Re: HHH(DDD) is correct to reject its input as non-halting --- EVIDENCE THAT I AM CORRECT

On 6/17/2025 5:09 AM, Mikko wrote:

On 2025-06-16 17:01:08 +0000, olcott said:
 

On 6/16/2025 6:37 AM, Mikko wrote:

On 2025-06-16 00:57:42 +0000, olcott said:
>

On 6/15/2025 6:44 PM, Richard Damon wrote:

On 6/15/25 4:10 PM, olcott wrote:

void DDD()
{
   HHH(DDD);
   return;
}
>
When I challenge anyone to show the details of exactly
how DDD correctly simulated by ANY simulating termination
analyzer HHH can possibly reach its own simulated "return"
statement final halt state they ignore this challenge.

>
And it seems you don't understand that the problem is that while, yes, if HHH does infact do a correct simulation, it will not reach a final state, that fact only applie *IF* HHH does that, and all the other HHHs which differ see different inputs.
>

>
*I should have said*

>
No, that is not how you should have said.
>

When one or more instructions of DDD are correctly
simulated by ANY simulating termination analyzer HHH
then DDD never reaches its simulated "return" statement
final halt state.

>
How does ANY simulating termination analyzer HHH differ form some
other simulating termination alalyzer?
>

>
I changed the evaluation from the HHH that I have coded
to every HHH that could possibly exist.

 Tnat means that you think the program
 int HHH((void *)x(void)) {
  return 1;
}
 when called with HHH(DDD) would return 0
 void DDD()
{
    HHH(DDD);
    return;
}
 to indicate that DDD does not halt?
 

Your HHH is not s simulating termination analyzer as required:
When one or more instructions of DDD are correctly
simulated by ANY simulating termination analyzer HHH
then DDD never reaches its simulated "return" statement
final halt state.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer