Re: How do simulating termination analyzers work?

On 2025-06-28 13:17:17 +0000, olcott said:

On 6/28/2025 6:41 AM, Mikko wrote:

On 2025-06-21 17:34:55 +0000, olcott said:
 

On 6/21/2025 4:52 AM, Mikko wrote:

On 2025-06-20 13:59:02 +0000, olcott said:
 

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

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

On 6/19/2025 4:21 AM, Fred. Zwarts wrote:

Op 18.jun.2025 om 15:46 schreef olcott:

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*
 

 At least it is true for all aborting ones, such as the one you presented in Halt7.c.

 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?

 

Yes, I confirmed many times that we can confirm this vacuous claim, because no such HHH exists. All of them fail to do a correct simulation up to the point where they can see whether the input specifies a halting program.

 if DDD correctly simulated by any simulating termination
analyzer HHH never aborts its simulation of DDD then

 that HHH is not interesting.

 *then the HP proofs are proved to be wrong*

 Does not follow. HHH and DDD are irrelevant to those proofs.

 void DDD()
{
   HHH(DDD);
   return;
}
 When I dumbed the original self-referential proof down
to HHH(DDD) everyone here proved that they did not even
understand what ordinary recursion is.

That you are dumb does not mean that others don't understand
ordinary recursion.
--
Mikko