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devel / comp.theory / Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.

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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Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.

<IvKdnUpgiu__PpH8nZ2dnUU7-I3NnZ2d@giganews.com>

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https://www.novabbs.com/devel/article-flat.php?id=19553&group=comp.theory#19553

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NNTP-Posting-Date: Thu, 05 Aug 2021 21:50:10 -0500
Subject: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct_and_forms_no_contradiction.
Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc> <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> <j8OdneamG91aK5f8nZ2dnUU7-fvNnZ2d@giganews.com> <87im0l2gc0.fsf@bsb.me.uk> <pu-dnVqdPMG4iJb8nZ2dnUU7-Q_NnZ2d@giganews.com> <87v94k23rn.fsf@bsb.me.uk> <Y5-dnbXu3ONoxpb8nZ2dnUU7-U3NnZ2d@giganews.com> <875ywj1qyk.fsf@bsb.me.uk> <IoydnYCO1pJUEJH8nZ2dnUU7-Q3NnZ2d@giganews.com> <87h7g3z4db.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Thu, 5 Aug 2021 21:50:09 -0500
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 by: olcott - Fri, 6 Aug 2021 02:50 UTC

On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> the Turing machine halting problem. Simply stated, the problem is:
>> given the description of a Turing machine M and an input w, does M,
>> when started in the initial configuration q0w, perform a computation
>> that eventually halts? Using an abbreviated way of talking about the
>> problem, we ask whether M applied to w, or simply (M, w), halts or
>> does not halt.
>> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>>
>> 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
>>
>> M refers to the Turing machine of the inputs not the Turing machine
>> that is being executed.
>
> M refers to the TM encoded as wM. The case in point -- the one you
> claimed to have something important to say about -- has wM = ⟨Ĥ⟩ and
> (consequently) M = Ĥ.
>

Yes the placement of this encoding of Ĥ in the specified machine
templates is crucial in that Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms
no contradiction.

> The key case is exactly the one where M refers to the TM being executed
> because the input, wM, is ⟨Ĥ⟩. Linz writes it out for you (though I'll
> switch to your notation):
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qy ∞ if Ĥ applied to ⟨Ĥ⟩ halts, and
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn if Ĥ applied to ⟨Ĥ⟩ does not halt.
>
> For obvious reasons (if not, ask), you do not have such an Ĥ.
>
>> if Machine_Of(wM) applied to wM does not halt
>
> Yes, Machine_Of(wM) is M. This is saying what Linz is saying though it
> a more wordy way.
>
>> So when Bill says that his identical twin brother is never going to
>> the store and then Bill goes to the store THIS IS NOT A CONTRADICTION.
>
> There is no contradiction. I've said this many times now. There is no
> contradiction (or paradox) with your Ĥ.
>
> Mind you, I don't see what your analogy is trying to tell me so I
> suspect you are wrong about something here. What are the two things
> that look identical but do different things here? Bill and his brother
> are... what... Ĥ and Ĥ or ⟨Ĥ⟩ and ⟨Ĥ⟩?
>
>> When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to Ĥ.qn THIS IS NOT A CONTRADICTION.
>
> <Sigh> How often do I have to say something for you take note of it?
> There is no contradiction or paradox with your Ĥ. Your Ĥ is simply not
> doing what Linz says it should be doing to be interesting. It's doing
> the first part of the spec: Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn but it's not meeting the
> second part because, obviously, Ĥ applied to ⟨Ĥ⟩ halts.
>
> If, 30 months ago, you'd said "I have two TMs H and Ĥ (as per Linz's
> construction) with this behaviour: Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn" no
> one would have cared. I don't know if you genuinely thought you had
> something interesting, but it 100% clear now that you never did.
>
> The false claim you made was that your H/Ĥ pair were not just
> constructed as in Linz, but met Linz's specifications, at least for the
> one key case.
>
>> If you don't want an honest dialogue then please say so.
>
> I can't imagine what you think I am being less than honest about. You,
> on the other hand have said things like:
>
> "Everyone has claimed that H on input pair (Ĥ, Ĥ) meeting the Linz
> specs does not exist. I now have a fully encoded pair of Turing
> Machines H / Ĥ proving them wrong."
>
> Now you may have been deluded or just mistaken when you said this, but
> you must see, now, that you don't have anything meeting Linz's
> specification. It's dishonest to keep saying (or implying) that you do.
>

--
Copyright 2021 Pete Olcott

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

Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.

<0_Sdnb6Qe8XGOZH8nZ2dnUU7-U3NnZ2d@giganews.com>

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NNTP-Posting-Date: Thu, 05 Aug 2021 21:54:19 -0500
Subject: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is corr
ect_and_forms_no_contradiction.
Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc> <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>
<zcadnTSOD5rtZ5f8nZ2dnUU7-T3NnZ2d@giganews.com> <877dh03l3c.fsf@bsb.me.uk>
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From: NoO...@NoWhere.com (olcott)
Date: Thu, 5 Aug 2021 21:54:18 -0500
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 by: olcott - Fri, 6 Aug 2021 02:54 UTC

On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>
>>>> The question is not: Does Ĥ halt on its input?
>>> Yes it is.
>>
>> The question is:
>> Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
>> The ansswer to this question is provably no!
>
> The question is: does Ĥ applied to ⟨Ĥ⟩ halt. It does:
>
>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn THIS IS NOT A CONTRADICTION
>
> Indeed. There is no contradiction. Just an Ĥ that does not meet Linz
> spec.
>

Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.
Because it is correct it meets the Linz spec.

if M applied to wM does not halt
means if the execution of the machine of the first ⟨Ĥ⟩ on its input of
the seocond ⟨Ĥ⟩ does not halt then ⊢* Ĥ.qn

I have proved that the execution of the machine of the first ⟨Ĥ⟩ on its
input of the second ⟨Ĥ⟩ does not halt.

--
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 ]

<OB1PI.137$3w6.14@fx26.iad>

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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
References: <20210719214640.00000dfc@reddwarf.jmc> <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>
<Rr6dnWKhH4ejIJf8nZ2dnUU7-Q3NnZ2d@giganews.com> <87czqt2ewt.fsf@bsb.me.uk>
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Fri, 6 Aug 2021 03:01 UTC

On 8/5/21 8:48 PM, olcott wrote:
> On 8/5/2021 5:15 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/4/2021 6:22 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/4/2021 2:22 PM, Ben Bacarisse wrote:
>>>>>> 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.
>>>>>
>>>>> It is not the first Ĥ that is being referred to it is *only* the
>>>>> machine represented by the input ⟨Ĥ⟩ that is being referred to.
>>>> The machine represented by ⟨Ĥ⟩ is Ĥ.  There is only one Ĥ being
>>>> discussed here -- yours.
>>>>
>>>>> That machine
>>>>> never reaches its final state.
>>>> Ĥ.q0 ⟨Ĥ⟩ transitions to Ĥ.qn, or so you've told us.  Are you changing
>>>> your story?  I'd like an answer.  It's not a hard question.
>>>
>>> Ĥ.q0 ⟨Ĥ⟩ transitions to Ĥ.qn
>>
>> OK, so you are not changing your story.  Linz tells us that
>>
>>    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>    if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>
>
> The input to H will be the description (encoded in some form) of M, say
> WM, as well as the input w. The requirement is then that, given any (WM,
> w), the Turing machine H will halt with either a yes or no answer.
> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>
> The M that is being referred below is the machine specified by WM
> H.q0 WM W ⊢* H.qn
> if M applied to wM does not halt
>
> The M that is being referred below is the machine specified by wM AKA ⟨Ĥ⟩
> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
> if M applied to wM does not halt
>
> When Ĥ.qx is a simulating halt decider the question becomes:
> Does the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ reach its final state.
>

NO.

The question is and always will be, does the Machine that the input is a
description of reach its Halting state in a finite number of steps.

I.e. is H^(H^), or M(w) a halting machine or not. What the aborted
simulation does means absolutely nothing.

If it was an UNABORTED simulation, then it NEVER reaching a final state
would mean something, or it the simulation reached a final state, but an
aborted simulation proves nothing.

Read what the condition were:

They final state was ALWAYS the ACTUAL machine applied to the input it
was given, that would be H^ (the actual machine) applied to the
representation of H^, and does that Halt or not, and even YOU have
agreed that this does halt.

Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.

<ptOdnX2yN-V0NZH8nZ2dnUU7-b_NnZ2d@giganews.com>

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References: <20210719214640.00000dfc@reddwarf.jmc> <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>
<Rr6dnWKhH4ejIJf8nZ2dnUU7-Q3NnZ2d@giganews.com> <87czqt2ewt.fsf@bsb.me.uk>
<8YKdnVBaxrJ_i5b8nZ2dnUU7-aPNnZ2d@giganews.com> <87y29g23sk.fsf@bsb.me.uk>
<no-dnYnca_3rxZb8nZ2dnUU7-YPNnZ2d@giganews.com> <87zgtvzgfl.fsf@bsb.me.uk>
<gZKdnY-eutttCZH8nZ2dnUU7-V3NnZ2d@giganews.com> <OB1PI.137$3w6.14@fx26.iad>
From: NoO...@NoWhere.com (olcott)
Date: Thu, 5 Aug 2021 22:13:44 -0500
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 by: olcott - Fri, 6 Aug 2021 03:13 UTC

On 8/5/2021 10:01 PM, Richard Damon wrote:
> On 8/5/21 8:48 PM, olcott wrote:
>> On 8/5/2021 5:15 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/4/2021 6:22 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/4/2021 2:22 PM, Ben Bacarisse wrote:
>>>>>>> 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.
>>>>>>
>>>>>> It is not the first Ĥ that is being referred to it is *only* the
>>>>>> machine represented by the input ⟨Ĥ⟩ that is being referred to.
>>>>> The machine represented by ⟨Ĥ⟩ is Ĥ.  There is only one Ĥ being
>>>>> discussed here -- yours.
>>>>>
>>>>>> That machine
>>>>>> never reaches its final state.
>>>>> Ĥ.q0 ⟨Ĥ⟩ transitions to Ĥ.qn, or so you've told us.  Are you changing
>>>>> your story?  I'd like an answer.  It's not a hard question.
>>>>
>>>> Ĥ.q0 ⟨Ĥ⟩ transitions to Ĥ.qn
>>>
>>> OK, so you are not changing your story.  Linz tells us that
>>>
>>>    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>    if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>>
>>
>> The input to H will be the description (encoded in some form) of M, say
>> WM, as well as the input w. The requirement is then that, given any (WM,
>> w), the Turing machine H will halt with either a yes or no answer.
>> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>>
>> The M that is being referred below is the machine specified by WM
>> H.q0 WM W ⊢* H.qn
>> if M applied to wM does not halt
>>
>> The M that is being referred below is the machine specified by wM AKA ⟨Ĥ⟩
>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>> if M applied to wM does not halt
>>
>> When Ĥ.qx is a simulating halt decider the question becomes:
>> Does the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ reach its final state.
>>
>
> NO.
>
> The question is and always will be, does the Machine that the input is a
> description of reach its Halting state in a finite number of steps.
Which can be summed up as: Does the Machine that the input is a
description of ever reach its Final state?

--
Copyright 2021 Pete Olcott

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

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.

<802PI.101$bS5.78@fx21.iad>

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References: <20210719214640.00000dfc@reddwarf.jmc> <87im0s0ydp.fsf@bsb.me.uk>
<Brqdnfehrf0Kj5n8nZ2dnUU7-X3NnZ2d@giganews.com> <87tukblgjy.fsf@bsb.me.uk>
<qtGdnfuXs4nFOZn8nZ2dnUU7-cnNnZ2d@giganews.com> <871r7ekugt.fsf@bsb.me.uk>
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Fri, 6 Aug 2021 03:29 UTC

On 8/5/21 10:13 PM, olcott wrote:
> On 8/5/2021 10:01 PM, Richard Damon wrote:
>> On 8/5/21 8:48 PM, olcott wrote:
>>> On 8/5/2021 5:15 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/4/2021 6:22 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 8/4/2021 2:22 PM, Ben Bacarisse wrote:
>>>>>>>> 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.
>>>>>>>
>>>>>>> It is not the first Ĥ that is being referred to it is *only* the
>>>>>>> machine represented by the input ⟨Ĥ⟩ that is being referred to.
>>>>>> The machine represented by ⟨Ĥ⟩ is Ĥ.  There is only one Ĥ being
>>>>>> discussed here -- yours.
>>>>>>
>>>>>>> That machine
>>>>>>> never reaches its final state.
>>>>>> Ĥ.q0 ⟨Ĥ⟩ transitions to Ĥ.qn, or so you've told us.  Are you changing
>>>>>> your story?  I'd like an answer.  It's not a hard question.
>>>>>
>>>>> Ĥ.q0 ⟨Ĥ⟩ transitions to Ĥ.qn
>>>>
>>>> OK, so you are not changing your story.  Linz tells us that
>>>>
>>>>     Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>     if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>
>>>
>>> The input to H will be the description (encoded in some form) of M, say
>>> WM, as well as the input w. The requirement is then that, given any (WM,
>>> w), the Turing machine H will halt with either a yes or no answer.
>>> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>>>
>>> The M that is being referred below is the machine specified by WM
>>> H.q0 WM W ⊢* H.qn
>>> if M applied to wM does not halt
>>>
>>> The M that is being referred below is the machine specified by wM AKA
>>> ⟨Ĥ⟩
>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>> if M applied to wM does not halt
>>>
>>> When Ĥ.qx is a simulating halt decider the question becomes:
>>> Does the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ reach its final state.
>>>
>>
>> NO.
>>
>> The question is and always will be, does the Machine that the input is a
>> description of reach its Halting state in a finite number of steps.
> Which can be summed up as: Does the Machine that the input is a
> description of ever reach its Final state?
>

Right, and we have shown that the H^(H^) that is the top level machine
(and thus can't be aborted) does reach its final state, and thus ALL
copies of H^(H^) will do the same. The fact that an aborted version
didn't get there YET, doesn't say that it wouldn't if you continued the
simulation of THAT EXACT MACHINE farther till it was ACTUALLY completted.

Thus non-Halting is the wrong answer.

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.

<982PI.4528$Fx8.3541@fx45.iad>

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Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
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Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc> <875yws36vt.fsf@bsb.me.uk>
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Fri, 6 Aug 2021 03:37 UTC

On 8/5/21 9:54 PM, olcott wrote:
> On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>  
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>
>>>>> The question is not: Does Ĥ halt on its input?
>>>> Yes it is.
>>>
>>> The question is:
>>> Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
>>> The ansswer to this question is provably no!
>>
>> The question is: does Ĥ applied to ⟨Ĥ⟩ halt.  It does:
>>
>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn  THIS IS NOT A CONTRADICTION
>>
>> Indeed.  There is no contradiction.  Just an Ĥ that does not meet Linz
>> spec.
>>
>
> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.
> Because it is correct it meets the Linz spec.
>
> if M applied to wM does not halt
> means if the execution of the machine of the first ⟨Ĥ⟩ on its input of
> the seocond ⟨Ĥ⟩ does not halt then ⊢* Ĥ.qn
>
> I have proved that the execution of the machine of the first ⟨Ĥ⟩ on its
> input of the second ⟨Ĥ⟩ does not halt.
>
>

Since M is H^, and H^ applied to H^, in other notations H^(<H^>), has
been PROVED and accepted by you, to Halt when run as the actual Turing
Machine

What you have shown is that the ABORTED simulation done by H of H^
applied to H^ didn't reach its final state in the limited number of
steps simulated, but from the trace of the actual run of H^(<H^>) we
know that that exact same computation with the exact same trace segment,
when it continues does reach that halting state.

That H^.q0 <H^> reached H^.qn is PROOF that H^(<H^>) is a Halting
computation, as H^.qn is a terminal Halting State off H^

We also have that since H^.qx is the equivalent of H.q0, and H^.qn is
the equivalent of H.qn, the state where H responds that H has decided
that its input is non-Halting, we also know that H(<H^>, <H^>) has
decided the H^(<H^>) is non-Halting, and since we have just also showed
that it actually is Halting (as the H^.qn is a halting state) we have
shown that H was WRONG.

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.

<rP2dnRpdyrC_LZH8nZ2dnUU7-e2dnZ2d@giganews.com>

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Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
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Newsgroups: comp.theory
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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Fri, 6 Aug 2021 03:44 UTC

On 8/5/2021 10:37 PM, Richard Damon wrote:
> On 8/5/21 9:54 PM, olcott wrote:
>> On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>
>>>>>> The question is not: Does Ĥ halt on its input?
>>>>> Yes it is.
>>>>
>>>> The question is:
>>>> Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
>>>> The ansswer to this question is provably no!
>>>
>>> The question is: does Ĥ applied to ⟨Ĥ⟩ halt.  It does:
>>>
>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn  THIS IS NOT A CONTRADICTION
>>>
>>> Indeed.  There is no contradiction.  Just an Ĥ that does not meet Linz
>>> spec.
>>>
>>
>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.
>> Because it is correct it meets the Linz spec.
>>
>> if M applied to wM does not halt
>> means if the execution of the machine of the first ⟨Ĥ⟩ on its input of
>> the seocond ⟨Ĥ⟩ does not halt then ⊢* Ĥ.qn
>>
>> I have proved that the execution of the machine of the first ⟨Ĥ⟩ on its
>> input of the second ⟨Ĥ⟩ does not halt.
>>
>>
>
> Since M is H^, and H^ applied to H^, in other notations H^(<H^>), has
> been PROVED and accepted by you, to Halt when run as the actual Turing
> Machine
>

The input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts.

--
Copyright 2021 Pete Olcott

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

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.

<seibjs$k2d$1@dont-email.me>

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From: news.x.r...@xoxy.net (Richard Damon)
Newsgroups: comp.theory
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct_and_forms_no_contradiction.
Date: Thu, 5 Aug 2021 22:51:22 -0500
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 by: Richard Damon - Fri, 6 Aug 2021 03:51 UTC

On 8/5/21 10:44 PM, olcott wrote:
> On 8/5/2021 10:37 PM, Richard Damon wrote:
>> On 8/5/21 9:54 PM, olcott wrote:
>>> On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>  
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>
>>>>>>> The question is not: Does Ĥ halt on its input?
>>>>>> Yes it is.
>>>>>
>>>>> The question is:
>>>>> Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
>>>>> The ansswer to this question is provably no!
>>>>
>>>> The question is: does Ĥ applied to ⟨Ĥ⟩ halt.  It does:
>>>>
>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn  THIS IS NOT A CONTRADICTION
>>>>
>>>> Indeed.  There is no contradiction.  Just an Ĥ that does not meet Linz
>>>> spec.
>>>>
>>>
>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.
>>> Because it is correct it meets the Linz spec.
>>>
>>> if M applied to wM does not halt
>>> means if the execution of the machine of the first ⟨Ĥ⟩ on its input of
>>> the seocond ⟨Ĥ⟩ does not halt then ⊢* Ĥ.qn
>>>
>>> I have proved that the execution of the machine of the first ⟨Ĥ⟩ on its
>>> input of the second ⟨Ĥ⟩ does not halt.
>>>
>>>
>>
>> Since M is H^, and H^ applied to H^, in other notations H^(<H^>), has
>> been PROVED and accepted by you, to Halt when run as the actual Turing
>> Machine
>>
>
> The input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts.
>

Category error.

Input don't halt or not, Machines Halt or not on an input.

The machine that is described by <H^> <H^> is H^(<H^>) which has been
accepted to be a Halting Computation, even by you, so your claim is false.

The fact that an H that can answer H(<H^>, <H^>) will never run its
simulation long enough to see it reach that terminal state just show the
error in H, not that H^(<H^>) is non-Halting.

Now, if you mean that H never halts when given <H^> <H^> as a input,
then that is something different, that just shows that in that case, H
has just proven that it fails to be a decider for that input.

Re: Black box halt decider is NOT a partial decider

<78bf063c-ac26-48d9-a24d-b0fc52f4fccfn@googlegroups.com>

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Subject: Re: Black box halt decider is NOT a partial decider
From: wyni...@gmail.com (wij)
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 by: wij - Fri, 6 Aug 2021 04:58 UTC

On Thursday, 5 August 2021 at 12:55:01 UTC+8, Chris M. Thomasson wrote:
> On 8/4/2021 9:53 PM, Chris M. Thomasson wrote:
> > On 7/23/2021 6:34 PM, Mike Terry wrote:
> >> On 24/07/2021 00:09, Chris M. Thomasson wrote:
> >>> On 7/23/2021 3:49 PM, André G. Isaak wrote:
> >>>> On 2021-07-23 16:28, Chris M. Thomasson wrote:
> >>>>
> >>>>> Sum[n=1..oo] 1/n^24 converges on 1.
> >>>>
> >>>> That's certainly news to me!
> >>>
> >>> https://www.wolframalpha.com/input/?i=Sum%5Bn%3D1..oo%5D+1%2Fn%5E24
> >>>
> >>> Bugger in Wolfram?
> >>>
> >>>
> >>
> >> The n=1 term is 1/1^24 = 1, and then there's all those other positive
> >> terms, so obviously the limit can't be 1. :)
> >>
> >> The Wolfram page is saying the limit is close to 1, which makes sense
> >> as all terms in the series apart from the first are quite small!
> >
> > close to 1 vs arbitrarily close to 1?
> >
> For instance, .999999999999999 is close to 1, however, its not
> arbitrarily close to 1 like .999... is

0.999...≠1 (except "0.999..." is ill-formed for analysis)

I created a New conversation to answer your question
https://groups.google.com/g/comp.theory/c/B2mIO2gTjEE

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.

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

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct and forms no contradiction.
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 by: Ben Bacarisse - Sun, 8 Aug 2021 00:34 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>
>>>>> The question is not: Does Ĥ halt on its input?
>>>> Yes it is.
>>>
>>> The question is:
>>> Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
>>> The ansswer to this question is provably no!
>> The question is: does Ĥ applied to ⟨Ĥ⟩ halt. It does:
>>
>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn THIS IS NOT A CONTRADICTION
>> Indeed. There is no contradiction. Just an Ĥ that does not meet Linz
>> spec.
>
> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.
> Because it is correct it meets the Linz spec.

I find it startling that you think that, but then it seems you don't yet
know what the key words mean:

> if M applied to wM does not halt
> means if the execution of the machine of the first ⟨Ĥ⟩ on its input of
> the seocond ⟨Ĥ⟩ does not halt then ⊢* Ĥ.qn

No. Would you like to know "what M applied to wM does not halt" means?
Do you need help to see that "Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn" is clearly a case of "M
applied to wM halts"?

--
Ben.

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.

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

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct and forms no contradiction.
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 by: Ben Bacarisse - Sun, 8 Aug 2021 00:34 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> the Turing machine halting problem. Simply stated, the problem is:
>>> given the description of a Turing machine M and an input w, does M,
>>> when started in the initial configuration q0w, perform a computation
>>> that eventually halts? Using an abbreviated way of talking about the
>>> problem, we ask whether M applied to w, or simply (M, w), halts or
>>> does not halt.
>>> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>>>
>>> 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
>>>
>>> M refers to the Turing machine of the inputs not the Turing machine
>>> that is being executed.
>> M refers to the TM encoded as wM. The case in point -- the one you
>> claimed to have something important to say about -- has wM = ⟨Ĥ⟩ and
>> (consequently) M = Ĥ.
>
> Yes the placement of this encoding of Ĥ in the specified machine
> templates is crucial in that Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms
> no contradiction.

There's no contradiction (for the umpteenth time). It just shows that
you've chosen to write H (and hence Ĥ) in such a way that Ĥ.q0 ⟨Ĥ⟩ ⊢*
Ĥ.qn when Linz says it shouldn't. No one cares about such TMs.

--
Ben.

Re: Page 6 conclusively proves that H(P,P)==0 is correct.

<87tuk0lqj9.fsf@bsb.me.uk>

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Newsgroups: comp.theory
Subject: Re: Page 6 conclusively proves that H(P,P)==0 is correct.
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 by: Ben Bacarisse - Sun, 8 Aug 2021 00:38 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/5/2021 9:35 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> The M that is being referred below is the machine specified by wM AKA ⟨Ĥ⟩
>> Yes, so M is Ĥ (what you wrote, rather redundantly, as Machine_Of(⟨Ĥ⟩).
>>
>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>> if M applied to wM does not halt
>> So
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>> if Ĥ applied to ⟨Ĥ⟩ does not halt
>> Linz even writes this out for you with the correct substitutions.
>>
>>> When Ĥ.qx is a simulating halt decider the question becomes:
>>> Does the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ reach its final state.
>> Your implementation choices for H (and thus Ĥ) do not change the
>> specification. You Ĥ does not meet it.
>> We can go round and round this loop forever, but you can't show that
>> your Ĥ does what it should to be interesting.
>
> That you ignore rather than point out errors in my reasoning really
> seems to prove that you don't want an honest dialogue.

I hope others can see the irony of the remark, attached as it is to a
paragraph that points out what's wrong with your reasoning.

> If you really do want an honest dialogue then we must go through the
> points on page 6 until we have mutual agreement. Page 6 conclusively
> proves that H(P,P)==0 is correct.

You keep posting undisputed claims that show that your Ĥ does not match
the specification in Linz. There's nothing more to say. If you don't
understand this, I might be able to help, but you are determined to
concentrate on the way in which H is wrong rather than the plain fact
that it is.

--
Ben.

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.

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<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>
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From: NoO...@NoWhere.com (olcott)
Date: Mon, 9 Aug 2021 17:32:05 -0500
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 by: olcott - Mon, 9 Aug 2021 22:32 UTC

On 8/7/2021 7:34 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>
>>>>>> The question is not: Does Ĥ halt on its input?
>>>>> Yes it is.
>>>>
>>>> The question is:
>>>> Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
>>>> The ansswer to this question is provably no!
>>> The question is: does Ĥ applied to ⟨Ĥ⟩ halt. It does:
>>>
>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn THIS IS NOT A CONTRADICTION
>>> Indeed. There is no contradiction. Just an Ĥ that does not meet Linz
>>> spec.
>>
>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.
>> Because it is correct it meets the Linz spec.
>
> I find it startling that you think that, but then it seems you don't yet
> know what the key words mean:
>
>> if M applied to wM does not halt
>> means if the execution of the machine of the first ⟨Ĥ⟩ on its input of
>> the seocond ⟨Ĥ⟩ does not halt then ⊢* Ĥ.qn
>
> No. Would you like to know "what M applied to wM does not halt" means?
> Do you need help to see that "Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn" is clearly a case of "M
> applied to wM halts"?
>

the Turing machine halting problem. Simply stated, the problem
is: given the description of a Turing machine M and an input w,
does M, when started in the initial configuration q0w, perform a
computation that eventually halts? (Linz:1990:317).

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn

When Ĥ is applied to ⟨Ĥ⟩ the description of the Turing Machine and its
input are specified as: ⟨Ĥ⟩ ⟨Ĥ⟩ for the embedded halt decider at Ĥ.qx.

When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn this is not a final state of the simulated
input it is a final state of the executed Ĥ.

--
Copyright 2021 Pete Olcott

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

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.

<2qkQI.22663$Fx8.8260@fx45.iad>

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Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct_and_forms_no_contradiction.
Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc> <87im0s0ydp.fsf@bsb.me.uk>
<Brqdnfehrf0Kj5n8nZ2dnUU7-X3NnZ2d@giganews.com> <87tukblgjy.fsf@bsb.me.uk>
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<w5edne8d06OkxpX8nZ2dnUU7-b_NnZ2d@giganews.com> <87bl6f5qvy.fsf@bsb.me.uk>
<1ZednUt-qbrXlZf8nZ2dnUU7-QnNnZ2d@giganews.com> <87tuk52h0e.fsf@bsb.me.uk>
<zcadnTSOD5rtZ5f8nZ2dnUU7-T3NnZ2d@giganews.com> <877dh03l3c.fsf@bsb.me.uk>
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<4JOdnRS2SLR7MYz8nZ2dnUU7-YnNnZ2d@giganews.com>
From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Tue, 10 Aug 2021 01:14 UTC

On 8/9/21 6:32 PM, olcott wrote:
> On 8/7/2021 7:34 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>   
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>
>>>>>>> The question is not: Does Ĥ halt on its input?
>>>>>> Yes it is.
>>>>>
>>>>> The question is:
>>>>> Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
>>>>> The ansswer to this question is provably no!
>>>> The question is: does Ĥ applied to ⟨Ĥ⟩ halt.  It does:
>>>>
>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn  THIS IS NOT A CONTRADICTION
>>>> Indeed.  There is no contradiction.  Just an Ĥ that does not meet Linz
>>>> spec.
>>>
>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.
>>> Because it is correct it meets the Linz spec.
>>
>> I find it startling that you think that, but then it seems you don't yet
>> know what the key words mean:
>>
>>> if M applied to wM does not halt
>>> means if the execution of the machine of the first ⟨Ĥ⟩ on its input of
>>> the seocond ⟨Ĥ⟩ does not halt then ⊢* Ĥ.qn
>>
>> No.  Would you like to know "what M applied to wM does not halt" means?
>> Do you need help to see that "Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn" is clearly a case of "M
>> applied to wM halts"?
>>
>
>      the Turing machine halting problem. Simply stated, the problem
>      is: given the description of a Turing machine M and an input w,
>      does M, when started in the initial configuration q0w, perform a
>      computation that eventually halts? (Linz:1990:317).
>
> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>
> When Ĥ is applied to ⟨Ĥ⟩ the description of the Turing Machine and its
> input are specified as: ⟨Ĥ⟩ ⟨Ĥ⟩ for the embedded halt decider at Ĥ.qx.
>
> When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn this is not a final state of the simulated
> input it is a final state of the executed Ĥ.
>
>

H^.qn IS the Halting state of the H^ machine that is being run, which is
what really counts.

The fact that the simulated version of H^ had its simulation aborted
means that the fact that this simulation didn't reach a halting state
does NOT prove that machine is non-halting. The fact that a copy of the
ACTUAL machine Halts, proves that it IS halting.

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.

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

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct and forms no contradiction.
Date: Wed, 11 Aug 2021 01:42:30 +0100
Organization: A noiseless patient Spider
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 by: Ben Bacarisse - Wed, 11 Aug 2021 00:42 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/7/2021 7:34 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>
>>>>>>> The question is not: Does Ĥ halt on its input?
>>>>>> Yes it is.
>>>>>
>>>>> The question is:
>>>>> Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
>>>>> The ansswer to this question is provably no!
>>>> The question is: does Ĥ applied to ⟨Ĥ⟩ halt. It does:
>>>>
>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn THIS IS NOT A CONTRADICTION
>>>> Indeed. There is no contradiction. Just an Ĥ that does not meet Linz
>>>> spec.
>>>
>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.
>>> Because it is correct it meets the Linz spec.
>> I find it startling that you think that, but then it seems you don't yet
>> know what the key words mean:
>>
>>> if M applied to wM does not halt
>>> means if the execution of the machine of the first ⟨Ĥ⟩ on its input of
>>> the seocond ⟨Ĥ⟩ does not halt then ⊢* Ĥ.qn
>> No. Would you like to know "what M applied to wM does not halt" means?
>> Do you need help to see that "Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn" is clearly a case of "M
>> applied to wM halts"?
>
> the Turing machine halting problem. Simply stated, the problem
> is: given the description of a Turing machine M and an input w,
> does M, when started in the initial configuration q0w, perform a
> computation that eventually halts? (Linz:1990:317).

Yes. I was offering to help you understand the key words in that text.

> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn

You've missed off the key lines yet again. Is that deliberate? They
are the lines that show you are wrong so I am suspicious that you keep
omitting them.

> When Ĥ is applied to ⟨Ĥ⟩ the description of the Turing Machine and its
> input are specified as: ⟨Ĥ⟩ ⟨Ĥ⟩ for the embedded halt decider at Ĥ.qx.

Ungrammatical.

> When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn this is not a final state of the simulated
> input it is a final state of the executed Ĥ.

Yes. You don't seem to know why that's wrong. If you'd like to know,
ask me questions. If you want to keep just saying things for the fun of
it, you are doing just fine without any help.

--
Ben.

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.

<brCdnZyB6OOdg478nZ2dnUU7-ePNnZ2d@giganews.com>

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Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct_and_forms_no_contradiction.
Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc> <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> <zcadnTSOD5rtZ5f8nZ2dnUU7-T3NnZ2d@giganews.com> <877dh03l3c.fsf@bsb.me.uk> <Z5adnd038KGXwJb8nZ2dnUU7-I_NnZ2d@giganews.com> <8735rn1qvj.fsf@bsb.me.uk> <goydnfCCIYUWE5H8nZ2dnUU7-e_NnZ2d@giganews.com> <87eeb7z4d1.fsf@bsb.me.uk> <0_Sdnb6Qe8XGOZH8nZ2dnUU7-U3NnZ2d@giganews.com> <87zgtslqpv.fsf@bsb.me.uk> <4JOdnRS2SLR7MYz8nZ2dnUU7-YnNnZ2d@giganews.com> <87zgtoizgp.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Tue, 10 Aug 2021 19:46:54 -0500
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 by: olcott - Wed, 11 Aug 2021 00:46 UTC

On 8/10/2021 7:42 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/7/2021 7:34 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>
>>>>>>>> The question is not: Does Ĥ halt on its input?
>>>>>>> Yes it is.
>>>>>>
>>>>>> The question is:
>>>>>> Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
>>>>>> The ansswer to this question is provably no!
>>>>> The question is: does Ĥ applied to ⟨Ĥ⟩ halt. It does:
>>>>>
>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn THIS IS NOT A CONTRADICTION
>>>>> Indeed. There is no contradiction. Just an Ĥ that does not meet Linz
>>>>> spec.
>>>>
>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.
>>>> Because it is correct it meets the Linz spec.
>>> I find it startling that you think that, but then it seems you don't yet
>>> know what the key words mean:
>>>
>>>> if M applied to wM does not halt
>>>> means if the execution of the machine of the first ⟨Ĥ⟩ on its input of
>>>> the seocond ⟨Ĥ⟩ does not halt then ⊢* Ĥ.qn
>>> No. Would you like to know "what M applied to wM does not halt" means?
>>> Do you need help to see that "Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn" is clearly a case of "M
>>> applied to wM halts"?
>>
>> the Turing machine halting problem. Simply stated, the problem
>> is: given the description of a Turing machine M and an input w,
>> does M, when started in the initial configuration q0w, perform a
>> computation that eventually halts? (Linz:1990:317).
>
> Yes. I was offering to help you understand the key words in that text.
>
>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>
> You've missed off the key lines yet again. Is that deliberate? They
> are the lines that show you are wrong so I am suspicious that you keep
> omitting them.
>
>> When Ĥ is applied to ⟨Ĥ⟩ the description of the Turing Machine and its
>> input are specified as: ⟨Ĥ⟩ ⟨Ĥ⟩ for the embedded halt decider at Ĥ.qx.
>
> Ungrammatical.
>
>> When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn this is not a final state of the simulated
>> input it is a final state of the executed Ĥ.
>
> Yes. You don't seem to know why that's wrong.

Because it is not wrong and you are lying and you can't possibly
correctly show otherwise.

> If you'd like to know,
> ask me questions. If you want to keep just saying things for the fun of
> it, you are doing just fine without any help.
>

--
Copyright 2021 Pete Olcott

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

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.

<Z6ednWt7SpIIv478nZ2dnUU7-WHNnZ2d@giganews.com>

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Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct_and_forms_no_contradiction.
Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc> <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> <zcadnTSOD5rtZ5f8nZ2dnUU7-T3NnZ2d@giganews.com> <877dh03l3c.fsf@bsb.me.uk> <Z5adnd038KGXwJb8nZ2dnUU7-I_NnZ2d@giganews.com> <8735rn1qvj.fsf@bsb.me.uk> <goydnfCCIYUWE5H8nZ2dnUU7-e_NnZ2d@giganews.com> <87eeb7z4d1.fsf@bsb.me.uk> <0_Sdnb6Qe8XGOZH8nZ2dnUU7-U3NnZ2d@giganews.com> <87zgtslqpv.fsf@bsb.me.uk> <4JOdnRS2SLR7MYz8nZ2dnUU7-YnNnZ2d@giganews.com> <87zgtoizgp.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Tue, 10 Aug 2021 20:06:27 -0500
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 by: olcott - Wed, 11 Aug 2021 01:06 UTC

On 8/10/2021 7:42 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/7/2021 7:34 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>
>>>>>>>> The question is not: Does Ĥ halt on its input?
>>>>>>> Yes it is.
>>>>>>
>>>>>> The question is:
>>>>>> Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
>>>>>> The ansswer to this question is provably no!
>>>>> The question is: does Ĥ applied to ⟨Ĥ⟩ halt. It does:
>>>>>
>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn THIS IS NOT A CONTRADICTION
>>>>> Indeed. There is no contradiction. Just an Ĥ that does not meet Linz
>>>>> spec.
>>>>
>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.
>>>> Because it is correct it meets the Linz spec.
>>> I find it startling that you think that, but then it seems you don't yet
>>> know what the key words mean:
>>>
>>>> if M applied to wM does not halt
>>>> means if the execution of the machine of the first ⟨Ĥ⟩ on its input of
>>>> the seocond ⟨Ĥ⟩ does not halt then ⊢* Ĥ.qn
>>> No. Would you like to know "what M applied to wM does not halt" means?
>>> Do you need help to see that "Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn" is clearly a case of "M
>>> applied to wM halts"?
>>
>> the Turing machine halting problem. Simply stated, the problem
>> is: given the description of a Turing machine M and an input w,
>> does M, when started in the initial configuration q0w, perform a
>> computation that eventually halts? (Linz:1990:317).
>
> Yes. I was offering to help you understand the key words in that text.
>
>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>
> You've missed off the key lines yet again. Is that deliberate? They
> are the lines that show you are wrong so I am suspicious that you keep
> omitting them.
>
>> When Ĥ is applied to ⟨Ĥ⟩ the description of the Turing Machine and its
>> input are specified as: ⟨Ĥ⟩ ⟨Ĥ⟩ for the embedded halt decider at Ĥ.qx.
>
> Ungrammatical.
>
>> When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn this is not a final state of the simulated
>> input it is a final state of the executed Ĥ.
>
> Yes. You don't seem to know why that's wrong.

What is your basis for believing that is wrong?

> If you'd like to know,
> ask me questions. If you want to keep just saying things for the fun of
> it, you are doing just fine without any help.
>

--
Copyright 2021 Pete Olcott

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

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.

<87pmukiwr5.fsf@bsb.me.uk>

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct and forms no contradiction.
Date: Wed, 11 Aug 2021 02:41:02 +0100
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 by: Ben Bacarisse - Wed, 11 Aug 2021 01:41 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/10/2021 7:42 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/7/2021 7:34 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>
>>>>>>>>> The question is not: Does Ĥ halt on its input?
>>>>>>>> Yes it is.
>>>>>>>
>>>>>>> The question is:
>>>>>>> Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
>>>>>>> The ansswer to this question is provably no!
>>>>>> The question is: does Ĥ applied to ⟨Ĥ⟩ halt. It does:
>>>>>>
>>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn THIS IS NOT A CONTRADICTION
>>>>>> Indeed. There is no contradiction. Just an Ĥ that does not meet Linz
>>>>>> spec.
>>>>>
>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.
>>>>> Because it is correct it meets the Linz spec.
>>>> I find it startling that you think that, but then it seems you don't yet
>>>> know what the key words mean:
>>>>
>>>>> if M applied to wM does not halt
>>>>> means if the execution of the machine of the first ⟨Ĥ⟩ on its input of
>>>>> the seocond ⟨Ĥ⟩ does not halt then ⊢* Ĥ.qn
>>>> No. Would you like to know "what M applied to wM does not halt" means?
>>>> Do you need help to see that "Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn" is clearly a case of "M
>>>> applied to wM halts"?
>>>
>>> the Turing machine halting problem. Simply stated, the problem
>>> is: given the description of a Turing machine M and an input w,
>>> does M, when started in the initial configuration q0w, perform a
>>> computation that eventually halts? (Linz:1990:317).
>> Yes. I was offering to help you understand the key words in that text.
>>
>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>> You've missed off the key lines yet again. Is that deliberate? They
>> are the lines that show you are wrong so I am suspicious that you keep
>> omitting them.
>>
>>> When Ĥ is applied to ⟨Ĥ⟩ the description of the Turing Machine and its
>>> input are specified as: ⟨Ĥ⟩ ⟨Ĥ⟩ for the embedded halt decider at Ĥ.qx.
>> Ungrammatical.
>>
>>> When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn this is not a final state of the simulated
>>> input it is a final state of the executed Ĥ.
>> Yes. You don't seem to know why that's wrong.
>
> What is your basis for believing that is wrong?

Ah, a question about what I'm saying. I can help there. The basis is
what Linz says about Ĥ. He says that (translating to your notation)

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn

should be the case "if Ĥ applied to ⟨Ĥ⟩ does not halt". But, as you can
see, your Ĥ does halt when applied to ⟨Ĥ⟩ (qn is a halting or final
state). Your Ĥ is not doing what it should in this one crucial case.

It is trivial to write a TM J from which a Ĵ can be derived that has the
properties of your Ĥ. Every student learning this material for the first
time should be able to come up with one a few minutes. There is nothing
interesting about a TM with

Ĵ.q0 ⟨Ĵ⟩ ⊢* Ĵ.qn
if Ĵ applied to ⟨Ĵ⟩ halts.

What you can't have (and you don't have) is a TM that behaves like
Linz's in this one particular case:

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

--
Ben.

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.

<5c6dnbn2gMMIsI78nZ2dnUU7-dnNnZ2d@giganews.com>

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NNTP-Posting-Date: Tue, 10 Aug 2021 20:53:25 -0500
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct_and_forms_no_contradiction.
Newsgroups: comp.theory
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From: NoO...@NoWhere.com (olcott)
Date: Tue, 10 Aug 2021 20:53:23 -0500
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 by: olcott - Wed, 11 Aug 2021 01:53 UTC

On 8/10/2021 8:41 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/10/2021 7:42 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/7/2021 7:34 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>
>>>>>>>>>> The question is not: Does Ĥ halt on its input?
>>>>>>>>> Yes it is.
>>>>>>>>
>>>>>>>> The question is:
>>>>>>>> Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
>>>>>>>> The ansswer to this question is provably no!
>>>>>>> The question is: does Ĥ applied to ⟨Ĥ⟩ halt. It does:
>>>>>>>
>>>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn THIS IS NOT A CONTRADICTION
>>>>>>> Indeed. There is no contradiction. Just an Ĥ that does not meet Linz
>>>>>>> spec.
>>>>>>
>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.
>>>>>> Because it is correct it meets the Linz spec.
>>>>> I find it startling that you think that, but then it seems you don't yet
>>>>> know what the key words mean:
>>>>>
>>>>>> if M applied to wM does not halt
>>>>>> means if the execution of the machine of the first ⟨Ĥ⟩ on its input of
>>>>>> the seocond ⟨Ĥ⟩ does not halt then ⊢* Ĥ.qn
>>>>> No. Would you like to know "what M applied to wM does not halt" means?
>>>>> Do you need help to see that "Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn" is clearly a case of "M
>>>>> applied to wM halts"?
>>>>
>>>> the Turing machine halting problem. Simply stated, the problem
>>>> is: given the description of a Turing machine M and an input w,
>>>> does M, when started in the initial configuration q0w, perform a
>>>> computation that eventually halts? (Linz:1990:317).
>>> Yes. I was offering to help you understand the key words in that text.
>>>
>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>> You've missed off the key lines yet again. Is that deliberate? They
>>> are the lines that show you are wrong so I am suspicious that you keep
>>> omitting them.
>>>
>>>> When Ĥ is applied to ⟨Ĥ⟩ the description of the Turing Machine and its
>>>> input are specified as: ⟨Ĥ⟩ ⟨Ĥ⟩ for the embedded halt decider at Ĥ.qx.
>>> Ungrammatical.
>>>
>>>> When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn this is not a final state of the simulated
>>>> input it is a final state of the executed Ĥ.
>>> Yes. You don't seem to know why that's wrong.
>>
>> What is your basis for believing that is wrong?
>
> Ah, a question about what I'm saying. I can help there. The basis is
> what Linz says about Ĥ. He says that (translating to your notation)
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>
> should be the case "if Ĥ applied to ⟨Ĥ⟩ does not halt". But, as you can
> see, your Ĥ does halt when applied to ⟨Ĥ⟩ (qn is a halting or final
> state). Your Ĥ is not doing what it should in this one crucial case.
>

the Turing machine halting problem. Simply stated, the problem
is: given the description of a Turing machine M and an input w,
does M, when started in the initial configuration q0w, perform a
computation that eventually halts? (Linz:1990:317).

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

But you are ignoring:
(a) The copying step.
(b) The point in the execution trace where the halt decider is located.
(c) That the halt decider only takes the description of a machine as input.
(d) That the halting question is applied to the machine of this
description: AKA the first param to: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩
(e) that the first param to: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts

> It is trivial to write a TM J from which a Ĵ can be derived that has the
> properties of your Ĥ. Every student learning this material for the first
> time should be able to come up with one a few minutes. There is nothing
> interesting about a TM with
>
> Ĵ.q0 ⟨Ĵ⟩ ⊢* Ĵ.qn
> if Ĵ applied to ⟨Ĵ⟩ halts.
>
> What you can't have (and you don't have) is a TM that behaves like
> Linz's in this one particular case:
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
> if Ĥ applied to ⟨Ĥ⟩ does not halt.
>

--
Copyright 2021 Pete Olcott

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

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.

<87eeb0iuo1.fsf@bsb.me.uk>

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https://www.novabbs.com/devel/article-flat.php?id=19655&group=comp.theory#19655

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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: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct and forms no contradiction.
Date: Wed, 11 Aug 2021 03:26:06 +0100
Organization: A noiseless patient Spider
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 by: Ben Bacarisse - Wed, 11 Aug 2021 02:26 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/10/2021 8:41 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/10/2021 7:42 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/7/2021 7:34 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>
>>>>>>>>>>> The question is not: Does Ĥ halt on its input?
>>>>>>>>>> Yes it is.
>>>>>>>>>
>>>>>>>>> The question is:
>>>>>>>>> Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
>>>>>>>>> The ansswer to this question is provably no!
>>>>>>>> The question is: does Ĥ applied to ⟨Ĥ⟩ halt. It does:
>>>>>>>>
>>>>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn THIS IS NOT A CONTRADICTION
>>>>>>>> Indeed. There is no contradiction. Just an Ĥ that does not meet Linz
>>>>>>>> spec.
>>>>>>>
>>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.
>>>>>>> Because it is correct it meets the Linz spec.
>>>>>> I find it startling that you think that, but then it seems you don't yet
>>>>>> know what the key words mean:
>>>>>>
>>>>>>> if M applied to wM does not halt
>>>>>>> means if the execution of the machine of the first ⟨Ĥ⟩ on its input of
>>>>>>> the seocond ⟨Ĥ⟩ does not halt then ⊢* Ĥ.qn
>>>>>> No. Would you like to know "what M applied to wM does not halt" means?
>>>>>> Do you need help to see that "Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn" is clearly a case of "M
>>>>>> applied to wM halts"?
>>>>>
>>>>> the Turing machine halting problem. Simply stated, the problem
>>>>> is: given the description of a Turing machine M and an input w,
>>>>> does M, when started in the initial configuration q0w, perform a
>>>>> computation that eventually halts? (Linz:1990:317).
>>>> Yes. I was offering to help you understand the key words in that text.
>>>>
>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>> You've missed off the key lines yet again. Is that deliberate? They
>>>> are the lines that show you are wrong so I am suspicious that you keep
>>>> omitting them.
>>>>
>>>>> When Ĥ is applied to ⟨Ĥ⟩ the description of the Turing Machine and its
>>>>> input are specified as: ⟨Ĥ⟩ ⟨Ĥ⟩ for the embedded halt decider at Ĥ.qx.
>>>> Ungrammatical.
>>>>
>>>>> When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn this is not a final state of the simulated
>>>>> input it is a final state of the executed Ĥ.
>>>> Yes. You don't seem to know why that's wrong.
>>>
>>> What is your basis for believing that is wrong?
>> Ah, a question about what I'm saying. I can help there. The basis is
>> what Linz says about Ĥ. He says that (translating to your notation)
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>> should be the case "if Ĥ applied to ⟨Ĥ⟩ does not halt". But, as you can
>> see, your Ĥ does halt when applied to ⟨Ĥ⟩ (qn is a halting or final
>> state). Your Ĥ is not doing what it should in this one crucial case.
>>
>
> the Turing machine halting problem. Simply stated, the problem
> is: given the description of a Turing machine M and an input w,
> does M, when started in the initial configuration q0w, perform a
> computation that eventually halts? (Linz:1990:317).

and so on. Same old stuff.

I'm sorry my explanation did not help at all. I'm happy to answer any
other questions you might have if you think it might help you understand
what I (and Linz) are saying.

--
Ben.

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.

<2-adncuMeNUVpY78nZ2dnUU7-fPNnZ2d@giganews.com>

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NNTP-Posting-Date: Tue, 10 Aug 2021 21:40:08 -0500
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct_and_forms_no_contradiction.
Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc> <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>
<zcadnTSOD5rtZ5f8nZ2dnUU7-T3NnZ2d@giganews.com> <877dh03l3c.fsf@bsb.me.uk>
<Z5adnd038KGXwJb8nZ2dnUU7-I_NnZ2d@giganews.com> <8735rn1qvj.fsf@bsb.me.uk>
<goydnfCCIYUWE5H8nZ2dnUU7-e_NnZ2d@giganews.com> <87eeb7z4d1.fsf@bsb.me.uk>
<0_Sdnb6Qe8XGOZH8nZ2dnUU7-U3NnZ2d@giganews.com> <87zgtslqpv.fsf@bsb.me.uk>
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From: NoO...@NoWhere.com (olcott)
Date: Tue, 10 Aug 2021 21:40:06 -0500
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 by: olcott - Wed, 11 Aug 2021 02:40 UTC

On 8/10/2021 9:26 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/10/2021 8:41 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/10/2021 7:42 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/7/2021 7:34 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>
>>>>>>>>>>>> The question is not: Does Ĥ halt on its input?
>>>>>>>>>>> Yes it is.
>>>>>>>>>>
>>>>>>>>>> The question is:
>>>>>>>>>> Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
>>>>>>>>>> The ansswer to this question is provably no!
>>>>>>>>> The question is: does Ĥ applied to ⟨Ĥ⟩ halt. It does:
>>>>>>>>>
>>>>>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn THIS IS NOT A CONTRADICTION
>>>>>>>>> Indeed. There is no contradiction. Just an Ĥ that does not meet Linz
>>>>>>>>> spec.
>>>>>>>>
>>>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.
>>>>>>>> Because it is correct it meets the Linz spec.
>>>>>>> I find it startling that you think that, but then it seems you don't yet
>>>>>>> know what the key words mean:
>>>>>>>
>>>>>>>> if M applied to wM does not halt
>>>>>>>> means if the execution of the machine of the first ⟨Ĥ⟩ on its input of
>>>>>>>> the seocond ⟨Ĥ⟩ does not halt then ⊢* Ĥ.qn
>>>>>>> No. Would you like to know "what M applied to wM does not halt" means?
>>>>>>> Do you need help to see that "Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn" is clearly a case of "M
>>>>>>> applied to wM halts"?
>>>>>>
>>>>>> the Turing machine halting problem. Simply stated, the problem
>>>>>> is: given the description of a Turing machine M and an input w,
>>>>>> does M, when started in the initial configuration q0w, perform a
>>>>>> computation that eventually halts? (Linz:1990:317).
>>>>> Yes. I was offering to help you understand the key words in that text.
>>>>>
>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>> You've missed off the key lines yet again. Is that deliberate? They
>>>>> are the lines that show you are wrong so I am suspicious that you keep
>>>>> omitting them.
>>>>>
>>>>>> When Ĥ is applied to ⟨Ĥ⟩ the description of the Turing Machine and its
>>>>>> input are specified as: ⟨Ĥ⟩ ⟨Ĥ⟩ for the embedded halt decider at Ĥ.qx.
>>>>> Ungrammatical.
>>>>>
>>>>>> When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn this is not a final state of the simulated
>>>>>> input it is a final state of the executed Ĥ.
>>>>> Yes. You don't seem to know why that's wrong.
>>>>
>>>> What is your basis for believing that is wrong?
>>> Ah, a question about what I'm saying. I can help there. The basis is
>>> what Linz says about Ĥ. He says that (translating to your notation)
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>> should be the case "if Ĥ applied to ⟨Ĥ⟩ does not halt". But, as you can
>>> see, your Ĥ does halt when applied to ⟨Ĥ⟩ (qn is a halting or final
>>> state). Your Ĥ is not doing what it should in this one crucial case.
>>>
>>
>> the Turing machine halting problem. Simply stated, the problem
>> is: given the description of a Turing machine M and an input w,
>> does M, when started in the initial configuration q0w, perform a
>> computation that eventually halts? (Linz:1990:317).
>
> and so on. Same old stuff.
>
> I'm sorry my explanation did not help at all. I'm happy to answer any
> other questions you might have if you think it might help you understand
> what I (and Linz) are saying.

I went point by point. If I am actually incorrect then you can go point
by point and point out each individual error step by step. Of course
everyone knows that this is impossible if I am totally correct.

- the Turing machine halting problem. Simply stated, the problem
- is: given the description of a Turing machine M and an input w,
- does M, when started in the initial configuration q0w, perform a
- computation that eventually halts? (Linz:1990:317).

The input to H will be the description (encoded in some form) of M, say
WM, as well as the input w. (Linz:1990:318)

H.q0 WM w ⊢* Ĥ.qn
becomes
H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn

Can you admit when you are wrong when you really are wrong?

--
Copyright 2021 Pete Olcott

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

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction. (fixed type)

<2-adncqMeNWXpI78nZ2dnUU7-fOdnZ2d@giganews.com>

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Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
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Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc> <87czqxa0zk.fsf@bsb.me.uk>
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<goydnfCCIYUWE5H8nZ2dnUU7-e_NnZ2d@giganews.com> <87eeb7z4d1.fsf@bsb.me.uk>
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From: NoO...@NoWhere.com (olcott)
Date: Tue, 10 Aug 2021 21:42:16 -0500
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 by: olcott - Wed, 11 Aug 2021 02:42 UTC

On 8/10/2021 9:26 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/10/2021 8:41 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/10/2021 7:42 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/7/2021 7:34 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>
>>>>>>>>>>>> The question is not: Does Ĥ halt on its input?
>>>>>>>>>>> Yes it is.
>>>>>>>>>>
>>>>>>>>>> The question is:
>>>>>>>>>> Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
>>>>>>>>>> The ansswer to this question is provably no!
>>>>>>>>> The question is: does Ĥ applied to ⟨Ĥ⟩ halt. It does:
>>>>>>>>>
>>>>>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn THIS IS NOT A CONTRADICTION
>>>>>>>>> Indeed. There is no contradiction. Just an Ĥ that does not meet Linz
>>>>>>>>> spec.
>>>>>>>>
>>>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.
>>>>>>>> Because it is correct it meets the Linz spec.
>>>>>>> I find it startling that you think that, but then it seems you don't yet
>>>>>>> know what the key words mean:
>>>>>>>
>>>>>>>> if M applied to wM does not halt
>>>>>>>> means if the execution of the machine of the first ⟨Ĥ⟩ on its input of
>>>>>>>> the seocond ⟨Ĥ⟩ does not halt then ⊢* Ĥ.qn
>>>>>>> No. Would you like to know "what M applied to wM does not halt" means?
>>>>>>> Do you need help to see that "Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn" is clearly a case of "M
>>>>>>> applied to wM halts"?
>>>>>>
>>>>>> the Turing machine halting problem. Simply stated, the problem
>>>>>> is: given the description of a Turing machine M and an input w,
>>>>>> does M, when started in the initial configuration q0w, perform a
>>>>>> computation that eventually halts? (Linz:1990:317).
>>>>> Yes. I was offering to help you understand the key words in that text.
>>>>>
>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>> You've missed off the key lines yet again. Is that deliberate? They
>>>>> are the lines that show you are wrong so I am suspicious that you keep
>>>>> omitting them.
>>>>>
>>>>>> When Ĥ is applied to ⟨Ĥ⟩ the description of the Turing Machine and its
>>>>>> input are specified as: ⟨Ĥ⟩ ⟨Ĥ⟩ for the embedded halt decider at Ĥ.qx.
>>>>> Ungrammatical.
>>>>>
>>>>>> When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn this is not a final state of the simulated
>>>>>> input it is a final state of the executed Ĥ.
>>>>> Yes. You don't seem to know why that's wrong.
>>>>
>>>> What is your basis for believing that is wrong?
>>> Ah, a question about what I'm saying. I can help there. The basis is
>>> what Linz says about Ĥ. He says that (translating to your notation)
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>> should be the case "if Ĥ applied to ⟨Ĥ⟩ does not halt". But, as you can
>>> see, your Ĥ does halt when applied to ⟨Ĥ⟩ (qn is a halting or final
>>> state). Your Ĥ is not doing what it should in this one crucial case.
>>>
>>
>> the Turing machine halting problem. Simply stated, the problem
>> is: given the description of a Turing machine M and an input w,
>> does M, when started in the initial configuration q0w, perform a
>> computation that eventually halts? (Linz:1990:317).
>
> and so on. Same old stuff.
>
> I'm sorry my explanation did not help at all. I'm happy to answer any
> other questions you might have if you think it might help you understand
> what I (and Linz) are saying.
>

I went point by point. If I am actually incorrect then you can go point
by point and point out each individual error step by step. Of course
everyone knows that this is impossible if I am totally correct.

- the Turing machine halting problem. Simply stated, the problem
- is: given the description of a Turing machine M and an input w,
- does M, when started in the initial configuration q0w, perform a
- computation that eventually halts? (Linz:1990:317).

The input to H will be the description (encoded in some form) of M, say
WM, as well as the input w. (Linz:1990:318)

H.q0 WM w ⊢* H.qn // fixed typo
becomes
H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn

Can you admit when you are wrong when you really are wrong?

--
Copyright 2021 Pete Olcott

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

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction. // fixed typo (link added)

<Js2dnYjKie1Bzo78nZ2dnUU7-XnNnZ2d@giganews.com>

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https://www.novabbs.com/devel/article-flat.php?id=19658&group=comp.theory#19658

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Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct_and_forms_no_contradiction._//_fixed_typo_(link_added
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Newsgroups: comp.theory
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From: NoO...@NoWhere.com (olcott)
Date: Tue, 10 Aug 2021 23:36:42 -0500
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 by: olcott - Wed, 11 Aug 2021 04:36 UTC

On 8/10/2021 9:26 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/10/2021 8:41 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/10/2021 7:42 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/7/2021 7:34 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>
>>>>>>>>>>>> The question is not: Does Ĥ halt on its input?
>>>>>>>>>>> Yes it is.
>>>>>>>>>>
>>>>>>>>>> The question is:
>>>>>>>>>> Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
>>>>>>>>>> The ansswer to this question is provably no!
>>>>>>>>> The question is: does Ĥ applied to ⟨Ĥ⟩ halt. It does:
>>>>>>>>>
>>>>>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn THIS IS NOT A CONTRADICTION
>>>>>>>>> Indeed. There is no contradiction. Just an Ĥ that does not meet Linz
>>>>>>>>> spec.
>>>>>>>>
>>>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.
>>>>>>>> Because it is correct it meets the Linz spec.
>>>>>>> I find it startling that you think that, but then it seems you don't yet
>>>>>>> know what the key words mean:
>>>>>>>
>>>>>>>> if M applied to wM does not halt
>>>>>>>> means if the execution of the machine of the first ⟨Ĥ⟩ on its input of
>>>>>>>> the seocond ⟨Ĥ⟩ does not halt then ⊢* Ĥ.qn
>>>>>>> No. Would you like to know "what M applied to wM does not halt" means?
>>>>>>> Do you need help to see that "Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn" is clearly a case of "M
>>>>>>> applied to wM halts"?
>>>>>>
>>>>>> the Turing machine halting problem. Simply stated, the problem
>>>>>> is: given the description of a Turing machine M and an input w,
>>>>>> does M, when started in the initial configuration q0w, perform a
>>>>>> computation that eventually halts? (Linz:1990:317).
>>>>> Yes. I was offering to help you understand the key words in that text.
>>>>>
>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>> You've missed off the key lines yet again. Is that deliberate? They
>>>>> are the lines that show you are wrong so I am suspicious that you keep
>>>>> omitting them.
>>>>>
>>>>>> When Ĥ is applied to ⟨Ĥ⟩ the description of the Turing Machine and its
>>>>>> input are specified as: ⟨Ĥ⟩ ⟨Ĥ⟩ for the embedded halt decider at Ĥ.qx.
>>>>> Ungrammatical.
>>>>>
>>>>>> When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn this is not a final state of the simulated
>>>>>> input it is a final state of the executed Ĥ.
>>>>> Yes. You don't seem to know why that's wrong.
>>>>
>>>> What is your basis for believing that is wrong?
>>> Ah, a question about what I'm saying. I can help there. The basis is
>>> what Linz says about Ĥ. He says that (translating to your notation)
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>> should be the case "if Ĥ applied to ⟨Ĥ⟩ does not halt". But, as you can
>>> see, your Ĥ does halt when applied to ⟨Ĥ⟩ (qn is a halting or final
>>> state). Your Ĥ is not doing what it should in this one crucial case.
>>>
>>
>> the Turing machine halting problem. Simply stated, the problem
>> is: given the description of a Turing machine M and an input w,
>> does M, when started in the initial configuration q0w, perform a
>> computation that eventually halts? (Linz:1990:317).
>
> and so on. Same old stuff.
>
> I'm sorry my explanation did not help at all. I'm happy to answer any
> other questions you might have if you think it might help you understand
> what I (and Linz) are saying.
>

I went point by point. If I am actually incorrect then you can go point
by point and point out each individual error step by step. Of course
everyone knows that this is impossible if I am totally correct.

- the Turing machine halting problem. Simply stated, the problem
- is: given the description of a Turing machine M and an input w,
- does M, when started in the initial configuration q0w, perform a
- computation that eventually halts? (Linz:1990:317).

The input to H will be the description (encoded in some form) of M, say
WM, as well as the input w. (Linz:1990:318)

H.q0 WM w ⊢* H.qn
becomes
H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn

Can you admit when you are wrong when you really are wrong?

http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf

--
Copyright 2021 Pete Olcott

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

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.

<00857ac3-32b7-46c5-a5a7-9cd6f6ec1561n@googlegroups.com>

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Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct an
d forms no contradiction.
From: malcolm....@gmail.com (Malcolm McLean)
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 by: Malcolm McLean - Wed, 11 Aug 2021 06:40 UTC

On Wednesday, 11 August 2021 at 02:41:04 UTC+1, Ben Bacarisse wrote:
>
> It is trivial to write a TM J from which a Ĵ can be derived that has the
> properties of your Ĥ. Every student learning this material for the first
> time should be able to come up with one a few minutes. There is nothing
> interesting about a TM with
>
> Ĵ.q0 ⟨Ĵ⟩ ⊢* Ĵ.qn
> if Ĵ applied to ⟨Ĵ⟩ halts.
>
It's trivial to write a J with these characteristics.
But that doesn't mean that every such J will be trivial or uninteresting. It just
won't be a counter-example to an established theorem.

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.

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

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct and forms no contradiction.
Date: Wed, 11 Aug 2021 11:48:11 +0100
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 by: Ben Bacarisse - Wed, 11 Aug 2021 10:48 UTC

Malcolm McLean <malcolm.arthur.mclean@gmail.com> writes:

> On Wednesday, 11 August 2021 at 02:41:04 UTC+1, Ben Bacarisse wrote:
>>
>> It is trivial to write a TM J from which a Ĵ can be derived that has the
>> properties of your Ĥ. Every student learning this material for the first
>> time should be able to come up with one a few minutes. There is nothing
>> interesting about a TM with
>>
>> Ĵ.q0 ⟨Ĵ⟩ ⊢* Ĵ.qn
>> if Ĵ applied to ⟨Ĵ⟩ halts.
>>
> It's trivial to write a J with these characteristics.
> But that doesn't mean that every such J will be trivial or
> uninteresting. It just won't be a counter-example to an established
> theorem.

That's an odd thing to point out, but of course it's true. Deciders for
all sort of sets might have this property: shortest paths through a
network, numbered digits of pi, integer factorisations... Some will and
some won't, but a very large proportion of interesting and non-trivial
TMs will have this property.

That's probably not what you meant though, but I don't know what you are
hinting at.

--
Ben.

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