Re: Who here understands that the last paragraph is Necessarily true?

Op 19.jul.2024 om 16:56 schreef olcott:

On 7/19/2024 1:40 AM, Fred. Zwarts wrote:

Op 18.jul.2024 om 16:18 schreef olcott:

On 7/18/2024 3:41 AM, Fred. Zwarts wrote:

Op 17.jul.2024 om 16:56 schreef olcott:

On 7/17/2024 9:32 AM, Fred. Zwarts wrote:

Op 17.jul.2024 om 16:20 schreef olcott:

On 7/17/2024 8:54 AM, Fred. Zwarts wrote:

Op 17.jul.2024 om 15:27 schreef olcott:

>
HHH is not allowed to report on the behavior of it actual self
in its own directly executed process. HHH is allowed to report on
the effect of the behavior of the simulation of itself simulating DDD.
>

>
But only on the effect of a correct simulation.

>
_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]
>
*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.
>

>
It is self evident that a program that aborts will halt.
The semantics of the x86 code of a halting program is also self-evident: it halts.
So, the aborting HHH, when simulated correctly, stops.
Dreaming of a HHH that does not abort is irrelevant.
>

>
That is all the dishonest dodge of the strawman deception.
HHH is required to halt by its design spec.
>
_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]
>
*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.
>

>
Dreaming of a HHH that does not halt, when we are talking about a HHH that aborts and halts is irrelevant. Therefore, the 'unless aborted' is irrelevant. The semantics of the x86 instructions are self-evident: HHH halts.

>
When you are hungry you remain hungry until you eat.
   Before HHH(DDD) aborts its emulation the directly
   executed DDD() cannot possibly halt.

>
No, but HHH would have halted when not aborted, because that is how it is programmed. That is the semantics of its x86 code.
>

 int main { DDD(); } calls HHH(DDD) that must abort the
emulation of its input or
HHH, emulated DDD and executed DDD never stop running.
 

Whether HHH *must* abort, or *not abort* is irrelevant. It aborts.
When HHH is coded to abort, the reason why is no longer relevant when the behaviour is analysed.
HHH aborts and halts, therefore it is incorrect to abort the simulated HHH, which would also abort and halt.
The semantics of the x86 language is the same for the direct execution and for the simulator. The semantics of x86 does not depend on who will interpret it.
HHH, which is coded to abort, cannot refuse to abort, which makes the simulation incorrect.
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.
Olcott could not point to an error, but prefers to ignore it. So, I will repeat it, until either an error is found, or olcott admits that the simulation is incorrect.