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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory,sci.logic,comp.ai.philosophy
Subject: Re: DDD correctly emulated by HHH is correctly rejected as
 non-halting V2
Date: Fri, 12 Jul 2024 22:52:13 +0200
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Op 12.jul.2024 om 16:56 schreef olcott:
> We stipulate that the only measure of a correct emulation is the
> semantics of the x86 programming language.
> 
> _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]
> 
> When N steps of DDD are emulated by HHH according to the
> semantics of the x86 language then N steps are emulated correctly.

But the following M steps are not simulated at all, making the as a 
whole invalid, because:
1) That contradicts your stipulation that the only measure of a correct 
emulation is the semantics of the x86 programming language. The x86 
language does not allow an abort halfway of a series of instructions.
2) Skipping the simulation of only some steps will make the behaviour 
different. The simulator must process the whole input, not only the 
first part.

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

Then it skips the last part of the simulation and that makes the 
simulation as a whole incorrect.

> HHH₂ two steps of DDD are correctly emulated by HHH.

Then it skips the last part of the simulation and that makes the 
simulation as a whole incorrect.

> HHH₃ three steps of DDD are correctly emulated by HHH.

Then it skips the last part of the simulation and that makes the 
simulation as a whole incorrect.

> ...
> HHH∞ The emulation of DDD by HHH never stops running.

Even that simulation is incorrect, because it does not end.

> 
> The above specifies the infinite set of every HHH/DDD pair
> where 1 to infinity steps of DDD are correctly emulated by HHH.

And where the HHH that abort skip the simulation of the last part of its 
input, making the simulation incorrect.

> 
> No DDD instance of each HHH/DDD pair ever reaches past its
> own machine address of 0000216b and halts.

Showing that none of these simulations was correct.
It only proves that no HHH in this set is able to simulate itself correctly.

> 
> Thus each HHH element of the above infinite set of HHH/DDD
> pairs is necessarily correct to reject its DDD as non-halting.
> 
No, the conclusion must be that none of the HHH in this set is able to 
simulate itself correctly and reach the end of the simulation.
They all miss the last part of the simulation and therefore, do not 
process the full input.
So, the abort was always premature and no conclusion about halting or 
non-halting is possible. For each of the HHH that abort, it makes no 
sense to dream of a HHH that does not abort, because that is not the one 
that this HHH is simulating.
Each of the HHH in this set only simulates itself, not another one of 
this set. Dreaming of another HHH in this set when simulating only 
itself, is irrelevant and does not make the simulation correct.

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 none of 
these simulations is correct, because none of them is able to reach the end.
For each HHH in this set we see that HHH cannot possibly simulate itself 
correctly.

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.
When it should decide about Finite_Recursion(5), the programmer starts 
to dream about an infinite set of Finite_Recursion(M), with M=0 to 
infinity, and decides that it must be an infinite recursion.