Re: HHH maps its input to the behavior specified by it --- never reaches its halt state ---

Op 09.aug.2024 om 15:41 schreef olcott:

On 8/9/2024 1:39 AM, Fred. Zwarts wrote:

Op 09.aug.2024 om 05:03 schreef olcott:

On 8/8/2024 9:52 PM, Richard Damon wrote:

On 8/8/24 9:15 AM, olcott wrote:

>
void DDD()
{
   HHH(DDD);
   return;
}
>
Each HHH of every HHH that can possibly exist definitely
*emulates zero to infinity instructions correctly* In
none of these cases does the emulated DDD ever reach
its "return" instruction halt state.
>
*There are no double-talk weasel words around this*
*There are no double-talk weasel words around this*
*There are no double-talk weasel words around this*
>
There is no need to show any execution trace at the x86 level
every expert in the C language sees that the emulated DDD
cannot possibly reaches its "return" instruction halt state.
>
Every rebuttal that anyone can possibly make is necessarily
erroneous because the first paragraph is a tautology.
>
>

>
Nope, it is a lie based on comfusing the behavior of DDD which is what "Halting" is.
>

>
Finally something besides
the strawman deception,
disagreeing with a tautology, or
pure ad hominem.
>
You must first agree with everything that I said above
before we can get to this last and final point that it
not actually directly referenced above.
>
*Two key facts*
(a) The "return" instruction is the halt state of DDD.
(b) DDD correctly emulated by any HHH never reaches this state.

There is no correct simulation of HHH by itself. HHH cannot possibly simulate itself correctly. A correct simulation of a halting program must reach this state.

 Try and show how it is incorrect.

We proved it many times, but it seems you do not understand it. It is unclear to me whether you are unable to understand it or unwilling to understand it. But I will go another mile.
(HHH is required to halt, so we don't need to consider the non-aborting HHH.)
1) HHH is programmed to abort after N cycles. When HHH simulates itself, the simulating HHH aborts after N cycles.
2) Note, we are looking at the HHH that aborts after N cycles, not the one that does not abort. Both the simulating and the simulated HHH use the same algorithm, namely, the algorithm to abort after N cycles.
3) At the moment that the simulating HHH aborts, the simulated HHH has done N-1 cycles. It still has one cycle to go, after which it would abort and return, after which DDD would return.
4) But the simulation cannot reach this end, because the simulating HHH aborts after N cycles, because that is how it is programmed, skipping in this way the simulation of the last cycle. (There is a last cycle, because that is how HHH is programmed: to abort after N cycles. We are not dreaming of another HHH that does not abort.)
5) Skipping the last cycle and the end of this halting program violates the purpose of the simulation, which makes the simulation incorrect.
Note that this proof holds for any N between 0 and infinity.
Now it is your turn. Not only repeat without evidence the claim that the simulation is correct, but prove it, by proving that no instructions of the halting program were skipped. (We don't need to prove that HHH is halting, because that is a requirement.)