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devel / comp.theory / Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ deceitful bastard ]

SubjectAuthor
* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ keyolcott
+- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Python
+- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Richard Damon
`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
 `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
  `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
   `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |+* Refuting the Peter Linz Halting Problem Proof --- Version(10) [Malcolm McLean
    ||`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    || `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [Malcolm McLean
    ||  `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    ||   +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    ||   |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    ||   | `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    ||   |  `- Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    ||   `- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Richard Damon
    |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    | +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [Richard Damon
    | |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    | | `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    | |  `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    | |   `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    | |    `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    | |     `- Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    | `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |  `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   | `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |  `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |   `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |    `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |     +- Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |     `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |      +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |      |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |      | `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |      |  `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |      |   `- Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |      `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |       `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |        |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        | `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |        |  `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |   +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |        |   |`- Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |   `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |        |    +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |+- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Python
    |   |        |    |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |        |    | `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |  `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |        |    |   `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |    `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |        |    |     `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |      +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [Jeff Barnett
    |   |        |    |      |`- Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |      +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |        |    |      |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |      | `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |        |    |      |  `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |      |   +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [Jeff Barnett
    |   |        |    |      |   |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |      |   | `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |        |    |      |   |  `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |      |   |   `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |        |    |      |   |    `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |      |   |     `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |        |    |      |   |      `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |      |   |       +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |        |    |      |   |       |`- Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |      |   |       `- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Richard Damon
    |   |        |    |      |   `- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Richard Damon
    |   |        |    |      `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |        |    |       `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |        `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |        |    |         `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |          +- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Richard Damon
    |   |        |    |          `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |        |    |           `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |            +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |        |    |            |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |            | +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [Dennis Bush
    |   |        |    |            | |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |            | | +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [Python
    |   |        |    |            | | |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |            | | | +- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Python
    |   |        |    |            | | | `- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Richard Damon
    |   |        |    |            | | `- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Richard Damon
    |   |        |    |            | +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |        |    |            | |`- Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |            | `- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Richard Damon
    |   |        |    |            `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |        |    |             +- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Malcolm McLean
    |   |        |    |             `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |              `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |        |    |               `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |                `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |        |    |                 `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [Andy Walker
    `- Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse

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Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ deceitful bastard ]

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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
deceitful bastard ]
From: dbush.mo...@gmail.com (Dennis Bush)
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 by: Dennis Bush - Sun, 10 Apr 2022 22:05 UTC

On Sunday, April 10, 2022 at 5:56:49 PM UTC-4, olcott wrote:
> On 4/10/2022 4:49 PM, André G. Isaak wrote:
> > On 2022-04-10 15:00, olcott wrote:
> >> On 4/10/2022 3:15 PM, olcott wrote:
> >>> On 4/10/2022 3:07 PM, André G. Isaak wrote:
> >
> >>>> I'm trying to get you to write using correct and coherent notation.
> >>>> That's one of the things you'll need to be able to do if you ever
> >>>> hope to publish. That involves remembering to always include
> >>>> conditions and using the same terms in your 'equations' as in your
> >>>> text.
> >>>>
> >>>> Not sure how that makes me a 'deceitful bastard'.
> >>>>
> >>>> André
> >>>>
> >>>
> >>> THAT you pretended to not know what I mean by embedded_H so that you
> >>> could artificially contrive a fake basis for rebuttal when no actual
> >>> basis for rebuttal exists makes you a deceitful bastard.
> >>
> >> IT IS THE CASE THAT the correctly simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to
> >> embedded_H never reaches its own final state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩
> >> under any condition what-so-ever therefore ⟨Ĥ0⟩ ⟨Ĥ1⟩ is proved to
> >> specify a non-halting sequence of configurations.
> >>
> >> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
> >> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn
> >
> > This is now the third reply you've made to the same post.
> >
> > That post didn't make any arguments whatsoever about your claims. It
> > simply pointed out that you are misusing your notation and urged you to
> > correct it.
> >
> THE NOTATION IS A STIPULATIVE DEFINITION THUS DISAGREEMENT IS INCORRECT.

A stipulative definition? You mean like this one?

---
A solution of the halting problem is a Turing machine
H, which for any <M> and <w>, performs the computation

H.q0 ⟨M⟩⟨w⟩ ⊦* H.qy

if M applied to <w> halts, and

H.q0 ⟨M⟩⟨w⟩ ⊦* H.qn

if M applied to <w> does not halt.
---

It is incorrect to disagree with stipulative definitions.
https://en.wikipedia.org/wiki/Stipulative_definition

Disagreeing with a stipulative definition is like disagreeing with
arithmetic.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ key missing piece in dialogue ][ back door ]

<BwI4K.247179$%uX7.104852@fx38.iad>

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https://www.novabbs.com/devel/article-flat.php?id=29900&group=comp.theory#29900

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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
key missing piece in dialogue ][ back door ]
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Sun, 10 Apr 2022 22:05 UTC

On 4/10/22 5:54 PM, olcott wrote:
> On 4/10/2022 4:41 PM, Ben wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 4/10/2022 11:14 AM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 4/10/2022 10:22 AM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 4/9/2022 7:14 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> On 4/9/2022 5:25 PM, Ben Bacarisse wrote:
>>>>>>>>
>>>>>>>>>> You state that for the H you are championing
>>>>>>>>>>
>>>>>>>>>>         Ĥ.q0 <Ĥ> ⊦* Ĥ.qx <Ĥ> <Ĥ> ⊦* Ĥ.qn.
>>>>>>>>>>
>>>>>>>>>> To see why this H is not a halt decider, you must answer the two
>>>>>>>>>> questions you are studiously avoiding.  What state does H <Ĥ> <Ĥ>
>>>>>>>>>> transition to, and what string must be passed to H for H to
>>>>>>>>>> tell us
>>>>>>>>>> whether or not Ĥ applied to <Ĥ> halts or not?
>>>>>>>>
>>>>>>>>> Is the input to embedded_H non-halting? YES
>>>>>>>> The only plausible meaning for whether a string, s, is
>>>>>>>> non-halting is
>>>>>>>> whether UTM(s) is non-halting.  UTM(<Ĥ> <Ĥ>) halts (according to
>>>>>>>> you).
>>>>>>>
>>>>>>> embedded_H contains a fully functional UTM.
>>>>>> So what?  It can contain a fully functional chess program for all
>>>>>> anyone
>>>>>> cares.  What matters is what "is the input to embedded_H non-halting"
>>>>>> means.  The only sane meaning is whether or not UTM(<Ĥ> <Ĥ>)
>>>>>> halts, and
>>>>>> it does (according to you).
>>>>>>
>>>>>>> The correctly simulated input to embedded_H would never reach its
>>>>>>> own
>>>>>>> final state whether or not aborted by embedded_H.
>>>>>>
>>>>>> UTM(<Ĥ> <Ĥ>) halts (according to you).  Are you retracing that?
>>>>>> Or are
>>>>>> you now saying both at once?  We know you are quite happy to claim
>>>>>> flat-out contradictory facts at the same time.
>>>>>
>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn
>>>> No once cares about this math poem.  You have invented some names to
>>>> cloak your mistakes, but unless ⟨Ĥ0⟩=⟨Ĥ1⟩=⟨Ĥ2⟩=⟨Ĥ⟩ it's junk.  What
>>>> matters is that
>>>>     Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy ⊦* oo  if UTM(⟨Ĥ⟩ ⟨Ĥ⟩) halts, and
>>>>     Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn        if UTM(⟨Ĥ⟩ ⟨Ĥ⟩) does not halt.
>>>>
>>>>> The simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to embedded_H never reaches its final
>>>>> state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩ under any condition what-so-ever
>>>> UTM(⟨Ĥ⟩ ⟨Ĥ⟩) halts.  But what string must be passed to H for H to tell
>>>
>>> The above means this:
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>
>> That's funny!  You really have no idea what this notation means, do you?
>>
>>> embedded_H is a simulating halt decider that has a full UTM embedded
>>> within it. As soon as it sees that the pure UTM simulation of its
>>> input would never reach the final state of this input it aborts this
>>> simulation and rejects this non-halting input.
>>
>> So you had no business writing those two junk lines, did you?  Or do you
>> really think that they are in some way compatible with that last
>> paragraph?  Probably neither.  I really think you see it much like
>> poetry.  Meanings are supposed to be intuited from unusual, often
>> metaphorical, juxtapositions of symbols.
>>
>
> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn
>
> THE CORRECT SIMULATION OF THE INPUT TO EMBEDDED_H NEVER REACHES ITS
> FINAL STATE
>
>
> CORRECTLY SIMULATED INPUT TO H(P,P)
>
> _P()
> [00000956](01)  55              push ebp
> [00000957](02)  8bec            mov ebp,esp
> [00000959](03)  8b4508          mov eax,[ebp+08]
> [0000095c](01)  50              push eax
> [0000095d](03)  8b4d08          mov ecx,[ebp+08]
> [00000960](01)  51              push ecx
> [00000961](05)  e8c0feffff      call 00000826
>
> NEVER GET PAST HERE
>
> [00000966](03)  83c408          add esp,+08
> [00000969](02)  85c0            test eax,eax
> [0000096b](02)  7402            jz 0000096f
> [0000096d](02)  ebfe            jmp 0000096d
> [0000096f](01)  5d              pop ebp
>
> TO GET TO HERE
> [00000970](01)  c3              ret
> Size in bytes:(0027) [00000970]
>
>
>
Which means that you are ADMITTING that your H never answers the
question H(P,P) and thus fails to be a decider, let alone being a
correct halt decider.

Remember, if ONE copy of H(P,P) doesn't answer, NO copies of H(P,P) can
answer if H is a computation.

IF you want to claim otherwise, you need to show an actual Turing
Machine that two identical copies given two identical copies of an input
can generate different results.

Please Try. That might be you eternal torment to work on that problem.

If you can't, it just shows that you fail to have even basic
understanding of what a Turing Machine is.

FAIL.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ deceitful bastard ]

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<87bkxb9tc9.fsf@bsb.me.uk> <h4ydnXCGgtZONs3_nZ2dnUU7_8zNnZ2d@giganews.com>
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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Sun, 10 Apr 2022 22:17 UTC

On 4/10/2022 5:05 PM, Dennis Bush wrote:
> On Sunday, April 10, 2022 at 5:56:49 PM UTC-4, olcott wrote:
>> On 4/10/2022 4:49 PM, André G. Isaak wrote:
>>> On 2022-04-10 15:00, olcott wrote:
>>>> On 4/10/2022 3:15 PM, olcott wrote:
>>>>> On 4/10/2022 3:07 PM, André G. Isaak wrote:
>>>
>>>>>> I'm trying to get you to write using correct and coherent notation.
>>>>>> That's one of the things you'll need to be able to do if you ever
>>>>>> hope to publish. That involves remembering to always include
>>>>>> conditions and using the same terms in your 'equations' as in your
>>>>>> text.
>>>>>>
>>>>>> Not sure how that makes me a 'deceitful bastard'.
>>>>>>
>>>>>> André
>>>>>>
>>>>>
>>>>> THAT you pretended to not know what I mean by embedded_H so that you
>>>>> could artificially contrive a fake basis for rebuttal when no actual
>>>>> basis for rebuttal exists makes you a deceitful bastard.
>>>>
>>>> IT IS THE CASE THAT the correctly simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to
>>>> embedded_H never reaches its own final state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩
>>>> under any condition what-so-ever therefore ⟨Ĥ0⟩ ⟨Ĥ1⟩ is proved to
>>>> specify a non-halting sequence of configurations.
>>>>
>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn
>>>
>>> This is now the third reply you've made to the same post.
>>>
>>> That post didn't make any arguments whatsoever about your claims. It
>>> simply pointed out that you are misusing your notation and urged you to
>>> correct it.
>>>
>> THE NOTATION IS A STIPULATIVE DEFINITION THUS DISAGREEMENT IS INCORRECT.
>
> A stipulative definition? You mean like this one?
>
> ---
> A solution of the halting problem is a Turing machine
> H, which for any <M> and <w>, performs the computation
>
> H.q0 ⟨M⟩⟨w⟩ ⊦* H.qy
>
> if M applied to <w> halts, and
>
> H.q0 ⟨M⟩⟨w⟩ ⊦* H.qn
>
> if M applied to <w> does not halt.
> ---
>

I show that the above definition is based on the false assumption that
(in some cases) a decider must compute the mapping of non-inputs to its
final accept or reject state. Because this directly contradicts the
definition of a decider in computer science we can know that it is
incorrect.

Here is its correction.

> H.q0 ⟨M⟩⟨w⟩ ⊦* H.qy
>
> if UTM ⟨M⟩⟨w⟩ halts, and // Ben originally encoded this
>
> H.q0 ⟨M⟩⟨w⟩ ⊦* H.qn
>
> if UTM ⟨M⟩⟨w⟩ does not halt. // Ben originally encoded this

This get trickier in this case:
Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn

We are asking if the H above was replaced by a UTM would its input halt?

--
Copyright 2022 Pete Olcott

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

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ deceitful bastard ]

<wJI4K.52280$rWve.3960@fx01.iad>

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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Sun, 10 Apr 2022 22:19 UTC

On 4/10/22 5:56 PM, olcott wrote:
> On 4/10/2022 4:49 PM, André G. Isaak wrote:
>> On 2022-04-10 15:00, olcott wrote:
>>> On 4/10/2022 3:15 PM, olcott wrote:
>>>> On 4/10/2022 3:07 PM, André G. Isaak wrote:
>>
>>>>> I'm trying to get you to write using correct and coherent notation.
>>>>> That's one of the things you'll need to be able to do if you ever
>>>>> hope to publish. That involves remembering to always include
>>>>> conditions and using the same terms in your 'equations' as in your
>>>>> text.
>>>>>
>>>>> Not sure how that makes me a 'deceitful bastard'.
>>>>>
>>>>> André
>>>>>
>>>>
>>>> THAT you pretended to not know what I mean by embedded_H so that you
>>>> could artificially contrive a fake basis for rebuttal when no actual
>>>> basis for rebuttal exists makes you a deceitful bastard.
>>>
>>> IT IS THE CASE THAT the correctly simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to
>>> embedded_H never reaches its own final state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩
>>> under any condition what-so-ever therefore ⟨Ĥ0⟩ ⟨Ĥ1⟩ is proved to
>>> specify a non-halting sequence of configurations.
>>>
>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn
>>
>> This is now the third reply you've made to the same post.
>>
>> That post didn't make any arguments whatsoever about your claims. It
>> simply pointed out that you are misusing your notation and urged you
>> to correct it.
>>
>
> THE NOTATION IS A STIPULATIVE DEFINITION THUS DISAGREEMENT IS INCORRECT.
>

Nope. To be a Stipulative Definition, you need to actually write a
DEFINITION of what you mean, not just use some sloppy notation that
isn't well defined.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ deceitful bastard ]

<0TI4K.571655$7F2.2835@fx12.iad>

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deceitful bastard ]
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 by: Richard Damon - Sun, 10 Apr 2022 22:29 UTC

On 4/10/22 6:17 PM, olcott wrote:
> On 4/10/2022 5:05 PM, Dennis Bush wrote:
>> On Sunday, April 10, 2022 at 5:56:49 PM UTC-4, olcott wrote:
>>> On 4/10/2022 4:49 PM, André G. Isaak wrote:
>>>> On 2022-04-10 15:00, olcott wrote:
>>>>> On 4/10/2022 3:15 PM, olcott wrote:
>>>>>> On 4/10/2022 3:07 PM, André G. Isaak wrote:
>>>>
>>>>>>> I'm trying to get you to write using correct and coherent notation.
>>>>>>> That's one of the things you'll need to be able to do if you ever
>>>>>>> hope to publish. That involves remembering to always include
>>>>>>> conditions and using the same terms in your 'equations' as in your
>>>>>>> text.
>>>>>>>
>>>>>>> Not sure how that makes me a 'deceitful bastard'.
>>>>>>>
>>>>>>> André
>>>>>>>
>>>>>>
>>>>>> THAT you pretended to not know what I mean by embedded_H so that you
>>>>>> could artificially contrive a fake basis for rebuttal when no actual
>>>>>> basis for rebuttal exists makes you a deceitful bastard.
>>>>>
>>>>> IT IS THE CASE THAT the correctly simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to
>>>>> embedded_H never reaches its own final state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩
>>>>> under any condition what-so-ever therefore ⟨Ĥ0⟩ ⟨Ĥ1⟩ is proved to
>>>>> specify a non-halting sequence of configurations.
>>>>>
>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn
>>>>
>>>> This is now the third reply you've made to the same post.
>>>>
>>>> That post didn't make any arguments whatsoever about your claims. It
>>>> simply pointed out that you are misusing your notation and urged you to
>>>> correct it.
>>>>
>>> THE NOTATION IS A STIPULATIVE DEFINITION THUS DISAGREEMENT IS INCORRECT.
>>
>> A stipulative definition?  You mean like this one?
>>
>> ---
>> A solution of the halting problem is a Turing machine
>> H, which for any <M> and <w>, performs the computation
>>
>>      H.q0 ⟨M⟩⟨w⟩ ⊦* H.qy
>>
>> if M applied to <w> halts, and
>>
>>      H.q0 ⟨M⟩⟨w⟩ ⊦* H.qn
>>
>> if M applied to <w> does not halt.
>> ---
>>
>
> I show that the above definition is based on the false assumption that
> (in some cases) a decider must compute the mapping of non-inputs to its
> final accept or reject state. Because this directly contradicts the
> definition of a decider in computer science we can know that it is
> incorrect.
>
> Here is its correction.
>
> >      H.q0 ⟨M⟩⟨w⟩ ⊦* H.qy
> >
> > if UTM ⟨M⟩⟨w⟩ halts, and      // Ben originally encoded this
> >
> >      H.q0 ⟨M⟩⟨w⟩ ⊦* H.qn
> >
> > if UTM ⟨M⟩⟨w⟩ does not halt. // Ben originally encoded this
>
> This get trickier in this case:
> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn
>
> We are asking if the H above was replaced by a UTM would its input halt?
>
>

WHO is asking about "if the H above was replace by a UTM"?

That isn't part of the definition of H.

NOWHERE does it ask about replacing itself with a UTM.

That is just part of the insanity of your POOP.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ deceitful bastard ]

<t2vm35$rth$1@dont-email.me>

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https://www.novabbs.com/devel/article-flat.php?id=29905&group=comp.theory#29905

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Path: i2pn2.org!i2pn.org!aioe.org!eternal-september.org!reader02.eternal-september.org!.POSTED!not-for-mail
From: agis...@gm.invalid (André G. Isaak)
Newsgroups: comp.theory
Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
deceitful bastard ]
Date: Sun, 10 Apr 2022 16:35:16 -0600
Organization: Christians and Atheists United Against Creeping Agnosticism
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 by: André G. Isaak - Sun, 10 Apr 2022 22:35 UTC

On 2022-04-10 15:56, olcott wrote:
> On 4/10/2022 4:49 PM, André G. Isaak wrote:
>> On 2022-04-10 15:00, olcott wrote:
>>> On 4/10/2022 3:15 PM, olcott wrote:
>>>> On 4/10/2022 3:07 PM, André G. Isaak wrote:
>>
>>>>> I'm trying to get you to write using correct and coherent notation.
>>>>> That's one of the things you'll need to be able to do if you ever
>>>>> hope to publish. That involves remembering to always include
>>>>> conditions and using the same terms in your 'equations' as in your
>>>>> text.
>>>>>
>>>>> Not sure how that makes me a 'deceitful bastard'.
>>>>>
>>>>> André
>>>>>
>>>>
>>>> THAT you pretended to not know what I mean by embedded_H so that you
>>>> could artificially contrive a fake basis for rebuttal when no actual
>>>> basis for rebuttal exists makes you a deceitful bastard.
>>>
>>> IT IS THE CASE THAT the correctly simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to
>>> embedded_H never reaches its own final state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩
>>> under any condition what-so-ever therefore ⟨Ĥ0⟩ ⟨Ĥ1⟩ is proved to
>>> specify a non-halting sequence of configurations.
>>>
>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn
>>
>> This is now the third reply you've made to the same post.
>>
>> That post didn't make any arguments whatsoever about your claims. It
>> simply pointed out that you are misusing your notation and urged you
>> to correct it.
>>
>
> THE NOTATION IS A STIPULATIVE DEFINITION THUS DISAGREEMENT IS INCORRECT.

If the notation is junk, then the definition is also junk.

That's like "stipulating" that

+×yz÷² = ±z+³

It's meaningless because the notation is meaningless, much like your
notation above.

This is meaningless:

Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy // what's the condition?
Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qn // what's the condition?

With no conditions specified, the above is just nonsense.

André

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

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ deceitful bastard ]

<2fmdne-EPaRPwc7_nZ2dnUU7_8zNnZ2d@giganews.com>

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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
deceitful bastard ]
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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Sun, 10 Apr 2022 22:40 UTC

On 4/10/2022 5:35 PM, André G. Isaak wrote:
> On 2022-04-10 15:56, olcott wrote:
>> On 4/10/2022 4:49 PM, André G. Isaak wrote:
>>> On 2022-04-10 15:00, olcott wrote:
>>>> On 4/10/2022 3:15 PM, olcott wrote:
>>>>> On 4/10/2022 3:07 PM, André G. Isaak wrote:
>>>
>>>>>> I'm trying to get you to write using correct and coherent
>>>>>> notation. That's one of the things you'll need to be able to do if
>>>>>> you ever hope to publish. That involves remembering to always
>>>>>> include conditions and using the same terms in your 'equations' as
>>>>>> in your text.
>>>>>>
>>>>>> Not sure how that makes me a 'deceitful bastard'.
>>>>>>
>>>>>> André
>>>>>>
>>>>>
>>>>> THAT you pretended to not know what I mean by embedded_H so that
>>>>> you could artificially contrive a fake basis for rebuttal when no
>>>>> actual basis for rebuttal exists makes you a deceitful bastard.
>>>>
>>>> IT IS THE CASE THAT the correctly simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to
>>>> embedded_H never reaches its own final state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩
>>>> under any condition what-so-ever therefore ⟨Ĥ0⟩ ⟨Ĥ1⟩ is proved to
>>>> specify a non-halting sequence of configurations.
>>>>
>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn
>>>
>>> This is now the third reply you've made to the same post.
>>>
>>> That post didn't make any arguments whatsoever about your claims. It
>>> simply pointed out that you are misusing your notation and urged you
>>> to correct it.
>>>
>>
>> THE NOTATION IS A STIPULATIVE DEFINITION THUS DISAGREEMENT IS INCORRECT.
>
> If the notation is junk, then the definition is also junk.
>
> That's like "stipulating" that
>
> +×yz÷² = ±z+³
>
> It's meaningless because the notation is meaningless, much like your
> notation above.
>
> This is meaningless:
>
> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy   // what's the condition?
> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qn   // what's the condition?
>
> With no conditions specified, the above is just nonsense.
>
> André
>

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would reach its final
state.

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never reach its
final state.

--
Copyright 2022 Pete Olcott

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

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ deceitful bastard ]

<171740c3-3f0d-4da0-98bd-9c296ebaa2f2n@googlegroups.com>

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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
deceitful bastard ]
From: dbush.mo...@gmail.com (Dennis Bush)
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 by: Dennis Bush - Sun, 10 Apr 2022 22:47 UTC

On Sunday, April 10, 2022 at 6:18:11 PM UTC-4, olcott wrote:
> On 4/10/2022 5:05 PM, Dennis Bush wrote:
> > On Sunday, April 10, 2022 at 5:56:49 PM UTC-4, olcott wrote:
> >> On 4/10/2022 4:49 PM, André G. Isaak wrote:
> >>> On 2022-04-10 15:00, olcott wrote:
> >>>> On 4/10/2022 3:15 PM, olcott wrote:
> >>>>> On 4/10/2022 3:07 PM, André G. Isaak wrote:
> >>>
> >>>>>> I'm trying to get you to write using correct and coherent notation..
> >>>>>> That's one of the things you'll need to be able to do if you ever
> >>>>>> hope to publish. That involves remembering to always include
> >>>>>> conditions and using the same terms in your 'equations' as in your
> >>>>>> text.
> >>>>>>
> >>>>>> Not sure how that makes me a 'deceitful bastard'.
> >>>>>>
> >>>>>> André
> >>>>>>
> >>>>>
> >>>>> THAT you pretended to not know what I mean by embedded_H so that you
> >>>>> could artificially contrive a fake basis for rebuttal when no actual
> >>>>> basis for rebuttal exists makes you a deceitful bastard.
> >>>>
> >>>> IT IS THE CASE THAT the correctly simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to
> >>>> embedded_H never reaches its own final state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩
> >>>> under any condition what-so-ever therefore ⟨Ĥ0⟩ ⟨Ĥ1⟩ is proved to
> >>>> specify a non-halting sequence of configurations.
> >>>>
> >>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
> >>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn
> >>>
> >>> This is now the third reply you've made to the same post.
> >>>
> >>> That post didn't make any arguments whatsoever about your claims. It
> >>> simply pointed out that you are misusing your notation and urged you to
> >>> correct it.
> >>>
> >> THE NOTATION IS A STIPULATIVE DEFINITION THUS DISAGREEMENT IS INCORRECT.
> >
> > A stipulative definition? You mean like this one?
> >
> > ---
> > A solution of the halting problem is a Turing machine
> > H, which for any <M> and <w>, performs the computation
> >
> > H.q0 ⟨M⟩⟨w⟩ ⊦* H.qy
> >
> > if M applied to <w> halts, and
> >
> > H.q0 ⟨M⟩⟨w⟩ ⊦* H.qn
> >
> > if M applied to <w> does not halt.
> > ---
> >
> I show that the above definition is based on the false assumption that
> (in some cases) a decider must compute the mapping of non-inputs to its
> final accept or reject state. Because this directly contradicts the
> definition of a decider in computer science we can know that it is
> incorrect.

As you yourself said:

It is incorrect to disagree with stipulative definitions.
https://en.wikipedia.org/wiki/Stipulative_definition

Disagreeing with a stipulative definition is like disagreeing with
arithmetic.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ deceitful bastard ]

<3OqdnRlHwJWJ_c7_nZ2dnUU7_8zNnZ2d@giganews.com>

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 by: olcott - Sun, 10 Apr 2022 22:54 UTC

On 4/10/2022 5:47 PM, Dennis Bush wrote:
> On Sunday, April 10, 2022 at 6:18:11 PM UTC-4, olcott wrote:
>> On 4/10/2022 5:05 PM, Dennis Bush wrote:
>>> On Sunday, April 10, 2022 at 5:56:49 PM UTC-4, olcott wrote:
>>>> On 4/10/2022 4:49 PM, André G. Isaak wrote:
>>>>> On 2022-04-10 15:00, olcott wrote:
>>>>>> On 4/10/2022 3:15 PM, olcott wrote:
>>>>>>> On 4/10/2022 3:07 PM, André G. Isaak wrote:
>>>>>
>>>>>>>> I'm trying to get you to write using correct and coherent notation.
>>>>>>>> That's one of the things you'll need to be able to do if you ever
>>>>>>>> hope to publish. That involves remembering to always include
>>>>>>>> conditions and using the same terms in your 'equations' as in your
>>>>>>>> text.
>>>>>>>>
>>>>>>>> Not sure how that makes me a 'deceitful bastard'.
>>>>>>>>
>>>>>>>> André
>>>>>>>>
>>>>>>>
>>>>>>> THAT you pretended to not know what I mean by embedded_H so that you
>>>>>>> could artificially contrive a fake basis for rebuttal when no actual
>>>>>>> basis for rebuttal exists makes you a deceitful bastard.
>>>>>>
>>>>>> IT IS THE CASE THAT the correctly simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to
>>>>>> embedded_H never reaches its own final state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩
>>>>>> under any condition what-so-ever therefore ⟨Ĥ0⟩ ⟨Ĥ1⟩ is proved to
>>>>>> specify a non-halting sequence of configurations.
>>>>>>
>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn
>>>>>
>>>>> This is now the third reply you've made to the same post.
>>>>>
>>>>> That post didn't make any arguments whatsoever about your claims. It
>>>>> simply pointed out that you are misusing your notation and urged you to
>>>>> correct it.
>>>>>
>>>> THE NOTATION IS A STIPULATIVE DEFINITION THUS DISAGREEMENT IS INCORRECT.
>>>
>>> A stipulative definition? You mean like this one?
>>>
>>> ---
>>> A solution of the halting problem is a Turing machine
>>> H, which for any <M> and <w>, performs the computation
>>>
>>> H.q0 ⟨M⟩⟨w⟩ ⊦* H.qy
>>>
>>> if M applied to <w> halts, and
>>>
>>> H.q0 ⟨M⟩⟨w⟩ ⊦* H.qn
>>>
>>> if M applied to <w> does not halt.
>>> ---
>>>
>> I show that the above definition is based on the false assumption that
>> (in some cases) a decider must compute the mapping of non-inputs to its
>> final accept or reject state. Because this directly contradicts the
>> definition of a decider in computer science we can know that it is
>> incorrect.
>
> As you yourself said:
>
> It is incorrect to disagree with stipulative definitions.
> https://en.wikipedia.org/wiki/Stipulative_definition
>
> Disagreeing with a stipulative definition is like disagreeing with
> arithmetic.

Until we drop into the coherence notion of truth, then when a set of
definitions results in incoherence one or more must be discarded until
coherence is restored.

When a textbook author disagrees with a fundamental principle of
computer science (how deciders work) the textbook author loses.

It is the case that a halt decider computes the mapping from its inputs
to its final accept or reject state on the basis of the actual behavior
specified by these inputs.

--
Copyright 2022 Pete Olcott

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

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ deceitful bastard ]

<t2vngt$526$1@dont-email.me>

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From: agis...@gm.invalid (André G. Isaak)
Newsgroups: comp.theory
Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
deceitful bastard ]
Date: Sun, 10 Apr 2022 16:59:40 -0600
Organization: Christians and Atheists United Against Creeping Agnosticism
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 by: André G. Isaak - Sun, 10 Apr 2022 22:59 UTC

On 2022-04-10 16:40, olcott wrote:
> On 4/10/2022 5:35 PM, André G. Isaak wrote:
>> On 2022-04-10 15:56, olcott wrote:
>>> On 4/10/2022 4:49 PM, André G. Isaak wrote:
>>>> On 2022-04-10 15:00, olcott wrote:
>>>>> On 4/10/2022 3:15 PM, olcott wrote:
>>>>>> On 4/10/2022 3:07 PM, André G. Isaak wrote:
>>>>
>>>>>>> I'm trying to get you to write using correct and coherent
>>>>>>> notation. That's one of the things you'll need to be able to do
>>>>>>> if you ever hope to publish. That involves remembering to always
>>>>>>> include conditions and using the same terms in your 'equations'
>>>>>>> as in your text.
>>>>>>>
>>>>>>> Not sure how that makes me a 'deceitful bastard'.
>>>>>>>
>>>>>>> André
>>>>>>>
>>>>>>
>>>>>> THAT you pretended to not know what I mean by embedded_H so that
>>>>>> you could artificially contrive a fake basis for rebuttal when no
>>>>>> actual basis for rebuttal exists makes you a deceitful bastard.
>>>>>
>>>>> IT IS THE CASE THAT the correctly simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to
>>>>> embedded_H never reaches its own final state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩
>>>>> under any condition what-so-ever therefore ⟨Ĥ0⟩ ⟨Ĥ1⟩ is proved to
>>>>> specify a non-halting sequence of configurations.
>>>>>
>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn
>>>>
>>>> This is now the third reply you've made to the same post.
>>>>
>>>> That post didn't make any arguments whatsoever about your claims. It
>>>> simply pointed out that you are misusing your notation and urged you
>>>> to correct it.
>>>>
>>>
>>> THE NOTATION IS A STIPULATIVE DEFINITION THUS DISAGREEMENT IS INCORRECT.
>>
>> If the notation is junk, then the definition is also junk.
>>
>> That's like "stipulating" that
>>
>> +×yz÷² = ±z+³
>>
>> It's meaningless because the notation is meaningless, much like your
>> notation above.
>>
>> This is meaningless:
>>
>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy   // what's the condition?
>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qn   // what's the condition?
>>
>> With no conditions specified, the above is just nonsense.
>>
>> André
>>
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
> If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would reach its final
> state.
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
> If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never reach its
> final state.

This is still nonsense.

For starters, you need to write either

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by H would reach its final
state.

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by H would never reach its
final state.

OR

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would reach its final
state.

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never reach its
final state.

Either use H consistently or use embedded_H consistently. You can't mix
and match

Second, H/embedded_H are supposedly simulating halt deciders, not "pure
simulators", so there can be no "pure simulation" by H/embedded_H
anymore than there can be a "pure simulation" by Adobe Photoshop.

You could say "if a pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ would never reach its
final state", but then, of course, you are asking about something which
isn't an input which according to you is disallowed.

André

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

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ deceitful bastard ]

<NkJ4K.692667$oF2.34561@fx10.iad>

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 by: Richard Damon - Sun, 10 Apr 2022 23:01 UTC

On 4/10/22 6:40 PM, olcott wrote:
> On 4/10/2022 5:35 PM, André G. Isaak wrote:
>> On 2022-04-10 15:56, olcott wrote:
>>> On 4/10/2022 4:49 PM, André G. Isaak wrote:
>>>> On 2022-04-10 15:00, olcott wrote:
>>>>> On 4/10/2022 3:15 PM, olcott wrote:
>>>>>> On 4/10/2022 3:07 PM, André G. Isaak wrote:
>>>>
>>>>>>> I'm trying to get you to write using correct and coherent
>>>>>>> notation. That's one of the things you'll need to be able to do
>>>>>>> if you ever hope to publish. That involves remembering to always
>>>>>>> include conditions and using the same terms in your 'equations'
>>>>>>> as in your text.
>>>>>>>
>>>>>>> Not sure how that makes me a 'deceitful bastard'.
>>>>>>>
>>>>>>> André
>>>>>>>
>>>>>>
>>>>>> THAT you pretended to not know what I mean by embedded_H so that
>>>>>> you could artificially contrive a fake basis for rebuttal when no
>>>>>> actual basis for rebuttal exists makes you a deceitful bastard.
>>>>>
>>>>> IT IS THE CASE THAT the correctly simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to
>>>>> embedded_H never reaches its own final state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩
>>>>> under any condition what-so-ever therefore ⟨Ĥ0⟩ ⟨Ĥ1⟩ is proved to
>>>>> specify a non-halting sequence of configurations.
>>>>>
>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn
>>>>
>>>> This is now the third reply you've made to the same post.
>>>>
>>>> That post didn't make any arguments whatsoever about your claims. It
>>>> simply pointed out that you are misusing your notation and urged you
>>>> to correct it.
>>>>
>>>
>>> THE NOTATION IS A STIPULATIVE DEFINITION THUS DISAGREEMENT IS INCORRECT.
>>
>> If the notation is junk, then the definition is also junk.
>>
>> That's like "stipulating" that
>>
>> +×yz÷² = ±z+³
>>
>> It's meaningless because the notation is meaningless, much like your
>> notation above.
>>
>> This is meaningless:
>>
>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy   // what's the condition?
>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qn   // what's the condition?
>>
>> With no conditions specified, the above is just nonsense.
>>
>> André
>>
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
> If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would reach its final
> state.

Nope, ERROR!! MISTAKE!! INCORRECT!! LIE!

The DEFINTION is "if H^ applied to <H^> Halts"

>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
> If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never reach its
> final state.
>
>

Nope, ERROR!! MISTAKE!! INCORRECT!! LIE!

The DEFINTION is "if H^ applied to <H^> Never Halts"

Altering the definition means you are no longer doing the Halting
Problem but just your POOP, which no one cares about.

Yes, you can change the "H^ applied to <H^>" to "UTM applied <H^> <H^>"

But to change it to "pure simulation of <H^> <H^> bu embedded_H" can
only be equivalent if embedded_H is DEFINED to be EQUIVALENT to a UTM,
which mean that it can NOT abort its simulation (DEFINITION)

This means the second case can NEVER be activated, as BY DEFINITION the
UTM simulation of a non-halting input never halts, so it can't 'go to H.Qn'

THIS IS ONE OF YOUR FUNDAMENTAL ERRORS.

FAIL.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ deceitful bastard ]

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 by: olcott - Sun, 10 Apr 2022 23:08 UTC

On 4/10/2022 5:59 PM, André G. Isaak wrote:
> On 2022-04-10 16:40, olcott wrote:
>> On 4/10/2022 5:35 PM, André G. Isaak wrote:
>>> On 2022-04-10 15:56, olcott wrote:
>>>> On 4/10/2022 4:49 PM, André G. Isaak wrote:
>>>>> On 2022-04-10 15:00, olcott wrote:
>>>>>> On 4/10/2022 3:15 PM, olcott wrote:
>>>>>>> On 4/10/2022 3:07 PM, André G. Isaak wrote:
>>>>>
>>>>>>>> I'm trying to get you to write using correct and coherent
>>>>>>>> notation. That's one of the things you'll need to be able to do
>>>>>>>> if you ever hope to publish. That involves remembering to always
>>>>>>>> include conditions and using the same terms in your 'equations'
>>>>>>>> as in your text.
>>>>>>>>
>>>>>>>> Not sure how that makes me a 'deceitful bastard'.
>>>>>>>>
>>>>>>>> André
>>>>>>>>
>>>>>>>
>>>>>>> THAT you pretended to not know what I mean by embedded_H so that
>>>>>>> you could artificially contrive a fake basis for rebuttal when no
>>>>>>> actual basis for rebuttal exists makes you a deceitful bastard.
>>>>>>
>>>>>> IT IS THE CASE THAT the correctly simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to
>>>>>> embedded_H never reaches its own final state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩
>>>>>> under any condition what-so-ever therefore ⟨Ĥ0⟩ ⟨Ĥ1⟩ is proved to
>>>>>> specify a non-halting sequence of configurations.
>>>>>>
>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn
>>>>>
>>>>> This is now the third reply you've made to the same post.
>>>>>
>>>>> That post didn't make any arguments whatsoever about your claims.
>>>>> It simply pointed out that you are misusing your notation and urged
>>>>> you to correct it.
>>>>>
>>>>
>>>> THE NOTATION IS A STIPULATIVE DEFINITION THUS DISAGREEMENT IS
>>>> INCORRECT.
>>>
>>> If the notation is junk, then the definition is also junk.
>>>
>>> That's like "stipulating" that
>>>
>>> +×yz÷² = ±z+³
>>>
>>> It's meaningless because the notation is meaningless, much like your
>>> notation above.
>>>
>>> This is meaningless:
>>>
>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy   // what's the condition?
>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qn   // what's the condition?
>>>
>>> With no conditions specified, the above is just nonsense.
>>>
>>> André
>>>
>>
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
>> If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would reach its final
>> state.
>>
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>> If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never reach its
>> final state.
>
>
> This is still nonsense.
>
Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would reach its
own final state.

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would never reach
its own final state.

This means that H pretends that it is only a UTM to see what its
simulated input would do in this case. If it would never reach its own
final state then H correctly rejects this input.

I don't see why this is so hard for people.
It is like I say that a dog is an animal and most people disagree on the
basis that they believe a dog is an office building.

--
Copyright 2022 Pete Olcott

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

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ deceitful bastard ]

<zrJ4K.417632$iK66.378416@fx46.iad>

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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
deceitful bastard ]
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Sun, 10 Apr 2022 23:08 UTC

On 4/10/22 6:54 PM, olcott wrote:
> On 4/10/2022 5:47 PM, Dennis Bush wrote:
>> On Sunday, April 10, 2022 at 6:18:11 PM UTC-4, olcott wrote:
>>> On 4/10/2022 5:05 PM, Dennis Bush wrote:
>>>> On Sunday, April 10, 2022 at 5:56:49 PM UTC-4, olcott wrote:
>>>>> On 4/10/2022 4:49 PM, André G. Isaak wrote:
>>>>>> On 2022-04-10 15:00, olcott wrote:
>>>>>>> On 4/10/2022 3:15 PM, olcott wrote:
>>>>>>>> On 4/10/2022 3:07 PM, André G. Isaak wrote:
>>>>>>
>>>>>>>>> I'm trying to get you to write using correct and coherent
>>>>>>>>> notation.
>>>>>>>>> That's one of the things you'll need to be able to do if you ever
>>>>>>>>> hope to publish. That involves remembering to always include
>>>>>>>>> conditions and using the same terms in your 'equations' as in your
>>>>>>>>> text.
>>>>>>>>>
>>>>>>>>> Not sure how that makes me a 'deceitful bastard'.
>>>>>>>>>
>>>>>>>>> André
>>>>>>>>>
>>>>>>>>
>>>>>>>> THAT you pretended to not know what I mean by embedded_H so that
>>>>>>>> you
>>>>>>>> could artificially contrive a fake basis for rebuttal when no
>>>>>>>> actual
>>>>>>>> basis for rebuttal exists makes you a deceitful bastard.
>>>>>>>
>>>>>>> IT IS THE CASE THAT the correctly simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to
>>>>>>> embedded_H never reaches its own final state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩
>>>>>>> under any condition what-so-ever therefore ⟨Ĥ0⟩ ⟨Ĥ1⟩ is proved to
>>>>>>> specify a non-halting sequence of configurations.
>>>>>>>
>>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
>>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn
>>>>>>
>>>>>> This is now the third reply you've made to the same post.
>>>>>>
>>>>>> That post didn't make any arguments whatsoever about your claims. It
>>>>>> simply pointed out that you are misusing your notation and urged
>>>>>> you to
>>>>>> correct it.
>>>>>>
>>>>> THE NOTATION IS A STIPULATIVE DEFINITION THUS DISAGREEMENT IS
>>>>> INCORRECT.
>>>>
>>>> A stipulative definition? You mean like this one?
>>>>
>>>> ---
>>>> A solution of the halting problem is a Turing machine
>>>> H, which for any <M> and <w>, performs the computation
>>>>
>>>> H.q0 ⟨M⟩⟨w⟩ ⊦* H.qy
>>>>
>>>> if M applied to <w> halts, and
>>>>
>>>> H.q0 ⟨M⟩⟨w⟩ ⊦* H.qn
>>>>
>>>> if M applied to <w> does not halt.
>>>> ---
>>>>
>>> I show that the above definition is based on the false assumption that
>>> (in some cases) a decider must compute the mapping of non-inputs to its
>>> final accept or reject state. Because this directly contradicts the
>>> definition of a decider in computer science we can know that it is
>>> incorrect.
>>
>> As you yourself said:
>>
>> It is incorrect to disagree with stipulative definitions.
>> https://en.wikipedia.org/wiki/Stipulative_definition
>>
>> Disagreeing with a stipulative definition is like disagreeing with
>> arithmetic.
>
> Until we drop into the coherence notion of truth, then when a set of
> definitions results in incoherence one or more must be discarded until
> coherence is restored.

Nope. FAIL.

If you can show actual incoherence, then you can show that the system as
defined is in error, and it becomes no longer a useful system,

This is what happened to Naive Set Theory when the set of all sets that
don't contain themselves showed a problem in it.

You HAVEN'T shown such an incoherence in Computation Theory, only that
you don't actually understand what Computation Theory is.

>
> When a textbook author disagrees with a fundamental principle of
> computer science (how deciders work) the textbook author loses.

Nope, you have just shown that YOU don't understand how deciders work.

You have shown ZERO understanding of the basics of the field, YOUR
claims that the textbook author is wrong is what loses.

>
> It is the case that a halt decider computes the mapping from its inputs
> to its final accept or reject state on the basis of the actual behavior
> specified by these inputs.
>

Just shows that you fail to understand even the basics of computer science.

H IS defined to map the input (in this case <H^> <H^>) on the bases of
the actual behavior specified by these inputs (in this case the behavior
of H^ applied to <H^>)

That fact that you can't comprehend this is YOUR problem, not the system.

FAIL.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ deceitful bastard ]

<t2vod6$9pl$1@dont-email.me>

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https://www.novabbs.com/devel/article-flat.php?id=29914&group=comp.theory#29914

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From: agis...@gm.invalid (André G. Isaak)
Newsgroups: comp.theory
Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
deceitful bastard ]
Date: Sun, 10 Apr 2022 17:14:45 -0600
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 by: André G. Isaak - Sun, 10 Apr 2022 23:14 UTC

On 2022-04-10 17:08, olcott wrote:
> On 4/10/2022 5:59 PM, André G. Isaak wrote:
>> On 2022-04-10 16:40, olcott wrote:
>>> On 4/10/2022 5:35 PM, André G. Isaak wrote:
>>>> On 2022-04-10 15:56, olcott wrote:
>>>>> On 4/10/2022 4:49 PM, André G. Isaak wrote:
>>>>>> On 2022-04-10 15:00, olcott wrote:
>>>>>>> On 4/10/2022 3:15 PM, olcott wrote:
>>>>>>>> On 4/10/2022 3:07 PM, André G. Isaak wrote:
>>>>>>
>>>>>>>>> I'm trying to get you to write using correct and coherent
>>>>>>>>> notation. That's one of the things you'll need to be able to do
>>>>>>>>> if you ever hope to publish. That involves remembering to
>>>>>>>>> always include conditions and using the same terms in your
>>>>>>>>> 'equations' as in your text.
>>>>>>>>>
>>>>>>>>> Not sure how that makes me a 'deceitful bastard'.
>>>>>>>>>
>>>>>>>>> André
>>>>>>>>>
>>>>>>>>
>>>>>>>> THAT you pretended to not know what I mean by embedded_H so that
>>>>>>>> you could artificially contrive a fake basis for rebuttal when
>>>>>>>> no actual basis for rebuttal exists makes you a deceitful bastard.
>>>>>>>
>>>>>>> IT IS THE CASE THAT the correctly simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to
>>>>>>> embedded_H never reaches its own final state of ⟨Ĥ0.qy⟩ or
>>>>>>> ⟨Ĥ0.qn⟩ under any condition what-so-ever therefore ⟨Ĥ0⟩ ⟨Ĥ1⟩ is
>>>>>>> proved to specify a non-halting sequence of configurations.
>>>>>>>
>>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
>>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn
>>>>>>
>>>>>> This is now the third reply you've made to the same post.
>>>>>>
>>>>>> That post didn't make any arguments whatsoever about your claims.
>>>>>> It simply pointed out that you are misusing your notation and
>>>>>> urged you to correct it.
>>>>>>
>>>>>
>>>>> THE NOTATION IS A STIPULATIVE DEFINITION THUS DISAGREEMENT IS
>>>>> INCORRECT.
>>>>
>>>> If the notation is junk, then the definition is also junk.
>>>>
>>>> That's like "stipulating" that
>>>>
>>>> +×yz÷² = ±z+³
>>>>
>>>> It's meaningless because the notation is meaningless, much like your
>>>> notation above.
>>>>
>>>> This is meaningless:
>>>>
>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy   // what's the condition?
>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qn   // what's the condition?
>>>>
>>>> With no conditions specified, the above is just nonsense.
>>>>
>>>> André
>>>>
>>>
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
>>> If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would reach its final
>>> state.
>>>
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>> If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never reach its
>>> final state.
>>
>>
>> This is still nonsense.
>>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
> If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would reach its
> own final state.
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
> If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would never reach
> its own final state.

And again you're still being inconsistent. You can either use H or use
embedded_H, but you can't mix the two.

> This means that H pretends that it is only a UTM to see what its
> simulated input would do in this case. If it would never reach its own
> final state then H correctly rejects this input.

A Turing Machine cannot "pretend" to be some different Turing Machine.

André

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

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ deceitful bastard ]

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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Sun, 10 Apr 2022 23:20 UTC

On 4/10/2022 6:14 PM, André G. Isaak wrote:
> On 2022-04-10 17:08, olcott wrote:
>> On 4/10/2022 5:59 PM, André G. Isaak wrote:
>>> On 2022-04-10 16:40, olcott wrote:
>>>> On 4/10/2022 5:35 PM, André G. Isaak wrote:
>>>>> On 2022-04-10 15:56, olcott wrote:
>>>>>> On 4/10/2022 4:49 PM, André G. Isaak wrote:
>>>>>>> On 2022-04-10 15:00, olcott wrote:
>>>>>>>> On 4/10/2022 3:15 PM, olcott wrote:
>>>>>>>>> On 4/10/2022 3:07 PM, André G. Isaak wrote:
>>>>>>>
>>>>>>>>>> I'm trying to get you to write using correct and coherent
>>>>>>>>>> notation. That's one of the things you'll need to be able to
>>>>>>>>>> do if you ever hope to publish. That involves remembering to
>>>>>>>>>> always include conditions and using the same terms in your
>>>>>>>>>> 'equations' as in your text.
>>>>>>>>>>
>>>>>>>>>> Not sure how that makes me a 'deceitful bastard'.
>>>>>>>>>>
>>>>>>>>>> André
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> THAT you pretended to not know what I mean by embedded_H so
>>>>>>>>> that you could artificially contrive a fake basis for rebuttal
>>>>>>>>> when no actual basis for rebuttal exists makes you a deceitful
>>>>>>>>> bastard.
>>>>>>>>
>>>>>>>> IT IS THE CASE THAT the correctly simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to
>>>>>>>> embedded_H never reaches its own final state of ⟨Ĥ0.qy⟩ or
>>>>>>>> ⟨Ĥ0.qn⟩ under any condition what-so-ever therefore ⟨Ĥ0⟩ ⟨Ĥ1⟩ is
>>>>>>>> proved to specify a non-halting sequence of configurations.
>>>>>>>>
>>>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
>>>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn
>>>>>>>
>>>>>>> This is now the third reply you've made to the same post.
>>>>>>>
>>>>>>> That post didn't make any arguments whatsoever about your claims.
>>>>>>> It simply pointed out that you are misusing your notation and
>>>>>>> urged you to correct it.
>>>>>>>
>>>>>>
>>>>>> THE NOTATION IS A STIPULATIVE DEFINITION THUS DISAGREEMENT IS
>>>>>> INCORRECT.
>>>>>
>>>>> If the notation is junk, then the definition is also junk.
>>>>>
>>>>> That's like "stipulating" that
>>>>>
>>>>> +×yz÷² = ±z+³
>>>>>
>>>>> It's meaningless because the notation is meaningless, much like
>>>>> your notation above.
>>>>>
>>>>> This is meaningless:
>>>>>
>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy   // what's the condition?
>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qn   // what's the condition?
>>>>>
>>>>> With no conditions specified, the above is just nonsense.
>>>>>
>>>>> André
>>>>>
>>>>
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
>>>> If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would reach its
>>>> final state.
>>>>
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>> If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never reach
>>>> its final state.
>>>
>>>
>>> This is still nonsense.
>>>
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
>> If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would reach its
>> own final state.
>>
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>> If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would never
>> reach its own final state.
>
> And again you're still being inconsistent. You can either use H or use
> embedded_H, but you can't mix the two.
>

Sure I can. I just did.

>> This means that H pretends that it is only a UTM to see what its
>> simulated input would do in this case. If it would never reach its own
>> final state then H correctly rejects this input.
>
> A Turing Machine cannot "pretend" to be some different Turing Machine.

It can perform a pure simulation of its input until this simulated input
matches a repeating behavior pattern that proves this input never
reaches its own final state.

It is like I say that a dog is an animal and everyone disagrees on the
basis that they believe that a dog is an office building.

--
Copyright 2022 Pete Olcott

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

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ deceitful bastard ]

<7256159c-d379-4cec-9730-0f958e7bb848n@googlegroups.com>

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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
deceitful bastard ]
From: dbush.mo...@gmail.com (Dennis Bush)
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 by: Dennis Bush - Sun, 10 Apr 2022 23:26 UTC

On Sunday, April 10, 2022 at 7:20:44 PM UTC-4, olcott wrote:
> On 4/10/2022 6:14 PM, André G. Isaak wrote:
> > On 2022-04-10 17:08, olcott wrote:
> >> On 4/10/2022 5:59 PM, André G. Isaak wrote:
> >>> On 2022-04-10 16:40, olcott wrote:
> >>>> On 4/10/2022 5:35 PM, André G. Isaak wrote:
> >>>>> On 2022-04-10 15:56, olcott wrote:
> >>>>>> On 4/10/2022 4:49 PM, André G. Isaak wrote:
> >>>>>>> On 2022-04-10 15:00, olcott wrote:
> >>>>>>>> On 4/10/2022 3:15 PM, olcott wrote:
> >>>>>>>>> On 4/10/2022 3:07 PM, André G. Isaak wrote:
> >>>>>>>
> >>>>>>>>>> I'm trying to get you to write using correct and coherent
> >>>>>>>>>> notation. That's one of the things you'll need to be able to
> >>>>>>>>>> do if you ever hope to publish. That involves remembering to
> >>>>>>>>>> always include conditions and using the same terms in your
> >>>>>>>>>> 'equations' as in your text.
> >>>>>>>>>>
> >>>>>>>>>> Not sure how that makes me a 'deceitful bastard'.
> >>>>>>>>>>
> >>>>>>>>>> André
> >>>>>>>>>>
> >>>>>>>>>
> >>>>>>>>> THAT you pretended to not know what I mean by embedded_H so
> >>>>>>>>> that you could artificially contrive a fake basis for rebuttal
> >>>>>>>>> when no actual basis for rebuttal exists makes you a deceitful
> >>>>>>>>> bastard.
> >>>>>>>>
> >>>>>>>> IT IS THE CASE THAT the correctly simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to
> >>>>>>>> embedded_H never reaches its own final state of ⟨Ĥ0..qy⟩ or
> >>>>>>>> ⟨Ĥ0.qn⟩ under any condition what-so-ever therefore ⟨Ĥ0⟩ ⟨Ĥ1⟩ is
> >>>>>>>> proved to specify a non-halting sequence of configurations.
> >>>>>>>>
> >>>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
> >>>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn
> >>>>>>>
> >>>>>>> This is now the third reply you've made to the same post.
> >>>>>>>
> >>>>>>> That post didn't make any arguments whatsoever about your claims.
> >>>>>>> It simply pointed out that you are misusing your notation and
> >>>>>>> urged you to correct it.
> >>>>>>>
> >>>>>>
> >>>>>> THE NOTATION IS A STIPULATIVE DEFINITION THUS DISAGREEMENT IS
> >>>>>> INCORRECT.
> >>>>>
> >>>>> If the notation is junk, then the definition is also junk.
> >>>>>
> >>>>> That's like "stipulating" that
> >>>>>
> >>>>> +×yz÷² = ±z+³
> >>>>>
> >>>>> It's meaningless because the notation is meaningless, much like
> >>>>> your notation above.
> >>>>>
> >>>>> This is meaningless:
> >>>>>
> >>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy // what's the condition?
> >>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qn // what's the condition?
> >>>>>
> >>>>> With no conditions specified, the above is just nonsense.
> >>>>>
> >>>>> André
> >>>>>
> >>>>
> >>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
> >>>> If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would reach its
> >>>> final state.
> >>>>
> >>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
> >>>> If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never reach
> >>>> its final state.
> >>>
> >>>
> >>> This is still nonsense.
> >>>
> >> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
> >> If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would reach its
> >> own final state.
> >>
> >> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
> >> If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would never
> >> reach its own final state.
> >
> > And again you're still being inconsistent. You can either use H or use
> > embedded_H, but you can't mix the two.
> >
> Sure I can. I just did.
> >> This means that H pretends that it is only a UTM to see what its
> >> simulated input would do in this case. If it would never reach its own
> >> final state then H correctly rejects this input.
> >
> > A Turing Machine cannot "pretend" to be some different Turing Machine.
> It can perform a pure simulation of its input until this simulated input
> matches a repeating behavior pattern that proves this input never
> reaches its own final state.

If that's the case, why does an actual UTM applied to the *same* input halt?

Hint: Because the result of an actual UTM applied to the input defines the correct answer, so H answers wrong.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ deceitful bastard ]

<hLJ4K.417634$iK66.193242@fx46.iad>

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deceitful bastard ]
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 by: Richard Damon - Sun, 10 Apr 2022 23:29 UTC

On 4/10/22 7:20 PM, olcott wrote:
> On 4/10/2022 6:14 PM, André G. Isaak wrote:
>> On 2022-04-10 17:08, olcott wrote:
>>> On 4/10/2022 5:59 PM, André G. Isaak wrote:
>>>> On 2022-04-10 16:40, olcott wrote:
>>>>> On 4/10/2022 5:35 PM, André G. Isaak wrote:
>>>>>> On 2022-04-10 15:56, olcott wrote:
>>>>>>> On 4/10/2022 4:49 PM, André G. Isaak wrote:
>>>>>>>> On 2022-04-10 15:00, olcott wrote:
>>>>>>>>> On 4/10/2022 3:15 PM, olcott wrote:
>>>>>>>>>> On 4/10/2022 3:07 PM, André G. Isaak wrote:
>>>>>>>>
>>>>>>>>>>> I'm trying to get you to write using correct and coherent
>>>>>>>>>>> notation. That's one of the things you'll need to be able to
>>>>>>>>>>> do if you ever hope to publish. That involves remembering to
>>>>>>>>>>> always include conditions and using the same terms in your
>>>>>>>>>>> 'equations' as in your text.
>>>>>>>>>>>
>>>>>>>>>>> Not sure how that makes me a 'deceitful bastard'.
>>>>>>>>>>>
>>>>>>>>>>> André
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> THAT you pretended to not know what I mean by embedded_H so
>>>>>>>>>> that you could artificially contrive a fake basis for rebuttal
>>>>>>>>>> when no actual basis for rebuttal exists makes you a deceitful
>>>>>>>>>> bastard.
>>>>>>>>>
>>>>>>>>> IT IS THE CASE THAT the correctly simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to
>>>>>>>>> embedded_H never reaches its own final state of ⟨Ĥ0.qy⟩ or
>>>>>>>>> ⟨Ĥ0.qn⟩ under any condition what-so-ever therefore ⟨Ĥ0⟩ ⟨Ĥ1⟩ is
>>>>>>>>> proved to specify a non-halting sequence of configurations.
>>>>>>>>>
>>>>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
>>>>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn
>>>>>>>>
>>>>>>>> This is now the third reply you've made to the same post.
>>>>>>>>
>>>>>>>> That post didn't make any arguments whatsoever about your
>>>>>>>> claims. It simply pointed out that you are misusing your
>>>>>>>> notation and urged you to correct it.
>>>>>>>>
>>>>>>>
>>>>>>> THE NOTATION IS A STIPULATIVE DEFINITION THUS DISAGREEMENT IS
>>>>>>> INCORRECT.
>>>>>>
>>>>>> If the notation is junk, then the definition is also junk.
>>>>>>
>>>>>> That's like "stipulating" that
>>>>>>
>>>>>> +×yz÷² = ±z+³
>>>>>>
>>>>>> It's meaningless because the notation is meaningless, much like
>>>>>> your notation above.
>>>>>>
>>>>>> This is meaningless:
>>>>>>
>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy   // what's the condition?
>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qn   // what's the condition?
>>>>>>
>>>>>> With no conditions specified, the above is just nonsense.
>>>>>>
>>>>>> André
>>>>>>
>>>>>
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
>>>>> If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would reach its
>>>>> final state.
>>>>>
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>>> If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never reach
>>>>> its final state.
>>>>
>>>>
>>>> This is still nonsense.
>>>>
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
>>> If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would reach
>>> its own final state.
>>>
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>> If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would never
>>> reach its own final state.
>>
>> And again you're still being inconsistent. You can either use H or use
>> embedded_H, but you can't mix the two.
>>
>
> Sure I can. I just did.

And prove you are being inconsistent (and thus a LIAR).

>
>>> This means that H pretends that it is only a UTM to see what its
>>> simulated input would do in this case. If it would never reach its
>>> own final state then H correctly rejects this input.
>>
>> A Turing Machine cannot "pretend" to be some different Turing Machine.
>
> It can perform a pure simulation of its input until this simulated input
> matches a repeating behavior pattern that proves this input never
> reaches its own final state.

Then it ISN'T performing a UTM simulation, PERIOD.

That is like saying you are immortal until you die.

UTMness, is an absolute, not a relative property. A UTM simulation is
ONLY a COMPLETE UTM Simulation that exactly replicates the behavior of a
machine.

You then have the fact that there exist NO such finite pattern that
PROVES that the simulation of <H^> <H^> will be non-halting, so by that
definition, your embedded_H was just defined to not abort and thus not
answer. FAIL.

>
> It is like I say that a dog is an animal and everyone disagrees on the
> basis that they believe that a dog is an office building.
>
>

Nope. You may THINK you have a dog, but you don't because apperently you
have never actually seen a dog before because you don't know what one
is. You keep on showing that you just don't understand the basics of
Computation Theory (or even just Logic), and don't actually know what
you are talking about.

Thus you have proved yourself to be LIAR, because you claim you do.

FAIL.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ deceitful bastard ]

<87pmlo4ht9.fsf@bsb.me.uk>

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https://www.novabbs.com/devel/article-flat.php?id=29920&group=comp.theory#29920

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From: ben.use...@bsb.me.uk (Ben)
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ deceitful bastard ]
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 by: Ben - Sun, 10 Apr 2022 23:52 UTC

Dennis Bush <dbush.mobile@gmail.com> writes:

> On Sunday, April 10, 2022 at 7:20:44 PM UTC-4, olcott wrote:
>> On 4/10/2022 6:14 PM, André G. Isaak wrote:
>> > On 2022-04-10 17:08, olcott wrote:

>> >> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>> >> If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would never
>> >> reach its own final state.
>> >
>> > And again you're still being inconsistent. You can either use H or use
>> > embedded_H, but you can't mix the two.
>> >
>> Sure I can. I just did.
>>
>> >> This means that H pretends that it is only a UTM to see what its
>> >> simulated input would do in this case. If it would never reach its own
>> >> final state then H correctly rejects this input.
>> >
>> > A Turing Machine cannot "pretend" to be some different Turing Machine.
>>
>> It can perform a pure simulation of its input until this simulated input
>> matches a repeating behavior pattern that proves this input never
>> reaches its own final state.
>
> If that's the case, why does an actual UTM applied to the *same* input
> halt?
>
> Hint: Because the result of an actual UTM applied to the input defines
> the correct answer, so H answers wrong.

Quite. The more detailed reason (in case you have spared yourself
looking at the history of this nonsense) is that the "pattern" being
looked for is not a non-halting pattern. It's a pattern that shows the
computation would not halt were it not for the fact that the
pattern-seeking code is about to halt.

PO has been trying to get this ruse past people for years. The biggest
mistake he made though was to be clear about it once. He sketched out a
Halts function that looped, single-stepping a simulation and checking
for whether the simulation "needs to be aborted". Halts returned false
when presented with the "hat" version of Halts, but PO deemed this
correct because of what would happen if line 15 (the "needs to be
aborted" check) were edited out!

One of the most useful words to look out for in PO's posts is "would".
It almost always signals some form of this ruse. TMs should be
specified by what is the case, not by what would be the case (leaving
some condition unstated).

--
Ben.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ correct halt deciding criteria ]

<zJqdnWmS9-al8s7_nZ2dnUU7_83NnZ2d@giganews.com>

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correct halt deciding criteria ]
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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Sun, 10 Apr 2022 23:58 UTC

On 4/10/2022 6:26 PM, Dennis Bush wrote:
> On Sunday, April 10, 2022 at 7:20:44 PM UTC-4, olcott wrote:
>> On 4/10/2022 6:14 PM, André G. Isaak wrote:
>>> On 2022-04-10 17:08, olcott wrote:
>>>> On 4/10/2022 5:59 PM, André G. Isaak wrote:
>>>>> On 2022-04-10 16:40, olcott wrote:
>>>>>> On 4/10/2022 5:35 PM, André G. Isaak wrote:
>>>>>>> On 2022-04-10 15:56, olcott wrote:
>>>>>>>> On 4/10/2022 4:49 PM, André G. Isaak wrote:
>>>>>>>>> On 2022-04-10 15:00, olcott wrote:
>>>>>>>>>> On 4/10/2022 3:15 PM, olcott wrote:
>>>>>>>>>>> On 4/10/2022 3:07 PM, André G. Isaak wrote:
>>>>>>>>>
>>>>>>>>>>>> I'm trying to get you to write using correct and coherent
>>>>>>>>>>>> notation. That's one of the things you'll need to be able to
>>>>>>>>>>>> do if you ever hope to publish. That involves remembering to
>>>>>>>>>>>> always include conditions and using the same terms in your
>>>>>>>>>>>> 'equations' as in your text.
>>>>>>>>>>>>
>>>>>>>>>>>> Not sure how that makes me a 'deceitful bastard'.
>>>>>>>>>>>>
>>>>>>>>>>>> André
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> THAT you pretended to not know what I mean by embedded_H so
>>>>>>>>>>> that you could artificially contrive a fake basis for rebuttal
>>>>>>>>>>> when no actual basis for rebuttal exists makes you a deceitful
>>>>>>>>>>> bastard.
>>>>>>>>>>
>>>>>>>>>> IT IS THE CASE THAT the correctly simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to
>>>>>>>>>> embedded_H never reaches its own final state of ⟨Ĥ0.qy⟩ or
>>>>>>>>>> ⟨Ĥ0.qn⟩ under any condition what-so-ever therefore ⟨Ĥ0⟩ ⟨Ĥ1⟩ is
>>>>>>>>>> proved to specify a non-halting sequence of configurations.
>>>>>>>>>>
>>>>>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
>>>>>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn
>>>>>>>>>
>>>>>>>>> This is now the third reply you've made to the same post.
>>>>>>>>>
>>>>>>>>> That post didn't make any arguments whatsoever about your claims.
>>>>>>>>> It simply pointed out that you are misusing your notation and
>>>>>>>>> urged you to correct it.
>>>>>>>>>
>>>>>>>>
>>>>>>>> THE NOTATION IS A STIPULATIVE DEFINITION THUS DISAGREEMENT IS
>>>>>>>> INCORRECT.
>>>>>>>
>>>>>>> If the notation is junk, then the definition is also junk.
>>>>>>>
>>>>>>> That's like "stipulating" that
>>>>>>>
>>>>>>> +×yz÷² = ±z+³
>>>>>>>
>>>>>>> It's meaningless because the notation is meaningless, much like
>>>>>>> your notation above.
>>>>>>>
>>>>>>> This is meaningless:
>>>>>>>
>>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy // what's the condition?
>>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qn // what's the condition?
>>>>>>>
>>>>>>> With no conditions specified, the above is just nonsense.
>>>>>>>
>>>>>>> André
>>>>>>>
>>>>>>
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
>>>>>> If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would reach its
>>>>>> final state.
>>>>>>
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>>>> If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never reach
>>>>>> its final state.
>>>>>
>>>>>
>>>>> This is still nonsense.
>>>>>
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
>>>> If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would reach its
>>>> own final state.
>>>>
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>> If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would never
>>>> reach its own final state.
>>>
>>> And again you're still being inconsistent. You can either use H or use
>>> embedded_H, but you can't mix the two.
>>>
>> Sure I can. I just did.
>>>> This means that H pretends that it is only a UTM to see what its
>>>> simulated input would do in this case. If it would never reach its own
>>>> final state then H correctly rejects this input.
>>>
>>> A Turing Machine cannot "pretend" to be some different Turing Machine.
>> It can perform a pure simulation of its input until this simulated input
>> matches a repeating behavior pattern that proves this input never
>> reaches its own final state.
>
> If that's the case, why does an actual UTM applied to the *same* input halt?
>
> Hint: Because the result of an actual UTM applied to the input defines the correct answer, so H answers wrong.

Intuitively that would seem to be true, this intuition is incorrect.

The ultimate definition of correct is the computation of the mapping of
the inputs to an accept or reject state on the basis of the behavior
that these inputs specify.

That simulated inputs to embedded_H would never reach their own final
state under any condition what-so-ever
IS THE ULTIMATE MEASURE OF THEIR HALTING BEHAVIOR
and conclusively proves they specify a non-halting sequence of
configurations.

--
Copyright 2022 Pete Olcott

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

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ key missing piece in dialogue ][ back door ]

<87k0bw4hgi.fsf@bsb.me.uk>

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https://www.novabbs.com/devel/article-flat.php?id=29923&group=comp.theory#29923

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From: ben.use...@bsb.me.uk (Ben)
Newsgroups: comp.theory
Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ key missing piece in dialogue ][ back door ]
Date: Mon, 11 Apr 2022 01:00:29 +0100
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 by: Ben - Mon, 11 Apr 2022 00:00 UTC

olcott <NoOne@NoWhere.com> writes:

> On 4/10/2022 4:41 PM, Ben wrote:
>> olcott <NoOne@NoWhere.com> writes:

>>> The above means this:
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>> That's funny! You really have no idea what this notation means, do you?
>>
>>> embedded_H is a simulating halt decider that has a full UTM embedded
>>> within it. As soon as it sees that the pure UTM simulation of its
>>> input would never reach the final state of this input it aborts this
>>> simulation and rejects this non-halting input.
>>
>> So you had no business writing those two junk lines, did you? Or do you
>> really think that they are in some way compatible with that last
>> paragraph? Probably neither. I really think you see it much like
>> poetry. Meanings are supposed to be intuited from unusual, often
>> metaphorical, juxtapositions of symbols.
>
> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn

Still junk.

--
Ben.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ deceitful bastard ]

<zJqdnWiS9-ZV8s7_nZ2dnUU7_81g4p2d@giganews.com>

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 by: olcott - Mon, 11 Apr 2022 00:01 UTC

On 4/10/2022 6:52 PM, Ben wrote:
> Dennis Bush <dbush.mobile@gmail.com> writes:
>
>> On Sunday, April 10, 2022 at 7:20:44 PM UTC-4, olcott wrote:
>>> On 4/10/2022 6:14 PM, André G. Isaak wrote:
>>>> On 2022-04-10 17:08, olcott wrote:
>
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>>> If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would never
>>>>> reach its own final state.
>>>>
>>>> And again you're still being inconsistent. You can either use H or use
>>>> embedded_H, but you can't mix the two.
>>>>
>>> Sure I can. I just did.
>>>
>>>>> This means that H pretends that it is only a UTM to see what its
>>>>> simulated input would do in this case. If it would never reach its own
>>>>> final state then H correctly rejects this input.
>>>>
>>>> A Turing Machine cannot "pretend" to be some different Turing Machine.
>>>
>>> It can perform a pure simulation of its input until this simulated input
>>> matches a repeating behavior pattern that proves this input never
>>> reaches its own final state.
>>
>> If that's the case, why does an actual UTM applied to the *same* input
>> halt?
>>
>> Hint: Because the result of an actual UTM applied to the input defines
>> the correct answer, so H answers wrong.
>
> Quite. The more detailed reason (in case you have spared yourself
> looking at the history of this nonsense) is that the "pattern" being
> looked for is not a non-halting pattern. It's a pattern that shows the
> computation would not halt were it not for the fact that the
> pattern-seeking code is about to halt.
>

THE SIMULATED INPUT CANNOT POSSIBLY REACH ITS OWN FINAL STATE
THIS SINGLE FACT BY ITSELF PROVES THAT THE INPUT IS CORRECTLY REJECTED

--
Copyright 2022 Pete Olcott

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

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ key missing piece in dialogue ][ back door ]

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 by: olcott - Mon, 11 Apr 2022 00:02 UTC

On 4/10/2022 7:00 PM, Ben wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 4/10/2022 4:41 PM, Ben wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>
>>>> The above means this:
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>> That's funny! You really have no idea what this notation means, do you?
>>>
>>>> embedded_H is a simulating halt decider that has a full UTM embedded
>>>> within it. As soon as it sees that the pure UTM simulation of its
>>>> input would never reach the final state of this input it aborts this
>>>> simulation and rejects this non-halting input.
>>>
>>> So you had no business writing those two junk lines, did you? Or do you
>>> really think that they are in some way compatible with that last
>>> paragraph? Probably neither. I really think you see it much like
>>> poetry. Meanings are supposed to be intuited from unusual, often
>>> metaphorical, juxtapositions of symbols.
>>
>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn
>
> Still junk.
>

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would reach its
own final state.

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would never reach
its own final state.

--
Copyright 2022 Pete Olcott

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

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ key missing piece in dialogue ][ back door ]

<87ee244h7c.fsf@bsb.me.uk>

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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ key missing piece in dialogue ][ back door ]
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 by: Ben - Mon, 11 Apr 2022 00:05 UTC

olcott <NoOne@NoWhere.com> writes:

> On 4/10/2022 4:18 PM, Ben wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 4/10/2022 10:52 AM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 4/9/2022 5:54 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 4/9/2022 7:20 AM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> On 4/8/2022 4:08 PM, Ben Bacarisse wrote:
>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>
>>>>>>>>>>> On 4/7/2022 8:14 PM, Ben Bacarisse wrote:
>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>>
>>>>>>>>>>>>> On 4/7/2022 6:37 PM, Ben Bacarisse wrote:
>>>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> On 4/7/2022 10:51 AM, Ben Bacarisse wrote:
>>>>>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> THIS PROVES THAT I AM CORRECT
>>>>>>>>>>>>>>>>> It is the case that the correctly simulated input to embedded_H can
>>>>>>>>>>>>>>>>> never possibly reach its own final state under any condition at all.
>>>>>>>>>>>>>>>>> Therefore embedded_H is necessarily correct to reject its input.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Yet you won't answer two simple questions! Why?
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Because I absolutely positively will not tolerate divergence from
>>>>>>>>>>>>>>> validating my 17 years worth of work.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> But you have no choice but to tolerate it. If someone wants to talk
>>>>>>>>>>>>>> about why you are wrong, they will do so.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> You are wrong (for the C version of H) because H(P,P) == false but P(P)
>>>>>>>>>>>>>> halts. You are wrong about your TM H because H <Ĥ> <Ĥ> transitions to
>>>>>>>>>>>>>> qn, but Ĥ applied to <Ĥ> is a halting computation. (Feel free to deny
>>>>>>>>>>>>>> any of these facts if the mood takes you.)
>>>>>>>>>>>>>
>>>>>>>>>>>>> If you believe (against the verified facts) that the simulated ⟨Ĥ0⟩
>>>>>>>>>>>>> reaches its final state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩...
>>>>>>>>>>>>
>>>>>>>>>>>> I believe what you've told me: that you claim that H(P,P)==false is
>>>>>>>>>>>> correct despite the fact that P(P) halts. That's wrong.
>>>>>>>>>>>
>>>>>>>>>>> If the input to H(P,P) cannot possibly reach its final state then this
>>>>>>>>>>> input is correctly rejected and nothing in the universe can possibly
>>>>>>>>>>> contradict this.
>>>>>>>>>>
>>>>>>>>>> Agreed facts: (1) H(P,P) == false, (2) P(P) halts. You don't dispute
>>>>>>>>>> either (indeed they come from you).
>>>>>>>> At least you don't contend these facts.
>>>>>>>>
>>>>>>>>>> Your new line in waffle is just an attempt to distract attention from a
>>>>>>>>>> very simple claim: that the wrong answer is the right one.
>>>>>>>>>
>>>>>>>>> Even Linz got this wrong because it is counter-intuitive.
>>>>>>>>>
>>>>>>>>> A halt decider must compute the mapping from its inputs (not any damn
>>>>>>>>> thing else in the universe) to its own final state on the basis of the
>>>>>>>>> behavior specified by these inputs
>>>>>>>>
>>>>>>>> That's not counter intuitive, it's basic. Everyone knows this, though
>>>>>>>> it took you a while to get round to it. A halt decider accepts or
>>>>>>>> rejects a string based on the behaviour of the computation specified by
>>>>>>>> that string. Of course, you never got as far in my exercises as
>>>>>>>> specifying any TM that decides something on the basis of behaviour, so
>>>>>>>> you really don't know how it's actually done. That was, I thought, the
>>>>>>>> whole point of the exercises -- to see how TMs are specified to decide
>>>>>>>> properties of computations.
>>>>>>>
>>>>>>> You have to actually pay attention to this,
>>>>>>
>>>>>> Flip, flop! Back to being wrong about TMs rather than being wrong about
>>>>>> your old C junk. These uncontested facts: (1) H(P,P) == false, (2) P(P)
>>>>>> halts are why your H and P are wrong.
>>>>>
>>>>> If you are able to break the problem down to it micro component parts
>>>>> and carefully analyze each of these separately instead of simply
>>>>> slipping down the slide of intuition then you can see that I am
>>>>> correct.
>>>>>
>>>>> If it is true that the correct simulation input to H(P,P) cannot
>>>>> possibly reach its own final state then
>>>>>
>>>>> The input to H(P,P) is non-halting then
>>>> There is no "input to H(P,P)".
>>>
>>> The correct simulation of the input to H
>> Better. I still would not call it "input" (since these are C functions)
>> but you've got the hang of what am saying. Well done.
>>
>>> cannot possibly ever reach it final state thus is a non-halting
>>> sequence of configurations even if everyone and everything in the
>>> universe disagrees.
>>
>> The truth is not determined by who does or does not agree with
>> something. But to find the truth of the matter you must first stop
>> talking literal nonsense. The arguments to H (what you call the
>> "input") are two pointers. What does simulating two pointers mean?
>> What you mean, I hope, is simulating calling the first pointer with the
>> second as it's argument. That simulation, according to you, will halt
>> (or "reach it's final state" in your flamboyant, sciencey, language).
>> It will halt because the direct call P(P) halts. Everything here halts
>> (according to you). That's why H is wrong.
>
> You simply are ignoring the actual execution trace that conclusively
> proves that the simulated input to H cannot possibly reach its final
> own state.

The traces that matter are the one of P(P) halting (you made the mistake
of posting it once), and the one of H(P,P) return false (you posted that
as well). You a free to retract any of these at any time, but until you
do, your H is wrong by your own supplied traces.

--
Ben.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ key missing piece in dialogue ][ back door ]

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 by: olcott - Mon, 11 Apr 2022 00:09 UTC

On 4/10/2022 7:05 PM, Ben wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 4/10/2022 4:18 PM, Ben wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 4/10/2022 10:52 AM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 4/9/2022 5:54 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 4/9/2022 7:20 AM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> On 4/8/2022 4:08 PM, Ben Bacarisse wrote:
>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>
>>>>>>>>>>>> On 4/7/2022 8:14 PM, Ben Bacarisse wrote:
>>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> On 4/7/2022 6:37 PM, Ben Bacarisse wrote:
>>>>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> On 4/7/2022 10:51 AM, Ben Bacarisse wrote:
>>>>>>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> THIS PROVES THAT I AM CORRECT
>>>>>>>>>>>>>>>>>> It is the case that the correctly simulated input to embedded_H can
>>>>>>>>>>>>>>>>>> never possibly reach its own final state under any condition at all.
>>>>>>>>>>>>>>>>>> Therefore embedded_H is necessarily correct to reject its input.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Yet you won't answer two simple questions! Why?
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Because I absolutely positively will not tolerate divergence from
>>>>>>>>>>>>>>>> validating my 17 years worth of work.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> But you have no choice but to tolerate it. If someone wants to talk
>>>>>>>>>>>>>>> about why you are wrong, they will do so.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> You are wrong (for the C version of H) because H(P,P) == false but P(P)
>>>>>>>>>>>>>>> halts. You are wrong about your TM H because H <Ĥ> <Ĥ> transitions to
>>>>>>>>>>>>>>> qn, but Ĥ applied to <Ĥ> is a halting computation. (Feel free to deny
>>>>>>>>>>>>>>> any of these facts if the mood takes you.)
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> If you believe (against the verified facts) that the simulated ⟨Ĥ0⟩
>>>>>>>>>>>>>> reaches its final state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩...
>>>>>>>>>>>>>
>>>>>>>>>>>>> I believe what you've told me: that you claim that H(P,P)==false is
>>>>>>>>>>>>> correct despite the fact that P(P) halts. That's wrong.
>>>>>>>>>>>>
>>>>>>>>>>>> If the input to H(P,P) cannot possibly reach its final state then this
>>>>>>>>>>>> input is correctly rejected and nothing in the universe can possibly
>>>>>>>>>>>> contradict this.
>>>>>>>>>>>
>>>>>>>>>>> Agreed facts: (1) H(P,P) == false, (2) P(P) halts. You don't dispute
>>>>>>>>>>> either (indeed they come from you).
>>>>>>>>> At least you don't contend these facts.
>>>>>>>>>
>>>>>>>>>>> Your new line in waffle is just an attempt to distract attention from a
>>>>>>>>>>> very simple claim: that the wrong answer is the right one.
>>>>>>>>>>
>>>>>>>>>> Even Linz got this wrong because it is counter-intuitive.
>>>>>>>>>>
>>>>>>>>>> A halt decider must compute the mapping from its inputs (not any damn
>>>>>>>>>> thing else in the universe) to its own final state on the basis of the
>>>>>>>>>> behavior specified by these inputs
>>>>>>>>>
>>>>>>>>> That's not counter intuitive, it's basic. Everyone knows this, though
>>>>>>>>> it took you a while to get round to it. A halt decider accepts or
>>>>>>>>> rejects a string based on the behaviour of the computation specified by
>>>>>>>>> that string. Of course, you never got as far in my exercises as
>>>>>>>>> specifying any TM that decides something on the basis of behaviour, so
>>>>>>>>> you really don't know how it's actually done. That was, I thought, the
>>>>>>>>> whole point of the exercises -- to see how TMs are specified to decide
>>>>>>>>> properties of computations.
>>>>>>>>
>>>>>>>> You have to actually pay attention to this,
>>>>>>>
>>>>>>> Flip, flop! Back to being wrong about TMs rather than being wrong about
>>>>>>> your old C junk. These uncontested facts: (1) H(P,P) == false, (2) P(P)
>>>>>>> halts are why your H and P are wrong.
>>>>>>
>>>>>> If you are able to break the problem down to it micro component parts
>>>>>> and carefully analyze each of these separately instead of simply
>>>>>> slipping down the slide of intuition then you can see that I am
>>>>>> correct.
>>>>>>
>>>>>> If it is true that the correct simulation input to H(P,P) cannot
>>>>>> possibly reach its own final state then
>>>>>>
>>>>>> The input to H(P,P) is non-halting then
>>>>> There is no "input to H(P,P)".
>>>>
>>>> The correct simulation of the input to H
>>> Better. I still would not call it "input" (since these are C functions)
>>> but you've got the hang of what am saying. Well done.
>>>
>>>> cannot possibly ever reach it final state thus is a non-halting
>>>> sequence of configurations even if everyone and everything in the
>>>> universe disagrees.
>>>
>>> The truth is not determined by who does or does not agree with
>>> something. But to find the truth of the matter you must first stop
>>> talking literal nonsense. The arguments to H (what you call the
>>> "input") are two pointers. What does simulating two pointers mean?
>>> What you mean, I hope, is simulating calling the first pointer with the
>>> second as it's argument. That simulation, according to you, will halt
>>> (or "reach it's final state" in your flamboyant, sciencey, language).
>>> It will halt because the direct call P(P) halts. Everything here halts
>>> (according to you). That's why H is wrong.
>>
>> You simply are ignoring the actual execution trace that conclusively
>> proves that the simulated input to H cannot possibly reach its final
>> own state.
>
> The traces that matter are the one of P(P) halting (you made the mistake
> of posting it once), and the one of H(P,P) return false (you posted that
> as well). You a free to retract any of these at any time, but until you
> do, your H is wrong by your own supplied traces.
>

_P()
[00000956](01) 55 push ebp
[00000957](02) 8bec mov ebp,esp
[00000959](03) 8b4508 mov eax,[ebp+08]
[0000095c](01) 50 push eax
[0000095d](03) 8b4d08 mov ecx,[ebp+08]
[00000960](01) 51 push ecx


Click here to read the complete article
Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ deceitful bastard ]

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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
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 by: Richard Damon - Mon, 11 Apr 2022 00:23 UTC

On 4/10/22 8:01 PM, olcott wrote:
> On 4/10/2022 6:52 PM, Ben wrote:
>> Dennis Bush <dbush.mobile@gmail.com> writes:
>>
>>> On Sunday, April 10, 2022 at 7:20:44 PM UTC-4, olcott wrote:
>>>> On 4/10/2022 6:14 PM, André G. Isaak wrote:
>>>>> On 2022-04-10 17:08, olcott wrote:
>>
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>>>> If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would never
>>>>>> reach its own final state.
>>>>>
>>>>> And again you're still being inconsistent. You can either use H or use
>>>>> embedded_H, but you can't mix the two.
>>>>>
>>>> Sure I can. I just did.
>>>>
>>>>>> This means that H pretends that it is only a UTM to see what its
>>>>>> simulated input would do in this case. If it would never reach its
>>>>>> own
>>>>>> final state then H correctly rejects this input.
>>>>>
>>>>> A Turing Machine cannot "pretend" to be some different Turing Machine.
>>>>
>>>> It can perform a pure simulation of its input until this simulated
>>>> input
>>>> matches a repeating behavior pattern that proves this input never
>>>> reaches its own final state.
>>>
>>> If that's the case, why does an actual UTM applied to the *same* input
>>> halt?
>>>
>>> Hint: Because the result of an actual UTM applied to the input defines
>>> the correct answer, so H answers wrong.
>>
>> Quite.  The more detailed reason (in case you have spared yourself
>> looking at the history of this nonsense) is that the "pattern" being
>> looked for is not a non-halting pattern.  It's a pattern that shows the
>> computation would not halt were it not for the fact that the
>> pattern-seeking code is about to halt.
>>
>
> THE SIMULATED INPUT CANNOT POSSIBLY REACH ITS OWN FINAL STATE
> THIS SINGLE FACT BY ITSELF PROVES THAT THE INPUT IS CORRECTLY REJECTED
>

Depends what you mean by "The Simulated Input".

The only path from the definition that gets to 'simulation' askes about
UTM applied to <H^> <H^>, and THAT has been shown to reach its final
state (if embedded_H applied to <H^> <H^> rejects its input and goes to
H.Qn, if it doesn't it doesn't matter if the input is non-halting,
because H never answers).

You seem to be stuck on the simulation by H / embedded_H, but if it
aborts its simulation, then it isn't a UTM, and thus its simulation is
irrelevant.

The fact you don't understand this just shows how little you understand
the field.

You just keep spouting off LIE that you claim something that is
blatantly false must be true, because you are just to stupid to
understand what you are talking about.

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