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devel / comp.theory / Re: H(P,P)==false is proven to be correct

SubjectAuthor
* On the halting problem (final)Mr Flibble
+- On the halting problem (final)Malcolm McLean
+- On the halting problem (final)Ben
+* On the halting problem (final)Heywood Jablome
|`- On the halting problem (final)Python
`* On the halting problem (final)olcott
 `* On the halting problem (final)Mr Flibble
  `* On the halting problem (final)olcott
   `* On the halting problem (final)wij
    `* On the halting problem (final)olcott
     `* On the halting problem (final)wij
      `* On the halting problem (final)olcott
       +* On the halting problem (final)Dennis Bush
       |`* On the halting problem (final)olcott
       | `* On the halting problem (final)Dennis Bush
       |  `* H(P,P)==false is proven to be correctolcott
       |   `* H(P,P)==false is proven to be correctDennis Bush
       |    `* H(P,P)==false is proven to be correctolcott
       |     +- H(P,P)==false is proven to be correctDennis Bush
       |     +- H(P,P)==false is proven to be correctRichard Damon
       |     `- H(P,P)==false is proven to be correctwij
       `- On the halting problem (final)wij

1
On the halting problem (final)

<20220508234650.000077b1@reddwarf.jmc>

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From: flib...@reddwarf.jmc (Mr Flibble)
Newsgroups: comp.theory
Subject: On the halting problem (final)
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 by: Mr Flibble - Sun, 8 May 2022 22:46 UTC

Based on the assumption that [Strachey, 1965] is not actually advocating
running a program (either through direct execution or by simulation) to
determine if that program halts Strachey's "Impossible Program" is
indeed impossible for the reason given (the contradiction).

The only open question in my mind is what it actually means for a
decider to evaluate the source code of itself in addition to the
source code of the program which would invoke the decider if it actually
was run.

I was wrong. Apologies for the noise.

/Flibble

Re: On the halting problem (final)

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Subject: Re: On the halting problem (final)
From: malcolm....@gmail.com (Malcolm McLean)
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 by: Malcolm McLean - Sun, 8 May 2022 22:59 UTC

On Sunday, 8 May 2022 at 23:46:52 UTC+1, Mr Flibble wrote:
> Based on the assumption that [Strachey, 1965] is not actually advocating
> running a program (either through direct execution or by simulation) to
> determine if that program halts Strachey's "Impossible Program" is
> indeed impossible for the reason given (the contradiction).
>
> The only open question in my mind is what it actually means for a
> decider to evaluate the source code of itself in addition to the
> source code of the program which would invoke the decider if it actually
> was run.
>
> I was wrong. Apologies for the noise.
>
Many programs can be run unproblematically on their own source. For
instnace if a word processor is used to edit its own source, no-one would
see anything paradoxical about that.

Re: On the halting problem (final)

<87k0avsi34.fsf@bsb.me.uk>

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From: ben.use...@bsb.me.uk (Ben)
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Subject: Re: On the halting problem (final)
Date: Mon, 09 May 2022 00:46:55 +0100
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 by: Ben - Sun, 8 May 2022 23:46 UTC

Mr Flibble <flibble@reddwarf.jmc> writes:

> Based on the assumption that [Strachey, 1965] is not actually advocating
> running a program (either through direct execution or by simulation) to
> determine if that program halts Strachey's "Impossible Program" is
> indeed impossible for the reason given (the contradiction).
>
> The only open question in my mind is what it actually means for a
> decider to evaluate the source code of itself in addition to the
> source code of the program which would invoke the decider if it actually
> was run.

Imagine a Lisp decider. This is a lisp function that always returns and
reports on some property of it's argument. For example

(defun is_big (exp) (> (size exp) 42))

with size a bit like this:

(defun size (exp)
(if (atom exp) 1
(+ (size (car exp)) (size (cdr exp)))))

Now, is_big can be called to see if anything is big or not, but that can
include "code" like this: (is_big '((lambda (x) (+ x 1)) 2)).

And there's nothing to stop us asking the question about a computation
involving is_big itself (is_big '(is_big ((lambda (x) (+ x 1)) 2))).

A lisp halts function would be similar. It might pull the expression
apart and examine it. It might partially evaluate it. But we know that
whatever it does, it can't always give the correct answer.

> I was wrong. Apologies for the noise.

Kudos.

--
Ben.

Re: On the halting problem (final)

<t59pg6$11hn$1@gioia.aioe.org>

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Newsgroups: comp.theory
Subject: Re: On the halting problem (final)
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 by: Heywood Jablome - Mon, 9 May 2022 01:07 UTC

On 5/8/2022 3:46 PM, Mr Flibble wrote:
> Based on the assumption that [Strachey, 1965] is not actually advocating
> running a program (either through direct execution or by simulation) to
> determine if that program halts Strachey's "Impossible Program" is
> indeed impossible for the reason given (the contradiction).
>
> The only open question in my mind is what it actually means for a
> decider to evaluate the source code of itself in addition to the
> source code of the program which would invoke the decider if it actually
> was run.
>
> I was wrong. Apologies for the noise.
>
> /Flibble
>

Post it 78,488,747 more times you retarded rectum load of Hitler penis.

Re: On the halting problem (final)

<t59re6$1gj3$1@gioia.aioe.org>

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Subject: Re: On the halting problem (final)
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 by: Python - Mon, 9 May 2022 01:40 UTC

Nym-shifting troll, aka Heywood Jablome wrote:
> On 5/8/2022 3:46 PM, Mr Flibble wrote:
>> Based on the assumption that [Strachey, 1965] is not actually advocating
>> running a program (either through direct execution or by simulation) to
>> determine if that program halts Strachey's "Impossible Program" is
>> indeed impossible for the reason given (the contradiction).
>>
>> The only open question in my mind is what it actually means for a
>> decider to evaluate the source code of itself in addition to the
>> source code of the program which would invoke the decider if it actually
>> was run.
>>
>> I was wrong. Apologies for the noise.
>>
>> /Flibble
>>
>
> Post it 78,488,747 more times you retarded rectum load of Hitler penis.

Everytime somewone wrong on Usenet admits he was wrong — what doesn't
happen often — the nym-shifting-troll lose a part of his body.

Re: On the halting problem (final)

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 by: olcott - Mon, 9 May 2022 15:57 UTC

On 5/8/2022 5:46 PM, Mr Flibble wrote:
> Based on the assumption that [Strachey, 1965] is not actually advocating
> running a program (either through direct execution or by simulation) to
> determine if that program halts Strachey's "Impossible Program" is
> indeed impossible for the reason given (the contradiction).
>
> The only open question in my mind is what it actually means for a
> decider to evaluate the source code of itself in addition to the
> source code of the program which would invoke the decider if it actually
> was run.
>
> I was wrong. Apologies for the noise.
>
> /Flibble
>

You were correct that the input never reaches its impossible part
because it is stuck in infinite recursion.

This only occurs if the halt decider H is a simulating halt decider.
When the input to H(P,P) does get stuck in infinite recursion H can see
this.

--
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: On the halting problem (final)

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From: flib...@reddwarf.jmc (Mr Flibble)
Newsgroups: comp.theory
Subject: Re: On the halting problem (final)
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 by: Mr Flibble - Mon, 9 May 2022 17:35 UTC

On Mon, 9 May 2022 10:57:39 -0500
olcott <NoOne@NoWhere.com> wrote:

> On 5/8/2022 5:46 PM, Mr Flibble wrote:
> > Based on the assumption that [Strachey, 1965] is not actually
> > advocating running a program (either through direct execution or by
> > simulation) to determine if that program halts Strachey's
> > "Impossible Program" is indeed impossible for the reason given (the
> > contradiction).
> >
> > The only open question in my mind is what it actually means for a
> > decider to evaluate the source code of itself in addition to the
> > source code of the program which would invoke the decider if it
> > actually was run.
> >
> > I was wrong. Apologies for the noise.
> >
> > /Flibble
> >
>
> You were correct that the input never reaches its impossible part
> because it is stuck in infinite recursion.

Have you not read my retraction? I was in error: there is no infinite
recursion.

>
> This only occurs if the halt decider H is a simulating halt decider.
> When the input to H(P,P) does get stuck in infinite recursion H can
> see this.

Simulation is an erroneous approach.

/Flibble

Re: On the halting problem (final)

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 by: olcott - Mon, 9 May 2022 17:44 UTC

On 5/9/2022 12:35 PM, Mr Flibble wrote:
> On Mon, 9 May 2022 10:57:39 -0500
> olcott <NoOne@NoWhere.com> wrote:
>
>> On 5/8/2022 5:46 PM, Mr Flibble wrote:
>>> Based on the assumption that [Strachey, 1965] is not actually
>>> advocating running a program (either through direct execution or by
>>> simulation) to determine if that program halts Strachey's
>>> "Impossible Program" is indeed impossible for the reason given (the
>>> contradiction).
>>>
>>> The only open question in my mind is what it actually means for a
>>> decider to evaluate the source code of itself in addition to the
>>> source code of the program which would invoke the decider if it
>>> actually was run.
>>>
>>> I was wrong. Apologies for the noise.
>>>
>>> /Flibble
>>>
>>
>> You were correct that the input never reaches its impossible part
>> because it is stuck in infinite recursion.
>
> Have you not read my retraction? I was in error: there is no infinite
> recursion.
>
>>
>> This only occurs if the halt decider H is a simulating halt decider.
>> When the input to H(P,P) does get stuck in infinite recursion H can
>> see this.
>
> Simulation is an erroneous approach.
>
> /Flibble
>

I have proven otherwise:

https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5

--
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: On the halting problem (final)

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Subject: Re: On the halting problem (final)
From: wynii...@gmail.com (wij)
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 by: wij - Mon, 9 May 2022 18:10 UTC

On Tuesday, 10 May 2022 at 01:44:28 UTC+8, olcott wrote:
> On 5/9/2022 12:35 PM, Mr Flibble wrote:
> > On Mon, 9 May 2022 10:57:39 -0500
> > olcott <No...@NoWhere.com> wrote:
> >
> >> On 5/8/2022 5:46 PM, Mr Flibble wrote:
> >>> Based on the assumption that [Strachey, 1965] is not actually
> >>> advocating running a program (either through direct execution or by
> >>> simulation) to determine if that program halts Strachey's
> >>> "Impossible Program" is indeed impossible for the reason given (the
> >>> contradiction).
> >>>
> >>> The only open question in my mind is what it actually means for a
> >>> decider to evaluate the source code of itself in addition to the
> >>> source code of the program which would invoke the decider if it
> >>> actually was run.
> >>>
> >>> I was wrong. Apologies for the noise.
> >>>
> >>> /Flibble
> >>>
> >>
> >> You were correct that the input never reaches its impossible part
> >> because it is stuck in infinite recursion.
> >
> > Have you not read my retraction? I was in error: there is no infinite
> > recursion.
> >
> >>
> >> This only occurs if the halt decider H is a simulating halt decider.
> >> When the input to H(P,P) does get stuck in infinite recursion H can
> >> see this.
> >
> > Simulation is an erroneous approach.
> >
> > /Flibble
> >
> I have proven otherwise:
>
> https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5
> --
> Copyright 2022 Pete Olcott
>
> "Talent hits a target no one else can hit;
> Genius hits a target no one else can see."
> Arthur Schopenhauer

According to GUR, your proof is wrong:
GUR says "No TM U can decide the property of a TM P if that property can be defied by TM P."

Re: On the halting problem (final)

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 by: olcott - Mon, 9 May 2022 18:13 UTC

On 5/9/2022 1:10 PM, wij wrote:
> On Tuesday, 10 May 2022 at 01:44:28 UTC+8, olcott wrote:
>> On 5/9/2022 12:35 PM, Mr Flibble wrote:
>>> On Mon, 9 May 2022 10:57:39 -0500
>>> olcott <No...@NoWhere.com> wrote:
>>>
>>>> On 5/8/2022 5:46 PM, Mr Flibble wrote:
>>>>> Based on the assumption that [Strachey, 1965] is not actually
>>>>> advocating running a program (either through direct execution or by
>>>>> simulation) to determine if that program halts Strachey's
>>>>> "Impossible Program" is indeed impossible for the reason given (the
>>>>> contradiction).
>>>>>
>>>>> The only open question in my mind is what it actually means for a
>>>>> decider to evaluate the source code of itself in addition to the
>>>>> source code of the program which would invoke the decider if it
>>>>> actually was run.
>>>>>
>>>>> I was wrong. Apologies for the noise.
>>>>>
>>>>> /Flibble
>>>>>
>>>>
>>>> You were correct that the input never reaches its impossible part
>>>> because it is stuck in infinite recursion.
>>>
>>> Have you not read my retraction? I was in error: there is no infinite
>>> recursion.
>>>
>>>>
>>>> This only occurs if the halt decider H is a simulating halt decider.
>>>> When the input to H(P,P) does get stuck in infinite recursion H can
>>>> see this.
>>>
>>> Simulation is an erroneous approach.
>>>
>>> /Flibble
>>>
>> I have proven otherwise:
>>
>> https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5
>> --
>> Copyright 2022 Pete Olcott
>>
>> "Talent hits a target no one else can hit;
>> Genius hits a target no one else can see."
>> Arthur Schopenhauer
>
> According to GUR, your proof is wrong:
> GUR says "No TM U can decide the property of a TM P if that property can be defied by TM P."

According to my proof GUR is wrong. You have to actually read it to see
this.

--
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: On the halting problem (final)

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Subject: Re: On the halting problem (final)
From: wynii...@gmail.com (wij)
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 by: wij - Mon, 9 May 2022 18:36 UTC

On Tuesday, 10 May 2022 at 02:13:22 UTC+8, olcott wrote:
> On 5/9/2022 1:10 PM, wij wrote:
> > On Tuesday, 10 May 2022 at 01:44:28 UTC+8, olcott wrote:
> >> On 5/9/2022 12:35 PM, Mr Flibble wrote:
> >>> On Mon, 9 May 2022 10:57:39 -0500
> >>> olcott <No...@NoWhere.com> wrote:
> >>>
> >>>> On 5/8/2022 5:46 PM, Mr Flibble wrote:
> >>>>> Based on the assumption that [Strachey, 1965] is not actually
> >>>>> advocating running a program (either through direct execution or by
> >>>>> simulation) to determine if that program halts Strachey's
> >>>>> "Impossible Program" is indeed impossible for the reason given (the
> >>>>> contradiction).
> >>>>>
> >>>>> The only open question in my mind is what it actually means for a
> >>>>> decider to evaluate the source code of itself in addition to the
> >>>>> source code of the program which would invoke the decider if it
> >>>>> actually was run.
> >>>>>
> >>>>> I was wrong. Apologies for the noise.
> >>>>>
> >>>>> /Flibble
> >>>>>
> >>>>
> >>>> You were correct that the input never reaches its impossible part
> >>>> because it is stuck in infinite recursion.
> >>>
> >>> Have you not read my retraction? I was in error: there is no infinite
> >>> recursion.
> >>>
> >>>>
> >>>> This only occurs if the halt decider H is a simulating halt decider.
> >>>> When the input to H(P,P) does get stuck in infinite recursion H can
> >>>> see this.
> >>>
> >>> Simulation is an erroneous approach.
> >>>
> >>> /Flibble
> >>>
> >> I have proven otherwise:
> >>
> >> https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5
> >> --
> >> Copyright 2022 Pete Olcott
> >>
> >> "Talent hits a target no one else can hit;
> >> Genius hits a target no one else can see."
> >> Arthur Schopenhauer
> >
> > According to GUR, your proof is wrong:
> > GUR says "No TM U can decide the property of a TM P if that property can be defied by TM P."
> According to my proof GUR is wrong. You have to actually read it to see
> this.
> --
> Copyright 2022 Pete Olcott
>
> "Talent hits a target no one else can hit;
> Genius hits a target no one else can see."
> Arthur Schopenhauer

Any halting decider is shown/proved not able to return a correct answer.

After so long, you still do not have and show us your POOH that can pass the
the test of the convention HP proof.

My GUR is more abstract which includes what the HP says. The basic idea is
proven. I did not fully write it verbosely because it is very 'intuitive'.

Note: I think you should show your POOH before refutation. It should have been a year.
You are not able to talk abstract things (Even a black cat could be a white dog in your mouth).

Re: On the halting problem (final)

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 by: olcott - Mon, 9 May 2022 20:21 UTC

On 5/9/2022 1:36 PM, wij wrote:
> On Tuesday, 10 May 2022 at 02:13:22 UTC+8, olcott wrote:
>> On 5/9/2022 1:10 PM, wij wrote:
>>> On Tuesday, 10 May 2022 at 01:44:28 UTC+8, olcott wrote:
>>>> On 5/9/2022 12:35 PM, Mr Flibble wrote:
>>>>> On Mon, 9 May 2022 10:57:39 -0500
>>>>> olcott <No...@NoWhere.com> wrote:
>>>>>
>>>>>> On 5/8/2022 5:46 PM, Mr Flibble wrote:
>>>>>>> Based on the assumption that [Strachey, 1965] is not actually
>>>>>>> advocating running a program (either through direct execution or by
>>>>>>> simulation) to determine if that program halts Strachey's
>>>>>>> "Impossible Program" is indeed impossible for the reason given (the
>>>>>>> contradiction).
>>>>>>>
>>>>>>> The only open question in my mind is what it actually means for a
>>>>>>> decider to evaluate the source code of itself in addition to the
>>>>>>> source code of the program which would invoke the decider if it
>>>>>>> actually was run.
>>>>>>>
>>>>>>> I was wrong. Apologies for the noise.
>>>>>>>
>>>>>>> /Flibble
>>>>>>>
>>>>>>
>>>>>> You were correct that the input never reaches its impossible part
>>>>>> because it is stuck in infinite recursion.
>>>>>
>>>>> Have you not read my retraction? I was in error: there is no infinite
>>>>> recursion.
>>>>>
>>>>>>
>>>>>> This only occurs if the halt decider H is a simulating halt decider.
>>>>>> When the input to H(P,P) does get stuck in infinite recursion H can
>>>>>> see this.
>>>>>
>>>>> Simulation is an erroneous approach.
>>>>>
>>>>> /Flibble
>>>>>
>>>> I have proven otherwise:
>>>>
>>>> https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5
>>>> --
>>>> Copyright 2022 Pete Olcott
>>>>
>>>> "Talent hits a target no one else can hit;
>>>> Genius hits a target no one else can see."
>>>> Arthur Schopenhauer
>>>
>>> According to GUR, your proof is wrong:
>>> GUR says "No TM U can decide the property of a TM P if that property can be defied by TM P."
>> According to my proof GUR is wrong. You have to actually read it to see
>> this.
>> --
>> Copyright 2022 Pete Olcott
>>
>> "Talent hits a target no one else can hit;
>> Genius hits a target no one else can see."
>> Arthur Schopenhauer
>
> Any halting decider is shown/proved not able to return a correct answer.

All deciders must compute the mapping from their inputs to an
accept/reject state on the basis of a property of these inputs thus

All halt deciders must compute the mapping from their inputs to an
accept/reject state on the basis of a the actual behavior specified by
these actual inputs.

The above two things are verifiable facts (that you are unaware of) and
my conclusion logically follows as a necessary consequence these facts.

H(P,P) is proven to correctly map its inputs to its own reject state on
the basis that the correct simulation of the input to H(P,P) would never
reach its final state at machine address [000009f0] whether or not its
simulation was aborted.

>
> After so long, you still do not have and show us your POOH that can pass the
> the test of the convention HP proof.
>
> My GUR is more abstract which includes what the HP says. The basic idea is
> proven. I did not fully write it verbosely because it is very 'intuitive'.
>
> Note: I think you should show your POOH before refutation. It should have been a year.
> You are not able to talk abstract things (Even a black cat could be a white dog in your mouth).

--
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: On the halting problem (final)

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Subject: Re: On the halting problem (final)
From: dbush.mo...@gmail.com (Dennis Bush)
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 by: Dennis Bush - Mon, 9 May 2022 20:37 UTC

On Monday, May 9, 2022 at 4:21:21 PM UTC-4, olcott wrote:
> On 5/9/2022 1:36 PM, wij wrote:
> > On Tuesday, 10 May 2022 at 02:13:22 UTC+8, olcott wrote:
> >> On 5/9/2022 1:10 PM, wij wrote:
> >>> On Tuesday, 10 May 2022 at 01:44:28 UTC+8, olcott wrote:
> >>>> On 5/9/2022 12:35 PM, Mr Flibble wrote:
> >>>>> On Mon, 9 May 2022 10:57:39 -0500
> >>>>> olcott <No...@NoWhere.com> wrote:
> >>>>>
> >>>>>> On 5/8/2022 5:46 PM, Mr Flibble wrote:
> >>>>>>> Based on the assumption that [Strachey, 1965] is not actually
> >>>>>>> advocating running a program (either through direct execution or by
> >>>>>>> simulation) to determine if that program halts Strachey's
> >>>>>>> "Impossible Program" is indeed impossible for the reason given (the
> >>>>>>> contradiction).
> >>>>>>>
> >>>>>>> The only open question in my mind is what it actually means for a
> >>>>>>> decider to evaluate the source code of itself in addition to the
> >>>>>>> source code of the program which would invoke the decider if it
> >>>>>>> actually was run.
> >>>>>>>
> >>>>>>> I was wrong. Apologies for the noise.
> >>>>>>>
> >>>>>>> /Flibble
> >>>>>>>
> >>>>>>
> >>>>>> You were correct that the input never reaches its impossible part
> >>>>>> because it is stuck in infinite recursion.
> >>>>>
> >>>>> Have you not read my retraction? I was in error: there is no infinite
> >>>>> recursion.
> >>>>>
> >>>>>>
> >>>>>> This only occurs if the halt decider H is a simulating halt decider.
> >>>>>> When the input to H(P,P) does get stuck in infinite recursion H can
> >>>>>> see this.
> >>>>>
> >>>>> Simulation is an erroneous approach.
> >>>>>
> >>>>> /Flibble
> >>>>>
> >>>> I have proven otherwise:
> >>>>
> >>>> https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5
> >>>> --
> >>>> Copyright 2022 Pete Olcott
> >>>>
> >>>> "Talent hits a target no one else can hit;
> >>>> Genius hits a target no one else can see."
> >>>> Arthur Schopenhauer
> >>>
> >>> According to GUR, your proof is wrong:
> >>> GUR says "No TM U can decide the property of a TM P if that property can be defied by TM P."
> >> According to my proof GUR is wrong. You have to actually read it to see
> >> this.
> >> --
> >> Copyright 2022 Pete Olcott
> >>
> >> "Talent hits a target no one else can hit;
> >> Genius hits a target no one else can see."
> >> Arthur Schopenhauer
> >
> > Any halting decider is shown/proved not able to return a correct answer..
> All deciders must compute the mapping from their inputs to an
> accept/reject state on the basis of a property of these inputs thus

And for the halting function, the property of the inputs is the representation of a turning machine and the input to that machine

>
> All halt deciders must compute the mapping from their inputs to an
> accept/reject state on the basis of a the actual behavior specified by
> these actual inputs.

And by definition, the actual behavior specified by the actual inputs is the behavior of the turing machine represented by the first input whose input is the second input. i.e. the the actual behavior specified by the actual inputs to H(P,P) is the behavior of P(P) *by the definition of the problem*.

>
> The above two things are verifiable facts (that you are unaware of) and
> my conclusion logically follows as a necessary consequence these facts.
>
> H(P,P) is proven to correctly map its inputs to its own reject state on
> the basis that the correct simulation of the input to H(P,P) would never
> reach its final state at machine address [000009f0] whether or not its
> simulation was aborted.

The fixed algorithm of H which we'll refer to as Ha, was designed to detect infinite simulation in Pn(Pn), where Hn simulates without aborting and Pn is built from Hn. Ha(Pa,Pa), where Pa calls Ha, does not perform a correct simulation of its input and is therefore not correct to reject. This is proved by Hb which simulates for k more steps than Ha. Hb(Pa,Pa) is able to simulate this input to its final state at machine address [000009f0]. Therefore Ha(Pa,Pa) did not perform a correct simulation of its input.

> >
> > After so long, you still do not have and show us your POOH that can pass the
> > the test of the convention HP proof.
> >
> > My GUR is more abstract which includes what the HP says. The basic idea is
> > proven. I did not fully write it verbosely because it is very 'intuitive'.
> >
> > Note: I think you should show your POOH before refutation. It should have been a year.
> > You are not able to talk abstract things (Even a black cat could be a white dog in your mouth).
> --
> 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: On the halting problem (final)

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Subject: Re: On the halting problem (final)
From: wynii...@gmail.com (wij)
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 by: wij - Mon, 9 May 2022 20:47 UTC

On Tuesday, 10 May 2022 at 04:21:21 UTC+8, olcott wrote:
> On 5/9/2022 1:36 PM, wij wrote:
> > On Tuesday, 10 May 2022 at 02:13:22 UTC+8, olcott wrote:
> >> On 5/9/2022 1:10 PM, wij wrote:
> >>> On Tuesday, 10 May 2022 at 01:44:28 UTC+8, olcott wrote:
> >>>> On 5/9/2022 12:35 PM, Mr Flibble wrote:
> >>>>> On Mon, 9 May 2022 10:57:39 -0500
> >>>>> olcott <No...@NoWhere.com> wrote:
> >>>>>
> >>>>>> On 5/8/2022 5:46 PM, Mr Flibble wrote:
> >>>>>>> Based on the assumption that [Strachey, 1965] is not actually
> >>>>>>> advocating running a program (either through direct execution or by
> >>>>>>> simulation) to determine if that program halts Strachey's
> >>>>>>> "Impossible Program" is indeed impossible for the reason given (the
> >>>>>>> contradiction).
> >>>>>>>
> >>>>>>> The only open question in my mind is what it actually means for a
> >>>>>>> decider to evaluate the source code of itself in addition to the
> >>>>>>> source code of the program which would invoke the decider if it
> >>>>>>> actually was run.
> >>>>>>>
> >>>>>>> I was wrong. Apologies for the noise.
> >>>>>>>
> >>>>>>> /Flibble
> >>>>>>>
> >>>>>>
> >>>>>> You were correct that the input never reaches its impossible part
> >>>>>> because it is stuck in infinite recursion.
> >>>>>
> >>>>> Have you not read my retraction? I was in error: there is no infinite
> >>>>> recursion.
> >>>>>
> >>>>>>
> >>>>>> This only occurs if the halt decider H is a simulating halt decider.
> >>>>>> When the input to H(P,P) does get stuck in infinite recursion H can
> >>>>>> see this.
> >>>>>
> >>>>> Simulation is an erroneous approach.
> >>>>>
> >>>>> /Flibble
> >>>>>
> >>>> I have proven otherwise:
> >>>>
> >>>> https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5
> >>>> --
> >>>> Copyright 2022 Pete Olcott
> >>>>
> >>>> "Talent hits a target no one else can hit;
> >>>> Genius hits a target no one else can see."
> >>>> Arthur Schopenhauer
> >>>
> >>> According to GUR, your proof is wrong:
> >>> GUR says "No TM U can decide the property of a TM P if that property can be defied by TM P."
> >> According to my proof GUR is wrong. You have to actually read it to see
> >> this.
> >> --
> >> Copyright 2022 Pete Olcott
> >>
> >> "Talent hits a target no one else can hit;
> >> Genius hits a target no one else can see."
> >> Arthur Schopenhauer
> >
> > Any halting decider is shown/proved not able to return a correct answer.
> All deciders must compute the mapping from their inputs to an
> accept/reject state on the basis of a property of these inputs thus
>
> All halt deciders must compute the mapping from their inputs to an
> accept/reject state on the basis of a the actual behavior specified by
> these actual inputs.
>
> The above two things are verifiable facts (that you are unaware of) and
> my conclusion logically follows as a necessary consequence these facts.
The HP (eventually) asks for fact to refute, not any logic or theory. No real H, no rebuttal.

Re: On the halting problem (final)

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 by: olcott - Mon, 9 May 2022 21:01 UTC

On 5/9/2022 3:37 PM, Dennis Bush wrote:
> On Monday, May 9, 2022 at 4:21:21 PM UTC-4, olcott wrote:
>> On 5/9/2022 1:36 PM, wij wrote:
>>> On Tuesday, 10 May 2022 at 02:13:22 UTC+8, olcott wrote:
>>>> On 5/9/2022 1:10 PM, wij wrote:
>>>>> On Tuesday, 10 May 2022 at 01:44:28 UTC+8, olcott wrote:
>>>>>> On 5/9/2022 12:35 PM, Mr Flibble wrote:
>>>>>>> On Mon, 9 May 2022 10:57:39 -0500
>>>>>>> olcott <No...@NoWhere.com> wrote:
>>>>>>>
>>>>>>>> On 5/8/2022 5:46 PM, Mr Flibble wrote:
>>>>>>>>> Based on the assumption that [Strachey, 1965] is not actually
>>>>>>>>> advocating running a program (either through direct execution or by
>>>>>>>>> simulation) to determine if that program halts Strachey's
>>>>>>>>> "Impossible Program" is indeed impossible for the reason given (the
>>>>>>>>> contradiction).
>>>>>>>>>
>>>>>>>>> The only open question in my mind is what it actually means for a
>>>>>>>>> decider to evaluate the source code of itself in addition to the
>>>>>>>>> source code of the program which would invoke the decider if it
>>>>>>>>> actually was run.
>>>>>>>>>
>>>>>>>>> I was wrong. Apologies for the noise.
>>>>>>>>>
>>>>>>>>> /Flibble
>>>>>>>>>
>>>>>>>>
>>>>>>>> You were correct that the input never reaches its impossible part
>>>>>>>> because it is stuck in infinite recursion.
>>>>>>>
>>>>>>> Have you not read my retraction? I was in error: there is no infinite
>>>>>>> recursion.
>>>>>>>
>>>>>>>>
>>>>>>>> This only occurs if the halt decider H is a simulating halt decider.
>>>>>>>> When the input to H(P,P) does get stuck in infinite recursion H can
>>>>>>>> see this.
>>>>>>>
>>>>>>> Simulation is an erroneous approach.
>>>>>>>
>>>>>>> /Flibble
>>>>>>>
>>>>>> I have proven otherwise:
>>>>>>
>>>>>> https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5
>>>>>> --
>>>>>> Copyright 2022 Pete Olcott
>>>>>>
>>>>>> "Talent hits a target no one else can hit;
>>>>>> Genius hits a target no one else can see."
>>>>>> Arthur Schopenhauer
>>>>>
>>>>> According to GUR, your proof is wrong:
>>>>> GUR says "No TM U can decide the property of a TM P if that property can be defied by TM P."
>>>> According to my proof GUR is wrong. You have to actually read it to see
>>>> this.
>>>> --
>>>> Copyright 2022 Pete Olcott
>>>>
>>>> "Talent hits a target no one else can hit;
>>>> Genius hits a target no one else can see."
>>>> Arthur Schopenhauer
>>>
>>> Any halting decider is shown/proved not able to return a correct answer.
>> All deciders must compute the mapping from their inputs to an
>> accept/reject state on the basis of a property of these inputs thus
>
> And for the halting function, the property of the inputs is the representation of a turning machine and the input to that machine

No not at all. There is no (indirect reference) of "representation" to
it. The input is a literal string that precisely specifies a sequence of
configurations. In this case it is x86 machine code.

>>
>> All halt deciders must compute the mapping from their inputs to an
>> accept/reject state on the basis of a the actual behavior specified by
>> these actual inputs.
>
> And by definition, the actual behavior specified by the actual inputs is the behavior of the turing machine represented by the first input whose input is the second input.

No not at all. There is no (indirect reference) of "representation" to
it. The input is a literal string that precisely specifies a sequence of
configurations. In this case it is x86 machine code.

> i.e. the the actual behavior specified by the actual inputs to H(P,P) is the behavior of P(P) *by the definition of the problem*.

No not at all. There is no (indirect reference) of "representation" to
it. The input is a literal string that precisely specifies a sequence of
configurations. In this case it is x86 machine code.

Either you are interpreting "representation" incorrectly or the
statement of the problem never noticed that it directly contradicts the
definition of a decider in some rare cases.

--
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: On the halting problem (final)

<de31660f-8385-4409-8949-0402e11d2dc2n@googlegroups.com>

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Subject: Re: On the halting problem (final)
From: dbush.mo...@gmail.com (Dennis Bush)
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 by: Dennis Bush - Mon, 9 May 2022 21:26 UTC

On Monday, May 9, 2022 at 5:01:47 PM UTC-4, olcott wrote:
> On 5/9/2022 3:37 PM, Dennis Bush wrote:
> > On Monday, May 9, 2022 at 4:21:21 PM UTC-4, olcott wrote:
> >> On 5/9/2022 1:36 PM, wij wrote:
> >>> On Tuesday, 10 May 2022 at 02:13:22 UTC+8, olcott wrote:
> >>>> On 5/9/2022 1:10 PM, wij wrote:
> >>>>> On Tuesday, 10 May 2022 at 01:44:28 UTC+8, olcott wrote:
> >>>>>> On 5/9/2022 12:35 PM, Mr Flibble wrote:
> >>>>>>> On Mon, 9 May 2022 10:57:39 -0500
> >>>>>>> olcott <No...@NoWhere.com> wrote:
> >>>>>>>
> >>>>>>>> On 5/8/2022 5:46 PM, Mr Flibble wrote:
> >>>>>>>>> Based on the assumption that [Strachey, 1965] is not actually
> >>>>>>>>> advocating running a program (either through direct execution or by
> >>>>>>>>> simulation) to determine if that program halts Strachey's
> >>>>>>>>> "Impossible Program" is indeed impossible for the reason given (the
> >>>>>>>>> contradiction).
> >>>>>>>>>
> >>>>>>>>> The only open question in my mind is what it actually means for a
> >>>>>>>>> decider to evaluate the source code of itself in addition to the
> >>>>>>>>> source code of the program which would invoke the decider if it
> >>>>>>>>> actually was run.
> >>>>>>>>>
> >>>>>>>>> I was wrong. Apologies for the noise.
> >>>>>>>>>
> >>>>>>>>> /Flibble
> >>>>>>>>>
> >>>>>>>>
> >>>>>>>> You were correct that the input never reaches its impossible part
> >>>>>>>> because it is stuck in infinite recursion.
> >>>>>>>
> >>>>>>> Have you not read my retraction? I was in error: there is no infinite
> >>>>>>> recursion.
> >>>>>>>
> >>>>>>>>
> >>>>>>>> This only occurs if the halt decider H is a simulating halt decider.
> >>>>>>>> When the input to H(P,P) does get stuck in infinite recursion H can
> >>>>>>>> see this.
> >>>>>>>
> >>>>>>> Simulation is an erroneous approach.
> >>>>>>>
> >>>>>>> /Flibble
> >>>>>>>
> >>>>>> I have proven otherwise:
> >>>>>>
> >>>>>> https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5
> >>>>>> --
> >>>>>> Copyright 2022 Pete Olcott
> >>>>>>
> >>>>>> "Talent hits a target no one else can hit;
> >>>>>> Genius hits a target no one else can see."
> >>>>>> Arthur Schopenhauer
> >>>>>
> >>>>> According to GUR, your proof is wrong:
> >>>>> GUR says "No TM U can decide the property of a TM P if that property can be defied by TM P."
> >>>> According to my proof GUR is wrong. You have to actually read it to see
> >>>> this.
> >>>> --
> >>>> Copyright 2022 Pete Olcott
> >>>>
> >>>> "Talent hits a target no one else can hit;
> >>>> Genius hits a target no one else can see."
> >>>> Arthur Schopenhauer
> >>>
> >>> Any halting decider is shown/proved not able to return a correct answer.
> >> All deciders must compute the mapping from their inputs to an
> >> accept/reject state on the basis of a property of these inputs thus
> >
> > And for the halting function, the property of the inputs is the representation of a turning machine and the input to that machine
> No not at all. There is no (indirect reference) of "representation" to
> it.
>

There is by the definition of the problem.

> The input is a literal string that precisely specifies a sequence of
> configurations. In this case it is x86 machine code.
> >>
> >> All halt deciders must compute the mapping from their inputs to an
> >> accept/reject state on the basis of a the actual behavior specified by
> >> these actual inputs.
> >
> > And by definition, the actual behavior specified by the actual inputs is the behavior of the turing machine represented by the first input whose input is the second input.
> No not at all. There is no (indirect reference) of "representation" to
> it. The input is a literal string that precisely specifies a sequence of
> configurations. In this case it is x86 machine code.
> > i.e. the the actual behavior specified by the actual inputs to H(P,P) is the behavior of P(P) *by the definition of the problem*.
> No not at all. There is no (indirect reference) of "representation" to
> it. The input is a literal string that precisely specifies a sequence of
> configurations. In this case it is x86 machine code.
>
> Either you are interpreting "representation" incorrectly or the
> statement of the problem never noticed that it directly contradicts the
> definition of a decider in some rare cases.

You mean this definition of a decider?

https://cs.stackexchange.com/a/84440

The one you "always accepted ... as the best one"?

All this says is that a decider maps input to accept/reject. It doesn't say anything about *how* that mapping occurs. In fact the post says "The term decider doesn't really have a standard meaning".

And the *definition* of the mapping performed by a halt decider:

H applied to <M> w goes to H.qy if and only if M applied to w halts, and
H applied to <M> w goes to H.qn if and only if M applied to w does not halt

In no way contradicts any of this. Not to mention a question like "does P(P) halt" (which is what the above is asking) can't be "wrong". And I'll bet that if the poster knew what you were proposing that he would agree that you're wrong. In fact he said of your definition: "You are looking for a decider whose language is the halting problem. There is no need for any special definitions".

So all your talk about "directly contradicts the definition of a decider" hold no merit.

H(P,P)==false is proven to be correct

<Z--dnd6-gL4VEuT_nZ2dnUU7_8zNnZ2d@giganews.com>

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 by: olcott - Mon, 9 May 2022 22:02 UTC

On 5/9/2022 4:26 PM, Dennis Bush wrote:
> On Monday, May 9, 2022 at 5:01:47 PM UTC-4, olcott wrote:
>> On 5/9/2022 3:37 PM, Dennis Bush wrote:
>>> On Monday, May 9, 2022 at 4:21:21 PM UTC-4, olcott wrote:
>>>> On 5/9/2022 1:36 PM, wij wrote:
>>>>> On Tuesday, 10 May 2022 at 02:13:22 UTC+8, olcott wrote:
>>>>>> On 5/9/2022 1:10 PM, wij wrote:
>>>>>>> On Tuesday, 10 May 2022 at 01:44:28 UTC+8, olcott wrote:
>>>>>>>> On 5/9/2022 12:35 PM, Mr Flibble wrote:
>>>>>>>>> On Mon, 9 May 2022 10:57:39 -0500
>>>>>>>>> olcott <No...@NoWhere.com> wrote:
>>>>>>>>>
>>>>>>>>>> On 5/8/2022 5:46 PM, Mr Flibble wrote:
>>>>>>>>>>> Based on the assumption that [Strachey, 1965] is not actually
>>>>>>>>>>> advocating running a program (either through direct execution or by
>>>>>>>>>>> simulation) to determine if that program halts Strachey's
>>>>>>>>>>> "Impossible Program" is indeed impossible for the reason given (the
>>>>>>>>>>> contradiction).
>>>>>>>>>>>
>>>>>>>>>>> The only open question in my mind is what it actually means for a
>>>>>>>>>>> decider to evaluate the source code of itself in addition to the
>>>>>>>>>>> source code of the program which would invoke the decider if it
>>>>>>>>>>> actually was run.
>>>>>>>>>>>
>>>>>>>>>>> I was wrong. Apologies for the noise.
>>>>>>>>>>>
>>>>>>>>>>> /Flibble
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> You were correct that the input never reaches its impossible part
>>>>>>>>>> because it is stuck in infinite recursion.
>>>>>>>>>
>>>>>>>>> Have you not read my retraction? I was in error: there is no infinite
>>>>>>>>> recursion.
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> This only occurs if the halt decider H is a simulating halt decider.
>>>>>>>>>> When the input to H(P,P) does get stuck in infinite recursion H can
>>>>>>>>>> see this.
>>>>>>>>>
>>>>>>>>> Simulation is an erroneous approach.
>>>>>>>>>
>>>>>>>>> /Flibble
>>>>>>>>>
>>>>>>>> I have proven otherwise:
>>>>>>>>
>>>>>>>> https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5
>>>>>>>> --
>>>>>>>> Copyright 2022 Pete Olcott
>>>>>>>>
>>>>>>>> "Talent hits a target no one else can hit;
>>>>>>>> Genius hits a target no one else can see."
>>>>>>>> Arthur Schopenhauer
>>>>>>>
>>>>>>> According to GUR, your proof is wrong:
>>>>>>> GUR says "No TM U can decide the property of a TM P if that property can be defied by TM P."
>>>>>> According to my proof GUR is wrong. You have to actually read it to see
>>>>>> this.
>>>>>> --
>>>>>> Copyright 2022 Pete Olcott
>>>>>>
>>>>>> "Talent hits a target no one else can hit;
>>>>>> Genius hits a target no one else can see."
>>>>>> Arthur Schopenhauer
>>>>>
>>>>> Any halting decider is shown/proved not able to return a correct answer.
>>>> All deciders must compute the mapping from their inputs to an
>>>> accept/reject state on the basis of a property of these inputs thus
>>>
>>> And for the halting function, the property of the inputs is the representation of a turning machine and the input to that machine
>> No not at all. There is no (indirect reference) of "representation" to
>> it.
>>
>
> There is by the definition of the problem.
>
>> The input is a literal string that precisely specifies a sequence of
>> configurations. In this case it is x86 machine code.
>>>>
>>>> All halt deciders must compute the mapping from their inputs to an
>>>> accept/reject state on the basis of a the actual behavior specified by
>>>> these actual inputs.
>>>
>>> And by definition, the actual behavior specified by the actual inputs is the behavior of the turing machine represented by the first input whose input is the second input.
>> No not at all. There is no (indirect reference) of "representation" to
>> it. The input is a literal string that precisely specifies a sequence of
>> configurations. In this case it is x86 machine code.
>>> i.e. the the actual behavior specified by the actual inputs to H(P,P) is the behavior of P(P) *by the definition of the problem*.
>> No not at all. There is no (indirect reference) of "representation" to
>> it. The input is a literal string that precisely specifies a sequence of
>> configurations. In this case it is x86 machine code.
>>
>> Either you are interpreting "representation" incorrectly or the
>> statement of the problem never noticed that it directly contradicts the
>> definition of a decider in some rare cases.
>
> You mean this definition of a decider?
>
> https://cs.stackexchange.com/a/84440
>
> The one you "always accepted ... as the best one"?
>
> All this says is that a decider maps input to accept/reject.
> It doesn't say anything about *how* that mapping occurs.

Now we are getting into nuances of meaning that are not commonly known.

It must do so in the basis of a property specified by its input finite
string.

For example a decider that decides whether or not its input has a string
length > 20 will simply base its decision on the actual length of the
actual input string.

Another decider may determine if the string contains: "the". Deciders
always decide on the basis of what the finite string actually specifies,
not what someone imagines that these strings "should" specify.

For the halt deciders this property is the actual behavior actually
specified by these finite strings. We already know that the correct
simulation of an input is the ultimate measure of the behavior of this
input.

As long as it can be verified that the simulating halt decider does
perform a correct simulation of its input we (and it) can base the halt
status decision on the behavior of this simulated input.

--
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: H(P,P)==false is proven to be correct

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Subject: Re: H(P,P)==false is proven to be correct
From: dbush.mo...@gmail.com (Dennis Bush)
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 by: Dennis Bush - Mon, 9 May 2022 22:49 UTC

On Monday, May 9, 2022 at 6:02:56 PM UTC-4, olcott wrote:
> On 5/9/2022 4:26 PM, Dennis Bush wrote:
> > On Monday, May 9, 2022 at 5:01:47 PM UTC-4, olcott wrote:
> >> On 5/9/2022 3:37 PM, Dennis Bush wrote:
> >>> On Monday, May 9, 2022 at 4:21:21 PM UTC-4, olcott wrote:
> >>>> On 5/9/2022 1:36 PM, wij wrote:
> >>>>> On Tuesday, 10 May 2022 at 02:13:22 UTC+8, olcott wrote:
> >>>>>> On 5/9/2022 1:10 PM, wij wrote:
> >>>>>>> On Tuesday, 10 May 2022 at 01:44:28 UTC+8, olcott wrote:
> >>>>>>>> On 5/9/2022 12:35 PM, Mr Flibble wrote:
> >>>>>>>>> On Mon, 9 May 2022 10:57:39 -0500
> >>>>>>>>> olcott <No...@NoWhere.com> wrote:
> >>>>>>>>>
> >>>>>>>>>> On 5/8/2022 5:46 PM, Mr Flibble wrote:
> >>>>>>>>>>> Based on the assumption that [Strachey, 1965] is not actually
> >>>>>>>>>>> advocating running a program (either through direct execution or by
> >>>>>>>>>>> simulation) to determine if that program halts Strachey's
> >>>>>>>>>>> "Impossible Program" is indeed impossible for the reason given (the
> >>>>>>>>>>> contradiction).
> >>>>>>>>>>>
> >>>>>>>>>>> The only open question in my mind is what it actually means for a
> >>>>>>>>>>> decider to evaluate the source code of itself in addition to the
> >>>>>>>>>>> source code of the program which would invoke the decider if it
> >>>>>>>>>>> actually was run.
> >>>>>>>>>>>
> >>>>>>>>>>> I was wrong. Apologies for the noise.
> >>>>>>>>>>>
> >>>>>>>>>>> /Flibble
> >>>>>>>>>>>
> >>>>>>>>>>
> >>>>>>>>>> You were correct that the input never reaches its impossible part
> >>>>>>>>>> because it is stuck in infinite recursion.
> >>>>>>>>>
> >>>>>>>>> Have you not read my retraction? I was in error: there is no infinite
> >>>>>>>>> recursion.
> >>>>>>>>>
> >>>>>>>>>>
> >>>>>>>>>> This only occurs if the halt decider H is a simulating halt decider.
> >>>>>>>>>> When the input to H(P,P) does get stuck in infinite recursion H can
> >>>>>>>>>> see this.
> >>>>>>>>>
> >>>>>>>>> Simulation is an erroneous approach.
> >>>>>>>>>
> >>>>>>>>> /Flibble
> >>>>>>>>>
> >>>>>>>> I have proven otherwise:
> >>>>>>>>
> >>>>>>>> https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5
> >>>>>>>> --
> >>>>>>>> Copyright 2022 Pete Olcott
> >>>>>>>>
> >>>>>>>> "Talent hits a target no one else can hit;
> >>>>>>>> Genius hits a target no one else can see."
> >>>>>>>> Arthur Schopenhauer
> >>>>>>>
> >>>>>>> According to GUR, your proof is wrong:
> >>>>>>> GUR says "No TM U can decide the property of a TM P if that property can be defied by TM P."
> >>>>>> According to my proof GUR is wrong. You have to actually read it to see
> >>>>>> this.
> >>>>>> --
> >>>>>> Copyright 2022 Pete Olcott
> >>>>>>
> >>>>>> "Talent hits a target no one else can hit;
> >>>>>> Genius hits a target no one else can see."
> >>>>>> Arthur Schopenhauer
> >>>>>
> >>>>> Any halting decider is shown/proved not able to return a correct answer.
> >>>> All deciders must compute the mapping from their inputs to an
> >>>> accept/reject state on the basis of a property of these inputs thus
> >>>
> >>> And for the halting function, the property of the inputs is the representation of a turning machine and the input to that machine
> >> No not at all. There is no (indirect reference) of "representation" to
> >> it.
> >>
> >
> > There is by the definition of the problem.
> >
> >> The input is a literal string that precisely specifies a sequence of
> >> configurations. In this case it is x86 machine code.
> >>>>
> >>>> All halt deciders must compute the mapping from their inputs to an
> >>>> accept/reject state on the basis of a the actual behavior specified by
> >>>> these actual inputs.
> >>>
> >>> And by definition, the actual behavior specified by the actual inputs is the behavior of the turing machine represented by the first input whose input is the second input.
> >> No not at all. There is no (indirect reference) of "representation" to
> >> it. The input is a literal string that precisely specifies a sequence of
> >> configurations. In this case it is x86 machine code.
> >>> i.e. the the actual behavior specified by the actual inputs to H(P,P) is the behavior of P(P) *by the definition of the problem*.
> >> No not at all. There is no (indirect reference) of "representation" to
> >> it. The input is a literal string that precisely specifies a sequence of
> >> configurations. In this case it is x86 machine code.
> >>
> >> Either you are interpreting "representation" incorrectly or the
> >> statement of the problem never noticed that it directly contradicts the
> >> definition of a decider in some rare cases.
> >
> > You mean this definition of a decider?
> >
> > https://cs.stackexchange.com/a/84440
> >
> > The one you "always accepted ... as the best one"?
> >
> > All this says is that a decider maps input to accept/reject.
> > It doesn't say anything about *how* that mapping occurs.
> Now we are getting into nuances of meaning that are not commonly known.
>
> It must do so in the basis of a property specified by its input finite
> string.

And for a halt decider H(P,P), the specified property *by definition* is the behavior of P(P)

>
> For example a decider that decides whether or not its input has a string
> length > 20 will simply base its decision on the actual length of the
> actual input string.
>
> Another decider may determine if the string contains: "the". Deciders
> always decide on the basis of what the finite string actually specifies,
> not what someone imagines that these strings "should" specify.
>
> For the halt deciders this property is the actual behavior actually
> specified by these finite strings.

Which by *the definition of the problem*, for a halt decider H(P,P), that property is the behavior of P(P)

> We already know that the correct
> simulation of an input is the ultimate measure of the behavior of this
> input.

And the correct simulation of the input to H(P,P) is by definition UTM(P,P) because it is equivalent to the behavior of the direct execution of P(P)

>
> As long as it can be verified that the simulating halt decider does
> perform a correct simulation of its input we (and it) can base the halt
> status decision on the behavior of this simulated input.

In the case of Ha(Pa,Pa) == false, we can verify that it does NOT do a correct simulation but instead aborts too soon. This is proved by UTM(Pa,Pa) which halts.

This is also proved by Hb which simulates for k more steps than Ha, and Hb(Pa,Pa) == true. And since Hb and Ha are both halt deciders (meaning that they compute the same function) and are given the same input, the fact that they give different answers means one must be wrong. And since Hb simulates this input to a final state, it is necessarily correct, proving Ha(Pa,Pa) == false wrong.

Re: H(P,P)==false is proven to be correct

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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Mon, 9 May 2022 23:00 UTC

On 5/9/2022 5:49 PM, Dennis Bush wrote:
> On Monday, May 9, 2022 at 6:02:56 PM UTC-4, olcott wrote:
>> On 5/9/2022 4:26 PM, Dennis Bush wrote:
>>> On Monday, May 9, 2022 at 5:01:47 PM UTC-4, olcott wrote:
>>>> On 5/9/2022 3:37 PM, Dennis Bush wrote:
>>>>> On Monday, May 9, 2022 at 4:21:21 PM UTC-4, olcott wrote:
>>>>>> On 5/9/2022 1:36 PM, wij wrote:
>>>>>>> On Tuesday, 10 May 2022 at 02:13:22 UTC+8, olcott wrote:
>>>>>>>> On 5/9/2022 1:10 PM, wij wrote:
>>>>>>>>> On Tuesday, 10 May 2022 at 01:44:28 UTC+8, olcott wrote:
>>>>>>>>>> On 5/9/2022 12:35 PM, Mr Flibble wrote:
>>>>>>>>>>> On Mon, 9 May 2022 10:57:39 -0500
>>>>>>>>>>> olcott <No...@NoWhere.com> wrote:
>>>>>>>>>>>
>>>>>>>>>>>> On 5/8/2022 5:46 PM, Mr Flibble wrote:
>>>>>>>>>>>>> Based on the assumption that [Strachey, 1965] is not actually
>>>>>>>>>>>>> advocating running a program (either through direct execution or by
>>>>>>>>>>>>> simulation) to determine if that program halts Strachey's
>>>>>>>>>>>>> "Impossible Program" is indeed impossible for the reason given (the
>>>>>>>>>>>>> contradiction).
>>>>>>>>>>>>>
>>>>>>>>>>>>> The only open question in my mind is what it actually means for a
>>>>>>>>>>>>> decider to evaluate the source code of itself in addition to the
>>>>>>>>>>>>> source code of the program which would invoke the decider if it
>>>>>>>>>>>>> actually was run.
>>>>>>>>>>>>>
>>>>>>>>>>>>> I was wrong. Apologies for the noise.
>>>>>>>>>>>>>
>>>>>>>>>>>>> /Flibble
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> You were correct that the input never reaches its impossible part
>>>>>>>>>>>> because it is stuck in infinite recursion.
>>>>>>>>>>>
>>>>>>>>>>> Have you not read my retraction? I was in error: there is no infinite
>>>>>>>>>>> recursion.
>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> This only occurs if the halt decider H is a simulating halt decider.
>>>>>>>>>>>> When the input to H(P,P) does get stuck in infinite recursion H can
>>>>>>>>>>>> see this.
>>>>>>>>>>>
>>>>>>>>>>> Simulation is an erroneous approach.
>>>>>>>>>>>
>>>>>>>>>>> /Flibble
>>>>>>>>>>>
>>>>>>>>>> I have proven otherwise:
>>>>>>>>>>
>>>>>>>>>> https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5
>>>>>>>>>> --
>>>>>>>>>> Copyright 2022 Pete Olcott
>>>>>>>>>>
>>>>>>>>>> "Talent hits a target no one else can hit;
>>>>>>>>>> Genius hits a target no one else can see."
>>>>>>>>>> Arthur Schopenhauer
>>>>>>>>>
>>>>>>>>> According to GUR, your proof is wrong:
>>>>>>>>> GUR says "No TM U can decide the property of a TM P if that property can be defied by TM P."
>>>>>>>> According to my proof GUR is wrong. You have to actually read it to see
>>>>>>>> this.
>>>>>>>> --
>>>>>>>> Copyright 2022 Pete Olcott
>>>>>>>>
>>>>>>>> "Talent hits a target no one else can hit;
>>>>>>>> Genius hits a target no one else can see."
>>>>>>>> Arthur Schopenhauer
>>>>>>>
>>>>>>> Any halting decider is shown/proved not able to return a correct answer.
>>>>>> All deciders must compute the mapping from their inputs to an
>>>>>> accept/reject state on the basis of a property of these inputs thus
>>>>>
>>>>> And for the halting function, the property of the inputs is the representation of a turning machine and the input to that machine
>>>> No not at all. There is no (indirect reference) of "representation" to
>>>> it.
>>>>
>>>
>>> There is by the definition of the problem.
>>>
>>>> The input is a literal string that precisely specifies a sequence of
>>>> configurations. In this case it is x86 machine code.
>>>>>>
>>>>>> All halt deciders must compute the mapping from their inputs to an
>>>>>> accept/reject state on the basis of a the actual behavior specified by
>>>>>> these actual inputs.
>>>>>
>>>>> And by definition, the actual behavior specified by the actual inputs is the behavior of the turing machine represented by the first input whose input is the second input.
>>>> No not at all. There is no (indirect reference) of "representation" to
>>>> it. The input is a literal string that precisely specifies a sequence of
>>>> configurations. In this case it is x86 machine code.
>>>>> i.e. the the actual behavior specified by the actual inputs to H(P,P) is the behavior of P(P) *by the definition of the problem*.
>>>> No not at all. There is no (indirect reference) of "representation" to
>>>> it. The input is a literal string that precisely specifies a sequence of
>>>> configurations. In this case it is x86 machine code.
>>>>
>>>> Either you are interpreting "representation" incorrectly or the
>>>> statement of the problem never noticed that it directly contradicts the
>>>> definition of a decider in some rare cases.
>>>
>>> You mean this definition of a decider?
>>>
>>> https://cs.stackexchange.com/a/84440
>>>
>>> The one you "always accepted ... as the best one"?
>>>
>>> All this says is that a decider maps input to accept/reject.
>>> It doesn't say anything about *how* that mapping occurs.
>> Now we are getting into nuances of meaning that are not commonly known.
>>
>> It must do so in the basis of a property specified by its input finite
>> string.
>
> And for a halt decider H(P,P), the specified property *by definition* is the behavior of P(P)
>
>>
>> For example a decider that decides whether or not its input has a string
>> length > 20 will simply base its decision on the actual length of the
>> actual input string.
>>
>> Another decider may determine if the string contains: "the". Deciders
>> always decide on the basis of what the finite string actually specifies,
>> not what someone imagines that these strings "should" specify.
>>
>> For the halt deciders this property is the actual behavior actually
>> specified by these finite strings.
>
> Which by *the definition of the problem*, for a halt decider H(P,P), that property is the behavior of P(P)
>
>> We already know that the correct
>> simulation of an input is the ultimate measure of the behavior of this
>> input.
>
> And the correct simulation of the input to H(P,P) is by definition UTM(P,P) because it is equivalent to the behavior of the direct execution of P(P)

void P(u32 x)
{ if (x86_emulate(x, x))
HERE: goto HERE;
return;
}

int main()
{ Output("Input_Halts = ", H((u32)P, (u32)P));
}

Yet the UTM must be embedded at the same place where H would be
embedded. The above P would never stop running.

--
Copyright 2022 Pete Olcott

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


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Re: H(P,P)==false is proven to be correct

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Subject: Re: H(P,P)==false is proven to be correct
From: dbush.mo...@gmail.com (Dennis Bush)
Injection-Date: Mon, 09 May 2022 23:32:10 +0000
Content-Type: text/plain; charset="UTF-8"
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 by: Dennis Bush - Mon, 9 May 2022 23:32 UTC

On Monday, May 9, 2022 at 7:00:22 PM UTC-4, olcott wrote:
> On 5/9/2022 5:49 PM, Dennis Bush wrote:
> > On Monday, May 9, 2022 at 6:02:56 PM UTC-4, olcott wrote:
> >> On 5/9/2022 4:26 PM, Dennis Bush wrote:
> >>> On Monday, May 9, 2022 at 5:01:47 PM UTC-4, olcott wrote:
> >>>> On 5/9/2022 3:37 PM, Dennis Bush wrote:
> >>>>> On Monday, May 9, 2022 at 4:21:21 PM UTC-4, olcott wrote:
> >>>>>> On 5/9/2022 1:36 PM, wij wrote:
> >>>>>>> On Tuesday, 10 May 2022 at 02:13:22 UTC+8, olcott wrote:
> >>>>>>>> On 5/9/2022 1:10 PM, wij wrote:
> >>>>>>>>> On Tuesday, 10 May 2022 at 01:44:28 UTC+8, olcott wrote:
> >>>>>>>>>> On 5/9/2022 12:35 PM, Mr Flibble wrote:
> >>>>>>>>>>> On Mon, 9 May 2022 10:57:39 -0500
> >>>>>>>>>>> olcott <No...@NoWhere.com> wrote:
> >>>>>>>>>>>
> >>>>>>>>>>>> On 5/8/2022 5:46 PM, Mr Flibble wrote:
> >>>>>>>>>>>>> Based on the assumption that [Strachey, 1965] is not actually
> >>>>>>>>>>>>> advocating running a program (either through direct execution or by
> >>>>>>>>>>>>> simulation) to determine if that program halts Strachey's
> >>>>>>>>>>>>> "Impossible Program" is indeed impossible for the reason given (the
> >>>>>>>>>>>>> contradiction).
> >>>>>>>>>>>>>
> >>>>>>>>>>>>> The only open question in my mind is what it actually means for a
> >>>>>>>>>>>>> decider to evaluate the source code of itself in addition to the
> >>>>>>>>>>>>> source code of the program which would invoke the decider if it
> >>>>>>>>>>>>> actually was run.
> >>>>>>>>>>>>>
> >>>>>>>>>>>>> I was wrong. Apologies for the noise.
> >>>>>>>>>>>>>
> >>>>>>>>>>>>> /Flibble
> >>>>>>>>>>>>>
> >>>>>>>>>>>>
> >>>>>>>>>>>> You were correct that the input never reaches its impossible part
> >>>>>>>>>>>> because it is stuck in infinite recursion.
> >>>>>>>>>>>
> >>>>>>>>>>> Have you not read my retraction? I was in error: there is no infinite
> >>>>>>>>>>> recursion.
> >>>>>>>>>>>
> >>>>>>>>>>>>
> >>>>>>>>>>>> This only occurs if the halt decider H is a simulating halt decider.
> >>>>>>>>>>>> When the input to H(P,P) does get stuck in infinite recursion H can
> >>>>>>>>>>>> see this.
> >>>>>>>>>>>
> >>>>>>>>>>> Simulation is an erroneous approach.
> >>>>>>>>>>>
> >>>>>>>>>>> /Flibble
> >>>>>>>>>>>
> >>>>>>>>>> I have proven otherwise:
> >>>>>>>>>>
> >>>>>>>>>> https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5
> >>>>>>>>>> --
> >>>>>>>>>> Copyright 2022 Pete Olcott
> >>>>>>>>>>
> >>>>>>>>>> "Talent hits a target no one else can hit;
> >>>>>>>>>> Genius hits a target no one else can see."
> >>>>>>>>>> Arthur Schopenhauer
> >>>>>>>>>
> >>>>>>>>> According to GUR, your proof is wrong:
> >>>>>>>>> GUR says "No TM U can decide the property of a TM P if that property can be defied by TM P."
> >>>>>>>> According to my proof GUR is wrong. You have to actually read it to see
> >>>>>>>> this.
> >>>>>>>> --
> >>>>>>>> Copyright 2022 Pete Olcott
> >>>>>>>>
> >>>>>>>> "Talent hits a target no one else can hit;
> >>>>>>>> Genius hits a target no one else can see."
> >>>>>>>> Arthur Schopenhauer
> >>>>>>>
> >>>>>>> Any halting decider is shown/proved not able to return a correct answer.
> >>>>>> All deciders must compute the mapping from their inputs to an
> >>>>>> accept/reject state on the basis of a property of these inputs thus
> >>>>>
> >>>>> And for the halting function, the property of the inputs is the representation of a turning machine and the input to that machine
> >>>> No not at all. There is no (indirect reference) of "representation" to
> >>>> it.
> >>>>
> >>>
> >>> There is by the definition of the problem.
> >>>
> >>>> The input is a literal string that precisely specifies a sequence of
> >>>> configurations. In this case it is x86 machine code.
> >>>>>>
> >>>>>> All halt deciders must compute the mapping from their inputs to an
> >>>>>> accept/reject state on the basis of a the actual behavior specified by
> >>>>>> these actual inputs.
> >>>>>
> >>>>> And by definition, the actual behavior specified by the actual inputs is the behavior of the turing machine represented by the first input whose input is the second input.
> >>>> No not at all. There is no (indirect reference) of "representation" to
> >>>> it. The input is a literal string that precisely specifies a sequence of
> >>>> configurations. In this case it is x86 machine code.
> >>>>> i.e. the the actual behavior specified by the actual inputs to H(P,P) is the behavior of P(P) *by the definition of the problem*.
> >>>> No not at all. There is no (indirect reference) of "representation" to
> >>>> it. The input is a literal string that precisely specifies a sequence of
> >>>> configurations. In this case it is x86 machine code.
> >>>>
> >>>> Either you are interpreting "representation" incorrectly or the
> >>>> statement of the problem never noticed that it directly contradicts the
> >>>> definition of a decider in some rare cases.
> >>>
> >>> You mean this definition of a decider?
> >>>
> >>> https://cs.stackexchange.com/a/84440
> >>>
> >>> The one you "always accepted ... as the best one"?
> >>>
> >>> All this says is that a decider maps input to accept/reject.
> >>> It doesn't say anything about *how* that mapping occurs.
> >> Now we are getting into nuances of meaning that are not commonly known..
> >>
> >> It must do so in the basis of a property specified by its input finite
> >> string.
> >
> > And for a halt decider H(P,P), the specified property *by definition* is the behavior of P(P)
> >
> >>
> >> For example a decider that decides whether or not its input has a string
> >> length > 20 will simply base its decision on the actual length of the
> >> actual input string.
> >>
> >> Another decider may determine if the string contains: "the". Deciders
> >> always decide on the basis of what the finite string actually specifies,
> >> not what someone imagines that these strings "should" specify.
> >>
> >> For the halt deciders this property is the actual behavior actually
> >> specified by these finite strings.
> >
> > Which by *the definition of the problem*, for a halt decider H(P,P), that property is the behavior of P(P)
> >
> >> We already know that the correct
> >> simulation of an input is the ultimate measure of the behavior of this
> >> input.
> >
> > And the correct simulation of the input to H(P,P) is by definition UTM(P,P) because it is equivalent to the behavior of the direct execution of P(P)
> void P(u32 x)
> {
> if (x86_emulate(x, x))
> HERE: goto HERE;
> return;
> }
>
> int main()
> {
> Output("Input_Halts = ", H((u32)P, (u32)P));
> }
>


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Re: H(P,P)==false is proven to be correct

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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Tue, 10 May 2022 00:26 UTC

On 5/9/22 7:00 PM, olcott wrote:
> On 5/9/2022 5:49 PM, Dennis Bush wrote:
>> On Monday, May 9, 2022 at 6:02:56 PM UTC-4, olcott wrote:
>>> On 5/9/2022 4:26 PM, Dennis Bush wrote:
>>>> On Monday, May 9, 2022 at 5:01:47 PM UTC-4, olcott wrote:
>>>>> On 5/9/2022 3:37 PM, Dennis Bush wrote:
>>>>>> On Monday, May 9, 2022 at 4:21:21 PM UTC-4, olcott wrote:
>>>>>>> On 5/9/2022 1:36 PM, wij wrote:
>>>>>>>> On Tuesday, 10 May 2022 at 02:13:22 UTC+8, olcott wrote:
>>>>>>>>> On 5/9/2022 1:10 PM, wij wrote:
>>>>>>>>>> On Tuesday, 10 May 2022 at 01:44:28 UTC+8, olcott wrote:
>>>>>>>>>>> On 5/9/2022 12:35 PM, Mr Flibble wrote:
>>>>>>>>>>>> On Mon, 9 May 2022 10:57:39 -0500
>>>>>>>>>>>> olcott <No...@NoWhere.com> wrote:
>>>>>>>>>>>>
>>>>>>>>>>>>> On 5/8/2022 5:46 PM, Mr Flibble wrote:
>>>>>>>>>>>>>> Based on the assumption that [Strachey, 1965] is not actually
>>>>>>>>>>>>>> advocating running a program (either through direct
>>>>>>>>>>>>>> execution or by
>>>>>>>>>>>>>> simulation) to determine if that program halts Strachey's
>>>>>>>>>>>>>> "Impossible Program" is indeed impossible for the reason
>>>>>>>>>>>>>> given (the
>>>>>>>>>>>>>> contradiction).
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> The only open question in my mind is what it actually
>>>>>>>>>>>>>> means for a
>>>>>>>>>>>>>> decider to evaluate the source code of itself in addition
>>>>>>>>>>>>>> to the
>>>>>>>>>>>>>> source code of the program which would invoke the decider
>>>>>>>>>>>>>> if it
>>>>>>>>>>>>>> actually was run.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> I was wrong. Apologies for the noise.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> /Flibble
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> You were correct that the input never reaches its
>>>>>>>>>>>>> impossible part
>>>>>>>>>>>>> because it is stuck in infinite recursion.
>>>>>>>>>>>>
>>>>>>>>>>>> Have you not read my retraction? I was in error: there is no
>>>>>>>>>>>> infinite
>>>>>>>>>>>> recursion.
>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> This only occurs if the halt decider H is a simulating halt
>>>>>>>>>>>>> decider.
>>>>>>>>>>>>> When the input to H(P,P) does get stuck in infinite
>>>>>>>>>>>>> recursion H can
>>>>>>>>>>>>> see this.
>>>>>>>>>>>>
>>>>>>>>>>>> Simulation is an erroneous approach.
>>>>>>>>>>>>
>>>>>>>>>>>> /Flibble
>>>>>>>>>>>>
>>>>>>>>>>> I have proven otherwise:
>>>>>>>>>>>
>>>>>>>>>>> https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5
>>>>>>>>>>>
>>>>>>>>>>> --
>>>>>>>>>>> Copyright 2022 Pete Olcott
>>>>>>>>>>>
>>>>>>>>>>> "Talent hits a target no one else can hit;
>>>>>>>>>>> Genius hits a target no one else can see."
>>>>>>>>>>> Arthur Schopenhauer
>>>>>>>>>>
>>>>>>>>>> According to GUR, your proof is wrong:
>>>>>>>>>> GUR says "No TM U can decide the property of a TM P if that
>>>>>>>>>> property can be defied by TM P."
>>>>>>>>> According to my proof GUR is wrong. You have to actually read
>>>>>>>>> it to see
>>>>>>>>> this.
>>>>>>>>> --
>>>>>>>>> Copyright 2022 Pete Olcott
>>>>>>>>>
>>>>>>>>> "Talent hits a target no one else can hit;
>>>>>>>>> Genius hits a target no one else can see."
>>>>>>>>> Arthur Schopenhauer
>>>>>>>>
>>>>>>>> Any halting decider is shown/proved not able to return a correct
>>>>>>>> answer.
>>>>>>> All deciders must compute the mapping from their inputs to an
>>>>>>> accept/reject state on the basis of a property of these inputs thus
>>>>>>
>>>>>> And for the halting function, the property of the inputs is the
>>>>>> representation of a turning machine and the input to that machine
>>>>> No not at all. There is no (indirect reference) of "representation" to
>>>>> it.
>>>>>
>>>>
>>>> There is by the definition of the problem.
>>>>
>>>>> The input is a literal string that precisely specifies a sequence of
>>>>> configurations. In this case it is x86 machine code.
>>>>>>>
>>>>>>> All halt deciders must compute the mapping from their inputs to an
>>>>>>> accept/reject state on the basis of a the actual behavior
>>>>>>> specified by
>>>>>>> these actual inputs.
>>>>>>
>>>>>> And by definition, the actual behavior specified by the actual
>>>>>> inputs is the behavior of the turing machine represented by the
>>>>>> first input whose input is the second input.
>>>>> No not at all. There is no (indirect reference) of "representation" to
>>>>> it. The input is a literal string that precisely specifies a
>>>>> sequence of
>>>>> configurations. In this case it is x86 machine code.
>>>>>> i.e. the the actual behavior specified by the actual inputs to
>>>>>> H(P,P) is the behavior of P(P) *by the definition of the problem*.
>>>>> No not at all. There is no (indirect reference) of "representation" to
>>>>> it. The input is a literal string that precisely specifies a
>>>>> sequence of
>>>>> configurations. In this case it is x86 machine code.
>>>>>
>>>>> Either you are interpreting "representation" incorrectly or the
>>>>> statement of the problem never noticed that it directly contradicts
>>>>> the
>>>>> definition of a decider in some rare cases.
>>>>
>>>> You mean this definition of a decider?
>>>>
>>>> https://cs.stackexchange.com/a/84440
>>>>
>>>> The one you "always accepted ... as the best one"?
>>>>
>>>> All this says is that a decider maps input to accept/reject.
>>>> It doesn't say anything about *how* that mapping occurs.
>>> Now we are getting into nuances of meaning that are not commonly known.
>>>
>>> It must do so in the basis of a property specified by its input finite
>>> string.
>>
>> And for a halt decider H(P,P), the specified property *by definition*
>> is the behavior of P(P)
>>
>>>
>>> For example a decider that decides whether or not its input has a string
>>> length > 20 will simply base its decision on the actual length of the
>>> actual input string.
>>>
>>> Another decider may determine if the string contains: "the". Deciders
>>> always decide on the basis of what the finite string actually specifies,
>>> not what someone imagines that these strings "should" specify.
>>>
>>> For the halt deciders this property is the actual behavior actually
>>> specified by these finite strings.
>>
>> Which by *the definition of the problem*, for a halt decider H(P,P),
>> that property is the behavior of P(P)
>>
>>> We already know that the correct
>>> simulation of an input is the ultimate measure of the behavior of this
>>> input.
>>
>> And the correct simulation of the input to H(P,P) is by definition
>> UTM(P,P) because it is equivalent to the behavior of the direct
>> execution of P(P)
>
>
> void P(u32 x)
> {
>   if (x86_emulate(x, x))
>     HERE: goto HERE;
>   return;
> }
>
> int main()
> {
>   Output("Input_Halts = ", H((u32)P, (u32)P));
> }
>
> Yet the UTM must be embedded at the same place where H would be
> embedded. The above P would never stop running.
>
>


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Subject: Re: H(P,P)==false is proven to be correct
From: wynii...@gmail.com (wij)
Injection-Date: Tue, 10 May 2022 15:25:35 +0000
Content-Type: text/plain; charset="UTF-8"
 by: wij - Tue, 10 May 2022 15:25 UTC

On Tuesday, 10 May 2022 at 07:00:22 UTC+8, olcott wrote:
> On 5/9/2022 5:49 PM, Dennis Bush wrote:
> > On Monday, May 9, 2022 at 6:02:56 PM UTC-4, olcott wrote:
> >> On 5/9/2022 4:26 PM, Dennis Bush wrote:
> >>> On Monday, May 9, 2022 at 5:01:47 PM UTC-4, olcott wrote:
> >>>> On 5/9/2022 3:37 PM, Dennis Bush wrote:
> >>>>> On Monday, May 9, 2022 at 4:21:21 PM UTC-4, olcott wrote:
> >>>>>> On 5/9/2022 1:36 PM, wij wrote:
> >>>>>>> On Tuesday, 10 May 2022 at 02:13:22 UTC+8, olcott wrote:
> >>>>>>>> On 5/9/2022 1:10 PM, wij wrote:
> >>>>>>>>> On Tuesday, 10 May 2022 at 01:44:28 UTC+8, olcott wrote:
> >>>>>>>>>> On 5/9/2022 12:35 PM, Mr Flibble wrote:
> >>>>>>>>>>> On Mon, 9 May 2022 10:57:39 -0500
> >>>>>>>>>>> olcott <No...@NoWhere.com> wrote:
> >>>>>>>>>>>
> >>>>>>>>>>>> On 5/8/2022 5:46 PM, Mr Flibble wrote:
> >>>>>>>>>>>>> Based on the assumption that [Strachey, 1965] is not actually
> >>>>>>>>>>>>> advocating running a program (either through direct execution or by
> >>>>>>>>>>>>> simulation) to determine if that program halts Strachey's
> >>>>>>>>>>>>> "Impossible Program" is indeed impossible for the reason given (the
> >>>>>>>>>>>>> contradiction).
> >>>>>>>>>>>>>
> >>>>>>>>>>>>> The only open question in my mind is what it actually means for a
> >>>>>>>>>>>>> decider to evaluate the source code of itself in addition to the
> >>>>>>>>>>>>> source code of the program which would invoke the decider if it
> >>>>>>>>>>>>> actually was run.
> >>>>>>>>>>>>>
> >>>>>>>>>>>>> I was wrong. Apologies for the noise.
> >>>>>>>>>>>>>
> >>>>>>>>>>>>> /Flibble
> >>>>>>>>>>>>>
> >>>>>>>>>>>>
> >>>>>>>>>>>> You were correct that the input never reaches its impossible part
> >>>>>>>>>>>> because it is stuck in infinite recursion.
> >>>>>>>>>>>
> >>>>>>>>>>> Have you not read my retraction? I was in error: there is no infinite
> >>>>>>>>>>> recursion.
> >>>>>>>>>>>
> >>>>>>>>>>>>
> >>>>>>>>>>>> This only occurs if the halt decider H is a simulating halt decider.
> >>>>>>>>>>>> When the input to H(P,P) does get stuck in infinite recursion H can
> >>>>>>>>>>>> see this.
> >>>>>>>>>>>
> >>>>>>>>>>> Simulation is an erroneous approach.
> >>>>>>>>>>>
> >>>>>>>>>>> /Flibble
> >>>>>>>>>>>
> >>>>>>>>>> I have proven otherwise:
> >>>>>>>>>>
> >>>>>>>>>> https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5
> >>>>>>>>>> --
> >>>>>>>>>> Copyright 2022 Pete Olcott
> >>>>>>>>>>
> >>>>>>>>>> "Talent hits a target no one else can hit;
> >>>>>>>>>> Genius hits a target no one else can see."
> >>>>>>>>>> Arthur Schopenhauer
> >>>>>>>>>
> >>>>>>>>> According to GUR, your proof is wrong:
> >>>>>>>>> GUR says "No TM U can decide the property of a TM P if that property can be defied by TM P."
> >>>>>>>> According to my proof GUR is wrong. You have to actually read it to see
> >>>>>>>> this.
> >>>>>>>> --
> >>>>>>>> Copyright 2022 Pete Olcott
> >>>>>>>>
> >>>>>>>> "Talent hits a target no one else can hit;
> >>>>>>>> Genius hits a target no one else can see."
> >>>>>>>> Arthur Schopenhauer
> >>>>>>>
> >>>>>>> Any halting decider is shown/proved not able to return a correct answer.
> >>>>>> All deciders must compute the mapping from their inputs to an
> >>>>>> accept/reject state on the basis of a property of these inputs thus
> >>>>>
> >>>>> And for the halting function, the property of the inputs is the representation of a turning machine and the input to that machine
> >>>> No not at all. There is no (indirect reference) of "representation" to
> >>>> it.
> >>>>
> >>>
> >>> There is by the definition of the problem.
> >>>
> >>>> The input is a literal string that precisely specifies a sequence of
> >>>> configurations. In this case it is x86 machine code.
> >>>>>>
> >>>>>> All halt deciders must compute the mapping from their inputs to an
> >>>>>> accept/reject state on the basis of a the actual behavior specified by
> >>>>>> these actual inputs.
> >>>>>
> >>>>> And by definition, the actual behavior specified by the actual inputs is the behavior of the turing machine represented by the first input whose input is the second input.
> >>>> No not at all. There is no (indirect reference) of "representation" to
> >>>> it. The input is a literal string that precisely specifies a sequence of
> >>>> configurations. In this case it is x86 machine code.
> >>>>> i.e. the the actual behavior specified by the actual inputs to H(P,P) is the behavior of P(P) *by the definition of the problem*.
> >>>> No not at all. There is no (indirect reference) of "representation" to
> >>>> it. The input is a literal string that precisely specifies a sequence of
> >>>> configurations. In this case it is x86 machine code.
> >>>>
> >>>> Either you are interpreting "representation" incorrectly or the
> >>>> statement of the problem never noticed that it directly contradicts the
> >>>> definition of a decider in some rare cases.
> >>>
> >>> You mean this definition of a decider?
> >>>
> >>> https://cs.stackexchange.com/a/84440
> >>>
> >>> The one you "always accepted ... as the best one"?
> >>>
> >>> All this says is that a decider maps input to accept/reject.
> >>> It doesn't say anything about *how* that mapping occurs.
> >> Now we are getting into nuances of meaning that are not commonly known.
> >>
> >> It must do so in the basis of a property specified by its input finite
> >> string.
> >
> > And for a halt decider H(P,P), the specified property *by definition* is the behavior of P(P)
> >
> >>
> >> For example a decider that decides whether or not its input has a string
> >> length > 20 will simply base its decision on the actual length of the
> >> actual input string.
> >>
> >> Another decider may determine if the string contains: "the". Deciders
> >> always decide on the basis of what the finite string actually specifies,
> >> not what someone imagines that these strings "should" specify.
> >>
> >> For the halt deciders this property is the actual behavior actually
> >> specified by these finite strings.
> >
> > Which by *the definition of the problem*, for a halt decider H(P,P), that property is the behavior of P(P)
> >
> >> We already know that the correct
> >> simulation of an input is the ultimate measure of the behavior of this
> >> input.
> >
> > And the correct simulation of the input to H(P,P) is by definition UTM(P,P) because it is equivalent to the behavior of the direct execution of P(P)
> void P(u32 x)
> {
> if (x86_emulate(x, x))
> HERE: goto HERE;
> return;
> }
>
> int main()
> {
> Output("Input_Halts = ", H((u32)P, (u32)P));
> }
>
> Yet the UTM must be embedded at the same place where H would be
> embedded. The above P would never stop running.
> --
> Copyright 2022 Pete Olcott
>
> "Talent hits a target no one else can hit;
> Genius hits a target no one else can see."
> Arthur Schopenhauer


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