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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory,sci.logic
Subject: Re: Simulating termination analyzers for dummies
Date: Tue, 18 Jun 2024 09:57:05 +0200
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Op 17.jun.2024 om 17:56 schreef olcott:
> On 6/17/2024 9:49 AM, Fred. Zwarts wrote:
>> Op 17.jun.2024 om 16:34 schreef olcott:
>>> On 6/17/2024 9:18 AM, Fred. Zwarts wrote:
>>>> Op 17.jun.2024 om 15:47 schreef olcott:
>>>>> On 6/17/2024 8:30 AM, Fred. Zwarts wrote:
>>>>>> Op 17.jun.2024 om 14:20 schreef olcott:
>>>>>>> On 6/17/2024 3:31 AM, Fred. Zwarts wrote:
>>>>>>>> Op 17.jun.2024 om 05:33 schreef olcott:
>>>>>>>>> To understand this analysis requires a sufficient knowledge of
>>>>>>>>> the C programming language and what an x86 emulator does.
>>>>>>>>>
>>>>>>>>> Unless every single detail is made 100% explicit false assumptions
>>>>>>>>> always slip though the cracks. This is why it must be examined at
>>>>>>>>> the C level before it is examined at the Turing Machine level.
>>>>>>>>>
>>>>>>>>> typedef void (*ptr)();
>>>>>>>>> int H0(ptr P);
>>>>>>>>>
>>>>>>>>> void Infinite_Loop()
>>>>>>>>> {
>>>>>>>>>    HERE: goto HERE;
>>>>>>>>> }
>>>>>>>>>
>>>>>>>>> void Infinite_Recursion()
>>>>>>>>> {
>>>>>>>>>    Infinite_Recursion();
>>>>>>>>> }
>>>>>>>>>
>>>>>>>>> void DDD()
>>>>>>>>> {
>>>>>>>>>    H0(DDD);
>>>>>>>>>    return;
>>>>>>>>> }
>>>>>>>>>
>>>>>>>>> int main()
>>>>>>>>> {
>>>>>>>>>    H0(Infinite_Loop);
>>>>>>>>>    H0(Infinite_Recursion);
>>>>>>>>>    H0(DDD);
>>>>>>>>> }
>>>>>>>>>
>>>>>>>>> Every C programmer that knows what an x86 emulator is knows 
>>>>>>>>> that when H0
>>>>>>>>> emulates the machine language of Infinite_Loop, 
>>>>>>>>> Infinite_Recursion, and
>>>>>>>>> DDD that it must abort these emulations so that itself can 
>>>>>>>>> terminate
>>>>>>>>> normally.
>>>>>>>>>
>>>>>>>>> When this is construed as non-halting criteria then simulating
>>>>>>>>> termination analyzer H0 is correct to reject these inputs as non-
>>>>>>>>> halting.
>>>>>>>>>
>>>>>>>>
>>>>>>>> For Infinite_Loop and Infinite_Recursion that might be true, 
>>>>>>>> because there the simulator processes the whole input.
>>>>>>>>
>>>>>>>> The H0 case is very different. For H0 there is indeed a false 
>>>>>>>> assumption, as you mentioned. Here H0 needs to simulate itself, 
>>>>>>>> but the simulation is never able to reach the final state of the 
>>>>>>>> simulated self. The abort is always one cycle too early, so that 
>>>>>>>> the simulating H0 misses the abort. Therefore this results in a 
>>>>>>>> false negative.
>>>>>>>> (Note that H0 should process its input, which includes the H0 
>>>>>>>> that aborts, not a non-input with an H that does not abort.)
>>>>>>>>
>>>>>>>> This results in a impossible dilemma for the programmer. It he 
>>>>>>>> creates a H that does not abort, it will not terminate. 
>>>>>>>
>>>>>>> *Therefore what I said is correct*
>>>>>>
>>>>>> No, that is not a logical conclusion. 
>>>>>
>>>>> Every C programmer that knows what an x86 emulator is knows
>>>>> that when H0 emulates the machine language of Infinite_Loop, 
>>>>> Infinite_Recursion, and DDD that it must abort these emulations
>>>>> so that itself can terminate normally.
>>>>
>>>> That might be correct.
>>>>>
>>>>> When this is construed as non-halting criteria then simulating
>>>>> termination analyzer H0 is correct to reject these inputs as non-
>>>>> halting.
>>>>
>>>> That is wrong. It only shows that H0 is unable to simulate itself. 
>>>> It tells nothing about the halting of the input.
>>>>
>>>>>
>>>>> *Too late you have already affirmed the words above*
>>>>> Affirming the first part necessitates the second part.
>>>>>
>>>> That is not logical. If a non-aborting program is wrong, it does not 
>>>> follow that a program that aborts is correct.
>>>> Please, think before you reply.
>>>>
>>>> So, I repeat:
>>>> The logical conclusion if both aborting and not aborting result in 
>>>> errors, is: a halt-decider cannot be based on such a simulation.
>>>
>>> Your view here is merely ignorant of the fact that deciders
>>> must report on the behavior specified by their inputs.
>>>
>>
>> It is incorrect to assume that a failing simulation is able to report 
>> about its input.
>> The simulation fails, because H0 is unable to simulate itself.
>>
> 
> There is no possible way for the call to H0 by DDD
> correctly simulated by any H0 to return to its caller.

We have not seen a proof for this claim and since H0 has contradictory 
requirements nobody else has ever provided a proof.
But OK, lets assume your are right. Simulated H0 is unable to simulate 
itself op to its final state and return to its caller, because it was 
aborted one cycle before it would return to its caller.

> 
> _DDD()
> [00001fd2] 55               push ebp
> [00001fd3] 8bec             mov ebp,esp
> [00001fd5] 68d21f0000       push 00001fd2 ; push address of DDD
> [00001fda] e8f3f9ffff       call 000019d2 ; call  H0
> [00001fdf] 83c404           add esp,+04
> [00001fe2] 5d               pop ebp
> [00001fe3] c3               ret
> Size in bytes:(0018) [00001fe3]
> 
> *THAT THIS IS OVER YOUR HEAD DOES NOT MEAN THAT I AM INCORRECT*
> DDD correctly simulated by H0 *is* the behavior that
> the finite string of x86 machine code of DDD specifies.
> 

It is such a simple fact that H0 cannot possibly correct. If it does not 
abort it does not return. If it does abort, it does not see the correct 
behaviour that was specified in the input, because it aborted its 
simulation one cycle too early to see the behaviour in the input.
H0 fails to do a correct simulation, because it does not process all of 
its input, but aborts its input before it can process the part of the 
input that also specifies behaviour. The final behaviour specified by 
the input is unreachable for the simulator, because it is unable to 
simulate itself.

If even such simple facts are over your head, you must be stuck in 
rebuttal mode very deeply.