Re: Liar detector: Fred, Richard, Joes and Alan --- Ben's agreement

Op 17.jul.2024 om 16:26 schreef olcott:

On 7/17/2024 9:08 AM, Fred. Zwarts wrote:

Op 17.jul.2024 om 15:39 schreef olcott:

On 7/17/2024 8:32 AM, Fred. Zwarts wrote:

Op 17.jul.2024 om 15:00 schreef olcott:

On 7/17/2024 1:43 AM, Mikko wrote:

On 2024-07-16 14:21:28 +0000, olcott said:

>
When simulated input DDD stops running {if and only if}
the simulation of this input DDD has been aborted this
necessitates that input DDD specifies non-halting behavior

>
DDD does not stop runnig unless it is completely exeuted.

>
_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]
>
DDD emulated by HHH according to the semantic meaning of
its x86 instructions never stop running unless aborted.

>
The 'unless aborted' is misleading, since we know that HHH *does* abort.

*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.

>
But HHH is programmed to abort after N cycles,

 _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 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

And it has been proved that this infinite amount of simulations are all incorrect, because they either do not halt, or the abort prematurely.

 No DDD of the infinite set of every HHH/DDD pair reaches
past its machine address 0000216b thus never halts. Every
HHH either aborts its simulation or DDD never stops running.
>

Proving that HHH cannot possibly simulate itself correctly.
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.