From Newsgroup: comp.ai.philosophy
Simulating (at least partial) halt decider Ĥ.embedded_H
either sees the repeating state of its input or not.
If it cannot possibly see the repeating state of its
input then we do know more about the halting problem
proof than we ever knew before, that the counter-example
input would be correctly decided as non-halting. We can
see that it is non-halting even if Ĥ.embedded_H cannot.
*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
(a) Ĥ copies its input ⟨Ĥ⟩
(b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
If Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ can see the repeating state then
it will abort its simulation and correctly transition
to Ĥ.qn on the basis that ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly reach
its own simulated final halt state of ⟨Ĥ.qn⟩.
Turing machine deciders only compute the mapping
from their inputs...
Thus the behavior of the input would overrule the
behavior of Ĥ applied to ⟨Ĥ⟩ in this case. If this
behavior cannot be overruled then that would seem
to indicate that embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly
see the repeating state.
I can't know that Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly
see the repeating state of its input until trying
every category of possible ways and failing.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
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