Re: How do simulating termination analyzers work?

On 6/20/25 10:08 AM, olcott wrote:

On 6/20/2025 4:29 AM, Fred. Zwarts wrote:

Op 20.jun.2025 om 02:13 schreef olcott:

On 6/19/2025 6:40 PM, Richard Damon wrote:

On 6/18/25 10:11 PM, olcott wrote:

On 6/18/2025 8:20 PM, Richard Damon wrote:

On 6/18/25 9:46 AM, olcott wrote:

On 6/18/2025 5:12 AM, Fred. Zwarts wrote:

Op 18.jun.2025 om 03:54 schreef olcott:

On 6/17/2025 8:19 PM, Richard Damon wrote:

On 6/17/25 4:34 PM, olcott wrote:

void Infinite_Recursion()
{
   Infinite_Recursion();
   return;
}
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void Infinite_Loop()
{
   HERE: goto HERE;
   return;
}
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void DDD()
{
   HHH(DDD);
   return;
}
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When it is understood that HHH does simulate itself
simulating DDD then any first year CS student knows
that when each of the above are correctly simulated
by HHH that none of them ever stop running unless aborted.

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WHich means that the code for HHH is part of the input, and thus there is just ONE HHH in existance at this time.
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Since that code aborts its simulation to return the answer that you claim, you are just lying that it did a correct simulation (which in this context means complete)
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*none of them ever stop running unless aborted*

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All of them do abort and their simulation does not need an abort.
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*It is not given that any of them abort*
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But it either does or it doesn't, and different HHHs give different DDD so you can't compare their behavior.
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My claim is that DDD correctly simulated by any
termination analyzer HHH that can possibly exist
will never stop running unless aborted.
*No one has ever been able to refute this*
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But the only HHH that DOES simulate any part of THIS DDD, is THIS HHH, and if it aborts to answer, it doesn't correctly simulate this input, so you LIE that it does.
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The only rebuttal to my actual point would be to
show how DDD simulated by any simulating termination
analyzer HHH would stop running if never aborted.
*Everything else is a dishonest*
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Wrong. There is another rebuttal.
It has been proven by world-class simulators that a correct simulation of exactly the same input reaches a natural end without an abort.

 Counter-factual
 if DDD correctly simulated by any simulating termination
analyzer HHH never aborts its simulation of DDD then
(a) DDD simulated by HHH never stops running.
(b) Directly executed HHH() never stops running.
(c) Directly executed DDD() never stops running.
(d) DDD correctly simulated by HHH1 never stops running.

And thus, if HHH correctly simulates its input. it never answers.
Thus, the HHH that you claim to be answering correctly, and that the input will be calling, doesn't behave that way.
You don't seem to understand that you get DIFFERENT input DDD when you change HHH, as you are trying to pull a fast one with using inconsistant definitions (which is just one of the forms of lying).

 

It has been proven also that no correct simulation of HHH by itself exists.

 *Every instruction that HHH emulates is emulated correctly*

Nope, not the last, since it aborted.
You like that LIE, just like your spiritual daddy.

 

Therefore, asking to correct an incorrect HHH is like asking to draw a square circle.
It is dishonest to ask for the impossible, in particular when this has been pointed out to you many times.
Not knowing something does not make your stupid. he resistance against learning from your errors, however, does.
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