Re: Anyone with sufficient knowledge of C knows that DD specifies non-terminating behavior to HHH

Op 09.feb.2025 om 16:15 schreef olcott:

On 2/9/2025 2:09 AM, Fred. Zwarts wrote:

Op 09.feb.2025 om 07:04 schreef olcott:

On 2/8/2025 3:49 PM, Fred. Zwarts wrote:

Op 08.feb.2025 om 15:43 schreef olcott:

On 2/8/2025 3:54 AM, Fred. Zwarts wrote:

Op 08.feb.2025 om 00:13 schreef olcott:

Experts in the C programming language will know that DD
correctly simulated by HHH cannot possibly reach its own
"if" statement.

>
Yes, it demonstrates the incapability of HHH to correctly determine the halting behaviour of DD
>

>
The finite string DD specifies non-terminating recursive
simulation to simulating termination analyzer HHH. This
makes HHH necessarily correct to reject its input as
non-halting.

>
The finite string defines one behaviour. This finite string, when given to an X86 processor shows halting behaviour. This finite string,when given to a world class simulator, shows halting behaviour. Only HHH fails to see this proven halting behaviour. So it proves the failure of HHH.
HHH aborts the simulation on unsound grounds one cycle before the simulation would terminate normally.
>

>
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);
}
>
https://www.researchgate.net/ publication/369971402_Simulating_Termination_Analyzer_H_is_Not_Fooled_by_Pathological_Input_D
>
https://github.com/plolcott/x86utm/blob/master/Halt7.c
has fully operational HHH and DD
>
The halting problem has always been a mathematical mapping
from finite strings to behaviors.

>
Yes. And the behaviour of this finite string has been proven to show halting behaviour. Only Olcott's HHH fails to see it.
His misunderstanding is that he thinks that the behaviour defined by the finite string depends on the simulator.

>
When DD calls HHH(DD) in recursive simulation it is a
verified fact that DD cannot possibly halt.

>
Which proves the failure of HHH. It does not reach the end of a halting program. All other methods show that DD halts.
>

>
Your comment only proves that you lack sufficient
understanding of the C programming language.
>

>
This is a proof of lack of logical reasoning.
>
Verified fact 1: DD halts

 Fallacy of equivocation error.
(a) All men are mortal
(b) No woman is a man
∴ No woman is mortal

Yes, the claim that DD does not halt is indeed such a fallacy:
(a) Direct execution and all simulators show that DD halts.
(b) My simulator is different
 > ∴ DD does not halt.

 The input to HHH(DD) cannot possibly terminate normally.
Referring to some other DD does not change this verfied fact.
 

That DD halts is a verified fact. Verified by direct execution and many simulators.
It is exactly the same finite string as used in HHH, otherwise HHH is simply talking about another finite string.
So, HHH is simply wrong if it claims that DD does not halt. The explanation is simple: There is a bug in the detection of the 'special condition' hat causes the abort of the simulation, which makes that the simulation is aborted one cycle before the normal termination of the simulation.