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

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.