Re: DDD correctly emulated by HHH cannot possibly halt

Op 12.jul.2024 om 16:51 schreef olcott:

On 7/12/2024 8:02 AM, Fred. Zwarts wrote:

Op 12.jul.2024 om 14:08 schreef olcott:

>
when 1 to ∞ steps of DDD are correctly emulated in
the infinite set of every HHH/DDD pair and no DDD
halts then we can say that DDD DOES NOT HALT.

>
If 1 to ∞ steps are simulated by HHH and none of these simulations is correct, then none of the simulations is correct.

*Vacuous "truth" is a form of deception*
 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 only those steps are simulated correctly. The steps that are skipped are not considered. So the simulation as a whole is incorrect.
Simulating only N steps of a program of N+1 steps makes the simulation incorrect, because it does not show the full behaviour, of the last step and the end of the program. The simulation does not process the full input that it is given.

 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 of further steps is skipped, making the simulation as a whole incorrect, because it does not show the full behaviour.

HHH₂ two steps of DDD are correctly emulated by HHH.

But the simulation of further steps is skipped, making the simulation as a whole incorrect, because it does not show the full behaviour.

HHH₃ three steps of DDD are correctly emulated by HHH.

But the simulation of further steps is skipped, making the simulation as a whole incorrect, because it does not show the full behaviour.

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

Which is also not a correct simulation, 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 last step is skipped, which makes the whole simulation incorrect. So none of the instances of HHH in this set was able to do a correct simulation.

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

Showing that the simulation of all instances is incorrect.

 Thus each HHH element of the above infinite set of HHH/DDD
pairs is necessarily correct to reject its DDD as non-halting.
 

No. None of these simulations is correct. Each HHH that aborts is shown to abort 1 step before the simulated HHH would reach its end. 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 to be it correct, or how many times you repeat that its is correct, it does not change the fact that none of these simulations is incorrect, 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 is required to halt, but it decides that HHH 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.