Re: How do simulating termination analyzers work?

Op 20.jun.2025 om 16:42 schreef olcott:

On 6/20/2025 4:35 AM, Fred. Zwarts wrote:

Op 19.jun.2025 om 17:13 schreef olcott:

On 6/19/2025 4:13 AM, Fred. Zwarts wrote:

Op 19.jun.2025 om 04:11 schreef olcott:

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;
}
>
void Infinite_Loop()
{
   HERE: goto HERE;
   return;
}
>
void DDD()
{
   HHH(DDD);
   return;
}
>
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.

>
WHich means that the code for HHH is part of the input, and thus there is just ONE HHH in existance at this time.
>
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)
>

>
*none of them ever stop running unless aborted*

>
All of them do abort and their simulation does not need an abort.
>

>
*It is not given that any of them abort*
>

>
>
>
But it either does or it doesn't, and different HHHs give different DDD so you can't compare their behavior.
>

>
My claim is that DDD correctly simulated by any
termination analyzer HHH that can possibly exist
will never stop running unless aborted.

>
A vacuous statement because no such termination analyser exist.

>
The above set of every termination analyzer HHH includes
those that get the wrong answer and those that never stop
running.

>
That is not the meaning of the words 'correctly simulated'.
None of them does a correct simulation.
>

 Every instruction of DDD emulated by HHH is emulated correctly.

A simulator that simulates only one instruction also correctly simulates   every instruction simulated. That does not mean that it is a correct simulation of the input.
Is even that too difficult for you?

 

>

The candidate you present

>
*Is each element of this infinite set*
My claim is that each of the above functions correctly
simulated by any termination analyzer HHH that can possibly
exist will never stop running unless aborted by HHH.
Can you affirm or correctly refute this?
>
>

>
The words 'correctly simulated' makes this a vacuous statement. There is no HHH that can correctly simulate itself.

 You say this only because you lack the technical competence
to verify that HHH does correctly emulate itself emulating DDD.
https://github.com/plolcott/x86utm/blob/master/Halt7.c
 

Again a claim without evidence.