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

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

It is the case that the copy of H (called embedded_H) at Ĥ.qx must abort
the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never stop running.

A simulating halt decider simulates N steps of its input until its input
halts on its own or it correctly determines that a pure simulation of
its input would never stop running.

https://www.liarparadox.org/Peter_Linz_HP_315-320.pdf
Linz, Peter 1990. An Introduction to Formal Languages and Automata. Lexington/Toronto: D. C. Heath and Company. (315-320)

--
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 1/13/2022 2:33 PM, olcott wrote:

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

It is the case that the copy of H (called embedded_H) at Ĥ.qx must abort
the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never stop running.

A simulating halt decider simulates N steps of its input until its input halts on its own or it correctly determines that a pure simulation of
its input would never stop running.

https://www.liarparadox.org/Peter_Linz_HP_315-320.pdf
Linz, Peter 1990. An Introduction to Formal Languages and Automata. Lexington/Toronto: D. C. Heath and Company. (315-320)

dbush said:
@polcott There is a point of agreement, namely that you've successfully
shown that a simulating H cannot simulate the P built from it to
completion. Your confusion is that you think this is the same as
not-halting. By definition, it is not.

It must be a point of mutual agreement or it does not count as a point
of agreement.

I have successfully shown that sequences of configurations that never
reach the final state of this sequence can be recognized as meeting the
Linz definition of non-halting computations.


--
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 1/13/2022 2:33 PM, olcott wrote:

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

It is the case that the copy of H (called embedded_H) at Ĥ.qx must abort
the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never stop running.

A simulating halt decider simulates N steps of its input until its input halts on its own or it correctly determines that a pure simulation of
its input would never stop running.

https://www.liarparadox.org/Peter_Linz_HP_315-320.pdf
Linz, Peter 1990. An Introduction to Formal Languages and Automata. Lexington/Toronto: D. C. Heath and Company. (315-320)

@polcott There is a point of agreement, namely that you've successfully
shown that a simulating H cannot simulate the P built from it to
completion. Your confusion is that you think this is the same as
not-halting. By definition, it is not. – dbush

Proof that my reviewers are not interested in an honest dialogue:
(1) There are never any points of mutual agreement.
(2) Whenever I prove my point reviewers always change the subject rather
than acknowledge that I proved my point.

A simulating halt decider continues to simulate N steps of its input
input until (a) The simulation ends on its own after reaching the final
state specified by the input. (b) An infinite execution pattern has been recognized.

It is ridiculously stupid to claim that the simulating halt decider
fails on the basis that it cannot simulate an input that specifies
infinite execution to completion.

H cannot simulate P to completion because P has no completion. All the
HP counter-example instance specify infinitely nested simulation to
every simulating halt decider.

--
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 1/14/2022 5:53 PM, Alex C wrote:

On Friday, January 14, 2022 at 7:49:26 PM UTC+1, olcott wrote:

On 1/14/2022 12:11 PM, Alex C wrote:

On Friday, January 14, 2022 at 5:59:13 PM UTC+1, olcott wrote:

On 1/14/2022 10:36 AM, Alex C wrote:

On Thursday, January 13, 2022 at 9:34:08 PM UTC+1, olcott wrote:

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

It is the case that the copy of H (called embedded_H) at Ĥ.qx must abort
the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never stop running.

A simulating halt decider simulates N steps of its input until its input >>>>>> halts on its own or it correctly determines that a pure simulation of >>>>>> its input would never stop running.

https://www.liarparadox.org/Peter_Linz_HP_315-320.pdf
Linz, Peter 1990. An Introduction to Formal Languages and Automata. >>>>>> Lexington/Toronto: D. C. Heath and Company. (315-320)

--
Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer

A Turing machine applied to some input reaches one final state or one infinite loop, not two different ones. This is a consequence of the definition of determinism.

In the case of Linz Ĥ applied to ⟨Ĥ⟩, embedded_H at Ĥ.qx correctly >>>> transitions to its final state of Ĥ.qn on the basis that a mathematical >>>> mapping between ⟨Ĥ⟩ ⟨Ĥ⟩ and Ĥ.qn is proven. It is proven on the basis
that the input ⟨Ĥ⟩ ⟨Ĥ⟩ specifies infinitely nested simulation to the
simulating halt decider of embedded_H.
--
Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer

Ĥ.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ??
is a question. A question that has only one answer. We can find the answer by passing the arguments ⟨Ĥ.q0⟩ and ⟨Ĥ⟩ to a UTM. Otherwise it is like saying:

2+2 = 4
2+2 = 5

That is not the question and you know it.
I can't really understand your motivation to lie.

This is the actual question: embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢*
and the answer is: Ĥ.qn.
--
Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer

if that is the question then why did you post the nonsense about Ĥ.q0 ⟨Ĥ⟩

If you are user253751 then you already know the full background.
Otherwise you need to study the 6 pages of Linz text on this link: https://www.liarparadox.org/Peter_Linz_HP_315-320.pdf

This is a simplification of the Linz syntax at the bottom of page 319
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

The copy of Linz H that is embedded at Ĥ.qx is called embedded_H.


--
Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

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

On 1/14/2022 5:53 PM, Alex C wrote:

On Friday, January 14, 2022 at 7:49:26 PM UTC+1, olcott wrote:

On 1/14/2022 12:11 PM, Alex C wrote:

On Friday, January 14, 2022 at 5:59:13 PM UTC+1, olcott wrote:

On 1/14/2022 10:36 AM, Alex C wrote:

On Thursday, January 13, 2022 at 9:34:08 PM UTC+1, olcott wrote:

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

It is the case that the copy of H (called embedded_H) at Ĥ.qx must abort
the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never stop running.

A simulating halt decider simulates N steps of its input until its input >>>>>> halts on its own or it correctly determines that a pure simulation of >>>>>> its input would never stop running.

https://www.liarparadox.org/Peter_Linz_HP_315-320.pdf
Linz, Peter 1990. An Introduction to Formal Languages and Automata. >>>>>> Lexington/Toronto: D. C. Heath and Company. (315-320)

--
Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer

A Turing machine applied to some input reaches one final state or one infinite loop, not two different ones. This is a consequence of the definition of determinism.

In the case of Linz Ĥ applied to ⟨Ĥ⟩, embedded_H at Ĥ.qx correctly >>>> transitions to its final state of Ĥ.qn on the basis that a mathematical >>>> mapping between ⟨Ĥ⟩ ⟨Ĥ⟩ and Ĥ.qn is proven. It is proven on the basis
that the input ⟨Ĥ⟩ ⟨Ĥ⟩ specifies infinitely nested simulation to the
simulating halt decider of embedded_H.
--
Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer

Ĥ.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ??
is a question. A question that has only one answer. We can find the answer by passing the arguments ⟨Ĥ.q0⟩ and ⟨Ĥ⟩ to a UTM. Otherwise it is like saying:

2+2 = 4
2+2 = 5

That is not the question and you know it.
I can't really understand your motivation to lie.

This is the actual question: embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢*
and the answer is: Ĥ.qn.
--
Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer

if that is the question then why did you post the nonsense about Ĥ.q0 ⟨Ĥ⟩

If you are user253751 then you already know the full background.
Otherwise you need to study the 6 pages of Linz text on this link: https://www.liarparadox.org/Peter_Linz_HP_315-320.pdf

This is a simplification of the Linz syntax at the bottom of page 319
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

The copy of Linz H that is embedded at Ĥ.qx is called embedded_H.


--
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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