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

Op 13.jul.2024 om 13:39 schreef olcott:

On 7/13/2024 3:15 AM, Fred. Zwarts wrote:

Op 13.jul.2024 om 01:19 schreef olcott:

On 7/12/2024 5:56 PM, Richard Damon wrote:

On 7/12/24 10:56 AM, olcott wrote:

We stipulate that the only measure of a correct emulation is the
semantics of the x86 programming language.

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Which means the only "correct emulation" that tells the behavior of the program at the input is a non-aborted one.
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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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When N steps of DDD are emulated by HHH according to the
semantics of the x86 language then N steps are emulated correctly.

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And thus HHH that do that know only the first N steps of the behavior of DDD, which continues per the definition of the x86 instruction set until the COMPLETE emulation (or direct execution) reaches a terminal instruction.
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When we examine the infinite set of every HHH/DDD pair such that:
HHH₁ one step of DDD is correctly emulated by HHH.
HHH₂ two steps of DDD are correctly emulated by HHH.
HHH₃ three steps of DDD are correctly emulated by HHH.
...
HHH∞ The emulation of DDD by HHH never stops running.

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And thus, the subset that only did a finite number of steps and aborted its emulation on a non-terminal instrucition only have partial knowledge of the behavior of their DDD, and by returning to their caller, they establish that behavior for ALL copies of that HHH, even the one that DDD calls, which shows that DDD will be halting, even though HHH stopped its observation of the input before it gets to that point.
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The above specifies the infinite set of every HHH/DDD pair
where 1 to infinity steps of DDD are correctly emulated by HHH.
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No DDD instance of each HHH/DDD pair ever reaches past its
own machine address of 0000216b and halts.

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Wrong. EVERY DDD of an HHH that simulated its input for only a finite number of steps WILL halt becuase it will reach its final return.
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The HHH that simulated it for only a finite number of steps, only learned that finite number of steps of the behaivor, and in EVERY case, when we look at the behavior past that point, which DOES occur per the definition of the x86 instruction set, as we have not reached a "termial" instruction that stops behavior, will see the HHH(DDD) that DDD called continuing to simulate its input to the point that this one was defined to stop, and then returns 0 to DDDD and then DDD returning and ending the behavior.
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You continue to stupidly confuse the PARTIAL observation that HHH does of the behavior of DDD by its PARTIAL emulation with the ACTUAL FULL behavior of DDD as defined by the full definition of the x86 insttuction set.
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Thus each HHH element of the above infinite set of HHH/DDD
pairs is necessarily correct to reject its DDD as non-halting.
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Nope.
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NONE Of them CORRECTLY rejected itS DDD as non-halting and you are shown to be ignorant of what you are talking about.
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The HHH that did a partial emulation got the wrong answer, because THEIR DDD will halt. and the HHH that doen't abort never get around to rejecting its DDD as non-halting.

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*Here is the gist of my proof it is irrefutable*
When no DDD of every HHH/DDD that can possibly exist
halts then each HHH that rejects its DDD as non-halting
is necessarily correct.
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*No double-talk and weasel words can overcome that*
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This is double talk, because no HHH can possibly exist that simulates itself correctly.

 Your definition of correct contradicts the semantics of
the x86 language making it wrong.
 

You have a wrong understanding of the semantics of the x86 language. You think that the x86 language specifies that skipping instructions do not change the behaviour of a program.
However, the x86 language does not allow an abort halfway the processing of the instructions, because that would change the behaviour of the program.
Your simulation does not follow the semantics of the x86 language, because it *does* stop the simulation halfway the simulation.
So it is wrong according to your own criteria.
DDD has nothing to do with it. 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 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 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.
Sipser would agree that this incorrect simulation cannot be used to detect a non-halting behaviour.