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computers / comp.theory / Re: Conquering the last rebuttal to H(P,P)==0 refutation of the halting problem proofs

Re: Conquering the last rebuttal to H(P,P)==0 refutation of the halting problem proofs

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Subject: Re: Conquering the last rebuttal to H(P,P)==0 refutation of the
halting problem proofs
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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Mon, 27 Jun 2022 18:41 UTC

On 6/27/2022 1:08 PM, Dennis Bush wrote:
> On Monday, June 27, 2022 at 1:59:58 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.
>>>
>> The semantics of the x86 language conclusively proves that H(P,P)
>> correctly determines/predicts that its complete and correct x86
>> emulation of its input would never reach the "ret" instruction of P.
>
> As previously stated, this is nonsense. Failure to explain *in detail* why the below is wrong will be taken as an admission that it is correct.
>
> 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.

I NEVER SAID that H does a complete and correct simulation of its input.

H correctly predicts the behavior of a complete and correct simulation
of its input.

I have corrected you and Richard on this far too many times.
I ended up blocking Richard which erased my copy of all of his messages.

THIS IS IMPOSSIBLY FALSE THUS IRREFUTABLE
Every simulating halt 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.

THIS IS IMPOSSIBLY FALSE THUS IRREFUTABLE
The semantics of the x86 language conclusively proves that H(P,P)
correctly determines/predicts that its complete and correct x86
emulation of its input would never reach the "ret" instruction of P.

--
Copyright 2022 Pete Olcott

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

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o Conquering the last rebuttal to H(P,P)==0 refutation of the halting

By: olcott on Mon, 27 Jun 2022

201olcott
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