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computers / comp.ai.philosophy / Re: Halting theorem refutation (V3)

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o Re: Halting theorem refutation (V3)olcott

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Re: Halting theorem refutation (V3)

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From: NoO...@NoWhere.com (olcott)
Subject: Re: Halting theorem refutation (V3)
Date: Thu, 20 May 2021 15:14:28 -0500
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 by: olcott - Thu, 20 May 2021 20:14 UTC

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn

The above is adapted from (Linz:1990:319).
It shows that Turing machine Ĥ copies its input at (q0) and begins
executing an embedded copy of the original halt decider with this input
at (qx).

The (qy) state indicates that the halt decider has determined that its
input would halt.

The appended (qa) and (qb) states cause Ĥ to infinitely loop if the halt
decider has determined that its input would halt.

The ((qn)) state indicates that the halt decider has decided that its
input would not halt.

Linz, Peter 1990. An Introduction to Formal Languages and Automata.
Lexington/Toronto: D. C. Heath and Company.

This halt deciding principle overcomes the pathological self-reference
error of the halting theorem:

(A) Every simulation of input P that never halts unless simulating halt
decider H stops simulating it <is> a non-halting computation. This
remains true even after H stops simulating P.

Any high school student would be able to analyze the basic logic of the
above and know that it is correct very easily.

*Basic logic of the above*
If X is the only cause of Y and not X then not Y.

X = H stops simulating P
Y = The simulation of P stops

~X = H never stops simulating P
~Y = The simulation of P never stops

(B) The simulation of: ([Ĥ][Ĥ]) by the simulating halt decider @ Ĥ.qx
never halts unless Ĥ.qx aborts this simulation.

(C) ∴ The simulating halt decider @ Ĥ.qx correctly decides its input
([Ĥ][Ĥ]) as non-halting.

http://www.liarparadox.org/Halting_problem_undecidability_and_infinitely_nested_simulation.pdf

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

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