XPost: comp.theory, sci.logic, sci.math

Does the Linz Ĥ applied to ⟨Ĥ⟩ correctly transition to its final reject state?

Let ⟨M⟩ describe a Turing machine M = (Q, Σ, Γ, δ, q₀, □, F), and let w
be any element of Σ⁺, A solution of the halting problem is a Turing
machine H, which for any ⟨M⟩ and w, performs the computation (Linz 1990:317)

H.q0 ⟨M⟩ w ⊢* H.qy ----- iff UTM( ⟨M⟩, w ) reaches the final state of M
H.q0 ⟨M⟩ w ⊢* H.qn ----- iff UTM( ⟨M⟩, w ) would never reach the final
state of M

RHS is a paraphrase of Ben Bacarisse encoding of my halt status
criterion measure.

The above criteria correctly determines the halt status of inputs with pathological self-reference (Olcott 2004). Simulating halt decider H
performs a pure simulation of its input as if it was a UTM unless and
until it detects an infinitely repeating pattern. Then it aborts the
simulation of its input and transitions to its final reject state.

The following simplifies the syntax for the definition of the Linz
Turing machine Ĥ, it is now a single machine with a single start state.
A copy of Linz H is embedded at Ĥ.qx.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Can the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H possibly transition
to ⟨Ĥ⟩.qn ?

Linz Ĥ applied to ⟨Ĥ⟩ does correctly transition to its final reject
state of Ĥ.qn because the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H
would never reach its own final state of ⟨Ĥ⟩.qn

When Ĥ is applied to ⟨Ĥ⟩
Ĥ copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then embedded_H simulates ⟨Ĥ1⟩ ⟨Ĥ2⟩

Then these steps would keep repeating:
Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩
Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates ⟨Ĥ4⟩ ⟨Ĥ5⟩...

The above repeating pattern shows that the correctly simulated input to embedded_H would never reach its final state of ⟨Ĥ⟩.qn conclusively proving that this simulated input never halts. This enables embedded_H
to abort its simulation and correctly transition to Ĥ.qn.

Because all simulating halt deciders are deciders they are only
accountable for computing the mapping from their input finite strings to
an accept or reject state on the basis of whether or not their correctly simulated input could ever reach its final state: ⟨Ĥ⟩⟨Ĥ⟩ ⊢* ⟨Ĥ⟩.qn.

embedded_H is only accountable for the behavior of its correctly
simulated input: ⟨Ĥ⟩ ⟨Ĥ⟩. embedded_H is not accountable for the behavior
of the computation that it is contained within: Ĥ applied to ⟨Ĥ⟩ because is it not actual input to embedded_H.



Halting problem undecidability and infinitely nested simulation (V3)

https://www.researchgate.net/publication/358009319_Halting_problem_undecidability_and_infinitely_nested_simulation_V3

--
Copyright 2021 Pete 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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XPost: comp.theory, sci.logic, sci.math

On 2/17/2022 8:50 PM, Ben Bacarisse wrote:

olcott <NoOne@NoWhere.com> writes:

H.q0 ⟨M⟩ w ⊢* H.qy ----- iff UTM( ⟨M⟩, w ) reaches the final state of M
H.q0 ⟨M⟩ w ⊢* H.qn ----- iff UTM( ⟨M⟩, w ) would never reach the final state of M

RHS is a paraphrase of Ben Bacarisse encoding of my halt status
criterion measure.

I never encoded your "halt status criterion measure" so you can not be paraphrasing such a thing here. I re-wrote Linz's condition in terms of
a UTM.

That is exactly the same thing that I am referring to.
That was a big help thanks.

Your "halt status criterion measure" has 'reject' (or false) as the
correct answer for at least one halting computation so you are not
talking about the halting problem.

My whole point is that the embedded copy of H in Ĥ now correctly
determines that halt status of Linz ⟨Ĥ⟩ ⟨Ĥ⟩ thus refuting all the conventional proofs as well as all proofs that can be reduced to the conventional proofs.

https://math.stackexchange.com/questions/749987/reducing-a-decidability-problem-to-the-halting-problem

This is completely rewritten today and has all my new material.

https://www.researchgate.net/publication/358009319_Halting_problem_undecidability_and_infinitely_nested_simulation_V3

Let ⟨M⟩ describe a Turing machine M = (Q, Σ, Γ, δ, q₀, □, F), and let
w be any element of Σ⁺, A solution of the halting problem is a Turing
machine H, which for any ⟨M⟩ and w, performs the computation (Linz
1990:317) ...

Neither are you are not talking about Turing machines because you assert
that the tape and the state transition function don't determine the transitions a machine makes.

I have no idea what you are talking about.

A simulating halt decider computes the mapping from a finite string
pair** to an accept or reject state entirely on the basis of the whether
or not its pure simulation of this pair would reach its own final state.

** (Turing machine description / finite string)

Unless you correct those huge mistakes you have nothing to say about the halting problem for Turing machines. You are discussing some other
problem about magic Olcott machines.

https://www.researchgate.net/publication/358009319_Halting_problem_undecidability_and_infinitely_nested_simulation_V3


--
Copyright 2021 Pete 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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