Re: How do simulating termination analyzers work?

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, and every HHH gets a diffent PROGRAM DDD, so "none of them" is a meaningless term for any of them.
You are just showing that you don't undertstand what you are talking about becuase you made yourself stupid.
Your argument begins with the equivalent of saying if 1 is equal to 2, then ...