Re: DDD correctly emulated by HHH is Correctly rejected as non-halting V2

Op 17.jul.2024 om 15:43 schreef olcott:

On 7/17/2024 8:35 AM, Fred. Zwarts wrote:

Op 17.jul.2024 om 15:02 schreef olcott:

On 7/17/2024 1:48 AM, Mikko wrote:

On 2024-07-16 15:57:04 +0000, olcott said:
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New slave_stack at:1038c4
Begin Local Halt Decider Simulation   Execution Trace Stored at:1138cc
[00002172][001138bc][001138c0] 55         push ebp      ; housekeeping
[00002173][001138bc][001138c0] 8bec       mov ebp,esp   ; housekeeping
[00002175][001138b8][00002172] 6872210000 push 00002172 ; push DDD
[0000217a][001138b4][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
New slave_stack at:14e2ec
[00002172][0015e2e4][0015e2e8] 55         push ebp      ; housekeeping
[00002173][0015e2e4][0015e2e8] 8bec       mov ebp,esp   ; housekeeping
[00002175][0015e2e0][00002172] 6872210000 push 00002172 ; push DDD
[0000217a][0015e2dc][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
Local Halt Decider: Infinite Recursion Detected Simulation Stopped

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The trace does not show that HHH returns so there is no basis to
think that HHH is a decider.
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The trace shows the data of the executed program of HHH that
does halt.

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It shows some of the data, not all, and in particular, not the halting.
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_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]
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DDD emulated by HHH according to the semantic meaning of
its x86 instructions never stop running unless aborted.
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You have shown that you do not understand the semantics of the x86 language.
HHH does abort and halt after N cycles,

That is counter-factual

No, it is a fact.

 When we examine the infinite set of every HHH/DDD pair such that:
HHH1  One step of DDD is correctly emulated by HHH

But the simulation is halted prematurely, so, the simulation as a whole is incorrect.

HHH2  Two steps of DDD are correctly emulated by HHH

But the simulation is halted prematurely, so, the simulation as a whole is incorrect.

HHH3  Three steps of DDD are correctly emulated by HHH

But the simulation is halted prematurely, so, the simulation as a whole is incorrect.

...

But the simulation for N steps (with N arbitrarily large) is halted prematurely, so, the simulation as a whole is incorrect.
Simulation invariant: An infinite amount of incorrect simulations.

HHH∞  The emulation of DDD by HHH never stops

Also an incorrect simulation, because the simulation never stops.

*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.

So, when not aborted it is incorrect and when aborted the simulation is also incorrect.
So, you showed an infinite amount of evidence that each HHH cannot possibly simulate itself correctly.
The semantics of the x86 language for a halting program is self-evident: it halts.
Any disagreement is either misleading, or stupid.
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.