Re: Defining a correct simulating halt decider

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Sujet : Re: Defining a correct simulating halt decider
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theory
Date : 03. Sep 2024, 00:45:46
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <04873adef20ff2198360813a66325ca855af089f@i2pn2.org>
References : 1 2
User-Agent : Mozilla Thunderbird
On 9/2/24 7:11 PM, olcott wrote:
On 9/2/2024 11:38 AM, olcott wrote:
A halt decider is a Turing machine that computes
the mapping from its finite string input to the
behavior that this finite string specifies.
>
If the finite string machine string machine
description specifies that it cannot possibly
reach its own final halt state then this machine
description specifies non-halting behavior.
>
A halt decider never ever computes the mapping
for the computation that itself is contained within.
>
Unless there is a pathological relationship between
the halt decider H and its input D the direct execution
of this input D will always have identical behavior to
D correctly simulated by simulating halt decider H.
>
*Simulating Termination Analyzer H Not Fooled by Pathological Input D*
https://www.researchgate.net/ publication/369971402_Simulating_Termination_Analyzer_H_is_Not_Fooled_by_Pathological_Input_D
>
A correct emulation of DDD by HHH only requires that HHH
emulate the instructions of DDD** including when DDD calls
HHH in recursive emulation such that HHH emulates itself
emulating DDD.
>
 ** According to the semantics of the x86 language.
 This prevents the correctly emulated** DDD from ever
reaching its final halt state no matter what HHH does.
 
Of course  not, since you have defined that your HHH actually DOES abort to return an answer, the CORRECT x86 emulation of the input DDD (which isn't what HHH does) steps through the instructions of HHH until it reaches the point where it aborts its emulation and return to DDD that returns.
WHat doesn't get there is the PARTIAL (and thus not correct) emulation done by your HHH.
You can only make your statement work if you stipulate that the HHH you have shown isn't the HHH that we have, but we have instead the HHH that never aborts, but then that HHH just fails to be a decider so is wrong also,
You don't get to have two different HHHs in the problem. That is just your LIE.
Sorry, you are just proving your utter stupidity and ignorance about what you are talking, and showing how you are just a pathological liar.

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4 Jul 25 o 

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