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On 9/2/2024 11:38 AM, olcott wrote: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.A halt decider is a Turing machine that computes** According to the semantics of the x86 language.
the mapping from its finite string input to the
behavior that this finite string specifies.
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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.
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A halt decider never ever computes the mapping
for the computation that itself is contained within.
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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.
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*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
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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.
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This prevents the correctly emulated** DDD from ever
reaching its final halt state no matter what HHH does.
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