Re: No decider is accountable for the computation that itself is contained within

Op 30.jul.2024 om 18:24 schreef olcott:

On 7/30/2024 2:24 AM, joes wrote:

Am Mon, 29 Jul 2024 15:32:44 -0500 schrieb olcott:

On 7/29/2024 3:17 PM, joes wrote:

Am Mon, 29 Jul 2024 11:32:00 -0500 schrieb olcott:

On 7/28/2024 3:40 AM, Mikko wrote:

On 2024-07-27 14:21:50 +0000, olcott said:

On 7/27/2024 2:46 AM, Mikko wrote:

On 2024-07-26 16:28:43 +0000, olcott said:

>

Halt deciders are not allowed to report on the behavior of the actual
computation that they themselves are contained within. They are only
allowed to compute the mapping from input finite strings.

What if the input is the same as the containing computation?

It always is except in the case where the decider is reporting on the TM
description that itself is contained within.

 

I don't understand. "The input is not the same as the containing
computation when deciding on the description of the containing
computation"?
>

 void DDD()
{
   HHH(DDD);
}
 The behavior of the correct emulation of the x86 machine
language input DDD to a emulating halt decider HHH is not
the same as behavior of the direct execution of DDD when
the x86 machine language of DDD is correctly emulated
by emulating halt decider HHH that calls HHH(DDD) (itself).
 

In fact, HHH cannot possibly simulate *itself* correctly. That is the difference. HHH deviates from the semantics of the x86 language by skipping the last few instructions of the simulation of itself.
This incorrect simulation does not show the behaviour of DDD, but it shows the error of 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.
It shows that 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 of a halting program.
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 incomplete and, therefore, incorrect.