Re: Simulation vs. Execution in the Halting Problem

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Sujet : Re: Simulation vs. Execution in the Halting Problem
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory
Date : 13. Jun 2025, 15:24:39
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <102hcb7$3fnmh$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
User-Agent : Mozilla Thunderbird
On 6/13/2025 4:11 AM, Fred. Zwarts wrote:
Op 12.jun.2025 om 17:51 schreef olcott:
On 6/12/2025 4:37 AM, Fred. Zwarts wrote:
Op 12.jun.2025 om 02:52 schreef olcott:
>
If you don't understand the difference between object
instances of OOP classes and the classes themselves
then you might not understand.
>
int main()
{
   D(); // calls H(D) and the parameter to this H(D)
}      // is not the caller of this instance of H(D)
>
The analyser should not decide about an instance, but about what is specified by the code.
void DDD()
{
   HHH(DDD);
   return;
}
>
The input to HHH(DDD) specifies a sequence of configurations
that cannot possibly reach their own "return" statement
final halt state.
 The input is a pointer to code that includes the code to abort and halt.
I m not going to correct you on this again.
I will simply stop looking at anything that you say.

That HHH is not able to reach that part of the code is a failure of HHH, not a property of the program specified in the input.
counter-factual and over-your-head.
It is a verified fact that unless the outermost HHH(DDD)
aborts its simulation of DDD then this simulated DDD, the
directly executed DDD() and the directly executed HHH()
would never stop running.
That you don't have enough technical skill to verify
this fact should cause you to have the position that
you don't know if this fact is verified. That you
take the position that I am wrong knowing that you
don't understand these things seems dishonest.

That HHH cannot reach it, does not change the specified code.
Within a correct execution trace unreachable code is
the same as non-existent code. That you don't know that
is merely a lack of sufficient technical competence on
your part.

The program aborts as world-class simulators en direct execution of exactly the same input prove.
 
int main()
{
   DDD(); // calls HHH(DDD) with a different instance of DDD
}

>
That you are insufficiently competent to see this is proven
by the fact that you have no rebuttal anchored in correct
reasoning.
>
 Ad hominem attacks only prove your lack of counter-arguments.
You say that I am wrong and cannot provide all of the details
showing exactly how and why I am wrong does not count as a
rebuttal. It only counts as incorrect intuition.
You prove your ignorance by the lack of details that
you provide in your purely dogmatic assertions that I
am wrong.
void DDD()
{
   HHH(DDD);
   return;
}
The input to HHH(DDD) specifies non-halting behavior
in that DDD correctly simulated by HHH cannot possibly
reach its own "return" statement final halt state.
Only a correct simulation of the input to HHH(DDD) by
HHH derives the actual behavior specified by this input.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

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