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Feb 21, 2023, 12:17:18 PMFeb 21

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When the ultimate measure of correct simulation is that the execution

trace of the simulated input exactly matches the behavior that the input

machine description specifies then:

*It is an easily verified fact that* Every counter-example input to the

halting theorem D cannot possibly reach its own simulated final state in

any finite number of steps when correctly simulated by simulating halt

decider H.

The halting theorem does not prove that a set of input pairs cannot be

divided into halting and not halting. It only proves that one criterion

measure for dividing these pairs does not always work.

*This same reasoning equally applies to the Turing machine based proofs*

int D(int (*x)())

{

int Halt_Status = H(x, x);

if (Halt_Status)

HERE: goto HERE;

return Halt_Status;

}

It is a verified fact that H correctly predicts that D correctly

simulated by H would never reach its own final state and terminate

normally, thus H does correctly decide halting for its input D.

Anyone with sufficient software engineering skill knows that

*D simulated by H cannot possibly correctly reach its ret instruction*

Everyone else lacks sufficient software engineering skill or lies

_D()

[00001d12] 55 push ebp

[00001d13] 8bec mov ebp,esp

[00001d15] 51 push ecx

[00001d16] 8b4508 mov eax,[ebp+08]

[00001d19] 50 push eax // push D

[00001d1a] 8b4d08 mov ecx,[ebp+08]

[00001d1d] 51 push ecx // push D

[00001d1e] e83ff8ffff call 00001562 // call H

[00001d23] 83c408 add esp,+08

[00001d26] 8945fc mov [ebp-04],eax

[00001d29] 837dfc00 cmp dword [ebp-04],+00

[00001d2d] 7402 jz 00001d31

[00001d2f] ebfe jmp 00001d2f

[00001d31] 8b45fc mov eax,[ebp-04]

[00001d34] 8be5 mov esp,ebp

[00001d36] 5d pop ebp

[00001d37] c3 ret

*Simulating Halt Deciders Defeat the Halting Theorem*

https://www.researchgate.net/publication/368568464_Simulating_Halt_Deciders_Defeat_the_Halting_Theorem

--

Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius

hits a target no one else can see." Arthur Schopenhauer

trace of the simulated input exactly matches the behavior that the input

machine description specifies then:

*It is an easily verified fact that* Every counter-example input to the

halting theorem D cannot possibly reach its own simulated final state in

any finite number of steps when correctly simulated by simulating halt

decider H.

The halting theorem does not prove that a set of input pairs cannot be

divided into halting and not halting. It only proves that one criterion

measure for dividing these pairs does not always work.

*This same reasoning equally applies to the Turing machine based proofs*

int D(int (*x)())

{

int Halt_Status = H(x, x);

if (Halt_Status)

HERE: goto HERE;

return Halt_Status;

}

It is a verified fact that H correctly predicts that D correctly

simulated by H would never reach its own final state and terminate

normally, thus H does correctly decide halting for its input D.

Anyone with sufficient software engineering skill knows that

*D simulated by H cannot possibly correctly reach its ret instruction*

Everyone else lacks sufficient software engineering skill or lies

_D()

[00001d12] 55 push ebp

[00001d13] 8bec mov ebp,esp

[00001d15] 51 push ecx

[00001d16] 8b4508 mov eax,[ebp+08]

[00001d19] 50 push eax // push D

[00001d1a] 8b4d08 mov ecx,[ebp+08]

[00001d1d] 51 push ecx // push D

[00001d1e] e83ff8ffff call 00001562 // call H

[00001d23] 83c408 add esp,+08

[00001d26] 8945fc mov [ebp-04],eax

[00001d29] 837dfc00 cmp dword [ebp-04],+00

[00001d2d] 7402 jz 00001d31

[00001d2f] ebfe jmp 00001d2f

[00001d31] 8b45fc mov eax,[ebp-04]

[00001d34] 8be5 mov esp,ebp

[00001d36] 5d pop ebp

[00001d37] c3 ret

*Simulating Halt Deciders Defeat the Halting Theorem*

https://www.researchgate.net/publication/368568464_Simulating_Halt_Deciders_Defeat_the_Halting_Theorem

--

Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius

hits a target no one else can see." Arthur Schopenhauer

Feb 21, 2023, 12:31:37 PMFeb 21

to

dishonest reviewers is that these dishonest reviewers change the subject

instead of providing any rebuttal to these points.

Feb 21, 2023, 6:53:21 PMFeb 21

to

On 2/21/23 12:17 PM, olcott wrote:

> When the ultimate measure of correct simulation is that the execution

> trace of the simulated input exactly matches the behavior that the input

> machine description specifies then:

And by that definition, a "Halt Decider" can "corrrectly" determine ANY
> When the ultimate measure of correct simulation is that the execution

> trace of the simulated input exactly matches the behavior that the input

> machine description specifies then:

input to be non-halting as it gets aborted before it can reach a final

state.

>

> *It is an easily verified fact that* Every counter-example input to the

> halting theorem D cannot possibly reach its own simulated final state in

> any finite number of steps when correctly simulated by simulating halt

> decider H.

>

> The halting theorem does not prove that a set of input pairs cannot be

> divided into halting and not halting. It only proves that one criterion

> measure for dividing these pairs does not always work.

that is basd on the actual behavior of the machine given.

>

> *This same reasoning equally applies to the Turing machine based proofs*

contains a copy of H.

>

> int D(int (*x)())

> {

> int Halt_Status = H(x, x);

> if (Halt_Status)

> HERE: goto HERE;

> return Halt_Status;

> }

>

> It is a verified fact that H correctly predicts that D correctly

> simulated by H would never reach its own final state and terminate

> normally, thus H does correctly decide halting for its input D.

>

> Anyone with sufficient software engineering skill knows that

> *D simulated by H cannot possibly correctly reach its ret instruction*

> Everyone else lacks sufficient software engineering skill or lies

>

> _D()

> [00001d12] 55 push ebp

> [00001d13] 8bec mov ebp,esp

> [00001d15] 51 push ecx

> [00001d16] 8b4508 mov eax,[ebp+08]

> [00001d19] 50 push eax // push D

> [00001d1a] 8b4d08 mov ecx,[ebp+08]

> [00001d1d] 51 push ecx // push D

> [00001d1e] e83ff8ffff call 00001562 // call H

> [00001d23] 83c408 add esp,+08

> [00001d26] 8945fc mov [ebp-04],eax

> [00001d29] 837dfc00 cmp dword [ebp-04],+00

> [00001d2d] 7402 jz 00001d31

> [00001d2f] ebfe jmp 00001d2f

> [00001d31] 8b45fc mov eax,[ebp-04]

> [00001d34] 8be5 mov esp,ebp

> [00001d36] 5d pop ebp

> [00001d37] c3 ret

>

> *Simulating Halt Deciders Defeat the Halting Theorem*

> https://www.researchgate.net/publication/368568464_Simulating_Halt_Deciders_Defeat_the_Halting_Theorem

>

>

it gets there, but an actual complete simulation of the input by a real

UTM will get there, showing that H is incorrect.

Feb 21, 2023, 6:54:12 PMFeb 21

to

to ignore the facts and just repeat your lies.

Feb 22, 2023, 12:06:07 PMFeb 22

to

On 2/22/2023 10:41 AM, Fritz Feldhase wrote:

> On Wednesday, February 22, 2023 at 4:18:56 AM UTC+1, olcott wrote:

>> On 2/21/2023 9:11 PM, Fritz Feldhase wrote:

>>>>

>>>> H(D,D) cannot report on the [halting] behavior of D(D)

>>>>

>>> Huh?! What?!

>>>

>> H(D,D) cannot possibly report on the behavior of non inputs.

>

> Huh?! Since when is your program/function D a "non input". What does this even mean?

>

> Just invented a new term (weasel word) to say something silly?

>

int sum (int x, int y) { return x + y; }

sum(3,4) cannot return the sum of 7 + 5 and would be wrong to do so.

Simulating halt decider H is not allowed to analyze the behavior of the

directly executed D(D) for the same reason that sum(3,4) is not allowed

to return the sum of 5 + 7.

>>

>

> Just a comment: H cannot correctly "determine" the halt status of D(D).

H is not allowed to "determine" the halt status of D(D).

All deciders compute the mapping from their inputs

All deciders compute the mapping from their inputs

All deciders compute the mapping from their inputs

*straw-man*

An intentionally misrepresented proposition that is set up because it is

easier to defeat than an opponent's real argument.

https://www.lexico.com/en/definition/straw_man

You continue to ignore and erase the proof that H does correctly

determine the halt status of D. *This is the straw-man deception*

Anyone with sufficient software engineering skill knows that

*D simulated by H cannot possibly correctly reach its ret instruction*

Everyone else lacks sufficient software engineering skill or lies

_D()

[00001d12] 55 push ebp

[00001d13] 8bec mov ebp,esp

[00001d15] 51 push ecx

[00001d16] 8b4508 mov eax,[ebp+08] // move 1st argument to eax

simulate D again endlessly or abort the entire recursive chain of

simulations before any of them pass machine address [00001d1e].

*Are you clueless about how the x86 language works*

[00001d16] mov eax,[ebp+08] // move 1st argument to eax

[00001d19] push eax // push D

[00001d1a] mov ecx,[ebp+08] // move 1st argument to ecx

[00001d1d] push ecx // push D

[00001d1e] call 00001562 // call H

This is the same thing as this C: H(D,D);

> On Wednesday, February 22, 2023 at 4:18:56 AM UTC+1, olcott wrote:

>> On 2/21/2023 9:11 PM, Fritz Feldhase wrote:

>>>>

>>>> H(D,D) cannot report on the [halting] behavior of D(D)

>>>>

>>> Huh?! What?!

>>>

>> H(D,D) cannot possibly report on the behavior of non inputs.

>

> Huh?! Since when is your program/function D a "non input". What does this even mean?

>

> Just invented a new term (weasel word) to say something silly?

>

int sum (int x, int y) { return x + y; }

sum(3,4) cannot return the sum of 7 + 5 and would be wrong to do so.

Simulating halt decider H is not allowed to analyze the behavior of the

directly executed D(D) for the same reason that sum(3,4) is not allowed

to return the sum of 5 + 7.

>>

>

> Just a comment: H cannot correctly "determine" the halt status of D(D).

H is not allowed to "determine" the halt status of D(D).

All deciders compute the mapping from their inputs

All deciders compute the mapping from their inputs

All deciders compute the mapping from their inputs

*straw-man*

An intentionally misrepresented proposition that is set up because it is

easier to defeat than an opponent's real argument.

https://www.lexico.com/en/definition/straw_man

You continue to ignore and erase the proof that H does correctly

determine the halt status of D. *This is the straw-man deception*

Anyone with sufficient software engineering skill knows that

*D simulated by H cannot possibly correctly reach its ret instruction*

Everyone else lacks sufficient software engineering skill or lies

_D()

[00001d12] 55 push ebp

[00001d13] 8bec mov ebp,esp

[00001d15] 51 push ecx

[00001d19] 50 push eax // push D

[00001d1a] 8b4d08 mov ecx,[ebp+08] // move 1st argument to ecx
[00001d1d] 51 push ecx // push D

[00001d1e] e83ff8ffff call 00001562 // call H

[00001d23] 83c408 add esp,+08

[00001d26] 8945fc mov [ebp-04],eax

[00001d29] 837dfc00 cmp dword [ebp-04],+00

[00001d2d] 7402 jz 00001d31

[00001d2f] ebfe jmp 00001d2f

[00001d31] 8b45fc mov eax,[ebp-04]

[00001d34] 8be5 mov esp,ebp

[00001d36] 5d pop ebp

[00001d37] c3 ret

When D correctly simulated by H reaches machine address [00001d1e] H can
[00001d1e] e83ff8ffff call 00001562 // call H

[00001d23] 83c408 add esp,+08

[00001d26] 8945fc mov [ebp-04],eax

[00001d29] 837dfc00 cmp dword [ebp-04],+00

[00001d2d] 7402 jz 00001d31

[00001d2f] ebfe jmp 00001d2f

[00001d31] 8b45fc mov eax,[ebp-04]

[00001d34] 8be5 mov esp,ebp

[00001d36] 5d pop ebp

[00001d37] c3 ret

simulate D again endlessly or abort the entire recursive chain of

simulations before any of them pass machine address [00001d1e].

*Are you clueless about how the x86 language works*

[00001d16] mov eax,[ebp+08] // move 1st argument to eax

[00001d19] push eax // push D

[00001d1a] mov ecx,[ebp+08] // move 1st argument to ecx

[00001d1d] push ecx // push D

[00001d1e] call 00001562 // call H

This is the same thing as this C: H(D,D);

Feb 22, 2023, 7:49:39 PMFeb 22

to

On 2/22/23 12:06 PM, olcott wrote:

> On 2/22/2023 10:41 AM, Fritz Feldhase wrote:

>> On Wednesday, February 22, 2023 at 4:18:56 AM UTC+1, olcott wrote:

>>> On 2/21/2023 9:11 PM, Fritz Feldhase wrote:

>>>>>

>>>>> H(D,D) cannot report on the [halting] behavior of D(D)

>>>>>

>>>> Huh?! What?!

>>>>

>>> H(D,D) cannot possibly report on the behavior of non inputs.

>>

>> Huh?! Since when is your program/function D a "non input". What does

>> this even mean?

>>

>> Just invented a new term (weasel word) to say something silly?

>>

>

> int sum (int x, int y) { return x + y; }

> sum(3,4) cannot return the sum of 7 + 5 and would be wrong to do so.

Right, just like Halt(D,D) can't return non-halting when D(D) Halts.
> On 2/22/2023 10:41 AM, Fritz Feldhase wrote:

>> On Wednesday, February 22, 2023 at 4:18:56 AM UTC+1, olcott wrote:

>>> On 2/21/2023 9:11 PM, Fritz Feldhase wrote:

>>>>>

>>>>> H(D,D) cannot report on the [halting] behavior of D(D)

>>>>>

>>>> Huh?! What?!

>>>>

>>> H(D,D) cannot possibly report on the behavior of non inputs.

>>

>> Huh?! Since when is your program/function D a "non input". What does

>> this even mean?

>>

>> Just invented a new term (weasel word) to say something silly?

>>

>

> int sum (int x, int y) { return x + y; }

> sum(3,4) cannot return the sum of 7 + 5 and would be wrong to do so.

>

> Simulating halt decider H is not allowed to analyze the behavior of the

> directly executed D(D) for the same reason that sum(3,4) is not allowed

> to return the sum of 5 + 7.

sum of3,3) is 3+4 just as Ha;lt(D,D) is supposed to answer if D(D) Halts.

>

> >>

> >

> > Just a comment: H cannot correctly "determine" the halt status of D(D).

>

> H is not allowed to "determine" the halt status of D(D).

determine means.

>

> All deciders compute the mapping from their inputs

> All deciders compute the mapping from their inputs

> All deciders compute the mapping from their inputs

>

> *straw-man*

> An intentionally misrepresented proposition that is set up because it is

> easier to defeat than an opponent's real argument.

> https://www.lexico.com/en/definition/straw_man

>

> You continue to ignore and erase the proof that H does correctly

> determine the halt status of D. *This is the straw-man deception*

>

Remember, teh DEFINITION of a Halt Decider comes from:

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.

So the answer/mapping that a Halt Decider is supposed to determine is

the behavior of the ACUTUAL machine.

Since D(D) Halts, the CORRECT answer for H(D,D) is Halting, so the fact

it says Non-Halting shows it to be incorrect as a Halt Decider.

It might "Correctly" answer some other question, but that just makes it

s correct POOP decider, not a Halt Decider.

>

>

> Anyone with sufficient software engineering skill knows that

> *D simulated by H cannot possibly correctly reach its ret instruction*

> Everyone else lacks sufficient software engineering skill or lies

>

It has also been shown that the CORRECT SIMULATION 0f this input by a

simulator that actually completes its job shows that it does halt.

This just shows that H always just gives up too soon, and thus its

conclusion is incorrect. You can argue whether this is due to an

incorrect simulation, or incorrrctly determining, but it IS INCORRECT.

> _D()

> [00001d12] 55 push ebp

> [00001d13] 8bec mov ebp,esp

> [00001d15] 51 push ecx

> [00001d16] 8b4508 mov eax,[ebp+08] // move 1st argument to eax

> [00001d19] 50 push eax // push D

> [00001d1a] 8b4d08 mov ecx,[ebp+08] // move 1st argument to ecx

> [00001d1d] 51 push ecx // push D

> [00001d1e] e83ff8ffff call 00001562 // call H

> [00001d23] 83c408 add esp,+08

> [00001d26] 8945fc mov [ebp-04],eax

> [00001d29] 837dfc00 cmp dword [ebp-04],+00

> [00001d2d] 7402 jz 00001d31

> [00001d2f] ebfe jmp 00001d2f

> [00001d31] 8b45fc mov eax,[ebp-04]

> [00001d34] 8be5 mov esp,ebp

> [00001d36] 5d pop ebp

> [00001d37] c3 ret

>

> When D correctly simulated by H reaches machine address [00001d1e] H can

> simulate D again endlessly or abort the entire recursive chain of

> simulations before any of them pass machine address [00001d1e].

arguements about what whould happen when it doesn't.

That is just the fallacy of using a false premise.

>

> *Are you clueless about how the x86 language works*

> [00001d16] mov eax,[ebp+08] // move 1st argument to eax

> [00001d19] push eax // push D

> [00001d1a] mov ecx,[ebp+08] // move 1st argument to ecx

> [00001d1d] push ecx // push D

> [00001d1e] call 00001562 // call H

> This is the same thing as this C: H(D,D);

>

>

So you know how to write a call instruction.

DO you understand what this does?

It calls an H that WILL abort its simulation and return 0, at least that

is what all the code you have provided that has an H that supposedly

"Correctly" returns 0 for H(D,D)

You don't seem to understand how computers actually work.

Since we have established that H(D,D) returns 0, the only thing that a

correct simulation of that code sequence can indicate is that eventually

we will get to the the next instruction with a 0 in the eax register.

Anything else is just proved to be an incorrect simulation.

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