On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:
On 2/27/26 6:09 AM, dart200 wrote:
On 2/27/26 2:51 AM, Tristan Wibberley wrote:
On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on >>>>>>>>>>>>> turing's original paper and can be reused anywhere... >>>>>>>>>>>>You think you can test all of them one by one? Don't tell me >>>>>>>>>>>> you think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude >>>>>>>>>> one by one.
that exactly what turing does in his proof: he defines a
comptuation
that enumerates out all the numbers, testing each one of they >>>>>>>>> represent
a "satisfactory"/"circle-free" machine, and adding that to
diagonal
across defined across computable numbers
it really would be a great exercise to carefully read p247 of >>>>>>>>> turing's
proof and produce the psuedo-code for the machine H, assuming that >>>>>>>>> machine D exists
I'll get to it sooner then, because it's mad. Are you sure he >>>>>>>> didn't
reason quantified over all but phrase it like a procedure for >>>>>>>> what he
the theory of computation is the theory of such procedures, and >>>>>>> understanding the diagonal procedure is critical to understanding >>>>>>> the *base* contradiction/paradox that the rest of his support for >>>>>>> godel's result is then built on
And focusing on what is said to be impossible and not changing the >>>>>> problem is also important.
The problem with the diagonal generation isn't the generation of
the diagonal itself, but effectively enumerating the enumeration
in the first place.
i don't see any indication that turing realized a difference there
Then you zre just showing your stupidity, because YOU can't tell the
difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable.
Here is is SPECIFICALLY talking about the effective enumeration of
the computable sequences.
He then points out that he can directly show that the "anti-
diagonal" of the (non-effectively computed) enumeration can't be
computed but that "This proof, although perfectly sound, has the
disadvantage that it may leave the reader with a feeling that 'there
must be something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being done.
I think he is refering he to the standard anti-diagonal arguement,
which shows that since for all n, position n differs from the value in
number n, there can not be any element that matches the anti-diagonal.
It is just a natural fact of countable infinity, something it seems
you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with turing's short
diagonal proof, but in writing this post i now find myself in a subtle,
yet entirely critical disagreement:
/let an be the n-th computable sequence, and let -an(m) be the m-th
figure in an. Let +# be the sequence with 1--an(m) as its n-th. figure. Since +# is computable, there exists a number K [== +#] such that 1--an(n)
= -aK(n) for all n. Putting n = K, we have 1 = 2-aK(K), i.e. 1 is even.
This is impossible/
the fallacy here is assuming that because the direct diagonal is
computable, that one can therefore compute the anti-diagonal using the direct diagonal. the abstract definition makes it look simple, but this ignores the complexities of self-referential analysis (like what turing details on the next page)
in both methods i have for rectifying the paradox found in the direct diagonal (either (1) filtering TMs or (2) using RTMs), neither can be
used to then compute the anti-diagonal
in (1) the algo to compute an inverse diagonal is filtered out like
turing's paradoxical variation of the direct diagonal would be, and
there is no analogous non-paradoxical variation that has a hard coded
value that is inverse to what it does return ... such a concept is
entirely nonsensical. a function can only return what it does, it can't
also return the inverse to what it returns eh???
in (2) the attempt to compute an inverse diagonal with RTMs just fails
for reasons u'd only understand by working thru the algo urself (p7 of
re: turing's diagonals)
the premise:
/Let +# be the sequence with 1--an(m) as its n-th/
= -aK(n) for all n/
\_(paa)_/->
is just not sufficient evidence that such +# is actually computable given the direct diagonal -an()
one cannot just assume that because the diagonal across computable
numbers is computable, therefore the anti-diagonal across computable
numbers is computable...
He doesn't. You are just showing your stupidity,
He is proving the enumeration is uncomputable, and without the
enumeration, you can't compute either of them.
neither method i have for fixing the diagonal computation across the
computable numbers can be used to compute the inverse diagonal
But your method still doesn't let you compute the enumeration, and
thus you can't actually compute the diagonal.
Remember, the problem definitions requires that the listing be a
COMPLETE listing of the computable numbers / machine that compute
computable numbers, in some definite order.
If your enumeration isn't complete, your diagonal isn't correct.
so while i agree with turing that the anti-diagonal is not
computable, i don't agree that the normal diagonal is not computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on all the other
machines, including his original H that doesn't use your "trick"
But instead, he can prove with a more obvious process, that the
Decider "D" that could be used to effectively enumerate the sequence
of machine that produce computable numbers can not esit.
Thus, he clearly knows the difference, but is pointing out that the
attempt to compute the diagonal clearly reveals the issue with
effectively enumerating the sequences.
well, he didn't consider that perhaps the proper algo for computing
the diagonal can avoid the paradox on itself ...
But it doesn't.
Your just don't understand that D just can't correctly decide on his
given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on turing's H,
because my response to this is that D does not need to decide correctly
on H to compute a diagonal
It doesn't matter that your new H doesn't get stuck on itself, it will
still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he doesn't specify what D (or H) needs to do when encountering the /undecidable input/ of
H, so therefore both D and H are an incomplete specifications of a machine
IF D is wrong by deciding it is not circle free, then your H will
compute the wrong diagonal, as the resulting version of his H WILL be
circle free (since it never tries to simulate itself) and thus DOES
produce an computable number that your computation misses.
Or, if that D is wrong by decing it IS circle free, then when you H
tries to process it, it will get stuck in the infinite loop.
The problem is that in stepping through the machines in order, you
WILL hit these actual machines built on your erroneous D (your D must
have this flaw, as no D without exists), and thus you will be wrong on
THAT input. IT doesn't matter if you get a good answer for yourself.
idk what he would have said about it, but prolly something more
substantial than just calling me ignorant repeatedly
I doubt it.
He likely would have gotten frustrated by your idiodic assertion of
bad logic. You would have likely been escorted out of the meeting as
showing you were unqualified and being a distraction.
Something that seems to be beyond your ignorant understanding.
interestingly: one can only fix the direct diagonal computation
H shows that *IF* you can make that enumeration, you can make the >>>>>> diagonal, and thus the anti-diagonal. The problem is you can't
make that enumeration, and assuming you can just shows unsoundness. >>>>>
like this
u can't do an analogous fix for the inverse/anti-diagonal
computation. it's not possible hard code a machine to return an
inverted value, a machine can only return what it does, not the
inverse of what it does...
so if we can filter out paradoxes from the enumeration, that will
leave a direct diagonal computation extant in that filtered (yet
still turing complete list), while any attempt to compute an
inverse diagonal will not be
But the problem is that "paradoxical machines" don't exist in
isolation, but only in relationship to a given machine trying to
decide them.
right. so if ur constructing a diagonal across computable numbers
then u only need to filter out paradoxes in regards to the classifier
that classifies them as a "satisfactory" number
Right, which he shows can not be done.
please do quote where turing shows we can't filter out such paradoxes...
(also why do always just make random assertions???)
any machine which *is not* "satisfactory" OR *is not* classifiable as
satisfactory by said classifier... can just be skipped
No, it can only skip those that are not satisfactory, not those that
are but it can not classify as such, or your enumeration will not be
complete, and thus just in error.
Thus, it needs to be able to correctly classify ALL machines (as it
will be asked about all machines as it counts through all the
descriptions) and thus Turing's H *WILL* be asked about.
similarly if u want to go a step further an filter out computable
numbers already included on this diagonal, any machine which either
*is* computably equivalent OR *is not* classifiable in regards to
*any* machine already the list... can just be skipped
Nope, you can't skip some machines, as you then might lose some of the
computable numbers.
see you can't compute a diagonal across *all* /machines/, with said
machines, but u can compute a diagonal across *all* /computable numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be done, since
ALL classifiers that attempt it will make an error, you can't do what
you want to do.
nah, (a) computing an enumeration of all /computable numbers/ is not the same thing as (b) computing the enumeration of all machines that compute computable numbers. (b) necessarily has duplicates while (a) does not
need them. turing's paper wrongly conflates (a) with (b)
i'm pretty sure (a) can be done with TMs
(b) probably can't be done with TMs
yes, i still do need to prove my thesis that for any paradoxical
machine, there exists a functionally equivalent machine without such
paradox
And the problem is that your "paradoxical" isn't actually a definable
property (let alone computable). Part of the problem is that if you
look at just a machine description, it doesn't (necessarily) tell you
about the use of an "interface" as that use of an interface can be
just inlined, leaving nothing "in the code" to show it exists.
i'm sorry, are you actually saying the machine description does not
describe what the machine does???
lol
His specified H, with an actually (incorrect) implementation of D
(which is all that CAN exist) will either be circle-free and thus
generate a number (but its D said it isn't, and thus omitted a valid
machine from the list) or it isn't circle-free, and fails to computa
a number, and thus should have been omitted from the list but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it is just built
on an assuption in error.
so despite turing's worries, the existence of a diagonal
computation does not actually then imply the existence of an anti-
diagonal computation, due the same particular self-referential
weirdness that stumped turing the first place
But there is no actuall SELF-REFERENCE, so your logic is just based
on ERROR.
Your attempt to REDEFINE self-reference to mean processing a number
that happens to represent yourself means that you you system "ALL"
doesn't actually mean ALL, and thus is just ill-defined.
i'm sorry, you have an issue with me labeling a number that directly
refers to yourself, as a "self-reference" ???
Sure, because it is just a number. The problem is that you still have
it's a *specific* number that has the currently running machine encoded
into it, it's not "just" a number whatever that means
problems with all the "equivalent" machines that have different numbers.
those are references to functionally equivalent machines, not self- references
It may let you filter out the simplest case used in the proofs, but
doesn't solve the actual problem, as the "Machine Number" doesn't
actually fully identify the problematic cases.
that's not actually true. you can't meaningfully paradox the paradox detector while filter *out* paradoxes to point of creating an actual
problem
let halts be a halting classifier that takes an input machines and
returns true/false whether it halts or not
let paradox_free be a paradox filtering classifier that takes an input classifier, and an input machine to search for paradoxes in regards to: paradox_free(classifier, machine)
for example if take our basic halting paradox:
-a und = () -> halts(und) loop()
then:
-a paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox classifier like such:
-a undp = () -> {
-a-a-a if ( paradox_free(halts, undp) == TRUE )
-a-a-a-a-a if ( halts(undp) ) loop()
-a }
here if paradox_free(halts,undp) -> FALSE, then the if case run a
halting paradox form making it /undecidable input/ to halts().
however, if paradox_free(halts,undp) -> TRUE, then that if case doesn't
run and clearly halts(undp) -> TRUE without issues
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within a machine is necessary for the particular input->output computation being done, so utilizing the return paradox_free(halts,undp) -> FALSE to filter out paradox_free() paradoxes as well as halts() paradoxes will suffice to produce a turing complete subset of machines that can be totally
classified by halts()
i think a major problem with the consensus perspective is an expectation
of a certain cleanliness in the logic as being the only possibility here
to compute what we want. TM's can't achieve that, but something slightly messier can.
dear future: u see this shit i'm trying work with???? sorry i didn't
manage to make progress any faster, but i'm literally dragging
toddlers kicking and screaming at this point jeez...
In other words, in your world "ALL" isn't a word that can actually be
used.
You don't understand that you need to use actual sound logic,
lol, i can only EfOA someone would engage in sound logic with me
This just shows that you real problem is you don't understand what
...which u won't find interesting, but i can't fix a lack of curiosity >>>>>
Efn+
expected was an inexpert audience, supposing that experts will >>>>>>>> recognise
the relevant mapping to universal quantification?
the actual problem is, and your world is just build on things that
are lies.
IT seems that fundamentally, your world doesn't actually have
"computations" as you don't understand the basic requirement that
they need to be fully defined in the actions they do.
On 3/3/26 12:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:
On 2/27/26 6:09 AM, dart200 wrote:
On 2/27/26 2:51 AM, Tristan Wibberley wrote:
On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on >>>>>>>>>>>>>> turing's original paper and can be reused anywhere... >>>>>>>>>>>>>You think you can test all of them one by one? Don't tell >>>>>>>>>>>>> me you think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude >>>>>>>>>>> one by one.
that exactly what turing does in his proof: he defines a
comptuation
that enumerates out all the numbers, testing each one of they >>>>>>>>>> represent
a "satisfactory"/"circle-free" machine, and adding that to >>>>>>>>>> diagonal
across defined across computable numbers
it really would be a great exercise to carefully read p247 of >>>>>>>>>> turing's
proof and produce the psuedo-code for the machine H, assuming >>>>>>>>>> that
machine D exists
I'll get to it sooner then, because it's mad. Are you sure he >>>>>>>>> didn't
reason quantified over all but phrase it like a procedure for >>>>>>>>> what he
the theory of computation is the theory of such procedures, and >>>>>>>> understanding the diagonal procedure is critical to
understanding the *base* contradiction/paradox that the rest of >>>>>>>> his support for godel's result is then built on
And focusing on what is said to be impossible and not changing
the problem is also important.
The problem with the diagonal generation isn't the generation of >>>>>>> the diagonal itself, but effectively enumerating the enumeration >>>>>>> in the first place.
i don't see any indication that turing realized a difference there
Then you zre just showing your stupidity, because YOU can't tell
the difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable.
Here is is SPECIFICALLY talking about the effective enumeration of
the computable sequences.
He then points out that he can directly show that the "anti-
diagonal" of the (non-effectively computed) enumeration can't be
computed but that "This proof, although perfectly sound, has the
disadvantage that it may leave the reader with a feeling that
'there must be something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being done.
I think he is refering he to the standard anti-diagonal arguement,
which shows that since for all n, position n differs from the value
in number n, there can not be any element that matches the anti-
diagonal.
It is just a natural fact of countable infinity, something it seems
you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with turing's short
diagonal proof, but in writing this post i now find myself in a
subtle, yet entirely critical disagreement:
/let an be the n-th computable sequence, and let -an(m) be the m-th
figure in an. Let +# be the sequence with 1--an(m) as its n-th. figure.
Since +# is computable, there exists a number K [== +#] such that 1-
-an(n) = -aK(n) for all n. Putting n = K, we have 1 = 2-aK(K), i.e. 1 is
even. This is impossible/
the fallacy here is assuming that because the direct diagonal is
computable, that one can therefore compute the anti-diagonal using the
direct diagonal. the abstract definition makes it look simple, but
this ignores the complexities of self-referential analysis (like what
turing details on the next page)
in both methods i have for rectifying the paradox found in the direct
diagonal (either (1) filtering TMs or (2) using RTMs), neither can be
used to then compute the anti-diagonal
in (1) the algo to compute an inverse diagonal is filtered out like
turing's paradoxical variation of the direct diagonal would be, and
there is no analogous non-paradoxical variation that has a hard coded
value that is inverse to what it does return ... such a concept is
entirely nonsensical. a function can only return what it does, it
can't also return the inverse to what it returns eh???
in (2) the attempt to compute an inverse diagonal with RTMs just fails
for reasons u'd only understand by working thru the algo urself (p7 of
re: turing's diagonals)
the premise:
/Let +# be the sequence with 1--an(m) as its n-th/
that sentence there, ben, from p246,
is the sentence of turing's paper /on computable numbers/ that i start
to diagree with,
that sentence just is wrong to imply anything about the computability of
a number
/there exists a number K [== +#] such that 1--an(n)
= -aK(n) for all n/
and that sentence is *only* /half-true/
it's *correct* in that can't be computed by a TM,
whoever a human with a TM could still write it down, but a human could
never pass that input entirely to a finite running machine eh???
so i kinda think the ct-thesis is actually cooked in a way
\_(paa)_/->
is just not sufficient evidence that such +# is actually computable
given the direct diagonal -an()
one cannot just assume that because the diagonal across computable
numbers is computable, therefore the anti-diagonal across computable
numbers is computable...
He doesn't. You are just showing your stupidity,
He is proving the enumeration is uncomputable, and without the
enumeration, you can't compute either of them.
neither method i have for fixing the diagonal computation across the
computable numbers can be used to compute the inverse diagonal
But your method still doesn't let you compute the enumeration, and
thus you can't actually compute the diagonal.
Remember, the problem definitions requires that the listing be a
COMPLETE listing of the computable numbers / machine that compute
computable numbers, in some definite order.
If your enumeration isn't complete, your diagonal isn't correct.
so while i agree with turing that the anti-diagonal is not
computable, i don't agree that the normal diagonal is not computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on all the other
machines, including his original H that doesn't use your "trick"
But instead, he can prove with a more obvious process, that the
Decider "D" that could be used to effectively enumerate the
sequence of machine that produce computable numbers can not esit.
Thus, he clearly knows the difference, but is pointing out that the >>>>> attempt to compute the diagonal clearly reveals the issue with
effectively enumerating the sequences.
well, he didn't consider that perhaps the proper algo for computing
the diagonal can avoid the paradox on itself ...
But it doesn't.
Your just don't understand that D just can't correctly decide on his
given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on turing's H,
because my response to this is that D does not need to decide
correctly on H to compute a diagonal
It doesn't matter that your new H doesn't get stuck on itself, it
will still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he doesn't
specify what D (or H) needs to do when encountering the /undecidable
input/ of H, so therefore both D and H are an incomplete
specifications of a machine
IF D is wrong by deciding it is not circle free, then your H will
compute the wrong diagonal, as the resulting version of his H WILL be
circle free (since it never tries to simulate itself) and thus DOES
produce an computable number that your computation misses.
Or, if that D is wrong by decing it IS circle free, then when you H
tries to process it, it will get stuck in the infinite loop.
The problem is that in stepping through the machines in order, you
WILL hit these actual machines built on your erroneous D (your D must
have this flaw, as no D without exists), and thus you will be wrong
on THAT input. IT doesn't matter if you get a good answer for yourself.
idk what he would have said about it, but prolly something more
substantial than just calling me ignorant repeatedly
I doubt it.
He likely would have gotten frustrated by your idiodic assertion of
bad logic. You would have likely been escorted out of the meeting as
showing you were unqualified and being a distraction.
Something that seems to be beyond your ignorant understanding.
interestingly: one can only fix the direct diagonal computation
H shows that *IF* you can make that enumeration, you can make the >>>>>>> diagonal, and thus the anti-diagonal. The problem is you can't
make that enumeration, and assuming you can just shows unsoundness. >>>>>>
like this
u can't do an analogous fix for the inverse/anti-diagonal
computation. it's not possible hard code a machine to return an
inverted value, a machine can only return what it does, not the
inverse of what it does...
so if we can filter out paradoxes from the enumeration, that will >>>>>> leave a direct diagonal computation extant in that filtered (yet
still turing complete list), while any attempt to compute an
inverse diagonal will not be
But the problem is that "paradoxical machines" don't exist in
isolation, but only in relationship to a given machine trying to
decide them.
right. so if ur constructing a diagonal across computable numbers
then u only need to filter out paradoxes in regards to the
classifier that classifies them as a "satisfactory" number
Right, which he shows can not be done.
please do quote where turing shows we can't filter out such paradoxes...
(also why do always just make random assertions???)
any machine which *is not* "satisfactory" OR *is not* classifiable
as satisfactory by said classifier... can just be skipped
No, it can only skip those that are not satisfactory, not those that
are but it can not classify as such, or your enumeration will not be
complete, and thus just in error.
Thus, it needs to be able to correctly classify ALL machines (as it
will be asked about all machines as it counts through all the
descriptions) and thus Turing's H *WILL* be asked about.
similarly if u want to go a step further an filter out computable
numbers already included on this diagonal, any machine which either
*is* computably equivalent OR *is not* classifiable in regards to
*any* machine already the list... can just be skipped
Nope, you can't skip some machines, as you then might lose some of
the computable numbers.
see you can't compute a diagonal across *all* /machines/, with said
machines, but u can compute a diagonal across *all* /computable
numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be done, since
ALL classifiers that attempt it will make an error, you can't do what
you want to do.
nah, (a) computing an enumeration of all /computable numbers/ is not
the same thing as (b) computing the enumeration of all machines that
compute computable numbers. (b) necessarily has duplicates while (a)
does not need them. turing's paper wrongly conflates (a) with (b)
i'm pretty sure (a) can be done with TMs
(b) probably can't be done with TMs
yes, i still do need to prove my thesis that for any paradoxical
machine, there exists a functionally equivalent machine without such
paradox
And the problem is that your "paradoxical" isn't actually a definable
property (let alone computable). Part of the problem is that if you
look at just a machine description, it doesn't (necessarily) tell you
about the use of an "interface" as that use of an interface can be
just inlined, leaving nothing "in the code" to show it exists.
i'm sorry, are you actually saying the machine description does not
describe what the machine does???
lol
His specified H, with an actually (incorrect) implementation of D
(which is all that CAN exist) will either be circle-free and thus
generate a number (but its D said it isn't, and thus omitted a
valid machine from the list) or it isn't circle-free, and fails to
computa a number, and thus should have been omitted from the list
but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it is just
built on an assuption in error.
so despite turing's worries, the existence of a diagonal
computation does not actually then imply the existence of an anti- >>>>>> diagonal computation, due the same particular self-referential
weirdness that stumped turing the first place
But there is no actuall SELF-REFERENCE, so your logic is just based >>>>> on ERROR.
Your attempt to REDEFINE self-reference to mean processing a number >>>>> that happens to represent yourself means that you you system "ALL"
doesn't actually mean ALL, and thus is just ill-defined.
i'm sorry, you have an issue with me labeling a number that directly
refers to yourself, as a "self-reference" ???
Sure, because it is just a number. The problem is that you still have
it's a *specific* number that has the currently running machine
encoded into it, it's not "just" a number whatever that means
problems with all the "equivalent" machines that have different numbers.
those are references to functionally equivalent machines, not self-
references
It may let you filter out the simplest case used in the proofs, but
doesn't solve the actual problem, as the "Machine Number" doesn't
actually fully identify the problematic cases.
that's not actually true. you can't meaningfully paradox the paradox
detector while filter *out* paradoxes to point of creating an actual
problem
let halts be a halting classifier that takes an input machines and
returns true/false whether it halts or not
let paradox_free be a paradox filtering classifier that takes an input
classifier, and an input machine to search for paradoxes in regards
to: paradox_free(classifier, machine)
for example if take our basic halting paradox:
-a-a und = () -> halts(und) loop()
then:
-a-a paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox classifier
like such:
-a-a undp = () -> {
-a-a-a-a if ( paradox_free(halts, undp) == TRUE )
-a-a-a-a-a-a if ( halts(undp) ) loop()
-a-a }
here if paradox_free(halts,undp) -> FALSE, then the if case run a
halting paradox form making it /undecidable input/ to halts().
however, if paradox_free(halts,undp) -> TRUE, then that if case
doesn't run and clearly halts(undp) -> TRUE without issues
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within a machine is
necessary for the particular input->output computation being done, so
utilizing the return paradox_free(halts,undp) -> FALSE to filter out
paradox_free() paradoxes as well as halts() paradoxes will suffice to
produce a turing complete subset of machines that can be totally
classified by halts()
i think a major problem with the consensus perspective is an
expectation of a certain cleanliness in the logic as being the only
possibility here to compute what we want. TM's can't achieve that, but
something slightly messier can.
dear future: u see this shit i'm trying work with???? sorry i didn't
manage to make progress any faster, but i'm literally dragging
toddlers kicking and screaming at this point jeez...
In other words, in your world "ALL" isn't a word that can actually be
used.
You don't understand that you need to use actual sound logic,
lol, i can only EfOA someone would engage in sound logic with me
This just shows that you real problem is you don't understand what
...which u won't find interesting, but i can't fix a lack of
curiosity
Efn+
expected was an inexpert audience, supposing that experts will >>>>>>>>> recognise
the relevant mapping to universal quantification?
the actual problem is, and your world is just build on things that
are lies.
IT seems that fundamentally, your world doesn't actually have
"computations" as you don't understand the basic requirement that
they need to be fully defined in the actions they do.
| Sysop: | Amessyroom |
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