From Newsgroup: comp.theory

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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From Newsgroup: comp.theory

On 8/20/2025 11:10 PM, olcott wrote:

Simulating (at least partial) halt decider Ĥ.embedded_H

False. The proof starts with the assumption that H is a total halt decider:

We assume the contrary, namely that there exists an algorithm,
and consequently some Turing machine H, that solves the halting problem

From there, through a series of truth-preserving operations, it is
concluded that it is possible to construct Ĥ from total halt decider H
to form a contradiction.
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From Newsgroup: comp.theory

On 8/20/25 11:10 PM, olcott wrote:

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.

The problem is the state isn't infinitely repeating if the ONE H thinks
it sees it.

Your problem is you lie to yourself as to what a program is.

*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.∞,

But only *IF* Ĥ.q0 ⟨Ĥ⟩ will halt, which it doesn't

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

But only **F* Ĥ.q0 ⟨Ĥ⟩ will run forever, which it doesn't.

Therefore, you can't make an H / Ĥ.embedded_H that meets the requirements.

(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⟩.

Not "correctly". as if it transitions to Ĥ.qn, then the input is halting
and didn't HAVE infinitely repeating state.

Turing machine deciders only compute the mapping
from their inputs...

But are required to compute the correct answer.

The the answer they compute doesn't match that, they are just wrong.

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.

Nope, just because you can't put together a logical argument doesn't
releive you of the requirement to actually prove your statement.

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.

And it isn't q quesiton about knowkedge, but of fact.

*DOES* the program described by the input halt when run.

THAT is the question.

All you are doing is PROVING that you don't understand between facts and knowledge, or the meaning of many other words you use, and you think
your stupidity is grounds to allow you to lie.

Sorry, all you have done is sealed you fate for eternity.

It seems your personal mental state prevents you from understanding the
nature of the things you want to talk about.
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