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

On 6/20/2025 3:35 AM, Mikko wrote:

On 2025-06-19 15:25:45 +0000, olcott said:
 

On 6/19/2025 3:21 AM, Mikko wrote:

On 2025-06-18 15:07:14 +0000, olcott said:
 

On 6/17/2025 8:27 PM, Richard Damon wrote:

On 6/17/25 10:46 AM, olcott wrote:

On 6/17/2025 4:33 AM, Fred. Zwarts wrote:

Op 16.jun.2025 om 19:01 schreef olcott:

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.

 And even a beginner can see that they all fail to reach the end of the simulation, even though the input is a pointer to code that includes the code to abort and halt.

 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.

 No, they understand that a pattern seen is a halting program (since you admit that DDD halts when run directly) can't be a pattern that proves the program is non-halting.

 You changed the subject from THIS EXACT POINT
*none of them ever stop running unless aborted*
(a) YES that is true
(b) No that is not true

 No, he did not. The paragraph responded to was about first year CS
students and what know, and so is the response.

 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?

 Now you are changed the topic.

 That is what I said (less clearly) all along.