Re: HHH(DD) does correctly reject its input as non-halting --- VERIFIED FACT

On 6/13/25 2:03 PM, olcott wrote:

On 6/13/2025 12:50 PM, Richard Damon wrote:

On 6/13/25 1:37 PM, olcott wrote:

On 6/13/2025 12:33 PM, Richard Damon wrote:

On 6/13/25 1:26 PM, olcott wrote:

On 6/13/2025 12:19 PM, Richard Damon wrote:

On 6/13/25 10:37 AM, olcott wrote:

On 6/13/2025 4:26 AM, Fred. Zwarts wrote:

Op 12.jun.2025 om 17:30 schreef olcott:
>
Even after many corrections, Olcott repeated his claims without learning anything from his previous errors.
Lack of knowledge does not make someone look stupid, but the resistance against learning does.
>

int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
>
It is a verified fact that DD() *is* one of the forms
of the counter-example input as such an input would
be encoded in C. Christopher Strachey wrote his in CPL.
>
// rec routine P
//   §L :if T[P] go to L
//     Return §
// https://academic.oup.com/comjnl/article/7/4/313/354243
void Strachey_P()
{
   L: if (HHH(Strachey_P)) goto L;
   return;
}
>
https://academic.oup.com/comjnl/article- abstract/7/4/313/354243? redirectedFrom=fulltext
>
It *is* a verified fact DD correctly simulated by HHH
cannot possibly reach its own "return" statement
final halt state.

>
Showing the failure of HHH to reach the end of the simulation.

>
The code of the input to HHH(DD) specifies
HHH simulates DD that calls HHH(DD)
HHH simulates DD that calls HHH(DD)
HHH simulates DD that calls HHH(DD)...

>
>
Then you are lying that HHH will abort and return 0.
>
That is your problem, you world is based on being able to just lie about what you want.
>

>
That you can't understand this is merely a lack
of sufficient tecnh9cal competence on your part.
>

>
No, it is merely a lack of honesty on your part.
>

That you continue to fail to show all of the details
of exactly how DD does reach its simulated "return"
statement final halt state proves that you know you
are not competent.
>
>

>
But DD DOES reach its final state

THE SIMULATED DD CANNOT POSSIBLY REACH ITS SIMULATED FINAL
YOU DAMNED JACKASS.
>

>
So you erroneously think. I have shown how it does.

You have never shown how DDD correctly emulated by
simulating termination analyzer HHH reaches its own
simulated final halt state.

And why should I?
I have instead shown that your HHH doesn't do what you claim, and thus your proof is just a lie.
I am just chopping down your strawman, rather than getting distracted into a fight with them.
The problem you have is that if you just say "the simulated input", that allows the simulation to be done by some other machine, especiailly if this one doesn't do it.
If you specify "the correct simulation of the input by this machine", if it doesn't DO such a correct simulation, because it aborts, then the clause is meaningless.
THus, you "proof" does have any ground to stand on, except the shifting sands built on your lies.

 I thought that I made a square circle once until
I added the unstated requirement that it must be
in the same two dimensional plane.
 

So?