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

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

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

<pu-dnVqdPMG4iJb8nZ2dnUU7-Q_NnZ2d@giganews.com>

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https://www.novabbs.com/devel/article-flat.php?id=19494&group=comp.theory#19494

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Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct ][ GIGO ]
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
References: <20210719214640.00000dfc@reddwarf.jmc> <871r7i6n2u.fsf@bsb.me.uk> <OqKdnROLKJ9CdJz8nZ2dnUU7-avNnZ2d@giganews.com> <87k0la542c.fsf@bsb.me.uk> <1NidnVPZ-NHDl5_8nZ2dnUU7-enNnZ2d@giganews.com> <87sfzw3ao1.fsf@bsb.me.uk> <7oKdnTjx4IC20p78nZ2dnUU7-TvNnZ2d@giganews.com> <875yws36vt.fsf@bsb.me.uk> <j66dnbdHrpV8_p78nZ2dnUU7-aXNnZ2d@giganews.com> <87im0s0ydp.fsf@bsb.me.uk> <Brqdnfehrf0Kj5n8nZ2dnUU7-X3NnZ2d@giganews.com> <87tukblgjy.fsf@bsb.me.uk> <qtGdnfuXs4nFOZn8nZ2dnUU7-cnNnZ2d@giganews.com> <871r7ekugt.fsf@bsb.me.uk> <K5-dndGZo_-VmJv8nZ2dnUU78QvNnZ2d@giganews.com> <87czqxa0zk.fsf@bsb.me.uk> <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>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 4 Aug 2021 17:31:01 -0500
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 by: olcott - Wed, 4 Aug 2021 22:31 UTC

On 8/4/2021 1:51 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/1/2021 11:00 AM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/1/2021 7:41 AM, Ben Bacarisse wrote:
>>>>> Malcolm McLean <malcolm.arthur.mclean@gmail.com> writes:
>>>>>
>>>>>> On Sunday, 1 August 2021 at 11:54:57 UTC+1, Ben Bacarisse wrote:
>>>>>>>
>>>>>>> Here we can see that Ĥ applied to ⟨Ĥ⟩ halts. You can call your Ĥ's
>>>>>>> behaviour "correct". You can call it anything you like. But it's not
>>>>>>> "as in Linz". It does not say anything about Linz's proof. It does not
>>>>>>> do anything people would call impossible or even interesting.
>>>>>>>
>>>>>> It seems to be established that H(H_Hat, H_Hat) returns "non-halting"
>>>>>> whilst H_Hat(H_Hat) halts. So all is as Linz says it must be and no
>>>>>> theorems are refuted. Which you would expect. If results were consistent
>>>>>> it would have to be some cheap trick.
>>>>> I case there is some confusion, I mean that PO's Ĥ is not an Ĥ as
>>>>> specified in Linz. Yes, everything is in accordance with the truth as
>>>>> laid out in Linz and, indeed, in any textbook.
>>>>> I point this out to PO because he brings it up. He keeps posting the
>>>>> specification of what an Ĥ, as Linz specifies it, would do:
>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>> if (and only if) M applied to wM does not halt.
>>>>> He claims (or used to claim) that his Ĥ meets this specification for at
>>>>> least the one case where wM == ⟨Ĥ⟩:
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>> if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>>> To remain relevant, he /must/ keep insisting that his Ĥ meets the
>>>>> requirements laid out in Linz, if only for this one key input.
>>>>>
>>>>
>>>> Ĥ[0].q0 is taken to mean Ĥ<sub>0</sub>.q0 which is the Turing machine.
>>>>
>>>> Ĥ[1].q0 is taken to mean Ĥ<sub>1</sub>.q0 which is the Turing machine
>>>> description input to Ĥ[0].q0
>>>>
>>>> Ĥ[2].q0 is taken to mean Ĥ<sub>2</sub>.q0 which is first copy of the
>>>> Turing machine description input to Ĥ[0].q0
>>>>
>>>> Ĥ[0].q0 ⟨Ĥ⟩ ⊢* Ĥ[0].qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ[0].qn
>>> Ĥ[0] is Ĥ so you are confirming, yet again, that
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>
>>>> It is neither a contradiction nor a paradox because there are three
>>>> different instances of Ĥ.
>>> I agree that this is neither a paradox nor a contradiction. It's just a
>>> fact derived form the logic of how your Ĥ is written (the majority of
>>> which you are keeping hidden from us).
>>>
>>>> Because the only reason that the first instance halts is that Ĥ[0].qx
>>>> correctly determines that its input cannot possibly ever reach its
>>>> final state of Ĥ[1].qn or Ĥ[1].qy whether or not the simulating halt
>>>> decider aborts its simulation of this input, we know with 100%
>>>> perfectly justified logical certainty that the input to Ĥ[0].qx never
>>>> halts.
>>> We know, since you keep telling us, that Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn. This clearly
>>> shows that Ĥ applied to ⟨Ĥ⟩ halts. You can see the final state right
>>
>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>> if M applied to wM halts, and
>>
>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>> if M applied to wM does not halt
>>
>> You are using the wrong Ĥ.
>
> First of all, let's be 100% clear: I am talking about what /your/ Ĥ
> does, based in the facts you have let slip about it.
>
>> Linz stipulates that wM is ⟨Ĥ⟩ and M is the underlying machine of this
>> ⟨Ĥ⟩ therefore M applied to wM means ⟨Ĥ⟩ applied to ⟨Ĥ⟩.
>
> No. How many years have you been staring at this one page from Linz?
> You still don't know what it says. Do ask me questions, if you'd like
> to know what the text you've been sure is wrong for 17 years really
> says.
>

....Turing machine halting problem.
Simply stated, the problem is:
given the description of a Turing machine M
given the description of a Turing machine M
given the description of a Turing machine M
given the description of a Turing machine M
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?

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

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

When ⟨M⟩ = ⟨Ĥ⟩: I have proved that Ĥ ⟨M⟩ transitions to Ĥ.qn because M
never reaches a final state.

--
Copyright 2021 Pete Olcott

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

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

<8YKdnVBaxrJ_i5b8nZ2dnUU7-aPNnZ2d@giganews.com>

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https://www.novabbs.com/devel/article-flat.php?id=19495&group=comp.theory#19495

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NNTP-Posting-Date: Wed, 04 Aug 2021 17:38:26 -0500
Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[
Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct
]
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
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From: NoO...@NoWhere.com (olcott)
Date: Wed, 4 Aug 2021 17:38:26 -0500
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 by: olcott - Wed, 4 Aug 2021 22:38 UTC

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. That machine
never reaches its final state.

Ĥ.q0 ⟨Ĥ[1]⟩ ⊢* Ĥ.qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ.qn
if Ĥ[1] applied to ⟨Ĥ[2]⟩ does not halt.

That you persistently ignore this distinction really seems to be a
diverge from an honest dialogue.

>> We can know that Machine_of(⟨Ĥ⟩) applied to ⟨Ĥ⟩ never reaches its
>
> Machine_of(⟨Ĥ⟩) = Ĥ.
>
>> final state whether or not the embedded halt decider at Ĥ.qx aborts it
>
> You keep telling me that Ĥ applied to ⟨Ĥ⟩ reaches a final state. You
> even tell me the final state.
>
>> simulation of Machine_of(⟨Ĥ⟩), therefore we know that Machine_of(⟨Ĥ⟩)
>> never halts. Therefore we know that M applied to wM does not halt.
>
> You keep telling me that Ĥ applied to ⟨Ĥ⟩ reaches the final state Ĥ.qn.
> Are you changing your mind after all this time? No, you are searching
> for some form of words that will make the wrong answer sound right.
>

--
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 ]

<87y29g23sk.fsf@bsb.me.uk>

 copy mid

https://www.novabbs.com/devel/article-flat.php?id=19496&group=comp.theory#19496

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider [
Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [
succinct ]
Followup-To: comp.theory
Date: Thu, 05 Aug 2021 00:22:51 +0100
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 by: Ben Bacarisse - Wed, 4 Aug 2021 23:22 UTC

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[1]⟩ ⊢* Ĥ.qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ.qn
> if Ĥ[1] applied to ⟨Ĥ[2]⟩ does not halt.
>
> That you persistently ignore this distinction really seems to be a
> diverge from an honest dialogue.

If the various strings you've chosen to number (badly) are not identical
then your "hat" construction is wrong. Linz's "hat" version makes an
exact copy.

You are free to write ⟨Ĥ[99]⟩ if you like, but I am also free to change
that back to ⟨Ĥ⟩ because they are identical strings. Anything true of

Ĥ.q0 ⟨Ĥ[1]⟩ ⊢* Ĥ.qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ.qn
if Ĥ[1] applied to ⟨Ĥ[2]⟩ does not halt.

is true of

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

But do say if it isn't. I'd like there to be another reason why your
claims are irrelevant -- the more the merrier. Please tell us that Ĥ[1]
!= Ĥ or that ⟨Ĥ[2]⟩ != ⟨Ĥ⟩.

>> You keep telling me that Ĥ applied to ⟨Ĥ⟩ reaches the final state Ĥ.qn.
>> Are you changing your mind after all this time?

You are not answering questions, are you?

--
Ben.

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

<87v94k23rn.fsf@bsb.me.uk>

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https://www.novabbs.com/devel/article-flat.php?id=19497&group=comp.theory#19497

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider [
Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [
succinct ][ GIGO ]
Followup-To: comp.theory
Date: Thu, 05 Aug 2021 00:23:24 +0100
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 by: Ben Bacarisse - Wed, 4 Aug 2021 23:23 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/4/2021 1:51 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/1/2021 11:00 AM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/1/2021 7:41 AM, Ben Bacarisse wrote:
>>>>>> Malcolm McLean <malcolm.arthur.mclean@gmail.com> writes:
>>>>>>
>>>>>>> On Sunday, 1 August 2021 at 11:54:57 UTC+1, Ben Bacarisse wrote:
>>>>>>>>
>>>>>>>> Here we can see that Ĥ applied to ⟨Ĥ⟩ halts. You can call your Ĥ's
>>>>>>>> behaviour "correct". You can call it anything you like. But it's not
>>>>>>>> "as in Linz". It does not say anything about Linz's proof. It does not
>>>>>>>> do anything people would call impossible or even interesting.
>>>>>>>>
>>>>>>> It seems to be established that H(H_Hat, H_Hat) returns "non-halting"
>>>>>>> whilst H_Hat(H_Hat) halts. So all is as Linz says it must be and no
>>>>>>> theorems are refuted. Which you would expect. If results were consistent
>>>>>>> it would have to be some cheap trick.
>>>>>> I case there is some confusion, I mean that PO's Ĥ is not an Ĥ as
>>>>>> specified in Linz. Yes, everything is in accordance with the truth as
>>>>>> laid out in Linz and, indeed, in any textbook.
>>>>>> I point this out to PO because he brings it up. He keeps posting the
>>>>>> specification of what an Ĥ, as Linz specifies it, would do:
>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>>> if (and only if) M applied to wM does not halt.
>>>>>> He claims (or used to claim) that his Ĥ meets this specification for at
>>>>>> least the one case where wM == ⟨Ĥ⟩:
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>> if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>>>> To remain relevant, he /must/ keep insisting that his Ĥ meets the
>>>>>> requirements laid out in Linz, if only for this one key input.
>>>>>>
>>>>>
>>>>> Ĥ[0].q0 is taken to mean Ĥ<sub>0</sub>.q0 which is the Turing machine.
>>>>>
>>>>> Ĥ[1].q0 is taken to mean Ĥ<sub>1</sub>.q0 which is the Turing machine
>>>>> description input to Ĥ[0].q0
>>>>>
>>>>> Ĥ[2].q0 is taken to mean Ĥ<sub>2</sub>.q0 which is first copy of the
>>>>> Turing machine description input to Ĥ[0].q0
>>>>>
>>>>> Ĥ[0].q0 ⟨Ĥ⟩ ⊢* Ĥ[0].qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ[0].qn
>>>> Ĥ[0] is Ĥ so you are confirming, yet again, that
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>
>>>>> It is neither a contradiction nor a paradox because there are three
>>>>> different instances of Ĥ.
>>>> I agree that this is neither a paradox nor a contradiction. It's just a
>>>> fact derived form the logic of how your Ĥ is written (the majority of
>>>> which you are keeping hidden from us).
>>>>
>>>>> Because the only reason that the first instance halts is that Ĥ[0].qx
>>>>> correctly determines that its input cannot possibly ever reach its
>>>>> final state of Ĥ[1].qn or Ĥ[1].qy whether or not the simulating halt
>>>>> decider aborts its simulation of this input, we know with 100%
>>>>> perfectly justified logical certainty that the input to Ĥ[0].qx never
>>>>> halts.
>>>> We know, since you keep telling us, that Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn. This clearly
>>>> shows that Ĥ applied to ⟨Ĥ⟩ halts. You can see the final state right
>>>
>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>> if M applied to wM halts, and
>>>
>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>> if M applied to wM does not halt
>>>
>>> You are using the wrong Ĥ.
>> First of all, let's be 100% clear: I am talking about what /your/ Ĥ
>> does, based in the facts you have let slip about it.
>>
>>> Linz stipulates that wM is ⟨Ĥ⟩ and M is the underlying machine of this
>>> ⟨Ĥ⟩ therefore M applied to wM means ⟨Ĥ⟩ applied to ⟨Ĥ⟩.
>> No. How many years have you been staring at this one page from Linz?
>> You still don't know what it says. Do ask me questions, if you'd like
>> to know what the text you've been sure is wrong for 17 years really
>> says.
>>
>
> ...Turing machine halting problem.
> Simply stated, the problem is:
> given the description of a Turing machine M
> given the description of a Turing machine M
> given the description of a Turing machine M
> given the description of a Turing machine M
> 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?
>
> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>
> Ĥ.q0 ⟨M⟩ ⊢* Ĥ.qx ⟨M⟩ ⟨M⟩ ⊢* Ĥ.qn
> if M applied to ⟨M⟩ does not halt
>
> When ⟨M⟩ = ⟨Ĥ⟩: I have proved that Ĥ ⟨M⟩ transitions to Ĥ.qn because M
> never reaches a final state.

This a schoolboy error. Why are you so scared to ask me questions? Do
you fear you might understand me?

--
Ben.

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

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Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[
Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct
][ GIGO ]
Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc> <871r7i6n2u.fsf@bsb.me.uk>
<OqKdnROLKJ9CdJz8nZ2dnUU7-avNnZ2d@giganews.com> <87k0la542c.fsf@bsb.me.uk>
<1NidnVPZ-NHDl5_8nZ2dnUU7-enNnZ2d@giganews.com> <87sfzw3ao1.fsf@bsb.me.uk>
<7oKdnTjx4IC20p78nZ2dnUU7-TvNnZ2d@giganews.com> <875yws36vt.fsf@bsb.me.uk>
<j66dnbdHrpV8_p78nZ2dnUU7-aXNnZ2d@giganews.com> <87im0s0ydp.fsf@bsb.me.uk>
<Brqdnfehrf0Kj5n8nZ2dnUU7-X3NnZ2d@giganews.com> <87tukblgjy.fsf@bsb.me.uk>
<qtGdnfuXs4nFOZn8nZ2dnUU7-cnNnZ2d@giganews.com> <871r7ekugt.fsf@bsb.me.uk>
<K5-dndGZo_-VmJv8nZ2dnUU78QvNnZ2d@giganews.com> <87czqxa0zk.fsf@bsb.me.uk>
<53d47ab9-818c-4f40-8e72-bdb76fa416een@googlegroups.com>
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Thu, 5 Aug 2021 00:05 UTC

On 8/4/21 4:31 PM, olcott wrote:
> On 8/4/2021 1:51 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/1/2021 11:00 AM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/1/2021 7:41 AM, Ben Bacarisse wrote:
>>>>>> Malcolm McLean <malcolm.arthur.mclean@gmail.com> writes:
>>>>>>
>>>>>>> On Sunday, 1 August 2021 at 11:54:57 UTC+1, Ben Bacarisse wrote:
>>>>>>>>
>>>>>>>> Here we can see that Ĥ applied to ⟨Ĥ⟩ halts. You can call your Ĥ's
>>>>>>>> behaviour "correct". You can call it anything you like. But it's
>>>>>>>> not
>>>>>>>> "as in Linz". It does not say anything about Linz's proof. It
>>>>>>>> does not
>>>>>>>> do anything people would call impossible or even interesting.
>>>>>>>>
>>>>>>> It seems to be established that H(H_Hat, H_Hat) returns
>>>>>>> "non-halting"
>>>>>>> whilst H_Hat(H_Hat) halts. So all is as Linz says it must be and no
>>>>>>> theorems are refuted. Which you would expect. If results were
>>>>>>> consistent
>>>>>>> it would have to be some cheap trick.
>>>>>> I case there is some confusion, I mean that PO's Ĥ is not an Ĥ as
>>>>>> specified in Linz.  Yes, everything is in accordance with the
>>>>>> truth as
>>>>>> laid out in Linz and, indeed, in any textbook.
>>>>>> I point this out to PO because he brings it up.  He keeps posting the
>>>>>> specification of what an Ĥ, as Linz specifies it, would do:
>>>>>>      Ĥ.q0 wM ⊢* Ĥ.qx wM wM  ⊢* Ĥ.qn
>>>>>>      if (and only if) M applied to wM does not halt.
>>>>>> He claims (or used to claim) that his Ĥ meets this specification
>>>>>> for at
>>>>>> least the one case where wM == ⟨Ĥ⟩:
>>>>>>      Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩  ⊢* Ĥ.qn
>>>>>>      if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>>>> To remain relevant, he /must/ keep insisting that his Ĥ meets the
>>>>>> requirements laid out in Linz, if only for this one key input.
>>>>>>
>>>>>
>>>>> Ĥ[0].q0 is taken to mean Ĥ<sub>0</sub>.q0 which is the Turing machine.
>>>>>
>>>>> Ĥ[1].q0 is taken to mean Ĥ<sub>1</sub>.q0 which is the Turing machine
>>>>> description input to Ĥ[0].q0
>>>>>
>>>>> Ĥ[2].q0 is taken to mean Ĥ<sub>2</sub>.q0 which is first copy of the
>>>>> Turing machine description input to Ĥ[0].q0
>>>>>
>>>>> Ĥ[0].q0 ⟨Ĥ⟩ ⊢* Ĥ[0].qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ[0].qn
>>>> Ĥ[0] is Ĥ so you are confirming, yet again, that
>>>>      Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>
>>>>> It is neither a contradiction nor a paradox because there are three
>>>>> different instances of Ĥ.
>>>> I agree that this is neither a paradox nor a contradiction.  It's
>>>> just a
>>>> fact derived form the logic of how your Ĥ is written (the majority of
>>>> which you are keeping hidden from us).
>>>>
>>>>> Because the only reason that the first instance halts is that Ĥ[0].qx
>>>>> correctly determines that its input cannot possibly ever reach its
>>>>> final state of Ĥ[1].qn or Ĥ[1].qy whether or not the simulating halt
>>>>> decider aborts its simulation of this input, we know with 100%
>>>>> perfectly justified logical certainty that the input to Ĥ[0].qx never
>>>>> halts.
>>>> We know, since you keep telling us, that Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn.  This
>>>> clearly
>>>> shows that Ĥ applied to ⟨Ĥ⟩ halts.  You can see the final state right
>>>
>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>> if M applied to wM halts, and
>>>
>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>> if M applied to wM does not halt
>>>
>>> You are using the wrong Ĥ.
>>
>> First of all, let's be 100% clear: I am talking about what /your/ Ĥ
>> does, based in the facts you have let slip about it.
>>
>>> Linz stipulates that wM is ⟨Ĥ⟩ and M is the underlying machine of this
>>> ⟨Ĥ⟩ therefore M applied to wM means ⟨Ĥ⟩ applied to ⟨Ĥ⟩.
>>
>> No.  How many years have you been staring at this one page from Linz?
>> You still don't know what it says.  Do ask me questions, if you'd like
>> to know what the text you've been sure is wrong for 17 years really
>> says.
>>
>
> ...Turing machine halting problem.
> Simply stated, the problem is:
> given the description of a Turing machine M
> given the description of a Turing machine M
> given the description of a Turing machine M
> given the description of a Turing machine M
> 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?
>
> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>
> Ĥ.q0 ⟨M⟩ ⊢* Ĥ.qx ⟨M⟩ ⟨M⟩ ⊢* Ĥ.qn
> if M applied to ⟨M⟩ does not halt
>
> When ⟨M⟩ = ⟨Ĥ⟩: I have proved that Ĥ ⟨M⟩ transitions to Ĥ.qn because M
> never reaches a final state.
>
>

Except that H^.qn is the state that H^ HALTS in because its copy of H
said it was non-halting.

H^ is NOT a decider, it qn state does NOT say that its input is
non-halting, it says that IT is Halting becuase its copy of H says (in
this case incorrectly) that it thinks the input is non-halting.

Arriving at H^.qn with the input = (H^) is the PROOF that H((H^),(H^))
is wrong.

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

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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> <87sfzw3ao1.fsf@bsb.me.uk>
<7oKdnTjx4IC20p78nZ2dnUU7-TvNnZ2d@giganews.com> <875yws36vt.fsf@bsb.me.uk>
<j66dnbdHrpV8_p78nZ2dnUU7-aXNnZ2d@giganews.com> <87im0s0ydp.fsf@bsb.me.uk>
<Brqdnfehrf0Kj5n8nZ2dnUU7-X3NnZ2d@giganews.com> <87tukblgjy.fsf@bsb.me.uk>
<qtGdnfuXs4nFOZn8nZ2dnUU7-cnNnZ2d@giganews.com> <871r7ekugt.fsf@bsb.me.uk>
<K5-dndGZo_-VmJv8nZ2dnUU78QvNnZ2d@giganews.com> <87czqxa0zk.fsf@bsb.me.uk>
<woudnXWBxPba95r8nZ2dnUU78ffNnZ2d@giganews.com> <87mtpz64sq.fsf@bsb.me.uk>
<w5edne8d06OkxpX8nZ2dnUU7-b_NnZ2d@giganews.com> <87bl6f5qvy.fsf@bsb.me.uk>
<xLednaPs_ZSXCZX8nZ2dnUU7-YnNnZ2d@giganews.com> <87o8af47y0.fsf@bsb.me.uk>
<NsudnY99rthDOJX8nZ2dnUU7-c_NnZ2d@giganews.com> <87zgtx2wxn.fsf@bsb.me.uk>
<Rr6dnWKhH4ejIJf8nZ2dnUU7-Q3NnZ2d@giganews.com> <87czqt2ewt.fsf@bsb.me.uk>
<8YKdnVBaxrJ_i5b8nZ2dnUU7-aPNnZ2d@giganews.com>
From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Thu, 5 Aug 2021 01:21 UTC

On 8/4/21 4:38 PM, olcott wrote:
> 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. That machine
> never reaches its final state.

Except that one copy of the Turing Machine P(I) acts exactly like EVERY
other copy of that same P(I).

Thus if one H^(H^) halts, then ALL H^(H^) are halting.

The fact that this instance had its simulation aborted before it got to
that Halting state just says that the simulator decided incorrectly that
it wouldn't halt and stop it too soon.
>
> Ĥ.q0 ⟨Ĥ[1]⟩ ⊢* Ĥ.qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ.qn
> if Ĥ[1] applied to ⟨Ĥ[2]⟩ does not halt.

FALSE.

>
> That you persistently ignore this distinction really seems to be a
> diverge from an honest dialogue.

That you miss the fundamentals of Turing Machines shows your (lack of)
understanding of the field.

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

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Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc>
<1NidnVPZ-NHDl5_8nZ2dnUU7-enNnZ2d@giganews.com> <87sfzw3ao1.fsf@bsb.me.uk>
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<qtGdnfuXs4nFOZn8nZ2dnUU7-cnNnZ2d@giganews.com> <871r7ekugt.fsf@bsb.me.uk>
<K5-dndGZo_-VmJv8nZ2dnUU78QvNnZ2d@giganews.com> <87czqxa0zk.fsf@bsb.me.uk>
<woudnXWBxPba95r8nZ2dnUU78ffNnZ2d@giganews.com> <87mtpz64sq.fsf@bsb.me.uk>
<w5edne8d06OkxpX8nZ2dnUU7-b_NnZ2d@giganews.com> <87bl6f5qvy.fsf@bsb.me.uk>
<1ZednUt-qbrXlZf8nZ2dnUU7-QnNnZ2d@giganews.com> <sed35j$od8$1@dont-email.me>
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Thu, 5 Aug 2021 01:29 UTC

On 8/3/21 10:55 PM, olcott wrote:

> I am not willing to talk about anything besides page 6 until we have
> mutual agreement on page 6.
>
> https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation

Errors in first Paragraph.

Because H acts like a pure simulator JSUT until it decides and then (if
it decides non-halting) aborts its simulation means that it NOT a pure
simulator, PERIOD.

DEFINITION: A Pure simulator is one that will accurate simulate the
machie until it reaches its final halting state. (A Pure simulator of a
non-hating computation is non-halting).

A given instance of H may not affect the copy of the machine it is
simuulating, but it DOES affect the machine it is part of, so H MUST
take that into account when simulating another copy of itself. To omit
that is to miss some of the behavior of the machine it is simulating and
thus is unsound.

THe replacement of a simulation of a simulator with the simulation of
the machihe it is simulating only is valid if you have a REALLY PURE
simulator, it does NOT apply if the simulator can ever abort its
simulation, thus the transformation being done is invalid.

Please provide what youy disagree about this, and PROOF that what I say
is wrong.

FAIR WARNING.

These error invalidate the whole rest of your paper

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

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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>
<Ob2dneXfOsPHVGD9nZ2dnUU78aHNnZ2d@brightview.co.uk>
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<871r7i6n2u.fsf@bsb.me.uk> <OqKdnROLKJ9CdJz8nZ2dnUU7-avNnZ2d@giganews.com>
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Thu, 5 Aug 2021 01:38 UTC

On 8/1/21 9:56 PM, olcott wrote:
> On 8/1/2021 5:54 AM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 7/31/2021 5:08 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 7/30/2021 2:58 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 7/30/2021 7:39 AM, Ben Bacarisse wrote:
>>>>>>
>>>>>>>> Any chance you will now say if
>>>>>>>>
>>>>>>>>>      Ĥ.qx(⟨Ĥ⟩, ⟨Ĥ⟩)
>>>>>>>>
>>>>>>>> transitions to Ĥ.qn or Ĥ.qy?  If you find this question difficult,
>>>>>>>> please ask for some help in understanding it.
>>>>>>>
>>>>>>> Ĥ.qx(⟨Ĥ⟩, ⟨Ĥ⟩) transitions to Ĥ.qn
>>>>>>
>>>>>> An answer.  Thank you.
>>>>>>
>>>>>>      Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>
>>>>>> For Ĥ to be "exactly and precisely as in Linz" this, then, is the
>>>>>> clause
>>>>>> that applies to your H and Ĥ:
>>>>>
>>>>> There is no H in the relevant last paragraph of the Linz proof that
>>>>> forms the basis for the Linz conclusion.
>>>> Distraction.  Everything you ignore below is about the proof and refers
>>>> only to Ĥ.
>>>>
>>>>>>>>>>>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>>>>>>>>>>>>>> if M applied to wM does not halt
>>>>>>
>>>>>> so Ĥ (M) applied to ⟨Ĥ⟩ (wM) does not halt, but you have just told me
>>>>>> that it does.  That is what this full (but abbreviated) state
>>>>>> transition
>>>>>> sequence means:
>>>>>>
>>>>>>      Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>
>>>>>> Which is it?
>>>>>
>>>>> Ĥ0.q0 copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then Ĥ0.qx simulates Ĥ1 with the
>>>>> ⟨Ĥ2⟩ copy then
>>>>> Ĥ1.q0 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then Ĥ1.qx simulates Ĥ2 with the
>>>>> ⟨Ĥ3⟩ copy then
>>>>> Ĥ2.q0 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then Ĥ2.qx simulates Ĥ3 with the
>>>>> ⟨Ĥ4⟩ copy then ...
>>>> This is an abuse of the notation (but I know what you mean).  There is
>>>> no Ĥ1 or Ĥ2.  If you think it helps to show which copy of ⟨Ĥ⟩ your
>>>> simulating "decider" is either running and/or currently looking at, you
>>>> need to come up with a notation that does that.
>>>
>>> A better notation is what I have in my PDF actual subscripts but
>>> people here tel me that their newsreader makes sure to totally ignore
>>> posts with HTML so that do even see the post at all.
>>
>> I am happy you have a notation you like.  Are you prepared to address
>> that fact that your H^ is not "as in Linz"?
>>
>>>> At least I know what
>>>> this "math poem" means, because you've been saying this "it's a
>>>> simulator until" stuff for years.
>>>>
>>>>> The outermost Ĥ0.qx correctly decides that its input: (⟨Ĥ1⟩, ⟨Ĥ2⟩)
>>>>> can't possibly ever reach its final state. Then it transitions to
>>>>> Ĥ0.qn causing the outermost Ĥ0 to halt.
>>>> Apart from the bad notation, yes.  All those copies and tests and
>>>> eventual deciding are neatly summed up in the last ⊢* Ĥ.qn of
>>>>     Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>
>>>>> Because the outermost Ĥ0.qx did not decide that Ĥ0 would never halt
>>>>> and it is self evident that its input: (⟨Ĥ1⟩, ⟨Ĥ2⟩) can't possibly
>>>>> ever reach its final state there is no contradiction or paradox and it
>>>>> decided correctly.
>>>> You are free to define "decide correctly" in any way you like provided
>>>> you are honest about it.  But you hooked people in by saying that
>>>> your Ĥ
>>>> is "exactly and precisely as in Linz", and you quoted, even now, what
>>>> Linz has to say about such TMs:
>>>>     Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>     if M applied to wM does not halt
>>>> This is your quote.  You brought it up.  You claimed your Ĥ was as Linz
>>>> states -- that Ĥ.q0 wM ⊢* Ĥ.qn if and only if M applied to wM does not
>>>> halt.  Linz makes no exceptions based on why the transitions from Ĥ.q0
>>>> wM to Ĥ.qn occur.  Linz does not say
>>>>     Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>     if M applied to wM does not halt or if M applied wM only halts
>>>>     because...
>>>> Are you now saying that your TM was not "as in Linz"?  (You should,
>>>> because you've admitted that elsewhere.)
>>>
>>> Ĥ[0] is to be interpreted to mean Ĥ<sub>0</sub>
>>>   [0] Means the actual Turing machine and not a TM description.
>>>   [1] Means the first TM description parameter
>>>   [2] Means a copy of the the first TM description parameter
>>>
>>> Now I am saying that when the actual unmodified Linz Ĥ is understood
>>> to have a UTM/Halt-Decider at Ĥ[0].qx that this Ĥ[0].qx does correctly
>>> decide that its input: (⟨Ĥ[1]⟩, ⟨Ĥ[2]⟩) can't possibly ever reach its
>>> final state of Ĥ[1].qn or Ĥ[2].qn, therefore we know that its input
>>> never halts therefore we know that a state transition from Ĥ[0].qx to
>>> Ĥ0.qn is necessarily correct.
>>
>> We all know you are declaring that to be correct.  Here's why your Ĥ is
>> not "as in Linz".  Linz requires that
>>
>>    Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>    if M applied to wM does not halt
>>
>> Any Ĥ that eventually transitions to Ĥ.qn on input wM must do so
>> if, and only if, the encoded M applied to wM does not halt.  But you've
>> given us a case where your Ĥ is not like this:
>>
>>    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
> *This did not show up in my other news server so I am pointing it again*
>
> And when we eliminate the fallacy of equivocation error we have
>
> Ĥ[0].q0 ⟨Ĥ⟩ ⊢* Ĥ[0].qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ[0].qn
>
> The way that you do it when your twin bother commits a crime that makes
> you guilty.

Except for Turing Machines, you do every thing your 'twin brother' does,
so if he commits a crime, so did you, so you ARE guilty.

>
> Ĥ[0].q0 ⟨Ĥ[1]⟩ only halts because the simulation of the
> the input to Ĥ[0].qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ was aborted.
>
> (a) The simulation of the input to Ĥ[0].qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩
> was aborted because neither ⟨Ĥ[1]⟩ nor ⟨Ĥ[2]⟩ could
> ever possibly reach their final states.
>
> (b) Because neither ⟨Ĥ[1]⟩ nor ⟨Ĥ[2]⟩ could ever possibly reach
> their final states we know that they never halt.
>
> (c) Because they never halt we know that Ĥ[0].qx correctly decided
> that its input never halts.
>
> This is the same as: X > Y & Y > Z therefore X > Z
> I do seem to have a correct chain of inference.
> I do not think that you can point to any error in
> this chain of inference.
>
> In an honest dialogue when you would disagree that a
> chain of inference derives a correct conclusion you
> would point to a specific error in this chain of inference.
>
> What is the error in (a)(b)(c) ?

Because that isn't the definition of Halting. The definition is what the
ACTUAL machine does, and by the definition of a UTM what the simulaion
by a REAL UTM would do. A simulator that aborts its simulation is NOT a
UTM -- PERIOD.

The fact that H^[1] is aborted before it reaches the state that H^[0] is
KNOWN AND ADMITTED to reach, we know that H^[1] WILL do exactly the same
thing if allowed to continue.

The fact that H make the error and aborts its simulation too soon
doesn't change this fact.

THe fact that UTM((H^[1]), (H^[2])) will halts (since this IS the
identical computation to H^[0]((H^[1]) since all (H^[i]) are identical.


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

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References: <20210719214640.00000dfc@reddwarf.jmc>
<OqKdnROLKJ9CdJz8nZ2dnUU7-avNnZ2d@giganews.com> <87k0la542c.fsf@bsb.me.uk>
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<GtmdnfBrysPrAZn8nZ2dnUU7-L_NnZ2d@giganews.com> <874kcakv3d.fsf@bsb.me.uk>
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Thu, 5 Aug 2021 01:43 UTC

On 8/4/21 8:38 AM, olcott wrote:
> On 8/3/2021 12:11 PM, André G. Isaak wrote:
>> On 2021-08-03 09:21, olcott wrote:
>>> On 8/3/2021 9:33 AM, Malcolm McLean wrote:
>>>> On Tuesday, 3 August 2021 at 14:00:28 UTC+1, olcott wrote:
>>>>> On 8/3/2021 12:29 AM, Malcolm McLean wrote:
>>>>>>
>>>>>> You're trying to claim that H_Hat<H_Hat> doens't really halt because
>>>>>> the recursion of H instances is terminated by H.
>>>>> No I have never been saying that. I am claiming that the input to
>>>>> H(P,P)
>>>>> never halts whether or not H terminates its simulation of this input.
>>>>>
>>>> We agree that P(P) halts.
>>>> So now you're drawing a distinction between P(P) and "the input to
>>>> H(P,P)".
>>>> ThIs is nonsense..
>>>
>>> Try and find any error in (a)(b)(c)(d) on page 6
>>>
>>> https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation
>>
>>
>>
>>
>> The error, which has been pointed out repeatedly, is in your (b).
>>
>> You claim "there are no control flow instructions in the execution
>> trace that would escape the infinite recursion", but there *are* flow
>> control instructions. But your trace is incomplete and skips over the
>> call to B82 where these flow control instructions reside.
>>
>
> I will rephrase that:
> There are no control flow instructions in P that break out of the
> infinite recursion. The control flow instructions in H that could abort
> the simulation of this infinite recursion would never allow P to reach
> its final state. Therefore we know that P never reaches its final state.
> Therefore because of the definition of halting that you provided we know
> that P never halts. Therefore we know that H(P,P)==0 is correct.
>

Yes, there are, since all the algorith of H is part of the algorithm of
P, the conditional within it are part of the algorithm of P

When these instructions abort the simulation of the copy of P that it is
looking at, the P that is using that H will get the answer and Halt.

>> The code at B82 is *part* of the computation performed by P. It *must*
>> be included in any trace of P which purports to be a complete and
>> honest trace.
>>
>
> On the basis of the above analysis we know for sure that this is
> irrelevant. Nothing that happens in H can possibly cause P to reach its
> final state. Nothing that happens in P can possibly cause P to reach its
> final state. Therefore P never reaches its final state. Therefore P
> never halts. Therefore H(P,)==0 is correct.
>

Wrong. Nothing in H, except by not returning the answer of non-halting,
can prevent the P that called that H from halting.

IF the outer H that is called by the actual machine P doesn't return
non-halting, then H never decides on H(H^,H) and thus it fails to be a
decider.

>> You don't seem to grasp the fact that *all* code executed by a
>> particular computation from the moment the computation begins to the
>> time it halts (assuming it halts) is part of that computation. It
>> doesn't matter whether the code in question is shared by some other
>> routine, or is operating system code or whatever. These are purely
>> artificial distinctions which play no role in the theory of computation.
>>
>> André
>>
>
> If the input to H(P,P) cannot possibly halt then H(P,P)==0 is correct.
>

Repeat after me, 'Inputs don't halt or not halt, Machine Halt or not-halt'

Since P(P) Halts, the right answer for H(P,P) is halting. PERIOD.

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

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References: <20210719214640.00000dfc@reddwarf.jmc> <7oKdnTjx4IC20p78nZ2dnUU7-TvNnZ2d@giganews.com> <875yws36vt.fsf@bsb.me.uk> <j66dnbdHrpV8_p78nZ2dnUU7-aXNnZ2d@giganews.com> <87im0s0ydp.fsf@bsb.me.uk> <Brqdnfehrf0Kj5n8nZ2dnUU7-X3NnZ2d@giganews.com> <87tukblgjy.fsf@bsb.me.uk> <qtGdnfuXs4nFOZn8nZ2dnUU7-cnNnZ2d@giganews.com> <871r7ekugt.fsf@bsb.me.uk> <K5-dndGZo_-VmJv8nZ2dnUU78QvNnZ2d@giganews.com> <87czqxa0zk.fsf@bsb.me.uk> <woudnXWBxPba95r8nZ2dnUU78ffNnZ2d@giganews.com> <87mtpz64sq.fsf@bsb.me.uk> <w5edne8d06OkxpX8nZ2dnUU7-b_NnZ2d@giganews.com> <87bl6f5qvy.fsf@bsb.me.uk> <1ZednUt-qbrXlZf8nZ2dnUU7-QnNnZ2d@giganews.com> <sed35j$od8$1@dont-email.me> <KcOdnSwOTcFujZf8nZ2dnUU7-W_NnZ2d@giganews.com> <sed4b4$tqd$1@dont-email.me> <PKGdnRIK9KVMi5f8nZ2dnUU78V3NnZ2d@giganews.com> <sed5fo$31j$1@dont-email.me> <INCdnRriUbgihJf8nZ2dnUU7-YmdnZ2d@giganews.com> <Ysedne93LONKgJf8nZ2dnUU7-XvNnZ2d@giganews.com> <sed904$iae$1@dont-email.me> <sv2dnSY1jKlPM5f8nZ2dnUU78eXNnZ2d@giganews.com>
From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Thu, 5 Aug 2021 02:05 UTC

On 8/4/21 9:14 AM, olcott wrote:
> On 8/4/2021 12:36 AM, André G. Isaak wrote:

>> That's not going to happen since your argument on page 6 is fallacious.
>>
>> Read my earlier message <sebtb9$plb$1@dont-email.me>
>>
>> I see no reason to repeat it.
>>
>
> You simply ignored my correction of your error.
> I responded to this message again with words that are more clear.
>
> Simply ignoring this response is sufficient proof that you are
> unmotivated to have an actual honest dialogue.
>
>

You mean your garbage of:
>
> Whether H aborts its simulation of P or not P never reaches its final state.
>
> The conditional instructions are merely the aspect of H aborting its simulation of P.
>
> Because P never reaches its final state whether H aborts its simulation of P or not we know that P never halts.

First statement is wrong.

If H abort its simulation of P(P), and returns the non-halting answer to
the P that called it, then that P Halts.

If H does not abort its simulation, then yes, THAT version of P is
non-halting but for that version of P, its H never answers, so that
doesn't matter. REMEMBER, the definition of P (which should be called
H^) is dependent of the definition of H, so each version of H needs to
be looked at independently.

Remember, Halting a proper of Turing Machines, not 'inputs'. If an input
is a rerpesentation of a Turing Machine, it is the behavior of the
Turing Machine that matters, not the partial simulation of it in the
decider.

Second statement is bogus, the conditional instructions in H are what
allows it to actually let H given an answer. To say that they don't
count is just being dishonest. Thats like saying your white cat is
really a black cat because you call him blackie.

Third statement is irrelevent, the definition of Halting is what the
ACTUAL machine does, not what an aborted simulation does, which in this
case has been shown, and admitted by your, that it is halt.

Yes, you can replace the running of the actual machine by a simulation
with a REAL UTM, and if we do that ALSO will halt. You can NOT make any
arguement about the non-halting if the simulation is aborted.

Try Again.

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

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Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct ]
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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> <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>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 4 Aug 2021 22:18:13 -0500
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 by: olcott - Thu, 5 Aug 2021 03:18 UTC

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 because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts.

>> Ĥ.q0 ⟨Ĥ[1]⟩ ⊢* Ĥ.qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ.qn
>> if Ĥ[1] applied to ⟨Ĥ[2]⟩ does not halt.
>>
>> That you persistently ignore this distinction really seems to be a
>> diverge from an honest dialogue.
>
> If the various strings you've chosen to number (badly) are not identical
> then your "hat" construction is wrong. Linz's "hat" version makes an
> exact copy.
>

When Bill says that his identical twin brother is not going to go to the
store, and then Bill goes to the store this does not make him a liar.

> You are free to write ⟨Ĥ[99]⟩ if you like, but I am also free to change
> that back to ⟨Ĥ⟩ because they are identical strings. Anything true of
>

Ĥ is not a string it is a Turing machine.
Turing machines are not identical to strings.
Ĥ halts the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ never reaches its final state
whether or not the halt decider at Ĥ.qx stop simulating it.

If you want an actual honest dialogue we must have closure on some of
the points. You must say what things you agree with and not merely
ignore those things that you agree with.

Do you understand that the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ never reaches its
final state whether or not the halt decider at Ĥ.qx stop simulating it?

If you disagree then to prove that you are not simply being disagreeable
you must proceed with a dialogue on this single point until we reach
mutual agreement. All other points will not be discussed until we reach
mutual agreement on this point.

> Ĥ.q0 ⟨Ĥ[1]⟩ ⊢* Ĥ.qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ.qn
> if Ĥ[1] applied to ⟨Ĥ[2]⟩ does not halt.
>
> is true of
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> if Ĥ applied to ⟨Ĥ⟩ does not halt.
>
> But do say if it isn't. I'd like there to be another reason why your
> claims are irrelevant -- the more the merrier. Please tell us that Ĥ[1]
> != Ĥ or that ⟨Ĥ[2]⟩ != ⟨Ĥ⟩.
>
>>> You keep telling me that Ĥ applied to ⟨Ĥ⟩ reaches the final state Ĥ.qn.
>>> Are you changing your mind after all this time?
>
> You are not answering questions, are you?
>

--
Copyright 2021 Pete Olcott

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

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

<Y5-dnbXu3ONoxpb8nZ2dnUU7-U3NnZ2d@giganews.com>

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Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct ][ GIGO ]
Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc> <87k0la542c.fsf@bsb.me.uk> <1NidnVPZ-NHDl5_8nZ2dnUU7-enNnZ2d@giganews.com> <87sfzw3ao1.fsf@bsb.me.uk> <7oKdnTjx4IC20p78nZ2dnUU7-TvNnZ2d@giganews.com> <875yws36vt.fsf@bsb.me.uk> <j66dnbdHrpV8_p78nZ2dnUU7-aXNnZ2d@giganews.com> <87im0s0ydp.fsf@bsb.me.uk> <Brqdnfehrf0Kj5n8nZ2dnUU7-X3NnZ2d@giganews.com> <87tukblgjy.fsf@bsb.me.uk> <qtGdnfuXs4nFOZn8nZ2dnUU7-cnNnZ2d@giganews.com> <871r7ekugt.fsf@bsb.me.uk> <K5-dndGZo_-VmJv8nZ2dnUU78QvNnZ2d@giganews.com> <87czqxa0zk.fsf@bsb.me.uk> <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>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 4 Aug 2021 22:33:08 -0500
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 by: olcott - Thu, 5 Aug 2021 03:33 UTC

On 8/4/2021 6:23 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/4/2021 1:51 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/1/2021 11:00 AM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/1/2021 7:41 AM, Ben Bacarisse wrote:
>>>>>>> Malcolm McLean <malcolm.arthur.mclean@gmail.com> writes:
>>>>>>>
>>>>>>>> On Sunday, 1 August 2021 at 11:54:57 UTC+1, Ben Bacarisse wrote:
>>>>>>>>>
>>>>>>>>> Here we can see that Ĥ applied to ⟨Ĥ⟩ halts. You can call your Ĥ's
>>>>>>>>> behaviour "correct". You can call it anything you like. But it's not
>>>>>>>>> "as in Linz". It does not say anything about Linz's proof. It does not
>>>>>>>>> do anything people would call impossible or even interesting.
>>>>>>>>>
>>>>>>>> It seems to be established that H(H_Hat, H_Hat) returns "non-halting"
>>>>>>>> whilst H_Hat(H_Hat) halts. So all is as Linz says it must be and no
>>>>>>>> theorems are refuted. Which you would expect. If results were consistent
>>>>>>>> it would have to be some cheap trick.
>>>>>>> I case there is some confusion, I mean that PO's Ĥ is not an Ĥ as
>>>>>>> specified in Linz. Yes, everything is in accordance with the truth as
>>>>>>> laid out in Linz and, indeed, in any textbook.
>>>>>>> I point this out to PO because he brings it up. He keeps posting the
>>>>>>> specification of what an Ĥ, as Linz specifies it, would do:
>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>>>> if (and only if) M applied to wM does not halt.
>>>>>>> He claims (or used to claim) that his Ĥ meets this specification for at
>>>>>>> least the one case where wM == ⟨Ĥ⟩:
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>> if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>>>>> To remain relevant, he /must/ keep insisting that his Ĥ meets the
>>>>>>> requirements laid out in Linz, if only for this one key input.
>>>>>>>
>>>>>>
>>>>>> Ĥ[0].q0 is taken to mean Ĥ<sub>0</sub>.q0 which is the Turing machine.
>>>>>>
>>>>>> Ĥ[1].q0 is taken to mean Ĥ<sub>1</sub>.q0 which is the Turing machine
>>>>>> description input to Ĥ[0].q0
>>>>>>
>>>>>> Ĥ[2].q0 is taken to mean Ĥ<sub>2</sub>.q0 which is first copy of the
>>>>>> Turing machine description input to Ĥ[0].q0
>>>>>>
>>>>>> Ĥ[0].q0 ⟨Ĥ⟩ ⊢* Ĥ[0].qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ[0].qn
>>>>> Ĥ[0] is Ĥ so you are confirming, yet again, that
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>
>>>>>> It is neither a contradiction nor a paradox because there are three
>>>>>> different instances of Ĥ.
>>>>> I agree that this is neither a paradox nor a contradiction. It's just a
>>>>> fact derived form the logic of how your Ĥ is written (the majority of
>>>>> which you are keeping hidden from us).
>>>>>
>>>>>> Because the only reason that the first instance halts is that Ĥ[0].qx
>>>>>> correctly determines that its input cannot possibly ever reach its
>>>>>> final state of Ĥ[1].qn or Ĥ[1].qy whether or not the simulating halt
>>>>>> decider aborts its simulation of this input, we know with 100%
>>>>>> perfectly justified logical certainty that the input to Ĥ[0].qx never
>>>>>> halts.
>>>>> We know, since you keep telling us, that Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn. This clearly
>>>>> shows that Ĥ applied to ⟨Ĥ⟩ halts. You can see the final state right
>>>>
>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>>> if M applied to wM halts, and
>>>>
>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>> if M applied to wM does not halt
>>>>
>>>> You are using the wrong Ĥ.
>>> First of all, let's be 100% clear: I am talking about what /your/ Ĥ
>>> does, based in the facts you have let slip about it.
>>>
>>>> Linz stipulates that wM is ⟨Ĥ⟩ and M is the underlying machine of this
>>>> ⟨Ĥ⟩ therefore M applied to wM means ⟨Ĥ⟩ applied to ⟨Ĥ⟩.
>>> No. How many years have you been staring at this one page from Linz?
>>> You still don't know what it says. Do ask me questions, if you'd like
>>> to know what the text you've been sure is wrong for 17 years really
>>> says.
>>>
>>
>> ...Turing machine halting problem.
>> Simply stated, the problem is:
>> given the description of a Turing machine M
>> given the description of a Turing machine M
>> given the description of a Turing machine M
>> given the description of a Turing machine M
>> 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?
>>
>> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>>
>> Ĥ.q0 ⟨M⟩ ⊢* Ĥ.qx ⟨M⟩ ⟨M⟩ ⊢* Ĥ.qn
>> if M applied to ⟨M⟩ does not halt
>>
>> When ⟨M⟩ = ⟨Ĥ⟩: I have proved that Ĥ ⟨M⟩ transitions to Ĥ.qn because M
>> never reaches a final state.
>
> This a schoolboy error. Why are you so scared to ask me questions? Do
> you fear you might understand me?
>

Speaking to me with denigration debases only yourself.

The halting problem is not about deciding whether or not a Turing
Machine halts, it is only about whether or not the description of a
Turing machine specifies a computation that reaches its final state.

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

--
Copyright 2021 Pete Olcott

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

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

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Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs_Count_]
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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Thu, 5 Aug 2021 03:37 UTC

On 8/4/2021 5:23 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/4/2021 1:37 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/2/2021 7:11 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/2/2021 2:10 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 8/1/2021 5:54 AM, Ben Bacarisse wrote:
>>>>>>>
>>>>>>>>> I am happy you have a notation you like. Are you prepared to address
>>>>>>>>> that fact that your H^ is not "as in Linz"?
>>>>>>>>
>>>>>>>> My Ĥ is exactly the Linz Ĥ with the additional elaboration that the
>>>>>>>> second wildcard state transition ⊢* is defined to be a simulating halt
>>>>>>>> decider.
>>>>>>> No. I've explained many times now why your Ĥ is not at all "the Linz
>>>>>>> Ĥ". Do you see any point in my doing so again?
>>>>> I suspect not. You certainly have not asked a single question that
>>>>> could help you to understand why your Ĥ is irrelevant.
>>>>>
>>>>>>>>> We all know you are declaring that to be correct. Here's why your Ĥ is
>>>>>>>>> not "as in Linz". Linz requires that
>>>>>>>>>
>>>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>>>>>> if M applied to wM does not halt
>>>>>>>>>
>>>>>>>>> Any Ĥ that eventually transitions to Ĥ.qn on input wM must do so
>>>>>>>>> if, and only if, the encoded M applied to wM does not halt. But you've
>>>>>>>>> given us a case where your Ĥ is not like this:
>>>>>>>>>
>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>
>>>>>>>> And when we eliminate the fallacy of equivocation error we have
>>>>>>>>
>>>>>>>> Ĥ[0].q0 ⟨Ĥ⟩ ⊢* Ĥ[0].qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ[0].qn
>>>>>>> The only thing I think you are equivocating on is pretending that Ĥ[0]
>>>>>>> is not Ĥ, and it doesn't look as if you've removed that error. I think
>>>>>>> you /should/ remove it so that you can be saying something of value
>>>>>>> about Ĥ.
>>>>>>
>>>>>> It is clear from the text that Linz does specify at least three
>>>>>> different instances of Ĥ, The TM the TMD input ⟨Ĥ⟩ and a copy of this
>>>>>> TMD input.
>>>>> Still equivocating. Either Ĥ[0] = Ĥ or you are wasting everyone's time.
>>>>> (This is mathematical equality. Ĥ is a tuple of sets. If Ĥ[0] is not
>>>>> exactly identical in every way to Ĥ then I don't care about it. Note
>>>>> that I'm not disputing your right, for ease of explanation, to give
>>>>> identical things more than one name. But the same permission allows me
>>>>> to use any of the names because they name the same thing.)
>>>>> You are wrong because
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>> where Linz requires that
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>> if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>>
>>>> You have the above incorrectly in these two
>>>> M refers to the Turing Machine described by wM.
>>>>
>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>>> if M applied to wM halts, and
>>>>
>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>> if M applied to wM does not halt
>>>>
>>>> This is *NOT* the Ĥ that is executed.
>>> I have no idea what you are objecting to. You can substitute Ĥ and ⟨Ĥ⟩
>>> into what Linz says, can't you? I did it above and you don't like it:
>>> For M = Ĥ and wM = ⟨Ĥ⟩ Linz tells us that
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>> if Ĥ applied to ⟨Ĥ⟩ does not halt.
>>> Your Ĥ is not like that. Its behaviour is boring and obvious and
>>> trivial.
>>
>> This is how it really is:
>> if ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt.
>> and ⟨Ĥ⟩ applied to ⟨Ĥ⟩ never reaches its final state.
>
> No. ⟨Ĥ⟩ is a string, not a TM. The TM that string represents is Ĥ and
> Ĥ applied to ⟨Ĥ⟩ does halt.
>

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?

The question is: Does the Turing Machine description ⟨Ĥ⟩ specify a
computation that reaches its final state on input ⟨Ĥ⟩?

Do you agree with this?

--
Copyright 2021 Pete Olcott

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

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

<GcJOI.2151$lK.2002@fx41.iad>

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Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[
Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inpu
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Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc> <87k0la542c.fsf@bsb.me.uk>
<1NidnVPZ-NHDl5_8nZ2dnUU7-enNnZ2d@giganews.com> <87sfzw3ao1.fsf@bsb.me.uk>
<7oKdnTjx4IC20p78nZ2dnUU7-TvNnZ2d@giganews.com> <875yws36vt.fsf@bsb.me.uk>
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<Brqdnfehrf0Kj5n8nZ2dnUU7-X3NnZ2d@giganews.com> <87tukblgjy.fsf@bsb.me.uk>
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<K5-dndGZo_-VmJv8nZ2dnUU78QvNnZ2d@giganews.com> <87czqxa0zk.fsf@bsb.me.uk>
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Thu, 5 Aug 2021 03:48 UTC

On 8/4/21 9:37 PM, olcott wrote:
> On 8/4/2021 5:23 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/4/2021 1:37 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/2/2021 7:11 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 8/2/2021 2:10 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> On 8/1/2021 5:54 AM, Ben Bacarisse wrote:
>>>>>>>>
>>>>>>>>>> I am happy you have a notation you like.  Are you prepared to
>>>>>>>>>> address
>>>>>>>>>> that fact that your H^ is not "as in Linz"?
>>>>>>>>>
>>>>>>>>> My Ĥ is exactly the Linz Ĥ with the additional elaboration that
>>>>>>>>> the
>>>>>>>>> second wildcard state transition ⊢* is defined to be a
>>>>>>>>> simulating halt
>>>>>>>>> decider.
>>>>>>>> No.  I've explained many times now why your Ĥ is not at all "the
>>>>>>>> Linz
>>>>>>>> Ĥ".  Do you see any point in my doing so again?
>>>>>> I suspect not.  You certainly have not asked a single question that
>>>>>> could help you to understand why your Ĥ is irrelevant.
>>>>>>
>>>>>>>>>> We all know you are declaring that to be correct.  Here's why
>>>>>>>>>> your Ĥ is
>>>>>>>>>> not "as in Linz".  Linz requires that
>>>>>>>>>>
>>>>>>>>>>        Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>>>>>>>        if M applied to wM does not halt
>>>>>>>>>>
>>>>>>>>>> Any Ĥ that eventually transitions to Ĥ.qn on input wM must do so
>>>>>>>>>> if, and only if, the encoded M applied to wM does not halt. 
>>>>>>>>>> But you've
>>>>>>>>>> given us a case where your Ĥ is not like this:
>>>>>>>>>>
>>>>>>>>>>        Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>
>>>>>>>>> And when we eliminate the fallacy of equivocation error we have
>>>>>>>>>
>>>>>>>>> Ĥ[0].q0 ⟨Ĥ⟩ ⊢* Ĥ[0].qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ[0].qn
>>>>>>>> The only thing I think you are equivocating on is pretending
>>>>>>>> that Ĥ[0]
>>>>>>>> is not Ĥ, and it doesn't look as if you've removed that error. 
>>>>>>>> I think
>>>>>>>> you /should/ remove it so that you can be saying something of value
>>>>>>>> about Ĥ.
>>>>>>>
>>>>>>> It is clear from the text that Linz does specify at least three
>>>>>>> different instances of Ĥ, The TM the TMD input ⟨Ĥ⟩ and a copy of
>>>>>>> this
>>>>>>> TMD input.
>>>>>> Still equivocating.  Either Ĥ[0] = Ĥ or you are wasting everyone's
>>>>>> time.
>>>>>> (This is mathematical equality.  Ĥ is a tuple of sets.  If Ĥ[0] is
>>>>>> not
>>>>>> exactly identical in every way to Ĥ then I don't care about it.  Note
>>>>>> that I'm not disputing your right, for ease of explanation, to give
>>>>>> identical things more than one name.  But the same permission
>>>>>> allows me
>>>>>> to use any of the names because they name the same thing.)
>>>>>> You are wrong because
>>>>>>      Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>> where Linz requires that
>>>>>>     Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>     if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>>>
>>>>> You have the above incorrectly in these two
>>>>> M refers to the Turing Machine described by wM.
>>>>>
>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>>>> if M applied to wM halts, and
>>>>>
>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>> if M applied to wM does not halt
>>>>>
>>>>> This is *NOT* the Ĥ that is executed.
>>>> I have no idea what you are objecting to.  You can substitute Ĥ and ⟨Ĥ⟩
>>>> into what Linz says, can't you?  I did it above and you don't like it:
>>>> For M = Ĥ and wM = ⟨Ĥ⟩ Linz tells us that
>>>>     Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>     if Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>> Your Ĥ is not like that.  Its behaviour is boring and obvious and
>>>> trivial.
>>>
>>> This is how it really is:
>>> if ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt.
>>> and ⟨Ĥ⟩ applied to ⟨Ĥ⟩ never reaches its final state.
>>
>> No.  ⟨Ĥ⟩ is a string, not a TM.  The TM that string represents is Ĥ and
>> Ĥ applied to ⟨Ĥ⟩ does halt.
>>
>
> 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?
>
> The question is: Does the Turing Machine description ⟨Ĥ⟩ specify a
> computation that reaches its final state on input ⟨Ĥ⟩?
>
> Do you agree with this?
>
>

The question is does the machine specified by description reach its
final state when given the specified input in a finite number of steps.

Description ONLY specify Computation via the Turing Machine they represent.

Maybe that is your confusion.

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

<AjJOI.687$qf5.450@fx07.iad>

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Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[
Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct
][ GIGO ]
Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc> <87k0la542c.fsf@bsb.me.uk>
<1NidnVPZ-NHDl5_8nZ2dnUU7-enNnZ2d@giganews.com> <87sfzw3ao1.fsf@bsb.me.uk>
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<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>
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<87v94k23rn.fsf@bsb.me.uk> <Y5-dnbXu3ONoxpb8nZ2dnUU7-U3NnZ2d@giganews.com>
From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Thu, 5 Aug 2021 03:56 UTC

On 8/4/21 9:33 PM, olcott wrote:
> On 8/4/2021 6:23 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/4/2021 1:51 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/1/2021 11:00 AM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 8/1/2021 7:41 AM, Ben Bacarisse wrote:
>>>>>>>> Malcolm McLean <malcolm.arthur.mclean@gmail.com> writes:
>>>>>>>>
>>>>>>>>> On Sunday, 1 August 2021 at 11:54:57 UTC+1, Ben Bacarisse wrote:
>>>>>>>>>>
>>>>>>>>>> Here we can see that Ĥ applied to ⟨Ĥ⟩ halts. You can call your
>>>>>>>>>> Ĥ's
>>>>>>>>>> behaviour "correct". You can call it anything you like. But
>>>>>>>>>> it's not
>>>>>>>>>> "as in Linz". It does not say anything about Linz's proof. It
>>>>>>>>>> does not
>>>>>>>>>> do anything people would call impossible or even interesting.
>>>>>>>>>>
>>>>>>>>> It seems to be established that H(H_Hat, H_Hat) returns
>>>>>>>>> "non-halting"
>>>>>>>>> whilst H_Hat(H_Hat) halts. So all is as Linz says it must be
>>>>>>>>> and no
>>>>>>>>> theorems are refuted. Which you would expect. If results were
>>>>>>>>> consistent
>>>>>>>>> it would have to be some cheap trick.
>>>>>>>> I case there is some confusion, I mean that PO's Ĥ is not an Ĥ as
>>>>>>>> specified in Linz.  Yes, everything is in accordance with the
>>>>>>>> truth as
>>>>>>>> laid out in Linz and, indeed, in any textbook.
>>>>>>>> I point this out to PO because he brings it up.  He keeps
>>>>>>>> posting the
>>>>>>>> specification of what an Ĥ, as Linz specifies it, would do:
>>>>>>>>       Ĥ.q0 wM ⊢* Ĥ.qx wM wM  ⊢* Ĥ.qn
>>>>>>>>       if (and only if) M applied to wM does not halt.
>>>>>>>> He claims (or used to claim) that his Ĥ meets this specification
>>>>>>>> for at
>>>>>>>> least the one case where wM == ⟨Ĥ⟩:
>>>>>>>>       Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩  ⊢* Ĥ.qn
>>>>>>>>       if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>>>>>> To remain relevant, he /must/ keep insisting that his Ĥ meets the
>>>>>>>> requirements laid out in Linz, if only for this one key input.
>>>>>>>>
>>>>>>>
>>>>>>> Ĥ[0].q0 is taken to mean Ĥ<sub>0</sub>.q0 which is the Turing
>>>>>>> machine.
>>>>>>>
>>>>>>> Ĥ[1].q0 is taken to mean Ĥ<sub>1</sub>.q0 which is the Turing
>>>>>>> machine
>>>>>>> description input to Ĥ[0].q0
>>>>>>>
>>>>>>> Ĥ[2].q0 is taken to mean Ĥ<sub>2</sub>.q0 which is first copy of the
>>>>>>> Turing machine description input to Ĥ[0].q0
>>>>>>>
>>>>>>> Ĥ[0].q0 ⟨Ĥ⟩ ⊢* Ĥ[0].qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ[0].qn
>>>>>> Ĥ[0] is Ĥ so you are confirming, yet again, that
>>>>>>       Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>
>>>>>>> It is neither a contradiction nor a paradox because there are three
>>>>>>> different instances of Ĥ.
>>>>>> I agree that this is neither a paradox nor a contradiction.  It's
>>>>>> just a
>>>>>> fact derived form the logic of how your Ĥ is written (the majority of
>>>>>> which you are keeping hidden from us).
>>>>>>
>>>>>>> Because the only reason that the first instance halts is that
>>>>>>> Ĥ[0].qx
>>>>>>> correctly determines that its input cannot possibly ever reach its
>>>>>>> final state of Ĥ[1].qn or Ĥ[1].qy whether or not the simulating halt
>>>>>>> decider aborts its simulation of this input, we know with 100%
>>>>>>> perfectly justified logical certainty that the input to Ĥ[0].qx
>>>>>>> never
>>>>>>> halts.
>>>>>> We know, since you keep telling us, that Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn.  This
>>>>>> clearly
>>>>>> shows that Ĥ applied to ⟨Ĥ⟩ halts.  You can see the final state right
>>>>>
>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>>>> if M applied to wM halts, and
>>>>>
>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>> if M applied to wM does not halt
>>>>>
>>>>> You are using the wrong Ĥ.
>>>> First of all, let's be 100% clear: I am talking about what /your/ Ĥ
>>>> does, based in the facts you have let slip about it.
>>>>
>>>>> Linz stipulates that wM is ⟨Ĥ⟩ and M is the underlying machine of this
>>>>> ⟨Ĥ⟩ therefore M applied to wM means ⟨Ĥ⟩ applied to ⟨Ĥ⟩.
>>>> No.  How many years have you been staring at this one page from Linz?
>>>> You still don't know what it says.  Do ask me questions, if you'd like
>>>> to know what the text you've been sure is wrong for 17 years really
>>>> says.
>>>>
>>>
>>> ...Turing machine halting problem.
>>> Simply stated, the problem is:
>>> given the description of a Turing machine M
>>> given the description of a Turing machine M
>>> given the description of a Turing machine M
>>> given the description of a Turing machine M
>>> 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?
>>>
>>> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>>>
>>> Ĥ.q0 ⟨M⟩ ⊢* Ĥ.qx ⟨M⟩ ⟨M⟩ ⊢* Ĥ.qn
>>> if M applied to ⟨M⟩ does not halt
>>>
>>> When ⟨M⟩ = ⟨Ĥ⟩: I have proved that Ĥ ⟨M⟩ transitions to Ĥ.qn because M
>>> never reaches a final state.
>>
>> This a schoolboy error.  Why are you so scared to ask me questions?  Do
>> you fear you might understand me?
>>
>
> Speaking to me with denigration debases only yourself.
>
> The halting problem is not about deciding whether or not a Turing
> Machine halts, it is only about whether or not the description of a
> Turing machine specifies a computation that reaches its final state.
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>

And a description only specifies a computation via the Turing Machine it
represents.

You seem to have uncoupled the concept of Turing Machine from this.

Computations are computed with Turing Machines which have a
representation. The representation does NOT specify directly a
Computation, only a Turing Machine.

Please note, a given Computation actually can be computed by a very
large number (infinite) of Turing Machines, and each Turing Machine can
be represented by a large number (infinite) of Representations, and how
you can build those representations is a function of the interpreter of
them. A given representation string only has meaning as it relates to
some interpreter (the UTM or decider that will process it) and might not
actually mean anything to some other machine.

The description can ONLY get back to the Computation via the Turing
Machine it represents.

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

<psJOI.67$WG5.22@fx38.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
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Thu, 5 Aug 2021 04:05 UTC

On 8/4/21 9:18 PM, olcott wrote:
> 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 because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts.

It goes to H^.qn because H decides that H (H^) (H^) will not halt, but H
is shown to be in error because H^ (H^) does halt.

>
>>> Ĥ.q0 ⟨Ĥ[1]⟩ ⊢* Ĥ.qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ.qn
>>> if Ĥ[1] applied to ⟨Ĥ[2]⟩ does not halt.
>>>
>>> That you persistently ignore this distinction really seems to be a
>>> diverge from an honest dialogue.
>>
>> If the various strings you've chosen to number (badly) are not identical
>> then your "hat" construction is wrong.  Linz's "hat" version makes an
>> exact copy.
>>
>
> When Bill says that his identical twin brother is not going to go to the
> store, and then Bill goes to the store this does not make him a liar.

Then this concept of Twin is not applicable to Turing Machines.

Alternate copies of a Turing Machine will ALWAYS behave xactly the same.

>
>> You are free to write ⟨Ĥ[99]⟩ if you like, but I am also free to change
>> that back to ⟨Ĥ⟩ because they are identical strings.  Anything true of
>>
>
> Ĥ is not a string it is a Turing machine.
> Turing machines are not identical to strings.
> Ĥ halts the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ never reaches its final state
> whether or not the halt decider at Ĥ.qx stop simulating it.

H^ (H^) Halt because H decides that it thinks that H^ (H^) will not
halt. The fact that H^ (H^) does halt means that H was wrong.

It doesn't matter that H never completed its simulation of H^ (H^) but
aborted it, the trace of H^ (H^) shows what it would have done if H had
simulated it a bit farther.

An aborted simulation NEVER proves non-Halting.

>
> If you want an actual honest dialogue we must have closure on some of
> the points. You must say what things you agree with and not merely
> ignore those things that you agree with.
>
> Do you understand that the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ never reaches its
> final state whether or not the halt decider at Ĥ.qx stop simulating it?

No, Aborted simulations don't prove anything, especially when the traces
you UNSOUND logic.

>
> If you disagree then to prove that you are not simply being disagreeable
> you must proceed with a dialogue on this single point until we reach
> mutual agreement. All other points will not be discussed until we reach
> mutual agreement on this point.

Maybe you should do the same, and try to point to an actual ERROR in the
statements people make to point out where YOU are in error. If you don't
think they are right, show the actual logical error in them. People are
mostly working directly off definitions, so it should be easy if you are
right.

Re: Black box halt decider is NOT a partial decider

<sefqsk$jns$3@gioia.aioe.org>

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https://www.novabbs.com/devel/article-flat.php?id=19512&group=comp.theory#19512

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Path: i2pn2.org!i2pn.org!aioe.org!ux6ld97kLXxG8kVFFLnoWg.user.46.165.242.75.POSTED!not-for-mail
From: chris.m....@gmail.com (Chris M. Thomasson)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider
Date: Wed, 4 Aug 2021 21:53:39 -0700
Organization: Aioe.org NNTP Server
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 by: Chris M. Thomasson - Thu, 5 Aug 2021 04:53 UTC

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?

Re: Black box halt decider is NOT a partial decider

<sefqv3$jns$4@gioia.aioe.org>

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From: chris.m....@gmail.com (Chris M. Thomasson)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider
Date: Wed, 4 Aug 2021 21:54:58 -0700
Organization: Aioe.org NNTP Server
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 by: Chris M. Thomasson - Thu, 5 Aug 2021 04:54 UTC

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

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

<20276f21-9bdc-4e64-9b20-1dc04bf8fa8an@googlegroups.com>

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Subject: Re: Black box halt decider is NOT a partial decider [ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct ]
From: malcolm....@gmail.com (Malcolm McLean)
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 by: Malcolm McLean - Thu, 5 Aug 2021 09:02 UTC

On Thursday, 5 August 2021 at 04:18:21 UTC+1, olcott wrote:
> On 8/4/2021 6:22 PM, Ben Bacarisse wrote:
> > olcott <No...@NoWhere.com> writes:
> >
> > If the various strings you've chosen to number (badly) are not identical
> > then your "hat" construction is wrong. Linz's "hat" version makes an
> > exact copy.
> >
> When Bill says that his identical twin brother is not going to go to the
> store, and then Bill goes to the store this does not make him a liar.
>
> > You are free to write ⟨Ĥ[99]⟩ if you like, but I am also free to change
> > that back to ⟨Ĥ⟩ because they are identical strings. Anything true of
> >
> Ĥ is not a string it is a Turing machine.
> Turing machines are not identical to strings.
> Ĥ halts the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ never reaches its final state
> whether or not the halt decider at Ĥ.qx stop simulating it.
>
> If you want an actual honest dialogue we must have closure on some of
> the points. You must say what things you agree with and not merely
> ignore those things that you agree with.
>
> Do you understand that the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ never reaches its
> final state whether or not the halt decider at Ĥ.qx stop simulating it?
>
H_Hat <H_Hat> halts. So if it is simulated by a UTM, the simulation
UTM <H_HAT><H_Hat> also halts.
But the halt decider embedded in H_Hat is not a UTM. It's a near UTM,
that has abort logic. Somewhere the abort logic must be triggered,
which causes H_Hat<H_Hat> to halt.
H<H_Hat><H_Hat> also triggers this abort logic, so H<H_Hat><H_Hat>
reports "false" (non-halting).
>
> If you disagree then to prove that you are not simply being disagreeable
> you must proceed with a dialogue on this single point until we reach
> mutual agreement. All other points will not be discussed until we reach
> mutual agreement on this point.
>
We agree that H_Hat<H_Hat> halts.
We agree that H <H_Hat> <H_Hat> reports "false" (non-halting).

Can't you see the obvious?

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

<lemdnTsHxYH0SZb8nZ2dnUU7-YfNnZ2d@giganews.com>

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Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct ]
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
References: <20210719214640.00000dfc@reddwarf.jmc> <875yws36vt.fsf@bsb.me.uk> <j66dnbdHrpV8_p78nZ2dnUU7-aXNnZ2d@giganews.com> <87im0s0ydp.fsf@bsb.me.uk> <Brqdnfehrf0Kj5n8nZ2dnUU7-X3NnZ2d@giganews.com> <87tukblgjy.fsf@bsb.me.uk> <qtGdnfuXs4nFOZn8nZ2dnUU7-cnNnZ2d@giganews.com> <871r7ekugt.fsf@bsb.me.uk> <K5-dndGZo_-VmJv8nZ2dnUU78QvNnZ2d@giganews.com> <87czqxa0zk.fsf@bsb.me.uk> <woudnXWBxPba95r8nZ2dnUU78ffNnZ2d@giganews.com> <87mtpz64sq.fsf@bsb.me.uk> <w5edne8d06OkxpX8nZ2dnUU7-b_NnZ2d@giganews.com> <87bl6f5qvy.fsf@bsb.me.uk> <xLednaPs_ZSXCZX8nZ2dnUU7-YnNnZ2d@giganews.com> <87o8af47y0.fsf@bsb.me.uk> <NsudnY99rthDOJX8nZ2dnUU7-c_NnZ2d@giganews.com> <87zgtx2wxn.fsf@bsb.me.uk> <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> <20276f21-9bdc-4e64-9b20-1dc04bf8fa8an@googlegroups.com>
From: NoO...@NoWhere.com (olcott)
Date: Thu, 5 Aug 2021 07:07:04 -0500
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 by: olcott - Thu, 5 Aug 2021 12:07 UTC

On 8/5/2021 4:02 AM, Malcolm McLean wrote:
> On Thursday, 5 August 2021 at 04:18:21 UTC+1, olcott wrote:
>> On 8/4/2021 6:22 PM, Ben Bacarisse wrote:
>>> olcott <No...@NoWhere.com> writes:
>>>
>>> If the various strings you've chosen to number (badly) are not identical
>>> then your "hat" construction is wrong. Linz's "hat" version makes an
>>> exact copy.
>>>
>> When Bill says that his identical twin brother is not going to go to the
>> store, and then Bill goes to the store this does not make him a liar.
>>
>>> You are free to write ⟨Ĥ[99]⟩ if you like, but I am also free to change
>>> that back to ⟨Ĥ⟩ because they are identical strings. Anything true of
>>>
>> Ĥ is not a string it is a Turing machine.
>> Turing machines are not identical to strings.
>> Ĥ halts the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ never reaches its final state
>> whether or not the halt decider at Ĥ.qx stop simulating it.
>>
>> If you want an actual honest dialogue we must have closure on some of
>> the points. You must say what things you agree with and not merely
>> ignore those things that you agree with.
>>
>> Do you understand that the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ never reaches its
>> final state whether or not the halt decider at Ĥ.qx stop simulating it?
>>
> H_Hat <H_Hat> halts. So if it is simulated by a UTM, the simulation
> UTM <H_HAT><H_Hat> also halts.

It is very easily proven that ⟨Ĥ⟩ on ⟨Ĥ⟩ cannot possibly ever reach its
final state. If it stops running without reaching is final state then
this does not count as halting.

> But the halt decider embedded in H_Hat is not a UTM. It's a near UTM,
> that has abort logic. Somewhere the abort logic must be triggered,
> which causes H_Hat<H_Hat> to halt.

If it stops running without reaching is final state then this does not
count as halting. It can not possibly reach its final state whether or
not its simulation is ever aborted, therefore it never halts.

> H<H_Hat><H_Hat> also triggers this abort logic, so H<H_Hat><H_Hat>
> reports "false" (non-halting).
>>
>> If you disagree then to prove that you are not simply being disagreeable
>> you must proceed with a dialogue on this single point until we reach
>> mutual agreement. All other points will not be discussed until we reach
>> mutual agreement on this point.
>>
> We agree that H_Hat<H_Hat> halts.
> We agree that H <H_Hat> <H_Hat> reports "false" (non-halting).
>
> Can't you see the obvious?
>

The mistake that you are making is counting stopping without reaching a
final state as halting, it is not.

--
Copyright 2021 Pete Olcott

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

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

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Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] ( Are you game ? )[ Only: H(P,P)==0 ][ André refuses an honest_dialogue_]
Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc> <OqKdnROLKJ9CdJz8nZ2dnUU7-avNnZ2d@giganews.com> <87k0la542c.fsf@bsb.me.uk> <x5mdnTC66uNJip_8nZ2dnUU7-aWdnZ2d@giganews.com> <87mtq438ge.fsf@bsb.me.uk> <PbednTcmR4_Mw578nZ2dnUU78WvNnZ2d@giganews.com> <87tukc12yh.fsf@bsb.me.uk> <e6OdnW_rdvusk5n8nZ2dnUU7-dPNnZ2d@giganews.com> <875ywrmwsr.fsf@bsb.me.uk> <GtmdnfBrysPrAZn8nZ2dnUU7-L_NnZ2d@giganews.com> <874kcakv3d.fsf@bsb.me.uk> <-M-dnfsXl4WvWpj8nZ2dnUU7-IGdnZ2d@giganews.com> <87im0pa1pp.fsf@bsb.me.uk> <6eOdnaoMUdZN_pr8nZ2dnUU7-cPNnZ2d@giganews.com> <87y29j6c5e.fsf@bsb.me.uk> <p8SdnV8dJu3wq5X8nZ2dnUU7-a3NnZ2d@giganews.com> <875ywn5qfr.fsf@bsb.me.uk> <hJmdnSWv8-zPCJX8nZ2dnUU7-T_NnZ2d@giganews.com> <3df616af-1e70-413f-8534-fb4ca6204c28n@googlegroups.com> <_ZudnRCgbJV5oJT8nZ2dnUU7-VXNnZ2d@giganews.com> <f5026a04-cbe0-4809-82d9-27ecaf1fd1afn@googlegroups.com> <3IadnZ2CTqZnw5T8nZ2dnUU7-VudnZ2d@giganews.com> <sebtb9$plb$1@dont-email.me> <Peqdncl6_7bIO5f8nZ2dnUU7-TXNnZ2d@giganews.com>
From: NoO...@NoWhere.com (olcott)
Date: Thu, 5 Aug 2021 09:04:05 -0500
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 by: olcott - Thu, 5 Aug 2021 14:04 UTC

On 8/4/2021 9:38 AM, olcott wrote:
> On 8/3/2021 12:11 PM, André G. Isaak wrote:
>> On 2021-08-03 09:21, olcott wrote:
>>> On 8/3/2021 9:33 AM, Malcolm McLean wrote:
>>>> On Tuesday, 3 August 2021 at 14:00:28 UTC+1, olcott wrote:
>>>>> On 8/3/2021 12:29 AM, Malcolm McLean wrote:
>>>>>>
>>>>>> You're trying to claim that H_Hat<H_Hat> doens't really halt because
>>>>>> the recursion of H instances is terminated by H.
>>>>> No I have never been saying that. I am claiming that the input to
>>>>> H(P,P)
>>>>> never halts whether or not H terminates its simulation of this input.
>>>>>
>>>> We agree that P(P) halts.
>>>> So now you're drawing a distinction between P(P) and "the input to
>>>> H(P,P)".
>>>> ThIs is nonsense..
>>>
>>> Try and find any error in (a)(b)(c)(d) on page 6
>>>
>>> https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation
>>
>>
>>
>>
>> The error, which has been pointed out repeatedly, is in your (b).
>>
>> You claim "there are no control flow instructions in the execution
>> trace that would escape the infinite recursion", but there *are* flow
>> control instructions. But your trace is incomplete and skips over the
>> call to B82 where these flow control instructions reside.
>>
>
> I will rephrase that:
> There are no control flow instructions in P that break out of the
> infinite recursion. The control flow instructions in H that could abort
> the simulation of this infinite recursion would never allow P to reach
> its final state. Therefore we know that P never reaches its final state.
> Therefore because of the definition of halting that you provided we know
> that P never halts. Therefore we know that H(P,P)==0 is correct.
>
>> The code at B82 is *part* of the computation performed by P. It *must*
>> be included in any trace of P which purports to be a complete and
>> honest trace.
>>
>
> On the basis of the above analysis we know for sure that this is
> irrelevant. Nothing that happens in H can possibly cause P to reach its
> final state. Nothing that happens in P can possibly cause P to reach its
> final state. Therefore P never reaches its final state. Therefore P
> never halts. Therefore H(P,)==0 is correct.
>
>> You don't seem to grasp the fact that *all* code executed by a
>> particular computation from the moment the computation begins to the
>> time it halts (assuming it halts) is part of that computation. It
>> doesn't matter whether the code in question is shared by some other
>> routine, or is operating system code or whatever. These are purely
>> artificial distinctions which play no role in the theory of computation.
>>
>> André
>>
>
> If the input to H(P,P) cannot possibly halt then H(P,P)==0 is correct.
>

André refuses an honest dialogue

--
Copyright 2021 Pete Olcott

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

H(P,P)==0 is proven to be correct [ André refuses an honest dialogue ]

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Subject: H(P,P)==0 is proven to be correct [ André_refuses_an_honest_dialogue_]
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References: <20210719214640.00000dfc@reddwarf.jmc> <OqKdnROLKJ9CdJz8nZ2dnUU7-avNnZ2d@giganews.com> <87k0la542c.fsf@bsb.me.uk> <x5mdnTC66uNJip_8nZ2dnUU7-aWdnZ2d@giganews.com> <87mtq438ge.fsf@bsb.me.uk> <PbednTcmR4_Mw578nZ2dnUU78WvNnZ2d@giganews.com> <87tukc12yh.fsf@bsb.me.uk> <e6OdnW_rdvusk5n8nZ2dnUU7-dPNnZ2d@giganews.com> <875ywrmwsr.fsf@bsb.me.uk> <GtmdnfBrysPrAZn8nZ2dnUU7-L_NnZ2d@giganews.com> <874kcakv3d.fsf@bsb.me.uk> <-M-dnfsXl4WvWpj8nZ2dnUU7-IGdnZ2d@giganews.com> <87im0pa1pp.fsf@bsb.me.uk> <6eOdnaoMUdZN_pr8nZ2dnUU7-cPNnZ2d@giganews.com> <87y29j6c5e.fsf@bsb.me.uk> <p8SdnV8dJu3wq5X8nZ2dnUU7-a3NnZ2d@giganews.com> <875ywn5qfr.fsf@bsb.me.uk> <hJmdnSWv8-zPCJX8nZ2dnUU7-T_NnZ2d@giganews.com> <3df616af-1e70-413f-8534-fb4ca6204c28n@googlegroups.com> <_ZudnRCgbJV5oJT8nZ2dnUU7-VXNnZ2d@giganews.com> <f5026a04-cbe0-4809-82d9-27ecaf1fd1afn@googlegroups.com> <3IadnZ2CTqZnw5T8nZ2dnUU7-VudnZ2d@giganews.com> <sebtb9$plb$1@dont-email.me> <Peqdncl6_7bIO5f8nZ2dnUU7-TXNnZ2d@giganews.com>
From: NoO...@NoWhere.com (olcott)
Date: Thu, 5 Aug 2021 09:07:08 -0500
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 by: olcott - Thu, 5 Aug 2021 14:07 UTC

On 8/4/2021 9:38 AM, olcott wrote:
> On 8/3/2021 12:11 PM, André G. Isaak wrote:
>> On 2021-08-03 09:21, olcott wrote:
>>> On 8/3/2021 9:33 AM, Malcolm McLean wrote:
>>>> On Tuesday, 3 August 2021 at 14:00:28 UTC+1, olcott wrote:
>>>>> On 8/3/2021 12:29 AM, Malcolm McLean wrote:
>>>>>>
>>>>>> You're trying to claim that H_Hat<H_Hat> doens't really halt because
>>>>>> the recursion of H instances is terminated by H.
>>>>> No I have never been saying that. I am claiming that the input to
>>>>> H(P,P)
>>>>> never halts whether or not H terminates its simulation of this input.
>>>>>
>>>> We agree that P(P) halts.
>>>> So now you're drawing a distinction between P(P) and "the input to
>>>> H(P,P)".
>>>> ThIs is nonsense..
>>>
>>> Try and find any error in (a)(b)(c)(d) on page 6
>>>
>>> https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation
>>
>>
>>
>>
>> The error, which has been pointed out repeatedly, is in your (b).
>>
>> You claim "there are no control flow instructions in the execution
>> trace that would escape the infinite recursion", but there *are* flow
>> control instructions. But your trace is incomplete and skips over the
>> call to B82 where these flow control instructions reside.
>>
>
> I will rephrase that:
> There are no control flow instructions in P that break out of the
> infinite recursion. The control flow instructions in H that could abort
> the simulation of this infinite recursion would never allow P to reach
> its final state. Therefore we know that P never reaches its final state.
> Therefore because of the definition of halting that you provided we know
> that P never halts. Therefore we know that H(P,P)==0 is correct.
>
>> The code at B82 is *part* of the computation performed by P. It *must*
>> be included in any trace of P which purports to be a complete and
>> honest trace.
>>
>
> On the basis of the above analysis we know for sure that this is
> irrelevant. Nothing that happens in H can possibly cause P to reach its
> final state. Nothing that happens in P can possibly cause P to reach its
> final state. Therefore P never reaches its final state. Therefore P
> never halts. Therefore H(P,)==0 is correct.
>
>> You don't seem to grasp the fact that *all* code executed by a
>> particular computation from the moment the computation begins to the
>> time it halts (assuming it halts) is part of that computation. It
>> doesn't matter whether the code in question is shared by some other
>> routine, or is operating system code or whatever. These are purely
>> artificial distinctions which play no role in the theory of computation.
>>
>> André
>>
>
> If the input to H(P,P) cannot possibly halt then H(P,P)==0 is correct.
>

André refuses an honest dialogue

--
Copyright 2021 Pete Olcott

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

Re: Black box halt decider is NOT a partial decider

<Eq2dnf_RVfcnapb8nZ2dnUU78cXNnZ2d@brightview.co.uk>

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References: <20210719214640.00000dfc@reddwarf.jmc>
<sd4pbc$f1b$1@gioia.aioe.org> <sd76r9$r63$3@dont-email.me>
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<sd8bim$1set$1@gioia.aioe.org> <87eebrlv2m.fsf@bsb.me.uk>
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From: news.dea...@darjeeling.plus.com (Mike Terry)
Date: Thu, 5 Aug 2021 15:37:45 +0100
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 by: Mike Terry - Thu, 5 Aug 2021 14:37 UTC

On 05/08/2021 05:54, 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

The page says:

sum_(n=1)^∞ 1/n^24 = (236364091 π^24)/201919571963756521875≈1.0000

Here, the ≈1.0000 means "approximately equal to", so in your terms
"close to". The LHS here is clearly irrational! (It is known that π is
transcendental.)

No two real numbers are "arbitrarily close" to each other. In
particular, 0.999... = 1, and the phrase "1 is arbitrarily close to 1"
is pointless. (So nobody uses "arbitrarily close" in that sense.)

Regards,
Mike.

Re: Black box halt decider is NOT a partial decider

<87mtpv27sp.fsf@bsb.me.uk>

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider
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 by: Ben Bacarisse - Thu, 5 Aug 2021 16:08 UTC

"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:

> For instance, .999999999999999 is close to 1, however, its not
> arbitrarily close to 1 like .999... is

The reals have the property that if x =/= y there are reals between x
and y. If 0.999... is only "arbitrarily close to 1" what reals lie
between it an 1? 0.999... == 1.

--
Ben.

Re: Black box halt decider is NOT a partial decider

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From: jbb...@notatt.com (Jeff Barnett)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider
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 by: Jeff Barnett - Thu, 5 Aug 2021 17:41 UTC

On 8/5/2021 10:08 AM, Ben Bacarisse wrote:
> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>
>> For instance, .999999999999999 is close to 1, however, its not
>> arbitrarily close to 1 like .999... is
>
> The reals have the property that if x =/= y there are reals between x
> and y. If 0.999... is only "arbitrarily close to 1" what reals lie
> between it an 1? 0.999... == 1.

Recently, I've been trying to remember the name of that property. I
think I once saw something where a set R was (totally) ordered by < with
the "in between" property, one could say that "R was dense in <". Even
if that phrasing is correct, I'd like to find a better sounding and more
familiar term for the property. Do you recall such a name or have any
suggestions?
--
Jeff Barnett

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