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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: Liar detector: Fred, Richard, Joes and Alan --- Ben's agreement
Date: Wed, 17 Jul 2024 16:08:18 +0200
Organization: A noiseless patient Spider
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Op 17.jul.2024 om 15:39 schreef olcott:
> On 7/17/2024 8:32 AM, Fred. Zwarts wrote:
>> Op 17.jul.2024 om 15:00 schreef olcott:
>>> On 7/17/2024 1:43 AM, Mikko wrote:
>>>> On 2024-07-16 14:21:28 +0000, olcott said:
>>>>>
>>>>> When simulated input DDD stops running {if and only if}
>>>>> the simulation of this input DDD has been aborted this
>>>>> necessitates that input DDD specifies non-halting behavior
>>>>
>>>> DDD does not stop runnig unless it is completely exeuted. 
>>>
>>> _DDD()
>>> [00002163] 55         push ebp      ; housekeeping
>>> [00002164] 8bec       mov ebp,esp   ; housekeeping
>>> [00002166] 6863210000 push 00002163 ; push DDD
>>> [0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
>>> [00002170] 83c404     add esp,+04
>>> [00002173] 5d         pop ebp
>>> [00002174] c3         ret
>>> Size in bytes:(0018) [00002174]
>>>
>>> DDD emulated by HHH according to the semantic meaning of
>>> its x86 instructions never stop running unless aborted.
>>
>> The 'unless aborted' is misleading, since we know that HHH *does* abort.
> *THIS IS SELF EVIDENT THUS DISAGREEMENT IS INCORRECT*
> DDD emulated by any pure function HHH according to the
> semantic meaning of its x86 instructions never stops
> running unless aborted.

But HHH is programmed to abort after N cycles, therefore, it stops 
running after N cycles.
Dreaming of HHH that does not abort is irrelevant.

The semantics of the x86 code of a halting program is self-evident: it 
halts. No abort needed.
Any denial displays a large misunderstanding of the x86 language.

DDD is a misleading and unneeded complication. It is easy to eliminate DDD:

        int main() {
          return HHH(main);
        }

This has the same problem. This proves that the problem is not in DDD, 
but in HHH, which halts when it aborts the simulation, but it decides 
that the simulation of itself does not halt.

HHH is simply unable to decide about finite recursions.

void Finite_Recursion (int N) {
   if (N > 0) Finite_Recursion (N - 1);
}

It decides after N recursions that there is an infinite recursion, which 
is incorrect.

Your HHH is programmed to abort the simulation after N cycles of 
recursive simulations. Therefore, it is incorrect to abort the 
simulation of HHH when the simulated HHH has performed only N-1 cycles, 
because that changes the behaviour of HHH.
Since the simulated HHH always runs one cycle behind the simulating HHH, 
it is clear that HHH can never simulate enough cycles for a correct 
simulation, as is required by the x86 language.
Therefore, the simulation is incorrect according to the criteria you 
stipulated.
The conclusion is simple:
HHH cannot possibly simulate itself correctly.

No matter how much you want it to be correct, or how many times you 
repeat that it is correct, it does not change the fact that such a 
simulation is incorrect, because it is unable to reach the end.
Your own claim that the simulated HHH does not reach its end confirms 
it. The trace you have shown also proves that HHH cannot reach the end 
of its own simulation. So, your own claims prove that it is true that 
HHH cannot possibly simulate itself up to the end, which makes the 
simulation incorrect.

Sipser would agree that this incorrect simulation cannot be used to 
detect a non-halting behaviour.