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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,comp.ai.philosophy,sci.logic
Subject: Halting Problem Proof ERROR
Date: Thu, 17 Jul 2025 09:44:23 -0500
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On 7/6/2025 11:02 AM, Alan Mackenzie wrote:
> [ Followup-To: set ]
>
> In comp.theory olcott <polcott333@gmail.com> wrote:
>> On 7/6/2025 5:16 AM, Alan Mackenzie wrote:
>>> olcott <polcott333@gmail.com> wrote:
>>>> On 7/5/2025 2:07 PM, Alan Mackenzie wrote:
>
>>>>> You lie. You don't have a proof. Many people in this group have pointed
>>>>> out lots of errors in various versions of your purported proof, which you
>>>>> just ignore. The section in Professor Linz's book you used to be so fond
>>>>> of citing will contain plenty of details, if only you would take the
>>>>> trouble to understand it (assuming you're capable of such understanding).
>
>>>> I have addressed ....
>
>>> Meaningless pompous word.
>
>>>> .... all of those details that you make sure to ignore so that you can
>>>> baselessly claim that I am wrong.
>
>>> I vaguely remember rolling my eyes at your hopeless lack of
>>> understanding. It was like watching a 7 year old trying to do calculus.
>>> The basic understanding was simply not there. Years later, it's still
>>> not there.
>
>>> And yes, you are wrong. The proofs of the halting theorem which involve
>>> constructing programs which purported halting deciders cannot decide
>>> correctly are correct.
>
>> Yet you cannot point to even one mistake because there are none.
>
> That's what I'm saying. Those proofs of the halting theorem are free
> from mistakes.
>
> More to the point, it is YOU who cannot point to any mistakes in them.
> They are valid proofs. Your work, if it contradicts those proofs (which
> isn't at all clear) can thus be dismissed without further consideration.
>
>>>> There cannot possibly be *AN ACTUAL INPUT* that does the
>>>> opposite of whatever its decider decides. All of the examples
>>>> of this have never been *ACTUAL INPUTS*
>
>>> That's so sloppily worded, it could mean almost anything.
>
>> The standard halting problem proof cannot even be constructed.
>
> It has been constructed, and is valid. But one would normally talk about
> formulating a proof, rather than constructing one.
>
> [ .... ]
>
>>>> No Turing machine can possibly take another directly executing
>>>> Turing machine as in input, thus removing these from the
>>>> domain of every halt decider.
>
>>> And that, too.
>
>>>> *Thus the requirement that HHH report on the behavior*
>>>> *of the directly executed DD has always been bogus*
>
>>> And that makes your hat trick.
>
>>>> Turing machine partial halt deciders compute the mapping
>>>> from their actual inputs to the actual behavior that these
>>>> inputs specify.
>
>>> And a fourth. There's some semblance of truth in there, but it's very
>>> confused.
>
>
>> It is not at all confused. I know exactly what it means.
>
> It's very confused to everybody but you, then.
>
>>> Sloppy wording is your technique to get people to go down to your level
>>> of discussion. That involves many posts trying just to tie you down to
>>> specific word meanings, and is very tiresome and unrewarding. I decline
>>> to get involved any further.
>
>
>> *Yet as I claimed you found no actual mistake*
>
> I've found plenty of actual mistakes. I was a software developer by
> profession.
>
>> Let me tell you the punchline so that you can
>> see why I said those things.
>
> Despite what I said last post, I will actually go to the trouble of
> analysing your sloppy expression.
>
>> Because directly executed Turing machines cannot
>> possibly be inputs to Turing machine deciders this
>> makes them outside of the domain of these deciders.
>
> It's entirely unclear what a "directly executed Turing machine" is. Most
> of the time turing machines are theoretical constructs used for proving
> theorems. They can be executed, but rarely are.
>
> It's unclear what you mean by a turing machine being an input to a turing
> machine. Read up about universal turing machines to get a bit of
> background.
>
>> When a partial halt decider is required to report
>> on the direct execution of a machine this requirement
>> is bogus.
>
> See above. That paragraph is meaningless.
>
>> This means that the behavior of DD() is none of the damn
>> business of HHH, thus does not contradict HHH(DD)==0.
>> *If you disagree this only proves that you do not understand*
>
> It's fully obscure what DD() and HHH mean, and thus impossible to
> affirm or contradict the meaningless "HHH(DD)==0".
>
>> HHH(DD) does correctly detect that DD simulated by HHH
>> according to the semantics pf the C programming language
>> cannot possibly reach its own "return"statement final
>> halt state.
>
> See above. By the way, people concerned with computation theory use
> turing machines, which are well-defined, simple, and powerful. They lack
> the complexity, ambiguity, and unsuitability for theoretical work of real
> world programming languages like C.
>
>> *If you disagree this only proves that you do not understand*
>
>> Any mindless idiot can disagree. Showing an error and proving
>> that it is an actual mistake requires much more than this.
>
> Indeed. All you have done is disagree with one of the proofs of the
> halting theorem. You have yet to show an error in it. That will be
> difficult, because there aren't any.
>
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ∞,
if M applied to WM halts, and
q0 WM ⊢* Ĥq0 Wm WM ⊢* Ĥ y1 qn y2,
if M applied to WM does not halt.
*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt.
<*Halting Problem Proof ERROR*>
Requires Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ to report on the
direct execution of Ĥ applied to ⟨Ĥ⟩ and thus not
⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by Ĥ.embedded_H.
No Turing Machine decider can ever report on the
behavior of anything that is not an input encoded
as a finite string.
Ĥ is not a finite string input to Ĥ.embedded_H
⟨Ĥ⟩ ⟨Ĥ⟩ are finite string inputs to Ĥ.embedded_H
</*Halting Problem Proof ERROR*>
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞
⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H reaches
its simulated final halt state of ⟨Ĥ.qn⟩, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
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