Re: key error in all the proofs --- Correction of Fred

Op 15.aug.2024 om 14:59 schreef olcott:

On 8/15/2024 3:20 AM, Fred. Zwarts wrote:

Op 14.aug.2024 om 23:08 schreef olcott:

On 8/14/2024 3:56 PM, Mike Terry wrote:

On 14/08/2024 18:45, olcott wrote:

On 8/14/2024 11:31 AM, joes wrote:

Am Wed, 14 Aug 2024 08:42:33 -0500 schrieb olcott:

On 8/14/2024 2:30 AM, Mikko wrote:

On 2024-08-13 13:30:08 +0000, olcott said:

On 8/13/2024 6:23 AM, Richard Damon wrote:

On 8/12/24 11:45 PM, olcott wrote:

>
*DDD correctly emulated by HHH cannot possibly reach its* *own
"return" instruction final halt state, thus never halts*
>

Which is only correct if HHH actuallly does a complete and correct
emulation, or the behavior DDD (but not the emulation of DDD by HHH)
will reach that return.
>

A complete emulation of a non-terminating input has always been a
contradiction in terms.
HHH correctly predicts that a correct and unlimited emulation of DDD
by HHH cannot possibly reach its own "return" instruction final halt
state.

>
That is not a meaningful prediction because a complete and unlimited
emulation of DDD by HHH never happens.
>

A complete emulation is not required to correctly predict that a
complete emulation would never halt.

What do we care about a complete simulation? HHH isn't doing one.
>

>
Please go read how Mike corrected you.
>

>
Lol, dude...  I mentioned nothing about complete/incomplete simulations.
>

>
*You corrected Joes most persistent error*
She made sure to ignore this correction.
>

But while we're here - a complete simulation of input D() would clearly halt.

>
_DDD()
[00002172] 55         push ebp      ; housekeeping
[00002173] 8bec       mov ebp,esp   ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404     add esp,+04
[00002182] 5d         pop ebp
[00002183] c3         ret
Size in bytes:(0018) [00002183]
>
A complete simulation *by HHH* remains stuck in
infinite recursion until aborted.

>
It is aborted, so the infinite recursion is just a dream.

 All simulating termination analyzers are required
to predict what the behavior would be when the
emulation is unlimited (never aborted) otherwise
they could never report on the behavior of this function:
 void Infinite_Loop()
{
   HERE: goto HERE;
}
 

Irrelevant when there is a halting input. The HHH that aborts halts, so forget about the HHH that does not halt.

Also something that you consistently ignore is that
HHH is not reporting on its own behavior. HHH is only
predicting whether or not an unlimited emulation of
DDD would reach the "return" instruction of DDD.
 void DDD()
{
   HHH(DDD);
   return;
}
 

And it is wrong, as is proved by the unlimited simulation of HHH by HHH1.
What you ignore is that DDD is completely irrelevant.
DDD is a misleading and unneeded complication. It is easy to eliminate DDD:
        int main() {
          return HHH(main);
        }
This has the same problem. This proves that the problem is not in DDD, but in HHH, which halts when it aborts the simulation, but it decides that the simulation of itself does not halt.
It shows that HHH cannot possibly simulate itself correctly.