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

On 8/13/24 11:25 AM, olcott wrote:

On 8/13/2024 9:40 AM, Fred. Zwarts wrote:

Op 13.aug.2024 om 15:04 schreef olcott:

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.

>
Contradiction in terminus.
A correct simulation is not possible.

 *YOU JUST DON'T GET THIS*
A simulation of N instructions of DDD by HHH according to
the semantics of the x86 language is stipulated to be correct.

Not a valid stipulation, you can not stipulate something to be "correct".
If you mean to stipulate a MEANING for corect, then you just locked your self out of using "correct" in the normal meaning, and thus locked you self out of proving anything.
Using your "stipulation" of correct, to mean that you are stipulating what will be considered to be a "Correct Simulation" for your previous statement, that make that statement FALSE, because ANY DDD that calls an HHH that will simulate for just N instructions and then return will be HALTING by simple inspection.
Yes, the simulation doesn't reach the final state, but partial simulation don't show what happens as a final state of the machine (or if it reaches on).
You keep repeating this error, proving that you seem to have a learning disability that prevents you from understanding the true meaning of words.
If you want to change your statement to be an actruism, then you need to say something like:  "It is true that no partial or complete correct simulaiton of DDD by HHH will ever reach the return instruction of DDD".
The key is you need simulation, not DDD to be the subject of the verb reach, so that is what you are actually talking about, and what reaches (or not) the final state.

 *YOU JUST DON'T GET THIS EITHER*
A correct simulation of N instructions of DDD by HHH is
sufficient to correctly predict the behavior of an unlimited
simulation.

Nopee, proven otherwise.

 *YOU JUST DON'T GET THIS EITHER*
Termination analyzers / halt deciders are only required
to correctly predict the behavior of their inputs.

Which is the behavior of the PROGRAM represented by the input, which is the behavior of the ACTUAL EXECUTION of that program.
Definitions, you know.

 *MOST JUST DON'T GET THIS*
Termination analyzers / halt deciders are only required
to correctly predict the behavior of their inputs, thus
the behavior of non-inputs is outside of their domain.
 

Right, but the behavior of the PROGRAM DDD, *IS* the behavior of the input.
You apparently just don't understand the basic definitins, because you CHOSE to not learn them.