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

Op 19.jul.2024 om 16:49 schreef olcott:

On 7/19/2024 4:14 AM, Mikko wrote:

On 2024-07-18 14:18:51 +0000, olcott said:
>

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.
>
After you eat you are no longer hungry.
   After HHH(DDD) aborts its emulation the directly
   executed DDD() halts.

>
If DDD does not halt it indicates that HHH is faulty. Therefore the
interesting question is whether DDD halts, not when DDD halts.
>

 *By your same reasoning*
If Infinite_Loop() does not halt HHH is faulty.
 In other words if Infinite_Loop()  is an actual infinite
loop then this is all the fault of HHH.
 void Infinite_Loop()
{
   HERE: goto HERE;
}
 int main()
{
   HHH(Infinite_Loop);
}
 

Irrelevant, because you do not understand the difference between Infinite_Loop and Finite_Recursion.
void Finite_Recursion (int N) {
   if (N > 0) Finite_Recursion (N - 1);
}
HHH is unable to recognise a finite recursion, such as in an aborting HHH.
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.
Olcott's 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 olcott wants it to be correct, or how many times olcott repeats that it is correct, it does not change the fact that such a simulation is incorrect, because it is unable to reach the end.
Olcott's own claim that the simulated HHH does not reach its end confirms it. The trace he has shown also proves that HHH cannot reach the end of its own simulation. So, his 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 HHH cannot possibly simulate itself correctly.