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devel / comp.theory / Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs Count ]

SubjectAuthor
* Black box halt decider is NOT a partial deciderMr Flibble
`* Black box halt decider is NOT a partial deciderChris M. Thomasson
 `* Black box halt decider is NOT a partial deciderDavid Brown
  `* Black box halt decider is NOT a partial deciderChris M. Thomasson
   +* Black box halt decider is NOT a partial deciderRichard Damon
   |`* Black box halt decider is NOT a partial deciderChris M. Thomasson
   | `* Black box halt decider is NOT a partial deciderRichard Damon
   |  `* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   +- Black box halt decider is NOT a partial deciderRichard Damon
   |   +* Black box halt decider is NOT a partial deciderBen Bacarisse
   |   |`* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | +* Black box halt decider is NOT a partial deciderBen Bacarisse
   |   | |`* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | | +- Black box halt decider is NOT a partial deciderRichard Damon
   |   | | `* Black box halt decider is NOT a partial deciderBen Bacarisse
   |   | |  `* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   +* Black box halt decider is NOT a partial deciderAndré G. Isaak
   |   | |   |`* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   | `* Black box halt decider is NOT a partial deciderMike Terry
   |   | |   |  `* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   |   `* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   |    +- Black box halt decider is NOT a partial deciderMike Terry
   |   | |   |    +* Black box halt decider is NOT a partial deciderBen Bacarisse
   |   | |   |    |+* Black box halt decider is NOT a partial deciderJeff Barnett
   |   | |   |    ||+- Black box halt decider is NOT a partial deciderJeff Barnett
   |   | |   |    ||`* Black box halt decider is NOT a partial deciderMike Terry
   |   | |   |    || +- Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   |    || `* Black box halt decider is NOT a partial deciderJeff Barnett
   |   | |   |    ||  `- Black box halt decider is NOT a partial deciderMike Terry
   |   | |   |    |`* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   |    | `* Black box halt decider is NOT a partial deciderBen Bacarisse
   |   | |   |    |  `* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   |    |   +- Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   |    |   `* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   |    |    `- Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   |    `- Black box halt decider is NOT a partial deciderwij
   |   | |   +* Black box halt decider is NOT a partial deciderRichard Damon
   |   | |   |`* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   | `* Black box halt decider is NOT a partial deciderRichard Damon
   |   | |   |  `- Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   `* Black box halt decider is NOT a partial deciderBen Bacarisse
   |   | |    +* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |    |`* Black box halt decider is NOT a partial deciderBen Bacarisse
   |   | |    | `* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |    |  `* Black box halt decider is NOT a partial deciderRichard Damon
   |   | |    |   `- Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |    `* Black box halt decider is NOT a partial deciderAndré G. Isaak
   |   | |     +* Black box halt decider is NOT a partial deciderBen Bacarisse
   |   | |     |+- Black box halt decider is NOT a partial deciderAndré G. Isaak
   |   | |     |`* Black box halt decider is NOT a partial deciderMike Terry
   |   | |     | +* Black box halt decider is NOT a partial deciderBen Bacarisse
   |   | |     | |+* Black box halt decider is NOT a partial deciderAndy Walker
   |   | |     | ||`* Black box halt decider is NOT a partial deciderMike Terry
   |   | |     | || +* Black box halt decider is NOT a partial deciderMalcolm McLean
   |   | |     | || |+* Black box halt decider is NOT a partial decider [ H(P,P)==0 is always correct ]olcott
   |   | |     | || ||`- Black box halt decider is NOT a partial decider [ H(P,P)==0 isRichard Damon
   |   | |     | || |+* Black box halt decider is NOT a partial decider [ H(P,P)==0 is always correct ]olcott
   |   | |     | || ||+- Black box halt decider is NOT a partial decider [ H(P,P)==0 isAndré G. Isaak
   |   | |     | || ||+* Black box halt decider is NOT a partial decider [ H(P,P)==0 isRichard Damon
   |   | |     | || |||`* Black box halt decider is NOT a partial decider [ H(P,P)==0 isMalcolm McLean
   |   | |     | || ||| `* Black box halt decider is NOT a partial decider [ H(P,P)==0 isRichard Damon
   |   | |     | || |||  `- Black box halt decider is NOT a partial decider [ H(P,P)==0 isJeff Barnett
   |   | |     | || ||`- Black box halt decider is NOT a partial decider [ H(P,P)==0 is always correct ]Ben Bacarisse
   |   | |     | || |+* Black box halt decider is NOT a partial deciderBen Bacarisse
   |   | |     | || ||`* Black box halt decider is NOT a partial deciderMalcolm McLean
   |   | |     | || || `* Black box halt decider is NOT a partial decider [ paradox ratherolcott
   |   | |     | || ||  +- Black box halt decider is NOT a partial decider [ paradox ratherRichard Damon
   |   | |     | || ||  `* Black box halt decider is NOT a partial decider [ paradox ratherAndré G. Isaak
   |   | |     | || ||   `* Black box halt decider is NOT a partial decider [ H refutes Rice's Theorem ]olcott
   |   | |     | || ||    +- Black box halt decider is NOT a partial decider [ H refutesRichard Damon
   |   | |     | || ||    `* Black box halt decider is NOT a partial decider [ H refutesAndré G. Isaak
   |   | |     | || ||     `* Black box halt decider is NOT a partial decider [ H refutes Rice's Theorem ]olcott
   |   | |     | || ||      +* Black box halt decider is NOT a partial decider [ H refutesAndré G. Isaak
   |   | |     | || ||      |`* Black box halt decider is NOT a partial decider [ H refutesolcott
   |   | |     | || ||      | `- Black box halt decider is NOT a partial decider [ H refutesRichard Damon
   |   | |     | || ||      `* Black box halt decider is NOT a partial decider [ H refutesJeff Barnett
   |   | |     | || ||       `* Black box halt decider is NOT a partial decider [ H refutesolcott
   |   | |     | || ||        `* Black box halt decider is NOT a partial decider [ H refutesAndré G. Isaak
   |   | |     | || ||         +* Black box halt decider is NOT a partial decider [ H refutesolcott
   |   | |     | || ||         |+- Black box halt decider is NOT a partial decider [ H refutesAndré G. Isaak
   |   | |     | || ||         |`- Black box halt decider is NOT a partial decider [ H refutesRichard Damon
   |   | |     | || ||         `* Black box halt decider is NOT a partial decider [ H refutesolcott
   |   | |     | || ||          +* Black box halt decider is NOT a partial decider [ H refutesAndré G. Isaak
   |   | |     | || ||          |`* Black box halt decider is NOT a partial decider [ H refutes Rice's Theorem ]olcott
   |   | |     | || ||          | `* Black box halt decider is NOT a partial decider [ H refutesAndré G. Isaak
   |   | |     | || ||          |  `* Black box halt decider is NOT a partial decider [ H refutesolcott
   |   | |     | || ||          |   +- Black box halt decider is NOT a partial decider [ H refutesAndré G. Isaak
   |   | |     | || ||          |   +- Black box halt decider is NOT a partial decider [ H refutesRichard Damon
   |   | |     | || ||          |   `* _Black_box_halt_decider_is_NOT_a_partial_decider_[_André_doesn't_know_Rice's_Theolcott
   |   | |     | || ||          |    +* _Black_box_halt_decider_is_NOT_a_partial_decider_[André G. Isaak
   |   | |     | || ||          |    |`* _Black_box_halt_decider_is_NOT_a_partial_decider_[olcott
   |   | |     | || ||          |    | +* _Black_box_halt_decider_is_NOT_a_partial_decider_[André G. Isaak
   |   | |     | || ||          |    | |`* _Black_box_halt_decider_is_NOT_a_partial_decider_Malcolm McLean
   |   | |     | || ||          |    | | `* _André_doesn't_know_Rice's_Theorem_[_Malcolm_]olcott
   |   | |     | || ||          |    | |  +* _André_doesn't_know_Rice's_Theorem_[_MalcRichard Damon
   |   | |     | || ||          |    | |  |`* _André_doesn't_know_Rice's_Theorem_[_Malcolcott
   |   | |     | || ||          |    | |  | `* _André_doesn't_know_Rice's_Theorem_[_MalcRichard Damon
   |   | |     | || ||          |    | |  |  `* _André_doesn't_know_Rice's_Theorem_[_Malcolm_](_attention_deficit_disorder_)olcott
   |   | |     | || ||          |    | |  |   `* _André_doesn't_know_Rice's_Theorem_[_MalcRichard Damon
   |   | |     | || ||          |    | |  |    `* _André_doesn't_know_Rice's_Theorem_[_Malcolcott
   |   | |     | || ||          |    | |  |     +- _André_doesn't_know_Rice's_Theorem_[_MalcRichard Damon
   |   | |     | || ||          |    | |  |     +* _André_doesn't_know_Rice's_Theorem_[_Malcolm_](_attention_deficit_disorder_)olcott
   |   | |     | || ||          |    | |  |     `* André doesn't know Rice's Theorem [ MalcolmBen Bacarisse
   |   | |     | || ||          |    | |  +* _André_doesn't_know_Rice's_Theorem_[_MalcAndré G. Isaak
   |   | |     | || ||          |    | |  `- _André_doesn't_know_Rice's_Theorem_[_MalcJeff Barnett
   |   | |     | || ||          |    | +- _Black_box_halt_decider_is_NOT_a_partial_decider_[Richard Damon
   |   | |     | || ||          |    | `* _Black_box_halt_decider_is_NOT_a_partial_decider_[_André_doesn't_know_Rice's_Theolcott
   |   | |     | || ||          |    `- _Black_box_halt_decider_is_NOT_a_partial_decider_[Richard Damon
   |   | |     | || ||          `- Black box halt decider is NOT a partial decider [ H refutesRichard Damon
   |   | |     | || |`* Black box halt decider is NOT a partial deciderMike Terry
   |   | |     | || `- Black box halt decider is NOT a partial deciderAndy Walker
   |   | |     | |`* Black box halt decider is NOT a partial deciderMike Terry
   |   | |     | `* Black box halt decider is NOT a partial deciderwij
   |   | |     `- Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | `* Black box halt decider is NOT a partial deciderRichard Damon
   |   `* Black box halt decider is NOT a partial deciderMalcolm McLean
   `* Black box halt decider is NOT a partial deciderJeff Barnett

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Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] ( Are you game ? )

<sed3c6$pco$1@dont-email.me>

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https://www.novabbs.com/devel/article-flat.php?id=19448&group=comp.theory#19448

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Path: i2pn2.org!i2pn.org!eternal-september.org!reader02.eternal-september.org!.POSTED!not-for-mail
From: agis...@gm.invalid (André G. Isaak)
Newsgroups: comp.theory
Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[
Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] ( Are you g
ame_?_)
Date: Tue, 3 Aug 2021 22:00:05 -0600
Organization: Christians and Atheists United Against Creeping Agnosticism
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 by: André G. Isaak - Wed, 4 Aug 2021 04:00 UTC

On 2021-08-03 21:30, olcott wrote:

> I have proven point-by-point that H(P,P)==0 is correct. That you keep
> changing the subject can't be tolerated within an honest dialogue.

No, you have not. All you've demonstrated is that whatever question it
is that your 'decider' is answering is unrelated to the actual halting
problem and that you don't really understand what that question is.

A halt decider *by definition* must decide whether the *independent*
computation represented by its input halts. The independent computation
P(P) as you have acknowledged, halts. The *only* correct answer a halt
decider can give is the one which actually conforms to the behaviour of
the independent computation described by the input.

André

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs Count ]

<KcOdnSwOTcFujZf8nZ2dnUU7-W_NnZ2d@giganews.com>

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https://www.novabbs.com/devel/article-flat.php?id=19449&group=comp.theory#19449

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Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs_Count_]
Newsgroups: comp.theory
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From: NoO...@NoWhere.com (olcott)
Date: Tue, 3 Aug 2021 23:00:50 -0500
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 by: olcott - Wed, 4 Aug 2021 04:00 UTC

On 8/3/2021 10:56 PM, André G. Isaak wrote:
> On 2021-08-03 21:23, olcott wrote:
>
>> The key point is that only the input to the halt decider is within the
>> scope of the halting problem.
>>
>> As long as the halt decider correctly decides its input what these
>> same programs do what they are not inputs to the halt decider make no
>> difference to the actual halting problem.
>
> You have this completely back-assward.
>
> How the TM described by the input to a halt decider behaves when they
> are run as *independent* computations (i.e. not as inputs to the halt
> decider) is the question a halt decider is supposed to answer by the
> very definition of a 'halt decider'. If the answer provided by the
> decider does not match the behavior of the *independent* computation,
> then that answer is wrong.
>
> Talking about how the description behaves "as an input" isn't even
> coherent. Inputs don't behave like anything. They are just data.
>
> André
>

In computability theory, the halting problem is the
problem of determining,

from a description of an arbitrary computer program and an input,
from a description of an arbitrary computer program and an input,
from a description of an arbitrary computer program and an input,
from a description of an arbitrary computer program and an input,

whether the program will
finish running, or continue to run forever.
https://en.wikipedia.org/wiki/Halting_problem

I have proved that the input to H(P,P) never halts whether or not it is
aborted by H, are you going to have an honest dialogue about the steps
of the proof or not?

--
Copyright 2021 Pete Olcott

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

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] ( Are you game ? )

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Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[
Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] ( Are you g
ame_?_)
Newsgroups: comp.theory
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From: NoO...@NoWhere.com (olcott)
Date: Tue, 3 Aug 2021 23:03:47 -0500
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 by: olcott - Wed, 4 Aug 2021 04:03 UTC

On 8/3/2021 11:00 PM, André G. Isaak wrote:
> On 2021-08-03 21:30, olcott wrote:
>
>> I have proven point-by-point that H(P,P)==0 is correct. That you keep
>> changing the subject can't be tolerated within an honest dialogue.
>
> No, you have not.

We can either have an honest dialogue about pages 4-6 or I will stop
talking to you.

https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation

> All you've demonstrated is that whatever question it
> is that your 'decider' is answering is unrelated to the actual halting
> problem and that you don't really understand what that question is.
>
> A halt decider *by definition* must decide whether the *independent*
> computation represented by its input halts. The independent computation
> P(P) as you have acknowledged, halts. The *only* correct answer a halt
> decider can give is the one which actually conforms to the behaviour of
> the independent computation described by the input.
>
> André
>

--
Copyright 2021 Pete Olcott

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

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] ( Are you game ? )

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Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[
Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] ( Are you g
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Wed, 4 Aug 2021 04:08 UTC

On 8/3/21 6:24 PM, olcott wrote:
>
> I corrected the Linz notation, Linz has two start states.
> Linz has no Ĥ.qx, that is my correction. It indicates H.q0.

I don't think his notation actually had two start states.

As I remember, he had a q0 as the start start for H^, from which we
duplicated the input and then we hit an internal state called H.q0 which
was where the copy of H that was in H^ began, and then it had copies of
ALL the states in H, ending in H.qn or H.qy, and then H.qy was modified
to form an infinite loop.

Note q0 != H.q0, H.q0 is NOT a 'start' state of H^, only the original
start state of the H machine that had all its states prefix with H. and
included (with qy being to become an infinite loop)

This if the H. as a scope operator on the state names.

Of course, if you recognized that, you would have to admit that H^
included a copy of all of H in it, so H when processing H^ would need to
actually look at those states, so maybe that it the source of your blind
spot.

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs Count ]

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Subject: Re: Black box halt decider is NOT a partial decider
[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs Count ]
From: wyni...@gmail.com (wij)
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 by: wij - Wed, 4 Aug 2021 04:12 UTC

On Wednesday, 4 August 2021 at 12:00:58 UTC+8, olcott wrote:
> On 8/3/2021 10:56 PM, André G. Isaak wrote:
> > On 2021-08-03 21:23, olcott wrote:
> >
> >> The key point is that only the input to the halt decider is within the
> >> scope of the halting problem.
> >>
> >> As long as the halt decider correctly decides its input what these
> >> same programs do what they are not inputs to the halt decider make no
> >> difference to the actual halting problem.
> >
> > You have this completely back-assward.
> >
> > How the TM described by the input to a halt decider behaves when they
> > are run as *independent* computations (i.e. not as inputs to the halt
> > decider) is the question a halt decider is supposed to answer by the
> > very definition of a 'halt decider'. If the answer provided by the
> > decider does not match the behavior of the *independent* computation,
> > then that answer is wrong.
> >
> > Talking about how the description behaves "as an input" isn't even
> > coherent. Inputs don't behave like anything. They are just data.
> >
> > André
> >
> In computability theory, the halting problem is the
> problem of determining,
>
> from a description of an arbitrary computer program and an input,
> from a description of an arbitrary computer program and an input,
> from a description of an arbitrary computer program and an input,
> from a description of an arbitrary computer program and an input,
>
> whether the program will
> finish running, or continue to run forever.
> https://en.wikipedia.org/wiki/Halting_problem
> I have proved that the input to H(P,P) never halts whether or not it is
> aborted by H, are you going to have an honest dialogue about the steps
> of the proof or not?
> --
> Copyright 2021 Pete Olcott
>
> "Great spirits have always encountered violent opposition from mediocre
> minds." Einstein

You are the one being dishonest, lying all the time.
For these month, your paper and your words keep changing. No one can reproduce
any useful thing from your paper or your words.

See GUR example 2:
https://groups.google.com/g/comp.theory/c/_tbCYyMox9M
U (halt decider) can be anything(e.g. Computer-God). No doubt CG can prove P halts or not
(Human could probably prove P halts or not, too).
BUT, if P can use the result, then the problem becomes undecidable.

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs Count ]

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Newsgroups: comp.theory
Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[
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 by: André G. Isaak - Wed, 4 Aug 2021 04:16 UTC

On 2021-08-03 22:00, olcott wrote:
> On 8/3/2021 10:56 PM, André G. Isaak wrote:
>> On 2021-08-03 21:23, olcott wrote:
>>
>>> The key point is that only the input to the halt decider is within
>>> the scope of the halting problem.
>>>
>>> As long as the halt decider correctly decides its input what these
>>> same programs do what they are not inputs to the halt decider make no
>>> difference to the actual halting problem.
>>
>> You have this completely back-assward.
>>
>> How the TM described by the input to a halt decider behaves when they
>> are run as *independent* computations (i.e. not as inputs to the halt
>> decider) is the question a halt decider is supposed to answer by the
>> very definition of a 'halt decider'. If the answer provided by the
>> decider does not match the behavior of the *independent* computation,
>> then that answer is wrong.
>>
>> Talking about how the description behaves "as an input" isn't even
>> coherent. Inputs don't behave like anything. They are just data.
>>
>> André
>>
>
>    In computability theory, the halting problem is the
>    problem of determining,
>
> from a description of an arbitrary computer program and an input,
> from a description of an arbitrary computer program and an input,
> from a description of an arbitrary computer program and an input,
> from a description of an arbitrary computer program and an input,
>
>    whether the program will
>    finish running, or continue to run forever.
>    https://en.wikipedia.org/wiki/Halting_problem
>
> I have proved that the input to H(P,P) never halts whether or not it is
> aborted by H, are you going to have an honest dialogue about the steps
> of the proof or not?

Yes, and when run H(P, P) the arbitrary program and input which it is
being given described P(P).

It is supposed to answer how that *program* will behave. And you have
acknowledged that that program *halts*. That behaviour is what *defines*
the correct answer to the question 'does P(P) halt'. And your H gets
this *wrong*.

The halting question is not concerned with how the input string behaves
inside your partial simulator. It is *only* concerned with the actual,
*independent* program P(P).

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs Count ]

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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Wed, 4 Aug 2021 04:25 UTC

On 8/3/2021 11:16 PM, André G. Isaak wrote:
> On 2021-08-03 22:00, olcott wrote:
>> On 8/3/2021 10:56 PM, André G. Isaak wrote:
>>> On 2021-08-03 21:23, olcott wrote:
>>>
>>>> The key point is that only the input to the halt decider is within
>>>> the scope of the halting problem.
>>>>
>>>> As long as the halt decider correctly decides its input what these
>>>> same programs do what they are not inputs to the halt decider make
>>>> no difference to the actual halting problem.
>>>
>>> You have this completely back-assward.
>>>
>>> How the TM described by the input to a halt decider behaves when they
>>> are run as *independent* computations (i.e. not as inputs to the halt
>>> decider) is the question a halt decider is supposed to answer by the
>>> very definition of a 'halt decider'. If the answer provided by the
>>> decider does not match the behavior of the *independent* computation,
>>> then that answer is wrong.
>>>
>>> Talking about how the description behaves "as an input" isn't even
>>> coherent. Inputs don't behave like anything. They are just data.
>>>
>>> André
>>>
>>
>>     In computability theory, the halting problem is the
>>     problem of determining,
>>
>> from a description of an arbitrary computer program and an input,
>> from a description of an arbitrary computer program and an input,
>> from a description of an arbitrary computer program and an input,
>> from a description of an arbitrary computer program and an input,
>>
>>     whether the program will
>>     finish running, or continue to run forever.
>>     https://en.wikipedia.org/wiki/Halting_problem
>>
>> I have proved that the input to H(P,P) never halts whether or not it
>> is aborted by H, are you going to have an honest dialogue about the
>> steps of the proof or not?
>
> Yes, and when run H(P, P) the arbitrary program and input which it is
> being given described P(P).
>
> It is supposed to answer how that *program* will behave. And you have
> acknowledged that that program *halts*. That behaviour is what *defines*
> the correct answer to the question 'does P(P) halt'. And your H gets
> this *wrong*.
>
> The halting question is not concerned with how the input string behaves
> inside your partial simulator. It is *only* concerned with the actual,
> *independent* program P(P).
>

So in other words you insist on not going through my steps (a)(b)(c)(d)
on page 6 and trying to find an error?

https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation

--
Copyright 2021 Pete Olcott

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

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs Count ]

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Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[
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Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc> <87pmv4ab6r.fsf@bsb.me.uk>
<JNadnQD-Ofr-SJz8nZ2dnUU7-XHNnZ2d@giganews.com> <871r7i6n2u.fsf@bsb.me.uk>
<OqKdnROLKJ9CdJz8nZ2dnUU7-avNnZ2d@giganews.com> <87k0la542c.fsf@bsb.me.uk>
<1NidnVPZ-NHDl5_8nZ2dnUU7-enNnZ2d@giganews.com> <87sfzw3ao1.fsf@bsb.me.uk>
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Wed, 4 Aug 2021 04:35 UTC

On 8/3/21 9:00 PM, olcott wrote:
> On 8/3/2021 10:56 PM, André G. Isaak wrote:
>> On 2021-08-03 21:23, olcott wrote:
>>
>>> The key point is that only the input to the halt decider is within
>>> the scope of the halting problem.
>>>
>>> As long as the halt decider correctly decides its input what these
>>> same programs do what they are not inputs to the halt decider make no
>>> difference to the actual halting problem.
>>
>> You have this completely back-assward.
>>
>> How the TM described by the input to a halt decider behaves when they
>> are run as *independent* computations (i.e. not as inputs to the halt
>> decider) is the question a halt decider is supposed to answer by the
>> very definition of a 'halt decider'. If the answer provided by the
>> decider does not match the behavior of the *independent* computation,
>> then that answer is wrong.
>>
>> Talking about how the description behaves "as an input" isn't even
>> coherent. Inputs don't behave like anything. They are just data.
>>
>> André
>>
>
>    In computability theory, the halting problem is the
>    problem of determining,
>
> from a description of an arbitrary computer program and an input,
> from a description of an arbitrary computer program and an input,
> from a description of an arbitrary computer program and an input,
> from a description of an arbitrary computer program and an input,
>
>    whether the program will
>    finish running, or continue to run forever.
>    https://en.wikipedia.org/wiki/Halting_problem
>
> I have proved that the input to H(P,P) never halts whether or not it is
> aborted by H, are you going to have an honest dialogue about the steps
> of the proof or not?
>
>

You just don't understand thw words do you.

H is given on its tape, the description of the machine P and its input I.

The job of H is from this description, which has been given to it on its
(H's) tape, determine whether the program (ie P) and its input (ie I)
when run will finish running or not.

ie, will P(I) Halt when run of run forever.

H's 'input' never does anything, it is just data. The job of H was to
determine what the machine which that input describes (ie P(I) ) will do.

I.e. H(H^,H^) needs to answer what H^(H^) does.

H^(H^) has been shown to Halt, and H(H^,H^) says it doesn't Halt.

Halt != Doesn't Halt, so H was WRONG.

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs Count ]

<sed5fo$31j$1@dont-email.me>

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From: agis...@gm.invalid (André G. Isaak)
Newsgroups: comp.theory
Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[
Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inpu
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Date: Tue, 3 Aug 2021 22:36:07 -0600
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 by: André G. Isaak - Wed, 4 Aug 2021 04:36 UTC

On 2021-08-03 22:25, olcott wrote:
> On 8/3/2021 11:16 PM, André G. Isaak wrote:
>> On 2021-08-03 22:00, olcott wrote:
>>> On 8/3/2021 10:56 PM, André G. Isaak wrote:
>>>> On 2021-08-03 21:23, olcott wrote:
>>>>
>>>>> The key point is that only the input to the halt decider is within
>>>>> the scope of the halting problem.
>>>>>
>>>>> As long as the halt decider correctly decides its input what these
>>>>> same programs do what they are not inputs to the halt decider make
>>>>> no difference to the actual halting problem.
>>>>
>>>> You have this completely back-assward.
>>>>
>>>> How the TM described by the input to a halt decider behaves when
>>>> they are run as *independent* computations (i.e. not as inputs to
>>>> the halt decider) is the question a halt decider is supposed to
>>>> answer by the very definition of a 'halt decider'. If the answer
>>>> provided by the decider does not match the behavior of the
>>>> *independent* computation, then that answer is wrong.
>>>>
>>>> Talking about how the description behaves "as an input" isn't even
>>>> coherent. Inputs don't behave like anything. They are just data.
>>>>
>>>> André
>>>>
>>>
>>>     In computability theory, the halting problem is the
>>>     problem of determining,
>>>
>>> from a description of an arbitrary computer program and an input,
>>> from a description of an arbitrary computer program and an input,
>>> from a description of an arbitrary computer program and an input,
>>> from a description of an arbitrary computer program and an input,
>>>
>>>     whether the program will
>>>     finish running, or continue to run forever.
>>>     https://en.wikipedia.org/wiki/Halting_problem
>>>
>>> I have proved that the input to H(P,P) never halts whether or not it
>>> is aborted by H, are you going to have an honest dialogue about the
>>> steps of the proof or not?
>>
>> Yes, and when run H(P, P) the arbitrary program and input which it is
>> being given described P(P).
>>
>> It is supposed to answer how that *program* will behave. And you have
>> acknowledged that that program *halts*. That behaviour is what
>> *defines* the correct answer to the question 'does P(P) halt'. And
>> your H gets this *wrong*.
>>
>> The halting question is not concerned with how the input string
>> behaves inside your partial simulator. It is *only* concerned with the
>> actual, *independent* program P(P).
>>
>
> So in other words you insist on not going through my steps (a)(b)(c)(d)
> on page 6 and trying to find an error?

I already pointed out that your (b) has major errors in an earlier post
which you disregarded. I see no reason to repeat that.

What I am pointing out here is that the halting status of P(P) is
definined *solely* in terms of the behaviour of the independent
computation P(P). Any claim about the halting status of P(P) which fails
to match this behaviour is wrong *by definition*.

André

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs Count ]

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Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc> <871r7i6n2u.fsf@bsb.me.uk> <OqKdnROLKJ9CdJz8nZ2dnUU7-avNnZ2d@giganews.com> <87k0la542c.fsf@bsb.me.uk> <1NidnVPZ-NHDl5_8nZ2dnUU7-enNnZ2d@giganews.com> <87sfzw3ao1.fsf@bsb.me.uk> <7oKdnTjx4IC20p78nZ2dnUU7-TvNnZ2d@giganews.com> <875yws36vt.fsf@bsb.me.uk> <j66dnbdHrpV8_p78nZ2dnUU7-aXNnZ2d@giganews.com> <87im0s0ydp.fsf@bsb.me.uk> <Brqdnfehrf0Kj5n8nZ2dnUU7-X3NnZ2d@giganews.com> <87tukblgjy.fsf@bsb.me.uk> <qtGdnfuXs4nFOZn8nZ2dnUU7-cnNnZ2d@giganews.com> <871r7ekugt.fsf@bsb.me.uk> <K5-dndGZo_-VmJv8nZ2dnUU78QvNnZ2d@giganews.com> <87czqxa0zk.fsf@bsb.me.uk> <woudnXWBxPba95r8nZ2dnUU78ffNnZ2d@giganews.com> <87mtpz64sq.fsf@bsb.me.uk> <w5edne8d06OkxpX8nZ2dnUU7-b_NnZ2d@giganews.com> <87bl6f5qvy.fsf@bsb.me.uk> <1ZednUt-qbrXlZf8nZ2dnUU7-QnNnZ2d@giganews.com> <sed35j$od8$1@dont-email.me> <KcOdnSwOTcFujZf8nZ2dnUU7-W_NnZ2d@giganews.com> <sed4b4$tqd$1@dont-email.me> <PKGdnRIK9KVMi5f8nZ2dnUU78V3NnZ2d@giganews.com> <sed5fo$31j$1@dont-email.me>
From: NoO...@NoWhere.com (olcott)
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 by: olcott - Wed, 4 Aug 2021 04:38 UTC

On 8/3/2021 11:36 PM, André G. Isaak wrote:
> On 2021-08-03 22:25, olcott wrote:
>> On 8/3/2021 11:16 PM, André G. Isaak wrote:
>>> On 2021-08-03 22:00, olcott wrote:
>>>> On 8/3/2021 10:56 PM, André G. Isaak wrote:
>>>>> On 2021-08-03 21:23, olcott wrote:
>>>>>
>>>>>> The key point is that only the input to the halt decider is within
>>>>>> the scope of the halting problem.
>>>>>>
>>>>>> As long as the halt decider correctly decides its input what these
>>>>>> same programs do what they are not inputs to the halt decider make
>>>>>> no difference to the actual halting problem.
>>>>>
>>>>> You have this completely back-assward.
>>>>>
>>>>> How the TM described by the input to a halt decider behaves when
>>>>> they are run as *independent* computations (i.e. not as inputs to
>>>>> the halt decider) is the question a halt decider is supposed to
>>>>> answer by the very definition of a 'halt decider'. If the answer
>>>>> provided by the decider does not match the behavior of the
>>>>> *independent* computation, then that answer is wrong.
>>>>>
>>>>> Talking about how the description behaves "as an input" isn't even
>>>>> coherent. Inputs don't behave like anything. They are just data.
>>>>>
>>>>> André
>>>>>
>>>>
>>>>     In computability theory, the halting problem is the
>>>>     problem of determining,
>>>>
>>>> from a description of an arbitrary computer program and an input,
>>>> from a description of an arbitrary computer program and an input,
>>>> from a description of an arbitrary computer program and an input,
>>>> from a description of an arbitrary computer program and an input,
>>>>
>>>>     whether the program will
>>>>     finish running, or continue to run forever.
>>>>     https://en.wikipedia.org/wiki/Halting_problem
>>>>
>>>> I have proved that the input to H(P,P) never halts whether or not it
>>>> is aborted by H, are you going to have an honest dialogue about the
>>>> steps of the proof or not?
>>>
>>> Yes, and when run H(P, P) the arbitrary program and input which it is
>>> being given described P(P).
>>>
>>> It is supposed to answer how that *program* will behave. And you have
>>> acknowledged that that program *halts*. That behaviour is what
>>> *defines* the correct answer to the question 'does P(P) halt'. And
>>> your H gets this *wrong*.
>>>
>>> The halting question is not concerned with how the input string
>>> behaves inside your partial simulator. It is *only* concerned with
>>> the actual, *independent* program P(P).
>>>
>>
>> So in other words you insist on not going through my steps
>> (a)(b)(c)(d) on page 6 and trying to find an error?
>
>
> I already pointed out that your (b) has major errors in an earlier post
> which you disregarded. I see no reason to repeat that.
>

I am not willing to talk about anything else.

> What I am pointing out here is that the halting status of P(P) is
> definined *solely* in terms of the behaviour of the independent
> computation P(P). Any claim about the halting status of P(P) which fails
> to match this behaviour is wrong *by definition*.
>
> André
>

--
Copyright 2021 Pete Olcott

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

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs Count ]

<Ysedne93LONKgJf8nZ2dnUU7-XvNnZ2d@giganews.com>

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https://www.novabbs.com/devel/article-flat.php?id=19459&group=comp.theory#19459

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Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[
Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inpu
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From: NoO...@NoWhere.com (olcott)
Date: Tue, 3 Aug 2021 23:55:51 -0500
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 by: olcott - Wed, 4 Aug 2021 04:55 UTC

On 8/3/2021 11:38 PM, olcott wrote:
> On 8/3/2021 11:36 PM, André G. Isaak wrote:
>> On 2021-08-03 22:25, olcott wrote:
>>> On 8/3/2021 11:16 PM, André G. Isaak wrote:
>>>> On 2021-08-03 22:00, olcott wrote:
>>>>> On 8/3/2021 10:56 PM, André G. Isaak wrote:
>>>>>> On 2021-08-03 21:23, olcott wrote:
>>>>>>
>>>>>>> The key point is that only the input to the halt decider is
>>>>>>> within the scope of the halting problem.
>>>>>>>
>>>>>>> As long as the halt decider correctly decides its input what
>>>>>>> these same programs do what they are not inputs to the halt
>>>>>>> decider make no difference to the actual halting problem.
>>>>>>
>>>>>> You have this completely back-assward.
>>>>>>
>>>>>> How the TM described by the input to a halt decider behaves when
>>>>>> they are run as *independent* computations (i.e. not as inputs to
>>>>>> the halt decider) is the question a halt decider is supposed to
>>>>>> answer by the very definition of a 'halt decider'. If the answer
>>>>>> provided by the decider does not match the behavior of the
>>>>>> *independent* computation, then that answer is wrong.
>>>>>>
>>>>>> Talking about how the description behaves "as an input" isn't even
>>>>>> coherent. Inputs don't behave like anything. They are just data.
>>>>>>
>>>>>> André
>>>>>>
>>>>>
>>>>>     In computability theory, the halting problem is the
>>>>>     problem of determining,
>>>>>
>>>>> from a description of an arbitrary computer program and an input,
>>>>> from a description of an arbitrary computer program and an input,
>>>>> from a description of an arbitrary computer program and an input,
>>>>> from a description of an arbitrary computer program and an input,
>>>>>
>>>>>     whether the program will
>>>>>     finish running, or continue to run forever.
>>>>>     https://en.wikipedia.org/wiki/Halting_problem
>>>>>
>>>>> I have proved that the input to H(P,P) never halts whether or not
>>>>> it is aborted by H, are you going to have an honest dialogue about
>>>>> the steps of the proof or not?
>>>>
>>>> Yes, and when run H(P, P) the arbitrary program and input which it
>>>> is being given described P(P).
>>>>
>>>> It is supposed to answer how that *program* will behave. And you
>>>> have acknowledged that that program *halts*. That behaviour is what
>>>> *defines* the correct answer to the question 'does P(P) halt'. And
>>>> your H gets this *wrong*.
>>>>
>>>> The halting question is not concerned with how the input string
>>>> behaves inside your partial simulator. It is *only* concerned with
>>>> the actual, *independent* program P(P).
>>>>
>>>
>>> So in other words you insist on not going through my steps
>>> (a)(b)(c)(d) on page 6 and trying to find an error?
>>
>>
>> I already pointed out that your (b) has major errors in an earlier
>> post which you disregarded. I see no reason to repeat that.
>>
>
> I am not willing to talk about anything else.

I am not willing to talk about anything besides page 6 until we have
mutual agreement on page 6.

https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation

>
>> What I am pointing out here is that the halting status of P(P) is
>> definined *solely* in terms of the behaviour of the independent
>> computation P(P). Any claim about the halting status of P(P) which
>> fails to match this behaviour is wrong *by definition*.
>>
>> André
>>
>
>

--
Copyright 2021 Pete Olcott

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

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs Count ]

<sed904$iae$1@dont-email.me>

 copy mid

https://www.novabbs.com/devel/article-flat.php?id=19460&group=comp.theory#19460

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Path: i2pn2.org!i2pn.org!eternal-september.org!reader02.eternal-september.org!.POSTED!not-for-mail
From: agis...@gm.invalid (André G. Isaak)
Newsgroups: comp.theory
Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[
Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inpu
ts_Count_]
Date: Tue, 3 Aug 2021 23:36:04 -0600
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 by: André G. Isaak - Wed, 4 Aug 2021 05:36 UTC

On 2021-08-03 22:55, olcott wrote:
> On 8/3/2021 11:38 PM, olcott wrote:
>> On 8/3/2021 11:36 PM, André G. Isaak wrote:
>>> On 2021-08-03 22:25, olcott wrote:
>>>> On 8/3/2021 11:16 PM, André G. Isaak wrote:
>>>>> On 2021-08-03 22:00, olcott wrote:
>>>>>> On 8/3/2021 10:56 PM, André G. Isaak wrote:
>>>>>>> On 2021-08-03 21:23, olcott wrote:
>>>>>>>
>>>>>>>> The key point is that only the input to the halt decider is
>>>>>>>> within the scope of the halting problem.
>>>>>>>>
>>>>>>>> As long as the halt decider correctly decides its input what
>>>>>>>> these same programs do what they are not inputs to the halt
>>>>>>>> decider make no difference to the actual halting problem.
>>>>>>>
>>>>>>> You have this completely back-assward.
>>>>>>>
>>>>>>> How the TM described by the input to a halt decider behaves when
>>>>>>> they are run as *independent* computations (i.e. not as inputs to
>>>>>>> the halt decider) is the question a halt decider is supposed to
>>>>>>> answer by the very definition of a 'halt decider'. If the answer
>>>>>>> provided by the decider does not match the behavior of the
>>>>>>> *independent* computation, then that answer is wrong.
>>>>>>>
>>>>>>> Talking about how the description behaves "as an input" isn't
>>>>>>> even coherent. Inputs don't behave like anything. They are just
>>>>>>> data.
>>>>>>>
>>>>>>> André
>>>>>>>
>>>>>>
>>>>>>     In computability theory, the halting problem is the
>>>>>>     problem of determining,
>>>>>>
>>>>>> from a description of an arbitrary computer program and an input,
>>>>>> from a description of an arbitrary computer program and an input,
>>>>>> from a description of an arbitrary computer program and an input,
>>>>>> from a description of an arbitrary computer program and an input,
>>>>>>
>>>>>>     whether the program will
>>>>>>     finish running, or continue to run forever.
>>>>>>     https://en.wikipedia.org/wiki/Halting_problem
>>>>>>
>>>>>> I have proved that the input to H(P,P) never halts whether or not
>>>>>> it is aborted by H, are you going to have an honest dialogue about
>>>>>> the steps of the proof or not?
>>>>>
>>>>> Yes, and when run H(P, P) the arbitrary program and input which it
>>>>> is being given described P(P).
>>>>>
>>>>> It is supposed to answer how that *program* will behave. And you
>>>>> have acknowledged that that program *halts*. That behaviour is what
>>>>> *defines* the correct answer to the question 'does P(P) halt'. And
>>>>> your H gets this *wrong*.
>>>>>
>>>>> The halting question is not concerned with how the input string
>>>>> behaves inside your partial simulator. It is *only* concerned with
>>>>> the actual, *independent* program P(P).
>>>>>
>>>>
>>>> So in other words you insist on not going through my steps
>>>> (a)(b)(c)(d) on page 6 and trying to find an error?
>>>
>>>
>>> I already pointed out that your (b) has major errors in an earlier
>>> post which you disregarded. I see no reason to repeat that.
>>>
>>
>> I am not willing to talk about anything else.
>
> I am not willing to talk about anything besides page 6 until we have
> mutual agreement on page 6.

That's not going to happen since your argument on page 6 is fallacious.

Read my earlier message <sebtb9$plb$1@dont-email.me>

I see no reason to repeat it.

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] ( Are you game ? )

<878s1h4k2k.fsf@bsb.me.uk>

 copy mid

https://www.novabbs.com/devel/article-flat.php?id=19470&group=comp.theory#19470

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Path: i2pn2.org!i2pn.org!eternal-september.org!reader02.eternal-september.org!.POSTED!not-for-mail
From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider [
Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] (
Are you game ? )
Date: Wed, 04 Aug 2021 10:48:19 +0100
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 by: Ben Bacarisse - Wed, 4 Aug 2021 09:48 UTC

Richard Damon <Richard@Damon-Family.org> writes:

> On 8/3/21 6:24 PM, olcott wrote:
>>
>> I corrected the Linz notation, Linz has two start states.
>> Linz has no Ĥ.qx, that is my correction. It indicates H.q0.
>
> I don't think his notation actually had two start states.

Indeed.

> As I remember, he had a q0 as the start start for H^, from which we
> duplicated the input and then we hit an internal state called H.q0 which
> was where the copy of H that was in H^ began,

Linz makes the mistake of not distinguishing these. His notation just
has the sate rather than any TM.state notation. He has H^ go from q0 to
q0 again after copying the input. This is not right.

PO's notation is a good one, but he does not use it! He prefixes all
the states in H^ with H^. He could have use H^.q0 and H.q0 to show that
H^ includes H's states as a subset. All of PO's configuration formulas
could have the prefixes dropped with no change in meaning.

> and then it had copies of
> ALL the states in H, ending in H.qn or H.qy, and then H.qy was modified
> to form an infinite loop.
>
> Note q0 != H.q0, H.q0 is NOT a 'start' state of H^, only the original
> start state of the H machine that had all its states prefix with H. and
> included (with qy being to become an infinite loop)

Using PO's notation to make that clear would have helped, but I think he
just got confused by the notion of one TM embedding another.

> This if the H. as a scope operator on the state names.
>
> Of course, if you recognized that, you would have to admit that H^
> included a copy of all of H in it, so H when processing H^ would need to
> actually look at those states, so maybe that it the source of your blind
> spot.

Or, as you say, it may have been a reluctance to admit what's really
going on.

--
Ben.

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct ]

<87zgtx2wxn.fsf@bsb.me.uk>

 copy mid

https://www.novabbs.com/devel/article-flat.php?id=19477&group=comp.theory#19477

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Path: i2pn2.org!i2pn.org!eternal-september.org!reader02.eternal-september.org!.POSTED!not-for-mail
From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider [
Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [
succinct ]
Date: Wed, 04 Aug 2021 13:53:24 +0100
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 by: Ben Bacarisse - Wed, 4 Aug 2021 12:53 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/2/2021 8:45 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/2/2021 7:11 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>
>> (I'm going to ignore errors that I've already pointed out.)
>>
>>> If you really do sincerely want an actual honest dialogue you would
>>> carefully work through all the steps to confirm that the input ⟨Ĥ⟩ ⟨Ĥ⟩
>>> to Ĥ.qx never halts.
>> If you want me to comment on that, write it without the errors. It has
>> two, both of which I've commented on before. If you are sincere, you
>> will want to write clearly and without errors.
>>
>>> Once we have mutual agreement that Ĥ.qx correctly decides that its
>>> input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts and Ĥ does halt, then we have the basis to
>>> go to the next step and resolve the actual paradox.
>> There is no paradox. That Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn when Ĥ applied to ⟨Ĥ⟩
>> clearly halts is not paradoxical. It's just not the sort of TM the
>> proof is talking about. And how could it be? The "Linz Ĥ" is as
>> illogical as a cat that is a dog.
>>
>>> As long as it is simply dismissed out-of-hand as a contradiction the
>>> paradox remains unresolved.
>> There is no contradiction or paradox. You Ĥ is just the wrong sort of
>> TM. The proof you want to "refute" is talking about this sort of Ĥ:
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>> if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>
> Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
> Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
> Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
> Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts

Maybe saying it a couple more times will help. After four times I can
tell you that it's still wrong. Maybe about a dozen more?

Whether what happens after Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ is correct or not is determined
by Linz, not by you. And you are clear that

Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn.

Linz is equally clear that this is wrong. You even helpfully keep
quoting Linz telling you it's wrong:

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.

You can repeat, again and again, that "Ĥ.qx correctly decides"
something, but you've already told the world that the computation halts
when is should not. Your Ĥ is not doing what it should to show a fault
in Linz's proof. It's doing something else that is of no interest to
anyone.

(It's possible that just can't see why it is that those two lines from
Linz are telling you that your Ĥ does is wrong. You've struggled with
what this notation really means, so it's possible. But reasonable
student would ask if they could not see that Linz is telling you your Ĥ
is wrong.)

> In computability theory, the halting problem is the problem of
> determining, from a description of an arbitrary computer program and
> an input, whether the program will finish running, or continue to run
> forever https://en.wikipedia.org/wiki/Halting_problem

Really? Well scan my tape and call me halting!

If you had had the two TMs you lied about having (sorry, used "poetic
license" to say you had) you could have run

Ĥ.q0 ⟨Ĥ⟩

and

H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩

and you would have seen, right away, that H(⟨Ĥ⟩ ⟨Ĥ⟩) is wrong about
Ĥ(⟨Ĥ⟩). Maybe you did and that's why you've spent 30 months waking back
that lie^H^H^H poetic license.

--
Ben.

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs Count ]

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Subject: Re: Black box halt decider is NOT a partial decider
[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs Count ]
From: malcolm....@gmail.com (Malcolm McLean)
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 by: Malcolm McLean - Wed, 4 Aug 2021 14:13 UTC

On Wednesday, 4 August 2021 at 05:55:59 UTC+1, olcott wrote:
>
> I am not willing to talk about anything besides page 6 until we have
> mutual agreement on page 6.
>
> https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation
>
We agree that P(P) halts and that H(P, P) reports false (non-halting).

I've suggested that maybe you are trying to argue that P(P)'s halting
doesn't count, because it is generated by the copy of H inside P.

You've rejected that idea, and insist that H(P,P) has correctly determined
that P(P) is non-halting.

I don't really think there's anything useful left to say.

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] ( Are you game ? )[ Only: H(P,P)==0 ]

<Peqdncl6_7bIO5f8nZ2dnUU7-TXNnZ2d@giganews.com>

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Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] ( Are you game ? )[ Only: H(P,P)==0 ]
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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Wed, 4 Aug 2021 14:38 UTC

On 8/3/2021 12:11 PM, André G. Isaak wrote:
> On 2021-08-03 09:21, olcott wrote:
>> On 8/3/2021 9:33 AM, Malcolm McLean wrote:
>>> On Tuesday, 3 August 2021 at 14:00:28 UTC+1, olcott wrote:
>>>> On 8/3/2021 12:29 AM, Malcolm McLean wrote:
>>>>>
>>>>> You're trying to claim that H_Hat<H_Hat> doens't really halt because
>>>>> the recursion of H instances is terminated by H.
>>>> No I have never been saying that. I am claiming that the input to
>>>> H(P,P)
>>>> never halts whether or not H terminates its simulation of this input.
>>>>
>>> We agree that P(P) halts.
>>> So now you're drawing a distinction between P(P) and "the input to
>>> H(P,P)".
>>> ThIs is nonsense..
>>
>> Try and find any error in (a)(b)(c)(d) on page 6
>>
>> https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation
>
>
>
> The error, which has been pointed out repeatedly, is in your (b).
>
> You claim "there are no control flow instructions in the execution trace
> that would escape the infinite recursion", but there *are* flow control
> instructions. But your trace is incomplete and skips over the call to
> B82 where these flow control instructions reside.
>

I will rephrase that:
There are no control flow instructions in P that break out of the
infinite recursion. The control flow instructions in H that could abort
the simulation of this infinite recursion would never allow P to reach
its final state. Therefore we know that P never reaches its final state.
Therefore because of the definition of halting that you provided we know
that P never halts. Therefore we know that H(P,P)==0 is correct.

> The code at B82 is *part* of the computation performed by P. It *must*
> be included in any trace of P which purports to be a complete and honest
> trace.
>

On the basis of the above analysis we know for sure that this is
irrelevant. Nothing that happens in H can possibly cause P to reach its
final state. Nothing that happens in P can possibly cause P to reach its
final state. Therefore P never reaches its final state. Therefore P
never halts. Therefore H(P,)==0 is correct.

> You don't seem to grasp the fact that *all* code executed by a
> particular computation from the moment the computation begins to the
> time it halts (assuming it halts) is part of that computation. It
> doesn't matter whether the code in question is shared by some other
> routine, or is operating system code or whatever. These are purely
> artificial distinctions which play no role in the theory of computation.
>
> André
>

If the input to H(P,P) cannot possibly halt then H(P,P)==0 is correct.

--
Copyright 2021 Pete Olcott

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

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs Count ]( Honest Dialogue )

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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Wed, 4 Aug 2021 15:11 UTC

On 8/4/2021 9:13 AM, Malcolm McLean wrote:
> On Wednesday, 4 August 2021 at 05:55:59 UTC+1, olcott wrote:
>>
>> I am not willing to talk about anything besides page 6 until we have
>> mutual agreement on page 6.
>>
>> https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation
>>
> We agree that P(P) halts and that H(P, P) reports false (non-halting).
>
> I've suggested that maybe you are trying to argue that P(P)'s halting
> doesn't count, because it is generated by the copy of H inside P.
>
> You've rejected that idea, and insist that H(P,P) has correctly determined
> that P(P) is non-halting.
>
> I don't really think there's anything useful left to say.
>

In computability theory, the halting problem is the problem
of determining, from a description of an arbitrary computer
program and an input, whether the program will finish running,
or continue to run forever.
https://en.wikipedia.org/wiki/Halting_problem

https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation

Because page 6 does prove that H(P,P) does correctly decide that its
input never halts and the halting problem only requires that H decide
its input correctly the fact that H(P,P)==0 is correct does directly
refute the halting problem counter-example conclusion.

That people are unwilling to critique page 6 only shows that they are
unwilling to have an actual honest dialogue.

--
Copyright 2021 Pete Olcott

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

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs Count ]

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From: NoO...@NoWhere.com (olcott)
Date: Wed, 4 Aug 2021 10:14:25 -0500
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 by: olcott - Wed, 4 Aug 2021 15:14 UTC

On 8/4/2021 12:36 AM, André G. Isaak wrote:
> On 2021-08-03 22:55, olcott wrote:
>> On 8/3/2021 11:38 PM, olcott wrote:
>>> On 8/3/2021 11:36 PM, André G. Isaak wrote:
>>>> On 2021-08-03 22:25, olcott wrote:
>>>>> On 8/3/2021 11:16 PM, André G. Isaak wrote:
>>>>>> On 2021-08-03 22:00, olcott wrote:
>>>>>>> On 8/3/2021 10:56 PM, André G. Isaak wrote:
>>>>>>>> On 2021-08-03 21:23, olcott wrote:
>>>>>>>>
>>>>>>>>> The key point is that only the input to the halt decider is
>>>>>>>>> within the scope of the halting problem.
>>>>>>>>>
>>>>>>>>> As long as the halt decider correctly decides its input what
>>>>>>>>> these same programs do what they are not inputs to the halt
>>>>>>>>> decider make no difference to the actual halting problem.
>>>>>>>>
>>>>>>>> You have this completely back-assward.
>>>>>>>>
>>>>>>>> How the TM described by the input to a halt decider behaves when
>>>>>>>> they are run as *independent* computations (i.e. not as inputs
>>>>>>>> to the halt decider) is the question a halt decider is supposed
>>>>>>>> to answer by the very definition of a 'halt decider'. If the
>>>>>>>> answer provided by the decider does not match the behavior of
>>>>>>>> the *independent* computation, then that answer is wrong.
>>>>>>>>
>>>>>>>> Talking about how the description behaves "as an input" isn't
>>>>>>>> even coherent. Inputs don't behave like anything. They are just
>>>>>>>> data.
>>>>>>>>
>>>>>>>> André
>>>>>>>>
>>>>>>>
>>>>>>>     In computability theory, the halting problem is the
>>>>>>>     problem of determining,
>>>>>>>
>>>>>>> from a description of an arbitrary computer program and an input,
>>>>>>> from a description of an arbitrary computer program and an input,
>>>>>>> from a description of an arbitrary computer program and an input,
>>>>>>> from a description of an arbitrary computer program and an input,
>>>>>>>
>>>>>>>     whether the program will
>>>>>>>     finish running, or continue to run forever.
>>>>>>>     https://en.wikipedia.org/wiki/Halting_problem
>>>>>>>
>>>>>>> I have proved that the input to H(P,P) never halts whether or not
>>>>>>> it is aborted by H, are you going to have an honest dialogue
>>>>>>> about the steps of the proof or not?
>>>>>>
>>>>>> Yes, and when run H(P, P) the arbitrary program and input which it
>>>>>> is being given described P(P).
>>>>>>
>>>>>> It is supposed to answer how that *program* will behave. And you
>>>>>> have acknowledged that that program *halts*. That behaviour is
>>>>>> what *defines* the correct answer to the question 'does P(P)
>>>>>> halt'. And your H gets this *wrong*.
>>>>>>
>>>>>> The halting question is not concerned with how the input string
>>>>>> behaves inside your partial simulator. It is *only* concerned with
>>>>>> the actual, *independent* program P(P).
>>>>>>
>>>>>
>>>>> So in other words you insist on not going through my steps
>>>>> (a)(b)(c)(d) on page 6 and trying to find an error?
>>>>
>>>>
>>>> I already pointed out that your (b) has major errors in an earlier
>>>> post which you disregarded. I see no reason to repeat that.
>>>>
>>>
>>> I am not willing to talk about anything else.
>>
>> I am not willing to talk about anything besides page 6 until we have
>> mutual agreement on page 6.
>
> That's not going to happen since your argument on page 6 is fallacious.
>
> Read my earlier message <sebtb9$plb$1@dont-email.me>
>
> I see no reason to repeat it.
>

You simply ignored my correction of your error.
I responded to this message again with words that are more clear.

Simply ignoring this response is sufficient proof that you are
unmotivated to have an actual honest dialogue.

--
Copyright 2021 Pete Olcott

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

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct ][ GIGO ]

<j8OdneamG91aK5f8nZ2dnUU7-fvNnZ2d@giganews.com>

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Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct ][ GIGO ]
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References: <20210719214640.00000dfc@reddwarf.jmc> <87pmv4ab6r.fsf@bsb.me.uk> <JNadnQD-Ofr-SJz8nZ2dnUU7-XHNnZ2d@giganews.com> <871r7i6n2u.fsf@bsb.me.uk> <OqKdnROLKJ9CdJz8nZ2dnUU7-avNnZ2d@giganews.com> <87k0la542c.fsf@bsb.me.uk> <1NidnVPZ-NHDl5_8nZ2dnUU7-enNnZ2d@giganews.com> <87sfzw3ao1.fsf@bsb.me.uk> <7oKdnTjx4IC20p78nZ2dnUU7-TvNnZ2d@giganews.com> <875yws36vt.fsf@bsb.me.uk> <j66dnbdHrpV8_p78nZ2dnUU7-aXNnZ2d@giganews.com> <87im0s0ydp.fsf@bsb.me.uk> <Brqdnfehrf0Kj5n8nZ2dnUU7-X3NnZ2d@giganews.com> <87tukblgjy.fsf@bsb.me.uk> <qtGdnfuXs4nFOZn8nZ2dnUU7-cnNnZ2d@giganews.com> <871r7ekugt.fsf@bsb.me.uk> <K5-dndGZo_-VmJv8nZ2dnUU78QvNnZ2d@giganews.com> <87czqxa0zk.fsf@bsb.me.uk> <53d47ab9-818c-4f40-8e72-bdb76fa416een@googlegroups.com> <87y29l8hhp.fsf@bsb.me.uk> <LZOdnR5aLooNKpv8nZ2dnUU7-SnNnZ2d@giganews.com> <87h7g988a6.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 4 Aug 2021 10:48:23 -0500
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 by: olcott - Wed, 4 Aug 2021 15:48 UTC

On 8/1/2021 11:00 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/1/2021 7:41 AM, Ben Bacarisse wrote:
>>> Malcolm McLean <malcolm.arthur.mclean@gmail.com> writes:
>>>
>>>> On Sunday, 1 August 2021 at 11:54:57 UTC+1, Ben Bacarisse wrote:
>>>>>
>>>>> Here we can see that Ĥ applied to ⟨Ĥ⟩ halts. You can call your Ĥ's
>>>>> behaviour "correct". You can call it anything you like. But it's not
>>>>> "as in Linz". It does not say anything about Linz's proof. It does not
>>>>> do anything people would call impossible or even interesting.
>>>>>
>>>> It seems to be established that H(H_Hat, H_Hat) returns "non-halting"
>>>> whilst H_Hat(H_Hat) halts. So all is as Linz says it must be and no
>>>> theorems are refuted. Which you would expect. If results were consistent
>>>> it would have to be some cheap trick.
>>> I case there is some confusion, I mean that PO's Ĥ is not an Ĥ as
>>> specified in Linz. Yes, everything is in accordance with the truth as
>>> laid out in Linz and, indeed, in any textbook.
>>> I point this out to PO because he brings it up. He keeps posting the
>>> specification of what an Ĥ, as Linz specifies it, would do:
>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>> if (and only if) M applied to wM does not halt.
>>> He claims (or used to claim) that his Ĥ meets this specification for at
>>> least the one case where wM == ⟨Ĥ⟩:
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>> if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>>> To remain relevant, he /must/ keep insisting that his Ĥ meets the
>>> requirements laid out in Linz, if only for this one key input.
>>>
>>
>> Ĥ[0].q0 is taken to mean Ĥ<sub>0</sub>.q0 which is the Turing machine.
>>
>> Ĥ[1].q0 is taken to mean Ĥ<sub>1</sub>.q0 which is the Turing machine
>> description input to Ĥ[0].q0
>>
>> Ĥ[2].q0 is taken to mean Ĥ<sub>2</sub>.q0 which is first copy of the
>> Turing machine description input to Ĥ[0].q0
>>
>> Ĥ[0].q0 ⟨Ĥ⟩ ⊢* Ĥ[0].qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ[0].qn
>
> Ĥ[0] is Ĥ so you are confirming, yet again, that
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>
>> It is neither a contradiction nor a paradox because there are three
>> different instances of Ĥ.
>
> I agree that this is neither a paradox nor a contradiction. It's just a
> fact derived form the logic of how your Ĥ is written (the majority of
> which you are keeping hidden from us).
>
>> Because the only reason that the first instance halts is that Ĥ[0].qx
>> correctly determines that its input cannot possibly ever reach its
>> final state of Ĥ[1].qn or Ĥ[1].qy whether or not the simulating halt
>> decider aborts its simulation of this input, we know with 100%
>> perfectly justified logical certainty that the input to Ĥ[0].qx never
>> halts.
>
> We know, since you keep telling us, that Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn. This clearly
> shows that Ĥ applied to ⟨Ĥ⟩ halts. You can see the final state right

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
if M applied to wM halts, and

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
if M applied to wM does not halt

You are using the wrong Ĥ. Linz stipulates that wM is ⟨Ĥ⟩ and M is the
underlying machine of this ⟨Ĥ⟩ therefore M applied to wM means ⟨Ĥ⟩
applied to ⟨Ĥ⟩. We can see that M never reaches its final state.

Because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ does specify infinitely nested
simulation we can know for sure that M never reaches its final state
whether or not Ĥ.qx aborts its simulation of M.

> there in the line you keep posting again and again. As far as I know,
> you have never disputed this fact.
>
> As you say, this is neither a paradox nor a contradiction. It just
> shows that Ĥ does not behave as Linz says it should, in the one case you
> have obsessed about for 17 years. Having an H (and thus an Ĥ) that is
> wrong (i.e. not "exactly and precisely as on Linz" as you once claimed)
> is trivial. It is not something that anyone (except Malcolm,
> apparently) would care about.
>

--
Copyright 2021 Pete Olcott

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

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct ]

<Rr6dnWKhH4ejIJf8nZ2dnUU7-Q3NnZ2d@giganews.com>

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Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct ]
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
References: <20210719214640.00000dfc@reddwarf.jmc> <87k0la542c.fsf@bsb.me.uk> <1NidnVPZ-NHDl5_8nZ2dnUU7-enNnZ2d@giganews.com> <87sfzw3ao1.fsf@bsb.me.uk> <7oKdnTjx4IC20p78nZ2dnUU7-TvNnZ2d@giganews.com> <875yws36vt.fsf@bsb.me.uk> <j66dnbdHrpV8_p78nZ2dnUU7-aXNnZ2d@giganews.com> <87im0s0ydp.fsf@bsb.me.uk> <Brqdnfehrf0Kj5n8nZ2dnUU7-X3NnZ2d@giganews.com> <87tukblgjy.fsf@bsb.me.uk> <qtGdnfuXs4nFOZn8nZ2dnUU7-cnNnZ2d@giganews.com> <871r7ekugt.fsf@bsb.me.uk> <K5-dndGZo_-VmJv8nZ2dnUU78QvNnZ2d@giganews.com> <87czqxa0zk.fsf@bsb.me.uk> <woudnXWBxPba95r8nZ2dnUU78ffNnZ2d@giganews.com> <87mtpz64sq.fsf@bsb.me.uk> <w5edne8d06OkxpX8nZ2dnUU7-b_NnZ2d@giganews.com> <87bl6f5qvy.fsf@bsb.me.uk> <xLednaPs_ZSXCZX8nZ2dnUU7-YnNnZ2d@giganews.com> <87o8af47y0.fsf@bsb.me.uk> <NsudnY99rthDOJX8nZ2dnUU7-c_NnZ2d@giganews.com> <87zgtx2wxn.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 4 Aug 2021 11:15:57 -0500
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 by: olcott - Wed, 4 Aug 2021 16:15 UTC

On 8/4/2021 7:53 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/2/2021 8:45 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/2/2021 7:11 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>
>>> (I'm going to ignore errors that I've already pointed out.)
>>>
>>>> If you really do sincerely want an actual honest dialogue you would
>>>> carefully work through all the steps to confirm that the input ⟨Ĥ⟩ ⟨Ĥ⟩
>>>> to Ĥ.qx never halts.
>>> If you want me to comment on that, write it without the errors. It has
>>> two, both of which I've commented on before. If you are sincere, you
>>> will want to write clearly and without errors.
>>>
>>>> Once we have mutual agreement that Ĥ.qx correctly decides that its
>>>> input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts and Ĥ does halt, then we have the basis to
>>>> go to the next step and resolve the actual paradox.
>>> There is no paradox. That Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn when Ĥ applied to ⟨Ĥ⟩
>>> clearly halts is not paradoxical. It's just not the sort of TM the
>>> proof is talking about. And how could it be? The "Linz Ĥ" is as
>>> illogical as a cat that is a dog.
>>>
>>>> As long as it is simply dismissed out-of-hand as a contradiction the
>>>> paradox remains unresolved.
>>> There is no contradiction or paradox. You Ĥ is just the wrong sort of
>>> TM. The proof you want to "refute" is talking about this sort of Ĥ:
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>> if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>>
>> Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>> Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>> Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>> Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>
> Maybe saying it a couple more times will help. After four times I can
> tell you that it's still wrong. Maybe about a dozen more?
>
> Whether what happens after Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ is correct or not is determined
> by Linz, not by you. And you are clear that
>
> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn.
>

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
if M applied to wM does not halt

As explained in complete detail below:
Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

because M applied to wM does not halt
where M is Machine_of(⟨Ĥ⟩) (1st param) above
and wM is ⟨Ĥ⟩ the second param above.

Because wM is referring to ⟨Ĥ⟩ and M is referring to the underlying
machine of ⟨Ĥ⟩ the last line above is translated to:
if Machine_of(⟨Ĥ⟩) applied to ⟨Ĥ⟩ does not halt

We can know that Machine_of(⟨Ĥ⟩) applied to ⟨Ĥ⟩ never reaches its final
state whether or not the embedded halt decider at Ĥ.qx aborts it
simulation of Machine_of(⟨Ĥ⟩), therefore we know that Machine_of(⟨Ĥ⟩)
never halts. Therefore we know that M applied to wM does not halt.

> Linz is equally clear that this is wrong. You even helpfully keep
> quoting Linz telling you it's wrong:
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>
> You can repeat, again and again, that "Ĥ.qx correctly decides"
> something, but you've already told the world that the computation halts
> when is should not. Your Ĥ is not doing what it should to show a fault
> in Linz's proof. It's doing something else that is of no interest to
> anyone.
>
> (It's possible that just can't see why it is that those two lines from
> Linz are telling you that your Ĥ does is wrong. You've struggled with
> what this notation really means, so it's possible. But reasonable
> student would ask if they could not see that Linz is telling you your Ĥ
> is wrong.)
>
>> In computability theory, the halting problem is the problem of
>> determining, from a description of an arbitrary computer program and
>> an input, whether the program will finish running, or continue to run
>> forever https://en.wikipedia.org/wiki/Halting_problem
>
> Really? Well scan my tape and call me halting!
>
> If you had had the two TMs you lied about having (sorry, used "poetic
> license" to say you had) you could have run
>
> Ĥ.q0 ⟨Ĥ⟩
>
> and
>
> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩
>
> and you would have seen, right away, that H(⟨Ĥ⟩ ⟨Ĥ⟩) is wrong about
> Ĥ(⟨Ĥ⟩). Maybe you did and that's why you've spent 30 months waking back
> that lie^H^H^H poetic license.
>

--
Copyright 2021 Pete Olcott

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

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs Count ]

<87tuk52h0e.fsf@bsb.me.uk>

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https://www.novabbs.com/devel/article-flat.php?id=19484&group=comp.theory#19484

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Path: i2pn2.org!i2pn.org!eternal-september.org!reader02.eternal-september.org!.POSTED!not-for-mail
From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider [
Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [
Only Inputs Count ]
Followup-To: comp.theory
Date: Wed, 04 Aug 2021 19:37:21 +0100
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 by: Ben Bacarisse - Wed, 4 Aug 2021 18:37 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/2/2021 7:11 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/2/2021 2:10 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/1/2021 5:54 AM, Ben Bacarisse wrote:
>>>>
>>>>>> I am happy you have a notation you like. Are you prepared to address
>>>>>> that fact that your H^ is not "as in Linz"?
>>>>>
>>>>> My Ĥ is exactly the Linz Ĥ with the additional elaboration that the
>>>>> second wildcard state transition ⊢* is defined to be a simulating halt
>>>>> decider.
>>>> No. I've explained many times now why your Ĥ is not at all "the Linz
>>>> Ĥ". Do you see any point in my doing so again?
>> I suspect not. You certainly have not asked a single question that
>> could help you to understand why your Ĥ is irrelevant.
>>
>>>>>> We all know you are declaring that to be correct. Here's why your Ĥ is
>>>>>> not "as in Linz". Linz requires that
>>>>>>
>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>>> if M applied to wM does not halt
>>>>>>
>>>>>> Any Ĥ that eventually transitions to Ĥ.qn on input wM must do so
>>>>>> if, and only if, the encoded M applied to wM does not halt. But you've
>>>>>> given us a case where your Ĥ is not like this:
>>>>>>
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>
>>>>> And when we eliminate the fallacy of equivocation error we have
>>>>>
>>>>> Ĥ[0].q0 ⟨Ĥ⟩ ⊢* Ĥ[0].qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ[0].qn
>>>> The only thing I think you are equivocating on is pretending that Ĥ[0]
>>>> is not Ĥ, and it doesn't look as if you've removed that error. I think
>>>> you /should/ remove it so that you can be saying something of value
>>>> about Ĥ.
>>>
>>> It is clear from the text that Linz does specify at least three
>>> different instances of Ĥ, The TM the TMD input ⟨Ĥ⟩ and a copy of this
>>> TMD input.
>> Still equivocating. Either Ĥ[0] = Ĥ or you are wasting everyone's time.
>> (This is mathematical equality. Ĥ is a tuple of sets. If Ĥ[0] is not
>> exactly identical in every way to Ĥ then I don't care about it. Note
>> that I'm not disputing your right, for ease of explanation, to give
>> identical things more than one name. But the same permission allows me
>> to use any of the names because they name the same thing.)
>> You are wrong because
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>> where Linz requires that
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>> if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>
> You have the above incorrectly in these two
> M refers to the Turing Machine described by wM.
>
> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
> if M applied to wM halts, and
>
> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
> if M applied to wM does not halt
>
> This is *NOT* the Ĥ that is executed.

I have no idea what you are objecting to. You can substitute Ĥ and ⟨Ĥ⟩
into what Linz says, can't you? I did it above and you don't like it:

For M = Ĥ and wM = ⟨Ĥ⟩ Linz tells us that

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt.

Your Ĥ is not like that. Its behaviour is boring and obvious and
trivial.

--
Ben.

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct ][ GIGO ]

<87im0l2gc0.fsf@bsb.me.uk>

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https://www.novabbs.com/devel/article-flat.php?id=19486&group=comp.theory#19486

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider [
Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [
succinct ][ GIGO ]
Followup-To: comp.theory
Date: Wed, 04 Aug 2021 19:51:59 +0100
Organization: A noiseless patient Spider
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 by: Ben Bacarisse - Wed, 4 Aug 2021 18:51 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/1/2021 11:00 AM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/1/2021 7:41 AM, Ben Bacarisse wrote:
>>>> Malcolm McLean <malcolm.arthur.mclean@gmail.com> writes:
>>>>
>>>>> On Sunday, 1 August 2021 at 11:54:57 UTC+1, Ben Bacarisse wrote:
>>>>>>
>>>>>> Here we can see that Ĥ applied to ⟨Ĥ⟩ halts. You can call your Ĥ's
>>>>>> behaviour "correct". You can call it anything you like. But it's not
>>>>>> "as in Linz". It does not say anything about Linz's proof. It does not
>>>>>> do anything people would call impossible or even interesting.
>>>>>>
>>>>> It seems to be established that H(H_Hat, H_Hat) returns "non-halting"
>>>>> whilst H_Hat(H_Hat) halts. So all is as Linz says it must be and no
>>>>> theorems are refuted. Which you would expect. If results were consistent
>>>>> it would have to be some cheap trick.
>>>> I case there is some confusion, I mean that PO's Ĥ is not an Ĥ as
>>>> specified in Linz. Yes, everything is in accordance with the truth as
>>>> laid out in Linz and, indeed, in any textbook.
>>>> I point this out to PO because he brings it up. He keeps posting the
>>>> specification of what an Ĥ, as Linz specifies it, would do:
>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>> if (and only if) M applied to wM does not halt.
>>>> He claims (or used to claim) that his Ĥ meets this specification for at
>>>> least the one case where wM == ⟨Ĥ⟩:
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>> if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>> To remain relevant, he /must/ keep insisting that his Ĥ meets the
>>>> requirements laid out in Linz, if only for this one key input.
>>>>
>>>
>>> Ĥ[0].q0 is taken to mean Ĥ<sub>0</sub>.q0 which is the Turing machine.
>>>
>>> Ĥ[1].q0 is taken to mean Ĥ<sub>1</sub>.q0 which is the Turing machine
>>> description input to Ĥ[0].q0
>>>
>>> Ĥ[2].q0 is taken to mean Ĥ<sub>2</sub>.q0 which is first copy of the
>>> Turing machine description input to Ĥ[0].q0
>>>
>>> Ĥ[0].q0 ⟨Ĥ⟩ ⊢* Ĥ[0].qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ[0].qn
>> Ĥ[0] is Ĥ so you are confirming, yet again, that
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>
>>> It is neither a contradiction nor a paradox because there are three
>>> different instances of Ĥ.
>> I agree that this is neither a paradox nor a contradiction. It's just a
>> fact derived form the logic of how your Ĥ is written (the majority of
>> which you are keeping hidden from us).
>>
>>> Because the only reason that the first instance halts is that Ĥ[0].qx
>>> correctly determines that its input cannot possibly ever reach its
>>> final state of Ĥ[1].qn or Ĥ[1].qy whether or not the simulating halt
>>> decider aborts its simulation of this input, we know with 100%
>>> perfectly justified logical certainty that the input to Ĥ[0].qx never
>>> halts.
>> We know, since you keep telling us, that Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn. This clearly
>> shows that Ĥ applied to ⟨Ĥ⟩ halts. You can see the final state right
>
> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
> if M applied to wM halts, and
>
> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
> if M applied to wM does not halt
>
> You are using the wrong Ĥ.

First of all, let's be 100% clear: I am talking about what /your/ Ĥ
does, based in the facts you have let slip about it.

> Linz stipulates that wM is ⟨Ĥ⟩ and M is the underlying machine of this
> ⟨Ĥ⟩ therefore M applied to wM means ⟨Ĥ⟩ applied to ⟨Ĥ⟩.

No. How many years have you been staring at this one page from Linz?
You still don't know what it says. Do ask me questions, if you'd like
to know what the text you've been sure is wrong for 17 years really
says.

--
Ben.

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct ]

<87czqt2ewt.fsf@bsb.me.uk>

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider [
Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [
succinct ]
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Date: Wed, 04 Aug 2021 20:22:42 +0100
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 by: Ben Bacarisse - Wed, 4 Aug 2021 19:22 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/4/2021 7:53 AM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/2/2021 8:45 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:

>>>>> As long as it is simply dismissed out-of-hand as a contradiction the
>>>>> paradox remains unresolved.
>>>>
>>>> There is no contradiction or paradox. You Ĥ is just the wrong sort of
>>>> TM. The proof you want to "refute" is talking about this sort of Ĥ:
>>>>
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>> if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>
>>> Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>>> Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>>> Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>>> Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>>
>> Maybe saying it a couple more times will help. After four times I can
>> tell you that it's still wrong. Maybe about a dozen more?
>>
>> Whether what happens after Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ is correct or not is determined
>> by Linz, not by you. And you are clear that
>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn.
>
> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
> if M applied to wM does not halt
>
> As explained in complete detail below:
> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Yes, please don't tell me the final state yet again. This is not been
in dispute for some time.

> because M applied to wM does not halt
> where M is Machine_of(⟨Ĥ⟩) (1st param) above
> and wM is ⟨Ĥ⟩ the second param above.
>
> Because wM is referring to ⟨Ĥ⟩ and M is referring to the underlying
> machine of ⟨Ĥ⟩ the last line above is translated to: if
> Machine_of(⟨Ĥ⟩) applied to ⟨Ĥ⟩ does not halt

That's convoluted. ⟨Ĥ⟩ is the encoding of Ĥ so to find out what Linz
expects from Ĥ applied to ⟨Ĥ⟩ we just substitute M = Ĥ and wM = ⟨Ĥ⟩ into
the above:

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt.

> We can know that Machine_of(⟨Ĥ⟩) applied to ⟨Ĥ⟩ never reaches its

Machine_of(⟨Ĥ⟩) = Ĥ.

> final state whether or not the embedded halt decider at Ĥ.qx aborts it

You keep telling me that Ĥ applied to ⟨Ĥ⟩ reaches a final state. You
even tell me the final state.

> simulation of Machine_of(⟨Ĥ⟩), therefore we know that Machine_of(⟨Ĥ⟩)
> never halts. Therefore we know that M applied to wM does not halt.

You keep telling me that Ĥ applied to ⟨Ĥ⟩ reaches the final state Ĥ.qn.
Are you changing your mind after all this time? No, you are searching
for some form of words that will make the wrong answer sound right.

--
Ben.

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs Count ]

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Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs_Count_]
Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc> <871r7i6n2u.fsf@bsb.me.uk> <OqKdnROLKJ9CdJz8nZ2dnUU7-avNnZ2d@giganews.com> <87k0la542c.fsf@bsb.me.uk> <1NidnVPZ-NHDl5_8nZ2dnUU7-enNnZ2d@giganews.com> <87sfzw3ao1.fsf@bsb.me.uk> <7oKdnTjx4IC20p78nZ2dnUU7-TvNnZ2d@giganews.com> <875yws36vt.fsf@bsb.me.uk> <j66dnbdHrpV8_p78nZ2dnUU7-aXNnZ2d@giganews.com> <87im0s0ydp.fsf@bsb.me.uk> <Brqdnfehrf0Kj5n8nZ2dnUU7-X3NnZ2d@giganews.com> <87tukblgjy.fsf@bsb.me.uk> <qtGdnfuXs4nFOZn8nZ2dnUU7-cnNnZ2d@giganews.com> <871r7ekugt.fsf@bsb.me.uk> <K5-dndGZo_-VmJv8nZ2dnUU78QvNnZ2d@giganews.com> <87czqxa0zk.fsf@bsb.me.uk> <woudnXWBxPba95r8nZ2dnUU78ffNnZ2d@giganews.com> <87mtpz64sq.fsf@bsb.me.uk> <w5edne8d06OkxpX8nZ2dnUU7-b_NnZ2d@giganews.com> <87bl6f5qvy.fsf@bsb.me.uk> <1ZednUt-qbrXlZf8nZ2dnUU7-QnNnZ2d@giganews.com> <87tuk52h0e.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 4 Aug 2021 15:37:02 -0500
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 by: olcott - Wed, 4 Aug 2021 20:37 UTC

On 8/4/2021 1:37 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/2/2021 7:11 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/2/2021 2:10 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/1/2021 5:54 AM, Ben Bacarisse wrote:
>>>>>
>>>>>>> I am happy you have a notation you like. Are you prepared to address
>>>>>>> that fact that your H^ is not "as in Linz"?
>>>>>>
>>>>>> My Ĥ is exactly the Linz Ĥ with the additional elaboration that the
>>>>>> second wildcard state transition ⊢* is defined to be a simulating halt
>>>>>> decider.
>>>>> No. I've explained many times now why your Ĥ is not at all "the Linz
>>>>> Ĥ". Do you see any point in my doing so again?
>>> I suspect not. You certainly have not asked a single question that
>>> could help you to understand why your Ĥ is irrelevant.
>>>
>>>>>>> We all know you are declaring that to be correct. Here's why your Ĥ is
>>>>>>> not "as in Linz". Linz requires that
>>>>>>>
>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>>>> if M applied to wM does not halt
>>>>>>>
>>>>>>> Any Ĥ that eventually transitions to Ĥ.qn on input wM must do so
>>>>>>> if, and only if, the encoded M applied to wM does not halt. But you've
>>>>>>> given us a case where your Ĥ is not like this:
>>>>>>>
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>
>>>>>> And when we eliminate the fallacy of equivocation error we have
>>>>>>
>>>>>> Ĥ[0].q0 ⟨Ĥ⟩ ⊢* Ĥ[0].qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ[0].qn
>>>>> The only thing I think you are equivocating on is pretending that Ĥ[0]
>>>>> is not Ĥ, and it doesn't look as if you've removed that error. I think
>>>>> you /should/ remove it so that you can be saying something of value
>>>>> about Ĥ.
>>>>
>>>> It is clear from the text that Linz does specify at least three
>>>> different instances of Ĥ, The TM the TMD input ⟨Ĥ⟩ and a copy of this
>>>> TMD input.
>>> Still equivocating. Either Ĥ[0] = Ĥ or you are wasting everyone's time.
>>> (This is mathematical equality. Ĥ is a tuple of sets. If Ĥ[0] is not
>>> exactly identical in every way to Ĥ then I don't care about it. Note
>>> that I'm not disputing your right, for ease of explanation, to give
>>> identical things more than one name. But the same permission allows me
>>> to use any of the names because they name the same thing.)
>>> You are wrong because
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>> where Linz requires that
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>> if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>>
>> You have the above incorrectly in these two
>> M refers to the Turing Machine described by wM.
>>
>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>> if M applied to wM halts, and
>>
>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>> if M applied to wM does not halt
>>
>> This is *NOT* the Ĥ that is executed.
>
> I have no idea what you are objecting to. You can substitute Ĥ and ⟨Ĥ⟩
> into what Linz says, can't you? I did it above and you don't like it:
>
> For M = Ĥ and wM = ⟨Ĥ⟩ Linz tells us that
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> if Ĥ applied to ⟨Ĥ⟩ does not halt.
>
> Your Ĥ is not like that. Its behaviour is boring and obvious and
> trivial.
>

This is how it really is:
if ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt.
and ⟨Ĥ⟩ applied to ⟨Ĥ⟩ never reaches its final state.

--
Copyright 2021 Pete Olcott

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

Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs Count ]

<877dh03l3c.fsf@bsb.me.uk>

 copy mid

https://www.novabbs.com/devel/article-flat.php?id=19493&group=comp.theory#19493

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Followup: comp.theory
Path: i2pn2.org!i2pn.org!eternal-september.org!reader02.eternal-september.org!.POSTED!not-for-mail
From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider [
Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [
Only Inputs Count ]
Followup-To: comp.theory
Date: Wed, 04 Aug 2021 23:23:51 +0100
Organization: A noiseless patient Spider
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 by: Ben Bacarisse - Wed, 4 Aug 2021 22:23 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/4/2021 1:37 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/2/2021 7:11 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/2/2021 2:10 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 8/1/2021 5:54 AM, Ben Bacarisse wrote:
>>>>>>
>>>>>>>> I am happy you have a notation you like. Are you prepared to address
>>>>>>>> that fact that your H^ is not "as in Linz"?
>>>>>>>
>>>>>>> My Ĥ is exactly the Linz Ĥ with the additional elaboration that the
>>>>>>> second wildcard state transition ⊢* is defined to be a simulating halt
>>>>>>> decider.
>>>>>> No. I've explained many times now why your Ĥ is not at all "the Linz
>>>>>> Ĥ". Do you see any point in my doing so again?
>>>> I suspect not. You certainly have not asked a single question that
>>>> could help you to understand why your Ĥ is irrelevant.
>>>>
>>>>>>>> We all know you are declaring that to be correct. Here's why your Ĥ is
>>>>>>>> not "as in Linz". Linz requires that
>>>>>>>>
>>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>>>>> if M applied to wM does not halt
>>>>>>>>
>>>>>>>> Any Ĥ that eventually transitions to Ĥ.qn on input wM must do so
>>>>>>>> if, and only if, the encoded M applied to wM does not halt. But you've
>>>>>>>> given us a case where your Ĥ is not like this:
>>>>>>>>
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>
>>>>>>> And when we eliminate the fallacy of equivocation error we have
>>>>>>>
>>>>>>> Ĥ[0].q0 ⟨Ĥ⟩ ⊢* Ĥ[0].qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ[0].qn
>>>>>> The only thing I think you are equivocating on is pretending that Ĥ[0]
>>>>>> is not Ĥ, and it doesn't look as if you've removed that error. I think
>>>>>> you /should/ remove it so that you can be saying something of value
>>>>>> about Ĥ.
>>>>>
>>>>> It is clear from the text that Linz does specify at least three
>>>>> different instances of Ĥ, The TM the TMD input ⟨Ĥ⟩ and a copy of this
>>>>> TMD input.
>>>> Still equivocating. Either Ĥ[0] = Ĥ or you are wasting everyone's time.
>>>> (This is mathematical equality. Ĥ is a tuple of sets. If Ĥ[0] is not
>>>> exactly identical in every way to Ĥ then I don't care about it. Note
>>>> that I'm not disputing your right, for ease of explanation, to give
>>>> identical things more than one name. But the same permission allows me
>>>> to use any of the names because they name the same thing.)
>>>> You are wrong because
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>> where Linz requires that
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>> if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>
>>> You have the above incorrectly in these two
>>> M refers to the Turing Machine described by wM.
>>>
>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>> if M applied to wM halts, and
>>>
>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>> if M applied to wM does not halt
>>>
>>> This is *NOT* the Ĥ that is executed.
>> I have no idea what you are objecting to. You can substitute Ĥ and ⟨Ĥ⟩
>> into what Linz says, can't you? I did it above and you don't like it:
>> For M = Ĥ and wM = ⟨Ĥ⟩ Linz tells us that
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>> if Ĥ applied to ⟨Ĥ⟩ does not halt.
>> Your Ĥ is not like that. Its behaviour is boring and obvious and
>> trivial.
>
> This is how it really is:
> if ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt.
> and ⟨Ĥ⟩ applied to ⟨Ĥ⟩ never reaches its final state.

No. ⟨Ĥ⟩ is a string, not a TM. The TM that string represents is Ĥ and
Ĥ applied to ⟨Ĥ⟩ does halt.

--
Ben.

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