Re: DDD simulated by HHH cannot possibly halt (Halting Problem)

On 4/13/25 3:46 PM, olcott wrote:

On 4/11/2025 2:57 AM, Mikko wrote:

On 2025-04-10 23:08:02 +0000, olcott said:
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On 4/10/2025 2:56 AM, Mikko wrote:

On 2025-04-09 20:35:30 +0000, olcott said:
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On 4/9/2025 1:58 PM, Fred. Zwarts wrote:

Op 09.apr.2025 om 19:29 schreef olcott:

>
On 4/8/2025 10:31 AM, Fred. Zwarts wrote:

Op 08.apr.2025 om 17:13 schreef olcott:

On 4/8/2025 2:45 AM, Fred. Zwarts wrote:

Op 08.apr.2025 om 06:33 schreef olcott:

>
typedef void (*ptr)();
int HHH(ptr P);
>
int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
>
int main()
{
   HHH(DD);
}
>
*Simulating termination analyzer Principle*
It is always correct for any simulating termination
analyzer to stop simulating and reject any input that
would otherwise prevent its own termination.

>
>
In this case there is nothing to prevent, because the finite string specifies a program that halts.

>
int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
>
This stuff is simply over-your-head.
HHH(DD) meets the above: *Simulating termination analyzer Principle*
Anyone with sufficient competence with the C programming language
will understand this.
>

Everyone with a little bit of C knowledge understands that if HHH returns with a value 0, then DDD halts.

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DDD CORRECTLY SIMULATED BY HHH
NOT ANY OTHER DAMN DDD IN THE UNIVERSE NITWIT.
>

If HHH would correctly simulate DD (and the functions called by DD) then the simulated HHH would return to DD and DD would halt.

>
Simply over your level of technical competence.
>

But HHH failed to complete the simulation of the halting program,

>
HHH is only required to report on the behavior of its
own correct simulation (meaning the according to the
semantics of the C programming language) and would be
incorrect to report on any other behavior.
>

because the programmer was dreaming of an infinite recursion.
>
If I didn't have to tell you this hundreds of times and you didn't
persist in the straw-man deception I would not have called you a nitwit.
>
>

>
I really think that you may simply be a troll playing head games.

>
It is not a good idea to think that everybody is just like you.

>
*Simulating termination analyzer Principle*
It is always correct for any simulating termination
analyzer to stop simulating and reject any input that
would otherwise prevent its own termination.

>
No proof of this principle has been shown so its use is not valid.
>

 The proof is self-evident.
Termination analyzers must halt (axiom) therefore
rejecting any input that would prevent the analyzer
from terminating is semantically entailed by this axiom.
 

That is self-evident only to an pathological liar, as it ignores the requirement that it answer must be correct.
Thanks for proving what you are.,