Re: Overview of proof that the input to HHH(DDD) specifies non-halting behavior --- Mike

On 8/14/2024 3:09 AM, Mikko wrote:

On 2024-08-13 13:04:17 +0000, olcott said:
 

On 8/13/2024 5:57 AM, Mikko wrote:

On 2024-08-13 01:43:49 +0000, olcott said:
>

We prove that the simulation is correct.
Then we prove that this simulation cannot possibly
reach its final halt state / ever stop running without being aborted.
The semantics of the x86 language conclusive proves this is true.
>
Thus when we measure the behavior specified by this finite
string by DDD correctly simulated/emulated by HHH it specifies
non-halting behavior.
>
https://www.researchgate.net/ publication/369971402_Simulating_Termination_Analyzer_H_is_Not_Fooled_by_Pathological_Input_D

>
Input to HHH(DDD) is DDD. If there is any other input then the proof is
not interesting.
>
The behviour specified by DDD on the first page of the linked article
is halting if HHH(DDD) halts. Otherwise HHH is not interesting.
>
Any proof of the false statement that "the input to HHH(DDD) specifies
non-halting behaviour" is either uninteresting or unsound.
>

>
void DDD()
{
   HHH(DDD);
   return;
}
>
It is true that DDD correctly emulated by any HHH cannot
possibly reach its own "return" instruction final halt state.

 If DDD does not halt then HHH does not 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]
The impossibility of DDD emulated by HHH
(according to the semantics of the x86 language)
to reach its own machine address [00002183] is
complete proof that DDD never halts.
This has nothing to do with whether or not HHH
halts.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer