>>>>>> therefore irrelevant.
So a paradoxical machine is one that a decider gets wrong? >>>>>>>>>>>>>> So this machine:
there is _no_ way to simulate a general ability to decide >>>>>>>>>>>>>>> on paradoxical turing machines, from within turing machines >>>>>>>>>>>>>>
void foo() { return; }
Is a paridoxical machine to this one:
int bar(void *p) { return 0; }
Because machine bar can't successfully report the status >>>>>>>>>>>>>> of machine foo. -a-aYes?
no. first, we went over this dud: constant return function >>>>>>>>>>>>> are not and will never qualify as a halting decider as they >>>>>>>>>>>>> do not output useful information in regards to the halting >>>>>>>>>>>>> status Efn+. the fact a machine happens to output the correct >>>>>>>>>>>>> answer does not qualify them as a halting decider, a >>>>>>>>>>>>> decider must output trustworthy information which is judged >>>>>>>>>>>>> across the entirety of the output space, not just single >>>>>>>>>>>>> instances
In other words, it's no different from *any* partial halt >>>>>>>>>>>> decider. That's not even mentioning ill-defined weasel words >>>>>>>>>>>> like "useful", "happens to", "trustworthy", and "judged". >>>>>>>>>>>
let me not mince words. the existing theory on the matter of >>>>>>>>>>> deciders is thus:
- a decider is not allowed a false positive or false
negative, and never allowed to diverge. we cannot build any >>>>>>>>>>> non-trivial decider (within turing machines)
a constant return 1 function, if use as a halting decider is >>>>>>>>>>> abound in false positives and therefore is completely
disqualified as being a halting decider.
- a recognizer is also not allowed false positives or false >>>>>>>>>>> negatives, it is however allowed to diverge on some
negatives, but never a positive. a halting recognizer is >>>>>>>>>>> allowed to diverge on some non- halting input, but never >>>>>>>>>>> halting input. this is sometimes called a partial decider, >>>>>>>>>>> but i'm going to label as it recognizer to align the rather >>>>>>>>>>> well known sipser on this, and because i have further
definitions to propose. we can build a halting recognizer, >>>>>>>>>>> but not a non-halting recognizer.
a constant return 1 function does not satisfy this for
halting either, as again it's abound in false positives. >>>>>>>>>>>
in my paper i propose two more types:
- a partial decider is still _not_ allowed false positive or >>>>>>>>>>> negatives. but it is allowed to diverge on some positives and >>>>>>>>>>> some negatives. we can be build withing turing machines both >>>>>>>>>>> halting and non-halting partial deciders
a constant return 1 function still does not satisfy this for >>>>>>>>>>> halting either
- a partial recognizer is still not allowed false positives, >>>>>>>>>>> but _does_ allow false negative ... but _only_ in the case >>>>>>>>>>> where returning positive would be a false positive. this >>>>>>>>>>> exception is allowed because a partial recognizer therefore >>>>>>>>>>> does not need to diverge, ever, and always halts. this can >>>>>>>>>>> also be built within turing machines for the halting and non- >>>>>>>>>>> halting problems.
a constant return 1 function again still does not satisfy >>>>>>>>>>> these requirements for halting
i just defined my terms here quite clearly. i suppose i'll thank >>>>>>> you for encouraging me to develop that, i'll may add it to my paper >>>>>>>
anyways, that was all covered in section -o5.2 which u never read >>>>>>>>>>>
i really don't know why i'm humoring you dud
second, an actual paradox is necessarily undecidable due to >>>>>>>>>>>>> the construction of the machine itself, the form is >>>>>>>>>>>>> generalized in -o3 of my paper. if it doesn't fit that form, >>>>>>>>>>>>> it's not a paradox, and is not undecidable in respect to >>>>>>>>>>>>> any classifier let alone in totality
Alright, let's satisfy your arbitrary requirement of a non- >>>>>>>>>>>> constant function and "useful":
void foo() {
-a-a-a-a puts("hello");
}
int bar(void *p) {
-a-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a-a return 1;
-a-a-a-a } else {
-a-a-a-a-a-a-a-a return 0;
-a-a-a-a }
}
Given that the first instruction of "foo" is not a "return" >>>>>>>>>>>> instruction, bar(foo) will return 0 even though foo() will >>>>>>>>>>>> halt.
Is the machine foo "undecidable" by machine bar?
No response to this?-a It goes directly to your definition of >>>>>>>>>> "undecidable" which seems core to your argument.
bar isn't a halting decider in the first place, no idea what >>>>>>>>> it's supposed to be
As noted above, it's a partial halt decider that reports halting >>>>>>>> if the first instruction is a "return" (0xc3 is the x86
instruction "ret") and reports non-halting otherwise.
Again, is the machine foo "undecidable" by machine bar?
It would be a shame to discard your whole argument just because >>>>>>>> your terms aren't well defined.
a partial halting decider is not allowed _any_ false positives or >>>>>>> false negatives, it must handle the entire input space of
machines without a _single_ false positive or false negative.
that's not even my assertion, that's the established consensus on >>>>>>> the matter
clearly bar would have a ton of false negatives and thereby be
disqualified as a partial halting decider
Alright, let's satisfy one more arbitrary requirement to see if we >>>>>> can get you to define your terms:
void foo() {
-a-a-a-a puts("hello");
}
int bar(void *p) {
-a-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a-a return 1;
-a-a-a-a } else {
-a-a-a-a-a-a-a-a while (1);
-a-a-a-a }
}
That takes care of the false negatives.-a So no more dodging the
point:
Is the machine foo "undecidable" by machine bar?
Failure to answer again will deem your term "undecidable" ill-
defined, and consequently the rest of your paper.
if i'm reading this correctly, bar now diverges (does not return)
on the halting input foo
i will clarify the definition for "partial decider": divergence is
only allowed when both positive and negative returns would be false >>>>> positive or false negative respectively (aka undecidable input to
the partial decider), so this would not quality as a partial
halting decider
So yet again you dodge the question on whether foo is "undecidable"
by machine bar.
bar is _not_ halting decider, partial or otherwise, so asking if the
halting semantics of foo is "undecidable" to bar is a category fallacy
wowee fallacy #8
Nope, you committed the fallacy of moving the goalposts after I
provided something that satisfied your arbitrary requirements in order
to dodge the question.
well if it isn't a decider than asking if an input is undecidable to it
is a category fallacy, no???
and ur just completely misinformed about conventional computing theory.
conventional computing theory only acknowledges total deciders, and recognizors, both of which _must_ return true for all positive input
without producing _any_ false positives. recognizers may diverge on some false input, but neither are allowed false negatives. that's the
established theory on the matter and if u disagree take it up with sipser.
i'm proposing two more, which still are _not_ allowed to produce false positives. partial deciders are allowed to diverge on some positive
input, whereas partial recognizer are allowed to produce false negatives (and never diverge),
and both only in the case where the input is
paradoxical (either return positive/negative would be false)
The exact code placed in function bar is irrelevant to the question.
there is no basis for this claim in conventional computing theory, nor
is what i'm proposing. classifiers for a particular semantic property
are not just random-ass algos that happens to output correctly for some machines, there are specific interface constraints they _must_ meet
That make that term ill-defined.
the term is quite well defined: if a classifier/decider is stuck
between returning a false positive and a false negative for a
particular input, then that input is undecidable to the classifier/
decider
Category error.-a There is no "stuck between" doing two things.-a A
classifier/decider does one thing and one thing only: what its fixed
instructions tell it to do.
handling input where responding either positive or negative would be
false
is a real input scenario that must be handled by any actual
machine attempting to accurately decide on other machines. i'm not describing what the machine is literally doing, i'm describing an
objective mathematical reality of the decider's output space, when faced with paradoxical input
Turing machines are defined by their instruction, not by where
instructions physically reside.
And not understanding that is your core mistake.-aThe same mistake made
by PO.
And because you whole paper is based on the use of that word, it is
reduced to meaninglessness.
a fallacy laden critique full of cognitive distortion and abuse is
never going to convince me dud
Without an answer, this deems your use of the term unclear, >>>>>>>>>> and therefore makes your entire argument null and void.
On 9/25/2026 2:19 PM, dart200 wrote:
On 9/25/26 5:16 AM, dbush wrote:
On 9/25/2026 12:13 AM, dart200 wrote:
On 9/24/26 6:00 PM, dbush wrote:
On 9/24/2026 7:13 PM, dart200 wrote:
On 9/24/26 1:54 PM, dbush wrote:Are you sure?
On 9/24/2026 4:09 PM, dart200 wrote:
On 9/24/26 11:23 AM, dbush wrote:
On 9/24/2026 2:03 PM, dart200 wrote:
On 9/24/26 5:37 AM, dbush wrote:
On 9/24/2026 5:38 AM, dart200 wrote:
On 9/23/26 7:08 PM, dbush wrote:
On 9/23/2026 9:35 PM, dart200 wrote:
On 9/23/26 12:43 PM, dbush wrote:
On 9/23/2026 3:17 PM, dart200 wrote:what rule??? lol deterministic only means running it >>>>>>>>>>>>>> always has the same result given the same input ... which >>>>>>>>>>>>>> a total decision algo run by the idealized agent certainly >>>>>>>>>>>>>> does
On 9/23/26 5:17 AM, dbush wrote:
So what you really did is break the rules of a >>>>>>>>>>>>>>>>> deterministic algorithm by using the side record and by >>>>>>>>>>>>>>>>> not simulating what you though you were simulating. >>>>>>>>>>>>>>
So what exactly are the inputs to the human agent?
the agent gets the machine description like any decision >>>>>>>>>>>> algo, that would be clear if u just read the paper
So you don't consider the side record as part of the human's >>>>>>>>>>> input. That makes it a non-deterministic algorithm, and >>>>>>>>>>> therefore irrelevant to church-turning.
As I said, you have a fundamental misunderstanding of the >>>>>>>>>>> problem, and you just demonstrated it quite clearly.
that ur quite sure of, even if the reason keep changing
I see you made no attempt to refute that you have a non-
deterministic algorithm that is therefore not applicable to >>>>>>>>> church- turning.-a Do you agree?
Still no response?
to clearly demonstrate that the sequence is produced, to
the side-record is nothing more than the aggregated output >>>>>>>>>>>> of the agent decision algo run across all turing machines. >>>>>>>>>>>> any particular n- th record does not influence any future >>>>>>>>>>>> records beyond n. or before n for that matter
You seem to think that the human's side record is not part >>>>>>>>>>>>> of the input. -a-aIf so, that makes it non-deterministic. >>>>>>>>>>>>
the algorithm is detailed in -o7.6 ... which u still >>>>>>>>>>>>>>>> haven't even opened
No deterministic algorithm, no refutation of church-turing >>>>>>>>>>>>>>>>
Not a deterministic algorithm, as I previously stated and >>>>>>>>>>>>>>> which you made no attempt to refute, and therefore >>>>>>>>>>>>>>> irrelevant to church- turning
u haven't even read the algo, so how are you going to >>>>>>>>>>>>>> explain why specifically it's non-deterministic ... ??? >>>>>>>>>>>>>
And if the human later reads it, that makes it an input.
i'm sorry, please explain to me why you think the agent needs >>>>>>>>>> to read from the side record after writing to it???
If the agent never reads the record, why does it need to write it? >>>>>>>>
demonstrate that the ability to objectively decide on any given >>>>>>>> turing machine is not confounded by the fact no turing machine >>>>>>>> itself can be a total machine decider
So it's just a log?
it is a computed sequence,
Meaning that if the side record doesn't exist the agent will
behave exactly the same?
yes beside the process of writing computed results to the side record >>>>>
7.3 "It is these recorded side effects that the idealized human
agent can exploit"
That implies the agent does in fact read the side record.
lol are you trying to educate me about the paper i wrote??? yes i'm
sure, as there is literally no need for the agent to read from the
side record, as no halting decision for any turing machine requires
doing so ... SO WHY WOULD HE???
Then anything the agent does can be accurately simulated by a turing
machine.
do you understand that repeating the unproven ct-thesis at me is not a
proof?
Do you understand that not showing a concrete example of a deterministic algorithm that a turing machine can't do is not a refutation?
the record is there to formalize the notion of computing a sequence,
not because it's used for later computations. and not because we
would actually want to compute the total halting set, that would be
as bad as busy beaver, so far beyond practical. it's to prove that
such computation exist in theory, so we know that proving so is
possible for any given machine
So nothing more than a log, and therefore irrelevant.
it's information that is written down and therefore definitively
computed.
And if a human executes a fixed deterministic sequence of steps using
only a machine description as input, a machine can run those same exact steps using a machine description as input and subsequently write down
that same information.
if that isn't meaningful to you, then perhaps ur not really that
interested in the theory of computing
So choose how you're wrong:
- The side record is not an input, so you have a non-
determinstic algorithm that is irrelevant to church-turning >>>>>>>>>>> - The side record is an input, and your abstraction is broken >>>>>>>>>>> with your simulator not actually simulating itself.
No response to this?
nothing written to the side record is used further in the
computation, so the agent never needs to read from it again
"Again".-a So it does read from the side record.
word-focus fallacy
Now you're just making things up.
You wouldn't have said "again" if the human agent *never* read from >>>>> the
do u honestly think this is a genuine line of inquiry?
Yes, as it means that either 1) you're trying to hide something, or
2) you have a very poor grasp of English.-a Both of which bring
everything you say into question.
i won't respond to abusive belittling
side record.-a In fact, you wouldn't have a whole section on it in
your paper, otherwise anything in it could simply be derived from
the agent's fixed algorithm.
You've basically just confirmed that you're lying about what the
agent does because you know you made a major fundamental mistake
and won't admit it.
bro unless u can tell me why the agent reads from the side record in
the future, then he does not, cause he doesn't need to, and this
line of trying to refute my paper is a failure
now we're at 4 named fallacies, 1 named cognitive distortion, and >>>>>> 1 named abuse tactic
Just because you gave the name of a fallacy doesn't mean it applies.
just because u can't identify forms of fallacies does not negate
their application. let's add #5 and #6 for ur latest post: bad-faith
interpretation + strawman
I explained why some of them don't apply and your response was "nope,
I gave the name of a fallacy, therefore it applies" without
explaining why I was wrong.
I guess that means you'd be OK with me just naming a fallacy when you
say something I don't like and you'll just accept it regardless of
what you have to say about it.
it would be incredible if u named even just one
talk about brain rot
for any location on the tape that a decider does attempt to write >>>>>> a decision, of which there are infinite locations on the tape, and >>>>>> infinite deciders writing to any location on the tape,
That makes it an input.
Which means that when und_cir_sr calls sim_cir_sr(und_cir_sr)
after calling Tdp, it's not simulating itself.
That also means the agent is disqualified from being a halt
decider/ recognizer, partial or otherwise.
So your abstraction is broken.
no one is changing the goal post as the goal post of a decider >>>>>>>>>> is handling _all_ possible input
ofc the code does that
To fix it, the fixed steps the human agent runs need >>>>>>>>>>>>>>>>>>> to be converted to a function which read/writes a >>>>>>>>>>>>>>>>>>> global variable, and that global variable need to be >>>>>>>>>>>>>>>>>>> set to the same value it had when und_cir_sr was >>>>>>>>>>>>>>>>>>> first called.
lol, turing machines don't have "global variables" >>>>>>>>>>>>>>>>>> dud, they just have a tape, and if the value is output >>>>>>>>>>>>>>>>>> to the tape anywhere, it can be picked out by a >>>>>>>>>>>>>>>>>> simulation and contradicted by a paradox
Strawman.-a "it can be picked out" means you're talking >>>>>>>>>>>>>>>>> about changing the code which means you're no longer >>>>>>>>>>>>>>>>> talking about the same machine.
that's incorrect. reading values from the tape during a >>>>>>>>>>>>>>>> step of the computation does not change the machine >>>>>>>>>>>>>>>> which is being simulated ...
No, but the code you've shown doesn't do that.-a So if you >>>>>>>>>>>>>>> change the code to do that, it's no longer the same machine. >>>>>>>>>>>>>>
Your code is reading the part of the tape where the >>>>>>>>>>>>> mechanized human agent is writing its side record?
No?
Then your code doesn't do that, and talking about what it >>>>>>>>>>>>> "can" do necessarily means changing the code to something >>>>>>>>>>>>> not being decided on.
u really don't get it: claiming a machine to be a total >>>>>>>>>>>> decider is subject to the full enumerations of machines, >>>>>>>>>>>> meaning if an input can exist, it will exist
That has nothing to do with the fact that you can't talk >>>>>>>>>>> about changing code and still claim it's the same machine. >>>>>>>>>>
*A* decider.-a There is no "can" or "can not", only "does" or >>>>>>>>> "does not".
there exists a machine that does a paradox for any location you >>>>>>>> can write a decision to the tape
You "can" write means changing the code of the decider, which is >>>>>>> not allowed.-a Otherwise you're not talking about the same decider. >>>>>>
there does exist infinite paradoxical machines which will read
from that specific bit and produce a paradox for it in regards to >>>>>> the written decision
there is _no_ way to produce a total halting decider for turing
machines, within turing machines
Which is basically Turing's proof.
so why are you trying to argue u can simulate the agent's
computation ... u can't do so within turing machines
If the agent is using *only* its fixed sequence of steps and a
description of the algorithm to decide on to make a decision, which
is what you now seem to be claiming, it can be exactly replicated.
again just continually begging the question that the ct-thesis is
true. u don't know have to proof of the assertion that all of
realizable computing is encapsulated within turing machines,
and u do not seem to be able to understand that u do not
You still haven't shown a concrete example otherwise, so until you do church-turing stands.
1) the agent does not read from the side record
Turing machines are defined by their instructions, not where they >>>>>>> physically reside.
there exists a paradox that contradicts any possible
location you could write a decision bit on a tape, and >>>>>>>>>>>> therefore there will exist a machine that defies the
decision, no matter where u try to write it
it is _not_ possible to simulate the side result created by >>>>>>>>>>>> the agent running a total decision algorithm.
1) you can't read the side result because your abstraction is >>>>>>>>>>> broken
2) the human agent isn't a halt decider, partial or
otherwise, because it takes an input that disqualifies it
No response to this?
neither of those are correct: false dichotomy
That wasn't an either/or.-a Both apply.
2) the agent does not utilize information from the side record
while computing the decision for any given input machine
neither apply
I don't believe you.
If the agent *never* uses the side record, that makes the side
record entirely irrelevant, and it means that any machine can use
the agent's
it's purpose is to demonstrate the existence of the computation, if
u don't care for the demonstration, ur free to remain in error
The existence of the computation is in the agent's algorithm which is
fixed.
algorithm to exactly reproduce what the agent does if it doesn't
use the
u keep asserting that, but you have not yet once told me how such a
simulation would handle it's own undecidability paradox: und_cir_sr
side data.-a You wouldn't have added a whole section to your paper
about the side record and made it the focus of your "refutation" if >>>>> the agent didn't use it.
turing machines don't utilize their own output to justify their
existence... they just output it. why would an agent need to?
i wonder if ur about to commit fallacy #7: special pleading
You're backtracking because you've been caught in a fundamental
mistake and won't admit it.
it is _not_ even possible to create a total decision
algorithm with turing machines, like seriously have u >>>>>>>>>>>> forgotten the point of turing's proof???
... getting "output" from a simulation requires reading >>>>>>>>>>>>>> the output from tape of the simulated machine, it can do >>>>>>>>>>>>>> that for any value, even ones that are specified to be >>>>>>>>>>>>>> "output"
and if a machine is simulating itself,
Which your code isn't doing because the side data isn't >>>>>>>>>>>>>>> the same when starting the simulation as it was when the >>>>>>>>>>>>>>> machine was invoked.
idk what u mean by "side data" there only data on the tape >>>>>>>>>>>>>> when it comes to turing machines, there is nothing on the >>>>>>>>>>>>>> side of data on the tape.
I'm referring specifically to what the mechanized version >>>>>>>>>>>>> of the human agent writes to the tape.
like in the case an undecidability paradox within >>>>>>>>>>>>>>>> computing, then this can be used to pick out values >>>>>>>>>>>>>>>> behind any form of data encapsulation that you might >>>>>>>>>>>>>>>> suggest to hide data away from a paradox. there's no >>>>>>>>>>>>>>>> where to hide output dud. there's no "global variable" >>>>>>>>>>>>>>>> that might prevent a paradox from being formed
"can be used" meaning it's not being used now, and >>>>>>>>>>>>>>> therefore irrelevant.
there is _no_ way to simulate a general ability to >>>>>>>>>>>>>>>> decide on paradoxical turing machines, from within >>>>>>>>>>>>>>>> turing machines
So a paradoxical machine is one that a decider gets >>>>>>>>>>>>>>> wrong? So this machine:
void foo() { return; }
Is a paridoxical machine to this one:
int bar(void *p) { return 0; }
Because machine bar can't successfully report the status >>>>>>>>>>>>>>> of machine foo. -a-aYes?
no. first, we went over this dud: constant return function >>>>>>>>>>>>>> are not and will never qualify as a halting decider as >>>>>>>>>>>>>> they do not output useful information in regards to the >>>>>>>>>>>>>> halting status Efn+. the fact a machine happens to output >>>>>>>>>>>>>> the correct answer does not qualify them as a halting >>>>>>>>>>>>>> decider, a decider must output trustworthy information >>>>>>>>>>>>>> which is judged across the entirety of the output space, >>>>>>>>>>>>>> not just single instances
In other words, it's no different from *any* partial halt >>>>>>>>>>>>> decider. That's not even mentioning ill-defined weasel >>>>>>>>>>>>> words like "useful", "happens to", "trustworthy", and >>>>>>>>>>>>> "judged".
dud if ur not here to effectively compute useful knowledge, >>>>>>>>>>>> idk what ur doing discussing the theory of computing.
let me not mince words. the existing theory on the matter of >>>>>>>>>>>> deciders is thus:
- a decider is not allowed a false positive or false
negative, and never allowed to diverge. we cannot build any >>>>>>>>>>>> non-trivial decider (within turing machines)
a constant return 1 function, if use as a halting decider is >>>>>>>>>>>> abound in false positives and therefore is completely >>>>>>>>>>>> disqualified as being a halting decider.
- a recognizer is also not allowed false positives or false >>>>>>>>>>>> negatives, it is however allowed to diverge on some
negatives, but never a positive. a halting recognizer is >>>>>>>>>>>> allowed to diverge on some non- halting input, but never >>>>>>>>>>>> halting input. this is sometimes called a partial decider, >>>>>>>>>>>> but i'm going to label as it recognizer to align the rather >>>>>>>>>>>> well known sipser on this, and because i have further >>>>>>>>>>>> definitions to propose. we can build a halting recognizer, >>>>>>>>>>>> but not a non-halting recognizer.
a constant return 1 function does not satisfy this for >>>>>>>>>>>> halting either, as again it's abound in false positives. >>>>>>>>>>>>
in my paper i propose two more types:
- a partial decider is still _not_ allowed false positive or >>>>>>>>>>>> negatives. but it is allowed to diverge on some positives >>>>>>>>>>>> and some negatives. we can be build withing turing machines >>>>>>>>>>>> both halting and non-halting partial deciders
a constant return 1 function still does not satisfy this for >>>>>>>>>>>> halting either
- a partial recognizer is still not allowed false positives, >>>>>>>>>>>> but _does_ allow false negative ... but _only_ in the case >>>>>>>>>>>> where returning positive would be a false positive. this >>>>>>>>>>>> exception is allowed because a partial recognizer therefore >>>>>>>>>>>> does not need to diverge, ever, and always halts. this can >>>>>>>>>>>> also be built within turing machines for the halting and >>>>>>>>>>>> non- halting problems.
a constant return 1 function again still does not satisfy >>>>>>>>>>>> these requirements for halting
i just defined my terms here quite clearly. i suppose i'll thank >>>>>>>> you for encouraging me to develop that, i'll may add it to my paper >>>>>>>>
anyways, that was all covered in section -o5.2 which u never >>>>>>>>>>>> read
i really don't know why i'm humoring you dud
second, an actual paradox is necessarily undecidable due >>>>>>>>>>>>>> to the construction of the machine itself, the form is >>>>>>>>>>>>>> generalized in -o3 of my paper. if it doesn't fit that >>>>>>>>>>>>>> form, it's not a paradox, and is not undecidable in >>>>>>>>>>>>>> respect to any classifier let alone in totality
Alright, let's satisfy your arbitrary requirement of a non- >>>>>>>>>>>>> constant function and "useful":
void foo() {
-a-a-a-a puts("hello");
}
int bar(void *p) {
-a-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a-a return 1;
-a-a-a-a } else {
-a-a-a-a-a-a-a-a return 0;
-a-a-a-a }
}
Given that the first instruction of "foo" is not a "return" >>>>>>>>>>>>> instruction, bar(foo) will return 0 even though foo() will >>>>>>>>>>>>> halt.
Is the machine foo "undecidable" by machine bar?
No response to this?-a It goes directly to your definition of >>>>>>>>>>> "undecidable" which seems core to your argument.
bar isn't a halting decider in the first place, no idea what >>>>>>>>>> it's supposed to be
As noted above, it's a partial halt decider that reports
halting if the first instruction is a "return" (0xc3 is the x86 >>>>>>>>> instruction "ret") and reports non-halting otherwise.
Again, is the machine foo "undecidable" by machine bar?
It would be a shame to discard your whole argument just because >>>>>>>>> your terms aren't well defined.
a partial halting decider is not allowed _any_ false positives >>>>>>>> or false negatives, it must handle the entire input space of
machines without a _single_ false positive or false negative. >>>>>>>> that's not even my assertion, that's the established consensus >>>>>>>> on the matter
clearly bar would have a ton of false negatives and thereby be >>>>>>>> disqualified as a partial halting decider
Alright, let's satisfy one more arbitrary requirement to see if >>>>>>> we can get you to define your terms:
void foo() {
-a-a-a-a puts("hello");
}
int bar(void *p) {
-a-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a-a return 1;
-a-a-a-a } else {
-a-a-a-a-a-a-a-a while (1);
-a-a-a-a }
}
That takes care of the false negatives.-a So no more dodging the >>>>>>> point:
Is the machine foo "undecidable" by machine bar?
Failure to answer again will deem your term "undecidable" ill-
defined, and consequently the rest of your paper.
if i'm reading this correctly, bar now diverges (does not return) >>>>>> on the halting input foo
i will clarify the definition for "partial decider": divergence is >>>>>> only allowed when both positive and negative returns would be
false positive or false negative respectively (aka undecidable
input to the partial decider), so this would not quality as a
partial halting decider
So yet again you dodge the question on whether foo is "undecidable" >>>>> by machine bar.
bar is _not_ halting decider, partial or otherwise, so asking if the
halting semantics of foo is "undecidable" to bar is a category fallacy >>>>
wowee fallacy #8
Nope, you committed the fallacy of moving the goalposts after I
provided something that satisfied your arbitrary requirements in
order to dodge the question.
well if it isn't a decider than asking if an input is undecidable to
it is a category fallacy, no???
It is a decider by the criteria everyone else but you uses.-a You're just trying to avoid the question.
and ur just completely misinformed about conventional computing theory.
conventional computing theory only acknowledges total deciders, and
recognizors, both of which _must_ return true for all positive input
without producing _any_ false positives. recognizers may diverge on
some false input, but neither are allowed false negatives. that's the
established theory on the matter and if u disagree take it up with
sipser.
i'm proposing two more, which still are _not_ allowed to produce false
positives. partial deciders are allowed to diverge on some positive
input, whereas partial recognizer are allowed to produce false
negatives (and never diverge), and both only in the case where the
input is paradoxical (either return positive/negative would be false)
No such thing, since all machines either halt or do not halt, and all deciders report either halting or non-halting for any given machine description.
The exact code placed in function bar is irrelevant to the question.
there is no basis for this claim in conventional computing theory, nor
is what i'm proposing. classifiers for a particular semantic property
are not just random-ass algos that happens to output correctly for
some machines, there are specific interface constraints they _must_ meet
There's no such thing as "happens to output correctly".-a Nowhere is
there any requirement that an algorithm has to use any particular steps
in order to map a mathematical function.
That make that term ill-defined.
the term is quite well defined: if a classifier/decider is stuck
between returning a false positive and a false negative for a
particular input, then that input is undecidable to the classifier/
decider
Category error.-a There is no "stuck between" doing two things.-a A
classifier/decider does one thing and one thing only: what its fixed
instructions tell it to do.
handling input where responding either positive or negative would be
false
No such thing.-a All machines either halt or do not halt, and a decider returns either one or the other (if it returns) for any given machine.
You still don't understand what defines a turing machine.
is a real input scenario that must be handled by any actual machine
attempting to accurately decide on other machines. i'm not describing
what the machine is literally doing, i'm describing an objective
mathematical reality of the decider's output space, when faced with
paradoxical input
Turing machines are defined by their instruction, not by where
instructions physically reside.
Your continued lack of response to this is telling.
And not understanding that is your core mistake.-aThe same mistake
made by PO.
And because you whole paper is based on the use of that word, it is >>>>> reduced to meaninglessness.
a fallacy laden critique full of cognitive distortion and abuse is
never going to convince me dud
Without an answer, this deems your use of the term unclear, >>>>>>>>>>> and therefore makes your entire argument null and void.
handling input where responding either positive or negative would be
false
No such thing.-a All machines either halt or do not halt, and a decider
returns either one or the other (if it returns) for any given machine.
You still don't understand what defines a turing machine.
ur not making even a semblance of sense to me anymore dud. yes ofc all machines either halt or not
that fact is orthogonal to the paradox of a decider having to deal with paradoxical input where returning true
causes the input machine to not
halt, and returning false
causes the input machine to halt. that is
literally the foundational paradox justifying the proof of
undecidability for the halting problem
On 9/25/2026 4:53 PM, dart200 wrote:
handling input where responding either positive or negative would be
false
No such thing.-a All machines either halt or do not halt, and a
decider returns either one or the other (if it returns) for any given
machine.
You still don't understand what defines a turing machine.
ur not making even a semblance of sense to me anymore dud. yes ofc all
machines either halt or not
Which means there's no such thing as returning 0 being incorrect AND returning 1 being incorrect.
that fact is orthogonal to the paradox of a decider having to deal
with paradoxical input where returning true
So we've defined decider X along with machine Y such that machine Y is implemented as making a call to decider X and X(Y) returns true and
machine Y does not halt.
causes the input machine to not halt, and returning false
Strawman.-a That's not decider X nor machine Y.
This is what I mean when I say you don't know what a Turing machine is.
And because your paper is based on an incorrect definition of a Turing machine, your whole argument breaks down.
causes the input machine to halt. that is literally the foundational
paradox justifying the proof of undecidability for the halting problem
On 9/25/26 2:40 PM, dbush wrote:
On 9/25/2026 4:53 PM, dart200 wrote:
handling input where responding either positive or negative would
be false
No such thing.-a All machines either halt or do not halt, and a
decider returns either one or the other (if it returns) for any
given machine.
You still don't understand what defines a turing machine.
ur not making even a semblance of sense to me anymore dud. yes ofc
all machines either halt or not
Which means there's no such thing as returning 0 being incorrect AND
returning 1 being incorrect.
it's literally _that_ paradox which proves undecidability ... are you disagreeing with Turing's proof of undecidability now???
that fact is orthogonal to the paradox of a decider having to deal
with paradoxical input where returning true
So we've defined decider X along with machine Y such that machine Y is
implemented as making a call to decider X and X(Y) returns true and
machine Y does not halt.
which makes X not a halting decider. it returned a false positive, and
no paradigm recognizes false positives as a valid specification for
halting deciders, partial or otherwise
causes the input machine to not halt, and returning false
Strawman.-a That's not decider X nor machine Y.
This is what I mean when I say you don't know what a Turing machine is.
And because your paper is based on an incorrect definition of a Turing
machine, your whole argument breaks down.
well that's certainly what u desperately hope for
causes the input machine to halt. that is literally the foundational
paradox justifying the proof of undecidability for the halting problem
On 9/25/2026 5:54 PM, dart200 wrote:
On 9/25/26 2:40 PM, dbush wrote:
On 9/25/2026 4:53 PM, dart200 wrote:
handling input where responding either positive or negative would >>>>>> be false
No such thing.-a All machines either halt or do not halt, and a
decider returns either one or the other (if it returns) for any
given machine.
You still don't understand what defines a turing machine.
ur not making even a semblance of sense to me anymore dud. yes ofc
all machines either halt or not
Which means there's no such thing as returning 0 being incorrect AND
returning 1 being incorrect.
it's literally _that_ paradox which proves undecidability ... are you
disagreeing with Turing's proof of undecidability now???
Not at all.-a Just with your interpretation of it.-a You seem to think you're talking about a single machine.-a You're not.
that fact is orthogonal to the paradox of a decider having to deal
with paradoxical input where returning true
So we've defined decider X along with machine Y such that machine Y
is implemented as making a call to decider X and X(Y) returns true
and machine Y does not halt.
which makes X not a halting decider. it returned a false positive, and
no paradigm recognizes false positives as a valid specification for
halting deciders, partial or otherwise
So we've at least defined machines X and Y.
causes the input machine to not halt, and returning false
Strawman.-a That's not decider X nor machine Y.
No response to this?-a This is the core of your error.
You're further proving that you don't know what a Turing machine is,
meaning your whole paper is based on a false premise.
This is what I mean when I say you don't know what a Turing machine is.
And because your paper is based on an incorrect definition of a
Turing machine, your whole argument breaks down.
well that's certainly what u desperately hope for
causes the input machine to halt. that is literally the foundational
paradox justifying the proof of undecidability for the halting problem
On 9/25/26 3:03 PM, dbush wrote:
On 9/25/2026 5:54 PM, dart200 wrote:
On 9/25/26 2:40 PM, dbush wrote:
On 9/25/2026 4:53 PM, dart200 wrote:
handling input where responding either positive or negative would >>>>>>> be false
No such thing.-a All machines either halt or do not halt, and a
decider returns either one or the other (if it returns) for any
given machine.
You still don't understand what defines a turing machine.
ur not making even a semblance of sense to me anymore dud. yes ofc
all machines either halt or not
Which means there's no such thing as returning 0 being incorrect AND
returning 1 being incorrect.
it's literally _that_ paradox which proves undecidability ... are you
disagreeing with Turing's proof of undecidability now???
Not at all.-a Just with your interpretation of it.-a You seem to think
you're talking about a single machine.-a You're not.
und = () -> {
-a if (halts(und))
-a-a-a loop
-a else
-a-a-a halt
}
what is halts(und) supposed to return?
that fact is orthogonal to the paradox of a decider having to deal
with paradoxical input where returning true
So we've defined decider X along with machine Y such that machine Y
is implemented as making a call to decider X and X(Y) returns true
and machine Y does not halt.
which makes X not a halting decider. it returned a false positive,
and no paradigm recognizes false positives as a valid specification
for halting deciders, partial or otherwise
So we've at least defined machines X and Y.
causes the input machine to not halt, and returning false
Strawman.-a That's not decider X nor machine Y.
No response to this?-a This is the core of your error.
the core of my error changes with every post u write
You're further proving that you don't know what a Turing machine is,
meaning your whole paper is based on a false premise.
i have no idea what ur talking about, that's for sure
This is what I mean when I say you don't know what a Turing machine is. >>>>
And because your paper is based on an incorrect definition of a
Turing machine, your whole argument breaks down.
well that's certainly what u desperately hope for
causes the input machine to halt. that is literally the
foundational paradox justifying the proof of undecidability for the >>>>> halting problem
On 9/24/2026 7:36 PM, Dude wrote:
On 9/24/2026 6:16 PM, dart200 wrote:
On 9/24/26 4:35 PM, Chris M. Thomasson wrote:You cannot solve the halting problem for every possible program.
On 9/24/2026 3:44 PM, dart200 wrote:
On 9/24/26 3:10 PM, Chris M. Thomasson wrote:
On 9/20/2026 10:39 PM, dart200 wrote:
On 9/20/26 7:51 PM, Chris M. Thomasson wrote:
On 9/20/2026 7:11 PM, dart200 wrote:
On 9/20/26 2:53 PM, Chris M. Thomasson wrote:
On 9/19/2026 2:38 PM, dart200 wrote:
[...]
If not, wtf! You remind me of PO. Sorry for that cut.
i don't care for ur fallacy by association brainrot
Sigh. I only said you kind of do remind me of the way PO dealt >>>>>>>>>> with the halting program.
and that statement has no bearing on the correctness of my
arguments, u brain-rotted boomer
You cannot predict a random number and you cannot solve the
halting problem. Sigh.
turing machines do not involve random numbers dud. if u knew
literally anything about basic computing theory u'd know that.
but u don't, so go back to ur fractals, eh?
Well, ponder on CSPRNG?
pseudo-random number generators are fully deterministic and
therefore entirely decidable in terms of whether it the machine
which produces it is circle-free or not (they are obviously). for
any decision ofc one needs a description of the machine including
any input (like a seed), and that combination can be used to decide
on the semantics of the described computation
when u say "random" that does not describe what can be produced by
a turing machine, as they can only compute pseudo-random sequences
that are inherently predictable, not truly random ones. if u meant
pseudo- random you should have said that...
seriously, bro: back to ur fractals, eh?
You have no idea how to get around the halting problem.
u have no idea what the halting problem even is, or the paradox that
makes it undecidable
and why anyone else isn't correcting u is just beyond me
But, you can get around it in practice by restricting your languages,
setting resource limits, or using automated parameters.
Alan Turing proved that no single program can correctly predict
whether any arbitrary code will finish running or loop forever.
However, real-world software engineering bypasses this theoretical
limit every day.
Where's Noah?
Say a program halts once all of its possible paths are hit?
On 9/25/2026 8:40 PM, dart200 wrote:
On 9/25/26 3:03 PM, dbush wrote:
On 9/25/2026 5:54 PM, dart200 wrote:
On 9/25/26 2:40 PM, dbush wrote:
On 9/25/2026 4:53 PM, dart200 wrote:
handling input where responding either positive or negative
would be false
No such thing.-a All machines either halt or do not halt, and a >>>>>>> decider returns either one or the other (if it returns) for any >>>>>>> given machine.
You still don't understand what defines a turing machine.
ur not making even a semblance of sense to me anymore dud. yes ofc >>>>>> all machines either halt or not
Which means there's no such thing as returning 0 being incorrect
AND returning 1 being incorrect.
it's literally _that_ paradox which proves undecidability ... are
you disagreeing with Turing's proof of undecidability now???
Not at all.-a Just with your interpretation of it.-a You seem to think
you're talking about a single machine.-a You're not.
und = () -> {
-a-a if (halts(und))
-a-a-a-a loop
-a-a else
-a-a-a-a halt
}
what is halts(und) supposed to return?
The machine und is not fully defined.-a All of its steps must be spelled
out before that can be determined.
that fact is orthogonal to the paradox of a decider having to deal >>>>>> with paradoxical input where returning true
So we've defined decider X along with machine Y such that machine Y >>>>> is implemented as making a call to decider X and X(Y) returns true
and machine Y does not halt.
which makes X not a halting decider. it returned a false positive,
and no paradigm recognizes false positives as a valid specification
for halting deciders, partial or otherwise
So we've at least defined machines X and Y.
causes the input machine to not halt, and returning false
Strawman.-a That's not decider X nor machine Y.
No response to this?-a This is the core of your error.
the core of my error changes with every post u write
You're further proving that you don't know what a Turing machine is,
meaning your whole paper is based on a false premise.
i have no idea what ur talking about, that's for sure
I'll make it simple for you:
Are these the same machine?
int foo(int x, int y)
{
-a-a-a return x*y;
}
int bar(int a, int b)
{
-a-a-a return a*b;
}
This is what I mean when I say you don't know what a Turing machine >>>>> is.
And because your paper is based on an incorrect definition of a
Turing machine, your whole argument breaks down.
well that's certainly what u desperately hope for
causes the input machine to halt. that is literally the
foundational paradox justifying the proof of undecidability for
the halting problem
On 2026-09-24 18:44, Mike Terry wrote:
On 25/09/2026 00:19, Andr|- G. Isaak wrote:
On 2026-09-24 14:09, dart200 wrote:I don't know - what you're describing seems to already have a term:
a partial halting decider is not allowed _any_ false positives or
false negatives, it must handle the entire input space of machines
without a _single_ false positive or false negative.
You're very confused here. If it is not allowed any false negatives
or false positives, then you're talking about a /total/ halt decider.
A /partial/ halt decider is something which correctly decides /some/
instances of the halting problem but which fails to correctly
decide /all/ instances. That's why that pesky word 'partial' is added.
it's a "decider".-a (But just not a "halt decider" since it gets some
inputs wrong by the halting criterion.)
The way "partial halt decider" [PHD] has typically been used /on this
newsgroup/ is to indicate a TM with three final "outcomes":
--a HALTS-a-a-a-a-a-a-a [input halts]
--a NEVERHALTS-a-a-a-a-a-a-a [input never halts]
--a PASS-a-a-a-a-a-a-a-a-a-a-a ["pass" - the PHD makes no halting decision] >>
So the PHD may always PASS for a given input, but when it does
indicate HALTS/NEVERHALTS that must /correctly/ match the input's
halting behaviour.
Quite a few posters over the years have casually suggested that "PHD"
is a well known term meaning [whatever - generally something like my
definition, rather than yours...] but I can't say I've seen any use of
the term in books I've used!-a Is it /really/ a standard term???-a If
"PHD" is more of a comp.theory common term, then posters should define
their usage for the term when they use it.
There are variations in how it might reasonably be defined, e.g. I
suspect requiring PASS to be an actual "decision" by the PHD [e.g.
transition to a 3rd TM halt state] is substantially different from
allowing a PHD to implicitly PASS by virtue of never halting.-a Since
this NG has often discussed concrete programs that run until they
detect a definite HALTS or NEVERHALTS pattern, my suggestion is to
allow PHDs to include programs that may never halt...-a [so PO's code
would then be classed a PHD if its bugs/logic-errors were fixed.]
I think you may be correct that this term really is one specific to this newsgroup rather than a term in wide usage.
On 9/24/26 7:38 PM, Dude wrote:
On 9/24/2026 4:37 PM, Chris M. Thomasson wrote:
On 9/24/2026 3:44 PM, dart200 wrote:It's a 5th grader response.
[...]
Oh my. Please refrain from calling yourself God, PO.......
You can use restricted programming languages or subsets (such as MISRA
C, SPARK, or Rocq) that are not fully Turing-complete.
By banning unbounded loops and wild recursion, you ensure that every
valid program in the language is guaranteed to finish
that's why u gotta read the paper dud!
what i show is that we don't need to restrict the computing power of the language, to gain total decidability of a language which can compute all that is computable! and prove that despite the computable enumerability
of that language, it is logicalluy resistant to the undecidability
paradoxes induced by diagonalization...
the logic required to enumerate all turing-computable sequences,
especially those involving self-references, is necessarily resistant to
and cannot be used to compute either a total anti-diagonal, or even just
a total diagonal. for every sequence that is computable, there must be
an anti-sequence that is also computable. the existence of (sequence, anti-sequence) pairs defining the set of turing computable sequences contraindicates both a total diagonal _and_ a total anti-diagonal
sequence from existing as a sequence that can be output by a turing
machine. it will neither be found in the enumeration of all sequences,
nor be definable using the logic required to enumerate those sequences.
the closest to a true diagonal that is turing computable is one that is
a direct diagonal across all sequences, except for producing anti-output
to it's anti-sequence. and the closed to an anti-diagonal that can be computed is one that is the anti-output for all other sequences except itself!
and no dud, u will not understand those words until u read the fucking paper! my next response will be to be read the fucking paper bro, i
swear to fucking god dude...
u are such a goddamn moron read the fucking paper Efy-Efy-Efy-
On 9/24/26 7:38 PM, Dude wrote:
On 9/24/2026 4:37 PM, Chris M. Thomasson wrote:
On 9/24/2026 3:44 PM, dart200 wrote:It's a 5th grader response.
[...]
Oh my. Please refrain from calling yourself God, PO.......
You can use restricted programming languages or subsets (such as MISRA
C, SPARK, or Rocq) that are not fully Turing-complete.
By banning unbounded loops and wild recursion, you ensure that every
valid program in the language is guaranteed to finish
that's why u gotta read the paper dud!
what i show is that we don't need to restrict the computing power of the language, to gain total decidability of a language which can compute all that is computable! and prove that despite the computable enumerability
of that language, it is logicalluy resistant to the undecidability
paradoxes induced by diagonalization...
the logic required to enumerate all turing-computable sequences,
especially those involving self-references, is necessarily resistant to
and cannot be used to compute either a total anti-diagonal, or even just
a total diagonal. for every sequence that is computable, there must be
an anti-sequence that is also computable. the existence of (sequence, anti-sequence) pairs defining the set of turing computable sequences contraindicates both a total diagonal _and_ a total anti-diagonal
sequence from existing as a sequence that can be output by a turing
machine. it will neither be found in the enumeration of all sequences,
nor be definable using the logic required to enumerate those sequences.
the closest to a true diagonal that is turing computable is one that is
a direct diagonal across all sequences, except for producing anti-output
to it's anti-sequence. and the closed to an anti-diagonal that can be computed is one that is the anti-output for all other sequences except itself!
and no dud, u will not understand those words until u read the fucking paper! my next response will be to be read the fucking paper bro, i
swear to fucking god dude...
u are such a goddamn moron read the fucking paper Efy-Efy-Efy-
On 9/25/26 7:24 PM, dbush wrote:
On 9/25/2026 8:40 PM, dart200 wrote:
On 9/25/26 3:03 PM, dbush wrote:
On 9/25/2026 5:54 PM, dart200 wrote:
On 9/25/26 2:40 PM, dbush wrote:
On 9/25/2026 4:53 PM, dart200 wrote:
handling input where responding either positive or negative >>>>>>>>> would be false
No such thing.-a All machines either halt or do not halt, and a >>>>>>>> decider returns either one or the other (if it returns) for any >>>>>>>> given machine.
You still don't understand what defines a turing machine.
ur not making even a semblance of sense to me anymore dud. yes
ofc all machines either halt or not
Which means there's no such thing as returning 0 being incorrect
AND returning 1 being incorrect.
it's literally _that_ paradox which proves undecidability ... are
you disagreeing with Turing's proof of undecidability now???
Not at all.-a Just with your interpretation of it.-a You seem to think >>>> you're talking about a single machine.-a You're not.
und = () -> {
-a-a if (halts(und))
-a-a-a-a loop
-a-a else
-a-a-a-a halt
}
what is halts(und) supposed to return?
The machine und is not fully defined.-a All of its steps must be
spelled out before that can be determined.
yeah and if we try to fully define it, by defining what halts(und) does there are three possible results:
- halts(und) diverges and doesn't return, causing und to also diverge.
this would be the case if halts is specified to be recognizer
- halts(und) returns false and then und halts. this would be the case if halts is specified to be a partial recognizer
- halts(und) returns true and then und diverges. this would make halts
not any type of classifier
there is no way to build a halt what would return an accurate decision
on what und() does when run. that is the paradoxical dilemma that
underpins all of undecidability within computing
that fact is orthogonal to the paradox of a decider having to
deal with paradoxical input where returning true
So we've defined decider X along with machine Y such that machine >>>>>> Y is implemented as making a call to decider X and X(Y) returns
true and machine Y does not halt.
which makes X not a halting decider. it returned a false positive,
and no paradigm recognizes false positives as a valid specification >>>>> for halting deciders, partial or otherwise
So we've at least defined machines X and Y.
causes the input machine to not halt, and returning false
Strawman.-a That's not decider X nor machine Y.
No response to this?-a This is the core of your error.
the core of my error changes with every post u write
You're further proving that you don't know what a Turing machine is,
meaning your whole paper is based on a false premise.
i have no idea what ur talking about, that's for sure
I'll make it simple for you:
Are these the same machine?
int foo(int x, int y)
{
-a-a-a-a return x*y;
}
int bar(int a, int b)
{
-a-a-a-a return a*b;
}
they should be the same machine, yes. what's the relevancy?
On 9/25/26 7:24 PM, dbush wrote:
On 9/25/2026 8:40 PM, dart200 wrote:
On 9/25/26 3:03 PM, dbush wrote:
On 9/25/2026 5:54 PM, dart200 wrote:
On 9/25/26 2:40 PM, dbush wrote:
On 9/25/2026 4:53 PM, dart200 wrote:
handling input where responding either positive or negative >>>>>>>> would be false
No such thing.-a All machines either halt or do not halt, and a >>>>>>> decider returns either one or the other (if it returns) for any >>>>>>> given machine.
You still don't understand what defines a turing machine.
ur not making even a semblance of sense to me anymore dud. yes ofc >>>>>> all machines either halt or not
Which means there's no such thing as returning 0 being incorrect
AND returning 1 being incorrect.
it's literally _that_ paradox which proves undecidability ... are
you disagreeing with Turing's proof of undecidability now???
Not at all.-a Just with your interpretation of it.-a You seem to think >>> you're talking about a single machine.-a You're not.
und = () -> {
-a-a if (halts(und))
-a-a-a-a loop
-a-a else
-a-a-a-a halt
}
what is halts(und) supposed to return?
The machine und is not fully defined.-a All of its steps must be spelled out before that can be determined.
yeah and if we try to fully define it, by defining what halts(und) does there are three possible results:
- halts(und) diverges and doesn't return, causing und to also diverge.
this would be the case if halts is specified to be recognizer
- halts(und) returns false and then und halts. this would be the case if halts is specified to be a partial recognizer
- halts(und) returns true and then und diverges. this would make halts
not any type of classifier
there is no way to build a halt what would return an accurate decision
on what und() does when run. that is the paradoxical dilemma that
underpins all of undecidability within computing
that fact is orthogonal to the paradox of a decider having to deal >>>>>> with paradoxical input where returning true
So we've defined decider X along with machine Y such that machine Y >>>>> is implemented as making a call to decider X and X(Y) returns true >>>>> and machine Y does not halt.
which makes X not a halting decider. it returned a false positive,
and no paradigm recognizes false positives as a valid specification >>>> for halting deciders, partial or otherwise
So we've at least defined machines X and Y.
causes the input machine to not halt, and returning false
Strawman.-a That's not decider X nor machine Y.
No response to this?-a This is the core of your error.
the core of my error changes with every post u write
You're further proving that you don't know what a Turing machine is,
meaning your whole paper is based on a false premise.
i have no idea what ur talking about, that's for sure
I'll make it simple for you:
Are these the same machine?
int foo(int x, int y)
{
-a-a-a return x*y;
}
int bar(int a, int b)
{
-a-a-a return a*b;
}
they should be the same machine, yes. what's the relevancy?
On 9/26/2026 1:33 AM, dart200 wrote:
On 9/25/26 7:24 PM, dbush wrote:
On 9/25/2026 8:40 PM, dart200 wrote:
On 9/25/26 3:03 PM, dbush wrote:
On 9/25/2026 5:54 PM, dart200 wrote:
On 9/25/26 2:40 PM, dbush wrote:
On 9/25/2026 4:53 PM, dart200 wrote:
handling input where responding either positive or negative >>>>>>>>>> would be false
No such thing.-a All machines either halt or do not halt, and a >>>>>>>>> decider returns either one or the other (if it returns) for any >>>>>>>>> given machine.
You still don't understand what defines a turing machine.
ur not making even a semblance of sense to me anymore dud. yes >>>>>>>> ofc all machines either halt or not
Which means there's no such thing as returning 0 being incorrect >>>>>>> AND returning 1 being incorrect.
it's literally _that_ paradox which proves undecidability ... are >>>>>> you disagreeing with Turing's proof of undecidability now???
Not at all.-a Just with your interpretation of it.-a You seem to
think you're talking about a single machine.-a You're not.
und = () -> {
-a-a if (halts(und))
-a-a-a-a loop
-a-a else
-a-a-a-a halt
}
what is halts(und) supposed to return?
The machine und is not fully defined.-a All of its steps must be
spelled out before that can be determined.
yeah and if we try to fully define it, by defining what halts(und)
does there are three possible results:
- halts(und) diverges and doesn't return, causing und to also diverge.
this would be the case if halts is specified to be recognizer
- halts(und) returns false and then und halts. this would be the case
if halts is specified to be a partial recognizer
- halts(und) returns true and then und diverges. this would make halts
not any type of classifier
Which means we're talking about 3 different deciders which give the
wrong answer to 3 different machines.
there is no way to build a halt what would return an accurate decision
on what und() does when run. that is the paradoxical dilemma that
underpins all of undecidability within computing
Not exactly.-a It is one way to show that any potential halt decider gets
at least one case wrong.-a But it doesn't have to be constructed that way.
that fact is orthogonal to the paradox of a decider having to >>>>>>>> deal with paradoxical input where returning true
So we've defined decider X along with machine Y such that machine >>>>>>> Y is implemented as making a call to decider X and X(Y) returns >>>>>>> true and machine Y does not halt.
which makes X not a halting decider. it returned a false positive, >>>>>> and no paradigm recognizes false positives as a valid
specification for halting deciders, partial or otherwise
So we've at least defined machines X and Y.
causes the input machine to not halt, and returning false
Strawman.-a That's not decider X nor machine Y.
No response to this?-a This is the core of your error.
the core of my error changes with every post u write
You're further proving that you don't know what a Turing machine
is, meaning your whole paper is based on a false premise.
i have no idea what ur talking about, that's for sure
I'll make it simple for you:
Are these the same machine?
int foo(int x, int y)
{
-a-a-a-a return x*y;
}
int bar(int a, int b)
{
-a-a-a-a return a*b;
}
they should be the same machine, yes. what's the relevancy?
Ah, so you DO understand that they're the same, defined by their actual instructions.-a Then you should be able to understand what follows.
Let's take a look at a candidate halt decider:
int x(void *p)
{
-a-a-a-a int result;
-a-a-a-a {
-a-a-a-a-a-a-a-a // machine bar
-a-a-a-a-a-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a-a-a-a-a-a result = 1;
-a-a-a-a-a-a-a-a } else {
-a-a-a-a-a-a-a-a-a-a-a-a result = 0;
-a-a-a-a-a-a-a-a }
-a-a-a }
-a-a-a return result;
}
The implementation is simple, but for now let's assume we haven't yet
done any analysis on it and are trying to determine what it gets right
and what it gets wrong.
The following machine is another way to build the "paradoxical" input,
and is in fact the method used in the Sipser proof that PO likes to quote:
void y()
{
-a-a-a-a // machine foo
-a-a-a-a void *p = y;
-a-a-a-a {
-a-a-a-a-a-a-a-a int result;
-a-a-a-a-a-a-a-a // machine bar
-a-a-a-a-a-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a-a-a-a-a-a result = 1;
-a-a-a-a-a-a-a-a } else {
-a-a-a-a-a-a-a-a-a-a-a-a result = 0;
-a-a-a-a-a-a-a-a }
-a-a-a }
-a-a-a if (result) while(1);
}
Since a Turing machine is defined by its instructions, we can build
another set of instructions that the first set gets wrong.
The common programming language paradigm of using functions as modular components tends to confuse programmers looking at computation theory as
to what actually composes an algorithm / Turing machine.
On 9/26/26 7:01 AM, dbush wrote:
On 9/26/2026 1:33 AM, dart200 wrote:
On 9/25/26 7:24 PM, dbush wrote:
On 9/25/2026 8:40 PM, dart200 wrote:
On 9/25/26 3:03 PM, dbush wrote:
On 9/25/2026 5:54 PM, dart200 wrote:
On 9/25/26 2:40 PM, dbush wrote:
On 9/25/2026 4:53 PM, dart200 wrote:
handling input where responding either positive or negative >>>>>>>>>>> would be false
No such thing. All machines either halt or do not halt, and a >>>>>>>>>> decider returns either one or the other (if it returns) for >>>>>>>>>> any given machine.
You still don't understand what defines a turing machine.
ur not making even a semblance of sense to me anymore dud. yes >>>>>>>>> ofc all machines either halt or not
Which means there's no such thing as returning 0 being incorrect >>>>>>>> AND returning 1 being incorrect.
it's literally _that_ paradox which proves undecidability ... are >>>>>>> you disagreeing with Turing's proof of undecidability now???
Not at all. Just with your interpretation of it. You seem to
think you're talking about a single machine. You're not.
und = () -> {
if (halts(und))
loop
else
halt
}
what is halts(und) supposed to return?
The machine und is not fully defined. All of its steps must be
spelled out before that can be determined.
yeah and if we try to fully define it, by defining what halts(und)
does there are three possible results:
- halts(und) diverges and doesn't return, causing und to also
diverge. this would be the case if halts is specified to be recognizer
- halts(und) returns false and then und halts. this would be the case
if halts is specified to be a partial recognizer
- halts(und) returns true and then und diverges. this would make
halts not any type of classifier
Which means we're talking about 3 different deciders which give the
wrong answer to 3 different machines.
but they all the face the same dilemma with respect to trying to compute
a halting decision
the dilemma is defined by the potential execution paths which can be
selected by the decider's output, not the actual path it does take. the paradox results from lack of a correct solution among all the potential paths. in fact adding/removing non-executed paths can add/remove the
dilemma, affecting how deciders react to it
there is no way to build a halt what would return an accurate
decision on what und() does when run. that is the paradoxical dilemma
that underpins all of undecidability within computing
Not exactly. It is one way to show that any potential halt decider
gets at least one case wrong. But it doesn't have to be constructed
that way.
all forms within computing result in a seemingly unsolvable dilemma
that fact is orthogonal to the paradox of a decider having to >>>>>>>>> deal with paradoxical input where returning true
So we've defined decider X along with machine Y such that
machine Y is implemented as making a call to decider X and X(Y) >>>>>>>> returns true and machine Y does not halt.
which makes X not a halting decider. it returned a false
positive, and no paradigm recognizes false positives as a valid
specification for halting deciders, partial or otherwise
So we've at least defined machines X and Y.
causes the input machine to not halt, and returning false
Strawman. That's not decider X nor machine Y.
No response to this? This is the core of your error.
the core of my error changes with every post u write
You're further proving that you don't know what a Turing machine
is, meaning your whole paper is based on a false premise.
i have no idea what ur talking about, that's for sure
I'll make it simple for you:
Are these the same machine?
int foo(int x, int y)
{
return x*y;
}
int bar(int a, int b)
{
return a*b;
}
they should be the same machine, yes. what's the relevancy?
Ah, so you DO understand that they're the same, defined by their
actual instructions. Then you should be able to understand what follows.
i'm well aware
Let's take a look at a candidate halt decider:
int x(void *p)
{
int result;
{
// machine bar
if (*(unsigned char *)p == 0xc3) {
result = 1;
} else {
result = 0;
}
}
return result;
}
The implementation is simple, but for now let's assume we haven't yet
done any analysis on it and are trying to determine what it gets right
and what it gets wrong.
The following machine is another way to build the "paradoxical" input,
and is in fact the method used in the Sipser proof that PO likes to
quote:
void y()
{
// machine foo
void *p = y;
{
int result;
// machine bar
if (*(unsigned char *)p == 0xc3) {
result = 1;
} else {
result = 0;
}
}
if (result) while(1);
}
Since a Turing machine is defined by its instructions, we can build
another set of instructions that the first set gets wrong.
sure and you could have just written it like
void y()
{
if (x(y)) while(1);
}
The common programming language paradigm of using functions as modular
components tends to confuse programmers looking at computation theory
as to what actually composes an algorithm / Turing machine.
the only thing i'm confused about is why u think this refutes anything
about my paper
On 9/24/26 3:21 PM, Chris M. Thomasson wrote:
On 9/24/2026 11:03 AM, dart200 wrote:
[...]
You really are PO! ;^o Woha!
i really am not polcott,
and this is really the best feedback u sheeple can give?
On 9/26/26 7:01 AM, dbush wrote:
On 9/26/2026 1:33 AM, dart200 wrote:
On 9/25/26 7:24 PM, dbush wrote:
On 9/25/2026 8:40 PM, dart200 wrote:
On 9/25/26 3:03 PM, dbush wrote:
On 9/25/2026 5:54 PM, dart200 wrote:
On 9/25/26 2:40 PM, dbush wrote:
On 9/25/2026 4:53 PM, dart200 wrote:
handling input where responding either positive or negative >>>>>>>>>>> would be false
No such thing.-a All machines either halt or do not halt, and a >>>>>>>>>> decider returns either one or the other (if it returns) for >>>>>>>>>> any given machine.
You still don't understand what defines a turing machine.
ur not making even a semblance of sense to me anymore dud. yes >>>>>>>>> ofc all machines either halt or not
Which means there's no such thing as returning 0 being incorrect >>>>>>>> AND returning 1 being incorrect.
it's literally _that_ paradox which proves undecidability ... are >>>>>>> you disagreeing with Turing's proof of undecidability now???
Not at all.-a Just with your interpretation of it.-a You seem to
think you're talking about a single machine.-a You're not.
und = () -> {
-a-a if (halts(und))
-a-a-a-a loop
-a-a else
-a-a-a-a halt
}
what is halts(und) supposed to return?
The machine und is not fully defined.-a All of its steps must be
spelled out before that can be determined.
yeah and if we try to fully define it, by defining what halts(und)
does there are three possible results:
- halts(und) diverges and doesn't return, causing und to also
diverge. this would be the case if halts is specified to be recognizer
- halts(und) returns false and then und halts. this would be the case
if halts is specified to be a partial recognizer
- halts(und) returns true and then und diverges. this would make
halts not any type of classifier
Which means we're talking about 3 different deciders which give the
wrong answer to 3 different machines.
but they all the face the same dilemma with respect to trying to compute
a halting decision
the dilemma is defined by the potential execution paths which can be selected by the decider's output, not the actual path it does take. the paradox results from lack of a correct solution among all the potential paths. in fact adding/removing non-executed paths can add/remove the dilemma, affecting how deciders react to it
there is no way to build a halt what would return an accurate
decision on what und() does when run. that is the paradoxical dilemma
that underpins all of undecidability within computing
Not exactly.-a It is one way to show that any potential halt decider
gets at least one case wrong.-a But it doesn't have to be constructed
that way.
all forms within computing result in a seemingly unsolvable dilemma
that fact is orthogonal to the paradox of a decider having to >>>>>>>>> deal with paradoxical input where returning true
So we've defined decider X along with machine Y such that
machine Y is implemented as making a call to decider X and X(Y) >>>>>>>> returns true and machine Y does not halt.
which makes X not a halting decider. it returned a false
positive, and no paradigm recognizes false positives as a valid >>>>>>> specification for halting deciders, partial or otherwise
So we've at least defined machines X and Y.
causes the input machine to not halt, and returning false
Strawman.-a That's not decider X nor machine Y.
No response to this?-a This is the core of your error.
the core of my error changes with every post u write
You're further proving that you don't know what a Turing machine
is, meaning your whole paper is based on a false premise.
i have no idea what ur talking about, that's for sure
I'll make it simple for you:
Are these the same machine?
int foo(int x, int y)
{
-a-a-a-a return x*y;
}
int bar(int a, int b)
{
-a-a-a-a return a*b;
}
they should be the same machine, yes. what's the relevancy?
Ah, so you DO understand that they're the same, defined by their
actual instructions.-a Then you should be able to understand what follows.
i'm well aware
Let's take a look at a candidate halt decider:
int x(void *p)
{
-a-a-a-a-a int result;
-a-a-a-a-a {
-a-a-a-a-a-a-a-a-a // machine bar
-a-a-a-a-a-a-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a-a-a-a-a-a-a result = 1;
-a-a-a-a-a-a-a-a-a } else {
-a-a-a-a-a-a-a-a-a-a-a-a-a result = 0;
-a-a-a-a-a-a-a-a-a }
-a-a-a-a }
-a-a-a-a return result;
}
The implementation is simple, but for now let's assume we haven't yet
done any analysis on it and are trying to determine what it gets right
and what it gets wrong.
The following machine is another way to build the "paradoxical" input,
and is in fact the method used in the Sipser proof that PO likes to
quote:
void y()
{
-a-a-a-a-a // machine foo
-a-a-a-a-a void *p = y;
-a-a-a-a-a {
-a-a-a-a-a-a-a-a-a int result;
-a-a-a-a-a-a-a-a-a // machine bar
-a-a-a-a-a-a-a-a-a if (*(unsigned char *)p == 0xc3) {
-a-a-a-a-a-a-a-a-a-a-a-a-a result = 1;
-a-a-a-a-a-a-a-a-a } else {
-a-a-a-a-a-a-a-a-a-a-a-a-a result = 0;
-a-a-a-a-a-a-a-a-a }
-a-a-a-a }
-a-a-a-a if (result) while(1);
}
Since a Turing machine is defined by its instructions, we can build
another set of instructions that the first set gets wrong.
sure and you could have just written it like
void y()
{
-a-a-a if (x(y)) while(1);
}
The common programming language paradigm of using functions as modular
components tends to confuse programmers looking at computation theory
as to what actually composes an algorithm / Turing machine.
the only thing i'm confused about is why u think this refutes anything
about my paper
On 9/24/2026 3:45 PM, dart200 wrote:
On 9/24/26 3:21 PM, Chris M. Thomasson wrote:
On 9/24/2026 11:03 AM, dart200 wrote:
[...]
You really are PO! ;^o Woha!
i really am not polcott,
and this is really the best feedback u sheeple can give?
You and him are VERY similar wrt the halting problem. Still not sure why.
On 9/24/2026 7:38 PM, Dude wrote:
On 9/24/2026 4:37 PM, Chris M. Thomasson wrote:
On 9/24/2026 3:44 PM, dart200 wrote:It's a 5th grader response.
[...]
Oh my. Please refrain from calling yourself God, PO.......
You can use restricted programming languages or subsets (such as MISRA
C, SPARK, or Rocq) that are not fully Turing-complete.
By banning unbounded loops and wild recursion, you ensure that every
valid program in the language is guaranteed to finish
lol You solve the halting problem by saying all must halt?
On 9/24/2026 7:43 PM, Chris M. Thomasson wrote:
On 9/24/2026 7:38 PM, Dude wrote:
On 9/24/2026 4:37 PM, Chris M. Thomasson wrote:
On 9/24/2026 3:44 PM, dart200 wrote:It's a 5th grader response.
[...]
Oh my. Please refrain from calling yourself God, PO.......
You can use restricted programming languages or subsets (such as
MISRA C, SPARK, or Rocq) that are not fully Turing-complete.
By banning unbounded loops and wild recursion, you ensure that every
valid program in the language is guaranteed to finish
lol You solve the halting problem by saying all must halt?
Nick came here to solve the Turing halting problem.
Turing proved that no algorithm can universally solve the halting
problem, establishing it as undecidable.
On 2026-09-26 19:45, dart200 wrote:
On 9/26/26 3:55 PM, dbush wrote:
What I don't get is why you think the agent writing its log to some
"private" area makes it special over a machine simulating the agent
coming up with the same decision and writing the same log to the tape.
it is representative of the fact u can't use a turing machine to
paradox _our_ innate ability to compute what any given turing machine
does...
Maybe you could expand upon this. What exactly do you mean by "_our_
innate ability to compute what any given turing machine does". Which
innate ability are you referring to? And since you claim to reject the Church-Turing Thesis, perhaps you could define *exactly* what the word 'compute' means to you (without mentioning TMs so as not to get tangled
up in Church-Turing).
For me, when we talk about 'computing' something we mean to solve
something using an algorithm, where an algorithm is purely mechanical sequence of deterministic steps which is guaranteed to lead one to the solution of the problem.
Put somewhat differently, an algorithm is a sequence of steps which, if followed, will always lead to the correct solution even when followed by someone (or something) who has absolutely no understanding of the actual problem being solved.
It's not clear to me that you share this definition.
Andr|-
in order to prove that paradox actually exist u need to compute what
the paradoxical machine does, to then show that it doesn't match the
output from the decider it is paradoxical in regards to.
On 9/24/2026 7:43 PM, Chris M. Thomasson wrote:
On 9/24/2026 7:38 PM, Dude wrote:
On 9/24/2026 4:37 PM, Chris M. Thomasson wrote:
On 9/24/2026 3:44 PM, dart200 wrote:It's a 5th grader response.
[...]
Oh my. Please refrain from calling yourself God, PO.......
You can use restricted programming languages or subsets (such as
MISRA C, SPARK, or Rocq) that are not fully Turing-complete.
By banning unbounded loops and wild recursion, you ensure that every
valid program in the language is guaranteed to finish
lol You solve the halting problem by saying all must halt?
Nick came here to solve the Turing halting problem.
Turing proved that no algorithm can universally solve the halting
problem, establishing it as undecidable.
On 9/26/26 7:36 PM, Andr|- G. Isaak wrote:
On 2026-09-26 19:45, dart200 wrote:
On 9/26/26 3:55 PM, dbush wrote:
What I don't get is why you think the agent writing its log to some
"private" area makes it special over a machine simulating the agent
coming up with the same decision and writing the same log to the tape.
it is representative of the fact u can't use a turing machine to
paradox _our_ innate ability to compute what any given turing machine
does...
Maybe you could expand upon this. What exactly do you mean by "_our_
innate ability to compute what any given turing machine does". Which
innate ability are you referring to? And since you claim to reject the
Church-Turing Thesis, perhaps you could define *exactly* what the word
'compute' means to you (without mentioning TMs so as not to get
tangled up in Church-Turing).
producing/writing down a binary (or any) sequence in a deterministic
manner with certainty. that is what computing most fundamentally is
For me, when we talk about 'computing' something we mean to solve
something using an algorithm, where an algorithm is purely mechanical
sequence of deterministic steps which is guaranteed to lead one to the
solution of the problem.
Put somewhat differently, an algorithm is a sequence of steps which,
if followed, will always lead to the correct solution even when
followed by someone (or something) who has absolutely no understanding
of the actual problem being solved.
i think this is still just another reiteration of the ct-thesis that tm-computing can encapsulate all of computing.
tm-computing is an
incredibly powerful tool for sure, but it's not able to compute all that
is computable, necessarily as a side effect of the power it does have
and if a man utilizes a level of intuition in a step while producing a sequence, i don't think that makes it not computing.
heck i might be
able to break down that step i suspect people will question (despite me giving demonstrations), into a more certain series of string transformations, which would be a stronger argument...
On 27/09/2026 04:52, Dude wrote:
On 9/24/2026 7:43 PM, Chris M. Thomasson wrote:
On 9/24/2026 7:38 PM, Dude wrote:Nick came here to solve the Turing halting problem.
On 9/24/2026 4:37 PM, Chris M. Thomasson wrote:
On 9/24/2026 3:44 PM, dart200 wrote:It's a 5th grader response.
[...]
Oh my. Please refrain from calling yourself God, PO.......
You can use restricted programming languages or subsets (such as
MISRA C, SPARK, or Rocq) that are not fully Turing-complete.
By banning unbounded loops and wild recursion, you ensure that every
valid program in the language is guaranteed to finish
lol You solve the halting problem by saying all must halt?
Turing proved that no algorithm can universally solve the halting
problem, establishing it as undecidable.
However, it is always possible to find a better partial solution.
On 9/24/26 4:35 PM, Chris M. Thomasson wrote:
On 9/24/2026 3:44 PM, dart200 wrote:
On 9/24/26 3:10 PM, Chris M. Thomasson wrote:
On 9/20/2026 10:39 PM, dart200 wrote:
On 9/20/26 7:51 PM, Chris M. Thomasson wrote:
On 9/20/2026 7:11 PM, dart200 wrote:
On 9/20/26 2:53 PM, Chris M. Thomasson wrote:
On 9/19/2026 2:38 PM, dart200 wrote:
[...]
If not, wtf! You remind me of PO. Sorry for that cut.
i don't care for ur fallacy by association brainrot
Sigh. I only said you kind of do remind me of the way PO dealt >>>>>>>> with the halting program.
and that statement has no bearing on the correctness of my
arguments, u brain-rotted boomer
You cannot predict a random number and you cannot solve the
halting problem. Sigh.
turing machines do not involve random numbers dud. if u knew
literally anything about basic computing theory u'd know that.
but u don't, so go back to ur fractals, eh?
Well, ponder on CSPRNG?
pseudo-random number generators are fully deterministic and therefore
entirely decidable in terms of whether it the machine which produces
it is circle-free or not (they are obviously). for any decision ofc
one needs a description of the machine including any input (like a
seed), and that combination can be used to decide on the semantics of
the described computation
when u say "random" that does not describe what can be produced by a
turing machine, as they can only compute pseudo-random sequences that
are inherently predictable, not truly random ones. if u meant pseudo-
random you should have said that...
seriously, bro: back to ur fractals, eh?
You have no idea how to get around the halting problem.
u have no idea what the halting problem even is, or the paradox that
makes it undecidable
and why anyone else isn't correcting u is just beyond me
On 9/26/26 3:49 PM, Chris M. Thomasson wrote:
On 9/24/2026 3:45 PM, dart200 wrote:
On 9/24/26 3:21 PM, Chris M. Thomasson wrote:
On 9/24/2026 11:03 AM, dart200 wrote:
[...]
You really are PO! ;^o Woha!
i really am not polcott,
and this is really the best feedback u sheeple can give?
You and him are VERY similar wrt the halting problem. Still not sure why.
bruh u make comments, and i don't really care because they never
demonstrate genuine consideration
On 9/24/26 3:21 PM, Chris M. Thomasson wrote:
On 9/24/2026 11:03 AM, dart200 wrote:
[...]
You really are PO! ;^o Woha!
i really am not polcott,
and this is really the best feedback u sheeple can give?
On 9/24/2026 7:43 PM, Chris M. Thomasson wrote:
On 9/24/2026 7:38 PM, Dude wrote:
On 9/24/2026 4:37 PM, Chris M. Thomasson wrote:
On 9/24/2026 3:44 PM, dart200 wrote:It's a 5th grader response.
[...]
Oh my. Please refrain from calling yourself God, PO.......
You can use restricted programming languages or subsets (such as
MISRA C, SPARK, or Rocq) that are not fully Turing-complete.
By banning unbounded loops and wild recursion, you ensure that every
valid program in the language is guaranteed to finish
lol You solve the halting problem by saying all must halt?
Nick came here to solve the Turing halting problem.
Turing proved that no algorithm can universally solve the halting
problem, establishing it as undecidable.
On 9/24/2026 6:16 PM, dart200 wrote:
On 9/24/26 4:35 PM, Chris M. Thomasson wrote:
On 9/24/2026 3:44 PM, dart200 wrote:
On 9/24/26 3:10 PM, Chris M. Thomasson wrote:
On 9/20/2026 10:39 PM, dart200 wrote:
On 9/20/26 7:51 PM, Chris M. Thomasson wrote:
On 9/20/2026 7:11 PM, dart200 wrote:
On 9/20/26 2:53 PM, Chris M. Thomasson wrote:
On 9/19/2026 2:38 PM, dart200 wrote:
[...]
If not, wtf! You remind me of PO. Sorry for that cut.
i don't care for ur fallacy by association brainrot
Sigh. I only said you kind of do remind me of the way PO dealt >>>>>>>>> with the halting program.
and that statement has no bearing on the correctness of my
arguments, u brain-rotted boomer
You cannot predict a random number and you cannot solve the
halting problem. Sigh.
turing machines do not involve random numbers dud. if u knew
literally anything about basic computing theory u'd know that.
but u don't, so go back to ur fractals, eh?
Well, ponder on CSPRNG?
pseudo-random number generators are fully deterministic and
therefore entirely decidable in terms of whether it the machine
which produces it is circle-free or not (they are obviously). for
any decision ofc one needs a description of the machine including
any input (like a seed), and that combination can be used to decide
on the semantics of the described computation
when u say "random" that does not describe what can be produced by a
turing machine, as they can only compute pseudo-random sequences
that are inherently predictable, not truly random ones. if u meant
pseudo- random you should have said that...
seriously, bro: back to ur fractals, eh?
You have no idea how to get around the halting problem.
u have no idea what the halting problem even is, or the paradox that
makes it undecidable
and why anyone else isn't correcting u is just beyond me
PO, its okay. We know way more about it. But, you already know about
that? Right? If not. WOW!
On 9/24/2026 3:45 PM, dart200 wrote:
On 9/24/26 3:21 PM, Chris M. Thomasson wrote:
You really are PO! ;^o Woha!
i really am not polcott,
Are you sure? You two are very much, alike.
On 9/28/26 2:33 AM, Mikko wrote:
On 27/09/2026 04:52, Dude wrote:
On 9/24/2026 7:43 PM, Chris M. Thomasson wrote:
On 9/24/2026 7:38 PM, Dude wrote:Nick came here to solve the Turing halting problem.
On 9/24/2026 4:37 PM, Chris M. Thomasson wrote:
On 9/24/2026 3:44 PM, dart200 wrote:It's a 5th grader response.
[...]
Oh my. Please refrain from calling yourself God, PO.......
You can use restricted programming languages or subsets (such as
MISRA C, SPARK, or Rocq) that are not fully Turing-complete.
By banning unbounded loops and wild recursion, you ensure that
every valid program in the language is guaranteed to finish
lol You solve the halting problem by saying all must halt?
Turing proved that no algorithm can universally solve the halting
problem, establishing it as undecidable.
However, it is always possible to find a better partial solution.
nah i have a proof against that
[ Followup-To: set ]
In comp.theory Chris M. Thomasson <chris.m.thomasson.1@gmail.com> wrote:
On 9/24/2026 3:45 PM, dart200 wrote:
On 9/24/26 3:21 PM, Chris M. Thomasson wrote:
[ .... ]
You really are PO! ;^o Woha!
i really am not polcott,
Are you sure? You two are very much, alike.
If you'd read any great deal of both of them, you would see straight
away they're not the same poster.
You've repeated that assertion so much it's in danger of becoming spam.
[ .... ]
On 9/28/26 3:00 PM, Chris M. Thomasson wrote:
On 9/24/2026 6:16 PM, dart200 wrote:
On 9/24/26 4:35 PM, Chris M. Thomasson wrote:
On 9/24/2026 3:44 PM, dart200 wrote:
On 9/24/26 3:10 PM, Chris M. Thomasson wrote:
On 9/20/2026 10:39 PM, dart200 wrote:
On 9/20/26 7:51 PM, Chris M. Thomasson wrote:
On 9/20/2026 7:11 PM, dart200 wrote:
On 9/20/26 2:53 PM, Chris M. Thomasson wrote:
On 9/19/2026 2:38 PM, dart200 wrote:
[...]
If not, wtf! You remind me of PO. Sorry for that cut.
i don't care for ur fallacy by association brainrot
Sigh. I only said you kind of do remind me of the way PO dealt >>>>>>>>>> with the halting program.
and that statement has no bearing on the correctness of my
arguments, u brain-rotted boomer
You cannot predict a random number and you cannot solve the
halting problem. Sigh.
turing machines do not involve random numbers dud. if u knew
literally anything about basic computing theory u'd know that.
but u don't, so go back to ur fractals, eh?
Well, ponder on CSPRNG?
pseudo-random number generators are fully deterministic and
therefore entirely decidable in terms of whether it the machine
which produces it is circle-free or not (they are obviously). for
any decision ofc one needs a description of the machine including
any input (like a seed), and that combination can be used to decide >>>>> on the semantics of the described computation
when u say "random" that does not describe what can be produced by
a turing machine, as they can only compute pseudo-random sequences
that are inherently predictable, not truly random ones. if u meant
pseudo- random you should have said that...
seriously, bro: back to ur fractals, eh?
You have no idea how to get around the halting problem.
u have no idea what the halting problem even is, or the paradox that
makes it undecidable
and why anyone else isn't correcting u is just beyond me
PO, its okay. We know way more about it. But, you already know about
that? Right? If not. WOW!
lol go back to ur fractals dud
On 09/24/2026 07:43 PM, Chris M. Thomasson wrote:
On 9/24/2026 7:36 PM, Dude wrote:
On 9/24/2026 6:16 PM, dart200 wrote:
On 9/24/26 4:35 PM, Chris M. Thomasson wrote:You cannot solve the halting problem for every possible program.
On 9/24/2026 3:44 PM, dart200 wrote:
On 9/24/26 3:10 PM, Chris M. Thomasson wrote:
On 9/20/2026 10:39 PM, dart200 wrote:
On 9/20/26 7:51 PM, Chris M. Thomasson wrote:
On 9/20/2026 7:11 PM, dart200 wrote:
On 9/20/26 2:53 PM, Chris M. Thomasson wrote:
On 9/19/2026 2:38 PM, dart200 wrote:
[...]
If not, wtf! You remind me of PO. Sorry for that cut. >>>>>>>>>>>>i don't care for ur fallacy by association brainrot
Sigh. I only said you kind of do remind me of the way PO dealt >>>>>>>>>>> with the halting program.
and that statement has no bearing on the correctness of my >>>>>>>>>> arguments, u brain-rotted boomer
You cannot predict a random number and you cannot solve the
halting problem. Sigh.
turing machines do not involve random numbers dud. if u knew
literally anything about basic computing theory u'd know that. >>>>>>>>
but u don't, so go back to ur fractals, eh?
Well, ponder on CSPRNG?
pseudo-random number generators are fully deterministic and
therefore entirely decidable in terms of whether it the machine
which produces it is circle-free or not (they are obviously). for
any decision ofc one needs a description of the machine including
any input (like a seed), and that combination can be used to decide >>>>>> on the semantics of the described computation
when u say "random" that does not describe what can be produced by >>>>>> a turing machine, as they can only compute pseudo-random sequences >>>>>> that are inherently predictable, not truly random ones. if u meant >>>>>> pseudo- random you should have said that...
seriously, bro: back to ur fractals, eh?
You have no idea how to get around the halting problem.
u have no idea what the halting problem even is, or the paradox that
makes it undecidable
and why anyone else isn't correcting u is just beyond me
But, you can get around it in practice by restricting your languages,
setting resource limits, or using automated parameters.
Alan Turing proved that no single program can correctly predict
whether any arbitrary code will finish running or loop forever.
However, real-world software engineering bypasses this theoretical
limit every day.
Where's Noah?
Say a program halts once all of its possible paths are hit?
In the finite it's all simple, and Turing's machine has at most
an _unbounded_ tape, which is a nice way of saying "not finite"
yet "each finite" since _quantifier disambiguation_ demands
distinction of differences for-any/for-each/for-every/for-all
when quantifying over the the infinite.
So, Chaitin's constant, or P(Halts), has a number of ways of
arriving at what it would be, whether it's approximately 0,
approximately 1, approximately 1/2, or as Chaitin arrives at
after a statistical account with "one standard deviation different
than the mean (of 1/2) about 0.85".
Anyways for any program of a bounded size, there is a program
of a bounded size that can determine whether it halts or not,
then those can be an infinite catalog or family of functions,
of static analysis, basically as a dirt-simple sketch of a
counter-example to the un-decide-ability of "Halts".
"Induction" is simply mathematical, what people make of it
is usually enough naively "empirical", which means simply enough "confirmation bias", anyways this was otherwise a useless thread.
On 28/09/2026 20:53, dart200 wrote:
On 9/28/26 2:33 AM, Mikko wrote:
On 27/09/2026 04:52, Dude wrote:
On 9/24/2026 7:43 PM, Chris M. Thomasson wrote:
On 9/24/2026 7:38 PM, Dude wrote:Nick came here to solve the Turing halting problem.
On 9/24/2026 4:37 PM, Chris M. Thomasson wrote:
On 9/24/2026 3:44 PM, dart200 wrote:It's a 5th grader response.
[...]
Oh my. Please refrain from calling yourself God, PO.......
You can use restricted programming languages or subsets (such as
MISRA C, SPARK, or Rocq) that are not fully Turing-complete.
By banning unbounded loops and wild recursion, you ensure that
every valid program in the language is guaranteed to finish
lol You solve the halting problem by saying all must halt?
Turing proved that no algorithm can universally solve the halting
problem, establishing it as undecidable.
However, it is always possible to find a better partial solution.
nah i have a proof against that
Every Turing machine is answered correctly by some partial decider.
Two partilal halt deciders can be combined to a better halt decider
that answers correctly about every Turing machine that at least one
of the component machines answers correctly. Therefore there is
always a way to construct a better partial halt deciders.
On 2026-09-28 01:12, dart200 wrote:
On 9/26/26 7:36 PM, Andr|- G. Isaak wrote:
On 2026-09-26 19:45, dart200 wrote:
On 9/26/26 3:55 PM, dbush wrote:
What I don't get is why you think the agent writing its log to some >>>>> "private" area makes it special over a machine simulating the agent >>>>> coming up with the same decision and writing the same log to the tape. >>>>it is representative of the fact u can't use a turing machine to
paradox _our_ innate ability to compute what any given turing
machine does...
Maybe you could expand upon this. What exactly do you mean by "_our_
innate ability to compute what any given turing machine does". Which
innate ability are you referring to? And since you claim to reject
the Church-Turing Thesis, perhaps you could define *exactly* what the
word 'compute' means to you (without mentioning TMs so as not to get
tangled up in Church-Turing).
producing/writing down a binary (or any) sequence in a deterministic
manner with certainty. that is what computing most fundamentally is
For me, when we talk about 'computing' something we mean to solve
something using an algorithm, where an algorithm is purely mechanical
sequence of deterministic steps which is guaranteed to lead one to
the solution of the problem.
Put somewhat differently, an algorithm is a sequence of steps which,
if followed, will always lead to the correct solution even when
followed by someone (or something) who has absolutely no
understanding of the actual problem being solved.
i think this is still just another reiteration of the ct-thesis that
tm-computing can encapsulate all of computing.
Since the description I gave doesn't even make mention of Turing
Machines, I'm not sure how it can be construed as a restatement of Church-Turing.
tm-computing is an incredibly powerful tool for sure, but it's not
able to compute all that is computable, necessarily as a side effect
of the power it does have
and if a man utilizes a level of intuition in a step while producing a
sequence, i don't think that makes it not computing.
'Intuition' clearly places something outside of the realm of algorithms
and thus outside the realm of computation.
heck i might be able to break down that step i suspect people will
question (despite me giving demonstrations), into a more certain
series of string transformations, which would be a stronger argument...
If you could reformulate 'intuition' as some deterministic sequence of mechanical steps, then we probably wouldn't want to call it 'intuition' anymore.
Andr|-
On 9/29/2026 2:43 AM, Alan Mackenzie wrote:
[ Followup-To: set ]
In comp.theory Chris M. Thomasson <chris.m.thomasson.1@gmail.com> wrote:
On 9/24/2026 3:45 PM, dart200 wrote:
On 9/24/26 3:21 PM, Chris M. Thomasson wrote:
[ .... ]
You really are PO! ;^o Woha!
i really am not polcott,
Are you sure? You two are very much, alike.
If you'd read any great deal of both of them, you would see straight
away they're not the same poster.
You've repeated that assertion so much it's in danger of becoming spam.
[ .... ]
Why do they [polcott and dart200] seem to act alike?
On 9/28/26 9:48 AM, Andr|- G. Isaak wrote:
On 2026-09-28 01:12, dart200 wrote:
On 9/26/26 7:36 PM, Andr|- G. Isaak wrote:
On 2026-09-26 19:45, dart200 wrote:
On 9/26/26 3:55 PM, dbush wrote:
What I don't get is why you think the agent writing its log to
some "private" area makes it special over a machine simulating the >>>>>> agent coming up with the same decision and writing the same log to >>>>>> the tape.
it is representative of the fact u can't use a turing machine to
paradox _our_ innate ability to compute what any given turing
machine does...
Maybe you could expand upon this. What exactly do you mean by "_our_
innate ability to compute what any given turing machine does". Which
innate ability are you referring to? And since you claim to reject
the Church-Turing Thesis, perhaps you could define *exactly* what
the word 'compute' means to you (without mentioning TMs so as not to
get tangled up in Church-Turing).
producing/writing down a binary (or any) sequence in a deterministic
manner with certainty. that is what computing most fundamentally is
For me, when we talk about 'computing' something we mean to solve
something using an algorithm, where an algorithm is purely
mechanical sequence of deterministic steps which is guaranteed to
lead one to the solution of the problem.
Put somewhat differently, an algorithm is a sequence of steps which,
if followed, will always lead to the correct solution even when
followed by someone (or something) who has absolutely no
understanding of the actual problem being solved.
i think this is still just another reiteration of the ct-thesis that
tm-computing can encapsulate all of computing.
Since the description I gave doesn't even make mention of Turing
Machines, I'm not sure how it can be construed as a restatement of
Church-Turing.
because ur assuming qualities of a turing machine, that the computer has zero fundamental understanding in any step of what it is computing. and
ur assuming that after almost a century of "propaganda" (turing's word
not mine) that turing's machines can effectively compute everything this
is "intuitive computable" by humans. u've lost perspective that
intuition can take place as part of a computation, just not a machine one
and honestly that part that requires intuition right now may in fact
have the kind of algorithm ur seeking ... i don't really have the time
left to do further intensive research without funding
tm-computing is an incredibly powerful tool for sure, but it's not
able to compute all that is computable, necessarily as a side effect
of the power it does have
and if a man utilizes a level of intuition in a step while producing
a sequence, i don't think that makes it not computing.
'Intuition' clearly places something outside of the realm of
algorithms and thus outside the realm of computation.
idk what writing down a sequence in a deterministic manner with
certainly is, except for computing the sequence
besides "impossible" (which is again asserting the ct-thesis) what would
u suggest that is???
heck i might be able to break down that step i suspect people will
question (despite me giving demonstrations), into a more certain
series of string transformations, which would be a stronger argument...
If you could reformulate 'intuition' as some deterministic sequence of
mechanical steps, then we probably wouldn't want to call it
'intuition' anymore.
Andr|-
On 9/28/2026 9:47 PM, dart200 wrote:
On 9/28/26 3:00 PM, Chris M. Thomasson wrote:
On 9/24/2026 6:16 PM, dart200 wrote:
On 9/24/26 4:35 PM, Chris M. Thomasson wrote:
On 9/24/2026 3:44 PM, dart200 wrote:
On 9/24/26 3:10 PM, Chris M. Thomasson wrote:
On 9/20/2026 10:39 PM, dart200 wrote:
On 9/20/26 7:51 PM, Chris M. Thomasson wrote:
On 9/20/2026 7:11 PM, dart200 wrote:
On 9/20/26 2:53 PM, Chris M. Thomasson wrote:
On 9/19/2026 2:38 PM, dart200 wrote:
[...]
If not, wtf! You remind me of PO. Sorry for that cut. >>>>>>>>>>>>i don't care for ur fallacy by association brainrot
Sigh. I only said you kind of do remind me of the way PO >>>>>>>>>>> dealt with the halting program.
and that statement has no bearing on the correctness of my >>>>>>>>>> arguments, u brain-rotted boomer
You cannot predict a random number and you cannot solve the >>>>>>>>> halting problem. Sigh.
turing machines do not involve random numbers dud. if u knew
literally anything about basic computing theory u'd know that. >>>>>>>>
but u don't, so go back to ur fractals, eh?
Well, ponder on CSPRNG?
pseudo-random number generators are fully deterministic and
therefore entirely decidable in terms of whether it the machine
which produces it is circle-free or not (they are obviously). for >>>>>> any decision ofc one needs a description of the machine including >>>>>> any input (like a seed), and that combination can be used to
decide on the semantics of the described computation
when u say "random" that does not describe what can be produced by >>>>>> a turing machine, as they can only compute pseudo-random sequences >>>>>> that are inherently predictable, not truly random ones. if u meant >>>>>> pseudo- random you should have said that...
seriously, bro: back to ur fractals, eh?
You have no idea how to get around the halting problem.
u have no idea what the halting problem even is, or the paradox that
makes it undecidable
and why anyone else isn't correcting u is just beyond me
PO, its okay. We know way more about it. But, you already know about
that? Right? If not. WOW!
lol go back to ur fractals dud
Remember my fuzzer code I made in AppleSoft BASIC? It only halts once
every path of the program under question has been hit. This is NOT
saying I solved the halting program, but its fun none the less, well according to me. ;^)
It fuzzes the program until all paths are hit. Once that occurs, it halts.
On 9/24/2026 6:16 PM, dart200 wrote:<SNIP>
On 9/24/26 4:35 PM, Chris M. Thomasson wrote:
seriously, bro: back to ur fractals, eh?
You have no idea how to get around the halting problem.
u have no idea what the halting problem even is, or the paradox that
makes it undecidable
and why anyone else isn't correcting u is just beyond me
PO, its okay. We know way more about it. But, you already know about
that? Right? If not. WOW!
On 9/30/2026 12:23 AM, Chris M. Thomasson wrote:
On 9/28/2026 9:47 PM, dart200 wrote:The main thing is that you halted the input of BASIC into a text editor
On 9/28/26 3:00 PM, Chris M. Thomasson wrote:
On 9/24/2026 6:16 PM, dart200 wrote:
On 9/24/26 4:35 PM, Chris M. Thomasson wrote:
On 9/24/2026 3:44 PM, dart200 wrote:
On 9/24/26 3:10 PM, Chris M. Thomasson wrote:
On 9/20/2026 10:39 PM, dart200 wrote:
On 9/20/26 7:51 PM, Chris M. Thomasson wrote:
On 9/20/2026 7:11 PM, dart200 wrote:
On 9/20/26 2:53 PM, Chris M. Thomasson wrote:
On 9/19/2026 2:38 PM, dart200 wrote:
[...]
If not, wtf! You remind me of PO. Sorry for that cut. >>>>>>>>>>>>>i don't care for ur fallacy by association brainrot
Sigh. I only said you kind of do remind me of the way PO >>>>>>>>>>>> dealt with the halting program.
and that statement has no bearing on the correctness of my >>>>>>>>>>> arguments, u brain-rotted boomer
You cannot predict a random number and you cannot solve the >>>>>>>>>> halting problem. Sigh.
turing machines do not involve random numbers dud. if u knew >>>>>>>>> literally anything about basic computing theory u'd know that. >>>>>>>>>
but u don't, so go back to ur fractals, eh?
Well, ponder on CSPRNG?
pseudo-random number generators are fully deterministic and
therefore entirely decidable in terms of whether it the machine >>>>>>> which produces it is circle-free or not (they are obviously). for >>>>>>> any decision ofc one needs a description of the machine including >>>>>>> any input (like a seed), and that combination can be used to
decide on the semantics of the described computation
when u say "random" that does not describe what can be produced >>>>>>> by a turing machine, as they can only compute pseudo-random
sequences that are inherently predictable, not truly random ones. >>>>>>> if u meant pseudo- random you should have said that...
seriously, bro: back to ur fractals, eh?
You have no idea how to get around the halting problem.
u have no idea what the halting problem even is, or the paradox
that makes it undecidable
and why anyone else isn't correcting u is just beyond me
PO, its okay. We know way more about it. But, you already know about
that? Right? If not. WOW!
lol go back to ur fractals dud
Remember my fuzzer code I made in AppleSoft BASIC? It only halts once
every path of the program under question has been hit. This is NOT
saying I solved the halting program, but its fun none the less, well
according to me. ;^)
for your own amusement, and you realized how boring it would be doing
that for a living somewhere in a dank basement for the rest of your
life. YMMV.
It fuzzes the program until all paths are hit. Once that occurs, it
halts.
Now you can write your paper and self-publish for review on Usenet. Now
get to work, Chris!
Chris M. Thomasson <chris.m.thomasson.1@gmail.com> wrote:
On 9/29/2026 2:43 AM, Alan Mackenzie wrote:
[ Followup-To: set ]
In comp.theory Chris M. Thomasson <chris.m.thomasson.1@gmail.com> wrote: >>>> On 9/24/2026 3:45 PM, dart200 wrote:
On 9/24/26 3:21 PM, Chris M. Thomasson wrote:
[ .... ]
You really are PO! ;^o Woha!
i really am not polcott,
Are you sure? You two are very much, alike.
If you'd read any great deal of both of them, you would see straight
away they're not the same poster.
You've repeated that assertion so much it's in danger of becoming spam.
[ .... ]
Why do they [polcott and dart200] seem to act alike?
What they have in common is a lack of respect for proven mathematics and
the achievements of great intellects of previous generations. Also, both
of their respective mathematical abilities are more meagre than they
believe; neither of them seems to understand proof by contradiction, for example.
Beyond that, they are different. Their posting mannerisms are different.
For example PO tends to write using full English words, correctly
punctuated, dart200 doesn't. PO writes in very short textual lines,
dart200 breaks up sentences by blank lines. Their uses of swear words differ. Only half of them try to divert discussion by citing
"fallacies". PO posts about his medical conditions, dart200 about his
family life.
And so on.
I'm not saying it would be impossible for a single poster to contrive
the two different personas we see, but that would take a lot of effort,
and why would anybody bother doing such a thing? The kick he'd get out
of deceiving Chris would be minimal compared with that effort.
On 9/26/2026 6:52 PM, Dude wrote:
On 9/24/2026 7:43 PM, Chris M. Thomasson wrote:
On 9/24/2026 7:38 PM, Dude wrote:Nick came here to solve the Turing halting problem.
On 9/24/2026 4:37 PM, Chris M. Thomasson wrote:
On 9/24/2026 3:44 PM, dart200 wrote:It's a 5th grader response.
[...]
Oh my. Please refrain from calling yourself God, PO.......
You can use restricted programming languages or subsets (such as
MISRA C, SPARK, or Rocq) that are not fully Turing-complete.
By banning unbounded loops and wild recursion, you ensure that every
valid program in the language is guaranteed to finish
lol You solve the halting problem by saying all must halt?
Turing proved that no algorithm can universally solve the halting
problem, establishing it as undecidable.
So, Nick is basically PO?
On 9/30/2026 9:36 AM, Dude wrote:
On 9/30/2026 12:23 AM, Chris M. Thomasson wrote:
On 9/28/2026 9:47 PM, dart200 wrote:The main thing is that you halted the input of BASIC into a text
On 9/28/26 3:00 PM, Chris M. Thomasson wrote:
On 9/24/2026 6:16 PM, dart200 wrote:
On 9/24/26 4:35 PM, Chris M. Thomasson wrote:
On 9/24/2026 3:44 PM, dart200 wrote:
On 9/24/26 3:10 PM, Chris M. Thomasson wrote:
On 9/20/2026 10:39 PM, dart200 wrote:
On 9/20/26 7:51 PM, Chris M. Thomasson wrote:
On 9/20/2026 7:11 PM, dart200 wrote:
On 9/20/26 2:53 PM, Chris M. Thomasson wrote:
On 9/19/2026 2:38 PM, dart200 wrote:
[...]
Sigh. I only said you kind of do remind me of the way PO >>>>>>>>>>>>> dealt with the halting program.If not, wtf! You remind me of PO. Sorry for that cut. >>>>>>>>>>>>>>i don't care for ur fallacy by association brainrot >>>>>>>>>>>>>
and that statement has no bearing on the correctness of my >>>>>>>>>>>> arguments, u brain-rotted boomer
You cannot predict a random number and you cannot solve the >>>>>>>>>>> halting problem. Sigh.
turing machines do not involve random numbers dud. if u knew >>>>>>>>>> literally anything about basic computing theory u'd know that. >>>>>>>>>>
but u don't, so go back to ur fractals, eh?
Well, ponder on CSPRNG?
pseudo-random number generators are fully deterministic and
therefore entirely decidable in terms of whether it the machine >>>>>>>> which produces it is circle-free or not (they are obviously). >>>>>>>> for any decision ofc one needs a description of the machine
including any input (like a seed), and that combination can be >>>>>>>> used to decide on the semantics of the described computation
when u say "random" that does not describe what can be produced >>>>>>>> by a turing machine, as they can only compute pseudo-random
sequences that are inherently predictable, not truly random
ones. if u meant pseudo- random you should have said that...
seriously, bro: back to ur fractals, eh?
You have no idea how to get around the halting problem.
u have no idea what the halting problem even is, or the paradox
that makes it undecidable
and why anyone else isn't correcting u is just beyond me
PO, its okay. We know way more about it. But, you already know
about that? Right? If not. WOW!
lol go back to ur fractals dud
Remember my fuzzer code I made in AppleSoft BASIC? It only halts once
every path of the program under question has been hit. This is NOT
saying I solved the halting program, but its fun none the less, well
according to me. ;^)
editor for your own amusement, and you realized how boring it would be
doing that for a living somewhere in a dank basement for the rest of
your life. YMMV.
Now you can write your paper and self-publish for review on Usenet.
It fuzzes the program until all paths are hit. Once that occurs, it
halts.
Now get to work, Chris!
;^) Fwiw, I wrote about it in this group. Just a fun way to fuzz a
program. It does not give a shit if it goes on forever or halts because
all possible paths were hit. It keeps a counter for every path it finds during static analysis on the program source code. So, it tries to fuzz
all on the conditionals. One counter per path. When all counters are non-zero, it says, we can halt for this accounting. That's all.
Actually, fuzzing is fairly helpful for testing things... :^)
Simply if we cannot hit all paths of a program, well, shit happens!
On 9/25/2026 11:28 AM, dart200 wrote:
[...]
You should read hours of PO nonsense as you are looking into a mirror.
On 9/30/26 5:47 AM, Alan Mackenzie wrote:
Chris M. Thomasson <chris.m.thomasson.1@gmail.com> wrote:
On 9/29/2026 2:43 AM, Alan Mackenzie wrote:
[ Followup-To: set ]
In comp.theory Chris M. Thomasson <chris.m.thomasson.1@gmail.com>
wrote:
On 9/24/2026 3:45 PM, dart200 wrote:
On 9/24/26 3:21 PM, Chris M. Thomasson wrote:
[ .... ]
You really are PO! ;^o Woha!
i really am not polcott,
Are you sure? You two are very much, alike.
If you'd read any great deal of both of them, you would see straight
away they're not the same poster.
You've repeated that assertion so much it's in danger of becoming spam.
[ .... ]
Why do they [polcott and dart200] seem to act alike?
What they have in common is a lack of respect for proven mathematics and
the achievements of great intellects of previous generations.-a Also, both
acting like the greats can never be wrong is when objectivity turns into propaganda, alan
for the record: turing wrote the most impactful math paper in human
history. certainly others were great as well, not discrediting them, but mechanical computing and the ability to imbue discrete logic into
objects, along with the sheer computational power of those objects, has clearly has the most significant impact on our progression as a species. narcissistic asshates scrambling for "muh AGI" have little appreciation
for just how miraculous computing already is...
and i'm _only_ improving on it u dipshit, nothing about my claims even remotely diminishes the the significance of his work.
the fact academics aren't even remotely open the idea truly exemplifies
how broken modern academia is. bureaucrats pushing for endless paper production and "innovation" for the sake of fame and ultimately
endorsements heavily distorts the academic mission of coherent and progressive knowledge production. we clearly aren't allowing for the flourishing of human intellect within academia anymore, and theoretical progression has mostly ground to a halt because of it
of their respective mathematical abilities are more meagre than they
believe; neither of them seems to understand proof by contradiction, for
example.
Beyond that, they are different.-a Their posting mannerisms are different. >> For example PO tends to write using full English words, correctly
punctuated, dart200 doesn't.-a PO writes in very short textual lines,
dart200 breaks up sentences by blank lines.-a Their uses of swear words
differ.-a Only half of them try to divert discussion by citing
"fallacies".-a PO posts about his medical conditions, dart200 about his
family life.
And so on.
I'm not saying it would be impossible for a single poster to contrive
the two different personas we see, but that would take a lot of effort,
and why would anybody bother doing such a thing?-a The kick he'd get out
of deceiving Chris would be minimal compared with that effort.
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