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

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

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

<3dde3011-0e46-45b0-87aa-1b7889506d0bn@googlegroups.com>

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Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct an
d forms no contradiction.
From: malcolm....@gmail.com (Malcolm McLean)
Injection-Date: Wed, 11 Aug 2021 11:09:48 +0000
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 by: Malcolm McLean - Wed, 11 Aug 2021 11:09 UTC

On Wednesday, 11 August 2021 at 11:48:14 UTC+1, Ben Bacarisse wrote:
> Malcolm McLean <malcolm.ar...@gmail.com> writes:
>
> > On Wednesday, 11 August 2021 at 02:41:04 UTC+1, Ben Bacarisse wrote:
> >>
> >> It is trivial to write a TM J from which a Ĵ can be derived that has the
> >> properties of your Ĥ. Every student learning this material for the first
> >> time should be able to come up with one a few minutes. There is nothing
> >> interesting about a TM with
> >>
> >> Ĵ.q0 ⟨Ĵ⟩ ⊢* Ĵ.qn
> >> if Ĵ applied to ⟨Ĵ⟩ halts.
> >>
> > It's trivial to write a J with these characteristics.
> > But that doesn't mean that every such J will be trivial or
> > uninteresting. It just won't be a counter-example to an established
> > theorem.
> That's an odd thing to point out, but of course it's true. Deciders for
> all sort of sets might have this property: shortest paths through a
> network, numbered digits of pi, integer factorisations... Some will and
> some won't, but a very large proportion of interesting and non-trivial
> TMs will have this property.
>
> That's probably not what you meant though, but I don't know what you are
> hinting at.
>
No, that's exactly what I meant.

You can't build a halt decider that decides every instance correctly. But you
can build an interesting TM that has lesser ambitions. Which is what some
people should maybe be focusing on.

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction. [ Is Ben a Liar or simply woefully ignorant? ]

<346dnYhkWPUNQ478nZ2dnUU7-UPNnZ2d@giganews.com>

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Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct_and_forms_no_contradiction._[_Is_Ben_a_Liar_or_simply
_woefully_ignorant?_]
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
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<zcadnTSOD5rtZ5f8nZ2dnUU7-T3NnZ2d@giganews.com> <877dh03l3c.fsf@bsb.me.uk>
<Z5adnd038KGXwJb8nZ2dnUU7-I_NnZ2d@giganews.com> <8735rn1qvj.fsf@bsb.me.uk>
<goydnfCCIYUWE5H8nZ2dnUU7-e_NnZ2d@giganews.com> <87eeb7z4d1.fsf@bsb.me.uk>
<0_Sdnb6Qe8XGOZH8nZ2dnUU7-U3NnZ2d@giganews.com> <87zgtslqpv.fsf@bsb.me.uk>
<4JOdnRS2SLR7MYz8nZ2dnUU7-YnNnZ2d@giganews.com> <87zgtoizgp.fsf@bsb.me.uk>
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From: NoO...@NoWhere.com (olcott)
Date: Wed, 11 Aug 2021 09:28:31 -0500
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 by: olcott - Wed, 11 Aug 2021 14:28 UTC

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

When the challenge to support one's assertion with reasoning is simply
ignored as you are ignoring it right now one can reasonably construe a
deceptive intent.

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

PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H

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

H.q0 WM w ⊢* H.qn
if M applied to W does not halt.

becomes

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

Pages of the Linz text to verify the above quotes in their full context:
http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf

M STILL REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO Ĥ.qx
M STILL REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO Ĥ.qx
M STILL REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO Ĥ.qx
M STILL REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO Ĥ.qx

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

When we know that M refers to the Turing machine specified by the first
wM then when Ĥ transitions to its final state of Ĥ.qn there is no direct
contradiction formed.

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

if M applied to wM does not halt (see above for definition of M)
means when the Turing machine of ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Ĥ.qx correctly transitions to its final state when the Ĥ.qx acts as a
UTM and simulates ⟨Ĥ⟩ ⟨Ĥ⟩ and determines that this input never halts.

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

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

--
Copyright 2021 Pete Olcott

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

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction. [ Is Ben a Liar or simply woefully ignorant? ]

<Ja-dneAl3poPeI78nZ2dnUU7-UvNnZ2d@giganews.com>

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Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct_and_forms_no_contradiction._[_Is_Ben_a_Liar_or_simply
_woefully_ignorant?_]
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<zcadnTSOD5rtZ5f8nZ2dnUU7-T3NnZ2d@giganews.com> <877dh03l3c.fsf@bsb.me.uk>
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From: NoO...@NoWhere.com (olcott)
Date: Wed, 11 Aug 2021 09:58:25 -0500
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 by: olcott - Wed, 11 Aug 2021 14:58 UTC

On 8/11/2021 9:28 AM, olcott wrote:
> On 8/10/2021 9:26 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/10/2021 8:41 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/10/2021 7:42 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 8/7/2021 7:34 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>
>>>>>>>>>>> On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>>
>>>>>>>>>>>>> The question is not: Does Ĥ halt on its input?
>>>>>>>>>>>> Yes it is.
>>>>>>>>>>>
>>>>>>>>>>> The question is:
>>>>>>>>>>> Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
>>>>>>>>>>> The ansswer to this question is provably no!
>>>>>>>>>> The question is: does Ĥ applied to ⟨Ĥ⟩ halt.  It does:
>>>>>>>>>>
>>>>>>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn  THIS IS NOT A CONTRADICTION
>>>>>>>>>> Indeed.  There is no contradiction.  Just an Ĥ that does not
>>>>>>>>>> meet Linz
>>>>>>>>>> spec.
>>>>>>>>>
>>>>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.
>>>>>>>>> Because it is correct it meets the Linz spec.
>>>>>>>> I find it startling that you think that, but then it seems you
>>>>>>>> don't yet
>>>>>>>> know what the key words mean:
>>>>>>>>
>>>>>>>>> if M applied to wM does not halt
>>>>>>>>> means if the execution of the machine of the first ⟨Ĥ⟩ on its
>>>>>>>>> input of
>>>>>>>>> the seocond ⟨Ĥ⟩ does not halt then ⊢* Ĥ.qn
>>>>>>>> No.  Would you like to know "what M applied to wM does not halt"
>>>>>>>> means?
>>>>>>>> Do you need help to see that "Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn" is clearly a
>>>>>>>> case of "M
>>>>>>>> applied to wM halts"?
>>>>>>>
>>>>>>>         the Turing machine halting problem. Simply stated, the
>>>>>>> problem
>>>>>>>         is: given the description of a Turing machine M and an
>>>>>>> input w,
>>>>>>>         does M, when started in the initial configuration q0w,
>>>>>>> perform a
>>>>>>>         computation that eventually halts? (Linz:1990:317).
>>>>>> Yes.  I was offering to help you understand the key words in that
>>>>>> text.
>>>>>>
>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>>> You've missed off the key lines yet again.  Is that deliberate?  They
>>>>>> are the lines that show you are wrong so I am suspicious that you
>>>>>> keep
>>>>>> omitting them.
>>>>>>
>>>>>>> When Ĥ is applied to ⟨Ĥ⟩ the description of the Turing Machine
>>>>>>> and its
>>>>>>> input are specified as: ⟨Ĥ⟩ ⟨Ĥ⟩ for the embedded halt decider at
>>>>>>> Ĥ.qx.
>>>>>> Ungrammatical.
>>>>>>
>>>>>>> When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn this is not a final state of the simulated
>>>>>>> input it is a final state of the executed Ĥ.
>>>>>> Yes.  You don't seem to know why that's wrong.
>>>>>
>>>>> What is your basis for believing that is wrong?
>>>> Ah, a question about what I'm saying.  I can help there.  The basis is
>>>> what Linz says about Ĥ.  He says that (translating to your notation)
>>>>     Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>> should be the case "if Ĥ applied to ⟨Ĥ⟩ does not halt".  But, as you
>>>> can
>>>> see, your Ĥ does halt when applied to ⟨Ĥ⟩ (qn is a halting or final
>>>> state).  Your Ĥ is not doing what it should in this one crucial case.
>>>>
>>>
>>>     the Turing machine halting problem. Simply stated, the problem
>>>     is: given the description of a Turing machine M and an input w,
>>>     does M, when started in the initial configuration q0w, perform a
>>>     computation that eventually halts? (Linz:1990:317).
>>
>> and so on.  Same old stuff.
>>
>
> When the challenge to support one's assertion with reasoning is simply
> ignored as you are ignoring it right now one can reasonably construe a
> deceptive intent.
>
> -- the Turing machine halting problem. Simply stated, the problem
> -- is: given the description of a Turing machine M and an input w,
> -- does M, when started in the initial configuration q0w, perform a
> -- computation that eventually halts? (Linz:1990:317).
>
> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>
> The input to H will be the description (encoded in some form) of M, say
> WM, as well as the input w. (Linz:1990:318)
>
> H.q0 WM w ⊢* H.qn
> if M applied to W does not halt.
>
>   becomes
>
> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
> if Ĥ applied to ⟨Ĥ⟩ does not halt.
>
>
> Pages of the Linz text to verify the above quotes in their full context:
> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>
> M STILL REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO Ĥ.qx
> M STILL REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO Ĥ.qx
> M STILL REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO Ĥ.qx
> M STILL REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO Ĥ.qx
>
> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
> if M applied to wM does not halt
>
> When we know that M refers to the Turing machine specified by the first
> wM then when Ĥ transitions to its final state of Ĥ.qn there is no direct
> contradiction formed.
>
> Can you admit when you are wrong when you really are wrong?
> Can you admit when you are wrong when you really are wrong?
> Can you admit when you are wrong when you really are wrong?
> Can you admit when you are wrong when you really are wrong?
>
> if M applied to wM does not halt (see above for definition of M)
> means when the Turing machine of ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt.
>
> Ĥ.q0 ⟨Ĥ⟩  ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> Ĥ.qx correctly transitions to its final state when the Ĥ.qx acts as a
> UTM and simulates ⟨Ĥ⟩ ⟨Ĥ⟩ and determines that this input never halts.
>
> https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation
>

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
if M applied to wM halts, and // M refers to the TM of the first wM
parameter to Ĥ.qx

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
if M applied to wM does not halt // M refers to the TM of the first wM
parameter to Ĥ.qx

--
Copyright 2021 Pete Olcott

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

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

<87o8a4ggzf.fsf@bsb.me.uk>

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct and forms no contradiction.
Followup-To: comp.theory
Date: Wed, 11 Aug 2021 16:04:36 +0100
Organization: A noiseless patient Spider
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 by: Ben Bacarisse - Wed, 11 Aug 2021 15:04 UTC

olcott <NoOne@NoWhere.com> writes:

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

You fail at the first hurdle. You can't hope to persuade anyone that an
add function with add(2, 3) == 9 is operating as specified simply by
detailing, step by step, exactly how you implement the wrong behaviour.
In your case, it's simply that Ĥ.q0 ⟨Ĥ⟩ transitions to Ĥ.qn (via Ĥ.qx
⟨Ĥ⟩ ⟨Ĥ⟩ as you pointlessly keep insisting) when Linz says it should not.

Your add function is entirely correct in that is does exactly what you
intend it to do. As far as I am concerned there is no significant error
in how you arrive at add(2, 3) == 9. The problem is that Linz says your
code should add numbers and not do whatever it is your code does with
100% correctness. Do you follow?

> Of course
> everyone knows that this is impossible if I am totally correct.

Then stop wasting time and try to publish! You'll need lots of time to
explain away why every editor simply laughs at the paper.

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

Yes, we all know that. It's exactly why your Ĥ does not meet Linz's
specification (for this case -- you don't claim to have a halt decider).

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

Yes. I am wrong all the time. In this case I'm having trouble working
out how I could be clearer about your Ĥ. Maybe if you didn't keep
removing the key text from Linz's explanations it might sink in?

--
Ben.

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

<XfGdnc2wcvv6dY78nZ2dnUU7-fednZ2d@giganews.com>

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Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc> <87mtpz64sq.fsf@bsb.me.uk>
<w5edne8d06OkxpX8nZ2dnUU7-b_NnZ2d@giganews.com> <87bl6f5qvy.fsf@bsb.me.uk>
<1ZednUt-qbrXlZf8nZ2dnUU7-QnNnZ2d@giganews.com> <87tuk52h0e.fsf@bsb.me.uk>
<zcadnTSOD5rtZ5f8nZ2dnUU7-T3NnZ2d@giganews.com> <877dh03l3c.fsf@bsb.me.uk>
<Z5adnd038KGXwJb8nZ2dnUU7-I_NnZ2d@giganews.com> <8735rn1qvj.fsf@bsb.me.uk>
<goydnfCCIYUWE5H8nZ2dnUU7-e_NnZ2d@giganews.com> <87eeb7z4d1.fsf@bsb.me.uk>
<0_Sdnb6Qe8XGOZH8nZ2dnUU7-U3NnZ2d@giganews.com> <87zgtslqpv.fsf@bsb.me.uk>
<4JOdnRS2SLR7MYz8nZ2dnUU7-YnNnZ2d@giganews.com> <87zgtoizgp.fsf@bsb.me.uk>
<Z6ednWt7SpIIv478nZ2dnUU7-WHNnZ2d@giganews.com> <87pmukiwr5.fsf@bsb.me.uk>
<5c6dnbn2gMMIsI78nZ2dnUU7-dnNnZ2d@giganews.com> <87eeb0iuo1.fsf@bsb.me.uk>
<2-adncuMeNUVpY78nZ2dnUU7-fPNnZ2d@giganews.com> <87o8a4ggzf.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 11 Aug 2021 10:10:30 -0500
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 by: olcott - Wed, 11 Aug 2021 15:10 UTC

On 8/11/2021 10:04 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/10/2021 9:26 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/10/2021 8:41 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/10/2021 7:42 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 8/7/2021 7:34 PM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>
>>>>>>>>>>>> On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
>>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>
>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> The question is not: Does Ĥ halt on its input?
>>>>>>>>>>>>> Yes it is.
>>>>>>>>>>>>
>>>>>>>>>>>> The question is:
>>>>>>>>>>>> Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
>>>>>>>>>>>> The ansswer to this question is provably no!
>>>>>>>>>>> The question is: does Ĥ applied to ⟨Ĥ⟩ halt. It does:
>>>>>>>>>>>
>>>>>>>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn THIS IS NOT A CONTRADICTION
>>>>>>>>>>> Indeed. There is no contradiction. Just an Ĥ that does not meet Linz
>>>>>>>>>>> spec.
>>>>>>>>>>
>>>>>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.
>>>>>>>>>> Because it is correct it meets the Linz spec.
>>>>>>>>> I find it startling that you think that, but then it seems you don't yet
>>>>>>>>> know what the key words mean:
>>>>>>>>>
>>>>>>>>>> if M applied to wM does not halt
>>>>>>>>>> means if the execution of the machine of the first ⟨Ĥ⟩ on its input of
>>>>>>>>>> the seocond ⟨Ĥ⟩ does not halt then ⊢* Ĥ.qn
>>>>>>>>> No. Would you like to know "what M applied to wM does not halt" means?
>>>>>>>>> Do you need help to see that "Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn" is clearly a case of "M
>>>>>>>>> applied to wM halts"?
>>>>>>>>
>>>>>>>> the Turing machine halting problem. Simply stated, the problem
>>>>>>>> is: given the description of a Turing machine M and an input w,
>>>>>>>> does M, when started in the initial configuration q0w, perform a
>>>>>>>> computation that eventually halts? (Linz:1990:317).
>>>>>>> Yes. I was offering to help you understand the key words in that text.
>>>>>>>
>>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>>>> You've missed off the key lines yet again. Is that deliberate? They
>>>>>>> are the lines that show you are wrong so I am suspicious that you keep
>>>>>>> omitting them.
>>>>>>>
>>>>>>>> When Ĥ is applied to ⟨Ĥ⟩ the description of the Turing Machine and its
>>>>>>>> input are specified as: ⟨Ĥ⟩ ⟨Ĥ⟩ for the embedded halt decider at Ĥ.qx.
>>>>>>> Ungrammatical.
>>>>>>>
>>>>>>>> When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn this is not a final state of the simulated
>>>>>>>> input it is a final state of the executed Ĥ.
>>>>>>> Yes. You don't seem to know why that's wrong.
>>>>>>
>>>>>> What is your basis for believing that is wrong?
>>>>> Ah, a question about what I'm saying. I can help there. The basis is
>>>>> what Linz says about Ĥ. He says that (translating to your notation)
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>> should be the case "if Ĥ applied to ⟨Ĥ⟩ does not halt". But, as you can
>>>>> see, your Ĥ does halt when applied to ⟨Ĥ⟩ (qn is a halting or final
>>>>> state). Your Ĥ is not doing what it should in this one crucial case.
>>>>>
>>>>
>>>> the Turing machine halting problem. Simply stated, the problem
>>>> is: given the description of a Turing machine M and an input w,
>>>> does M, when started in the initial configuration q0w, perform a
>>>> computation that eventually halts? (Linz:1990:317).
>>> and so on. Same old stuff.
>>> I'm sorry my explanation did not help at all. I'm happy to answer any
>>> other questions you might have if you think it might help you understand
>>> what I (and Linz) are saying.
>>
>> I went point by point. If I am actually incorrect then you can go
>> point by point and point out each individual error step by step.
>
> You fail at the first hurdle. You can't hope to persuade anyone that an
> add function with add(2, 3) == 9 is operating as specified simply by
> detailing, step by step, exactly how you implement the wrong behaviour.
> In your case, it's simply that Ĥ.q0 ⟨Ĥ⟩ transitions to Ĥ.qn (via Ĥ.qx
> ⟨Ĥ⟩ ⟨Ĥ⟩ as you pointlessly keep insisting) when Linz says it should not.
>
> Your add function is entirely correct in that is does exactly what you
> intend it to do. As far as I am concerned there is no significant error
> in how you arrive at add(2, 3) == 9. The problem is that Linz says your
> code should add numbers and not do whatever it is your code does with
> 100% correctness. Do you follow?
>
>> Of course
>> everyone knows that this is impossible if I am totally correct.
>
> Then stop wasting time and try to publish! You'll need lots of time to
> explain away why every editor simply laughs at the paper.
>
>> H.q0 WM w ⊢* Ĥ.qn
>> becomes
>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>
> Yes, we all know that. It's exactly why your Ĥ does not meet Linz's
> specification (for this case -- you don't claim to have a halt decider).
>
>> Can you admit when you are wrong when you really are wrong?
>
> Yes. I am wrong all the time. In this case I'm having trouble working
> out how I could be clearer about your Ĥ. Maybe if you didn't keep
> removing the key text from Linz's explanations it might sink in?
>

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
if M applied to wM halts, and // M refers to the TM of the
first wM parameter to Ĥ.qx

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
if M applied to wM does not halt // M refers to the TM of the first
wM parameter to Ĥ.qx

--
Copyright 2021 Pete Olcott

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

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

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Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct_and_forms_no_contradiction.
Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc> <87mtpz64sq.fsf@bsb.me.uk>
<w5edne8d06OkxpX8nZ2dnUU7-b_NnZ2d@giganews.com> <87bl6f5qvy.fsf@bsb.me.uk>
<1ZednUt-qbrXlZf8nZ2dnUU7-QnNnZ2d@giganews.com> <87tuk52h0e.fsf@bsb.me.uk>
<zcadnTSOD5rtZ5f8nZ2dnUU7-T3NnZ2d@giganews.com> <877dh03l3c.fsf@bsb.me.uk>
<Z5adnd038KGXwJb8nZ2dnUU7-I_NnZ2d@giganews.com> <8735rn1qvj.fsf@bsb.me.uk>
<goydnfCCIYUWE5H8nZ2dnUU7-e_NnZ2d@giganews.com> <87eeb7z4d1.fsf@bsb.me.uk>
<0_Sdnb6Qe8XGOZH8nZ2dnUU7-U3NnZ2d@giganews.com> <87zgtslqpv.fsf@bsb.me.uk>
<4JOdnRS2SLR7MYz8nZ2dnUU7-YnNnZ2d@giganews.com> <87zgtoizgp.fsf@bsb.me.uk>
<Z6ednWt7SpIIv478nZ2dnUU7-WHNnZ2d@giganews.com> <87pmukiwr5.fsf@bsb.me.uk>
<5c6dnbn2gMMIsI78nZ2dnUU7-dnNnZ2d@giganews.com> <87eeb0iuo1.fsf@bsb.me.uk>
<2-adncuMeNUVpY78nZ2dnUU7-fPNnZ2d@giganews.com> <87o8a4ggzf.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 11 Aug 2021 10:15:06 -0500
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 by: olcott - Wed, 11 Aug 2021 15:15 UTC

On 8/11/2021 10:04 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/10/2021 9:26 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/10/2021 8:41 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/10/2021 7:42 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 8/7/2021 7:34 PM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>
>>>>>>>>>>>> On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
>>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>
>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> The question is not: Does Ĥ halt on its input?
>>>>>>>>>>>>> Yes it is.
>>>>>>>>>>>>
>>>>>>>>>>>> The question is:
>>>>>>>>>>>> Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
>>>>>>>>>>>> The ansswer to this question is provably no!
>>>>>>>>>>> The question is: does Ĥ applied to ⟨Ĥ⟩ halt. It does:
>>>>>>>>>>>
>>>>>>>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn THIS IS NOT A CONTRADICTION
>>>>>>>>>>> Indeed. There is no contradiction. Just an Ĥ that does not meet Linz
>>>>>>>>>>> spec.
>>>>>>>>>>
>>>>>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.
>>>>>>>>>> Because it is correct it meets the Linz spec.
>>>>>>>>> I find it startling that you think that, but then it seems you don't yet
>>>>>>>>> know what the key words mean:
>>>>>>>>>
>>>>>>>>>> if M applied to wM does not halt
>>>>>>>>>> means if the execution of the machine of the first ⟨Ĥ⟩ on its input of
>>>>>>>>>> the seocond ⟨Ĥ⟩ does not halt then ⊢* Ĥ.qn
>>>>>>>>> No. Would you like to know "what M applied to wM does not halt" means?
>>>>>>>>> Do you need help to see that "Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn" is clearly a case of "M
>>>>>>>>> applied to wM halts"?
>>>>>>>>
>>>>>>>> the Turing machine halting problem. Simply stated, the problem
>>>>>>>> is: given the description of a Turing machine M and an input w,
>>>>>>>> does M, when started in the initial configuration q0w, perform a
>>>>>>>> computation that eventually halts? (Linz:1990:317).
>>>>>>> Yes. I was offering to help you understand the key words in that text.
>>>>>>>
>>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>>>> You've missed off the key lines yet again. Is that deliberate? They
>>>>>>> are the lines that show you are wrong so I am suspicious that you keep
>>>>>>> omitting them.
>>>>>>>
>>>>>>>> When Ĥ is applied to ⟨Ĥ⟩ the description of the Turing Machine and its
>>>>>>>> input are specified as: ⟨Ĥ⟩ ⟨Ĥ⟩ for the embedded halt decider at Ĥ.qx.
>>>>>>> Ungrammatical.
>>>>>>>
>>>>>>>> When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn this is not a final state of the simulated
>>>>>>>> input it is a final state of the executed Ĥ.
>>>>>>> Yes. You don't seem to know why that's wrong.
>>>>>>
>>>>>> What is your basis for believing that is wrong?
>>>>> Ah, a question about what I'm saying. I can help there. The basis is
>>>>> what Linz says about Ĥ. He says that (translating to your notation)
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>> should be the case "if Ĥ applied to ⟨Ĥ⟩ does not halt". But, as you can
>>>>> see, your Ĥ does halt when applied to ⟨Ĥ⟩ (qn is a halting or final
>>>>> state). Your Ĥ is not doing what it should in this one crucial case.
>>>>>
>>>>
>>>> the Turing machine halting problem. Simply stated, the problem
>>>> is: given the description of a Turing machine M and an input w,
>>>> does M, when started in the initial configuration q0w, perform a
>>>> computation that eventually halts? (Linz:1990:317).
>>> and so on. Same old stuff.
>>> I'm sorry my explanation did not help at all. I'm happy to answer any
>>> other questions you might have if you think it might help you understand
>>> what I (and Linz) are saying.
>>
>> I went point by point. If I am actually incorrect then you can go
>> point by point and point out each individual error step by step.
>
> You fail at the first hurdle. You can't hope to persuade anyone that an
> add function with add(2, 3) == 9 is operating as specified simply by
> detailing, step by step, exactly how you implement the wrong behaviour.
> In your case, it's simply that Ĥ.q0 ⟨Ĥ⟩ transitions to Ĥ.qn (via Ĥ.qx
> ⟨Ĥ⟩ ⟨Ĥ⟩ as you pointlessly keep insisting) when Linz says it should not.
>
> Your add function is entirely correct in that is does exactly what you
> intend it to do. As far as I am concerned there is no significant error
> in how you arrive at add(2, 3) == 9. The problem is that Linz says your
> code should add numbers and not do whatever it is your code does with
> 100% correctness. Do you follow?
>
>> Of course
>> everyone knows that this is impossible if I am totally correct.
>
> Then stop wasting time and try to publish! You'll need lots of time to
> explain away why every editor simply laughs at the paper.
>
>> H.q0 WM w ⊢* Ĥ.qn
>> becomes
>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>
> Yes, we all know that. It's exactly why your Ĥ does not meet Linz's
> specification (for this case -- you don't claim to have a halt decider).
>
>> Can you admit when you are wrong when you really are wrong?
>
> Yes. I am wrong all the time. In this case I'm having trouble working
> out how I could be clearer about your Ĥ. Maybe if you didn't keep
> removing the key text from Linz's explanations it might sink in?
>

Pages of the Linz text to verify the above quotes in their full context:
http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf

PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H

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

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
if M applied to wM halts, and // M refers to the TM of the first wM
parameter to Ĥ.qx

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
if M applied to wM does not halt // M refers to the TM of the first wM
parameter to Ĥ.qx

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

--
Copyright 2021 Pete Olcott

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

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction. [ Is Ben a Liar or simply woefully ignorant? ]

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

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct and forms no contradiction. [ Is Ben a Liar or simply woefully
ignorant? ]
Followup-To: comp.theory
Date: Wed, 11 Aug 2021 17:10:16 +0100
Organization: A noiseless patient Spider
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 by: Ben Bacarisse - Wed, 11 Aug 2021 16:10 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/11/2021 9:28 AM, olcott wrote:

Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no
contradiction. [ Is Ben a Liar or simply woefully ignorant? ]

This is a scumbag move. It's also cowardly and disingenuous --
pretending to be coy about your scumbag opinions. I'll leave it
unedited as it says everything readers need to know about your
character.

>> When the challenge to support one's assertion with reasoning is simply
>> ignored as you are ignoring it right now one can reasonably construe
>> a deceptive intent.

A scumbag could construe it that way. Reasonable people could come to
all sorts of other conclusions.

>> H.q0 WM w ⊢* H.qn
>> if M applied to W does not halt.
if M applied to MW does not halt.

(typo corrected)

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

Yes. After a lot a pressing (and I mean lots, over several years!) you
eventually admitted that H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ does indeed transition to H.qn.
You also keep telling us that Ĥ applied to ⟨Ĥ⟩ halts. That's why your H
(and its associated Ĥ) are wrong but for some reason you can't see this.

You will plainly state that H rejects the string "⟨Ĥ⟩ ⟨Ĥ⟩" which it
should do if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt. And you
will, time and time again, show us exactly how Ĥ applied to ⟨Ĥ⟩ halts.

This should be the end of the matter, but apparently your stating facts
that show that H and Ĥ are wrong does not mean you know that H and Ĥ are
wrong. I don't think I know any way to make progress on this.

>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>> if M applied to wM does not halt
....
>> if M applied to wM does not halt (see above for definition of
>> M) means when the Turing machine of ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt.

Of course. We all know that. Unless you are pulling a fast one. "the
Turing machine of ⟨Ĥ⟩" is just Ĥ. Is there a reason you are not simply
saying "when Ĥ applied to ⟨Ĥ⟩ does not halt"?

>> Ĥ.q0 ⟨Ĥ⟩  ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>> Ĥ.qx correctly transitions to its final state when the Ĥ.qx acts as a
>> UTM and simulates ⟨Ĥ⟩ ⟨Ĥ⟩ and determines that this input never halts.

Yes, we know the ruse: the computation would not halt if it were not the
computation that it is. You've been trying to pull off this trick ever
since the infamous "it wouldn't halt if line 15 was commented out"
admission. Your Ĥ, however, not being a UTM, has the property that

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

when it should not. Ĥ.q0 ⟨Ĥ⟩ should eventually transition to qn only
"if Ĥ applied to ⟨Ĥ⟩ does not halt" (or, as you rather suspiciously
write "if the Turing machine of ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt"). But
you've told us, time and time again, that Ĥ applied to ⟨Ĥ⟩ halts. You
keep showing us the summary description of it's configuration sequence.
You keep showing us the final state it transitions to. It's so obvious,
someone would need about 16 years misunderstanding Turing machines to
avoid seeing it.

> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
> if M applied to wM halts, and // M refers to the TM of the first wM
> parameter to Ĥ.qx
>
> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
> if M applied to wM does not halt // M refers to the TM of the first wM
> parameter to Ĥ.qx

The case you care about has M = Ĥ and wM = ⟨Ĥ⟩ as you've written it out
above.

--
Ben.

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction. [ Is Ben a Liar or simply woefully ignorant? ]

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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Wed, 11 Aug 2021 16:52 UTC

On 8/11/2021 11:10 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/11/2021 9:28 AM, olcott wrote:
>
> Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no
> contradiction. [ Is Ben a Liar or simply woefully ignorant? ]
>
> This is a scumbag move. It's also cowardly and disingenuous --
> pretending to be coy about your scumbag opinions. I'll leave it
> unedited as it says everything readers need to know about your
> character.
>
>>> When the challenge to support one's assertion with reasoning is simply
>>> ignored as you are ignoring it right now one can reasonably construe
>>> a deceptive intent.
>
> A scumbag could construe it that way. Reasonable people could come to
> all sorts of other conclusions.
>
>>> H.q0 WM w ⊢* H.qn
>>> if M applied to W does not halt.
> if M applied to MW does not halt.
>
> (typo corrected)
>
>>>   becomes
>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>> if Ĥ applied to ⟨Ĥ⟩ does not halt.
>
> Yes. After a lot a pressing (and I mean lots, over several years!) you
> eventually admitted that H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ does indeed transition to H.qn.

You are saying that in a misleading way.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩⟨Ĥ⟩ ⊢* Ĥ.qn
The Turing machine of Ĥ halts on input ⟨Ĥ⟩ because the simulating halt
decider at Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decides that its simulation of ⟨Ĥ⟩ on
input ⟨Ĥ⟩ never halts.

> You also keep telling us that Ĥ applied to ⟨Ĥ⟩ halts. That's why your H
> (and its associated Ĥ) are wrong but for some reason you can't see this.
>
> You will plainly state that H rejects the string "⟨Ĥ⟩ ⟨Ĥ⟩" which it
> should do if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt. And you
> will, time and time again, show us exactly how Ĥ applied to ⟨Ĥ⟩ halts.
>

The Ĥ of the first parameter to Ĥ.qx ⟨Ĥ⟩⟨Ĥ⟩ never halts
thus the Ĥ of the Turing machine Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩⟨Ĥ⟩ ⊢* Ĥ.qn
correctly transitions to its final state of Ĥ.qn

The full Linz proof is included at the end of the paper with the portion
of the Linz text that proves M refers to the first parameter WM to Ĥ.qx
⟨Ĥ⟩⟨Ĥ⟩ highlighted in YELLOW.

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 // see highlighted portion of Linz text
to confirm:
if M applied to wM does not halt // M refers to the TM of the first
wM parameter to Ĥ.qx

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

> This should be the end of the matter, but apparently your stating facts
> that show that H and Ĥ are wrong does not mean you know that H and Ĥ are
> wrong. I don't think I know any way to make progress on this.
>
>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>> if M applied to wM does not halt
> ...
>>> if M applied to wM does not halt (see above for definition of
>>> M) means when the Turing machine of ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt.
>
> Of course. We all know that. Unless you are pulling a fast one. "the
> Turing machine of ⟨Ĥ⟩" is just Ĥ. Is there a reason you are not simply
> saying "when Ĥ applied to ⟨Ĥ⟩ does not halt"?
>

The simulation of the parameter to Ĥ on input ⟨Ĥ⟩ does not halt on the
copy of this parameter.

I can prove what I am saying with Linz text, you cannot prove what you
are saying at all.

PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H

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

H.q0 WM w ⊢* H.qn
if M applied to W does not halt.

--
Copyright 2021 Pete Olcott

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

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction. [ Is Ben a Liar or simply woefully ignorant? ]

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From: NoO...@NoWhere.com (olcott)
Date: Wed, 11 Aug 2021 14:53:11 -0500
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 by: olcott - Wed, 11 Aug 2021 19:53 UTC

On 8/11/2021 11:10 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/11/2021 9:28 AM, olcott wrote:
>
> Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no
> contradiction. [ Is Ben a Liar or simply woefully ignorant? ]
>
> This is a scumbag move. It's also cowardly and disingenuous --
> pretending to be coy about your scumbag opinions. I'll leave it
> unedited as it says everything readers need to know about your
> character.
>
>>> When the challenge to support one's assertion with reasoning is simply
>>> ignored as you are ignoring it right now one can reasonably construe
>>> a deceptive intent.
>
> A scumbag could construe it that way. Reasonable people could come to
> all sorts of other conclusions.
>
>>> H.q0 WM w ⊢* H.qn
>>> if M applied to W does not halt.
> if M applied to MW does not halt.
>
> (typo corrected)
>
>>>   becomes
>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>> if Ĥ applied to ⟨Ĥ⟩ does not halt.
>
> Yes. After a lot a pressing (and I mean lots, over several years!) you
> eventually admitted that H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ does indeed transition to H.qn.
> You also keep telling us that Ĥ applied to ⟨Ĥ⟩ halts. That's why your H
> (and its associated Ĥ) are wrong but for some reason you can't see this.
>
> You will plainly state that H rejects the string "⟨Ĥ⟩ ⟨Ĥ⟩" which it
> should do if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt. And you
> will, time and time again, show us exactly how Ĥ applied to ⟨Ĥ⟩ halts.
>
> This should be the end of the matter, but apparently your stating facts
> that show that H and Ĥ are wrong does not mean you know that H and Ĥ are
> wrong. I don't think I know any way to make progress on this.
>
>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>> if M applied to wM does not halt
> ...
>>> if M applied to wM does not halt (see above for definition of
>>> M) means when the Turing machine of ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt.
>
> Of course. We all know that. Unless you are pulling a fast one. "the
> Turing machine of ⟨Ĥ⟩" is just Ĥ. Is there a reason you are not simply
> saying "when Ĥ applied to ⟨Ĥ⟩ does not halt"?
>
>>> Ĥ.q0 ⟨Ĥ⟩  ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>> Ĥ.qx correctly transitions to its final state when the Ĥ.qx acts as a
>>> UTM and simulates ⟨Ĥ⟩ ⟨Ĥ⟩ and determines that this input never halts.
>
> Yes, we know the ruse: the computation would not halt if it were not the
> computation that it is. You've been trying to pull off this trick ever
> since the infamous "it wouldn't halt if line 15 was commented out"
> admission. Your Ĥ, however, not being a UTM, has the property that
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
> when it should not. Ĥ.q0 ⟨Ĥ⟩ should eventually transition to qn only
> "if Ĥ applied to ⟨Ĥ⟩ does not halt" (or, as you rather suspiciously
> write "if the Turing machine of ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt"). But
> you've told us, time and time again, that Ĥ applied to ⟨Ĥ⟩ halts. You
> keep showing us the summary description of it's configuration sequence.
> You keep showing us the final state it transitions to. It's so obvious,
> someone would need about 16 years misunderstanding Turing machines to
> avoid seeing it.
>
>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>> if M applied to wM halts, and // M refers to the TM of the first wM
>> parameter to Ĥ.qx
>>
>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>> if M applied to wM does not halt // M refers to the TM of the first wM
>> parameter to Ĥ.qx
>
> The case you care about has M = Ĥ and wM = ⟨Ĥ⟩ as you've written it out
> above.
>

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 // see highlighted portion of Linz text
to confirm:
if M applied to wM does not halt // M refers to the TM of the first
wM parameter to Ĥ.qx

Turing machine Ĥ is applied to its input ⟨Ĥ⟩. It copies this input such
that this input and the copy of this input become the first and second
parameters to the simulating halt decider at Ĥ.qx. When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩
decides that the simulation of its first parameter on the input of its
second parameter never halt it correctly transitions to its own final
state of Ĥ.qn.

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

--
Copyright 2021 Pete Olcott

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

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

<87v94bfus2.fsf@bsb.me.uk>

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct and forms no contradiction.
Followup-To: comp.theory
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 by: Ben Bacarisse - Wed, 11 Aug 2021 23:04 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/11/2021 10:04 AM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:

>>> I went point by point. If I am actually incorrect then you can go
>>> point by point and point out each individual error step by step.
>> You fail at the first hurdle. You can't hope to persuade anyone that an
>> add function with add(2, 3) == 9 is operating as specified simply by
>> detailing, step by step, exactly how you implement the wrong behaviour.
>> In your case, it's simply that Ĥ.q0 ⟨Ĥ⟩ transitions to Ĥ.qn (via Ĥ.qx
>> ⟨Ĥ⟩ ⟨Ĥ⟩ as you pointlessly keep insisting) when Linz says it should not.
>> Your add function is entirely correct in that is does exactly what you
>> intend it to do. As far as I am concerned there is no significant error
>> in how you arrive at add(2, 3) == 9. The problem is that Linz says your
>> code should add numbers and not do whatever it is your code does with
>> 100% correctness. Do you follow?
>>
>>> Of course
>>> everyone knows that this is impossible if I am totally correct.
>> Then stop wasting time and try to publish! You'll need lots of time to
>> explain away why every editor simply laughs at the paper.
>>
>>> H.q0 WM w ⊢* Ĥ.qn
>>> becomes
>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>> Yes, we all know that. It's exactly why your Ĥ does not meet Linz's
>> specification (for this case -- you don't claim to have a halt decider).
>>
>>> Can you admit when you are wrong when you really are wrong?
>> Yes. I am wrong all the time. In this case I'm having trouble working
>> out how I could be clearer about your Ĥ. Maybe if you didn't keep
>> removing the key text from Linz's explanations it might sink in?
>>
>
> Pages of the Linz text to verify the above quotes in their full context:
> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>
> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H

I see you are in "paste the same text" again mode. If you think I can
help in any way, do let me know.

--
Ben.

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

<zKudnTxkSrW1xYn8nZ2dnUU7-bnNnZ2d@giganews.com>

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Newsgroups: comp.theory
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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Wed, 11 Aug 2021 23:07 UTC

On 8/11/2021 11:10 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/11/2021 9:28 AM, olcott wrote:
>
> Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no
> contradiction. [ Is Ben a Liar or simply woefully ignorant? ]
>
> This is a scumbag move. It's also cowardly and disingenuous --
> pretending to be coy about your scumbag opinions. I'll leave it
> unedited as it says everything readers need to know about your
> character.
>
>>> When the challenge to support one's assertion with reasoning is simply
>>> ignored as you are ignoring it right now one can reasonably construe
>>> a deceptive intent.
>
> A scumbag could construe it that way. Reasonable people could come to
> all sorts of other conclusions.
>
>>> H.q0 WM w ⊢* H.qn
>>> if M applied to W does not halt.
> if M applied to MW does not halt.
>
> (typo corrected)
>
>>>   becomes
>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>> if Ĥ applied to ⟨Ĥ⟩ does not halt.
>
> Yes. After a lot a pressing (and I mean lots, over several years!) you
> eventually admitted that H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ does indeed transition to H.qn.
> You also keep telling us that Ĥ applied to ⟨Ĥ⟩ halts. That's why your H
> (and its associated Ĥ) are wrong but for some reason you can't see this.
>
> You will plainly state that H rejects the string "⟨Ĥ⟩ ⟨Ĥ⟩" which it
> should do if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt. And you
> will, time and time again, show us exactly how Ĥ applied to ⟨Ĥ⟩ halts.
>
> This should be the end of the matter, but apparently your stating facts
> that show that H and Ĥ are wrong does not mean you know that H and Ĥ are
> wrong. I don't think I know any way to make progress on this.
>
>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>> if M applied to wM does not halt
> ...
>>> if M applied to wM does not halt (see above for definition of
>>> M) means when the Turing machine of ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt.
>
> Of course. We all know that. Unless you are pulling a fast one. "the
> Turing machine of ⟨Ĥ⟩" is just Ĥ. Is there a reason you are not simply
> saying "when Ĥ applied to ⟨Ĥ⟩ does not halt"?
>
>>> Ĥ.q0 ⟨Ĥ⟩  ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>> Ĥ.qx correctly transitions to its final state when the Ĥ.qx acts as a
>>> UTM and simulates ⟨Ĥ⟩ ⟨Ĥ⟩ and determines that this input never halts.
>
> Yes, we know the ruse: the computation would not halt if it were not the
> computation that it is. You've been trying to pull off this trick ever
> since the infamous "it wouldn't halt if line 15 was commented out"
> admission. Your Ĥ, however, not being a UTM, has the property that
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
> when it should not. Ĥ.q0 ⟨Ĥ⟩ should eventually transition to qn only
> "if Ĥ applied to ⟨Ĥ⟩ does not halt" (or, as you rather suspiciously
> write

> "if the Turing machine of ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt").

We can see that the above never halts.
In other words we can see that the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts.

There is an infinite cycle from Ĥ.qx to Ĥ.q0:

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

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

Therefore when Ĥ.qx on input ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to its own final state
of Ĥ.qn this is the correct never halting decision for its input because
its input never halts. It input never halts because its input does
specify infinitely nested simulation.

It does not matter that Ĥ on input ⟨Ĥ⟩ halts, the verifiable fact that
its input never halts proves that its halt decider did correctly decide
the "impossible" input. To meet the halting problem challenge a halt
decider only need decide its input correctly. Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ does decide
its input correctly, therefore the challenge has been met.

Turing machine Ĥ is applied to its input ⟨Ĥ⟩. It copies this input such
that this input and the copy of this input become the first and second
parameters to the simulating halt decider at Ĥ.qx. When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩
decides that the simulation of its first parameter on the input of its
second parameter never halt it correctly transitions to its own final
state of Ĥ.qn.

--
Copyright 2021 Pete Olcott

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

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

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Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
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Newsgroups: comp.theory
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From: NoO...@NoWhere.com (olcott)
Date: Wed, 11 Aug 2021 18:16:22 -0500
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 by: olcott - Wed, 11 Aug 2021 23:16 UTC

On 8/11/2021 6:04 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/11/2021 10:04 AM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>
>>>> I went point by point. If I am actually incorrect then you can go
>>>> point by point and point out each individual error step by step.
>>> You fail at the first hurdle. You can't hope to persuade anyone that an
>>> add function with add(2, 3) == 9 is operating as specified simply by
>>> detailing, step by step, exactly how you implement the wrong behaviour.
>>> In your case, it's simply that Ĥ.q0 ⟨Ĥ⟩ transitions to Ĥ.qn (via Ĥ.qx
>>> ⟨Ĥ⟩ ⟨Ĥ⟩ as you pointlessly keep insisting) when Linz says it should not.
>>> Your add function is entirely correct in that is does exactly what you
>>> intend it to do. As far as I am concerned there is no significant error
>>> in how you arrive at add(2, 3) == 9. The problem is that Linz says your
>>> code should add numbers and not do whatever it is your code does with
>>> 100% correctness. Do you follow?
>>>
>>>> Of course
>>>> everyone knows that this is impossible if I am totally correct.
>>> Then stop wasting time and try to publish! You'll need lots of time to
>>> explain away why every editor simply laughs at the paper.
>>>
>>>> H.q0 WM w ⊢* Ĥ.qn
>>>> becomes
>>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>> Yes, we all know that. It's exactly why your Ĥ does not meet Linz's
>>> specification (for this case -- you don't claim to have a halt decider).
>>>
>>>> Can you admit when you are wrong when you really are wrong?
>>> Yes. I am wrong all the time. In this case I'm having trouble working
>>> out how I could be clearer about your Ĥ. Maybe if you didn't keep
>>> removing the key text from Linz's explanations it might sink in?
>>>
>>
>> Pages of the Linz text to verify the above quotes in their full context:
>> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>>
>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>
> I see you are in "paste the same text" again mode. If you think I can
> help in any way, do let me know.
>

I just noticed that you acknowledged that M refers to the TM represented
by the first input parameter to Ĥ.qx wM wM so we can move to the next point:

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 // see highlighted portion of Linz text
to confirm:
if M applied to wM does not halt // M refers to the TM of the first wM
parameter to Ĥ.qx

Here is the overview of what I am claiming:
Turing machine Ĥ is applied to its input ⟨Ĥ⟩. It copies this input such
that this input and the copy of this input become the first and second
parameters to the simulating halt decider at Ĥ.qx. When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩
decides that the simulation of its first parameter on the input of its
second parameter never halt it correctly transitions to its own final
state of Ĥ.qn.

It is easy to see that when the halt decider at Ĥ.qx is a simulating
halt decider that there is an infinite cycle from Ĥ.qx to Ĥ.q0 that
prevents the input from every reaching its final state.

When we define halting as reaching a final state so that we can
unequivocally divide computations that complete from those that are
aborted, then we know that the input to Ĥ.qx never halts, thus making
its halt status decison correct.

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

--
Copyright 2021 Pete Olcott

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

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

<sf1mm1$7o8$1@dont-email.me>

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Path: i2pn2.org!i2pn.org!eternal-september.org!reader02.eternal-september.org!.POSTED!not-for-mail
From: jbb...@notatt.com (Jeff Barnett)
Newsgroups: comp.theory
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct_and_forms_no_contradiction.
Date: Wed, 11 Aug 2021 17:32:12 -0600
Organization: A noiseless patient Spider
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 by: Jeff Barnett - Wed, 11 Aug 2021 23:32 UTC

On 8/11/2021 5:04 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>
> I see you are in "paste the same text" again mode. If you think I can
> help in any way, do let me know.

That's not nice. Aren't you afraid that he'll develop carpal tunnel
syndrome? That along with all of his other deficiencies will surely do
him in. Perhaps we can invent shortcuts a la LaTeX macros and font
switches to help him cut down the strain. He wont be insulted and will
take to it as a pig to mud. Just think, he'll have yet another notation
to misuse and abuse.
--
Jeff Barnett

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

<yPSdnar5BfFpwon8nZ2dnUU7-WmdnZ2d@giganews.com>

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Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct_and_forms_no_contradiction.
Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc> <87bl6f5qvy.fsf@bsb.me.uk>
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From: NoO...@NoWhere.com (olcott)
Date: Wed, 11 Aug 2021 18:40:36 -0500
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 by: olcott - Wed, 11 Aug 2021 23:40 UTC

On 8/11/2021 6:32 PM, Jeff Barnett wrote:
> On 8/11/2021 5:04 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>
>> I see you are in "paste the same text" again mode.  If you think I can
>> help in any way, do let me know.
>
> That's not nice. Aren't you afraid that he'll develop carpal tunnel
> syndrome? That along with all of his other deficiencies will surely do
> him in. Perhaps we can invent shortcuts a la LaTeX macros and font
> switches to help him cut down the strain. He wont be insulted and will
> take to it as a pig to mud. Just think, he'll have yet another notation
> to misuse and abuse.

Ben finally acknowledged the point that I was making:

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

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

// M refers to the TM of the first wM parameter to Ĥ.qx

My proof requires a whole inference chain that cannot proceed to the
next point until the current point is accepted as correct.

--
Copyright 2021 Pete Olcott

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

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction. [ Is Ben a Liar or simply woefully ignorant? ]

<87pmujfs4o.fsf@bsb.me.uk>

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Followup: comp.theory
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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct and forms no contradiction. [ Is Ben a Liar or simply woefully
ignorant? ]
Followup-To: comp.theory
Date: Thu, 12 Aug 2021 01:01:27 +0100
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 by: Ben Bacarisse - Thu, 12 Aug 2021 00:01 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/11/2021 11:10 AM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/11/2021 9:28 AM, olcott wrote:
>> Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no
>> contradiction. [ Is Ben a Liar or simply woefully ignorant? ]
>>
>> This is a scumbag move. It's also cowardly and disingenuous --
>> pretending to be coy about your scumbag opinions. I'll leave it
>> unedited as it says everything readers need to know about your
>> character.

This is still a scumbag move.

>>>> When the challenge to support one's assertion with reasoning is simply
>>>> ignored as you are ignoring it right now one can reasonably construe
>>>> a deceptive intent.
>> A scumbag could construe it that way. Reasonable people could come to
>> all sorts of other conclusions.
>>
>>>> H.q0 WM w ⊢* H.qn
>>>> if M applied to W does not halt.
>> if M applied to MW does not halt.
>> (typo corrected)
>>
>>>>   becomes
>>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>> if Ĥ applied to ⟨Ĥ⟩ does not halt.
>>
>> Yes. After a lot a pressing (and I mean lots, over several years!) you
>> eventually admitted that H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ does indeed transition to H.qn.
>> You also keep telling us that Ĥ applied to ⟨Ĥ⟩ halts. That's why your H
>> (and its associated Ĥ) are wrong but for some reason you can't see this.
>>
>> You will plainly state that H rejects the string "⟨Ĥ⟩ ⟨Ĥ⟩" which it
>> should do if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt. And you
>> will, time and time again, show us exactly how Ĥ applied to ⟨Ĥ⟩ halts.
>>
>> This should be the end of the matter, but apparently your stating facts
>> that show that H and Ĥ are wrong does not mean you know that H and Ĥ are
>> wrong. I don't think I know any way to make progress on this.

You just read past this explanation of why your H and Ĥ are wrong and
you decided that you understood everything I'd said, did you? And you
also decided that nothing in these three paragraphs needs to be
challenged?

There's not much more I can add if you do indeed understand and accept
these remarks of mine. But, as I keep saying, I'm happy to answer any
questions you may have about my explanation.

--
Ben.

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction. [ Is Ben a Liar or simply woefully ignorant? ]

<TdOdnXoxb90094n8nZ2dnUU7-VHNnZ2d@giganews.com>

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<goydnfCCIYUWE5H8nZ2dnUU7-e_NnZ2d@giganews.com> <87eeb7z4d1.fsf@bsb.me.uk>
<0_Sdnb6Qe8XGOZH8nZ2dnUU7-U3NnZ2d@giganews.com> <87zgtslqpv.fsf@bsb.me.uk>
<4JOdnRS2SLR7MYz8nZ2dnUU7-YnNnZ2d@giganews.com> <87zgtoizgp.fsf@bsb.me.uk>
<Z6ednWt7SpIIv478nZ2dnUU7-WHNnZ2d@giganews.com> <87pmukiwr5.fsf@bsb.me.uk>
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From: NoO...@NoWhere.com (olcott)
Date: Wed, 11 Aug 2021 19:26:14 -0500
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 by: olcott - Thu, 12 Aug 2021 00:26 UTC

On 8/11/2021 7:01 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/11/2021 11:10 AM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/11/2021 9:28 AM, olcott wrote:
>>> Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no
>>> contradiction. [ Is Ben a Liar or simply woefully ignorant? ]
>>>
>>> This is a scumbag move. It's also cowardly and disingenuous --
>>> pretending to be coy about your scumbag opinions. I'll leave it
>>> unedited as it says everything readers need to know about your
>>> character.
>
> This is still a scumbag move.
>
>>>>> When the challenge to support one's assertion with reasoning is simply
>>>>> ignored as you are ignoring it right now one can reasonably construe
>>>>> a deceptive intent.
>>> A scumbag could construe it that way. Reasonable people could come to
>>> all sorts of other conclusions.
>>>
>>>>> H.q0 WM w ⊢* H.qn
>>>>> if M applied to W does not halt.
>>> if M applied to MW does not halt.
>>> (typo corrected)
>>>
>>>>>   becomes
>>>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>>> if Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>
>>> Yes. After a lot a pressing (and I mean lots, over several years!) you
>>> eventually admitted that H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ does indeed transition to H.qn.
>>> You also keep telling us that Ĥ applied to ⟨Ĥ⟩ halts. That's why your H
>>> (and its associated Ĥ) are wrong but for some reason you can't see this.
>>>
>>> You will plainly state that H rejects the string "⟨Ĥ⟩ ⟨Ĥ⟩" which it
>>> should do if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt. And you
>>> will, time and time again, show us exactly how Ĥ applied to ⟨Ĥ⟩ halts.
>>>
>>> This should be the end of the matter, but apparently your stating facts
>>> that show that H and Ĥ are wrong does not mean you know that H and Ĥ are
>>> wrong. I don't think I know any way to make progress on this.
>
> You just read past this explanation of why your H and Ĥ are wrong and
> you decided that you understood everything I'd said, did you? And you
> also decided that nothing in these three paragraphs needs to be
> challenged?
>

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

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

// M refers to the TM of the first wM parameter to Ĥ.qx

Now that you accept that the above is true we can move on to the next
point. My proof must proceed exactly one point at a time an cannot
possibly move to the next point until the current point is fully accepted.

That you believe that the fact that Ĥ applied to ⟨Ĥ⟩ transitions to its
final state of Ĥ.qn and halts nullifies the fact that Ĥ.qx wM wM does
correctly decide that its input never halts is the next point.

It is easier to discuss this with the H(P,P) model of the exact same
thing because with the H(P,P) model we can examine every single detail.

_P()
[00000d02](01) 55 push ebp
[00000d03](02) 8bec mov ebp,esp
[00000d05](03) 8b4508 mov eax,[ebp+08]
[00000d08](01) 50 push eax // push 2nd Param
[00000d09](03) 8b4d08 mov ecx,[ebp+08]
[00000d0c](01) 51 push ecx // push 1st Param
[00000d0d](05) e870feffff call 00000b82 // call H
[00000d12](03) 83c408 add esp,+08
[00000d15](02) 85c0 test eax,eax
[00000d17](02) 7402 jz 00000d1b
[00000d19](02) ebfe jmp 00000d19
[00000d1b](01) 5d pop ebp
[00000d1c](01) c3 ret
Size in bytes:(0027) [00000d1c]

....[00000d0d][00101829][00000d12] e870feffff call 00000b82 // call H

Begin Local Halt Decider Simulation at Machine Address:d02
....[00000d02][002118f1][002118f5] 55 push ebp
....[00000d03][002118f1][002118f5] 8bec mov ebp,esp
....[00000d05][002118f1][002118f5] 8b4508 mov eax,[ebp+08]
....[00000d08][002118ed][00000d02] 50 push eax // push P
....[00000d09][002118ed][00000d02] 8b4d08 mov ecx,[ebp+08]
....[00000d0c][002118e9][00000d02] 51 push ecx // push P
....[00000d0d][002118e5][00000d12] e870feffff call 00000b82 // call H

....[00000d02][0025c319][0025c31d] 55 push ebp
....[00000d03][0025c319][0025c31d] 8bec mov ebp,esp
....[00000d05][0025c319][0025c31d] 8b4508 mov eax,[ebp+08]
....[00000d08][0025c315][00000d02] 50 push eax // push P
....[00000d09][0025c315][00000d02] 8b4d08 mov ecx,[ebp+08]
....[00000d0c][0025c311][00000d02] 51 push ecx // push P
....[00000d0d][0025c30d][00000d12] e870feffff call 00000b82 // call H
Local Halt Decider: Infinite Recursion Detected Simulation Stopped

It is very obvious to people that know the x86 language very well that
while H acts as a pure simulator of P(P) that P cannot possibly stop
running.

The same thing applies to the infinite cycle from Ĥ.qx to Ĥ.q0 while the
simulating halt decider at Ĥ.qx acts as a UTM.

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

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

// M refers to the TM of the first wM parameter to Ĥ.qx

> There's not much more I can add if you do indeed understand and accept
> these remarks of mine. But, as I keep saying, I'm happy to answer any
> questions you may have about my explanation.
>

--
Copyright 2021 Pete Olcott

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

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction. [ Is Ben a Liar or simply woefully ignorant? ]

<878s17fqji.fsf@bsb.me.uk>

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Path: i2pn2.org!i2pn.org!eternal-september.org!reader02.eternal-september.org!.POSTED!not-for-mail
From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct and forms no contradiction. [ Is Ben a Liar or simply woefully
ignorant? ]
Followup-To: comp.theory
Date: Thu, 12 Aug 2021 01:35:45 +0100
Organization: A noiseless patient Spider
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 by: Ben Bacarisse - Thu, 12 Aug 2021 00:35 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/11/2021 11:10 AM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/11/2021 9:28 AM, olcott wrote:
>> Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no
>> contradiction. [ Is Ben a Liar or simply woefully ignorant? ]
>> This is a scumbag move. It's also cowardly and disingenuous --
>> pretending to be coy about your scumbag opinions. I'll leave it
>> unedited as it says everything readers need to know about your
>> character.
>>
>>>> When the challenge to support one's assertion with reasoning is simply
>>>> ignored as you are ignoring it right now one can reasonably construe
>>>> a deceptive intent.
>> A scumbag could construe it that way. Reasonable people could come to
>> all sorts of other conclusions.
>>
>>>> H.q0 WM w ⊢* H.qn
>>>> if M applied to W does not halt.
>> if M applied to MW does not halt.
>> (typo corrected)
>>
>>>>   becomes
>>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>> if Ĥ applied to ⟨Ĥ⟩ does not halt.
>> Yes. After a lot a pressing (and I mean lots, over several years!) you
>> eventually admitted that H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ does indeed transition to H.qn.
>
> You are saying that in a misleading way.

Curious. I agree with what say but it's misleading. Why? Did you not
continually resist saying whether the TM you (falsely, as it happened)
claimed to have accepted or rejected the key input? When, in your
recollection did you first say which it was without any equivocation?

> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩⟨Ĥ⟩ ⊢* Ĥ.qn
> The Turing machine of Ĥ halts on input ⟨Ĥ⟩ because the simulating halt
> decider at Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decides that its simulation of ⟨Ĥ⟩
> on input ⟨Ĥ⟩ never halts.

Garbled. What is "The Turing machine of Ĥ"? You can't mean Ĥ or your
would have just said Ĥ.

>> You also keep telling us that Ĥ applied to ⟨Ĥ⟩ halts. That's why your H
>> (and its associated Ĥ) are wrong but for some reason you can't see this.
>> You will plainly state that H rejects the string "⟨Ĥ⟩ ⟨Ĥ⟩" which it
>> should do if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt. And you
>> will, time and time again, show us exactly how Ĥ applied to ⟨Ĥ⟩ halts.
>
> The Ĥ of the first parameter to Ĥ.qx ⟨Ĥ⟩⟨Ĥ⟩ never halts

Garbled. But who cares? Your H rejects the string "⟨Ĥ⟩ ⟨Ĥ⟩", but your
Ĥ applied to ⟨Ĥ⟩ halts. This is not what Linz requires of your H/Ĥ. End
of story. If you want to retract ether of these statements about your
H/Ĥ, please do, but until you do your H/Ĥ are of no interest as far as
the proof is concerned.

--
Ben.

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction. [ Ben accepts one point? ]

<hMqdnWbwOMFh8In8nZ2dnUU7-dmdnZ2d@giganews.com>

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Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct_and_forms_no_contradiction._[_Ben_accepts_one_point? ]
Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc> <1ZednUt-qbrXlZf8nZ2dnUU7-QnNnZ2d@giganews.com> <87tuk52h0e.fsf@bsb.me.uk> <zcadnTSOD5rtZ5f8nZ2dnUU7-T3NnZ2d@giganews.com> <877dh03l3c.fsf@bsb.me.uk> <Z5adnd038KGXwJb8nZ2dnUU7-I_NnZ2d@giganews.com> <8735rn1qvj.fsf@bsb.me.uk> <goydnfCCIYUWE5H8nZ2dnUU7-e_NnZ2d@giganews.com> <87eeb7z4d1.fsf@bsb.me.uk> <0_Sdnb6Qe8XGOZH8nZ2dnUU7-U3NnZ2d@giganews.com> <87zgtslqpv.fsf@bsb.me.uk> <4JOdnRS2SLR7MYz8nZ2dnUU7-YnNnZ2d@giganews.com> <87zgtoizgp.fsf@bsb.me.uk> <Z6ednWt7SpIIv478nZ2dnUU7-WHNnZ2d@giganews.com> <87pmukiwr5.fsf@bsb.me.uk> <5c6dnbn2gMMIsI78nZ2dnUU7-dnNnZ2d@giganews.com> <87eeb0iuo1.fsf@bsb.me.uk> <346dnYhkWPUNQ478nZ2dnUU7-UPNnZ2d@giganews.com> <Ja-dneAl3poPeI78nZ2dnUU7-UvNnZ2d@giganews.com> <87fsvggdxz.fsf@bsb.me.uk> <HISdnXl3NPSvnYn8nZ2dnUU7-eHNnZ2d@giganews.com> <878s17fqji.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 11 Aug 2021 19:40:27 -0500
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 by: olcott - Thu, 12 Aug 2021 00:40 UTC

On 8/11/2021 7:35 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/11/2021 11:10 AM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/11/2021 9:28 AM, olcott wrote:
>>> Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no
>>> contradiction. [ Is Ben a Liar or simply woefully ignorant? ]
>>> This is a scumbag move. It's also cowardly and disingenuous --
>>> pretending to be coy about your scumbag opinions. I'll leave it
>>> unedited as it says everything readers need to know about your
>>> character.
>>>
>>>>> When the challenge to support one's assertion with reasoning is simply
>>>>> ignored as you are ignoring it right now one can reasonably construe
>>>>> a deceptive intent.
>>> A scumbag could construe it that way. Reasonable people could come to
>>> all sorts of other conclusions.
>>>
>>>>> H.q0 WM w ⊢* H.qn
>>>>> if M applied to W does not halt.
>>> if M applied to MW does not halt.
>>> (typo corrected)
>>>
>>>>>   becomes
>>>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>>> if Ĥ applied to ⟨Ĥ⟩ does not halt.
>>> Yes. After a lot a pressing (and I mean lots, over several years!) you
>>> eventually admitted that H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ does indeed transition to H.qn.
>>
>> You are saying that in a misleading way.
>
> Curious. I agree with what say but it's misleading. Why? Did you not
> continually resist saying whether the TM you (falsely, as it happened)
> claimed to have accepted or rejected the key input? When, in your
> recollection did you first say which it was without any equivocation?
>
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩⟨Ĥ⟩ ⊢* Ĥ.qn
>> The Turing machine of Ĥ halts on input ⟨Ĥ⟩ because the simulating halt
>> decider at Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decides that its simulation of ⟨Ĥ⟩
>> on input ⟨Ĥ⟩ never halts.
>
> Garbled. What is "The Turing machine of Ĥ"? You can't mean Ĥ or your
> would have just said Ĥ.
>
>>> You also keep telling us that Ĥ applied to ⟨Ĥ⟩ halts. That's why your H
>>> (and its associated Ĥ) are wrong but for some reason you can't see this.
>>> You will plainly state that H rejects the string "⟨Ĥ⟩ ⟨Ĥ⟩" which it
>>> should do if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt. And you
>>> will, time and time again, show us exactly how Ĥ applied to ⟨Ĥ⟩ halts.
>>
>> The Ĥ of the first parameter to Ĥ.qx ⟨Ĥ⟩⟨Ĥ⟩ never halts
>
> Garbled. But who cares? Your H rejects the string "⟨Ĥ⟩ ⟨Ĥ⟩", but your
> Ĥ applied to ⟨Ĥ⟩ halts. This is not what Linz requires of your H/Ĥ. End
> of story. If you want to retract ether of these statements about your
> H/Ĥ, please do, but until you do your H/Ĥ are of no interest as far as
> the proof is concerned.
>

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

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

// M refers to the TM of the first wM parameter to Ĥ.qx

Now that you accept that the above is true we can move on to the next
point. My proof must proceed exactly one point at a time an cannot
possibly move to the next point until the current point is fully accepted.

That you believe that the fact that Ĥ applied to ⟨Ĥ⟩ transitions to its
final state of Ĥ.qn and halts nullifies the fact that Ĥ.qx wM wM does
correctly decide that its input never halts is the next point.

--
Copyright 2021 Pete Olcott

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

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction. [ Ben accepts one point? ] (typo fixed )

<yuGdneaGkeN-84n8nZ2dnUU7-cnNnZ2d@giganews.com>

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Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
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Newsgroups: comp.theory
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<goydnfCCIYUWE5H8nZ2dnUU7-e_NnZ2d@giganews.com> <87eeb7z4d1.fsf@bsb.me.uk>
<0_Sdnb6Qe8XGOZH8nZ2dnUU7-U3NnZ2d@giganews.com> <87zgtslqpv.fsf@bsb.me.uk>
<4JOdnRS2SLR7MYz8nZ2dnUU7-YnNnZ2d@giganews.com> <87zgtoizgp.fsf@bsb.me.uk>
<Z6ednWt7SpIIv478nZ2dnUU7-WHNnZ2d@giganews.com> <87pmukiwr5.fsf@bsb.me.uk>
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From: NoO...@NoWhere.com (olcott)
Date: Wed, 11 Aug 2021 19:44:18 -0500
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 by: olcott - Thu, 12 Aug 2021 00:44 UTC

On 8/11/2021 7:35 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/11/2021 11:10 AM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/11/2021 9:28 AM, olcott wrote:
>>> Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no
>>> contradiction. [ Is Ben a Liar or simply woefully ignorant? ]
>>> This is a scumbag move. It's also cowardly and disingenuous --
>>> pretending to be coy about your scumbag opinions. I'll leave it
>>> unedited as it says everything readers need to know about your
>>> character.
>>>
>>>>> When the challenge to support one's assertion with reasoning is simply
>>>>> ignored as you are ignoring it right now one can reasonably construe
>>>>> a deceptive intent.
>>> A scumbag could construe it that way. Reasonable people could come to
>>> all sorts of other conclusions.
>>>
>>>>> H.q0 WM w ⊢* H.qn
>>>>> if M applied to W does not halt.
>>> if M applied to MW does not halt.
>>> (typo corrected)
>>>
>>>>>   becomes
>>>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>>> if Ĥ applied to ⟨Ĥ⟩ does not halt.
>>> Yes. After a lot a pressing (and I mean lots, over several years!) you
>>> eventually admitted that H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ does indeed transition to H.qn.
>>
>> You are saying that in a misleading way.
>
> Curious. I agree with what say but it's misleading. Why? Did you not
> continually resist saying whether the TM you (falsely, as it happened)
> claimed to have accepted or rejected the key input? When, in your
> recollection did you first say which it was without any equivocation?
>
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩⟨Ĥ⟩ ⊢* Ĥ.qn
>> The Turing machine of Ĥ halts on input ⟨Ĥ⟩ because the simulating halt
>> decider at Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decides that its simulation of ⟨Ĥ⟩
>> on input ⟨Ĥ⟩ never halts.
>
> Garbled. What is "The Turing machine of Ĥ"? You can't mean Ĥ or your
> would have just said Ĥ.
>
>>> You also keep telling us that Ĥ applied to ⟨Ĥ⟩ halts. That's why your H
>>> (and its associated Ĥ) are wrong but for some reason you can't see this.
>>> You will plainly state that H rejects the string "⟨Ĥ⟩ ⟨Ĥ⟩" which it
>>> should do if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt. And you
>>> will, time and time again, show us exactly how Ĥ applied to ⟨Ĥ⟩ halts.
>>
>> The Ĥ of the first parameter to Ĥ.qx ⟨Ĥ⟩⟨Ĥ⟩ never halts
>
> Garbled. But who cares? Your H rejects the string "⟨Ĥ⟩ ⟨Ĥ⟩", but your
> Ĥ applied to ⟨Ĥ⟩ halts. This is not what Linz requires of your H/Ĥ. End
> of story. If you want to retract ether of these statements about your
> H/Ĥ, please do, but until you do your H/Ĥ are of no interest as far as
> the proof is concerned.
>

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

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

// M refers to the TM of the first wM parameter to Ĥ.qx

Now that you accept that the above is true we can move on to the next
point. My proof must proceed exactly one point at a time an cannot
possibly move to the next point until the current point is fully accepted.

That you believe that the fact that Ĥ applied to ⟨Ĥ⟩ transitions to its
final state of Ĥ.qn and halts nullifies the fact that Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ does
correctly decide that its input never halts is the next point.

--
Copyright 2021 Pete Olcott

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

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

<8735rffoh2.fsf@bsb.me.uk>

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct and forms no contradiction.
Followup-To: comp.theory
Date: Thu, 12 Aug 2021 02:20:25 +0100
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 by: Ben Bacarisse - Thu, 12 Aug 2021 01:20 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/11/2021 6:04 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/11/2021 10:04 AM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>
>>>>> I went point by point. If I am actually incorrect then you can go
>>>>> point by point and point out each individual error step by step.
>>>> You fail at the first hurdle. You can't hope to persuade anyone that an
>>>> add function with add(2, 3) == 9 is operating as specified simply by
>>>> detailing, step by step, exactly how you implement the wrong behaviour.
>>>> In your case, it's simply that Ĥ.q0 ⟨Ĥ⟩ transitions to Ĥ.qn (via Ĥ.qx
>>>> ⟨Ĥ⟩ ⟨Ĥ⟩ as you pointlessly keep insisting) when Linz says it should not.
>>>> Your add function is entirely correct in that is does exactly what you
>>>> intend it to do. As far as I am concerned there is no significant error
>>>> in how you arrive at add(2, 3) == 9. The problem is that Linz says your
>>>> code should add numbers and not do whatever it is your code does with
>>>> 100% correctness. Do you follow?
>>>>
>>>>> Of course
>>>>> everyone knows that this is impossible if I am totally correct.
>>>> Then stop wasting time and try to publish! You'll need lots of time to
>>>> explain away why every editor simply laughs at the paper.
>>>>
>>>>> H.q0 WM w ⊢* Ĥ.qn
>>>>> becomes
>>>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>> Yes, we all know that. It's exactly why your Ĥ does not meet Linz's
>>>> specification (for this case -- you don't claim to have a halt decider).
>>>>
>>>>> Can you admit when you are wrong when you really are wrong?
>>>> Yes. I am wrong all the time. In this case I'm having trouble working
>>>> out how I could be clearer about your Ĥ. Maybe if you didn't keep
>>>> removing the key text from Linz's explanations it might sink in?
>>>>
>>>
>>> Pages of the Linz text to verify the above quotes in their full context:
>>> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>>>
>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>> I see you are in "paste the same text" again mode. If you think I can
>> help in any way, do let me know.
>
> I just noticed that you acknowledged that M refers to the TM
> represented by the first input parameter to Ĥ.qx wM wM

Did I? Oh dear. It's garbage. Where did I "acknowledge" it? I'd like
to go back and point out that it's mathematical junk.

> so we can move to the next point:
>
> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
> if M applied to wM halts, and

This is just a metaphorical math poem. You've taken a mathematical
statement and substituted for only some on the occurrences of a
variable. That is a schoolboy error.

Maybe you confused agreement with "can't be bothered to go of on a
tangent about a math poem". I certainly let many meaningless things you
write slip by unremarked upon. There are not enough hours in the day to
point out all your mistakes.

> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn // see highlighted portion of Linz text
> to confirm:
> if M applied to wM does not halt // M refers to the TM of the first wM
> parameter to Ĥ.qx

Linz requires that

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

Unlike your garbage, this makes sense because all occurrences of M and
the derived wM have now been substituted for actual values -- your Ĥ and
its encoding.

> Here is the overview of what I am claiming:
> Turing machine Ĥ is applied to its input ⟨Ĥ⟩.

Ĥ.q0 ⟨Ĥ⟩

> It copies this input such that this input and the copy of this input
> become the first and second parameters to the simulating halt decider
> at Ĥ.qx.

⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩

> When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ decides that the simulation of its first parameter
> on the input of its second parameter never halt it correctly
> transitions to its own final state of Ĥ.qn.

⊢* Ĥ.qn

None of which (other than your poor use of technical terms) has been in
doubt for months.

> It is easy to see that when the halt decider at Ĥ.qx is a simulating
> halt decider that there is an infinite cycle from Ĥ.qx to Ĥ.q0 that
> prevents the input from every reaching its final state.

If Ĥ were not written as you have chosen to write it -- if it were
instead a pure simulator, then

⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* ∞

would be the case. Mathematicians would avoid this messy language and
the possibility of confusion by not re-using the name Ĥ for both your
partial decider and one based on pure simulation.

Also, you are wrong about the cycles of states, but that's because you
have no clue how a TM would simulate another TM. It's a rabbit hole we
shouldn't go down, because (a) I don't think you could ever understand
how a UTM really works, and (b) it's irrelevant since I agree that there
would be infinite execution if Ĥ were not as it is but were, instead, a
pure simulator.

> When we define halting as reaching a final state

Ĥ applied to ⟨Ĥ⟩ halts. The computation reaches the final state Ĥ.qn.

> so that we can unequivocally divide computations that complete from
> those that are aborted,

So this is today's equivocation. TM computations either halt or they
don't halt. None of them "abort".

> then we know that the input to Ĥ.qx never
> halts, thus making its halt status decison correct.

Garbled. Inputs don't halt or not halt. Your Ĥ does not meet Linz's
spec because Ĥ applied to ⟨Ĥ⟩ halts when it should not. You don't get
to say it does because it halts in what you think of as special way.

--
Ben.

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction. [ Is Ben a Liar or simply woefully ignorant? ]

<87wnore9j3.fsf@bsb.me.uk>

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct and forms no contradiction. [ Is Ben a Liar or simply woefully
ignorant? ]
Followup-To: comp.theory
Date: Thu, 12 Aug 2021 02:28:32 +0100
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 by: Ben Bacarisse - Thu, 12 Aug 2021 01:28 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/11/2021 7:01 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/11/2021 11:10 AM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/11/2021 9:28 AM, olcott wrote:
>>>> Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no
>>>> contradiction. [ Is Ben a Liar or simply woefully ignorant? ]
>>>>
>>>> This is a scumbag move. It's also cowardly and disingenuous --
>>>> pretending to be coy about your scumbag opinions. I'll leave it
>>>> unedited as it says everything readers need to know about your
>>>> character.
>>
>> This is still a scumbag move.

Still a scumbag move.

>>>>>> When the challenge to support one's assertion with reasoning is simply
>>>>>> ignored as you are ignoring it right now one can reasonably construe
>>>>>> a deceptive intent.
>>>> A scumbag could construe it that way. Reasonable people could come to
>>>> all sorts of other conclusions.
>>>>
>>>>>> H.q0 WM w ⊢* H.qn
>>>>>> if M applied to W does not halt.
>>>> if M applied to MW does not halt.
>>>> (typo corrected)
>>>>
>>>>>>   becomes
>>>>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>>>> if Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>>
>>>> Yes. After a lot a pressing (and I mean lots, over several years!) you
>>>> eventually admitted that H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ does indeed transition to H.qn.
>>>> You also keep telling us that Ĥ applied to ⟨Ĥ⟩ halts. That's why your H
>>>> (and its associated Ĥ) are wrong but for some reason you can't see this.
>>>>
>>>> You will plainly state that H rejects the string "⟨Ĥ⟩ ⟨Ĥ⟩" which it
>>>> should do if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt. And you
>>>> will, time and time again, show us exactly how Ĥ applied to ⟨Ĥ⟩ halts.
>>>>
>>>> This should be the end of the matter, but apparently your stating facts
>>>> that show that H and Ĥ are wrong does not mean you know that H and Ĥ are
>>>> wrong. I don't think I know any way to make progress on this.
>>
>> You just read past this explanation of why your H and Ĥ are wrong and
>> you decided that you understood everything I'd said, did you? And you
>> also decided that nothing in these three paragraphs needs to be
>> challenged?

Nothing about this of course.

> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
> if M applied to wM halts, and
>
> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
> if M applied to wM does not halt
>
> // M refers to the TM of the first wM parameter to Ĥ.qx
>
> Now that you accept that the above is true

The above is junk. Please point me at the post that made you think I
agree with it. I may have a done so by accident (you post similar
texts, some of which are, rather surprisingly, correct) and I'd like to
post a follow-up correcting this impression.

--
Ben.

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

<87r1eze981.fsf@bsb.me.uk>

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct and forms no contradiction.
Followup-To: comp.theory
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 by: Ben Bacarisse - Thu, 12 Aug 2021 01:35 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/11/2021 11:10 AM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/11/2021 9:28 AM, olcott wrote:

>>>> H.q0 WM w ⊢* H.qn
>>>> if M applied to W does not halt.
>> if M applied to MW does not halt.
>> (typo corrected)
>>
>>>>   becomes
>>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>> if Ĥ applied to ⟨Ĥ⟩ does not halt.
>> Yes. After a lot a pressing (and I mean lots, over several years!) you
>> eventually admitted that H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ does indeed transition to H.qn.
>> You also keep telling us that Ĥ applied to ⟨Ĥ⟩ halts. That's why your H
>> (and its associated Ĥ) are wrong but for some reason you can't see this.
>> You will plainly state that H rejects the string "⟨Ĥ⟩ ⟨Ĥ⟩" which it
>> should do if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt. And you
>> will, time and time again, show us exactly how Ĥ applied to ⟨Ĥ⟩ halts.
>> This should be the end of the matter, but apparently your stating facts
>> that show that H and Ĥ are wrong does not mean you know that H and Ĥ are
>> wrong. I don't think I know any way to make progress on this.
>>
>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>> if M applied to wM does not halt
>> ...
>>>> if M applied to wM does not halt (see above for definition of
>>>> M) means when the Turing machine of ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt.
>> Of course. We all know that. Unless you are pulling a fast one. "the
>> Turing machine of ⟨Ĥ⟩" is just Ĥ. Is there a reason you are not simply
>> saying "when Ĥ applied to ⟨Ĥ⟩ does not halt"?

I didn't expect an answer. You may be reserving this phrase as a
get-out clause for later, so saying what you mean too early will block
that escape route.

>>>> Ĥ.q0 ⟨Ĥ⟩  ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>> Ĥ.qx correctly transitions to its final state when the Ĥ.qx acts as a
>>>> UTM and simulates ⟨Ĥ⟩ ⟨Ĥ⟩ and determines that this input never halts.
>> Yes, we know the ruse: the computation would not halt if it were not the
>> computation that it is. You've been trying to pull off this trick ever
>> since the infamous "it wouldn't halt if line 15 was commented out"
>> admission. Your Ĥ, however, not being a UTM, has the property that
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>> when it should not. Ĥ.q0 ⟨Ĥ⟩ should eventually transition to qn only
>> "if Ĥ applied to ⟨Ĥ⟩ does not halt" (or, as you rather suspiciously
>> write "if the Turing machine of ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt").
>
> We can see that the above never halts.

We can see that it halts. It's right there in a fact that has been
undisputed for months:

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

--
Ben.

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

<BaednXQRjpr9HIn8nZ2dnUU7-SvNnZ2d@giganews.com>

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Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct_and_forms_no_contradiction.
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
References: <20210719214640.00000dfc@reddwarf.jmc> <87tuk52h0e.fsf@bsb.me.uk> <zcadnTSOD5rtZ5f8nZ2dnUU7-T3NnZ2d@giganews.com> <877dh03l3c.fsf@bsb.me.uk> <Z5adnd038KGXwJb8nZ2dnUU7-I_NnZ2d@giganews.com> <8735rn1qvj.fsf@bsb.me.uk> <goydnfCCIYUWE5H8nZ2dnUU7-e_NnZ2d@giganews.com> <87eeb7z4d1.fsf@bsb.me.uk> <0_Sdnb6Qe8XGOZH8nZ2dnUU7-U3NnZ2d@giganews.com> <87zgtslqpv.fsf@bsb.me.uk> <4JOdnRS2SLR7MYz8nZ2dnUU7-YnNnZ2d@giganews.com> <87zgtoizgp.fsf@bsb.me.uk> <Z6ednWt7SpIIv478nZ2dnUU7-WHNnZ2d@giganews.com> <87pmukiwr5.fsf@bsb.me.uk> <5c6dnbn2gMMIsI78nZ2dnUU7-dnNnZ2d@giganews.com> <87eeb0iuo1.fsf@bsb.me.uk> <2-adncuMeNUVpY78nZ2dnUU7-fPNnZ2d@giganews.com> <87o8a4ggzf.fsf@bsb.me.uk> <XfGdncywcvvmdI78nZ2dnUU7-fednZ2d@giganews.com> <87v94bfus2.fsf@bsb.me.uk> <GKSdnVVNYtzbx4n8nZ2dnUU7-THNnZ2d@giganews.com> <8735rffoh2.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 11 Aug 2021 21:03:11 -0500
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 by: olcott - Thu, 12 Aug 2021 02:03 UTC

On 8/11/2021 8:20 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/11/2021 6:04 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/11/2021 10:04 AM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>>>> I went point by point. If I am actually incorrect then you can go
>>>>>> point by point and point out each individual error step by step.
>>>>> You fail at the first hurdle. You can't hope to persuade anyone that an
>>>>> add function with add(2, 3) == 9 is operating as specified simply by
>>>>> detailing, step by step, exactly how you implement the wrong behaviour.
>>>>> In your case, it's simply that Ĥ.q0 ⟨Ĥ⟩ transitions to Ĥ.qn (via Ĥ.qx
>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩ as you pointlessly keep insisting) when Linz says it should not.
>>>>> Your add function is entirely correct in that is does exactly what you
>>>>> intend it to do. As far as I am concerned there is no significant error
>>>>> in how you arrive at add(2, 3) == 9. The problem is that Linz says your
>>>>> code should add numbers and not do whatever it is your code does with
>>>>> 100% correctness. Do you follow?
>>>>>
>>>>>> Of course
>>>>>> everyone knows that this is impossible if I am totally correct.
>>>>> Then stop wasting time and try to publish! You'll need lots of time to
>>>>> explain away why every editor simply laughs at the paper.
>>>>>
>>>>>> H.q0 WM w ⊢* Ĥ.qn
>>>>>> becomes
>>>>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>>> Yes, we all know that. It's exactly why your Ĥ does not meet Linz's
>>>>> specification (for this case -- you don't claim to have a halt decider).
>>>>>
>>>>>> Can you admit when you are wrong when you really are wrong?
>>>>> Yes. I am wrong all the time. In this case I'm having trouble working
>>>>> out how I could be clearer about your Ĥ. Maybe if you didn't keep
>>>>> removing the key text from Linz's explanations it might sink in?
>>>>>
>>>>
>>>> Pages of the Linz text to verify the above quotes in their full context:
>>>> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>>>>
>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>> I see you are in "paste the same text" again mode. If you think I can
>>> help in any way, do let me know.
>>
>> I just noticed that you acknowledged that M refers to the TM
>> represented by the first input parameter to Ĥ.qx wM wM
>
> Did I? Oh dear. It's garbage. Where did I "acknowledge" it? I'd like
> to go back and point out that it's mathematical junk.
>
>> so we can move to the next point:
>>
>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>> if M applied to wM halts, and
>
> This is just a metaphorical math poem. You've taken a mathematical
> statement and substituted for only some on the occurrences of a
> variable. That is a schoolboy error.
>

That seems to be a very stupid thing to say when it only clarifies and
corrects Linz.

> Maybe you confused agreement with "can't be bothered to go of on a
> tangent about a math poem". I certainly let many meaningless things you
> write slip by unremarked upon. There are not enough hours in the day to
> point out all your mistakes.
>
>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn // see highlighted portion of Linz text
>> to confirm:
>> if M applied to wM does not halt // M refers to the TM of the first wM
>> parameter to Ĥ.qx
>
> Linz requires that
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> if, and only if, Ĥ applied to ⟨Ĥ⟩ does not halt.
>

Sure and the Ĥ being referred to is the one that the first param to Ĥ.qx
⟨Ĥ⟩ ⟨Ĥ⟩ specifies.

> Unlike your garbage, this makes sense because all occurrences of M and
> the derived wM have now been substituted for actual values -- your Ĥ and
> its encoding.
>
>> Here is the overview of what I am claiming:
>> Turing machine Ĥ is applied to its input ⟨Ĥ⟩.
>
> Ĥ.q0 ⟨Ĥ⟩
>
>> It copies this input such that this input and the copy of this input
>> become the first and second parameters to the simulating halt decider
>> at Ĥ.qx.
>
> ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩
>
>> When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ decides that the simulation of its first parameter
>> on the input of its second parameter never halt it correctly
>> transitions to its own final state of Ĥ.qn.
>
> ⊢* Ĥ.qn
>
> None of which (other than your poor use of technical terms) has been in
> doubt for months.
>

You sure acted like M only referred to the first Ĥ shown below:
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> if, and only if, Ĥ applied to ⟨Ĥ⟩ does not halt.

>> It is easy to see that when the halt decider at Ĥ.qx is a simulating
>> halt decider that there is an infinite cycle from Ĥ.qx to Ĥ.q0 that
>> prevents the input from every reaching its final state.
>
> If Ĥ were not written as you have chosen to write it -- if it were
> instead a pure simulator, then
>
> ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* ∞
>

This: Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
specifies an infinite cycle from Ĥ.qx to Ĥ.q0 while the simulating halt
decider at Ĥ.qx acts as a UTM.

> would be the case. Mathematicians would avoid this messy language and
> the possibility of confusion by not re-using the name Ĥ for both your
> partial decider and one based on pure simulation.
>

Since it is the very same machine in two different modes of operation
this seems really stupid.

This <is> the same Ĥ that Linz specifies with the extra detail of how
this wildcard state transition works: ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

> Also, you are wrong about the cycles of states, but that's because you
> have no clue how a TM would simulate another TM.

No one ever investigates this because their paper on the subject would
be millions of pages long if it included a full listing of the source
code. If we add the single feature of random access memory TMs would be
100,000-fold less tedious.

> It's a rabbit hole we
> shouldn't go down, because (a) I don't think you could ever understand
> how a UTM really works, and (b) it's irrelevant since I agree that there
> would be infinite execution if Ĥ were not as it is but were, instead, a
> pure simulator.
>

Great. This is a big breakthrough.

>> When we define halting as reaching a final state
>
> Ĥ applied to ⟨Ĥ⟩ halts. The computation reaches the final state Ĥ.qn.
>

Yes. The halting problem is not about programs that halt. It is about
inputs to a halt decider. The input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts.

>> so that we can unequivocally divide computations that complete from
>> those that are aborted,
>
> So this is today's equivocation. TM computations either halt or they
> don't halt. None of them "abort".
>

I am concurrently perfecting my understanding of the technical terms via
Linz. A TM can also be said to halt if there is no transition out of the
current state. Ultimately I only need a clear and unequivocal way to
divide computations that reach their natural conclusion from ones that
have had their simulation aborted.

>> then we know that the input to Ĥ.qx never
>> halts, thus making its halt status decison correct.
>
> Garbled. Inputs don't halt or not halt. Your Ĥ does not meet Linz's
> spec because Ĥ applied to ⟨Ĥ⟩ halts when it should not. You don't get
> to say it does because it halts in what you think of as special way.
>

That is like saying because we know the liar paradox: "this sentence is
not true" is not true then it must be true.

My purpose of creating the x86utm operating system was so that every
single detail of the halting problem counter-examples can be completely
examined.


Click here to read the complete article
Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction. [ Ben's key param agreement ]

<fN2dneG14c8TH4n8nZ2dnUU7-RfNnZ2d@giganews.com>

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Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct_and_forms_no_contradiction._[_Ben's_key_param_agree
ment_]
Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc> <87tuk52h0e.fsf@bsb.me.uk>
<zcadnTSOD5rtZ5f8nZ2dnUU7-T3NnZ2d@giganews.com> <877dh03l3c.fsf@bsb.me.uk>
<Z5adnd038KGXwJb8nZ2dnUU7-I_NnZ2d@giganews.com> <8735rn1qvj.fsf@bsb.me.uk>
<goydnfCCIYUWE5H8nZ2dnUU7-e_NnZ2d@giganews.com> <87eeb7z4d1.fsf@bsb.me.uk>
<0_Sdnb6Qe8XGOZH8nZ2dnUU7-U3NnZ2d@giganews.com> <87zgtslqpv.fsf@bsb.me.uk>
<4JOdnRS2SLR7MYz8nZ2dnUU7-YnNnZ2d@giganews.com> <87zgtoizgp.fsf@bsb.me.uk>
<Z6ednWt7SpIIv478nZ2dnUU7-WHNnZ2d@giganews.com> <87pmukiwr5.fsf@bsb.me.uk>
<5c6dnbn2gMMIsI78nZ2dnUU7-dnNnZ2d@giganews.com> <87eeb0iuo1.fsf@bsb.me.uk>
<2-adncuMeNUVpY78nZ2dnUU7-fPNnZ2d@giganews.com> <87o8a4ggzf.fsf@bsb.me.uk>
<XfGdncywcvvmdI78nZ2dnUU7-fednZ2d@giganews.com> <87v94bfus2.fsf@bsb.me.uk>
<GKSdnVVNYtzbx4n8nZ2dnUU7-THNnZ2d@giganews.com> <8735rffoh2.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 11 Aug 2021 21:08:14 -0500
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 by: olcott - Thu, 12 Aug 2021 02:08 UTC

On 8/11/2021 8:20 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/11/2021 6:04 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/11/2021 10:04 AM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>>>> I went point by point. If I am actually incorrect then you can go
>>>>>> point by point and point out each individual error step by step.
>>>>> You fail at the first hurdle. You can't hope to persuade anyone that an
>>>>> add function with add(2, 3) == 9 is operating as specified simply by
>>>>> detailing, step by step, exactly how you implement the wrong behaviour.
>>>>> In your case, it's simply that Ĥ.q0 ⟨Ĥ⟩ transitions to Ĥ.qn (via Ĥ.qx
>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩ as you pointlessly keep insisting) when Linz says it should not.
>>>>> Your add function is entirely correct in that is does exactly what you
>>>>> intend it to do. As far as I am concerned there is no significant error
>>>>> in how you arrive at add(2, 3) == 9. The problem is that Linz says your
>>>>> code should add numbers and not do whatever it is your code does with
>>>>> 100% correctness. Do you follow?
>>>>>
>>>>>> Of course
>>>>>> everyone knows that this is impossible if I am totally correct.
>>>>> Then stop wasting time and try to publish! You'll need lots of time to
>>>>> explain away why every editor simply laughs at the paper.
>>>>>
>>>>>> H.q0 WM w ⊢* Ĥ.qn
>>>>>> becomes
>>>>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>>> Yes, we all know that. It's exactly why your Ĥ does not meet Linz's
>>>>> specification (for this case -- you don't claim to have a halt decider).
>>>>>
>>>>>> Can you admit when you are wrong when you really are wrong?
>>>>> Yes. I am wrong all the time. In this case I'm having trouble working
>>>>> out how I could be clearer about your Ĥ. Maybe if you didn't keep
>>>>> removing the key text from Linz's explanations it might sink in?
>>>>>
>>>>
>>>> Pages of the Linz text to verify the above quotes in their full context:
>>>> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>>>>
>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>> I see you are in "paste the same text" again mode. If you think I can
>>> help in any way, do let me know.
>>
>> I just noticed that you acknowledged that M refers to the TM
>> represented by the first input parameter to Ĥ.qx wM wM
>
> Did I? Oh dear. It's garbage. Where did I "acknowledge" it? I'd like
> to go back and point out that it's mathematical junk.
>
>> so we can move to the next point:
>>
>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>> if M applied to wM halts, and
>
> This is just a metaphorical math poem. You've taken a mathematical
> statement and substituted for only some on the occurrences of a
> variable. That is a schoolboy error.
>
> Maybe you confused agreement with "can't be bothered to go of on a
> tangent about a math poem". I certainly let many meaningless things you
> write slip by unremarked upon. There are not enough hours in the day to
> point out all your mistakes.
>
>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn // see highlighted portion of Linz text
>> to confirm:
>> if M applied to wM does not halt // M refers to the TM of the first wM
>> parameter to Ĥ.qx
>
> Linz requires that
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> if, and only if, Ĥ applied to ⟨Ĥ⟩ does not halt.
>
> Unlike your garbage, this makes sense because all occurrences of M and
> the derived wM have now been substituted for actual values -- your Ĥ and
> its encoding.
>
>> Here is the overview of what I am claiming:
>> Turing machine Ĥ is applied to its input ⟨Ĥ⟩.
>
> Ĥ.q0 ⟨Ĥ⟩
>
>> It copies this input such that this input and the copy of this input
>> become the first and second parameters to the simulating halt decider
>> at Ĥ.qx.
>
> ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩
>
>> When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ decides that the simulation of its first parameter
>> on the input of its second parameter never halt it correctly
>> transitions to its own final state of Ĥ.qn.
>
> ⊢* Ĥ.qn
>
> None of which (other than your poor use of technical terms) has been in
> doubt for months.

--
Copyright 2021 Pete Olcott

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

Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction. [ Halting Problem is solved for Ĥ on ⟨Ĥ⟩ ]

<4NednYbjduF0Hon8nZ2dnUU7-QXNnZ2d@giganews.com>

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Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct_and_forms_no_contradiction._[_Halting_Problem_is_solv
ed for Ĥ on ⟨Ĥ⟩ ]
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References: <20210719214640.00000dfc@reddwarf.jmc>
<1ZednUt-qbrXlZf8nZ2dnUU7-QnNnZ2d@giganews.com> <87tuk52h0e.fsf@bsb.me.uk>
<zcadnTSOD5rtZ5f8nZ2dnUU7-T3NnZ2d@giganews.com> <877dh03l3c.fsf@bsb.me.uk>
<Z5adnd038KGXwJb8nZ2dnUU7-I_NnZ2d@giganews.com> <8735rn1qvj.fsf@bsb.me.uk>
<goydnfCCIYUWE5H8nZ2dnUU7-e_NnZ2d@giganews.com> <87eeb7z4d1.fsf@bsb.me.uk>
<0_Sdnb6Qe8XGOZH8nZ2dnUU7-U3NnZ2d@giganews.com> <87zgtslqpv.fsf@bsb.me.uk>
<4JOdnRS2SLR7MYz8nZ2dnUU7-YnNnZ2d@giganews.com> <87zgtoizgp.fsf@bsb.me.uk>
<Z6ednWt7SpIIv478nZ2dnUU7-WHNnZ2d@giganews.com> <87pmukiwr5.fsf@bsb.me.uk>
<5c6dnbn2gMMIsI78nZ2dnUU7-dnNnZ2d@giganews.com> <87eeb0iuo1.fsf@bsb.me.uk>
<346dnYhkWPUNQ478nZ2dnUU7-UPNnZ2d@giganews.com>
<Ja-dneAl3poPeI78nZ2dnUU7-UvNnZ2d@giganews.com> <87fsvggdxz.fsf@bsb.me.uk>
<zKudnTxkSrW1xYn8nZ2dnUU7-bnNnZ2d@giganews.com> <87r1eze981.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 11 Aug 2021 21:14:00 -0500
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 by: olcott - Thu, 12 Aug 2021 02:14 UTC

On 8/11/2021 8:35 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/11/2021 11:10 AM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/11/2021 9:28 AM, olcott wrote:
>
>>>>> H.q0 WM w ⊢* H.qn
>>>>> if M applied to W does not halt.
>>> if M applied to MW does not halt.
>>> (typo corrected)
>>>
>>>>>   becomes
>>>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>>> if Ĥ applied to ⟨Ĥ⟩ does not halt.
>>> Yes. After a lot a pressing (and I mean lots, over several years!) you
>>> eventually admitted that H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ does indeed transition to H.qn.
>>> You also keep telling us that Ĥ applied to ⟨Ĥ⟩ halts. That's why your H
>>> (and its associated Ĥ) are wrong but for some reason you can't see this.
>>> You will plainly state that H rejects the string "⟨Ĥ⟩ ⟨Ĥ⟩" which it
>>> should do if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt. And you
>>> will, time and time again, show us exactly how Ĥ applied to ⟨Ĥ⟩ halts.
>>> This should be the end of the matter, but apparently your stating facts
>>> that show that H and Ĥ are wrong does not mean you know that H and Ĥ are
>>> wrong. I don't think I know any way to make progress on this.
>>>
>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>> if M applied to wM does not halt
>>> ...
>>>>> if M applied to wM does not halt (see above for definition of
>>>>> M) means when the Turing machine of ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt.
>>> Of course. We all know that. Unless you are pulling a fast one. "the
>>> Turing machine of ⟨Ĥ⟩" is just Ĥ. Is there a reason you are not simply
>>> saying "when Ĥ applied to ⟨Ĥ⟩ does not halt"?
>
> I didn't expect an answer. You may be reserving this phrase as a
> get-out clause for later, so saying what you mean too early will block
> that escape route.
>
>>>>> Ĥ.q0 ⟨Ĥ⟩  ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>> Ĥ.qx correctly transitions to its final state when the Ĥ.qx acts as a
>>>>> UTM and simulates ⟨Ĥ⟩ ⟨Ĥ⟩ and determines that this input never halts.
>>> Yes, we know the ruse: the computation would not halt if it were not the
>>> computation that it is. You've been trying to pull off this trick ever
>>> since the infamous "it wouldn't halt if line 15 was commented out"
>>> admission. Your Ĥ, however, not being a UTM, has the property that
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>> when it should not. Ĥ.q0 ⟨Ĥ⟩ should eventually transition to qn only
>>> "if Ĥ applied to ⟨Ĥ⟩ does not halt" (or, as you rather suspiciously
>>> write "if the Turing machine of ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt").
>>
>> We can see that the above never halts.
>
> We can see that it halts. It's right there in a fact that has been
> undisputed for months:
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>

The above Ĥ halts only because Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decides that its
input never halts.

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

Thus the halting problem is solved for this input because the halting
problem only applies to inputs. It does not apply to computations that
are not inputs.

--
Copyright 2021 Pete Olcott

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

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