Re: Proving the: Simulating termination analyzer Principle

On 06/04/2025 01:35, olcott wrote:

On 4/5/2025 5:31 PM, Richard Heathfield wrote:

On 05/04/2025 23:20, olcott wrote:

On 4/5/2025 4:58 PM, Richard Heathfield wrote:

>
<snip>
>

hp(arg candidate, arg testdata)
{
   if(terminates(candidate(testdata)))
   {
     while(forever);
   }
   else
   {
     halt;
   }
}
>
We then invoke the program:
>
hp(hp, hp)
>
and try to predict what terminates() will report, and of course the answer is that we don't know, because neither does terminates(). The function cannot be written.
>

>
Understanding my simpler example was a mandatory
prerequisite

>
No, it wasn't.
>
Understanding my example isn't mandatory either, which is just as well where you're concerned.
>

int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}

>
That's fine, but it does beg the HHH() question. You are handwaving it for the same reason I am, which is that it can't be written. The difference between us is that I know it and you don't.
>

 HHH(DDD) is isomorphic to HHH(DD),

Irrelevant.

yet failing
to understand that HHH(DDD) meets the
*Simulating termination analyzer Principle*
prevents the significance of this from being seen.

It has no significance.
There are only two possibilities: either it always gives the right answer or it doesn't. If it gives the wrong answer, it's of no interest.
If it is claimed always to give the right answer, it becomes possible (as shown above in the chevrons) to write a program for which it will not be able to work out the right answer - reductio ad absurdum.
Your 'principle' doesn't matter a jot.
--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within