Re: How do simulating termination analyzers work?

On 6/18/2025 6:01 AM, Richard Damon wrote:

On 6/17/25 9:54 PM, olcott wrote:

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*
*none of them ever stop running unless aborted*
*none of them ever stop running unless aborted*
>
Do you agree or can you refute THIS EXACT POINT?
Do you agree or can you refute THIS EXACT POINT?
Do you agree or can you refute THIS EXACT POINT?
>
>

 How about the fact that if they abort, they never did a correct simulation,

*You are not addressing THE EXACT POINT*
*When HHH never aborts any of the above functions then*
(a) None of the functions ever stops running.
(b) Each of the above functions stops running anyway.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer