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

On 4/10/25 7:20 PM, olcott wrote:

On 4/10/2025 3:49 AM, Fred. Zwarts wrote:

Op 09.apr.2025 om 21:20 schreef olcott:

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.

>
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.

>
No way, José:
By "correctly simulate" I mean a simulation or an emulation
of DD by HHH that obeys the semantics of the C or the x86utm
programming languages respectfully. This expressly includes
that HHH simulates/emulates itself simulating/emulating DD.

 

That would make HHH to never halt.

 *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.
 The only rebuttal to this is to stupidly reject the notion
that all deciders must always halt.
 HHH detects this pattern through mathematical induction
of n steps of DD emulated/simulated by HHH. The criteria
is that the emulated/simulated DD fails to ever reach its
own “return instruction” final halt state.
 

But you don't, as you don't have a proper logic system to work on your programs (which don't seem to really be programs)
Which only applies if the Termination Analyzer actually does a correct simulation, or evalutates what a correct simultion would do.
It also requires that the input be correctly Simulated.
You have already established that if the Analyzer actually correctly simulates this input, then it can't return an answer, as this input (if correctly simulated because it includes the code of the HHH that it calls) when correctly simulated never returns because that emulator can never reach the final state of the input.
Thus, since it must halt, it can't be a correct simulation, and thus the results of correctly simulating this input must be based on the actua; behavior of a real correct simulator given this EXACT same input, which you have shown with HHH1, that it will halt.
So, the problem with your application of that principle, is thst it is self-contradictory and thus invalid, as Any simulator trying to follow it can't correctly simulate the input.
Either the input just can't be simulated, as it isn't a program because it doesn't include the code of the HHH that it calls, and thus your principle fails because it assumes a false premise, or the input is a program because it includes that code, and thus from the previous discussion, HHH doesn't do a correct simulation, and the correct simulation is shown to halt.
There is no problem with the fact that decider must always halt, but the correlary of that requirement is that no decider can be an always correct simulator, in particular, it can never be a correct simulator for a non-halting input, and the fact that it "decides" to abort it simulation can not be used a proof that the input is non-halting when correctly simulated, but you need to look at the actually correct simulation of the input.
Thus your phrase "Correctly Simulated by HHH" isn't a valid condition, as it doesn't happen.
Sorry, you are just showing that you don't understand what you are taloing about, and don't care, because you are just a hypocritical pathological liar.