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

On 4/13/2025 5:02 PM, olcott wrote:

On 4/13/2025 2:54 PM, dbush wrote:

On 4/13/2025 3:52 PM, olcott wrote:

On 4/13/2025 4:18 AM, joes wrote:

Am Wed, 09 Apr 2025 15:35:30 -0500 schrieb 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:

>

*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;
}
HHH(DD) meets the above: *Simulating termination analyzer Principle*
>

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.
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.

>

Such as what actually happens when you run it?
>

>
Why do you insist on dishonestly ignoring the pathological
self-reference difference?
>

>
Why do you insist on dishonestly ignoring that the input to HHH halts when executed directly?

 Because this stupidly ignores the difference
of pathological self-reference.
 

But I want to know if any arbitrary algorithm X with input Y will halt when executed directly.  If I had an H that could tell me that in *all* possible cases, I could show that all truths are provable.
If you're not interested in such an H, then you have no interest in making all true statements provable.