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On 6/3/2025 10:58 PM, olcott wrote:
On 6/3/2025 9:46 PM, dbush wrote:
On 6/3/2025 10:34 PM, olcott wrote:
On 6/3/2025 9:12 PM, dbush wrote:
Given any algorithm (i.e. a fixed immutable sequence of
instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that computes
the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed
directly
Yes there is no algorithm that does that
Excellent!
Let The Record Show
That Peter Olcott
Has *EXPLICITLY* admitted
That no algorithm H exists that meets the above requirements, which
is precisely the theorem that the halting problem proofs prove.
In the exact same way that there is no set of all set
that contain themselves. ZFC did not solve Russell's
Paradox as much as it showed that Russell's Paradox
was anchored in an incoherent foundation, now called
naive set theory.
Which arose because the axioms of naive set theory created a contradiction.
In contrast, the axioms of computation theory do *not* create a contradiction. It simply follows from those axioms that no H exists the meets the above requirements, which is a completely valid conclusion.
On 6/3/2025 10:02 PM, dbush wrote:
On 6/3/2025 10:58 PM, olcott wrote:
On 6/3/2025 9:46 PM, dbush wrote:
On 6/3/2025 10:34 PM, olcott wrote:
On 6/3/2025 9:12 PM, dbush wrote:
Given any algorithm (i.e. a fixed immutable sequence of
instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that computes >>>>>> the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly >>>>>> (<X>,Y) maps to 0 if and only if X(Y) does not halt when executed >>>>>> directly
Yes there is no algorithm that does that
Excellent!
Let The Record Show
That Peter Olcott
Has *EXPLICITLY* admitted
That no algorithm H exists that meets the above requirements, which
is precisely the theorem that the halting problem proofs prove.
In the exact same way that there is no set of all set
that contain themselves. ZFC did not solve Russell's
Paradox as much as it showed that Russell's Paradox
was anchored in an incoherent foundation, now called
naive set theory.
Which arose because the axioms of naive set theory created a
contradiction.
Likewise with halt deciders that are required to report
on the behavior of directly executed Turing machines.
Directly executed Turing machines are outside of the
domain of every Turing machine decider.
In contrast, the axioms of computation theory do *not* create a
contradiction. It simply follows from those axioms that no H exists
the meets the above requirements, which is a completely valid conclusion.
*Claude.ai seems to be the smartest bot about computation* https://claude.ai/share/48aab578-aec3-44a5-8bb3-6851e0f8b02e
On 7/4/25 4:43 PM, olcott wrote:
On 6/3/2025 10:02 PM, dbush wrote:
On 6/3/2025 10:58 PM, olcott wrote:
On 6/3/2025 9:46 PM, dbush wrote:
On 6/3/2025 10:34 PM, olcott wrote:
On 6/3/2025 9:12 PM, dbush wrote:
Given any algorithm (i.e. a fixed immutable sequence of
instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that computes >>>>>>> the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly >>>>>>> (<X>,Y) maps to 0 if and only if X(Y) does not halt when executed >>>>>>> directly
Yes there is no algorithm that does that
Excellent!
Let The Record Show
That Peter Olcott
Has *EXPLICITLY* admitted
That no algorithm H exists that meets the above requirements, which >>>>> is precisely the theorem that the halting problem proofs prove.
In the exact same way that there is no set of all set
that contain themselves. ZFC did not solve Russell's
Paradox as much as it showed that Russell's Paradox
was anchored in an incoherent foundation, now called
naive set theory.
Which arose because the axioms of naive set theory created a
contradiction.
Likewise with halt deciders that are required to report
on the behavior of directly executed Turing machines.
And what is the CONTRADICTION?
The result is just some things are not computable.
On 7/4/25 4:43 PM, olcott wrote:
On 6/3/2025 10:02 PM, dbush wrote:
On 6/3/2025 10:58 PM, olcott wrote:
On 6/3/2025 9:46 PM, dbush wrote:
On 6/3/2025 10:34 PM, olcott wrote:
On 6/3/2025 9:12 PM, dbush wrote:
Given any algorithm (i.e. a fixed immutable sequence of
instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that computes >>>>>>> the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly >>>>>>> (<X>,Y) maps to 0 if and only if X(Y) does not halt when executed >>>>>>> directly
Yes there is no algorithm that does that
Excellent!
Let The Record Show
That Peter Olcott
Has *EXPLICITLY* admitted
That no algorithm H exists that meets the above requirements, which >>>>> is precisely the theorem that the halting problem proofs prove.
In the exact same way that there is no set of all set
that contain themselves. ZFC did not solve Russell's
Paradox as much as it showed that Russell's Paradox
was anchored in an incoherent foundation, now called
naive set theory.
Which arose because the axioms of naive set theory created a
contradiction.
Likewise with halt deciders that are required to report
on the behavior of directly executed Turing machines.
And what is the CONTRADICTION?
The result is just some things are not computable.
Directly executed Turing machines are outside of the
domain of every Turing machine decider.
Then so is mathematics, as "numbers" can't be given to Turing Machines,
only representations of them.
By the exact same idea that we can represent a number by a finite--
string, we can express the algorithm, and input, of a Turing Machine as
a finite string, and thus can talk about what it will do.
In contrast, the axioms of computation theory do *not* create a
contradiction. It simply follows from those axioms that no H exists
the meets the above requirements, which is a completely valid
conclusion.
*Claude.ai seems to be the smartest bot about computation*
https://claude.ai/share/48aab578-aec3-44a5-8bb3-6851e0f8b02e
Which you just continue to lie to, so proving that you are just a pathological liar.
On 7/4/2025 3:53 PM, Richard Damon wrote:
On 7/4/25 4:43 PM, olcott wrote:
On 6/3/2025 10:02 PM, dbush wrote:
On 6/3/2025 10:58 PM, olcott wrote:
On 6/3/2025 9:46 PM, dbush wrote:
On 6/3/2025 10:34 PM, olcott wrote:In the exact same way that there is no set of all set
On 6/3/2025 9:12 PM, dbush wrote:
Given any algorithm (i.e. a fixed immutable sequence of
instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that
computes the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly >>>>>>>> (<X>,Y) maps to 0 if and only if X(Y) does not halt when
executed directly
Yes there is no algorithm that does that
Excellent!
Let The Record Show
That Peter Olcott
Has *EXPLICITLY* admitted
That no algorithm H exists that meets the above requirements,
which is precisely the theorem that the halting problem proofs prove. >>>>>
that contain themselves. ZFC did not solve Russell's
Paradox as much as it showed that Russell's Paradox
was anchored in an incoherent foundation, now called
naive set theory.
Which arose because the axioms of naive set theory created a
contradiction.
Likewise with halt deciders that are required to report
on the behavior of directly executed Turing machines.
And what is the CONTRADICTION?
The result is just some things are not computable.
Directly executed Turing machines are outside of the
domain of every Turing machine decider.
Then so is mathematics, as "numbers" can't be given to Turing
Machines, only representations of them.
Numbers always work the same way so it makes no difference.
*HHH(DDD)==0 and HHH1(DDD)==1 are both correct* https://claude.ai/share/da9b8e3f-eb16-42ca-a9e8-913f4b88202c
When we compare DDD emulated by HHH and DDD emulated
by HHH1 SIDE-BY-SIDE. (Mike didn't do it this way).
*The difference is when*
HHH begins to simulate itself simulating DDD and
HHH1 NEVER begins to simulate itself simulating DDD.
On 7/4/2025 3:53 PM, Richard Damon wrote:
On 7/4/25 4:43 PM, olcott wrote:
On 6/3/2025 10:02 PM, dbush wrote:
On 6/3/2025 10:58 PM, olcott wrote:
On 6/3/2025 9:46 PM, dbush wrote:
On 6/3/2025 10:34 PM, olcott wrote:In the exact same way that there is no set of all set
On 6/3/2025 9:12 PM, dbush wrote:
Given any algorithm (i.e. a fixed immutable sequence of
instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that
computes the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly >>>>>>>> (<X>,Y) maps to 0 if and only if X(Y) does not halt when
executed directly
Yes there is no algorithm that does that
Excellent!
Let The Record Show
That Peter Olcott
Has *EXPLICITLY* admitted
That no algorithm H exists that meets the above requirements,
which is precisely the theorem that the halting problem proofs prove. >>>>>
that contain themselves. ZFC did not solve Russell's
Paradox as much as it showed that Russell's Paradox
was anchored in an incoherent foundation, now called
naive set theory.
Which arose because the axioms of naive set theory created a
contradiction.
Likewise with halt deciders that are required to report
on the behavior of directly executed Turing machines.
And what is the CONTRADICTION?
The result is just some things are not computable.
The result is that there cannot possibly be
an *ACTUAL INPUT* that does the opposite of
whatever its partial halt decider decides
thus the HP proof fails before it begins.
On 7/4/2025 3:53 PM, Richard Damon wrote:
On 7/4/25 4:43 PM, olcott wrote:
On 6/3/2025 10:02 PM, dbush wrote:
On 6/3/2025 10:58 PM, olcott wrote:
On 6/3/2025 9:46 PM, dbush wrote:
On 6/3/2025 10:34 PM, olcott wrote:In the exact same way that there is no set of all set
On 6/3/2025 9:12 PM, dbush wrote:
Given any algorithm (i.e. a fixed immutable sequence of
instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that
computes the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly >>>>>>>> (<X>,Y) maps to 0 if and only if X(Y) does not halt when
executed directly
Yes there is no algorithm that does that
Excellent!
Let The Record Show
That Peter Olcott
Has *EXPLICITLY* admitted
That no algorithm H exists that meets the above requirements,
which is precisely the theorem that the halting problem proofs prove. >>>>>
that contain themselves. ZFC did not solve Russell's
Paradox as much as it showed that Russell's Paradox
was anchored in an incoherent foundation, now called
naive set theory.
Which arose because the axioms of naive set theory created a
contradiction.
Likewise with halt deciders that are required to report
on the behavior of directly executed Turing machines.
And what is the CONTRADICTION?
The result is just some things are not computable.
Directly executed Turing machines are outside of the
domain of every Turing machine decider.
Then so is mathematics, as "numbers" can't be given to Turing
Machines, only representations of them.
Numbers always work the same way so it makes no difference.
*HHH(DDD)==0 and HHH1(DDD)==1 are both correct* https://claude.ai/share/da9b8e3f-eb16-42ca-a9e8-913f4b88202c
When we compare DDD emulated by HHH and DDD emulated
by HHH1 SIDE-BY-SIDE. (Mike didn't do it this way).
*The difference is when*
HHH begins to simulate itself simulating DDD and
HHH1 NEVER begins to simulate itself simulating DDD.
HHH doesn't actually abort its simulation of DDD until
after has simulated many hundreds of simulated instructions
later. HHH simulates itself simulating DDD until DDD calls
HHH(DDD) again.
By the exact same idea that we can represent a number by a finite
string, we can express the algorithm, and input, of a Turing Machine
as a finite string, and thus can talk about what it will do.
In contrast, the axioms of computation theory do *not* create a
contradiction. It simply follows from those axioms that no H exists >>>> the meets the above requirements, which is a completely valid
conclusion.
*Claude.ai seems to be the smartest bot about computation*
https://claude.ai/share/48aab578-aec3-44a5-8bb3-6851e0f8b02e
Which you just continue to lie to, so proving that you are just a
pathological liar.
On 7/4/25 4:43 PM, olcott wrote:
On 6/3/2025 10:02 PM, dbush wrote:
On 6/3/2025 10:58 PM, olcott wrote:Likewise with halt deciders that are required to report on the behavior
On 6/3/2025 9:46 PM, dbush wrote:
On 6/3/2025 10:34 PM, olcott wrote:
On 6/3/2025 9:12 PM, dbush wrote:
Yes there is no algorithm that does that
Given any algorithm (i.e. a fixed immutable sequence of
instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that computes >>>>>>> the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly >>>>>>> (<X>,Y) maps to 0 if and only if X(Y) does not halt when executed >>>>>>> directly
Excellent!
Let The Record Show
That Peter Olcott
Has *EXPLICITLY* admitted
That no algorithm H exists that meets the above requirements, which
is precisely the theorem that the halting problem proofs prove.
In the exact same way that there is no set of all set that contain
themselves. ZFC did not solve Russell's Paradox as much as it showed
that Russell's Paradox was anchored in an incoherent foundation, now
called naive set theory.
Which arose because the axioms of naive set theory created a
contradiction.
of directly executed Turing machines.
And what is the CONTRADICTION?
The result is just some things are not computable.
Directly executed Turing machines are outside of the domain of every
Turing machine decider.
Then so is mathematics, as "numbers" can't be given to Turing Machines,
only representations of them.
By the exact same idea that we can represent a number by a finite
string, we can express the algorithm, and input, of a Turing Machine as
a finite string, and thus can talk about what it will do.
Which you just continue to lie to, so proving that you are just a pathological liar.In contrast, the axioms of computation theory do *not* create a
contradiction. It simply follows from those axioms that no H exists
the meets the above requirements, which is a completely valid
conclusion.
*Claude.ai seems to be the smartest bot about computation*
https://claude.ai/share/48aab578-aec3-44a5-8bb3-6851e0f8b02e
On Fri, 04 Jul 2025 16:53:50 -0400, Richard Damon wrote:
On 7/4/25 4:43 PM, olcott wrote:
On 6/3/2025 10:02 PM, dbush wrote:
On 6/3/2025 10:58 PM, olcott wrote:Likewise with halt deciders that are required to report on the behavior
On 6/3/2025 9:46 PM, dbush wrote:
On 6/3/2025 10:34 PM, olcott wrote:
On 6/3/2025 9:12 PM, dbush wrote:
Yes there is no algorithm that does that
Given any algorithm (i.e. a fixed immutable sequence of
instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that computes >>>>>>>> the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly >>>>>>>> (<X>,Y) maps to 0 if and only if X(Y) does not halt when executed >>>>>>>> directly
Excellent!
Let The Record Show
That Peter Olcott
Has *EXPLICITLY* admitted
That no algorithm H exists that meets the above requirements, which >>>>>> is precisely the theorem that the halting problem proofs prove.
In the exact same way that there is no set of all set that contain
themselves. ZFC did not solve Russell's Paradox as much as it showed >>>>> that Russell's Paradox was anchored in an incoherent foundation, now >>>>> called naive set theory.
Which arose because the axioms of naive set theory created a
contradiction.
of directly executed Turing machines.
And what is the CONTRADICTION?
The result is just some things are not computable.
Directly executed Turing machines are outside of the domain of every
Turing machine decider.
Then so is mathematics, as "numbers" can't be given to Turing Machines,
only representations of them.
By the exact same idea that we can represent a number by a finite
string, we can express the algorithm, and input, of a Turing Machine as
a finite string, and thus can talk about what it will do.
Which you just continue to lie to, so proving that you are just aIn contrast, the axioms of computation theory do *not* create a
contradiction. It simply follows from those axioms that no H exists
the meets the above requirements, which is a completely valid
conclusion.
*Claude.ai seems to be the smartest bot about computation*
https://claude.ai/share/48aab578-aec3-44a5-8bb3-6851e0f8b02e
pathological liar.
It is YOU who is the pathological liar: the recursive self reference in
the classical halting problem proofs is a category error just as in
Russell's Paradox. All halting problems based on such an erroneous construction are thus refuted, here and now, by me, Mr Flibble.
/Flibble
Op 05.jul.2025 om 00:15 schreef olcott:
On 7/4/2025 3:53 PM, Richard Damon wrote:Misleading words. Both simulators are given the same input. Both are simulating the same input, which includes HHH, not HHH1. So, there is no difference.
On 7/4/25 4:43 PM, olcott wrote:
On 6/3/2025 10:02 PM, dbush wrote:
On 6/3/2025 10:58 PM, olcott wrote:
On 6/3/2025 9:46 PM, dbush wrote:
On 6/3/2025 10:34 PM, olcott wrote:
On 6/3/2025 9:12 PM, dbush wrote:
Given any algorithm (i.e. a fixed immutable sequence of
instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that
computes the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly >>>>>>>>> (<X>,Y) maps to 0 if and only if X(Y) does not halt when
executed directly
Yes there is no algorithm that does that
Excellent!
Let The Record Show
That Peter Olcott
Has *EXPLICITLY* admitted
That no algorithm H exists that meets the above requirements,
which is precisely the theorem that the halting problem proofs
prove.
In the exact same way that there is no set of all set
that contain themselves. ZFC did not solve Russell's
Paradox as much as it showed that Russell's Paradox
was anchored in an incoherent foundation, now called
naive set theory.
Which arose because the axioms of naive set theory created a
contradiction.
Likewise with halt deciders that are required to report
on the behavior of directly executed Turing machines.
And what is the CONTRADICTION?
The result is just some things are not computable.
Directly executed Turing machines are outside of the
domain of every Turing machine decider.
Then so is mathematics, as "numbers" can't be given to Turing
Machines, only representations of them.
Numbers always work the same way so it makes no difference.
*HHH(DDD)==0 and HHH1(DDD)==1 are both correct*
https://claude.ai/share/da9b8e3f-eb16-42ca-a9e8-913f4b88202c
When we compare DDD emulated by HHH and DDD emulated
by HHH1 SIDE-BY-SIDE. (Mike didn't do it this way).
*The difference is when*
HHH begins to simulate itself simulating DDD and
HHH1 NEVER begins to simulate itself simulating DDD.
If there is a difference, you could point to the first instruction that--
has a different result in the simulations. But you can't.
On 7/4/25 6:11 PM, olcott wrote:
On 7/4/2025 3:53 PM, Richard Damon wrote:
On 7/4/25 4:43 PM, olcott wrote:
On 6/3/2025 10:02 PM, dbush wrote:
On 6/3/2025 10:58 PM, olcott wrote:
On 6/3/2025 9:46 PM, dbush wrote:
On 6/3/2025 10:34 PM, olcott wrote:
On 6/3/2025 9:12 PM, dbush wrote:
Given any algorithm (i.e. a fixed immutable sequence of
instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that
computes the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly >>>>>>>>> (<X>,Y) maps to 0 if and only if X(Y) does not halt when
executed directly
Yes there is no algorithm that does that
Excellent!
Let The Record Show
That Peter Olcott
Has *EXPLICITLY* admitted
That no algorithm H exists that meets the above requirements,
which is precisely the theorem that the halting problem proofs
prove.
In the exact same way that there is no set of all set
that contain themselves. ZFC did not solve Russell's
Paradox as much as it showed that Russell's Paradox
was anchored in an incoherent foundation, now called
naive set theory.
Which arose because the axioms of naive set theory created a
contradiction.
Likewise with halt deciders that are required to report
on the behavior of directly executed Turing machines.
And what is the CONTRADICTION?
The result is just some things are not computable.
The result is that there cannot possibly be
an *ACTUAL INPUT* that does the opposite of
whatever its partial halt decider decides
thus the HP proof fails before it begins.
Sure there is.
On 7/4/25 6:15 PM, olcott wrote:
On 7/4/2025 3:53 PM, Richard Damon wrote:
On 7/4/25 4:43 PM, olcott wrote:
On 6/3/2025 10:02 PM, dbush wrote:
On 6/3/2025 10:58 PM, olcott wrote:
On 6/3/2025 9:46 PM, dbush wrote:
On 6/3/2025 10:34 PM, olcott wrote:
On 6/3/2025 9:12 PM, dbush wrote:
Given any algorithm (i.e. a fixed immutable sequence of
instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that
computes the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly >>>>>>>>> (<X>,Y) maps to 0 if and only if X(Y) does not halt when
executed directly
Yes there is no algorithm that does that
Excellent!
Let The Record Show
That Peter Olcott
Has *EXPLICITLY* admitted
That no algorithm H exists that meets the above requirements,
which is precisely the theorem that the halting problem proofs
prove.
In the exact same way that there is no set of all set
that contain themselves. ZFC did not solve Russell's
Paradox as much as it showed that Russell's Paradox
was anchored in an incoherent foundation, now called
naive set theory.
Which arose because the axioms of naive set theory created a
contradiction.
Likewise with halt deciders that are required to report
on the behavior of directly executed Turing machines.
And what is the CONTRADICTION?
The result is just some things are not computable.
Directly executed Turing machines are outside of the
domain of every Turing machine decider.
Then so is mathematics, as "numbers" can't be given to Turing
Machines, only representations of them.
Numbers always work the same way so it makes no difference.
So do programs. (When they are programs)
*HHH(DDD)==0 and HHH1(DDD)==1 are both correct*
https://claude.ai/share/da9b8e3f-eb16-42ca-a9e8-913f4b88202c
Based on your LIE:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0
that the pattern HHH used was an actual non-termination pattern
On 7/5/2025 8:14 AM, Richard Damon wrote:
On 7/4/25 6:11 PM, olcott wrote:
On 7/4/2025 3:53 PM, Richard Damon wrote:
On 7/4/25 4:43 PM, olcott wrote:
On 6/3/2025 10:02 PM, dbush wrote:
On 6/3/2025 10:58 PM, olcott wrote:
On 6/3/2025 9:46 PM, dbush wrote:
On 6/3/2025 10:34 PM, olcott wrote:
On 6/3/2025 9:12 PM, dbush wrote:
Given any algorithm (i.e. a fixed immutable sequence of
instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that
computes the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed >>>>>>>>>> directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when
executed directly
Yes there is no algorithm that does that
Excellent!
Let The Record Show
That Peter Olcott
Has *EXPLICITLY* admitted
That no algorithm H exists that meets the above requirements, >>>>>>>> which is precisely the theorem that the halting problem proofs >>>>>>>> prove.
In the exact same way that there is no set of all set
that contain themselves. ZFC did not solve Russell's
Paradox as much as it showed that Russell's Paradox
was anchored in an incoherent foundation, now called
naive set theory.
Which arose because the axioms of naive set theory created a
contradiction.
Likewise with halt deciders that are required to report
on the behavior of directly executed Turing machines.
And what is the CONTRADICTION?
The result is just some things are not computable.
The result is that there cannot possibly be
an *ACTUAL INPUT* that does the opposite of
whatever its partial halt decider decides
thus the HP proof fails before it begins.
Sure there is.
In order to have an honest dialogue you must pay
100% complete attention to every single word.
You can't just erase one of the words that I said
and then form a rebuttal on that basis.
Directly executed Turing machines have always been
outside of the domain of every Turing machine based
decider.
On 7/5/2025 8:20 AM, Richard Damon wrote:
On 7/4/25 6:15 PM, olcott wrote:
On 7/4/2025 3:53 PM, Richard Damon wrote:
On 7/4/25 4:43 PM, olcott wrote:
On 6/3/2025 10:02 PM, dbush wrote:
On 6/3/2025 10:58 PM, olcott wrote:
On 6/3/2025 9:46 PM, dbush wrote:
On 6/3/2025 10:34 PM, olcott wrote:
On 6/3/2025 9:12 PM, dbush wrote:
Given any algorithm (i.e. a fixed immutable sequence of
instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that
computes the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed >>>>>>>>>> directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when
executed directly
Yes there is no algorithm that does that
Excellent!
Let The Record Show
That Peter Olcott
Has *EXPLICITLY* admitted
That no algorithm H exists that meets the above requirements, >>>>>>>> which is precisely the theorem that the halting problem proofs >>>>>>>> prove.
In the exact same way that there is no set of all set
that contain themselves. ZFC did not solve Russell's
Paradox as much as it showed that Russell's Paradox
was anchored in an incoherent foundation, now called
naive set theory.
Which arose because the axioms of naive set theory created a
contradiction.
Likewise with halt deciders that are required to report
on the behavior of directly executed Turing machines.
And what is the CONTRADICTION?
The result is just some things are not computable.
Directly executed Turing machines are outside of the
domain of every Turing machine decider.
Then so is mathematics, as "numbers" can't be given to Turing
Machines, only representations of them.
Numbers always work the same way so it makes no difference.
So do programs. (When they are programs)
*HHH(DDD)==0 and HHH1(DDD)==1 are both correct*
https://claude.ai/share/da9b8e3f-eb16-42ca-a9e8-913f4b88202c
Based on your LIE:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0
that the pattern HHH used was an actual non-termination pattern
Since it proves that it figured out all of the details
of this non-terminating pattern itself its agreement
is not biased by my generic definition of the notion
of simulating termination analyzer.
HHH simulating DDD:
- Enter DDD()
- Call HHH(DDD)
- Pattern detected: infinite recursion of HHH simulating DDD
- Simulation aborted, return 0
On 7/5/25 12:26 PM, olcott wrote:
On 7/5/2025 8:14 AM, Richard Damon wrote:
On 7/4/25 6:11 PM, olcott wrote:
On 7/4/2025 3:53 PM, Richard Damon wrote:
On 7/4/25 4:43 PM, olcott wrote:
On 6/3/2025 10:02 PM, dbush wrote:
On 6/3/2025 10:58 PM, olcott wrote:
On 6/3/2025 9:46 PM, dbush wrote:
On 6/3/2025 10:34 PM, olcott wrote:
On 6/3/2025 9:12 PM, dbush wrote:
Given any algorithm (i.e. a fixed immutable sequence of >>>>>>>>>>> instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that >>>>>>>>>>> computes the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed >>>>>>>>>>> directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when >>>>>>>>>>> executed directly
Yes there is no algorithm that does that
Excellent!
Let The Record Show
That Peter Olcott
Has *EXPLICITLY* admitted
That no algorithm H exists that meets the above requirements, >>>>>>>>> which is precisely the theorem that the halting problem proofs >>>>>>>>> prove.
In the exact same way that there is no set of all set
that contain themselves. ZFC did not solve Russell's
Paradox as much as it showed that Russell's Paradox
was anchored in an incoherent foundation, now called
naive set theory.
Which arose because the axioms of naive set theory created a
contradiction.
Likewise with halt deciders that are required to report
on the behavior of directly executed Turing machines.
And what is the CONTRADICTION?
The result is just some things are not computable.
The result is that there cannot possibly be
an *ACTUAL INPUT* that does the opposite of
whatever its partial halt decider decides
thus the HP proof fails before it begins.
Sure there is.
In order to have an honest dialogue you must pay
100% complete attention to every single word.
You can't just erase one of the words that I said
and then form a rebuttal on that basis.
Directly executed Turing machines have always been
outside of the domain of every Turing machine based
decider.
Nope.
Your refusal to providee a source is your admission that you are just a liar.
Remember, The DEFINITION of a Halt Deicder is that it is to be a decider that decides if the program represented by its input will halt when run.
On 7/5/2025 7:44 PM, Richard Damon wrote:
On 7/5/25 12:26 PM, olcott wrote:
On 7/5/2025 8:14 AM, Richard Damon wrote:
On 7/4/25 6:11 PM, olcott wrote:
On 7/4/2025 3:53 PM, Richard Damon wrote:
On 7/4/25 4:43 PM, olcott wrote:
On 6/3/2025 10:02 PM, dbush wrote:
On 6/3/2025 10:58 PM, olcott wrote:
On 6/3/2025 9:46 PM, dbush wrote:
On 6/3/2025 10:34 PM, olcott wrote:
On 6/3/2025 9:12 PM, dbush wrote:
Given any algorithm (i.e. a fixed immutable sequence of >>>>>>>>>>>> instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that >>>>>>>>>>>> computes the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed >>>>>>>>>>>> directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when >>>>>>>>>>>> executed directly
Yes there is no algorithm that does that
Excellent!
Let The Record Show
That Peter Olcott
Has *EXPLICITLY* admitted
That no algorithm H exists that meets the above requirements, >>>>>>>>>> which is precisely the theorem that the halting problem proofs >>>>>>>>>> prove.
In the exact same way that there is no set of all set
that contain themselves. ZFC did not solve Russell's
Paradox as much as it showed that Russell's Paradox
was anchored in an incoherent foundation, now called
naive set theory.
Which arose because the axioms of naive set theory created a
contradiction.
Likewise with halt deciders that are required to report
on the behavior of directly executed Turing machines.
And what is the CONTRADICTION?
The result is just some things are not computable.
The result is that there cannot possibly be
an *ACTUAL INPUT* that does the opposite of
whatever its partial halt decider decides
thus the HP proof fails before it begins.
Sure there is.
In order to have an honest dialogue you must pay
100% complete attention to every single word.
You can't just erase one of the words that I said
and then form a rebuttal on that basis.
Directly executed Turing machines have always been
outside of the domain of every Turing machine based
decider.
Nope.
Your refusal to providee a source is your admission that you are just
a liar.
Remember, The DEFINITION of a Halt Deicder is that it is to be a
decider that decides if the program represented by its input will halt
when run.
It has never been the program represented by its input
it has always been the behavior specified by its input.
This is the key mistake that no one noticed in 90 years.
On 7/5/25 10:43 PM, olcott wrote:
On 7/5/2025 7:44 PM, Richard Damon wrote:
On 7/5/25 12:26 PM, olcott wrote:
On 7/5/2025 8:14 AM, Richard Damon wrote:
On 7/4/25 6:11 PM, olcott wrote:
On 7/4/2025 3:53 PM, Richard Damon wrote:
On 7/4/25 4:43 PM, olcott wrote:
On 6/3/2025 10:02 PM, dbush wrote:
On 6/3/2025 10:58 PM, olcott wrote:
On 6/3/2025 9:46 PM, dbush wrote:
On 6/3/2025 10:34 PM, olcott wrote:
On 6/3/2025 9:12 PM, dbush wrote:
Given any algorithm (i.e. a fixed immutable sequence of >>>>>>>>>>>>> instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that >>>>>>>>>>>>> computes the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed >>>>>>>>>>>>> directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when >>>>>>>>>>>>> executed directly
Yes there is no algorithm that does that
Excellent!
Let The Record Show
That Peter Olcott
Has *EXPLICITLY* admitted
That no algorithm H exists that meets the above requirements, >>>>>>>>>>> which is precisely the theorem that the halting problem >>>>>>>>>>> proofs prove.
In the exact same way that there is no set of all set
that contain themselves. ZFC did not solve Russell's
Paradox as much as it showed that Russell's Paradox
was anchored in an incoherent foundation, now called
naive set theory.
Which arose because the axioms of naive set theory created a >>>>>>>>> contradiction.
Likewise with halt deciders that are required to report
on the behavior of directly executed Turing machines.
And what is the CONTRADICTION?
The result is just some things are not computable.
The result is that there cannot possibly be
an *ACTUAL INPUT* that does the opposite of
whatever its partial halt decider decides
thus the HP proof fails before it begins.
Sure there is.
In order to have an honest dialogue you must pay
100% complete attention to every single word.
You can't just erase one of the words that I said
and then form a rebuttal on that basis.
Directly executed Turing machines have always been
outside of the domain of every Turing machine based
decider.
Nope.
Your refusal to providee a source is your admission that you are just
a liar.
Remember, The DEFINITION of a Halt Deicder is that it is to be a
decider that decides if the program represented by its input will
halt when run.
It has never been the program represented by its input
it has always been the behavior specified by its input.
This is the key mistake that no one noticed in 90 years.
Really?
In computability theory, the halting problem is the problem of
determining, from a description of an arbitrary computer program and an input, whether the program will finish running, or continue to run forever.
Sounds like the program and its representation.
The Program comes first, and THAT is what the Halting Mapping is based on.
The finite string representation is the implementation detail for giving
it to the decider.
It seems you don't even understand the 101 level terms.
You are just proving how stupid and ignorant you are. A self-made
stupidity and ignorance, because you are afraid the truth will brainwash you, so you preemptively brainwashed yourself to be immune to the truth.
This just make you a pathological liar.
On 7/5/2025 2:39 AM, Fred. Zwarts wrote:
Op 05.jul.2025 om 00:15 schreef olcott:
On 7/4/2025 3:53 PM, Richard Damon wrote:Misleading words. Both simulators are given the same input. Both are
On 7/4/25 4:43 PM, olcott wrote:
On 6/3/2025 10:02 PM, dbush wrote:
On 6/3/2025 10:58 PM, olcott wrote:
On 6/3/2025 9:46 PM, dbush wrote:
On 6/3/2025 10:34 PM, olcott wrote:
On 6/3/2025 9:12 PM, dbush wrote:
Given any algorithm (i.e. a fixed immutable sequence of
instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that
computes the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed >>>>>>>>>> directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when
executed directly
Yes there is no algorithm that does that
Excellent!
Let The Record Show
That Peter Olcott
Has *EXPLICITLY* admitted
That no algorithm H exists that meets the above requirements, >>>>>>>> which is precisely the theorem that the halting problem proofs >>>>>>>> prove.
In the exact same way that there is no set of all set
that contain themselves. ZFC did not solve Russell's
Paradox as much as it showed that Russell's Paradox
was anchored in an incoherent foundation, now called
naive set theory.
Which arose because the axioms of naive set theory created a
contradiction.
Likewise with halt deciders that are required to report
on the behavior of directly executed Turing machines.
And what is the CONTRADICTION?
The result is just some things are not computable.
Directly executed Turing machines are outside of the
domain of every Turing machine decider.
Then so is mathematics, as "numbers" can't be given to Turing
Machines, only representations of them.
Numbers always work the same way so it makes no difference.
*HHH(DDD)==0 and HHH1(DDD)==1 are both correct*
https://claude.ai/share/da9b8e3f-eb16-42ca-a9e8-913f4b88202c
When we compare DDD emulated by HHH and DDD emulated
by HHH1 SIDE-BY-SIDE. (Mike didn't do it this way).
*The difference is when*
HHH begins to simulate itself simulating DDD and
HHH1 NEVER begins to simulate itself simulating DDD.
simulating the same input, which includes HHH, not HHH1. So, there is
no difference.
*Its just over your head*
HHH(DDD) Does simulate itself simulating DDD
thus causing recursive simulation that cannot
possibly reach the simulated final halt state of DDD.
On 7/5/2025 10:02 PM, Richard Damon wrote:
On 7/5/25 10:43 PM, olcott wrote:
On 7/5/2025 7:44 PM, Richard Damon wrote:
On 7/5/25 12:26 PM, olcott wrote:
On 7/5/2025 8:14 AM, Richard Damon wrote:
On 7/4/25 6:11 PM, olcott wrote:
On 7/4/2025 3:53 PM, Richard Damon wrote:
On 7/4/25 4:43 PM, olcott wrote:
On 6/3/2025 10:02 PM, dbush wrote:
On 6/3/2025 10:58 PM, olcott wrote:
On 6/3/2025 9:46 PM, dbush wrote:
On 6/3/2025 10:34 PM, olcott wrote:
On 6/3/2025 9:12 PM, dbush wrote:
Given any algorithm (i.e. a fixed immutable sequence of >>>>>>>>>>>>>> instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that >>>>>>>>>>>>>> computes the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed >>>>>>>>>>>>>> directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when >>>>>>>>>>>>>> executed directly
Yes there is no algorithm that does that
Excellent!
Let The Record Show
That Peter Olcott
Has *EXPLICITLY* admitted
That no algorithm H exists that meets the above
requirements, which is precisely the theorem that the >>>>>>>>>>>> halting problem proofs prove.
In the exact same way that there is no set of all set
that contain themselves. ZFC did not solve Russell's
Paradox as much as it showed that Russell's Paradox
was anchored in an incoherent foundation, now called
naive set theory.
Which arose because the axioms of naive set theory created a >>>>>>>>>> contradiction.
Likewise with halt deciders that are required to report
on the behavior of directly executed Turing machines.
And what is the CONTRADICTION?
The result is just some things are not computable.
The result is that there cannot possibly be
an *ACTUAL INPUT* that does the opposite of
whatever its partial halt decider decides
thus the HP proof fails before it begins.
Sure there is.
In order to have an honest dialogue you must pay
100% complete attention to every single word.
You can't just erase one of the words that I said
and then form a rebuttal on that basis.
Directly executed Turing machines have always been
outside of the domain of every Turing machine based
decider.
Nope.
Your refusal to providee a source is your admission that you are
just a liar.
Remember, The DEFINITION of a Halt Deicder is that it is to be a
decider that decides if the program represented by its input will
halt when run.
It has never been the program represented by its input
it has always been the behavior specified by its input.
This is the key mistake that no one noticed in 90 years.
Really?
In computability theory, the halting problem is the problem of
determining, from a description of an arbitrary computer program and
an input, whether the program will finish running, or continue to run
forever.
Sounds like the program and its representation.
With pathological self-reference the directly
executed machine will not have the same
behavior as the correctly simulated machine
specification.