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

On 2025-06-23 08:24:58 +0000, Fred. Zwarts said:

Op 22.jun.2025 om 21:27 schreef olcott:

On 6/22/2025 11:01 AM, Fred. Zwarts wrote:

Op 20.jun.2025 om 16:53 schreef olcott:

On 6/20/2025 4:42 AM, Fred. Zwarts wrote:

Op 19.jun.2025 om 17:23 schreef olcott:

On 6/19/2025 3:55 AM, Fred. Zwarts wrote:

Op 18.jun.2025 om 17:41 schreef olcott:

On 6/18/2025 4:36 AM, Fred. Zwarts wrote:

Op 17.jun.2025 om 16:36 schreef olcott:

On 6/17/2025 4:28 AM, Fred. Zwarts wrote:

Op 17.jun.2025 om 00:26 schreef olcott:

On 6/16/2025 3:53 AM, Fred. Zwarts wrote:

Op 15.jun.2025 om 22:10 schreef olcott:

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.

 It seems very difficult for you to read.
We clearly stated that the challenge is improper.

 Are you too stupid to understand that dogmatic
assertions that are utterly bereft of any supporting
reasoning DO NOT COUNT AS REBUTTALS ???

 No, you are too stupid to realise that challenging for a recipe to draw a square circle does not count as a proof that square circles exist.
 

 Claiming that I made a mistake with no ability to
show this mistake is DISHONEST.
 

 Indeed, but irrelevant,

 That alternative is that you are dishonest.
When you claim that I am wrong and have
no ability to show how and where I am wrong
this would seem to make you a liar.
 No one has ever even attempted to show the details
of how this is not correct:
 void DDD()
{
   HHH(DDD);
   return;
}
 When one or more instructions of DDD are correctly
simulated by ANY simulating termination analyzer HHH
then this correctly simulated DDD never reaches its
simulated "return" statement final halt state.

 Indeed, HHH fails to reach the end of the simulation, even though the end is only one cycle further from the point where it gave up the simulation.
 

 That is counter-factual and over-your-head.
 

 No evidence presented for this claim. Dreaming again?
Even a beginner understands that when HHH has code to abort and halt, the simulated HHH runs one cycle behind the simulating HHH, so that when the simulating HHH aborts, the simulated HHH is only one cycle away from the same point.

 Proving that you do not understand what unreachable code is.
First year CS students and EE majors may not understand this.
All CS graduates would understand this.

 That you do not understand what I write makes it difficult for you to learn from your errors.
It is not that difficult. Try again and pay full attention to it.
Even a beginner understands that when HHH has code to abort and halt,
the simulated HHH runs one cycle behind the simulating HHH, so that when the simulating HHH aborts, the simulated HHH is only one cycle away from the same point.

 Yes this is factual.
 *This is only ordinary computer programming with*
*no theory of computation computer science required*
 Every simulated HHH remains one cycle behind its simulator
no matter how deep the recursive simulations go. This means
that the outermost directly executed HHH reaches its abort
criteria first.

 And it fails to see that the simulated HHH would reach exactly the same abort criteria one cycle later.
In this way, it misses the fact that it is simulating an HHH that would abort and halt.
 

 void Infinite_Loop()
{
  HERE: goto HERE;
  printf("Fred Zwarts can't understand this is never reached\n");
}
 

Another claim without any evidence.
 Olcott does not understand that his HHH does not see an infinite loop.

Olcott's HHH does see an infinite recursion. More specifically, it
hallucinates one, as the infinite recursion that HHH sees does not
exist.
--
Mikko