*This is the outline of the complete refutation*
*of the only rebuttal that anyone has left*

(a) The complete and correct x86 emulation of the input to H(P,P) by H
never reaches the "ret" instruction of P.

(b) The direct execution of P(P) does reach its "ret" instruction.

The above two are proven to be verified facts entirely on the basis of
the semantics of the x86 language.

A halt decider must compute the mapping from its inputs to an accept or
reject state on the basis of the actual behavior that is actually
specified by these inputs.

P(P) is provably not the actual behavior of the actual input.

void P(u32 x)
{ if (H(x, x))
HERE: goto HERE;
return;
}

int main()
{ P(P);
}

_P()
[000011f0](01) 55 push ebp
[000011f1](02) 8bec mov ebp,esp
[000011f3](03) 8b4508 mov eax,[ebp+08]
[000011f6](01) 50 push eax
[000011f7](03) 8b4d08 mov ecx,[ebp+08]
[000011fa](01) 51 push ecx
[000011fb](05) e820feffff call 00001020

[00001200](03) 83c408 add esp,+08
[00001203](02) 85c0 test eax,eax
[00001205](02) 7402 jz 00001209
[00001207](02) ebfe jmp 00001207
[00001209](01) 5d pop ebp
[0000120a](01) c3 ret
Size in bytes:(0027) [0000120a]

_main()
[00001210](01) 55 push ebp
[00001211](02) 8bec mov ebp,esp
[00001213](05) 68f0110000 push 000011f0
[00001218](05) e8d3ffffff call 000011f0
[0000121d](03) 83c404 add esp,+04
[00001220](02) 33c0 xor eax,eax
[00001222](01) 5d pop ebp
[00001223](01) c3 ret
Size in bytes:(0020) [00001223]

machine stack stack machine assembly
address address data code language
======== ======== ======== ========= =============
[00001210][00101fba][00000000] 55 push ebp
[00001211][00101fba][00000000] 8bec mov ebp,esp
[00001213][00101fb6][000011f0] 68f0110000 push 000011f0 // push P
[00001218][00101fb2][0000121d] e8d3ffffff call 000011f0 // call P
[000011f0][00101fae][00101fba] 55 push ebp // enter executed P
[000011f1][00101fae][00101fba] 8bec mov ebp,esp
[000011f3][00101fae][00101fba] 8b4508 mov eax,[ebp+08]
[000011f6][00101faa][000011f0] 50 push eax // push P
[000011f7][00101faa][000011f0] 8b4d08 mov ecx,[ebp+08]
[000011fa][00101fa6][000011f0] 51 push ecx // push P
[000011fb][00101fa2][00001200] e820feffff call 00001020 // call H

Begin Simulation Execution Trace Stored at:21206e
Address_of_H:1020
[000011f0][0021205a][0021205e] 55 push ebp // enter emulated P
[000011f1][0021205a][0021205e] 8bec mov ebp,esp
[000011f3][0021205a][0021205e] 8b4508 mov eax,[ebp+08]
[000011f6][00212056][000011f0] 50 push eax // push P
[000011f7][00212056][000011f0] 8b4d08 mov ecx,[ebp+08]
[000011fa][00212052][000011f0] 51 push ecx // push P
[000011fb][0021204e][00001200] e820feffff call 00001020 // call emulated H
Infinitely Recursive Simulation Detected Simulation Stopped

H knows its own machine address and on this basis it can easily
examine its stored execution_trace of P (see above) to determine:
(a) P is calling H with the same arguments that H was called with.
(b) No instructions in P could possibly escape this otherwise infinitely
recursive emulation.
(c) H aborts its emulation of P before its call to H is emulated.

[00001200][00101fae][00101fba] 83c408 add esp,+08 // return to
executed P
[00001203][00101fae][00101fba] 85c0 test eax,eax
[00001205][00101fae][00101fba] 7402 jz 00001209
[00001209][00101fb2][0000121d] 5d pop ebp
[0000120a][00101fb6][000011f0] c3 ret // return from
executed P
[0000121d][00101fba][00000000] 83c404 add esp,+04
[00001220][00101fba][00000000] 33c0 xor eax,eax
[00001222][00101fbe][00100000] 5d pop ebp
[00001223][00101fc2][00000000] c3 ret // ret from main
Number of Instructions Executed(878) / 67 = 13 pages

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

On Monday, June 27, 2022 at 7:11:21 AM UTC-4, olcott wrote:
> *This is the outline of the complete refutation*
> *of the only rebuttal that anyone has left*
>
> (a) The complete and correct x86 emulation of the input to H(P,P) by H
> never reaches the "ret" instruction of P.

It is known that H aborts its simulation and returns 0. That means that H *by definition* does not perform a complete and correct emulation. And since H (if it was constructed correctly) is a pure function of its inputs, it will ALWAYS return the same return for a given input.

To state that H aborts *and* does a complete and correct simulation is simply nonsense.

What *does* perform a complete and correct emulation of the input to H(P,P) is UTM(P,P) which halts, therefore H(P,P)==0 is wrong.

>
> (b) The direct execution of P(P) does reach its "ret" instruction.
>
> The above two are proven to be verified facts entirely on the basis of
> the semantics of the x86 language.

A nonsense statement can't be a "fact", verified or otherwise.

>
> A halt decider must compute the mapping from its inputs to an accept or
> reject state on the basis of the actual behavior that is actually
> specified by these inputs.

Which for H(P,P) is BY DEFINITION is the behavior of P(P). The halting problem says that H MUST implement the following mapping (i.e. the halting function):

H(x,y)==1 if and only if x(y) halts, and
H(x,y)==0 if and only if x(y) does not halt.

H does not perform this mapping.

>
> P(P) is provably not the actual behavior of the actual input.

It is BY DEFINITION. So if H comes up with something different that means it did something wrong, specifically it aborted a halting computation too soon.

>
> void P(u32 x)
> {
> if (H(x, x))
> HERE: goto HERE;
> return;
> }
>
> int main()
> {
> P(P);
> }
>
> _P()
> [000011f0](01) 55 push ebp
> [000011f1](02) 8bec mov ebp,esp
> [000011f3](03) 8b4508 mov eax,[ebp+08]
> [000011f6](01) 50 push eax
> [000011f7](03) 8b4d08 mov ecx,[ebp+08]
> [000011fa](01) 51 push ecx
> [000011fb](05) e820feffff call 00001020
>
> [00001200](03) 83c408 add esp,+08
> [00001203](02) 85c0 test eax,eax
> [00001205](02) 7402 jz 00001209
> [00001207](02) ebfe jmp 00001207
> [00001209](01) 5d pop ebp
> [0000120a](01) c3 ret
> Size in bytes:(0027) [0000120a]
>
> _main()
> [00001210](01) 55 push ebp
> [00001211](02) 8bec mov ebp,esp
> [00001213](05) 68f0110000 push 000011f0
> [00001218](05) e8d3ffffff call 000011f0
> [0000121d](03) 83c404 add esp,+04
> [00001220](02) 33c0 xor eax,eax
> [00001222](01) 5d pop ebp
> [00001223](01) c3 ret
> Size in bytes:(0020) [00001223]
>
> machine stack stack machine assembly
> address address data code language
> ======== ======== ======== ========= =============
> [00001210][00101fba][00000000] 55 push ebp
> [00001211][00101fba][00000000] 8bec mov ebp,esp
> [00001213][00101fb6][000011f0] 68f0110000 push 000011f0 // push P
> [00001218][00101fb2][0000121d] e8d3ffffff call 000011f0 // call P
> [000011f0][00101fae][00101fba] 55 push ebp // enter executed P
> [000011f1][00101fae][00101fba] 8bec mov ebp,esp
> [000011f3][00101fae][00101fba] 8b4508 mov eax,[ebp+08]
> [000011f6][00101faa][000011f0] 50 push eax // push P
> [000011f7][00101faa][000011f0] 8b4d08 mov ecx,[ebp+08]
> [000011fa][00101fa6][000011f0] 51 push ecx // push P
> [000011fb][00101fa2][00001200] e820feffff call 00001020 // call H
>
> Begin Simulation Execution Trace Stored at:21206e
> Address_of_H:1020
> [000011f0][0021205a][0021205e] 55 push ebp // enter emulated P
> [000011f1][0021205a][0021205e] 8bec mov ebp,esp
> [000011f3][0021205a][0021205e] 8b4508 mov eax,[ebp+08]
> [000011f6][00212056][000011f0] 50 push eax // push P
> [000011f7][00212056][000011f0] 8b4d08 mov ecx,[ebp+08]
> [000011fa][00212052][000011f0] 51 push ecx // push P
> [000011fb][0021204e][00001200] e820feffff call 00001020 // call emulated H
> Infinitely Recursive Simulation Detected Simulation Stopped
>
> H knows its own machine address and on this basis it can easily
> examine its stored execution_trace of P (see above) to determine:
> (a) P is calling H with the same arguments that H was called with.
> (b) No instructions in P could possibly escape this otherwise infinitely
> recursive emulation.

FALSE. P contains these instructions:

if (Decide_Halting(&execution_trace, &decoded, code_end, &master_state,
&slave_state, &slave_stack, Address_of_H, P, I))
goto END_OF_CODE;
return 0; // Does not halt
END_OF_CODE:
return 1; // Input has normally terminated

That *can* and *do* prevent infinite recursive emulation in a *correct* and *complete* emulation

> (c) H aborts its emulation of P before its call to H is emulated.

And in doing so it fails to see that the call to H, since H is a computation, will aborts *its* simulation of P and return 0 to the outer P being emulated, which would cause it to halt, making H(P,P)==0 wrong.

>
> [00001200][00101fae][00101fba] 83c408 add esp,+08 // return to
> executed P
> [00001203][00101fae][00101fba] 85c0 test eax,eax
> [00001205][00101fae][00101fba] 7402 jz 00001209
> [00001209][00101fb2][0000121d] 5d pop ebp
> [0000120a][00101fb6][000011f0] c3 ret // return from
> executed P
> [0000121d][00101fba][00000000] 83c404 add esp,+04
> [00001220][00101fba][00000000] 33c0 xor eax,eax
> [00001222][00101fbe][00100000] 5d pop ebp
> [00001223][00101fc2][00000000] c3 ret // ret from main
> Number of Instructions Executed(878) / 67 = 13 pages
>
>

I've stated in detail why you're wrong. Now you need to explain in detail why I'm wrong.

Failure to do so, which includes simply restating your original point, will be taken as admission that what I've stated is correct.

*This is the outline of the complete refutation*
*of the only rebuttal that anyone has left*

(a) The complete and correct x86 emulation of the input to H(P,P) by H
never reaches the "ret" instruction of P.

(b) The direct execution of P(P) does reach its "ret" instruction.

The above two are proven to be verified facts entirely on the basis of
the semantics of the x86 language.

A halt decider must compute the mapping from its inputs to an accept or
reject state on the basis of the actual behavior that is actually
specified by these inputs.

P(P) is provably not the actual behavior of the actual input.

void P(u32 x)
{ if (H(x, x))
HERE: goto HERE;
return;
}

int main()
{ Output("Input_Halts = ", H1((u32)P, (u32)P));
}

_P()
[0000163a](01) 55 push ebp
[0000163b](02) 8bec mov ebp,esp
[0000163d](03) 8b4508 mov eax,[ebp+08]
[00001640](01) 50 push eax
[00001641](03) 8b4d08 mov ecx,[ebp+08]
[00001644](01) 51 push ecx
[00001645](05) e8f0fdffff call 0000143a // call H
[0000164a](03) 83c408 add esp,+08
[0000164d](02) 85c0 test eax,eax
[0000164f](02) 7402 jz 00001653
[00001651](02) ebfe jmp 00001651
[00001653](01) 5d pop ebp
[00001654](01) c3 ret
Size in bytes:(0027) [00001654]

_main()
[0000165a](01) 55 push ebp
[0000165b](02) 8bec mov ebp,esp
[0000165d](05) 683a160000 push 0000163a // push P
[00001662](05) 683a160000 push 0000163a // push P
[00001667](05) e8cef9ffff call 0000103a // call H1
[0000166c](03) 83c408 add esp,+08
[0000166f](01) 50 push eax // push return value
[00001670](05) 689b040000 push 0000049b // "Input_Halts = "
[00001675](05) e870eeffff call 000004ea // call Output
[0000167a](03) 83c408 add esp,+08
[0000167d](02) 33c0 xor eax,eax
[0000167f](01) 5d pop ebp
[00001680](01) c3 ret
Size in bytes:(0039) [00001680]

machine stack stack machine assembly
address address data code language
======== ======== ======== ========= =============
[0000165a][001026a9][00000000] 55 push ebp
[0000165b][001026a9][00000000] 8bec mov ebp,esp
[0000165d][001026a5][0000163a] 683a160000 push 0000163a // push P
[00001662][001026a1][0000163a] 683a160000 push 0000163a // push P
[00001667][0010269d][0000166c] e8cef9ffff call 0000103a // call H1

H1: Begin Simulation Execution Trace Stored at:21275d
Address_of_H1:103a
[0000163a][00212749][0021274d] 55 push ebp
[0000163b][00212749][0021274d] 8bec mov ebp,esp
[0000163d][00212749][0021274d] 8b4508 mov eax,[ebp+08]
[00001640][00212745][0000163a] 50 push eax // push P
[00001641][00212745][0000163a] 8b4d08 mov ecx,[ebp+08]
[00001644][00212741][0000163a] 51 push ecx // push P
[00001645][0021273d][0000164a] e8f0fdffff call 0000143a // call H

H: Begin Simulation Execution Trace Stored at:2285c5
Address_of_H:143a
[0000163a][002285b1][002285b5] 55 push ebp
[0000163b][002285b1][002285b5] 8bec mov ebp,esp
[0000163d][002285b1][002285b5] 8b4508 mov eax,[ebp+08]
[00001640][002285ad][0000163a] 50 push eax // push P
[00001641][002285ad][0000163a] 8b4d08 mov ecx,[ebp+08]
[00001644][002285a9][0000163a] 51 push ecx // push P
[00001645][002285a5][0000164a] e8f0fdffff call 0000143a // call H
H: Infinitely Recursive Simulation Detected Simulation Stopped

[0000164a][00212749][0021274d] 83c408 add esp,+08
[0000164d][00212749][0021274d] 85c0 test eax,eax
[0000164f][00212749][0021274d] 7402 jz 00001653
[00001653][0021274d][0000111e] 5d pop ebp
[00001654][00212751][0000163a] c3 ret
H1: End Simulation Input Terminated Normally

[0000166c][001026a9][00000000] 83c408 add esp,+08
[0000166f][001026a5][00000001] 50 push eax // return value
[00001670][001026a1][0000049b] 689b040000 push 0000049b // "Input_Halts = "
[00001675][001026a1][0000049b] e870eeffff call 000004ea // call Output
Input_Halts = 1
[0000167a][001026a9][00000000] 83c408 add esp,+08
[0000167d][001026a9][00000000] 33c0 xor eax,eax
[0000167f][001026ad][00100000] 5d pop ebp
[00001680][001026b1][00000004] c3 ret
Number of Instructions Executed(409590) == 6113 Pages

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

On 6/27/2022 6:57 AM, Dennis Bush wrote:
> On Monday, June 27, 2022 at 7:11:21 AM UTC-4, olcott wrote:
>> *This is the outline of the complete refutation*
>> *of the only rebuttal that anyone has left*
>>
>> (a) The complete and correct x86 emulation of the input to H(P,P) by H
>> never reaches the "ret" instruction of P.
>
> It is known that H aborts its simulation and returns 0. That means that H *by definition* does not perform a complete and correct emulation. And since H (if it was constructed correctly) is a pure function of its inputs, it will ALWAYS return the same return for a given input.
>
> To state that H aborts *and* does a complete and correct simulation is simply nonsense.
>
> What *does* perform a complete and correct emulation of the input to H(P,P) is UTM(P,P) which halts, therefore H(P,P)==0 is wrong.
>
>>
>> (b) The direct execution of P(P) does reach its "ret" instruction.
>>
>> The above two are proven to be verified facts entirely on the basis of
>> the semantics of the x86 language.
>
> A nonsense statement can't be a "fact", verified or otherwise.
>
>>
>> A halt decider must compute the mapping from its inputs to an accept or
>> reject state on the basis of the actual behavior that is actually
>> specified by these inputs.
>
> Which for H(P,P) is BY DEFINITION is the behavior of P(P). The halting problem says that H MUST implement the following mapping (i.e. the halting function):
>
> H(x,y)==1 if and only if x(y) halts, and
> H(x,y)==0 if and only if x(y) does not halt.
>
> H does not perform this mapping.
>
>>
>> P(P) is provably not the actual behavior of the actual input.
>
> It is BY DEFINITION. So if H comes up with something different that means it did something wrong, specifically it aborted a halting computation too soon.
>
>>
>> void P(u32 x)
>> {
>> if (H(x, x))
>> HERE: goto HERE;
>> return;
>> }
>>
>> int main()
>> {
>> P(P);
>> }
>>
>> _P()
>> [000011f0](01) 55 push ebp
>> [000011f1](02) 8bec mov ebp,esp
>> [000011f3](03) 8b4508 mov eax,[ebp+08]
>> [000011f6](01) 50 push eax
>> [000011f7](03) 8b4d08 mov ecx,[ebp+08]
>> [000011fa](01) 51 push ecx
>> [000011fb](05) e820feffff call 00001020
>>
>> [00001200](03) 83c408 add esp,+08
>> [00001203](02) 85c0 test eax,eax
>> [00001205](02) 7402 jz 00001209
>> [00001207](02) ebfe jmp 00001207
>> [00001209](01) 5d pop ebp
>> [0000120a](01) c3 ret
>> Size in bytes:(0027) [0000120a]
>>
>> _main()
>> [00001210](01) 55 push ebp
>> [00001211](02) 8bec mov ebp,esp
>> [00001213](05) 68f0110000 push 000011f0
>> [00001218](05) e8d3ffffff call 000011f0
>> [0000121d](03) 83c404 add esp,+04
>> [00001220](02) 33c0 xor eax,eax
>> [00001222](01) 5d pop ebp
>> [00001223](01) c3 ret
>> Size in bytes:(0020) [00001223]
>>
>> machine stack stack machine assembly
>> address address data code language
>> ======== ======== ======== ========= =============
>> [00001210][00101fba][00000000] 55 push ebp
>> [00001211][00101fba][00000000] 8bec mov ebp,esp
>> [00001213][00101fb6][000011f0] 68f0110000 push 000011f0 // push P
>> [00001218][00101fb2][0000121d] e8d3ffffff call 000011f0 // call P
>> [000011f0][00101fae][00101fba] 55 push ebp // enter executed P
>> [000011f1][00101fae][00101fba] 8bec mov ebp,esp
>> [000011f3][00101fae][00101fba] 8b4508 mov eax,[ebp+08]
>> [000011f6][00101faa][000011f0] 50 push eax // push P
>> [000011f7][00101faa][000011f0] 8b4d08 mov ecx,[ebp+08]
>> [000011fa][00101fa6][000011f0] 51 push ecx // push P
>> [000011fb][00101fa2][00001200] e820feffff call 00001020 // call H
>>
>> Begin Simulation Execution Trace Stored at:21206e
>> Address_of_H:1020
>> [000011f0][0021205a][0021205e] 55 push ebp // enter emulated P
>> [000011f1][0021205a][0021205e] 8bec mov ebp,esp
>> [000011f3][0021205a][0021205e] 8b4508 mov eax,[ebp+08]
>> [000011f6][00212056][000011f0] 50 push eax // push P
>> [000011f7][00212056][000011f0] 8b4d08 mov ecx,[ebp+08]
>> [000011fa][00212052][000011f0] 51 push ecx // push P
>> [000011fb][0021204e][00001200] e820feffff call 00001020 // call emulated H
>> Infinitely Recursive Simulation Detected Simulation Stopped
>>
>> H knows its own machine address and on this basis it can easily
>> examine its stored execution_trace of P (see above) to determine:
>> (a) P is calling H with the same arguments that H was called with.
>> (b) No instructions in P could possibly escape this otherwise infinitely
>> recursive emulation.
>
> FALSE. P contains these instructions:
>
> if (Decide_Halting(&execution_trace, &decoded, code_end, &master_state,
> &slave_state, &slave_stack, Address_of_H, P, I))
> goto END_OF_CODE;
> return 0; // Does not halt
> END_OF_CODE:
> return 1; // Input has normally terminated
>
> That *can* and *do* prevent infinite recursive emulation in a *correct* and *complete* emulation
>
>> (c) H aborts its emulation of P before its call to H is emulated.
>
> And in doing so it fails to see that the call to H, since H is a computation, will aborts *its* simulation of P and return 0 to the outer P being emulated, which would cause it to halt, making H(P,P)==0 wrong.
>
>>
>> [00001200][00101fae][00101fba] 83c408 add esp,+08 // return to
>> executed P
>> [00001203][00101fae][00101fba] 85c0 test eax,eax
>> [00001205][00101fae][00101fba] 7402 jz 00001209
>> [00001209][00101fb2][0000121d] 5d pop ebp
>> [0000120a][00101fb6][000011f0] c3 ret // return from
>> executed P
>> [0000121d][00101fba][00000000] 83c404 add esp,+04
>> [00001220][00101fba][00000000] 33c0 xor eax,eax
>> [00001222][00101fbe][00100000] 5d pop ebp
>> [00001223][00101fc2][00000000] c3 ret // ret from main
>> Number of Instructions Executed(878) / 67 = 13 pages
>>
>>
>
> I've stated in detail why you're wrong. Now you need to explain in detail why I'm wrong.
>
> Failure to do so, which includes simply restating your original point, will be taken as admission that what I've stated is correct.

H correctly predicts that its complete and correct x86 emulation of its
input never reaches the the "ret" instruction of P.

This is the actual behavior of the input to H(P,P).

P(P) is not the actual behavior of the input to H(P,P).

Never reaching the "ret" instruction means that P does not halt.

A halt decider must compute the mapping from its inputs to an accept or
reject state on the basis of the actual behavior that is actually
specified by these inputs.

Therefore H(P,P)==0 is correct.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

On Monday, June 27, 2022 at 8:14:31 AM UTC-4, olcott wrote:
> On 6/27/2022 6:57 AM, Dennis Bush wrote:
> > On Monday, June 27, 2022 at 7:11:21 AM UTC-4, olcott wrote:
> >> *This is the outline of the complete refutation*
> >> *of the only rebuttal that anyone has left*
> >>
> >> (a) The complete and correct x86 emulation of the input to H(P,P) by H
> >> never reaches the "ret" instruction of P.
> >
> > It is known that H aborts its simulation and returns 0. That means that H *by definition* does not perform a complete and correct emulation. And since H (if it was constructed correctly) is a pure function of its inputs, it will ALWAYS return the same return for a given input.
> >
> > To state that H aborts *and* does a complete and correct simulation is simply nonsense.
> >
> > What *does* perform a complete and correct emulation of the input to H(P,P) is UTM(P,P) which halts, therefore H(P,P)==0 is wrong.
> >
> >>
> >> (b) The direct execution of P(P) does reach its "ret" instruction.
> >>
> >> The above two are proven to be verified facts entirely on the basis of
> >> the semantics of the x86 language.
> >
> > A nonsense statement can't be a "fact", verified or otherwise.
> >
> >>
> >> A halt decider must compute the mapping from its inputs to an accept or
> >> reject state on the basis of the actual behavior that is actually
> >> specified by these inputs.
> >
> > Which for H(P,P) is BY DEFINITION is the behavior of P(P). The halting problem says that H MUST implement the following mapping (i.e. the halting function):
> >
> > H(x,y)==1 if and only if x(y) halts, and
> > H(x,y)==0 if and only if x(y) does not halt.
> >
> > H does not perform this mapping.
> >
> >>
> >> P(P) is provably not the actual behavior of the actual input.
> >
> > It is BY DEFINITION. So if H comes up with something different that means it did something wrong, specifically it aborted a halting computation too soon.
> >
> >>
> >> void P(u32 x)
> >> {
> >> if (H(x, x))
> >> HERE: goto HERE;
> >> return;
> >> }
> >>
> >> int main()
> >> {
> >> P(P);
> >> }
> >>
> >> _P()
> >> [000011f0](01) 55 push ebp
> >> [000011f1](02) 8bec mov ebp,esp
> >> [000011f3](03) 8b4508 mov eax,[ebp+08]
> >> [000011f6](01) 50 push eax
> >> [000011f7](03) 8b4d08 mov ecx,[ebp+08]
> >> [000011fa](01) 51 push ecx
> >> [000011fb](05) e820feffff call 00001020
> >>
> >> [00001200](03) 83c408 add esp,+08
> >> [00001203](02) 85c0 test eax,eax
> >> [00001205](02) 7402 jz 00001209
> >> [00001207](02) ebfe jmp 00001207
> >> [00001209](01) 5d pop ebp
> >> [0000120a](01) c3 ret
> >> Size in bytes:(0027) [0000120a]
> >>
> >> _main()
> >> [00001210](01) 55 push ebp
> >> [00001211](02) 8bec mov ebp,esp
> >> [00001213](05) 68f0110000 push 000011f0
> >> [00001218](05) e8d3ffffff call 000011f0
> >> [0000121d](03) 83c404 add esp,+04
> >> [00001220](02) 33c0 xor eax,eax
> >> [00001222](01) 5d pop ebp
> >> [00001223](01) c3 ret
> >> Size in bytes:(0020) [00001223]
> >>
> >> machine stack stack machine assembly
> >> address address data code language
> >> ======== ======== ======== ========= =============
> >> [00001210][00101fba][00000000] 55 push ebp
> >> [00001211][00101fba][00000000] 8bec mov ebp,esp
> >> [00001213][00101fb6][000011f0] 68f0110000 push 000011f0 // push P
> >> [00001218][00101fb2][0000121d] e8d3ffffff call 000011f0 // call P
> >> [000011f0][00101fae][00101fba] 55 push ebp // enter executed P
> >> [000011f1][00101fae][00101fba] 8bec mov ebp,esp
> >> [000011f3][00101fae][00101fba] 8b4508 mov eax,[ebp+08]
> >> [000011f6][00101faa][000011f0] 50 push eax // push P
> >> [000011f7][00101faa][000011f0] 8b4d08 mov ecx,[ebp+08]
> >> [000011fa][00101fa6][000011f0] 51 push ecx // push P
> >> [000011fb][00101fa2][00001200] e820feffff call 00001020 // call H
> >>
> >> Begin Simulation Execution Trace Stored at:21206e
> >> Address_of_H:1020
> >> [000011f0][0021205a][0021205e] 55 push ebp // enter emulated P
> >> [000011f1][0021205a][0021205e] 8bec mov ebp,esp
> >> [000011f3][0021205a][0021205e] 8b4508 mov eax,[ebp+08]
> >> [000011f6][00212056][000011f0] 50 push eax // push P
> >> [000011f7][00212056][000011f0] 8b4d08 mov ecx,[ebp+08]
> >> [000011fa][00212052][000011f0] 51 push ecx // push P
> >> [000011fb][0021204e][00001200] e820feffff call 00001020 // call emulated H
> >> Infinitely Recursive Simulation Detected Simulation Stopped
> >>
> >> H knows its own machine address and on this basis it can easily
> >> examine its stored execution_trace of P (see above) to determine:
> >> (a) P is calling H with the same arguments that H was called with.
> >> (b) No instructions in P could possibly escape this otherwise infinitely
> >> recursive emulation.
> >
> > FALSE. P contains these instructions:
> >
> > if (Decide_Halting(&execution_trace, &decoded, code_end, &master_state,
> > &slave_state, &slave_stack, Address_of_H, P, I))
> > goto END_OF_CODE;
> > return 0; // Does not halt
> > END_OF_CODE:
> > return 1; // Input has normally terminated
> >
> > That *can* and *do* prevent infinite recursive emulation in a *correct* and *complete* emulation
> >
> >> (c) H aborts its emulation of P before its call to H is emulated.
> >
> > And in doing so it fails to see that the call to H, since H is a computation, will aborts *its* simulation of P and return 0 to the outer P being emulated, which would cause it to halt, making H(P,P)==0 wrong.
> >
> >>
> >> [00001200][00101fae][00101fba] 83c408 add esp,+08 // return to
> >> executed P
> >> [00001203][00101fae][00101fba] 85c0 test eax,eax
> >> [00001205][00101fae][00101fba] 7402 jz 00001209
> >> [00001209][00101fb2][0000121d] 5d pop ebp
> >> [0000120a][00101fb6][000011f0] c3 ret // return from
> >> executed P
> >> [0000121d][00101fba][00000000] 83c404 add esp,+04
> >> [00001220][00101fba][00000000] 33c0 xor eax,eax
> >> [00001222][00101fbe][00100000] 5d pop ebp
> >> [00001223][00101fc2][00000000] c3 ret // ret from main
> >> Number of Instructions Executed(878) / 67 = 13 pages
> >>
> >>
> >
> > I've stated in detail why you're wrong. Now you need to explain in detail why I'm wrong.
> >
> > Failure to do so, which includes simply restating your original point, will be taken as admission that what I've stated is correct.
> H correctly predicts that its complete and correct x86 emulation of its
> input never reaches the the "ret" instruction of P.

Repeating your original point.

>
> This is the actual behavior of the input to H(P,P).
>
> P(P) is not the actual behavior of the input to H(P,P).
>

Again, repeating your original point.

> Never reaching the "ret" instruction means that P does not halt.
> A halt decider must compute the mapping from its inputs to an accept or
> reject state on the basis of the actual behavior that is actually
> specified by these inputs.
> Therefore H(P,P)==0 is correct.

And again, repeating your original point.

So no explanation as to why I'm wrong, just a repetition of your original points.

That means you accept that what I've said is true, and that H(P,P)==0 is wrong.

On Monday, 27 June 2022 at 12:11:21 UTC+1, olcott wrote:
> *This is the outline of the complete refutation*
> *of the only rebuttal that anyone has left*
>
> (a) The complete and correct x86 emulation of the input to H(P,P) by H
> never reaches the "ret" instruction of P.
>
> (b) The direct execution of P(P) does reach its "ret" instruction.
>
Well that's quite extraordinary.
>
> The above two are proven to be verified facts entirely on the basis of
> the semantics of the x86 language.
>
You don't really understand what it means to "prove" something. What you
mean is you've established that this is the case, to your satisfaction. But by far
the most likely explanation is a mistake in your analysis. The next most likely
explanation is a fault in the computer. The notion that the correct emulation
of a piece of code doesn't match its run time behaviour is absolutely the last
thing we should consider, after all other explanations have been exhausted
and we are driven to this conclusion.

On 6/27/2022 7:26 AM, Dennis Bush wrote:
> On Monday, June 27, 2022 at 8:14:31 AM UTC-4, olcott wrote:
>> On 6/27/2022 6:57 AM, Dennis Bush wrote:
>>> On Monday, June 27, 2022 at 7:11:21 AM UTC-4, olcott wrote:
>>>> *This is the outline of the complete refutation*
>>>> *of the only rebuttal that anyone has left*
>>>>
>>>> (a) The complete and correct x86 emulation of the input to H(P,P) by H
>>>> never reaches the "ret" instruction of P.
>>>
>>> It is known that H aborts its simulation and returns 0. That means that H *by definition* does not perform a complete and correct emulation. And since H (if it was constructed correctly) is a pure function of its inputs, it will ALWAYS return the same return for a given input.
>>>
>>> To state that H aborts *and* does a complete and correct simulation is simply nonsense.
>>>
>>> What *does* perform a complete and correct emulation of the input to H(P,P) is UTM(P,P) which halts, therefore H(P,P)==0 is wrong.
>>>
>>>>
>>>> (b) The direct execution of P(P) does reach its "ret" instruction.
>>>>
>>>> The above two are proven to be verified facts entirely on the basis of
>>>> the semantics of the x86 language.
>>>
>>> A nonsense statement can't be a "fact", verified or otherwise.
>>>
>>>>
>>>> A halt decider must compute the mapping from its inputs to an accept or
>>>> reject state on the basis of the actual behavior that is actually
>>>> specified by these inputs.
>>>
>>> Which for H(P,P) is BY DEFINITION is the behavior of P(P). The halting problem says that H MUST implement the following mapping (i.e. the halting function):
>>>
>>> H(x,y)==1 if and only if x(y) halts, and
>>> H(x,y)==0 if and only if x(y) does not halt.
>>>
>>> H does not perform this mapping.
>>>
>>>>
>>>> P(P) is provably not the actual behavior of the actual input.
>>>
>>> It is BY DEFINITION. So if H comes up with something different that means it did something wrong, specifically it aborted a halting computation too soon.
>>>
>>>>
>>>> void P(u32 x)
>>>> {
>>>> if (H(x, x))
>>>> HERE: goto HERE;
>>>> return;
>>>> }
>>>>
>>>> int main()
>>>> {
>>>> P(P);
>>>> }
>>>>
>>>> _P()
>>>> [000011f0](01) 55 push ebp
>>>> [000011f1](02) 8bec mov ebp,esp
>>>> [000011f3](03) 8b4508 mov eax,[ebp+08]
>>>> [000011f6](01) 50 push eax
>>>> [000011f7](03) 8b4d08 mov ecx,[ebp+08]
>>>> [000011fa](01) 51 push ecx
>>>> [000011fb](05) e820feffff call 00001020
>>>>
>>>> [00001200](03) 83c408 add esp,+08
>>>> [00001203](02) 85c0 test eax,eax
>>>> [00001205](02) 7402 jz 00001209
>>>> [00001207](02) ebfe jmp 00001207
>>>> [00001209](01) 5d pop ebp
>>>> [0000120a](01) c3 ret
>>>> Size in bytes:(0027) [0000120a]
>>>>
>>>> _main()
>>>> [00001210](01) 55 push ebp
>>>> [00001211](02) 8bec mov ebp,esp
>>>> [00001213](05) 68f0110000 push 000011f0
>>>> [00001218](05) e8d3ffffff call 000011f0
>>>> [0000121d](03) 83c404 add esp,+04
>>>> [00001220](02) 33c0 xor eax,eax
>>>> [00001222](01) 5d pop ebp
>>>> [00001223](01) c3 ret
>>>> Size in bytes:(0020) [00001223]
>>>>
>>>> machine stack stack machine assembly
>>>> address address data code language
>>>> ======== ======== ======== ========= =============
>>>> [00001210][00101fba][00000000] 55 push ebp
>>>> [00001211][00101fba][00000000] 8bec mov ebp,esp
>>>> [00001213][00101fb6][000011f0] 68f0110000 push 000011f0 // push P
>>>> [00001218][00101fb2][0000121d] e8d3ffffff call 000011f0 // call P
>>>> [000011f0][00101fae][00101fba] 55 push ebp // enter executed P
>>>> [000011f1][00101fae][00101fba] 8bec mov ebp,esp
>>>> [000011f3][00101fae][00101fba] 8b4508 mov eax,[ebp+08]
>>>> [000011f6][00101faa][000011f0] 50 push eax // push P
>>>> [000011f7][00101faa][000011f0] 8b4d08 mov ecx,[ebp+08]
>>>> [000011fa][00101fa6][000011f0] 51 push ecx // push P
>>>> [000011fb][00101fa2][00001200] e820feffff call 00001020 // call H
>>>>
>>>> Begin Simulation Execution Trace Stored at:21206e
>>>> Address_of_H:1020
>>>> [000011f0][0021205a][0021205e] 55 push ebp // enter emulated P
>>>> [000011f1][0021205a][0021205e] 8bec mov ebp,esp
>>>> [000011f3][0021205a][0021205e] 8b4508 mov eax,[ebp+08]
>>>> [000011f6][00212056][000011f0] 50 push eax // push P
>>>> [000011f7][00212056][000011f0] 8b4d08 mov ecx,[ebp+08]
>>>> [000011fa][00212052][000011f0] 51 push ecx // push P
>>>> [000011fb][0021204e][00001200] e820feffff call 00001020 // call emulated H
>>>> Infinitely Recursive Simulation Detected Simulation Stopped
>>>>
>>>> H knows its own machine address and on this basis it can easily
>>>> examine its stored execution_trace of P (see above) to determine:
>>>> (a) P is calling H with the same arguments that H was called with.
>>>> (b) No instructions in P could possibly escape this otherwise infinitely
>>>> recursive emulation.
>>>
>>> FALSE. P contains these instructions:
>>>
>>> if (Decide_Halting(&execution_trace, &decoded, code_end, &master_state,
>>> &slave_state, &slave_stack, Address_of_H, P, I))
>>> goto END_OF_CODE;
>>> return 0; // Does not halt
>>> END_OF_CODE:
>>> return 1; // Input has normally terminated
>>>
>>> That *can* and *do* prevent infinite recursive emulation in a *correct* and *complete* emulation
>>>
>>>> (c) H aborts its emulation of P before its call to H is emulated.
>>>
>>> And in doing so it fails to see that the call to H, since H is a computation, will aborts *its* simulation of P and return 0 to the outer P being emulated, which would cause it to halt, making H(P,P)==0 wrong.
>>>
>>>>
>>>> [00001200][00101fae][00101fba] 83c408 add esp,+08 // return to
>>>> executed P
>>>> [00001203][00101fae][00101fba] 85c0 test eax,eax
>>>> [00001205][00101fae][00101fba] 7402 jz 00001209
>>>> [00001209][00101fb2][0000121d] 5d pop ebp
>>>> [0000120a][00101fb6][000011f0] c3 ret // return from
>>>> executed P
>>>> [0000121d][00101fba][00000000] 83c404 add esp,+04
>>>> [00001220][00101fba][00000000] 33c0 xor eax,eax
>>>> [00001222][00101fbe][00100000] 5d pop ebp
>>>> [00001223][00101fc2][00000000] c3 ret // ret from main
>>>> Number of Instructions Executed(878) / 67 = 13 pages
>>>>
>>>>
>>>
>>> I've stated in detail why you're wrong. Now you need to explain in detail why I'm wrong.
>>>
>>> Failure to do so, which includes simply restating your original point, will be taken as admission that what I've stated is correct.
>> H correctly predicts that its complete and correct x86 emulation of its
>> input never reaches the the "ret" instruction of P.
>
> Repeating your original point.
>
>>
>> This is the actual behavior of the input to H(P,P).
>>
>> P(P) is not the actual behavior of the input to H(P,P).
>>
>
> Again, repeating your original point.
>
>> Never reaching the "ret" instruction means that P does not halt.
>> A halt decider must compute the mapping from its inputs to an accept or
>> reject state on the basis of the actual behavior that is actually
>> specified by these inputs.
>> Therefore H(P,P)==0 is correct.
>
> And again, repeating your original point.
>
> So no explanation as to why I'm wrong, just a repetition of your original points.
>
> That means you accept that what I've said is true, and that H(P,P)==0 is wrong.

You rebuttal is based on rejecting the verified fact that a simulating
halt decider correctly predicts that its correctly simulated input will
never reach the final state "ret" instruction of this input.

On Monday, June 27, 2022 at 8:34:22 AM UTC-4, olcott wrote:
> On 6/27/2022 7:26 AM, Dennis Bush wrote:
> > On Monday, June 27, 2022 at 8:14:31 AM UTC-4, olcott wrote:
> >> On 6/27/2022 6:57 AM, Dennis Bush wrote:
> >>> On Monday, June 27, 2022 at 7:11:21 AM UTC-4, olcott wrote:
> >>>> *This is the outline of the complete refutation*
> >>>> *of the only rebuttal that anyone has left*
> >>>>
> >>>> (a) The complete and correct x86 emulation of the input to H(P,P) by H
> >>>> never reaches the "ret" instruction of P.
> >>>
> >>> It is known that H aborts its simulation and returns 0. That means that H *by definition* does not perform a complete and correct emulation. And since H (if it was constructed correctly) is a pure function of its inputs, it will ALWAYS return the same return for a given input.
> >>>
> >>> To state that H aborts *and* does a complete and correct simulation is simply nonsense.
> >>>
> >>> What *does* perform a complete and correct emulation of the input to H(P,P) is UTM(P,P) which halts, therefore H(P,P)==0 is wrong.
> >>>
> >>>>
> >>>> (b) The direct execution of P(P) does reach its "ret" instruction.
> >>>>
> >>>> The above two are proven to be verified facts entirely on the basis of
> >>>> the semantics of the x86 language.
> >>>
> >>> A nonsense statement can't be a "fact", verified or otherwise.
> >>>
> >>>>
> >>>> A halt decider must compute the mapping from its inputs to an accept or
> >>>> reject state on the basis of the actual behavior that is actually
> >>>> specified by these inputs.
> >>>
> >>> Which for H(P,P) is BY DEFINITION is the behavior of P(P). The halting problem says that H MUST implement the following mapping (i.e. the halting function):
> >>>
> >>> H(x,y)==1 if and only if x(y) halts, and
> >>> H(x,y)==0 if and only if x(y) does not halt.
> >>>
> >>> H does not perform this mapping.
> >>>
> >>>>
> >>>> P(P) is provably not the actual behavior of the actual input.
> >>>
> >>> It is BY DEFINITION. So if H comes up with something different that means it did something wrong, specifically it aborted a halting computation too soon.
> >>>
> >>>>
> >>>> void P(u32 x)
> >>>> {
> >>>> if (H(x, x))
> >>>> HERE: goto HERE;
> >>>> return;
> >>>> }
> >>>>
> >>>> int main()
> >>>> {
> >>>> P(P);
> >>>> }
> >>>>
> >>>> _P()
> >>>> [000011f0](01) 55 push ebp
> >>>> [000011f1](02) 8bec mov ebp,esp
> >>>> [000011f3](03) 8b4508 mov eax,[ebp+08]
> >>>> [000011f6](01) 50 push eax
> >>>> [000011f7](03) 8b4d08 mov ecx,[ebp+08]
> >>>> [000011fa](01) 51 push ecx
> >>>> [000011fb](05) e820feffff call 00001020
> >>>>
> >>>> [00001200](03) 83c408 add esp,+08
> >>>> [00001203](02) 85c0 test eax,eax
> >>>> [00001205](02) 7402 jz 00001209
> >>>> [00001207](02) ebfe jmp 00001207
> >>>> [00001209](01) 5d pop ebp
> >>>> [0000120a](01) c3 ret
> >>>> Size in bytes:(0027) [0000120a]
> >>>>
> >>>> _main()
> >>>> [00001210](01) 55 push ebp
> >>>> [00001211](02) 8bec mov ebp,esp
> >>>> [00001213](05) 68f0110000 push 000011f0
> >>>> [00001218](05) e8d3ffffff call 000011f0
> >>>> [0000121d](03) 83c404 add esp,+04
> >>>> [00001220](02) 33c0 xor eax,eax
> >>>> [00001222](01) 5d pop ebp
> >>>> [00001223](01) c3 ret
> >>>> Size in bytes:(0020) [00001223]
> >>>>
> >>>> machine stack stack machine assembly
> >>>> address address data code language
> >>>> ======== ======== ======== ========= =============
> >>>> [00001210][00101fba][00000000] 55 push ebp
> >>>> [00001211][00101fba][00000000] 8bec mov ebp,esp
> >>>> [00001213][00101fb6][000011f0] 68f0110000 push 000011f0 // push P
> >>>> [00001218][00101fb2][0000121d] e8d3ffffff call 000011f0 // call P
> >>>> [000011f0][00101fae][00101fba] 55 push ebp // enter executed P
> >>>> [000011f1][00101fae][00101fba] 8bec mov ebp,esp
> >>>> [000011f3][00101fae][00101fba] 8b4508 mov eax,[ebp+08]
> >>>> [000011f6][00101faa][000011f0] 50 push eax // push P
> >>>> [000011f7][00101faa][000011f0] 8b4d08 mov ecx,[ebp+08]
> >>>> [000011fa][00101fa6][000011f0] 51 push ecx // push P
> >>>> [000011fb][00101fa2][00001200] e820feffff call 00001020 // call H
> >>>>
> >>>> Begin Simulation Execution Trace Stored at:21206e
> >>>> Address_of_H:1020
> >>>> [000011f0][0021205a][0021205e] 55 push ebp // enter emulated P
> >>>> [000011f1][0021205a][0021205e] 8bec mov ebp,esp
> >>>> [000011f3][0021205a][0021205e] 8b4508 mov eax,[ebp+08]
> >>>> [000011f6][00212056][000011f0] 50 push eax // push P
> >>>> [000011f7][00212056][000011f0] 8b4d08 mov ecx,[ebp+08]
> >>>> [000011fa][00212052][000011f0] 51 push ecx // push P
> >>>> [000011fb][0021204e][00001200] e820feffff call 00001020 // call emulated H
> >>>> Infinitely Recursive Simulation Detected Simulation Stopped
> >>>>
> >>>> H knows its own machine address and on this basis it can easily
> >>>> examine its stored execution_trace of P (see above) to determine:
> >>>> (a) P is calling H with the same arguments that H was called with.
> >>>> (b) No instructions in P could possibly escape this otherwise infinitely
> >>>> recursive emulation.
> >>>
> >>> FALSE. P contains these instructions:
> >>>
> >>> if (Decide_Halting(&execution_trace, &decoded, code_end, &master_state,
> >>> &slave_state, &slave_stack, Address_of_H, P, I))
> >>> goto END_OF_CODE;
> >>> return 0; // Does not halt
> >>> END_OF_CODE:
> >>> return 1; // Input has normally terminated
> >>>
> >>> That *can* and *do* prevent infinite recursive emulation in a *correct* and *complete* emulation
> >>>
> >>>> (c) H aborts its emulation of P before its call to H is emulated.
> >>>
> >>> And in doing so it fails to see that the call to H, since H is a computation, will aborts *its* simulation of P and return 0 to the outer P being emulated, which would cause it to halt, making H(P,P)==0 wrong.
> >>>
> >>>>
> >>>> [00001200][00101fae][00101fba] 83c408 add esp,+08 // return to
> >>>> executed P
> >>>> [00001203][00101fae][00101fba] 85c0 test eax,eax
> >>>> [00001205][00101fae][00101fba] 7402 jz 00001209
> >>>> [00001209][00101fb2][0000121d] 5d pop ebp
> >>>> [0000120a][00101fb6][000011f0] c3 ret // return from
> >>>> executed P
> >>>> [0000121d][00101fba][00000000] 83c404 add esp,+04
> >>>> [00001220][00101fba][00000000] 33c0 xor eax,eax
> >>>> [00001222][00101fbe][00100000] 5d pop ebp
> >>>> [00001223][00101fc2][00000000] c3 ret // ret from main
> >>>> Number of Instructions Executed(878) / 67 = 13 pages
> >>>>
> >>>>
> >>>
> >>> I've stated in detail why you're wrong. Now you need to explain in detail why I'm wrong.
> >>>
> >>> Failure to do so, which includes simply restating your original point, will be taken as admission that what I've stated is correct.
> >> H correctly predicts that its complete and correct x86 emulation of its
> >> input never reaches the the "ret" instruction of P.
> >
> > Repeating your original point.
> >
> >>
> >> This is the actual behavior of the input to H(P,P).
> >>
> >> P(P) is not the actual behavior of the input to H(P,P).
> >>
> >
> > Again, repeating your original point.
> >
> >> Never reaching the "ret" instruction means that P does not halt.
> >> A halt decider must compute the mapping from its inputs to an accept or
> >> reject state on the basis of the actual behavior that is actually
> >> specified by these inputs.
> >> Therefore H(P,P)==0 is correct.
> >
> > And again, repeating your original point.
> >
> > So no explanation as to why I'm wrong, just a repetition of your original points.
> >
> > That means you accept that what I've said is true, and that H(P,P)==0 is wrong.
> You rebuttal is based on rejecting the verified fact that a simulating
> halt decider correctly predicts that its correctly simulated input will
> never reach the final state "ret" instruction of this input.

On 6/27/2022 7:28 AM, Malcolm McLean wrote:
> On Monday, 27 June 2022 at 12:11:21 UTC+1, olcott wrote:
>> *This is the outline of the complete refutation*
>> *of the only rebuttal that anyone has left*
>>
>> (a) The complete and correct x86 emulation of the input to H(P,P) by H
>> never reaches the "ret" instruction of P.
>>
>> (b) The direct execution of P(P) does reach its "ret" instruction.
>>
> Well that's quite extraordinary.
>>
>> The above two are proven to be verified facts entirely on the basis of
>> the semantics of the x86 language.
>>
> You don't really understand what it means to "prove" something. What you
> mean is you've established that this is the case, to your satisfaction. But by far
> the most likely explanation is a mistake in your analysis.

In other words you disagree with the semantics of the x86 language.
Not a smart thing to do.

> The next most likely
> explanation is a fault in the computer. The notion that the correct emulation
> of a piece of code doesn't match its run time behaviour is absolutely the last
> thing we should consider, after all other explanations have been exhausted
> and we are driven to this conclusion.

In other words when you look at the code below that conclusively proves
my point you just don't understand the code so you guess that I must be
wrong.

void P(u32 x)
{ if (H(x, x))
HERE: goto HERE;
return;
}

int main()
{ Output("Input_Halts = ", H1((u32)P, (u32)P));
}

_P()
[0000163a](01) 55 push ebp
[0000163b](02) 8bec mov ebp,esp
[0000163d](03) 8b4508 mov eax,[ebp+08]
[00001640](01) 50 push eax
[00001641](03) 8b4d08 mov ecx,[ebp+08]
[00001644](01) 51 push ecx
[00001645](05) e8f0fdffff call 0000143a // call H
[0000164a](03) 83c408 add esp,+08
[0000164d](02) 85c0 test eax,eax
[0000164f](02) 7402 jz 00001653
[00001651](02) ebfe jmp 00001651
[00001653](01) 5d pop ebp
[00001654](01) c3 ret
Size in bytes:(0027) [00001654]

_main()
[0000165a](01) 55 push ebp
[0000165b](02) 8bec mov ebp,esp
[0000165d](05) 683a160000 push 0000163a // push P
[00001662](05) 683a160000 push 0000163a // push P
[00001667](05) e8cef9ffff call 0000103a // call H1
[0000166c](03) 83c408 add esp,+08
[0000166f](01) 50 push eax // push return value
[00001670](05) 689b040000 push 0000049b // "Input_Halts = "
[00001675](05) e870eeffff call 000004ea // call Output
[0000167a](03) 83c408 add esp,+08
[0000167d](02) 33c0 xor eax,eax
[0000167f](01) 5d pop ebp
[00001680](01) c3 ret
Size in bytes:(0039) [00001680]

machine stack stack machine assembly
address address data code language
======== ======== ======== ========= =============
[0000165a][001026a9][00000000] 55 push ebp
[0000165b][001026a9][00000000] 8bec mov ebp,esp
[0000165d][001026a5][0000163a] 683a160000 push 0000163a // push P
[00001662][001026a1][0000163a] 683a160000 push 0000163a // push P
[00001667][0010269d][0000166c] e8cef9ffff call 0000103a // call H1

H1: Begin Simulation Execution Trace Stored at:21275d
Address_of_H1:103a
[0000163a][00212749][0021274d] 55 push ebp
[0000163b][00212749][0021274d] 8bec mov ebp,esp
[0000163d][00212749][0021274d] 8b4508 mov eax,[ebp+08]
[00001640][00212745][0000163a] 50 push eax // push P
[00001641][00212745][0000163a] 8b4d08 mov ecx,[ebp+08]
[00001644][00212741][0000163a] 51 push ecx // push P
[00001645][0021273d][0000164a] e8f0fdffff call 0000143a // call H

H: Begin Simulation Execution Trace Stored at:2285c5
Address_of_H:143a
[0000163a][002285b1][002285b5] 55 push ebp
[0000163b][002285b1][002285b5] 8bec mov ebp,esp
[0000163d][002285b1][002285b5] 8b4508 mov eax,[ebp+08]
[00001640][002285ad][0000163a] 50 push eax // push P
[00001641][002285ad][0000163a] 8b4d08 mov ecx,[ebp+08]
[00001644][002285a9][0000163a] 51 push ecx // push P
[00001645][002285a5][0000164a] e8f0fdffff call 0000143a // call H
H: Infinitely Recursive Simulation Detected Simulation Stopped

[0000164a][00212749][0021274d] 83c408 add esp,+08
[0000164d][00212749][0021274d] 85c0 test eax,eax
[0000164f][00212749][0021274d] 7402 jz 00001653
[00001653][0021274d][0000111e] 5d pop ebp
[00001654][00212751][0000163a] c3 ret
H1: End Simulation Input Terminated Normally

[0000166c][001026a9][00000000] 83c408 add esp,+08
[0000166f][001026a5][00000001] 50 push eax // return value
[00001670][001026a1][0000049b] 689b040000 push 0000049b // "Input_Halts = "
[00001675][001026a1][0000049b] e870eeffff call 000004ea // call Output
Input_Halts = 1
[0000167a][001026a9][00000000] 83c408 add esp,+08
[0000167d][001026a9][00000000] 33c0 xor eax,eax
[0000167f][001026ad][00100000] 5d pop ebp
[00001680][001026b1][00000004] c3 ret
Number of Instructions Executed(409590) == 6113 Pages

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

On 6/27/2022 7:40 AM, Dennis Bush wrote:
> On Monday, June 27, 2022 at 8:34:22 AM UTC-4, olcott wrote:
>> On 6/27/2022 7:26 AM, Dennis Bush wrote:
>>> On Monday, June 27, 2022 at 8:14:31 AM UTC-4, olcott wrote:
>>>> On 6/27/2022 6:57 AM, Dennis Bush wrote:
>>>>> On Monday, June 27, 2022 at 7:11:21 AM UTC-4, olcott wrote:
>>>>>> *This is the outline of the complete refutation*
>>>>>> *of the only rebuttal that anyone has left*
>>>>>>
>>>>>> (a) The complete and correct x86 emulation of the input to H(P,P) by H
>>>>>> never reaches the "ret" instruction of P.
>>>>>
>>>>> It is known that H aborts its simulation and returns 0. That means that H *by definition* does not perform a complete and correct emulation. And since H (if it was constructed correctly) is a pure function of its inputs, it will ALWAYS return the same return for a given input.
>>>>>
>>>>> To state that H aborts *and* does a complete and correct simulation is simply nonsense.
>>>>>
>>>>> What *does* perform a complete and correct emulation of the input to H(P,P) is UTM(P,P) which halts, therefore H(P,P)==0 is wrong.
>>>>>
>>>>>>
>>>>>> (b) The direct execution of P(P) does reach its "ret" instruction.
>>>>>>
>>>>>> The above two are proven to be verified facts entirely on the basis of
>>>>>> the semantics of the x86 language.
>>>>>
>>>>> A nonsense statement can't be a "fact", verified or otherwise.
>>>>>
>>>>>>
>>>>>> A halt decider must compute the mapping from its inputs to an accept or
>>>>>> reject state on the basis of the actual behavior that is actually
>>>>>> specified by these inputs.
>>>>>
>>>>> Which for H(P,P) is BY DEFINITION is the behavior of P(P). The halting problem says that H MUST implement the following mapping (i.e. the halting function):
>>>>>
>>>>> H(x,y)==1 if and only if x(y) halts, and
>>>>> H(x,y)==0 if and only if x(y) does not halt.
>>>>>
>>>>> H does not perform this mapping.
>>>>>
>>>>>>
>>>>>> P(P) is provably not the actual behavior of the actual input.
>>>>>
>>>>> It is BY DEFINITION. So if H comes up with something different that means it did something wrong, specifically it aborted a halting computation too soon.
>>>>>
>>>>>>
>>>>>> void P(u32 x)
>>>>>> {
>>>>>> if (H(x, x))
>>>>>> HERE: goto HERE;
>>>>>> return;
>>>>>> }
>>>>>>
>>>>>> int main()
>>>>>> {
>>>>>> P(P);
>>>>>> }
>>>>>>
>>>>>> _P()
>>>>>> [000011f0](01) 55 push ebp
>>>>>> [000011f1](02) 8bec mov ebp,esp
>>>>>> [000011f3](03) 8b4508 mov eax,[ebp+08]
>>>>>> [000011f6](01) 50 push eax
>>>>>> [000011f7](03) 8b4d08 mov ecx,[ebp+08]
>>>>>> [000011fa](01) 51 push ecx
>>>>>> [000011fb](05) e820feffff call 00001020
>>>>>>
>>>>>> [00001200](03) 83c408 add esp,+08
>>>>>> [00001203](02) 85c0 test eax,eax
>>>>>> [00001205](02) 7402 jz 00001209
>>>>>> [00001207](02) ebfe jmp 00001207
>>>>>> [00001209](01) 5d pop ebp
>>>>>> [0000120a](01) c3 ret
>>>>>> Size in bytes:(0027) [0000120a]
>>>>>>
>>>>>> _main()
>>>>>> [00001210](01) 55 push ebp
>>>>>> [00001211](02) 8bec mov ebp,esp
>>>>>> [00001213](05) 68f0110000 push 000011f0
>>>>>> [00001218](05) e8d3ffffff call 000011f0
>>>>>> [0000121d](03) 83c404 add esp,+04
>>>>>> [00001220](02) 33c0 xor eax,eax
>>>>>> [00001222](01) 5d pop ebp
>>>>>> [00001223](01) c3 ret
>>>>>> Size in bytes:(0020) [00001223]
>>>>>>
>>>>>> machine stack stack machine assembly
>>>>>> address address data code language
>>>>>> ======== ======== ======== ========= =============
>>>>>> [00001210][00101fba][00000000] 55 push ebp
>>>>>> [00001211][00101fba][00000000] 8bec mov ebp,esp
>>>>>> [00001213][00101fb6][000011f0] 68f0110000 push 000011f0 // push P
>>>>>> [00001218][00101fb2][0000121d] e8d3ffffff call 000011f0 // call P
>>>>>> [000011f0][00101fae][00101fba] 55 push ebp // enter executed P
>>>>>> [000011f1][00101fae][00101fba] 8bec mov ebp,esp
>>>>>> [000011f3][00101fae][00101fba] 8b4508 mov eax,[ebp+08]
>>>>>> [000011f6][00101faa][000011f0] 50 push eax // push P
>>>>>> [000011f7][00101faa][000011f0] 8b4d08 mov ecx,[ebp+08]
>>>>>> [000011fa][00101fa6][000011f0] 51 push ecx // push P
>>>>>> [000011fb][00101fa2][00001200] e820feffff call 00001020 // call H
>>>>>>
>>>>>> Begin Simulation Execution Trace Stored at:21206e
>>>>>> Address_of_H:1020
>>>>>> [000011f0][0021205a][0021205e] 55 push ebp // enter emulated P
>>>>>> [000011f1][0021205a][0021205e] 8bec mov ebp,esp
>>>>>> [000011f3][0021205a][0021205e] 8b4508 mov eax,[ebp+08]
>>>>>> [000011f6][00212056][000011f0] 50 push eax // push P
>>>>>> [000011f7][00212056][000011f0] 8b4d08 mov ecx,[ebp+08]
>>>>>> [000011fa][00212052][000011f0] 51 push ecx // push P
>>>>>> [000011fb][0021204e][00001200] e820feffff call 00001020 // call emulated H
>>>>>> Infinitely Recursive Simulation Detected Simulation Stopped
>>>>>>
>>>>>> H knows its own machine address and on this basis it can easily
>>>>>> examine its stored execution_trace of P (see above) to determine:
>>>>>> (a) P is calling H with the same arguments that H was called with.
>>>>>> (b) No instructions in P could possibly escape this otherwise infinitely
>>>>>> recursive emulation.
>>>>>
>>>>> FALSE. P contains these instructions:
>>>>>
>>>>> if (Decide_Halting(&execution_trace, &decoded, code_end, &master_state,
>>>>> &slave_state, &slave_stack, Address_of_H, P, I))
>>>>> goto END_OF_CODE;
>>>>> return 0; // Does not halt
>>>>> END_OF_CODE:
>>>>> return 1; // Input has normally terminated
>>>>>
>>>>> That *can* and *do* prevent infinite recursive emulation in a *correct* and *complete* emulation
>>>>>
>>>>>> (c) H aborts its emulation of P before its call to H is emulated.
>>>>>
>>>>> And in doing so it fails to see that the call to H, since H is a computation, will aborts *its* simulation of P and return 0 to the outer P being emulated, which would cause it to halt, making H(P,P)==0 wrong.
>>>>>
>>>>>>
>>>>>> [00001200][00101fae][00101fba] 83c408 add esp,+08 // return to
>>>>>> executed P
>>>>>> [00001203][00101fae][00101fba] 85c0 test eax,eax
>>>>>> [00001205][00101fae][00101fba] 7402 jz 00001209
>>>>>> [00001209][00101fb2][0000121d] 5d pop ebp
>>>>>> [0000120a][00101fb6][000011f0] c3 ret // return from
>>>>>> executed P
>>>>>> [0000121d][00101fba][00000000] 83c404 add esp,+04
>>>>>> [00001220][00101fba][00000000] 33c0 xor eax,eax
>>>>>> [00001222][00101fbe][00100000] 5d pop ebp
>>>>>> [00001223][00101fc2][00000000] c3 ret // ret from main
>>>>>> Number of Instructions Executed(878) / 67 = 13 pages
>>>>>>
>>>>>>
>>>>>
>>>>> I've stated in detail why you're wrong. Now you need to explain in detail why I'm wrong.
>>>>>
>>>>> Failure to do so, which includes simply restating your original point, will be taken as admission that what I've stated is correct.
>>>> H correctly predicts that its complete and correct x86 emulation of its
>>>> input never reaches the the "ret" instruction of P.
>>>
>>> Repeating your original point.
>>>
>>>>
>>>> This is the actual behavior of the input to H(P,P).
>>>>
>>>> P(P) is not the actual behavior of the input to H(P,P).
>>>>
>>>
>>> Again, repeating your original point.
>>>
>>>> Never reaching the "ret" instruction means that P does not halt.
>>>> A halt decider must compute the mapping from its inputs to an accept or
>>>> reject state on the basis of the actual behavior that is actually
>>>> specified by these inputs.
>>>> Therefore H(P,P)==0 is correct.
>>>
>>> And again, repeating your original point.
>>>
>>> So no explanation as to why I'm wrong, just a repetition of your original points.
>>>
>>> That means you accept that what I've said is true, and that H(P,P)==0 is wrong.
>> You rebuttal is based on rejecting the verified fact that a simulating
>> halt decider correctly predicts that its correctly simulated input will
>> never reach the final state "ret" instruction of this input.
>
> FALSE. The correctly simulated input to H(P,P) is UTM(P,P) which halts, therefore H(P,P)==0 is wrong.

On Monday, June 27, 2022 at 8:50:05 AM UTC-4, olcott wrote:
> On 6/27/2022 7:40 AM, Dennis Bush wrote:
> > On Monday, June 27, 2022 at 8:34:22 AM UTC-4, olcott wrote:
> >> On 6/27/2022 7:26 AM, Dennis Bush wrote:
> >>> On Monday, June 27, 2022 at 8:14:31 AM UTC-4, olcott wrote:
> >>>> On 6/27/2022 6:57 AM, Dennis Bush wrote:
> >>>>> On Monday, June 27, 2022 at 7:11:21 AM UTC-4, olcott wrote:
> >>>>>> *This is the outline of the complete refutation*
> >>>>>> *of the only rebuttal that anyone has left*
> >>>>>>
> >>>>>> (a) The complete and correct x86 emulation of the input to H(P,P) by H
> >>>>>> never reaches the "ret" instruction of P.
> >>>>>
> >>>>> It is known that H aborts its simulation and returns 0. That means that H *by definition* does not perform a complete and correct emulation. And since H (if it was constructed correctly) is a pure function of its inputs, it will ALWAYS return the same return for a given input.
> >>>>>
> >>>>> To state that H aborts *and* does a complete and correct simulation is simply nonsense.
> >>>>>
> >>>>> What *does* perform a complete and correct emulation of the input to H(P,P) is UTM(P,P) which halts, therefore H(P,P)==0 is wrong.
> >>>>>
> >>>>>>
> >>>>>> (b) The direct execution of P(P) does reach its "ret" instruction.
> >>>>>>
> >>>>>> The above two are proven to be verified facts entirely on the basis of
> >>>>>> the semantics of the x86 language.
> >>>>>
> >>>>> A nonsense statement can't be a "fact", verified or otherwise.
> >>>>>
> >>>>>>
> >>>>>> A halt decider must compute the mapping from its inputs to an accept or
> >>>>>> reject state on the basis of the actual behavior that is actually
> >>>>>> specified by these inputs.
> >>>>>
> >>>>> Which for H(P,P) is BY DEFINITION is the behavior of P(P). The halting problem says that H MUST implement the following mapping (i.e. the halting function):
> >>>>>
> >>>>> H(x,y)==1 if and only if x(y) halts, and
> >>>>> H(x,y)==0 if and only if x(y) does not halt.
> >>>>>
> >>>>> H does not perform this mapping.
> >>>>>
> >>>>>>
> >>>>>> P(P) is provably not the actual behavior of the actual input.
> >>>>>
> >>>>> It is BY DEFINITION. So if H comes up with something different that means it did something wrong, specifically it aborted a halting computation too soon.
> >>>>>
> >>>>>>
> >>>>>> void P(u32 x)
> >>>>>> {
> >>>>>> if (H(x, x))
> >>>>>> HERE: goto HERE;
> >>>>>> return;
> >>>>>> }
> >>>>>>
> >>>>>> int main()
> >>>>>> {
> >>>>>> P(P);
> >>>>>> }
> >>>>>>
> >>>>>> _P()
> >>>>>> [000011f0](01) 55 push ebp
> >>>>>> [000011f1](02) 8bec mov ebp,esp
> >>>>>> [000011f3](03) 8b4508 mov eax,[ebp+08]
> >>>>>> [000011f6](01) 50 push eax
> >>>>>> [000011f7](03) 8b4d08 mov ecx,[ebp+08]
> >>>>>> [000011fa](01) 51 push ecx
> >>>>>> [000011fb](05) e820feffff call 00001020
> >>>>>>
> >>>>>> [00001200](03) 83c408 add esp,+08
> >>>>>> [00001203](02) 85c0 test eax,eax
> >>>>>> [00001205](02) 7402 jz 00001209
> >>>>>> [00001207](02) ebfe jmp 00001207
> >>>>>> [00001209](01) 5d pop ebp
> >>>>>> [0000120a](01) c3 ret
> >>>>>> Size in bytes:(0027) [0000120a]
> >>>>>>
> >>>>>> _main()
> >>>>>> [00001210](01) 55 push ebp
> >>>>>> [00001211](02) 8bec mov ebp,esp
> >>>>>> [00001213](05) 68f0110000 push 000011f0
> >>>>>> [00001218](05) e8d3ffffff call 000011f0
> >>>>>> [0000121d](03) 83c404 add esp,+04
> >>>>>> [00001220](02) 33c0 xor eax,eax
> >>>>>> [00001222](01) 5d pop ebp
> >>>>>> [00001223](01) c3 ret
> >>>>>> Size in bytes:(0020) [00001223]
> >>>>>>
> >>>>>> machine stack stack machine assembly
> >>>>>> address address data code language
> >>>>>> ======== ======== ======== ========= =============
> >>>>>> [00001210][00101fba][00000000] 55 push ebp
> >>>>>> [00001211][00101fba][00000000] 8bec mov ebp,esp
> >>>>>> [00001213][00101fb6][000011f0] 68f0110000 push 000011f0 // push P
> >>>>>> [00001218][00101fb2][0000121d] e8d3ffffff call 000011f0 // call P
> >>>>>> [000011f0][00101fae][00101fba] 55 push ebp // enter executed P
> >>>>>> [000011f1][00101fae][00101fba] 8bec mov ebp,esp
> >>>>>> [000011f3][00101fae][00101fba] 8b4508 mov eax,[ebp+08]
> >>>>>> [000011f6][00101faa][000011f0] 50 push eax // push P
> >>>>>> [000011f7][00101faa][000011f0] 8b4d08 mov ecx,[ebp+08]
> >>>>>> [000011fa][00101fa6][000011f0] 51 push ecx // push P
> >>>>>> [000011fb][00101fa2][00001200] e820feffff call 00001020 // call H
> >>>>>>
> >>>>>> Begin Simulation Execution Trace Stored at:21206e
> >>>>>> Address_of_H:1020
> >>>>>> [000011f0][0021205a][0021205e] 55 push ebp // enter emulated P
> >>>>>> [000011f1][0021205a][0021205e] 8bec mov ebp,esp
> >>>>>> [000011f3][0021205a][0021205e] 8b4508 mov eax,[ebp+08]
> >>>>>> [000011f6][00212056][000011f0] 50 push eax // push P
> >>>>>> [000011f7][00212056][000011f0] 8b4d08 mov ecx,[ebp+08]
> >>>>>> [000011fa][00212052][000011f0] 51 push ecx // push P
> >>>>>> [000011fb][0021204e][00001200] e820feffff call 00001020 // call emulated H
> >>>>>> Infinitely Recursive Simulation Detected Simulation Stopped
> >>>>>>
> >>>>>> H knows its own machine address and on this basis it can easily
> >>>>>> examine its stored execution_trace of P (see above) to determine:
> >>>>>> (a) P is calling H with the same arguments that H was called with.
> >>>>>> (b) No instructions in P could possibly escape this otherwise infinitely
> >>>>>> recursive emulation.
> >>>>>
> >>>>> FALSE. P contains these instructions:
> >>>>>
> >>>>> if (Decide_Halting(&execution_trace, &decoded, code_end, &master_state,
> >>>>> &slave_state, &slave_stack, Address_of_H, P, I))
> >>>>> goto END_OF_CODE;
> >>>>> return 0; // Does not halt
> >>>>> END_OF_CODE:
> >>>>> return 1; // Input has normally terminated
> >>>>>
> >>>>> That *can* and *do* prevent infinite recursive emulation in a *correct* and *complete* emulation
> >>>>>
> >>>>>> (c) H aborts its emulation of P before its call to H is emulated.
> >>>>>
> >>>>> And in doing so it fails to see that the call to H, since H is a computation, will aborts *its* simulation of P and return 0 to the outer P being emulated, which would cause it to halt, making H(P,P)==0 wrong.
> >>>>>
> >>>>>>
> >>>>>> [00001200][00101fae][00101fba] 83c408 add esp,+08 // return to
> >>>>>> executed P
> >>>>>> [00001203][00101fae][00101fba] 85c0 test eax,eax
> >>>>>> [00001205][00101fae][00101fba] 7402 jz 00001209
> >>>>>> [00001209][00101fb2][0000121d] 5d pop ebp
> >>>>>> [0000120a][00101fb6][000011f0] c3 ret // return from
> >>>>>> executed P
> >>>>>> [0000121d][00101fba][00000000] 83c404 add esp,+04
> >>>>>> [00001220][00101fba][00000000] 33c0 xor eax,eax
> >>>>>> [00001222][00101fbe][00100000] 5d pop ebp
> >>>>>> [00001223][00101fc2][00000000] c3 ret // ret from main
> >>>>>> Number of Instructions Executed(878) / 67 = 13 pages
> >>>>>>
> >>>>>>
> >>>>>
> >>>>> I've stated in detail why you're wrong. Now you need to explain in detail why I'm wrong.
> >>>>>
> >>>>> Failure to do so, which includes simply restating your original point, will be taken as admission that what I've stated is correct.
> >>>> H correctly predicts that its complete and correct x86 emulation of its
> >>>> input never reaches the the "ret" instruction of P.
> >>>
> >>> Repeating your original point.
> >>>
> >>>>
> >>>> This is the actual behavior of the input to H(P,P).
> >>>>
> >>>> P(P) is not the actual behavior of the input to H(P,P).
> >>>>
> >>>
> >>> Again, repeating your original point.
> >>>
> >>>> Never reaching the "ret" instruction means that P does not halt.
> >>>> A halt decider must compute the mapping from its inputs to an accept or
> >>>> reject state on the basis of the actual behavior that is actually
> >>>> specified by these inputs.
> >>>> Therefore H(P,P)==0 is correct.
> >>>
> >>> And again, repeating your original point.
> >>>
> >>> So no explanation as to why I'm wrong, just a repetition of your original points.
> >>>
> >>> That means you accept that what I've said is true, and that H(P,P)==0 is wrong.
> >> You rebuttal is based on rejecting the verified fact that a simulating
> >> halt decider correctly predicts that its correctly simulated input will
> >> never reach the final state "ret" instruction of this input.
> >
> > FALSE. The correctly simulated input to H(P,P) is UTM(P,P) which halts, therefore H(P,P)==0 is wrong.
> First of all I am not talking about UTM's.
> I am talking about x86 emulation.
> Try to encode what you mean, here is my guess:
>
> void P(u32 x)
> {
> if (emulate_x86(x, x))
> HERE: goto HERE;
> return;
> }
>
> int main()
> {
> Output("Input_Halts = ", H((u32)P, (u32)P));
> }

On 6/27/2022 7:58 AM, Dennis Bush wrote:
> On Monday, June 27, 2022 at 8:50:05 AM UTC-4, olcott wrote:
>> On 6/27/2022 7:40 AM, Dennis Bush wrote:
>>> On Monday, June 27, 2022 at 8:34:22 AM UTC-4, olcott wrote:
>>>> On 6/27/2022 7:26 AM, Dennis Bush wrote:
>>>>> On Monday, June 27, 2022 at 8:14:31 AM UTC-4, olcott wrote:
>>>>>> On 6/27/2022 6:57 AM, Dennis Bush wrote:
>>>>>>> On Monday, June 27, 2022 at 7:11:21 AM UTC-4, olcott wrote:
>>>>>>>> *This is the outline of the complete refutation*
>>>>>>>> *of the only rebuttal that anyone has left*
>>>>>>>>
>>>>>>>> (a) The complete and correct x86 emulation of the input to H(P,P) by H
>>>>>>>> never reaches the "ret" instruction of P.
>>>>>>>
>>>>>>> It is known that H aborts its simulation and returns 0. That means that H *by definition* does not perform a complete and correct emulation. And since H (if it was constructed correctly) is a pure function of its inputs, it will ALWAYS return the same return for a given input.
>>>>>>>
>>>>>>> To state that H aborts *and* does a complete and correct simulation is simply nonsense.
>>>>>>>
>>>>>>> What *does* perform a complete and correct emulation of the input to H(P,P) is UTM(P,P) which halts, therefore H(P,P)==0 is wrong.
>>>>>>>
>>>>>>>>
>>>>>>>> (b) The direct execution of P(P) does reach its "ret" instruction.
>>>>>>>>
>>>>>>>> The above two are proven to be verified facts entirely on the basis of
>>>>>>>> the semantics of the x86 language.
>>>>>>>
>>>>>>> A nonsense statement can't be a "fact", verified or otherwise.
>>>>>>>
>>>>>>>>
>>>>>>>> A halt decider must compute the mapping from its inputs to an accept or
>>>>>>>> reject state on the basis of the actual behavior that is actually
>>>>>>>> specified by these inputs.
>>>>>>>
>>>>>>> Which for H(P,P) is BY DEFINITION is the behavior of P(P). The halting problem says that H MUST implement the following mapping (i.e. the halting function):
>>>>>>>
>>>>>>> H(x,y)==1 if and only if x(y) halts, and
>>>>>>> H(x,y)==0 if and only if x(y) does not halt.
>>>>>>>
>>>>>>> H does not perform this mapping.
>>>>>>>
>>>>>>>>
>>>>>>>> P(P) is provably not the actual behavior of the actual input.
>>>>>>>
>>>>>>> It is BY DEFINITION. So if H comes up with something different that means it did something wrong, specifically it aborted a halting computation too soon.
>>>>>>>
>>>>>>>>
>>>>>>>> void P(u32 x)
>>>>>>>> {
>>>>>>>> if (H(x, x))
>>>>>>>> HERE: goto HERE;
>>>>>>>> return;
>>>>>>>> }
>>>>>>>>
>>>>>>>> int main()
>>>>>>>> {
>>>>>>>> P(P);
>>>>>>>> }
>>>>>>>>
>>>>>>>> _P()
>>>>>>>> [000011f0](01) 55 push ebp
>>>>>>>> [000011f1](02) 8bec mov ebp,esp
>>>>>>>> [000011f3](03) 8b4508 mov eax,[ebp+08]
>>>>>>>> [000011f6](01) 50 push eax
>>>>>>>> [000011f7](03) 8b4d08 mov ecx,[ebp+08]
>>>>>>>> [000011fa](01) 51 push ecx
>>>>>>>> [000011fb](05) e820feffff call 00001020
>>>>>>>>
>>>>>>>> [00001200](03) 83c408 add esp,+08
>>>>>>>> [00001203](02) 85c0 test eax,eax
>>>>>>>> [00001205](02) 7402 jz 00001209
>>>>>>>> [00001207](02) ebfe jmp 00001207
>>>>>>>> [00001209](01) 5d pop ebp
>>>>>>>> [0000120a](01) c3 ret
>>>>>>>> Size in bytes:(0027) [0000120a]
>>>>>>>>
>>>>>>>> _main()
>>>>>>>> [00001210](01) 55 push ebp
>>>>>>>> [00001211](02) 8bec mov ebp,esp
>>>>>>>> [00001213](05) 68f0110000 push 000011f0
>>>>>>>> [00001218](05) e8d3ffffff call 000011f0
>>>>>>>> [0000121d](03) 83c404 add esp,+04
>>>>>>>> [00001220](02) 33c0 xor eax,eax
>>>>>>>> [00001222](01) 5d pop ebp
>>>>>>>> [00001223](01) c3 ret
>>>>>>>> Size in bytes:(0020) [00001223]
>>>>>>>>
>>>>>>>> machine stack stack machine assembly
>>>>>>>> address address data code language
>>>>>>>> ======== ======== ======== ========= =============
>>>>>>>> [00001210][00101fba][00000000] 55 push ebp
>>>>>>>> [00001211][00101fba][00000000] 8bec mov ebp,esp
>>>>>>>> [00001213][00101fb6][000011f0] 68f0110000 push 000011f0 // push P
>>>>>>>> [00001218][00101fb2][0000121d] e8d3ffffff call 000011f0 // call P
>>>>>>>> [000011f0][00101fae][00101fba] 55 push ebp // enter executed P
>>>>>>>> [000011f1][00101fae][00101fba] 8bec mov ebp,esp
>>>>>>>> [000011f3][00101fae][00101fba] 8b4508 mov eax,[ebp+08]
>>>>>>>> [000011f6][00101faa][000011f0] 50 push eax // push P
>>>>>>>> [000011f7][00101faa][000011f0] 8b4d08 mov ecx,[ebp+08]
>>>>>>>> [000011fa][00101fa6][000011f0] 51 push ecx // push P
>>>>>>>> [000011fb][00101fa2][00001200] e820feffff call 00001020 // call H
>>>>>>>>
>>>>>>>> Begin Simulation Execution Trace Stored at:21206e
>>>>>>>> Address_of_H:1020
>>>>>>>> [000011f0][0021205a][0021205e] 55 push ebp // enter emulated P
>>>>>>>> [000011f1][0021205a][0021205e] 8bec mov ebp,esp
>>>>>>>> [000011f3][0021205a][0021205e] 8b4508 mov eax,[ebp+08]
>>>>>>>> [000011f6][00212056][000011f0] 50 push eax // push P
>>>>>>>> [000011f7][00212056][000011f0] 8b4d08 mov ecx,[ebp+08]
>>>>>>>> [000011fa][00212052][000011f0] 51 push ecx // push P
>>>>>>>> [000011fb][0021204e][00001200] e820feffff call 00001020 // call emulated H
>>>>>>>> Infinitely Recursive Simulation Detected Simulation Stopped
>>>>>>>>
>>>>>>>> H knows its own machine address and on this basis it can easily
>>>>>>>> examine its stored execution_trace of P (see above) to determine:
>>>>>>>> (a) P is calling H with the same arguments that H was called with.
>>>>>>>> (b) No instructions in P could possibly escape this otherwise infinitely
>>>>>>>> recursive emulation.
>>>>>>>
>>>>>>> FALSE. P contains these instructions:
>>>>>>>
>>>>>>> if (Decide_Halting(&execution_trace, &decoded, code_end, &master_state,
>>>>>>> &slave_state, &slave_stack, Address_of_H, P, I))
>>>>>>> goto END_OF_CODE;
>>>>>>> return 0; // Does not halt
>>>>>>> END_OF_CODE:
>>>>>>> return 1; // Input has normally terminated
>>>>>>>
>>>>>>> That *can* and *do* prevent infinite recursive emulation in a *correct* and *complete* emulation
>>>>>>>
>>>>>>>> (c) H aborts its emulation of P before its call to H is emulated.
>>>>>>>
>>>>>>> And in doing so it fails to see that the call to H, since H is a computation, will aborts *its* simulation of P and return 0 to the outer P being emulated, which would cause it to halt, making H(P,P)==0 wrong.
>>>>>>>
>>>>>>>>
>>>>>>>> [00001200][00101fae][00101fba] 83c408 add esp,+08 // return to
>>>>>>>> executed P
>>>>>>>> [00001203][00101fae][00101fba] 85c0 test eax,eax
>>>>>>>> [00001205][00101fae][00101fba] 7402 jz 00001209
>>>>>>>> [00001209][00101fb2][0000121d] 5d pop ebp
>>>>>>>> [0000120a][00101fb6][000011f0] c3 ret // return from
>>>>>>>> executed P
>>>>>>>> [0000121d][00101fba][00000000] 83c404 add esp,+04
>>>>>>>> [00001220][00101fba][00000000] 33c0 xor eax,eax
>>>>>>>> [00001222][00101fbe][00100000] 5d pop ebp
>>>>>>>> [00001223][00101fc2][00000000] c3 ret // ret from main
>>>>>>>> Number of Instructions Executed(878) / 67 = 13 pages
>>>>>>>>
>>>>>>>>
>>>>>>>
>>>>>>> I've stated in detail why you're wrong. Now you need to explain in detail why I'm wrong.
>>>>>>>
>>>>>>> Failure to do so, which includes simply restating your original point, will be taken as admission that what I've stated is correct.
>>>>>> H correctly predicts that its complete and correct x86 emulation of its
>>>>>> input never reaches the the "ret" instruction of P.
>>>>>
>>>>> Repeating your original point.
>>>>>
>>>>>>
>>>>>> This is the actual behavior of the input to H(P,P).
>>>>>>
>>>>>> P(P) is not the actual behavior of the input to H(P,P).
>>>>>>
>>>>>
>>>>> Again, repeating your original point.
>>>>>
>>>>>> Never reaching the "ret" instruction means that P does not halt.
>>>>>> A halt decider must compute the mapping from its inputs to an accept or
>>>>>> reject state on the basis of the actual behavior that is actually
>>>>>> specified by these inputs.
>>>>>> Therefore H(P,P)==0 is correct.
>>>>>
>>>>> And again, repeating your original point.
>>>>>
>>>>> So no explanation as to why I'm wrong, just a repetition of your original points.
>>>>>
>>>>> That means you accept that what I've said is true, and that H(P,P)==0 is wrong.
>>>> You rebuttal is based on rejecting the verified fact that a simulating
>>>> halt decider correctly predicts that its correctly simulated input will
>>>> never reach the final state "ret" instruction of this input.
>>>
>>> FALSE. The correctly simulated input to H(P,P) is UTM(P,P) which halts, therefore H(P,P)==0 is wrong.
>> First of all I am not talking about UTM's.
>> I am talking about x86 emulation.
>> Try to encode what you mean, here is my guess:
>>
>> void P(u32 x)
>> {
>> if (emulate_x86(x, x))
>> HERE: goto HERE;
>> return;
>> }
>>
>> int main()
>> {
>> Output("Input_Halts = ", H((u32)P, (u32)P));
>> }
>
> P is a computation and therefore it can't be changed, nor can anything that it calls. I mean UTM(P,P). The simplest way to implement this is:
>
> void UTM(u32 x, u32 y)
> {
> x(y)
> }
>
> H cannot by definition perform a correct and complete emulation if it aborts, so the input must be passed to something that does, i.e. a UTM, for the source of truth.
>

On 6/27/2022 4:11 AM, olcott wrote:
> *This is the outline of the complete refutation*
> *of the only rebuttal that anyone has left*
>

Post it 213,558 more times you impotent blithering wankmaggot.

On Monday, June 27, 2022 at 9:28:22 AM UTC-4, olcott wrote:
> On 6/27/2022 7:58 AM, Dennis Bush wrote:
> > On Monday, June 27, 2022 at 8:50:05 AM UTC-4, olcott wrote:
> >> On 6/27/2022 7:40 AM, Dennis Bush wrote:
> >>> On Monday, June 27, 2022 at 8:34:22 AM UTC-4, olcott wrote:
> >>>> On 6/27/2022 7:26 AM, Dennis Bush wrote:
> >>>>> On Monday, June 27, 2022 at 8:14:31 AM UTC-4, olcott wrote:
> >>>>>> On 6/27/2022 6:57 AM, Dennis Bush wrote:
> >>>>>>> On Monday, June 27, 2022 at 7:11:21 AM UTC-4, olcott wrote:
> >>>>>>>> *This is the outline of the complete refutation*
> >>>>>>>> *of the only rebuttal that anyone has left*
> >>>>>>>>
> >>>>>>>> (a) The complete and correct x86 emulation of the input to H(P,P) by H
> >>>>>>>> never reaches the "ret" instruction of P.
> >>>>>>>
> >>>>>>> It is known that H aborts its simulation and returns 0. That means that H *by definition* does not perform a complete and correct emulation. And since H (if it was constructed correctly) is a pure function of its inputs, it will ALWAYS return the same return for a given input.
> >>>>>>>
> >>>>>>> To state that H aborts *and* does a complete and correct simulation is simply nonsense.
> >>>>>>>
> >>>>>>> What *does* perform a complete and correct emulation of the input to H(P,P) is UTM(P,P) which halts, therefore H(P,P)==0 is wrong.
> >>>>>>>
> >>>>>>>>
> >>>>>>>> (b) The direct execution of P(P) does reach its "ret" instruction.
> >>>>>>>>
> >>>>>>>> The above two are proven to be verified facts entirely on the basis of
> >>>>>>>> the semantics of the x86 language.
> >>>>>>>
> >>>>>>> A nonsense statement can't be a "fact", verified or otherwise.
> >>>>>>>
> >>>>>>>>
> >>>>>>>> A halt decider must compute the mapping from its inputs to an accept or
> >>>>>>>> reject state on the basis of the actual behavior that is actually
> >>>>>>>> specified by these inputs.
> >>>>>>>
> >>>>>>> Which for H(P,P) is BY DEFINITION is the behavior of P(P). The halting problem says that H MUST implement the following mapping (i.e. the halting function):
> >>>>>>>
> >>>>>>> H(x,y)==1 if and only if x(y) halts, and
> >>>>>>> H(x,y)==0 if and only if x(y) does not halt.
> >>>>>>>
> >>>>>>> H does not perform this mapping.
> >>>>>>>
> >>>>>>>>
> >>>>>>>> P(P) is provably not the actual behavior of the actual input.
> >>>>>>>
> >>>>>>> It is BY DEFINITION. So if H comes up with something different that means it did something wrong, specifically it aborted a halting computation too soon.
> >>>>>>>
> >>>>>>>>
> >>>>>>>> void P(u32 x)
> >>>>>>>> {
> >>>>>>>> if (H(x, x))
> >>>>>>>> HERE: goto HERE;
> >>>>>>>> return;
> >>>>>>>> }
> >>>>>>>>
> >>>>>>>> int main()
> >>>>>>>> {
> >>>>>>>> P(P);
> >>>>>>>> }
> >>>>>>>>
> >>>>>>>> _P()
> >>>>>>>> [000011f0](01) 55 push ebp
> >>>>>>>> [000011f1](02) 8bec mov ebp,esp
> >>>>>>>> [000011f3](03) 8b4508 mov eax,[ebp+08]
> >>>>>>>> [000011f6](01) 50 push eax
> >>>>>>>> [000011f7](03) 8b4d08 mov ecx,[ebp+08]
> >>>>>>>> [000011fa](01) 51 push ecx
> >>>>>>>> [000011fb](05) e820feffff call 00001020
> >>>>>>>>
> >>>>>>>> [00001200](03) 83c408 add esp,+08
> >>>>>>>> [00001203](02) 85c0 test eax,eax
> >>>>>>>> [00001205](02) 7402 jz 00001209
> >>>>>>>> [00001207](02) ebfe jmp 00001207
> >>>>>>>> [00001209](01) 5d pop ebp
> >>>>>>>> [0000120a](01) c3 ret
> >>>>>>>> Size in bytes:(0027) [0000120a]
> >>>>>>>>
> >>>>>>>> _main()
> >>>>>>>> [00001210](01) 55 push ebp
> >>>>>>>> [00001211](02) 8bec mov ebp,esp
> >>>>>>>> [00001213](05) 68f0110000 push 000011f0
> >>>>>>>> [00001218](05) e8d3ffffff call 000011f0
> >>>>>>>> [0000121d](03) 83c404 add esp,+04
> >>>>>>>> [00001220](02) 33c0 xor eax,eax
> >>>>>>>> [00001222](01) 5d pop ebp
> >>>>>>>> [00001223](01) c3 ret
> >>>>>>>> Size in bytes:(0020) [00001223]
> >>>>>>>>
> >>>>>>>> machine stack stack machine assembly
> >>>>>>>> address address data code language
> >>>>>>>> ======== ======== ======== ========= =============
> >>>>>>>> [00001210][00101fba][00000000] 55 push ebp
> >>>>>>>> [00001211][00101fba][00000000] 8bec mov ebp,esp
> >>>>>>>> [00001213][00101fb6][000011f0] 68f0110000 push 000011f0 // push P
> >>>>>>>> [00001218][00101fb2][0000121d] e8d3ffffff call 000011f0 // call P
> >>>>>>>> [000011f0][00101fae][00101fba] 55 push ebp // enter executed P
> >>>>>>>> [000011f1][00101fae][00101fba] 8bec mov ebp,esp
> >>>>>>>> [000011f3][00101fae][00101fba] 8b4508 mov eax,[ebp+08]
> >>>>>>>> [000011f6][00101faa][000011f0] 50 push eax // push P
> >>>>>>>> [000011f7][00101faa][000011f0] 8b4d08 mov ecx,[ebp+08]
> >>>>>>>> [000011fa][00101fa6][000011f0] 51 push ecx // push P
> >>>>>>>> [000011fb][00101fa2][00001200] e820feffff call 00001020 // call H
> >>>>>>>>
> >>>>>>>> Begin Simulation Execution Trace Stored at:21206e
> >>>>>>>> Address_of_H:1020
> >>>>>>>> [000011f0][0021205a][0021205e] 55 push ebp // enter emulated P
> >>>>>>>> [000011f1][0021205a][0021205e] 8bec mov ebp,esp
> >>>>>>>> [000011f3][0021205a][0021205e] 8b4508 mov eax,[ebp+08]
> >>>>>>>> [000011f6][00212056][000011f0] 50 push eax // push P
> >>>>>>>> [000011f7][00212056][000011f0] 8b4d08 mov ecx,[ebp+08]
> >>>>>>>> [000011fa][00212052][000011f0] 51 push ecx // push P
> >>>>>>>> [000011fb][0021204e][00001200] e820feffff call 00001020 // call emulated H
> >>>>>>>> Infinitely Recursive Simulation Detected Simulation Stopped
> >>>>>>>>
> >>>>>>>> H knows its own machine address and on this basis it can easily
> >>>>>>>> examine its stored execution_trace of P (see above) to determine:
> >>>>>>>> (a) P is calling H with the same arguments that H was called with.
> >>>>>>>> (b) No instructions in P could possibly escape this otherwise infinitely
> >>>>>>>> recursive emulation.
> >>>>>>>
> >>>>>>> FALSE. P contains these instructions:
> >>>>>>>
> >>>>>>> if (Decide_Halting(&execution_trace, &decoded, code_end, &master_state,
> >>>>>>> &slave_state, &slave_stack, Address_of_H, P, I))
> >>>>>>> goto END_OF_CODE;
> >>>>>>> return 0; // Does not halt
> >>>>>>> END_OF_CODE:
> >>>>>>> return 1; // Input has normally terminated
> >>>>>>>
> >>>>>>> That *can* and *do* prevent infinite recursive emulation in a *correct* and *complete* emulation
> >>>>>>>
> >>>>>>>> (c) H aborts its emulation of P before its call to H is emulated..
> >>>>>>>
> >>>>>>> And in doing so it fails to see that the call to H, since H is a computation, will aborts *its* simulation of P and return 0 to the outer P being emulated, which would cause it to halt, making H(P,P)==0 wrong.
> >>>>>>>
> >>>>>>>>
> >>>>>>>> [00001200][00101fae][00101fba] 83c408 add esp,+08 // return to
> >>>>>>>> executed P
> >>>>>>>> [00001203][00101fae][00101fba] 85c0 test eax,eax
> >>>>>>>> [00001205][00101fae][00101fba] 7402 jz 00001209
> >>>>>>>> [00001209][00101fb2][0000121d] 5d pop ebp
> >>>>>>>> [0000120a][00101fb6][000011f0] c3 ret // return from
> >>>>>>>> executed P
> >>>>>>>> [0000121d][00101fba][00000000] 83c404 add esp,+04
> >>>>>>>> [00001220][00101fba][00000000] 33c0 xor eax,eax
> >>>>>>>> [00001222][00101fbe][00100000] 5d pop ebp
> >>>>>>>> [00001223][00101fc2][00000000] c3 ret // ret from main
> >>>>>>>> Number of Instructions Executed(878) / 67 = 13 pages
> >>>>>>>>
> >>>>>>>>
> >>>>>>>
> >>>>>>> I've stated in detail why you're wrong. Now you need to explain in detail why I'm wrong.
> >>>>>>>
> >>>>>>> Failure to do so, which includes simply restating your original point, will be taken as admission that what I've stated is correct.
> >>>>>> H correctly predicts that its complete and correct x86 emulation of its
> >>>>>> input never reaches the the "ret" instruction of P.
> >>>>>
> >>>>> Repeating your original point.
> >>>>>
> >>>>>>
> >>>>>> This is the actual behavior of the input to H(P,P).
> >>>>>>
> >>>>>> P(P) is not the actual behavior of the input to H(P,P).
> >>>>>>
> >>>>>
> >>>>> Again, repeating your original point.
> >>>>>
> >>>>>> Never reaching the "ret" instruction means that P does not halt.
> >>>>>> A halt decider must compute the mapping from its inputs to an accept or
> >>>>>> reject state on the basis of the actual behavior that is actually
> >>>>>> specified by these inputs.
> >>>>>> Therefore H(P,P)==0 is correct.
> >>>>>
> >>>>> And again, repeating your original point.
> >>>>>
> >>>>> So no explanation as to why I'm wrong, just a repetition of your original points.
> >>>>>
> >>>>> That means you accept that what I've said is true, and that H(P,P)==0 is wrong.
> >>>> You rebuttal is based on rejecting the verified fact that a simulating
> >>>> halt decider correctly predicts that its correctly simulated input will
> >>>> never reach the final state "ret" instruction of this input.
> >>>
> >>> FALSE. The correctly simulated input to H(P,P) is UTM(P,P) which halts, therefore H(P,P)==0 is wrong.
> >> First of all I am not talking about UTM's.
> >> I am talking about x86 emulation.
> >> Try to encode what you mean, here is my guess:
> >>
> >> void P(u32 x)
> >> {
> >> if (emulate_x86(x, x))
> >> HERE: goto HERE;
> >> return;
> >> }
> >>
> >> int main()
> >> {
> >> Output("Input_Halts = ", H((u32)P, (u32)P));
> >> }
> >
> > P is a computation and therefore it can't be changed, nor can anything that it calls. I mean UTM(P,P). The simplest way to implement this is:
> >
> > void UTM(u32 x, u32 y)
> > {
> > x(y)
> > }
> >
> > H cannot by definition perform a correct and complete emulation if it aborts, so the input must be passed to something that does, i.e. a UTM, for the source of truth.
> >
> A simulating halt decider correctly simulates its input until it
> correctly determines that this simulated input would never reach its
> final state.

On 6/27/2022 12:03 PM, Dennis Bush wrote:
> On Monday, June 27, 2022 at 9:28:22 AM UTC-4, olcott wrote:
>> On 6/27/2022 7:58 AM, Dennis Bush wrote:
>>> On Monday, June 27, 2022 at 8:50:05 AM UTC-4, olcott wrote:
>>>> On 6/27/2022 7:40 AM, Dennis Bush wrote:
>>>>> On Monday, June 27, 2022 at 8:34:22 AM UTC-4, olcott wrote:
>>>>>> On 6/27/2022 7:26 AM, Dennis Bush wrote:
>>>>>>> On Monday, June 27, 2022 at 8:14:31 AM UTC-4, olcott wrote:
>>>>>>>> On 6/27/2022 6:57 AM, Dennis Bush wrote:
>>>>>>>>> On Monday, June 27, 2022 at 7:11:21 AM UTC-4, olcott wrote:
>>>>>>>>>> *This is the outline of the complete refutation*
>>>>>>>>>> *of the only rebuttal that anyone has left*
>>>>>>>>>>
>>>>>>>>>> (a) The complete and correct x86 emulation of the input to H(P,P) by H
>>>>>>>>>> never reaches the "ret" instruction of P.
>>>>>>>>>
>>>>>>>>> It is known that H aborts its simulation and returns 0. That means that H *by definition* does not perform a complete and correct emulation. And since H (if it was constructed correctly) is a pure function of its inputs, it will ALWAYS return the same return for a given input.
>>>>>>>>>
>>>>>>>>> To state that H aborts *and* does a complete and correct simulation is simply nonsense.
>>>>>>>>>
>>>>>>>>> What *does* perform a complete and correct emulation of the input to H(P,P) is UTM(P,P) which halts, therefore H(P,P)==0 is wrong.
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> (b) The direct execution of P(P) does reach its "ret" instruction.
>>>>>>>>>>
>>>>>>>>>> The above two are proven to be verified facts entirely on the basis of
>>>>>>>>>> the semantics of the x86 language.
>>>>>>>>>
>>>>>>>>> A nonsense statement can't be a "fact", verified or otherwise.
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> A halt decider must compute the mapping from its inputs to an accept or
>>>>>>>>>> reject state on the basis of the actual behavior that is actually
>>>>>>>>>> specified by these inputs.
>>>>>>>>>
>>>>>>>>> Which for H(P,P) is BY DEFINITION is the behavior of P(P). The halting problem says that H MUST implement the following mapping (i.e. the halting function):
>>>>>>>>>
>>>>>>>>> H(x,y)==1 if and only if x(y) halts, and
>>>>>>>>> H(x,y)==0 if and only if x(y) does not halt.
>>>>>>>>>
>>>>>>>>> H does not perform this mapping.
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> P(P) is provably not the actual behavior of the actual input.
>>>>>>>>>
>>>>>>>>> It is BY DEFINITION. So if H comes up with something different that means it did something wrong, specifically it aborted a halting computation too soon.
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> void P(u32 x)
>>>>>>>>>> {
>>>>>>>>>> if (H(x, x))
>>>>>>>>>> HERE: goto HERE;
>>>>>>>>>> return;
>>>>>>>>>> }
>>>>>>>>>>
>>>>>>>>>> int main()
>>>>>>>>>> {
>>>>>>>>>> P(P);
>>>>>>>>>> }
>>>>>>>>>>
>>>>>>>>>> _P()
>>>>>>>>>> [000011f0](01) 55 push ebp
>>>>>>>>>> [000011f1](02) 8bec mov ebp,esp
>>>>>>>>>> [000011f3](03) 8b4508 mov eax,[ebp+08]
>>>>>>>>>> [000011f6](01) 50 push eax
>>>>>>>>>> [000011f7](03) 8b4d08 mov ecx,[ebp+08]
>>>>>>>>>> [000011fa](01) 51 push ecx
>>>>>>>>>> [000011fb](05) e820feffff call 00001020
>>>>>>>>>>
>>>>>>>>>> [00001200](03) 83c408 add esp,+08
>>>>>>>>>> [00001203](02) 85c0 test eax,eax
>>>>>>>>>> [00001205](02) 7402 jz 00001209
>>>>>>>>>> [00001207](02) ebfe jmp 00001207
>>>>>>>>>> [00001209](01) 5d pop ebp
>>>>>>>>>> [0000120a](01) c3 ret
>>>>>>>>>> Size in bytes:(0027) [0000120a]
>>>>>>>>>>
>>>>>>>>>> _main()
>>>>>>>>>> [00001210](01) 55 push ebp
>>>>>>>>>> [00001211](02) 8bec mov ebp,esp
>>>>>>>>>> [00001213](05) 68f0110000 push 000011f0
>>>>>>>>>> [00001218](05) e8d3ffffff call 000011f0
>>>>>>>>>> [0000121d](03) 83c404 add esp,+04
>>>>>>>>>> [00001220](02) 33c0 xor eax,eax
>>>>>>>>>> [00001222](01) 5d pop ebp
>>>>>>>>>> [00001223](01) c3 ret
>>>>>>>>>> Size in bytes:(0020) [00001223]
>>>>>>>>>>
>>>>>>>>>> machine stack stack machine assembly
>>>>>>>>>> address address data code language
>>>>>>>>>> ======== ======== ======== ========= =============
>>>>>>>>>> [00001210][00101fba][00000000] 55 push ebp
>>>>>>>>>> [00001211][00101fba][00000000] 8bec mov ebp,esp
>>>>>>>>>> [00001213][00101fb6][000011f0] 68f0110000 push 000011f0 // push P
>>>>>>>>>> [00001218][00101fb2][0000121d] e8d3ffffff call 000011f0 // call P
>>>>>>>>>> [000011f0][00101fae][00101fba] 55 push ebp // enter executed P
>>>>>>>>>> [000011f1][00101fae][00101fba] 8bec mov ebp,esp
>>>>>>>>>> [000011f3][00101fae][00101fba] 8b4508 mov eax,[ebp+08]
>>>>>>>>>> [000011f6][00101faa][000011f0] 50 push eax // push P
>>>>>>>>>> [000011f7][00101faa][000011f0] 8b4d08 mov ecx,[ebp+08]
>>>>>>>>>> [000011fa][00101fa6][000011f0] 51 push ecx // push P
>>>>>>>>>> [000011fb][00101fa2][00001200] e820feffff call 00001020 // call H
>>>>>>>>>>
>>>>>>>>>> Begin Simulation Execution Trace Stored at:21206e
>>>>>>>>>> Address_of_H:1020
>>>>>>>>>> [000011f0][0021205a][0021205e] 55 push ebp // enter emulated P
>>>>>>>>>> [000011f1][0021205a][0021205e] 8bec mov ebp,esp
>>>>>>>>>> [000011f3][0021205a][0021205e] 8b4508 mov eax,[ebp+08]
>>>>>>>>>> [000011f6][00212056][000011f0] 50 push eax // push P
>>>>>>>>>> [000011f7][00212056][000011f0] 8b4d08 mov ecx,[ebp+08]
>>>>>>>>>> [000011fa][00212052][000011f0] 51 push ecx // push P
>>>>>>>>>> [000011fb][0021204e][00001200] e820feffff call 00001020 // call emulated H
>>>>>>>>>> Infinitely Recursive Simulation Detected Simulation Stopped
>>>>>>>>>>
>>>>>>>>>> H knows its own machine address and on this basis it can easily
>>>>>>>>>> examine its stored execution_trace of P (see above) to determine:
>>>>>>>>>> (a) P is calling H with the same arguments that H was called with.
>>>>>>>>>> (b) No instructions in P could possibly escape this otherwise infinitely
>>>>>>>>>> recursive emulation.
>>>>>>>>>
>>>>>>>>> FALSE. P contains these instructions:
>>>>>>>>>
>>>>>>>>> if (Decide_Halting(&execution_trace, &decoded, code_end, &master_state,
>>>>>>>>> &slave_state, &slave_stack, Address_of_H, P, I))
>>>>>>>>> goto END_OF_CODE;
>>>>>>>>> return 0; // Does not halt
>>>>>>>>> END_OF_CODE:
>>>>>>>>> return 1; // Input has normally terminated
>>>>>>>>>
>>>>>>>>> That *can* and *do* prevent infinite recursive emulation in a *correct* and *complete* emulation
>>>>>>>>>
>>>>>>>>>> (c) H aborts its emulation of P before its call to H is emulated.
>>>>>>>>>
>>>>>>>>> And in doing so it fails to see that the call to H, since H is a computation, will aborts *its* simulation of P and return 0 to the outer P being emulated, which would cause it to halt, making H(P,P)==0 wrong.
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> [00001200][00101fae][00101fba] 83c408 add esp,+08 // return to
>>>>>>>>>> executed P
>>>>>>>>>> [00001203][00101fae][00101fba] 85c0 test eax,eax
>>>>>>>>>> [00001205][00101fae][00101fba] 7402 jz 00001209
>>>>>>>>>> [00001209][00101fb2][0000121d] 5d pop ebp
>>>>>>>>>> [0000120a][00101fb6][000011f0] c3 ret // return from
>>>>>>>>>> executed P
>>>>>>>>>> [0000121d][00101fba][00000000] 83c404 add esp,+04
>>>>>>>>>> [00001220][00101fba][00000000] 33c0 xor eax,eax
>>>>>>>>>> [00001222][00101fbe][00100000] 5d pop ebp
>>>>>>>>>> [00001223][00101fc2][00000000] c3 ret // ret from main
>>>>>>>>>> Number of Instructions Executed(878) / 67 = 13 pages
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> I've stated in detail why you're wrong. Now you need to explain in detail why I'm wrong.
>>>>>>>>>
>>>>>>>>> Failure to do so, which includes simply restating your original point, will be taken as admission that what I've stated is correct.
>>>>>>>> H correctly predicts that its complete and correct x86 emulation of its
>>>>>>>> input never reaches the the "ret" instruction of P.
>>>>>>>
>>>>>>> Repeating your original point.
>>>>>>>
>>>>>>>>
>>>>>>>> This is the actual behavior of the input to H(P,P).
>>>>>>>>
>>>>>>>> P(P) is not the actual behavior of the input to H(P,P).
>>>>>>>>
>>>>>>>
>>>>>>> Again, repeating your original point.
>>>>>>>
>>>>>>>> Never reaching the "ret" instruction means that P does not halt.
>>>>>>>> A halt decider must compute the mapping from its inputs to an accept or
>>>>>>>> reject state on the basis of the actual behavior that is actually
>>>>>>>> specified by these inputs.
>>>>>>>> Therefore H(P,P)==0 is correct.
>>>>>>>
>>>>>>> And again, repeating your original point.
>>>>>>>
>>>>>>> So no explanation as to why I'm wrong, just a repetition of your original points.
>>>>>>>
>>>>>>> That means you accept that what I've said is true, and that H(P,P)==0 is wrong.
>>>>>> You rebuttal is based on rejecting the verified fact that a simulating
>>>>>> halt decider correctly predicts that its correctly simulated input will
>>>>>> never reach the final state "ret" instruction of this input.
>>>>>
>>>>> FALSE. The correctly simulated input to H(P,P) is UTM(P,P) which halts, therefore H(P,P)==0 is wrong.
>>>> First of all I am not talking about UTM's.
>>>> I am talking about x86 emulation.
>>>> Try to encode what you mean, here is my guess:
>>>>
>>>> void P(u32 x)
>>>> {
>>>> if (emulate_x86(x, x))
>>>> HERE: goto HERE;
>>>> return;
>>>> }
>>>>
>>>> int main()
>>>> {
>>>> Output("Input_Halts = ", H((u32)P, (u32)P));
>>>> }
>>>
>>> P is a computation and therefore it can't be changed, nor can anything that it calls. I mean UTM(P,P). The simplest way to implement this is:
>>>
>>> void UTM(u32 x, u32 y)
>>> {
>>> x(y)
>>> }
>>>
>>> H cannot by definition perform a correct and complete emulation if it aborts, so the input must be passed to something that does, i.e. a UTM, for the source of truth.
>>>
>> A simulating halt decider correctly simulates its input until it
>> correctly determines that this simulated input would never reach its
>> final state.
>
> But H isn't such a decider, because the simulated input does reach a final state when correctly simulated, i.e. by UTM(P,P), therefore H(P,P)==0 is wrong.
>

On 6/27/2022 12:03 PM, Dennis Bush wrote:
> On Monday, June 27, 2022 at 9:28:22 AM UTC-4, olcott wrote:
>> On 6/27/2022 7:58 AM, Dennis Bush wrote:
>>> On Monday, June 27, 2022 at 8:50:05 AM UTC-4, olcott wrote:
>>>> On 6/27/2022 7:40 AM, Dennis Bush wrote:
>>>>> On Monday, June 27, 2022 at 8:34:22 AM UTC-4, olcott wrote:
>>>>>> On 6/27/2022 7:26 AM, Dennis Bush wrote:
>>>>>>> On Monday, June 27, 2022 at 8:14:31 AM UTC-4, olcott wrote:
>>>>>>>> On 6/27/2022 6:57 AM, Dennis Bush wrote:
>>>>>>>>> On Monday, June 27, 2022 at 7:11:21 AM UTC-4, olcott wrote:
>>>>>>>>>> *This is the outline of the complete refutation*
>>>>>>>>>> *of the only rebuttal that anyone has left*
>>>>>>>>>>
>>>>>>>>>> (a) The complete and correct x86 emulation of the input to H(P,P) by H
>>>>>>>>>> never reaches the "ret" instruction of P.
>>>>>>>>>
>>>>>>>>> It is known that H aborts its simulation and returns 0. That means that H *by definition* does not perform a complete and correct emulation. And since H (if it was constructed correctly) is a pure function of its inputs, it will ALWAYS return the same return for a given input.
>>>>>>>>>
>>>>>>>>> To state that H aborts *and* does a complete and correct simulation is simply nonsense.
>>>>>>>>>
>>>>>>>>> What *does* perform a complete and correct emulation of the input to H(P,P) is UTM(P,P) which halts, therefore H(P,P)==0 is wrong.
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> (b) The direct execution of P(P) does reach its "ret" instruction.
>>>>>>>>>>
>>>>>>>>>> The above two are proven to be verified facts entirely on the basis of
>>>>>>>>>> the semantics of the x86 language.
>>>>>>>>>
>>>>>>>>> A nonsense statement can't be a "fact", verified or otherwise.
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> A halt decider must compute the mapping from its inputs to an accept or
>>>>>>>>>> reject state on the basis of the actual behavior that is actually
>>>>>>>>>> specified by these inputs.
>>>>>>>>>
>>>>>>>>> Which for H(P,P) is BY DEFINITION is the behavior of P(P). The halting problem says that H MUST implement the following mapping (i.e. the halting function):
>>>>>>>>>
>>>>>>>>> H(x,y)==1 if and only if x(y) halts, and
>>>>>>>>> H(x,y)==0 if and only if x(y) does not halt.
>>>>>>>>>
>>>>>>>>> H does not perform this mapping.
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> P(P) is provably not the actual behavior of the actual input.
>>>>>>>>>
>>>>>>>>> It is BY DEFINITION. So if H comes up with something different that means it did something wrong, specifically it aborted a halting computation too soon.
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> void P(u32 x)
>>>>>>>>>> {
>>>>>>>>>> if (H(x, x))
>>>>>>>>>> HERE: goto HERE;
>>>>>>>>>> return;
>>>>>>>>>> }
>>>>>>>>>>
>>>>>>>>>> int main()
>>>>>>>>>> {
>>>>>>>>>> P(P);
>>>>>>>>>> }
>>>>>>>>>>
>>>>>>>>>> _P()
>>>>>>>>>> [000011f0](01) 55 push ebp
>>>>>>>>>> [000011f1](02) 8bec mov ebp,esp
>>>>>>>>>> [000011f3](03) 8b4508 mov eax,[ebp+08]
>>>>>>>>>> [000011f6](01) 50 push eax
>>>>>>>>>> [000011f7](03) 8b4d08 mov ecx,[ebp+08]
>>>>>>>>>> [000011fa](01) 51 push ecx
>>>>>>>>>> [000011fb](05) e820feffff call 00001020
>>>>>>>>>>
>>>>>>>>>> [00001200](03) 83c408 add esp,+08
>>>>>>>>>> [00001203](02) 85c0 test eax,eax
>>>>>>>>>> [00001205](02) 7402 jz 00001209
>>>>>>>>>> [00001207](02) ebfe jmp 00001207
>>>>>>>>>> [00001209](01) 5d pop ebp
>>>>>>>>>> [0000120a](01) c3 ret
>>>>>>>>>> Size in bytes:(0027) [0000120a]
>>>>>>>>>>
>>>>>>>>>> _main()
>>>>>>>>>> [00001210](01) 55 push ebp
>>>>>>>>>> [00001211](02) 8bec mov ebp,esp
>>>>>>>>>> [00001213](05) 68f0110000 push 000011f0
>>>>>>>>>> [00001218](05) e8d3ffffff call 000011f0
>>>>>>>>>> [0000121d](03) 83c404 add esp,+04
>>>>>>>>>> [00001220](02) 33c0 xor eax,eax
>>>>>>>>>> [00001222](01) 5d pop ebp
>>>>>>>>>> [00001223](01) c3 ret
>>>>>>>>>> Size in bytes:(0020) [00001223]
>>>>>>>>>>
>>>>>>>>>> machine stack stack machine assembly
>>>>>>>>>> address address data code language
>>>>>>>>>> ======== ======== ======== ========= =============
>>>>>>>>>> [00001210][00101fba][00000000] 55 push ebp
>>>>>>>>>> [00001211][00101fba][00000000] 8bec mov ebp,esp
>>>>>>>>>> [00001213][00101fb6][000011f0] 68f0110000 push 000011f0 // push P
>>>>>>>>>> [00001218][00101fb2][0000121d] e8d3ffffff call 000011f0 // call P
>>>>>>>>>> [000011f0][00101fae][00101fba] 55 push ebp // enter executed P
>>>>>>>>>> [000011f1][00101fae][00101fba] 8bec mov ebp,esp
>>>>>>>>>> [000011f3][00101fae][00101fba] 8b4508 mov eax,[ebp+08]
>>>>>>>>>> [000011f6][00101faa][000011f0] 50 push eax // push P
>>>>>>>>>> [000011f7][00101faa][000011f0] 8b4d08 mov ecx,[ebp+08]
>>>>>>>>>> [000011fa][00101fa6][000011f0] 51 push ecx // push P
>>>>>>>>>> [000011fb][00101fa2][00001200] e820feffff call 00001020 // call H
>>>>>>>>>>
>>>>>>>>>> Begin Simulation Execution Trace Stored at:21206e
>>>>>>>>>> Address_of_H:1020
>>>>>>>>>> [000011f0][0021205a][0021205e] 55 push ebp // enter emulated P
>>>>>>>>>> [000011f1][0021205a][0021205e] 8bec mov ebp,esp
>>>>>>>>>> [000011f3][0021205a][0021205e] 8b4508 mov eax,[ebp+08]
>>>>>>>>>> [000011f6][00212056][000011f0] 50 push eax // push P
>>>>>>>>>> [000011f7][00212056][000011f0] 8b4d08 mov ecx,[ebp+08]
>>>>>>>>>> [000011fa][00212052][000011f0] 51 push ecx // push P
>>>>>>>>>> [000011fb][0021204e][00001200] e820feffff call 00001020 // call emulated H
>>>>>>>>>> Infinitely Recursive Simulation Detected Simulation Stopped
>>>>>>>>>>
>>>>>>>>>> H knows its own machine address and on this basis it can easily
>>>>>>>>>> examine its stored execution_trace of P (see above) to determine:
>>>>>>>>>> (a) P is calling H with the same arguments that H was called with.
>>>>>>>>>> (b) No instructions in P could possibly escape this otherwise infinitely
>>>>>>>>>> recursive emulation.
>>>>>>>>>
>>>>>>>>> FALSE. P contains these instructions:
>>>>>>>>>
>>>>>>>>> if (Decide_Halting(&execution_trace, &decoded, code_end, &master_state,
>>>>>>>>> &slave_state, &slave_stack, Address_of_H, P, I))
>>>>>>>>> goto END_OF_CODE;
>>>>>>>>> return 0; // Does not halt
>>>>>>>>> END_OF_CODE:
>>>>>>>>> return 1; // Input has normally terminated
>>>>>>>>>
>>>>>>>>> That *can* and *do* prevent infinite recursive emulation in a *correct* and *complete* emulation
>>>>>>>>>
>>>>>>>>>> (c) H aborts its emulation of P before its call to H is emulated.
>>>>>>>>>
>>>>>>>>> And in doing so it fails to see that the call to H, since H is a computation, will aborts *its* simulation of P and return 0 to the outer P being emulated, which would cause it to halt, making H(P,P)==0 wrong.
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> [00001200][00101fae][00101fba] 83c408 add esp,+08 // return to
>>>>>>>>>> executed P
>>>>>>>>>> [00001203][00101fae][00101fba] 85c0 test eax,eax
>>>>>>>>>> [00001205][00101fae][00101fba] 7402 jz 00001209
>>>>>>>>>> [00001209][00101fb2][0000121d] 5d pop ebp
>>>>>>>>>> [0000120a][00101fb6][000011f0] c3 ret // return from
>>>>>>>>>> executed P
>>>>>>>>>> [0000121d][00101fba][00000000] 83c404 add esp,+04
>>>>>>>>>> [00001220][00101fba][00000000] 33c0 xor eax,eax
>>>>>>>>>> [00001222][00101fbe][00100000] 5d pop ebp
>>>>>>>>>> [00001223][00101fc2][00000000] c3 ret // ret from main
>>>>>>>>>> Number of Instructions Executed(878) / 67 = 13 pages
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> I've stated in detail why you're wrong. Now you need to explain in detail why I'm wrong.
>>>>>>>>>
>>>>>>>>> Failure to do so, which includes simply restating your original point, will be taken as admission that what I've stated is correct.
>>>>>>>> H correctly predicts that its complete and correct x86 emulation of its
>>>>>>>> input never reaches the the "ret" instruction of P.
>>>>>>>
>>>>>>> Repeating your original point.
>>>>>>>
>>>>>>>>
>>>>>>>> This is the actual behavior of the input to H(P,P).
>>>>>>>>
>>>>>>>> P(P) is not the actual behavior of the input to H(P,P).
>>>>>>>>
>>>>>>>
>>>>>>> Again, repeating your original point.
>>>>>>>
>>>>>>>> Never reaching the "ret" instruction means that P does not halt.
>>>>>>>> A halt decider must compute the mapping from its inputs to an accept or
>>>>>>>> reject state on the basis of the actual behavior that is actually
>>>>>>>> specified by these inputs.
>>>>>>>> Therefore H(P,P)==0 is correct.
>>>>>>>
>>>>>>> And again, repeating your original point.
>>>>>>>
>>>>>>> So no explanation as to why I'm wrong, just a repetition of your original points.
>>>>>>>
>>>>>>> That means you accept that what I've said is true, and that H(P,P)==0 is wrong.
>>>>>> You rebuttal is based on rejecting the verified fact that a simulating
>>>>>> halt decider correctly predicts that its correctly simulated input will
>>>>>> never reach the final state "ret" instruction of this input.
>>>>>
>>>>> FALSE. The correctly simulated input to H(P,P) is UTM(P,P) which halts, therefore H(P,P)==0 is wrong.
>>>> First of all I am not talking about UTM's.
>>>> I am talking about x86 emulation.
>>>> Try to encode what you mean, here is my guess:
>>>>
>>>> void P(u32 x)
>>>> {
>>>> if (emulate_x86(x, x))
>>>> HERE: goto HERE;
>>>> return;
>>>> }
>>>>
>>>> int main()
>>>> {
>>>> Output("Input_Halts = ", H((u32)P, (u32)P));
>>>> }
>>>
>>> P is a computation and therefore it can't be changed, nor can anything that it calls. I mean UTM(P,P). The simplest way to implement this is:
>>>
>>> void UTM(u32 x, u32 y)
>>> {
>>> x(y)
>>> }
>>>
>>> H cannot by definition perform a correct and complete emulation if it aborts, so the input must be passed to something that does, i.e. a UTM, for the source of truth.
>>>
>> A simulating halt decider correctly simulates its input until it
>> correctly determines that this simulated input would never reach its
>> final state.
>
> But H isn't such a decider, because the simulated input does reach a final state when correctly simulated, i.e. by UTM(P,P), therefore H(P,P)==0 is wrong.
>

On Monday, June 27, 2022 at 1:20:39 PM UTC-4, olcott wrote:
> On 6/27/2022 12:03 PM, Dennis Bush wrote:
> > On Monday, June 27, 2022 at 9:28:22 AM UTC-4, olcott wrote:
> >> On 6/27/2022 7:58 AM, Dennis Bush wrote:
> >>> On Monday, June 27, 2022 at 8:50:05 AM UTC-4, olcott wrote:
> >>>> On 6/27/2022 7:40 AM, Dennis Bush wrote:
> >>>>> On Monday, June 27, 2022 at 8:34:22 AM UTC-4, olcott wrote:
> >>>>>> On 6/27/2022 7:26 AM, Dennis Bush wrote:
> >>>>>>> On Monday, June 27, 2022 at 8:14:31 AM UTC-4, olcott wrote:
> >>>>>>>> On 6/27/2022 6:57 AM, Dennis Bush wrote:
> >>>>>>>>> On Monday, June 27, 2022 at 7:11:21 AM UTC-4, olcott wrote:
> >>>>>>>>>> *This is the outline of the complete refutation*
> >>>>>>>>>> *of the only rebuttal that anyone has left*
> >>>>>>>>>>
> >>>>>>>>>> (a) The complete and correct x86 emulation of the input to H(P,P) by H
> >>>>>>>>>> never reaches the "ret" instruction of P.
> >>>>>>>>>
> >>>>>>>>> It is known that H aborts its simulation and returns 0. That means that H *by definition* does not perform a complete and correct emulation. And since H (if it was constructed correctly) is a pure function of its inputs, it will ALWAYS return the same return for a given input.
> >>>>>>>>>
> >>>>>>>>> To state that H aborts *and* does a complete and correct simulation is simply nonsense.
> >>>>>>>>>
> >>>>>>>>> What *does* perform a complete and correct emulation of the input to H(P,P) is UTM(P,P) which halts, therefore H(P,P)==0 is wrong.
> >>>>>>>>>
> >>>>>>>>>>
> >>>>>>>>>> (b) The direct execution of P(P) does reach its "ret" instruction.
> >>>>>>>>>>
> >>>>>>>>>> The above two are proven to be verified facts entirely on the basis of
> >>>>>>>>>> the semantics of the x86 language.
> >>>>>>>>>
> >>>>>>>>> A nonsense statement can't be a "fact", verified or otherwise.
> >>>>>>>>>
> >>>>>>>>>>
> >>>>>>>>>> A halt decider must compute the mapping from its inputs to an accept or
> >>>>>>>>>> reject state on the basis of the actual behavior that is actually
> >>>>>>>>>> specified by these inputs.
> >>>>>>>>>
> >>>>>>>>> Which for H(P,P) is BY DEFINITION is the behavior of P(P). The halting problem says that H MUST implement the following mapping (i.e. the halting function):
> >>>>>>>>>
> >>>>>>>>> H(x,y)==1 if and only if x(y) halts, and
> >>>>>>>>> H(x,y)==0 if and only if x(y) does not halt.
> >>>>>>>>>
> >>>>>>>>> H does not perform this mapping.
> >>>>>>>>>
> >>>>>>>>>>
> >>>>>>>>>> P(P) is provably not the actual behavior of the actual input.
> >>>>>>>>>
> >>>>>>>>> It is BY DEFINITION. So if H comes up with something different that means it did something wrong, specifically it aborted a halting computation too soon.
> >>>>>>>>>
> >>>>>>>>>>
> >>>>>>>>>> void P(u32 x)
> >>>>>>>>>> {
> >>>>>>>>>> if (H(x, x))
> >>>>>>>>>> HERE: goto HERE;
> >>>>>>>>>> return;
> >>>>>>>>>> }
> >>>>>>>>>>
> >>>>>>>>>> int main()
> >>>>>>>>>> {
> >>>>>>>>>> P(P);
> >>>>>>>>>> }
> >>>>>>>>>>
> >>>>>>>>>> _P()
> >>>>>>>>>> [000011f0](01) 55 push ebp
> >>>>>>>>>> [000011f1](02) 8bec mov ebp,esp
> >>>>>>>>>> [000011f3](03) 8b4508 mov eax,[ebp+08]
> >>>>>>>>>> [000011f6](01) 50 push eax
> >>>>>>>>>> [000011f7](03) 8b4d08 mov ecx,[ebp+08]
> >>>>>>>>>> [000011fa](01) 51 push ecx
> >>>>>>>>>> [000011fb](05) e820feffff call 00001020
> >>>>>>>>>>
> >>>>>>>>>> [00001200](03) 83c408 add esp,+08
> >>>>>>>>>> [00001203](02) 85c0 test eax,eax
> >>>>>>>>>> [00001205](02) 7402 jz 00001209
> >>>>>>>>>> [00001207](02) ebfe jmp 00001207
> >>>>>>>>>> [00001209](01) 5d pop ebp
> >>>>>>>>>> [0000120a](01) c3 ret
> >>>>>>>>>> Size in bytes:(0027) [0000120a]
> >>>>>>>>>>
> >>>>>>>>>> _main()
> >>>>>>>>>> [00001210](01) 55 push ebp
> >>>>>>>>>> [00001211](02) 8bec mov ebp,esp
> >>>>>>>>>> [00001213](05) 68f0110000 push 000011f0
> >>>>>>>>>> [00001218](05) e8d3ffffff call 000011f0
> >>>>>>>>>> [0000121d](03) 83c404 add esp,+04
> >>>>>>>>>> [00001220](02) 33c0 xor eax,eax
> >>>>>>>>>> [00001222](01) 5d pop ebp
> >>>>>>>>>> [00001223](01) c3 ret
> >>>>>>>>>> Size in bytes:(0020) [00001223]
> >>>>>>>>>>
> >>>>>>>>>> machine stack stack machine assembly
> >>>>>>>>>> address address data code language
> >>>>>>>>>> ======== ======== ======== ========= =============
> >>>>>>>>>> [00001210][00101fba][00000000] 55 push ebp
> >>>>>>>>>> [00001211][00101fba][00000000] 8bec mov ebp,esp
> >>>>>>>>>> [00001213][00101fb6][000011f0] 68f0110000 push 000011f0 // push P
> >>>>>>>>>> [00001218][00101fb2][0000121d] e8d3ffffff call 000011f0 // call P
> >>>>>>>>>> [000011f0][00101fae][00101fba] 55 push ebp // enter executed P
> >>>>>>>>>> [000011f1][00101fae][00101fba] 8bec mov ebp,esp
> >>>>>>>>>> [000011f3][00101fae][00101fba] 8b4508 mov eax,[ebp+08]
> >>>>>>>>>> [000011f6][00101faa][000011f0] 50 push eax // push P
> >>>>>>>>>> [000011f7][00101faa][000011f0] 8b4d08 mov ecx,[ebp+08]
> >>>>>>>>>> [000011fa][00101fa6][000011f0] 51 push ecx // push P
> >>>>>>>>>> [000011fb][00101fa2][00001200] e820feffff call 00001020 // call H
> >>>>>>>>>>
> >>>>>>>>>> Begin Simulation Execution Trace Stored at:21206e
> >>>>>>>>>> Address_of_H:1020
> >>>>>>>>>> [000011f0][0021205a][0021205e] 55 push ebp // enter emulated P
> >>>>>>>>>> [000011f1][0021205a][0021205e] 8bec mov ebp,esp
> >>>>>>>>>> [000011f3][0021205a][0021205e] 8b4508 mov eax,[ebp+08]
> >>>>>>>>>> [000011f6][00212056][000011f0] 50 push eax // push P
> >>>>>>>>>> [000011f7][00212056][000011f0] 8b4d08 mov ecx,[ebp+08]
> >>>>>>>>>> [000011fa][00212052][000011f0] 51 push ecx // push P
> >>>>>>>>>> [000011fb][0021204e][00001200] e820feffff call 00001020 // call emulated H
> >>>>>>>>>> Infinitely Recursive Simulation Detected Simulation Stopped
> >>>>>>>>>>
> >>>>>>>>>> H knows its own machine address and on this basis it can easily
> >>>>>>>>>> examine its stored execution_trace of P (see above) to determine:
> >>>>>>>>>> (a) P is calling H with the same arguments that H was called with.
> >>>>>>>>>> (b) No instructions in P could possibly escape this otherwise infinitely
> >>>>>>>>>> recursive emulation.
> >>>>>>>>>
> >>>>>>>>> FALSE. P contains these instructions:
> >>>>>>>>>
> >>>>>>>>> if (Decide_Halting(&execution_trace, &decoded, code_end, &master_state,
> >>>>>>>>> &slave_state, &slave_stack, Address_of_H, P, I))
> >>>>>>>>> goto END_OF_CODE;
> >>>>>>>>> return 0; // Does not halt
> >>>>>>>>> END_OF_CODE:
> >>>>>>>>> return 1; // Input has normally terminated
> >>>>>>>>>
> >>>>>>>>> That *can* and *do* prevent infinite recursive emulation in a *correct* and *complete* emulation
> >>>>>>>>>
> >>>>>>>>>> (c) H aborts its emulation of P before its call to H is emulated.
> >>>>>>>>>
> >>>>>>>>> And in doing so it fails to see that the call to H, since H is a computation, will aborts *its* simulation of P and return 0 to the outer P being emulated, which would cause it to halt, making H(P,P)==0 wrong.
> >>>>>>>>>
> >>>>>>>>>>
> >>>>>>>>>> [00001200][00101fae][00101fba] 83c408 add esp,+08 // return to
> >>>>>>>>>> executed P
> >>>>>>>>>> [00001203][00101fae][00101fba] 85c0 test eax,eax
> >>>>>>>>>> [00001205][00101fae][00101fba] 7402 jz 00001209
> >>>>>>>>>> [00001209][00101fb2][0000121d] 5d pop ebp
> >>>>>>>>>> [0000120a][00101fb6][000011f0] c3 ret // return from
> >>>>>>>>>> executed P
> >>>>>>>>>> [0000121d][00101fba][00000000] 83c404 add esp,+04
> >>>>>>>>>> [00001220][00101fba][00000000] 33c0 xor eax,eax
> >>>>>>>>>> [00001222][00101fbe][00100000] 5d pop ebp
> >>>>>>>>>> [00001223][00101fc2][00000000] c3 ret // ret from main
> >>>>>>>>>> Number of Instructions Executed(878) / 67 = 13 pages
> >>>>>>>>>>
> >>>>>>>>>>
> >>>>>>>>>
> >>>>>>>>> I've stated in detail why you're wrong. Now you need to explain in detail why I'm wrong.
> >>>>>>>>>
> >>>>>>>>> Failure to do so, which includes simply restating your original point, will be taken as admission that what I've stated is correct.
> >>>>>>>> H correctly predicts that its complete and correct x86 emulation of its
> >>>>>>>> input never reaches the the "ret" instruction of P.
> >>>>>>>
> >>>>>>> Repeating your original point.
> >>>>>>>
> >>>>>>>>
> >>>>>>>> This is the actual behavior of the input to H(P,P).
> >>>>>>>>
> >>>>>>>> P(P) is not the actual behavior of the input to H(P,P).
> >>>>>>>>
> >>>>>>>
> >>>>>>> Again, repeating your original point.
> >>>>>>>
> >>>>>>>> Never reaching the "ret" instruction means that P does not halt.
> >>>>>>>> A halt decider must compute the mapping from its inputs to an accept or
> >>>>>>>> reject state on the basis of the actual behavior that is actually
> >>>>>>>> specified by these inputs.
> >>>>>>>> Therefore H(P,P)==0 is correct.
> >>>>>>>
> >>>>>>> And again, repeating your original point.
> >>>>>>>
> >>>>>>> So no explanation as to why I'm wrong, just a repetition of your original points.
> >>>>>>>
> >>>>>>> That means you accept that what I've said is true, and that H(P,P)==0 is wrong.
> >>>>>> You rebuttal is based on rejecting the verified fact that a simulating
> >>>>>> halt decider correctly predicts that its correctly simulated input will
> >>>>>> never reach the final state "ret" instruction of this input.
> >>>>>
> >>>>> FALSE. The correctly simulated input to H(P,P) is UTM(P,P) which halts, therefore H(P,P)==0 is wrong.
> >>>> First of all I am not talking about UTM's.
> >>>> I am talking about x86 emulation.
> >>>> Try to encode what you mean, here is my guess:
> >>>>
> >>>> void P(u32 x)
> >>>> {
> >>>> if (emulate_x86(x, x))
> >>>> HERE: goto HERE;
> >>>> return;
> >>>> }
> >>>>
> >>>> int main()
> >>>> {
> >>>> Output("Input_Halts = ", H((u32)P, (u32)P));
> >>>> }
> >>>
> >>> P is a computation and therefore it can't be changed, nor can anything that it calls. I mean UTM(P,P). The simplest way to implement this is:
> >>>
> >>> void UTM(u32 x, u32 y)
> >>> {
> >>> x(y)
> >>> }
> >>>
> >>> H cannot by definition perform a correct and complete emulation if it aborts, so the input must be passed to something that does, i.e. a UTM, for the source of truth.
> >>>
> >> A simulating halt decider correctly simulates its input until it
> >> correctly determines that this simulated input would never reach its
> >> final state.
> >
> > But H isn't such a decider, because the simulated input does reach a final state when correctly simulated, i.e. by UTM(P,P), therefore H(P,P)==0 is wrong.
> >
> Every simulating decider that correctly simulates its input until it
> correctly determines that this simulated input would never reach its
> final state correctly determines the halt status of this input.