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devel / comp.theory / Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versus logic ]

SubjectAuthor
* H(P,P)==0 is correct for every simulating halt decider H --- V2olcott
+* H(P,P)==0 is correct for every simulating halt decider H --- V2Ben Bacarisse
|`* H(P,P)==0 is correct for every simulating halt decider H --- V2olcott
| +* H(P,P)==0 is correct for every simulating halt decider H --- V2Ben Bacarisse
| |`* H(P,P)==0 is correct for every simulating halt decider H --- V2olcott
| | +- H(P,P)==0 is correct for every simulating halt decider H --- V2Richard Damon
| | `* H(P,P)==0 is correct for every simulating halt decider H --- V2Ben Bacarisse
| |  `* H(P,P)==0 is correct for every simulating halt decider H --- V2olcott
| |   +* H(P,P)==0 is correct for every simulating halt decider H --- V2Ben Bacarisse
| |   |+* H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versolcott
| |   ||`* H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versBen Bacarisse
| |   || `* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   ||  +* H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versBen Bacarisse
| |   ||  |`* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   ||  | +* H(P,P)==0 is correct for every simulating halt decider H --- V2 [André G. Isaak
| |   ||  | |`* H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versolcott
| |   ||  | | +* H(P,P)==0 is correct for every simulating halt decider H --- V2 [André G. Isaak
| |   ||  | | |`* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   ||  | | | +* H(P,P)==0 is correct for every simulating halt decider H --- V2 [André G. Isaak
| |   ||  | | | |`* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   ||  | | | | +* H(P,P)==0 is correct for every simulating halt decider H --- V2 [André G. Isaak
| |   ||  | | | | |`* H(P,P)==0 is correct for every simulating halt decider H --- V2 [ impossibly incolcott
| |   ||  | | | | | +- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  | | | | | `* H(P,P)==0 is correct for every simulating halt decider H --- V2 [André G. Isaak
| |   ||  | | | | |  `* H(P,P)==0 is correct for every simulating halt decider H --- V2 [ impossibly incolcott
| |   ||  | | | | |   +* H(P,P)==0 is correct for every simulating halt decider H --- V2 [André G. Isaak
| |   ||  | | | | |   |`* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   ||  | | | | |   | +- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  | | | | |   | +* H(P,P)==0 is correct for every simulating halt decider H --- V2 [André G. Isaak
| |   ||  | | | | |   | |`* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   ||  | | | | |   | | +* H(P,P)==0 is correct for every simulating halt decider H --- V2 [André G. Isaak
| |   ||  | | | | |   | | |+* H(P,P)==0 is correct for every simulating halt decider H --- V2 [Mike Terry
| |   ||  | | | | |   | | ||+* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   ||  | | | | |   | | |||`- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  | | | | |   | | ||`* H(P,P)==0 is correct for every simulating halt decider H --- V2 [Mike Terry
| |   ||  | | | | |   | | || +* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   ||  | | | | |   | | || |`- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  | | | | |   | | || `- H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   ||  | | | | |   | | |+* H(P,P)==0 is correct for every simulating halt decider H --- V2 [Malcolm McLean
| |   ||  | | | | |   | | ||`* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   ||  | | | | |   | | || `- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  | | | | |   | | |`* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   ||  | | | | |   | | | `* H(P,P)==0 is correct for every simulating halt decider H --- V2 [André G. Isaak
| |   ||  | | | | |   | | |  `* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   ||  | | | | |   | | |   +* H(P,P)==0 is correct for every simulating halt decider H --- V2 [André G. Isaak
| |   ||  | | | | |   | | |   |`* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   ||  | | | | |   | | |   | `- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  | | | | |   | | |   `- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  | | | | |   | | `- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  | | | | |   | `- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  | | | | |   `- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  | | | | `- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  | | | `- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  | | `- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  | +- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  | `* H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versBen Bacarisse
| |   ||  |  `* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   ||  |   +* H(P,P)==0 is correct for every simulating halt decider H --- V2 [André G. Isaak
| |   ||  |   |`* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   ||  |   | +* H(P,P)==0 is correct for every simulating halt decider H --- V2 [ updated criterBen Bacarisse
| |   ||  |   | |`* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   ||  |   | | `- H(P,P)==0 is correct for every simulating halt decider H --- V2 [ updated criterBen Bacarisse
| |   ||  |   | +* H(P,P)==0 is correct for every simulating halt decider H --- V2 [André G. Isaak
| |   ||  |   | |`* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   ||  |   | | +* H(P,P)==0 is correct for every simulating halt decider H --- V2 [André G. Isaak
| |   ||  |   | | |`* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   ||  |   | | | +* H(P,P)==0 is correct for every simulating halt decider H --- V2 [André G. Isaak
| |   ||  |   | | | |+* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   ||  |   | | | ||`* H(P,P)==0 is correct for every simulating halt decider H --- V2 [André G. Isaak
| |   ||  |   | | | || `* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   ||  |   | | | ||  +* H(P,P)==0 is correct for every simulating halt decider H --- V2 [André G. Isaak
| |   ||  |   | | | ||  |`* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   ||  |   | | | ||  | +* H(P,P)==0 is correct for every simulating halt decider H --- V2 [André G. Isaak
| |   ||  |   | | | ||  | |`* H(P,P)==0 is correct for every simulating halt decider H --- V2 [ complete proofolcott
| |   ||  |   | | | ||  | | +- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  |   | | | ||  | | `- H(P,P)==0 is correct for every simulating halt decider H --- V2 [André G. Isaak
| |   ||  |   | | | ||  | `* H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  |   | | | ||  |  `* H(P,P)==0 is correct for every simulating halt decider H --- V2 [ complete proofolcott
| |   ||  |   | | | ||  |   `- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  |   | | | ||  `- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  |   | | | |`- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  |   | | | `- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  |   | | `- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  |   | `- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  |   +* H(P,P)==0 is correct for every simulating halt decider H --- V2 [ updated criterBen Bacarisse
| |   ||  |   |`* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   ||  |   | +- H(P,P)==0 is correct for every simulating halt decider H --- V2 [André G. Isaak
| |   ||  |   | +- H(P,P)==0 is correct for every simulating halt decider H --- V2 [ updated criterBen Bacarisse
| |   ||  |   | `- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  |   `- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   ||  `- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   |`* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   | +* H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versBen Bacarisse
| |   | |`* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   | | `* H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versBen Bacarisse
| |   | |  `* H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   | |   +* H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versBen Bacarisse
| |   | |   |`- H(P,P)==0 is correct for every simulating halt decider H --- V2 [olcott
| |   | |   `* H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   | |    `* H(P,P)==0 is correct for every simulating halt decider H --- V2 [André G. Isaak
| |   | |     `- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   | `- H(P,P)==0 is correct for every simulating halt decider H --- V2 [Richard Damon
| |   +* H(P,P)==0 is correct for every simulating halt decider H --- V2André G. Isaak
| |   `- H(P,P)==0 is correct for every simulating halt decider H --- V2Richard Damon
| `- H(P,P)==0 is correct for every simulating halt decider H --- V2Richard Damon
+- H(P,P)==0 is correct for every simulating halt decider H --- V2Richard Damon
`* H(P,P)==0 is correct for every simulating halt decider H --- V2Malcolm McLean

Pages:123456
Re: H(P,P)==0 is correct for every simulating halt decider H --- V2

<hL2dnRDG7N7dBh38nZ2dnUU7-cmdnZ2d@giganews.com>

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 by: olcott - Tue, 2 Nov 2021 01:58 UTC

On 11/1/2021 8:31 PM, Jeff Barnett wrote:
> On 11/1/2021 3:22 PM, olcott wrote:
>> On 11/1/2021 3:33 PM, Jeff Barnett wrote:
>>> On 11/1/2021 10:46 AM, olcott wrote:
>>>> On 11/1/2021 11:31 AM, Jeff Barnett wrote:
>>>>> On 11/1/2021 6:46 AM, Malcolm McLean wrote:
>>>>>> On Monday, 1 November 2021 at 03:45:47 UTC, olcott wrote:
>>>>>
>>>>>       <SNIP POOP>
>>>>>
>>>>>> I can see where the confusion lies. If H never aborts, it
>>>>>> simulates P forever.
>>>>>> If it aborts, then it aborts because it has detected that
>>>>>> otherwise P would
>>>>>> simulate forever.
>>>>>
>>>>> Malcolm, note well: "confusion" does not lie; it's the permanently
>>>>> self-bamboozled PO who lies.
>>>>>
>>>>>        <SNIP>
>>>>
>>>> The part that you quoted proves that I am correct, yet because you
>>>> are clueless about these things you don't see this. You have never
>>>> ever provided any reply what-so-ever (that I remember) that was
>>>> anything at all besides denigration. Ad Hominem is your only tool.
>>>>
>>>> Ad Hominem
>>>> (Attacking the person): This fallacy occurs when, instead of
>>>> addressing someone's argument or position, you irrelevantly attack
>>>> the person or some aspect of the person who is making the argument.
>>>> https://www.txstate.edu/philosophy/resources/fallacy-definitions/Ad-Hominem.html
>>>
>>>
>>>
>>>
>>> So this is your refutation
>>
>> There is nothing to refute all that you ever have is denigration.
>
> I see you cut all of my message except for the first line. I also
> noticed that you didn't expand the newsgroup list to the normal ones.
> Why not? Would it embarrass you? Some hints about how to face your
> unhappy life and improve your existence? You really are a befuddled
> buffoon choosing to live with the label of proud ignoramus tattooed on
> your forehead. Unfortunately nothing you can or will say could change
> your now permanent reputation.

You apparently simply choose to be a jackass at least towards me.
This will be quite embarrassing for you when my work is accepted as
correct.

--
Copyright 2021 Pete Olcott

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

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2

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 by: Richard Damon - Tue, 2 Nov 2021 10:43 UTC

On 11/1/21 8:11 PM, olcott wrote:
> On 11/1/2021 5:38 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>>> || Here's the key question: do you still assert that H(P,P) == false is
>>>> || the "correct" answer even though P(P) halts?
>>
>> To which you replied (but have for some reason cut from this post)
>>
>>>> | Yes that is the correct answer even though P(P) halts.
>>
>>> H(P,P) reports that its input never halts.
>>
>> H(P,P) should report on whether P(P) halts.  Stating that the wrong
>> answer is the right one is not how mathematics is done.
>>
>
>
> H1(P,P) is computationally equivalent to P(P).
> H(P,P) is not computationally equivalent to P(P).
>
>

Incorrect. Doesn't match the definition of 'Computationally Equivalent'.

Computations (a pairing of an Algorithm and an input) are NEVER
'Computationally Equivalent', that term is defined between Computational
Structures and between Algorithms/Turing Machines, not Computations.

You are falling for the trap that is part of 'First Principles' of not
knowing the established terminology of the field, because you haven't
studied it, but just guessing what words mean and trying to borrow
principles you just don't understand.

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2

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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Tue, 2 Nov 2021 10:44 UTC

On 11/2/21 12:20 AM, olcott wrote:
> On 11/1/2021 11:08 PM, Jeff Barnett wrote:
>> On 11/1/2021 8:04 PM, olcott wrote:
>>> On 11/1/2021 8:31 PM, Jeff Barnett wrote:
>>>> On 11/1/2021 3:22 PM, olcott wrote:
>>>>> On 11/1/2021 3:33 PM, Jeff Barnett wrote:
>>>>>> On 11/1/2021 10:46 AM, olcott wrote:
>>>>>>> On 11/1/2021 11:31 AM, Jeff Barnett wrote:
>>>>>>>> On 11/1/2021 6:46 AM, Malcolm McLean wrote:
>>>>>>>>> On Monday, 1 November 2021 at 03:45:47 UTC, olcott wrote:
>>>>>>>>
>>>>>>>>       <SNIP POOP>
>>>>>>>>
>>>>>>>>> I can see where the confusion lies. If H never aborts, it
>>>>>>>>> simulates P forever.
>>>>>>>>> If it aborts, then it aborts because it has detected that
>>>>>>>>> otherwise P would
>>>>>>>>> simulate forever.
>>>>>>>>
>>>>>>>> Malcolm, note well: "confusion" does not lie; it's the
>>>>>>>> permanently self-bamboozled PO who lies.
>>>>>>>>
>>>>>>>>        <SNIP>
>>>>>>>
>>>>>>> The part that you quoted proves that I am correct, yet because
>>>>>>> you are clueless about these things you don't see this. You have
>>>>>>> never ever provided any reply what-so-ever (that I remember) that
>>>>>>> was anything at all besides denigration. Ad Hominem is your only
>>>>>>> tool.
>>>>>>>
>>>>>>> Ad Hominem
>>>>>>> (Attacking the person): This fallacy occurs when, instead of
>>>>>>> addressing someone's argument or position, you irrelevantly
>>>>>>> attack the person or some aspect of the person who is making the
>>>>>>> argument.
>>>>>>> https://www.txstate.edu/philosophy/resources/fallacy-definitions/Ad-Hominem.html
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>> So this is your refutation
>>>>>
>>>>> There is nothing to refute all that you ever have is denigration.
>>>>
>>>> I see you cut all of my message except for the first line. I also
>>>> noticed that you didn't expand the newsgroup list to the normal
>>>> ones. Why not? Would it embarrass you? Some hints about how to face
>>>> your unhappy life and improve your existence? You really are a
>>>> befuddled buffoon choosing to live with the label of proud ignoramus
>>>> tattooed on your forehead. Unfortunately nothing you can or will say
>>>> could change your now permanent reputation.
>>>
>>> You apparently simply choose to be a jackass at least towards me.
>>> This will be quite embarrassing for you when my work is accepted as
>>> correct.
>>
>> You correct on any of the many topics in this newsgroup!!??!? Don't
>> make me laugh. The only notable scientific thing to your credit is the
>> record for longest consecutive string of thinking errors. You've been
>> at it for more than a decade and haven't been right once. Not Ad
>> Homiem; fact! Don't you just love these Latin phrases?
>
> ad hominem
> You attacked your opponent's character or personal traits in an attempt
> to undermine their argument. https://yourlogicalfallacyis.com/ad-hominem
>

Isn't that one of your favorite tactics, calling your 'opponent' an
Idiot because they don't accept the POOP you are talking about as the
Halting Problem?

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2

<87lf26pueb.fsf@bsb.me.uk>

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: H(P,P)==0 is correct for every simulating halt decider H --- V2
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 by: Ben Bacarisse - Tue, 2 Nov 2021 13:19 UTC

olcott <NoOne@NoWhere.com> writes:

> On 11/1/2021 5:38 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>>> || Here's the key question: do you still assert that H(P,P) == false is
>>>> || the "correct" answer even though P(P) halts?
>> To which you replied (but have for some reason cut from this post)
>>
>>>> | Yes that is the correct answer even though P(P) halts.
>>
>>> H(P,P) reports that its input never halts.
>> H(P,P) should report on whether P(P) halts. Stating that the wrong
>> answer is the right one is not how mathematics is done.
>
> H1(P,P) is computationally equivalent to P(P).
> H(P,P) is not computationally equivalent to P(P).

H(P,P) == false is wrong if P(P) halts. This comes from the definition
of the problem you have been studying for more than 14 years. I think
it's time you did something else. It seems such a waste...

--
Ben.

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2

<xsudnaztFImi1Bz8nZ2dnUU7-fHNnZ2d@giganews.com>

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 by: olcott - Tue, 2 Nov 2021 14:20 UTC

On 11/2/2021 8:19 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 11/1/2021 5:38 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>>> || Here's the key question: do you still assert that H(P,P) == false is
>>>>> || the "correct" answer even though P(P) halts?
>>> To which you replied (but have for some reason cut from this post)
>>>
>>>>> | Yes that is the correct answer even though P(P) halts.
>>>
>>>> H(P,P) reports that its input never halts.
>>> H(P,P) should report on whether P(P) halts. Stating that the wrong
>>> answer is the right one is not how mathematics is done.
>>
>> H1(P,P) is computationally equivalent to P(P).
>> H(P,P) is not computationally equivalent to P(P).
>
> H(P,P) == false is wrong if P(P) halts.

The input to H(P,P) really does never halt this has been conclusively
proven. That you keep ignoring the verified fact that the input to
H(P,P) has been totally proven to never reaches its final state sure
seems to be an irrational break from reality to me.

It really seems that P(P) should have behavior consistent with H(P,P).
Intuitively they seem to be the same computation. When they don't have
behavior that is consistent with each other this merely proves that
intuition was wrong. Intuition is not infallible. Running the actual
code does infallibly show what the code actually does.

Since a halt decider is supposed to report what its input actually does
we still have H1(P,P) that does just that.

The whole issue with the halting problem proofs is that an input was
intentionally designed to thwart the halt decider. This has always been
understood to be an input that does the opposite of whatever the halt
decider decides.

One of the two halt deciders does get an answer that is consistent with
the behavior of its input.

> This comes from the definition
> of the problem you have been studying for more than 14 years. I think
> it's time you did something else. It seems such a waste...
>

H1(P,P) is computationally equivalent to P(P) and and reports that P(P)
halts. This is the same as applying H to ⟨Ĥ⟩ ⟨Ĥ⟩ which was something
that Linz "proved" was impossible.

--
Copyright 2021 Pete Olcott

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

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2

<slrj5q$1ov$1@dont-email.me>

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From: polco...@gmail.com (olcott)
Newsgroups: comp.theory
Subject: Re: H(P,P)==0 is correct for every simulating halt decider H --- V2
Date: Tue, 2 Nov 2021 09:49:28 -0500
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 by: olcott - Tue, 2 Nov 2021 14:49 UTC

On 11/1/2021 11:08 PM, Jeff Barnett wrote:
> On 11/1/2021 8:04 PM, olcott wrote:
>> On 11/1/2021 8:31 PM, Jeff Barnett wrote:
>>> On 11/1/2021 3:22 PM, olcott wrote:
>>>> On 11/1/2021 3:33 PM, Jeff Barnett wrote:
>>>>> On 11/1/2021 10:46 AM, olcott wrote:
>>>>>> On 11/1/2021 11:31 AM, Jeff Barnett wrote:
>>>>>>> On 11/1/2021 6:46 AM, Malcolm McLean wrote:
>>>>>>>> On Monday, 1 November 2021 at 03:45:47 UTC, olcott wrote:
>>>>>>>
>>>>>>>       <SNIP POOP>
>>>>>>>
>>>>>>>> I can see where the confusion lies. If H never aborts, it
>>>>>>>> simulates P forever.
>>>>>>>> If it aborts, then it aborts because it has detected that
>>>>>>>> otherwise P would
>>>>>>>> simulate forever.
>>>>>>>
>>>>>>> Malcolm, note well: "confusion" does not lie; it's the
>>>>>>> permanently self-bamboozled PO who lies.
>>>>>>>
>>>>>>>        <SNIP>
>>>>>>
>>>>>> The part that you quoted proves that I am correct, yet because you
>>>>>> are clueless about these things you don't see this. You have never
>>>>>> ever provided any reply what-so-ever (that I remember) that was
>>>>>> anything at all besides denigration. Ad Hominem is your only tool.
>>>>>>
>>>>>> Ad Hominem
>>>>>> (Attacking the person): This fallacy occurs when, instead of
>>>>>> addressing someone's argument or position, you irrelevantly attack
>>>>>> the person or some aspect of the person who is making the argument.
>>>>>> https://www.txstate.edu/philosophy/resources/fallacy-definitions/Ad-Hominem.html
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>> So this is your refutation
>>>>
>>>> There is nothing to refute all that you ever have is denigration.
>>>
>>> I see you cut all of my message except for the first line. I also
>>> noticed that you didn't expand the newsgroup list to the normal ones.
>>> Why not? Would it embarrass you? Some hints about how to face your
>>> unhappy life and improve your existence? You really are a befuddled
>>> buffoon choosing to live with the label of proud ignoramus tattooed
>>> on your forehead. Unfortunately nothing you can or will say could
>>> change your now permanent reputation.
>>
>> You apparently simply choose to be a jackass at least towards me.
>> This will be quite embarrassing for you when my work is accepted as
>> correct.
>
> You correct on any of the many topics in this newsgroup!!??!? Don't make
> me laugh. The only notable scientific thing to your credit is the record
> for longest consecutive string of thinking errors. You've been at it for
> more than a decade and haven't been right once. Not Ad Homiem; fact!
> Don't you just love these Latin phrases?

ad hominem
You attacked your opponent's character or personal traits in an attempt
to undermine their argument. https://yourlogicalfallacyis.com/ad-hominem

--
Copyright 2021 Pete Olcott "Great spirits have always encountered
violent opposition from mediocre minds." Einstein

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2

<874k8upp9k.fsf@bsb.me.uk>

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: H(P,P)==0 is correct for every simulating halt decider H --- V2
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 by: Ben Bacarisse - Tue, 2 Nov 2021 15:10 UTC

olcott <NoOne@NoWhere.com> writes:

> On 11/2/2021 8:19 AM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 11/1/2021 5:38 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>>> || Here's the key question: do you still assert that H(P,P) == false is
>>>>>> || the "correct" answer even though P(P) halts?
>>>> To which you replied (but have for some reason cut from this post)
>>>>
>>>>>> | Yes that is the correct answer even though P(P) halts.
>>>>
>>>>> H(P,P) reports that its input never halts.
>>>> H(P,P) should report on whether P(P) halts. Stating that the wrong
>>>> answer is the right one is not how mathematics is done.
>>>
>>> H1(P,P) is computationally equivalent to P(P).
>>> H(P,P) is not computationally equivalent to P(P).
>> H(P,P) == false is wrong if P(P) halts.
>
> The input to H(P,P) really does never halt this has been conclusively
> proven.

P(P) halts. H(P,P) == false is the wrong answer if P(P) halts.

> That you keep ignoring the verified fact that the input to H(P,P) has
> been totally proven to never reaches its final state sure seems to be
> an irrational break from reality to me.

You want me to get sucked into discussing your other errors, but really
only one counts: H(P,P) == false is the wrong answer if P(P) halts.

> It really seems that P(P) should have behavior consistent with H(P,P).

H(P,P) should report the halting status of P(P). I don't know if that's
what you mean. If that is what you mean, why not use my much simpler
words? And if it isn't, you're wrong.

> Intuitively they seem to be the same computation.

Nonsense. No one thinks they are the same computation.

> When they don't have behavior that is consistent with each other this
> merely proves that intuition was wrong. Intuition is not
> infallible. Running the actual code does infallibly show what the code
> actually does.

I get my facts from you. That P(P) halts is a fact you stated and have
never retracted. Likewise the fact that H(P,P) == false. I can't run either
because you are hiding the code, so I have to take what you tell me as fact.

> Since a halt decider is supposed to report what its input actually
> does we still have H1(P,P) that does just that.

What matters is that H(P,P) is wrong. Other functions can of course get
the right answer.

> The whole issue with the halting problem proofs is that an input was
> intentionally designed to thwart the halt decider.

Yes.

> This has always been understood to be an input that does the opposite
> of whatever the halt decider decides.

Almost. This has been understood to be an input that /represents a
computation/ that does the opposite. The input itself does nothing (or,
if it does, what it does it irrelevant).

> One of the two halt deciders does get an answer that is consistent
> with the behavior of its input.

As you, I think, finally acknowledged, every instance of the halting
problem has a correct yes/no answer, so there is always another trivial
function that gets the right answer.

What matters is that H(P,P) == false is wrong because P(P) halts.

>> This comes from the definition
>> of the problem you have been studying for more than 14 years. I think
>> it's time you did something else. It seems such a waste...
>
> H1(P,P) is computationally equivalent to P(P) and and reports that
> P(P) halts.

It's H that's wrong. H(P,P) == false is wrong because P(P) halts. You
have stated as fact everything we need to know that H is wrong about
this one case, the one case you said your had TMs fully encoded to
decide three years ago. Not a sign of them has been seen, of course. I
doubt they ever existed.

> This is the same as applying H to ⟨Ĥ⟩ ⟨Ĥ⟩ which was something
> that Linz "proved" was impossible.

No he does not. But here's the kicker: if Linz's proof is wrong surely
you can produce an H that is right about H(P,P)? You'd only have to
spend all this time trying to justify the wrong answer if, as Linz and
others prove, no code can give the right answer.

Three years ago you boasted that you had Turing machines (though you
later said this was "poetic license" for C-like code) that did something
everyone thought was impossible. That's what we want to see! Code that
gives H(P,P) == false when P(P) halts is trivial and not interesting.

--
Ben.

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2

<slrm7v$pog$1@dont-email.me>

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Newsgroups: comp.theory
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 by: André G. Isaak - Tue, 2 Nov 2021 15:41 UTC

On 2021-11-02 08:20, olcott wrote:
> On 11/2/2021 8:19 AM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 11/1/2021 5:38 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>>> || Here's the key question: do you still assert that H(P,P) ==
>>>>>> false is
>>>>>> || the "correct" answer even though P(P) halts?
>>>> To which you replied (but have for some reason cut from this post)
>>>>
>>>>>> | Yes that is the correct answer even though P(P) halts.
>>>>
>>>>> H(P,P) reports that its input never halts.
>>>> H(P,P) should report on whether P(P) halts.  Stating that the wrong
>>>> answer is the right one is not how mathematics is done.
>>>
>>> H1(P,P) is computationally equivalent to P(P).
>>> H(P,P) is not computationally equivalent to P(P).
>>
>> H(P,P) == false is wrong if P(P) halts.
>
> The input to H(P,P) really does never halt this has been conclusively
> proven. That you keep ignoring the verified fact that the input to
> H(P,P) has been totally proven to never reaches its final state sure
> seems to be an irrational break from reality to me.

The input to a halt decider is supposed to be a *description* of a
Turing Machine and an input string, and the halt decider is supposed to
determine whether the Turing Machine *represented* by that description
halts on that input.

> It really seems that P(P) should have behavior consistent with H(P,P).

It doesn't just seem that way. That's the definition of what a Halt
Decider is supposed to do.

> Intuitively they seem to be the same computation. When they don't have

No, P(P) and H(P, P) do *not* intuitively seem to be the same
computation. The former does whatever it does. The latter is supposed to
report on the behaviour of the former. How are those two very different
things the "same" computation?

> behavior that is consistent with each other this merely proves that
> intuition was wrong. Intuition is not infallible. Running the actual
> code does infallibly show what the code actually does.
>
> Since a halt decider is supposed to report what its input actually does
> we still have H1(P,P) that does just that.

"Inputs" don't do anything. It is supposed to report what the machine
represented by its input does. That would be P(P) which halts.

> The whole issue with the halting problem proofs is that an input was
> intentionally designed to thwart the halt decider. This has always been
> understood to be an input that does the opposite of whatever the halt
> decider decides.
>
> One of the two halt deciders does get an answer that is consistent with
> the behavior of its input.

And the fact that they get two different answers means they cannot both
be halt deciders. At least one of them is answering a different question
from the one required of a halt decider.

>> This comes from the definition
>> of the problem you have been studying for more than 14 years.  I think
>> it's time you did something else.  It seems such a waste...
>>
>
> H1(P,P) is computationally equivalent to P(P)

No, it isn't.

> and and reports that P(P)
> halts. This is the same as applying H to ⟨Ĥ⟩ ⟨Ĥ⟩ which was something
> that Linz "proved" was impossible.

No, it isn't, since Ĥ is derived from H, not from H1. Unless H1 can
determine what Ĥ1(Ĥ1) does it does not do what Linz claimed was impossible.

André

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

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versus logic ]

<t_mdnSlILtJIwhz8nZ2dnUU7-KnNnZ2d@giganews.com>

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 by: olcott - Tue, 2 Nov 2021 15:57 UTC

On 11/2/2021 10:10 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 11/2/2021 8:19 AM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 11/1/2021 5:38 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>>> || Here's the key question: do you still assert that H(P,P) == false is
>>>>>>> || the "correct" answer even though P(P) halts?
>>>>> To which you replied (but have for some reason cut from this post)
>>>>>
>>>>>>> | Yes that is the correct answer even though P(P) halts.
>>>>>
>>>>>> H(P,P) reports that its input never halts.
>>>>> H(P,P) should report on whether P(P) halts. Stating that the wrong
>>>>> answer is the right one is not how mathematics is done.
>>>>
>>>> H1(P,P) is computationally equivalent to P(P).
>>>> H(P,P) is not computationally equivalent to P(P).
>>> H(P,P) == false is wrong if P(P) halts.
>>
>> The input to H(P,P) really does never halt this has been conclusively
>> proven.
>
> P(P) halts. H(P,P) == false is the wrong answer if P(P) halts.
>
>> That you keep ignoring the verified fact that the input to H(P,P) has
>> been totally proven to never reaches its final state sure seems to be
>> an irrational break from reality to me.
>
> You want me to get sucked into discussing your other errors, but really
> only one counts: H(P,P) == false is the wrong answer if P(P) halts.
>

As long as H(P,P)==0 is correct none of my other "errors" are of any
consequence what-so-ever. I have conclusively proved that the input to
H(P,P) cannot possibly ever reach its final state for every possible
encoding of simulating halt decider H.

// Simplified Linz Ĥ (Linz:1990:319)
// Strachey(1965) CPL translated to C
void P(u32 x)
{ if (H(x, x))
HERE: goto HERE;
}

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

_P()
[00000c36](01) 55 push ebp
[00000c37](02) 8bec mov ebp,esp
[00000c39](03) 8b4508 mov eax,[ebp+08] // 2nd Param
[00000c3c](01) 50 push eax
[00000c3d](03) 8b4d08 mov ecx,[ebp+08] // 1st Param
[00000c40](01) 51 push ecx
[00000c41](05) e820fdffff call 00000966 // call H
[00000c46](03) 83c408 add esp,+08
[00000c49](02) 85c0 test eax,eax
[00000c4b](02) 7402 jz 00000c4f
[00000c4d](02) ebfe jmp 00000c4d
[00000c4f](01) 5d pop ebp
[00000c50](01) c3 ret
Size in bytes:(0027) [00000c50]

machine stack stack machine assembly
address address data code language
======== ======== ======== ========= =============
Begin Local Halt Decider Simulation at Machine Address:c36
[00000c36][002117ca][002117ce] 55 push ebp
[00000c37][002117ca][002117ce] 8bec mov ebp,esp
[00000c39][002117ca][002117ce] 8b4508 mov eax,[ebp+08]
[00000c3c][002117c6][00000c36] 50 push eax // push P
[00000c3d][002117c6][00000c36] 8b4d08 mov ecx,[ebp+08]
[00000c40][002117c2][00000c36] 51 push ecx // push P
[00000c41][002117be][00000c46] e820fdffff call 00000966 // call H(P,P)

There are only two possible behaviors for every possible encoding of
simulating halt decider H and only one of these behaviors is correct:

(1) Continue to act as a pure simulator that never aborts the simulation
of its input. This causes the first seven instructions of P to
infinitely repeat. (You might not know the x86 language well enough to
see this).

(2) Abort the simulation of P at one of the machine addresses of P
between 0xc36 to 0xc41 at some point in the simulation of P.

Those are the only two possibilities for every possible encoding of
simulating halt decider H.

Because we applied categorically exhaustive reasoning to H(P,P) we know
that there are no gaps in this reasoning. Every possible encoding of
simulating halt decider H either aborts its simulation of its input or
fails to abort the simulation of its input. There is no third option.

Since for every possible encoding of simulating halt decider H for every
possible abort versus not abort behavior of H(P,P) the input never
reaches its final state we know know with logically justified complete
certainty that the input to H(P,P) never halts thus H(P,P)==0 is correct.

The fact that correct logic contradicts your intuition merely proves
that your intuition is incorrect.

--
Copyright 2021 Pete Olcott

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

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2

<slrnra$87g$1@dont-email.me>

 copy mid

https://www.novabbs.com/devel/article-flat.php?id=22891&group=comp.theory#22891

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From: polco...@gmail.com (olcott)
Newsgroups: comp.theory
Subject: Re: H(P,P)==0 is correct for every simulating halt decider H --- V2
Date: Tue, 2 Nov 2021 11:09:10 -0500
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 by: olcott - Tue, 2 Nov 2021 16:09 UTC

On 11/2/2021 10:41 AM, André G. Isaak wrote:
> On 2021-11-02 08:20, olcott wrote:
>> On 11/2/2021 8:19 AM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 11/1/2021 5:38 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>>> || Here's the key question: do you still assert that H(P,P) ==
>>>>>>> false is
>>>>>>> || the "correct" answer even though P(P) halts?
>>>>> To which you replied (but have for some reason cut from this post)
>>>>>
>>>>>>> | Yes that is the correct answer even though P(P) halts.
>>>>>
>>>>>> H(P,P) reports that its input never halts.
>>>>> H(P,P) should report on whether P(P) halts.  Stating that the wrong
>>>>> answer is the right one is not how mathematics is done.
>>>>
>>>> H1(P,P) is computationally equivalent to P(P).
>>>> H(P,P) is not computationally equivalent to P(P).
>>>
>>> H(P,P) == false is wrong if P(P) halts.
>>
>> The input to H(P,P) really does never halt this has been conclusively
>> proven. That you keep ignoring the verified fact that the input to
>> H(P,P) has been totally proven to never reaches its final state sure
>> seems to be an irrational break from reality to me.
>
> The input to a halt decider is supposed to be a *description* of a
> Turing Machine and an input string, and the halt decider is supposed to
> determine whether the Turing Machine *represented* by that description
> halts on that input.
>

Therefore if the input never reaches its final state when simulated by a
simulating halt decider in pure simulation mode we can conclude that the
input to the halt decider definitely never halts even though an
intuitively similar computation does halt.

The Linz H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
has the same non pathological relationship to its input as H1(P,P).

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The Linz Ĥ.qx applied to ⟨Ĥ⟩ ⟨Ĥ⟩
has the same pathological relationship to its input as H(P,P).

>> It really seems that P(P) should have behavior consistent with H(P,P).
>
> It doesn't just seem that way. That's the definition of what a Halt
> Decider is supposed to do.
>
>> Intuitively they seem to be the same computation. When they don't have
>
> No, P(P) and H(P, P) do *not* intuitively seem to be the same
> computation. The former does whatever it does. The latter is supposed to
> report on the behaviour of the former. How are those two very different
> things the "same" computation?
>
>> behavior that is consistent with each other this merely proves that
>> intuition was wrong. Intuition is not infallible. Running the actual
>> code does infallibly show what the code actually does.
>>
>> Since a halt decider is supposed to report what its input actually
>> does we still have H1(P,P) that does just that.
>
> "Inputs" don't do anything. It is supposed to report what the machine
> represented by its input does. That would be P(P) which halts.
>
>> The whole issue with the halting problem proofs is that an input was
>> intentionally designed to thwart the halt decider. This has always
>> been understood to be an input that does the opposite of whatever the
>> halt decider decides.
>>
>> One of the two halt deciders does get an answer that is consistent
>> with the behavior of its input.
>
> And the fact that they get two different answers means they cannot both
> be halt deciders. At least one of them is answering a different question
> from the one required of a halt decider.
>
>>> This comes from the definition
>>> of the problem you have been studying for more than 14 years.  I think
>>> it's time you did something else.  It seems such a waste...
>>>
>>
>> H1(P,P) is computationally equivalent to P(P)
>
> No, it isn't.
>
>> and and reports that P(P) halts. This is the same as applying H to ⟨Ĥ⟩
>> ⟨Ĥ⟩ which was something that Linz "proved" was impossible.
>
> No, it isn't, since Ĥ is derived from H, not from H1. Unless H1 can
> determine what Ĥ1(Ĥ1) does it does not do what Linz claimed was impossible.
>
> André
>

--
Copyright 2021 Pete Olcott "Great spirits have always encountered
violent opposition from mediocre minds." Einstein

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versus logic ]

<slro2c$87g$2@dont-email.me>

 copy mid

https://www.novabbs.com/devel/article-flat.php?id=22892&group=comp.theory#22892

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From: polco...@gmail.com (olcott)
Newsgroups: comp.theory
Subject: Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [
intuition versus logic ]
Date: Tue, 2 Nov 2021 11:12:58 -0500
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 by: olcott - Tue, 2 Nov 2021 16:12 UTC

On 11/2/2021 10:10 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 11/2/2021 8:19 AM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 11/1/2021 5:38 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>>> || Here's the key question: do you still assert that H(P,P) == false is
>>>>>>> || the "correct" answer even though P(P) halts?
>>>>> To which you replied (but have for some reason cut from this post)
>>>>>
>>>>>>> | Yes that is the correct answer even though P(P) halts.
>>>>>
>>>>>> H(P,P) reports that its input never halts.
>>>>> H(P,P) should report on whether P(P) halts. Stating that the wrong
>>>>> answer is the right one is not how mathematics is done.
>>>>
>>>> H1(P,P) is computationally equivalent to P(P).
>>>> H(P,P) is not computationally equivalent to P(P).
>>> H(P,P) == false is wrong if P(P) halts.
>>
>> The input to H(P,P) really does never halt this has been conclusively
>> proven.
>
> P(P) halts. H(P,P) == false is the wrong answer if P(P) halts.
>
>> That you keep ignoring the verified fact that the input to H(P,P) has
>> been totally proven to never reaches its final state sure seems to be
>> an irrational break from reality to me.
>
> You want me to get sucked into discussing your other errors, but really
> only one counts: H(P,P) == false is the wrong answer if P(P) halts.
As long as H(P,P)==0 is correct none of my other "errors" are of any
consequence what-so-ever. I have conclusively proved that the input to
H(P,P) cannot possibly ever reach its final state for every possible
encoding of simulating halt decider H.

// Simplified Linz Ĥ (Linz:1990:319)
// Strachey(1965) CPL translated to C
void P(u32 x)
{ if (H(x, x))
HERE: goto HERE;
}

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

_P()
[00000c36](01) 55 push ebp
[00000c37](02) 8bec mov ebp,esp
[00000c39](03) 8b4508 mov eax,[ebp+08] // 2nd Param
[00000c3c](01) 50 push eax
[00000c3d](03) 8b4d08 mov ecx,[ebp+08] // 1st Param
[00000c40](01) 51 push ecx
[00000c41](05) e820fdffff call 00000966 // call H
[00000c46](03) 83c408 add esp,+08
[00000c49](02) 85c0 test eax,eax
[00000c4b](02) 7402 jz 00000c4f
[00000c4d](02) ebfe jmp 00000c4d
[00000c4f](01) 5d pop ebp
[00000c50](01) c3 ret
Size in bytes:(0027) [00000c50]

machine stack stack machine assembly
address address data code language
======== ======== ======== ========= =============
Begin Local Halt Decider Simulation at Machine Address:c36
[00000c36][002117ca][002117ce] 55 push ebp
[00000c37][002117ca][002117ce] 8bec mov ebp,esp
[00000c39][002117ca][002117ce] 8b4508 mov eax,[ebp+08]
[00000c3c][002117c6][00000c36] 50 push eax // push P
[00000c3d][002117c6][00000c36] 8b4d08 mov ecx,[ebp+08]
[00000c40][002117c2][00000c36] 51 push ecx // push P
[00000c41][002117be][00000c46] e820fdffff call 00000966 // call H(P,P)

There are only two possible behaviors for every possible encoding of
simulating halt decider H and only one of these behaviors is correct:

(1) Continue to act as a pure simulator that never aborts the simulation
of its input. This causes the first seven instructions of P to
infinitely repeat. (You might not know the x86 language well enough to
see this).

(2) Abort the simulation of P at one of the machine addresses of P
between 0xc36 to 0xc41 at some point in the simulation of P.

Those are the only two possibilities for every possible encoding of
simulating halt decider H.

Because we applied categorically exhaustive reasoning to H(P,P) we know
that there are no gaps in this reasoning. Every possible encoding of
simulating halt decider H either aborts its simulation of its input or
fails to abort the simulation of its input. There is no third option.

Since for every possible encoding of simulating halt decider H for every
possible abort versus not abort behavior of H(P,P) the input never
reaches its final state we know know with logically justified complete
certainty that the input to H(P,P) never halts thus H(P,P)==0 is correct.

The fact that correct logic contradicts your intuition merely proves
that your intuition is incorrect.

--
Copyright 2021 Pete Olcott "Great spirits have always encountered
violent opposition from mediocre minds." Einstein

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versus logic ]

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versus logic ]
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 by: Ben Bacarisse - Tue, 2 Nov 2021 16:15 UTC

olcott <NoOne@NoWhere.com> writes:

> On 11/2/2021 10:10 AM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 11/2/2021 8:19 AM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 11/1/2021 5:38 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>>> || Here's the key question: do you still assert that H(P,P) == false is
>>>>>>>> || the "correct" answer even though P(P) halts?
>>>>>> To which you replied (but have for some reason cut from this post)
>>>>>>
>>>>>>>> | Yes that is the correct answer even though P(P) halts.
>>>>>>
>>>>>>> H(P,P) reports that its input never halts.
>>>>>> H(P,P) should report on whether P(P) halts. Stating that the wrong
>>>>>> answer is the right one is not how mathematics is done.
>>>>>
>>>>> H1(P,P) is computationally equivalent to P(P).
>>>>> H(P,P) is not computationally equivalent to P(P).
>>>> H(P,P) == false is wrong if P(P) halts.
>>>
>>> The input to H(P,P) really does never halt this has been conclusively
>>> proven.
>>
>> P(P) halts. H(P,P) == false is the wrong answer if P(P) halts.
>>
>>> That you keep ignoring the verified fact that the input to H(P,P) has
>>> been totally proven to never reaches its final state sure seems to be
>>> an irrational break from reality to me.
>>
>> You want me to get sucked into discussing your other errors, but really
>> only one counts: H(P,P) == false is the wrong answer if P(P) halts.
>
> As long as H(P,P)==0 is correct none of my other "errors" are of any
> consequence what-so-ever.

That's why I said one error really count: H(P,P)==0 is not correct
because P(P) halts. How is it that you can keep ignoring this?

And because those two undisputed facts are all the matters, I've cut the
rest. You like an analogy don't you? It's as if you've admitted that
your 'add' function has add("2","2") == 5 and that 2+2 == 4, but instead
of addressing this obvious problem, you keep talking about what "the
input to add does" and you keep showing traces of how or why 'add'
arrives at the wrong answer.

--
Ben.

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versus logic ]

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Subject: Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [
intuition versus logic ]
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 by: olcott - Tue, 2 Nov 2021 16:25 UTC

On 11/2/2021 11:15 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 11/2/2021 10:10 AM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 11/2/2021 8:19 AM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 11/1/2021 5:38 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>>> || Here's the key question: do you still assert that H(P,P) == false is
>>>>>>>>> || the "correct" answer even though P(P) halts?
>>>>>>> To which you replied (but have for some reason cut from this post)
>>>>>>>
>>>>>>>>> | Yes that is the correct answer even though P(P) halts.
>>>>>>>
>>>>>>>> H(P,P) reports that its input never halts.
>>>>>>> H(P,P) should report on whether P(P) halts. Stating that the wrong
>>>>>>> answer is the right one is not how mathematics is done.
>>>>>>
>>>>>> H1(P,P) is computationally equivalent to P(P).
>>>>>> H(P,P) is not computationally equivalent to P(P).
>>>>> H(P,P) == false is wrong if P(P) halts.
>>>>
>>>> The input to H(P,P) really does never halt this has been conclusively
>>>> proven.
>>>
>>> P(P) halts. H(P,P) == false is the wrong answer if P(P) halts.
>>>
>>>> That you keep ignoring the verified fact that the input to H(P,P) has
>>>> been totally proven to never reaches its final state sure seems to be
>>>> an irrational break from reality to me.
>>>
>>> You want me to get sucked into discussing your other errors, but really
>>> only one counts: H(P,P) == false is the wrong answer if P(P) halts.
>>
>> As long as H(P,P)==0 is correct none of my other "errors" are of any
>> consequence what-so-ever.
>
> That's why I said one error really count: H(P,P)==0 is not correct
> because P(P) halts. How is it that you can keep ignoring this?
>

It is a verified fact that for every possible (abort / do not abort)
behavior of every possible encoding of simulating halt decider H that
the input to H(P,P) never reaches its final state.

It is like I am telling there are no integers between 1 and 2 and you
just don't believe it. If in all the possibilities in the universe the
input to H(P,P) does not halt then we know for sure that the input to
H(P,P) dopes not halt.

It seems to be intuitively true that H(P,P) should report that its input
halts because P(P) halts. This intuition is conclusively proved to be
incorrect. The gist of the intuition is satisfied when H1(P,P) reports
that its input does halt.

> And because those two undisputed facts are all the matters, I've cut the
> rest. You like an analogy don't you? It's as if you've admitted that
> your 'add' function has add("2","2") == 5 and that 2+2 == 4, but instead
> of addressing this obvious problem, you keep talking about what "the
> input to add does" and you keep showing traces of how or why 'add'
> arrives at the wrong answer.
>

--
Copyright 2021 Pete Olcott

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

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versus logic ]

<87sfweo4ze.fsf@bsb.me.uk>

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versus logic ]
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 by: Ben Bacarisse - Tue, 2 Nov 2021 17:13 UTC

olcott <polcott2@gmail.com> writes:

> On 11/2/2021 10:10 AM, Ben Bacarisse wrote:

>> You want me to get sucked into discussing your other errors, but really
>> only one counts: H(P,P) == false is the wrong answer if P(P) halts.
>>
> As long as H(P,P)==0 is correct none of my other "errors" are of any
> consequence what-so-ever.

Only one error really matters, and that's the fact that H(P,P)==0 is not
correct if P(P) halts, so I'm not sure why you keep posting other stuff
unless it's an attempt to draw attention away from this rather obvious
problem.

--
Ben.

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versus logic ]

<87mtmmo4gh.fsf@bsb.me.uk>

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versus logic ]
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 by: Ben Bacarisse - Tue, 2 Nov 2021 17:25 UTC

olcott <NoOne@NoWhere.com> writes:

> On 11/2/2021 11:15 AM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:

>>> As long as H(P,P)==0 is correct none of my other "errors" are of any
>>> consequence what-so-ever.
>>
>> That's why I said one error really count: H(P,P)==0 is not correct
>> because P(P) halts. How is it that you can keep ignoring this?
>
> It is a verified fact that for every possible (abort / do not abort)
> behavior of every possible encoding of simulating halt decider H that
> the input to H(P,P) never reaches its final state.

H(P,P)==0 is wrong because P(P) halts. You keep trying to explain some
other decision problem that you think H is getting right. For the
halting problem -- the one you've been "studying" for more than 14 years
-- H(P,P)==0 is only correct if P(P) does not halt and you've told us
that is does.

> It is like I am telling there are no integers between 1 and 2 and you
> just don't believe it.

No, its like you are tell me that H(P,P)==false is the right answer from
a halt decider when P(P) is a halting computation. In fact it's very
much like that. Almost exactly like that in fact.

> It seems to be intuitively true that H(P,P) should report that its
> input halts because P(P) halts.

No. I have no intuition about what you even mean because inputs don't
do anything. What is true by definition (no intuition required) is that
H(P,P) should be false only if P(P) does not halt.

> This intuition

I don't have that intuition. What "the input" does is meaningless.

>> And because those two undisputed facts are all the matters, I've cut the
>> rest. You like an analogy don't you? It's as if you've admitted that
>> your 'add' function has add("2","2") == 5 and that 2+2 == 4, but instead
>> of addressing this obvious problem, you keep talking about what "the
>> input to add does" and you keep showing traces of how or why 'add'
>> arrives at the wrong answer.

--
Ben.

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versus logic ]

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From: polco...@gmail.com (olcott)
Newsgroups: comp.theory
Subject: Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [
intuition versus logic ]
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 by: olcott - Tue, 2 Nov 2021 17:29 UTC

On 11/2/2021 12:13 PM, Ben Bacarisse wrote:
> olcott <polcott2@gmail.com> writes:
>
>> On 11/2/2021 10:10 AM, Ben Bacarisse wrote:
>
>>> You want me to get sucked into discussing your other errors, but really
>>> only one counts: H(P,P) == false is the wrong answer if P(P) halts.
>>>
>> As long as H(P,P)==0 is correct none of my other "errors" are of any
>> consequence what-so-ever.
>
> Only one error really matters, and that's the fact that H(P,P)==0 is not
> correct if P(P) halts,

As long as the input to H(P,P) cannot possibly halt for every possible
encoding of simulating halt decider H then H(P,P)==0 can't possibly be
incorrect NO MATTER WHAT.

If we have a black cat then it is utterly impossible that we do not have
a cat.

H1(P,P) reports that P(P) halts.
H(P,P) is an entirely different computation than the above two no matter
how much you simply believe otherwise.

> so I'm not sure why you keep posting other stuff
> unless it's an attempt to draw attention away from this rather obvious
> problem.
>

--
Copyright 2021 Pete Olcott "Great spirits have always encountered
violent opposition from mediocre minds." Einstein

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versus logic ]

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From: polco...@gmail.com (olcott)
Newsgroups: comp.theory
Subject: Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [
intuition versus logic ]
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 by: olcott - Tue, 2 Nov 2021 17:32 UTC

On 11/2/2021 12:25 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 11/2/2021 11:15 AM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>
>>>> As long as H(P,P)==0 is correct none of my other "errors" are of any
>>>> consequence what-so-ever.
>>>
>>> That's why I said one error really count: H(P,P)==0 is not correct
>>> because P(P) halts. How is it that you can keep ignoring this?
>>
>> It is a verified fact that for every possible (abort / do not abort)
>> behavior of every possible encoding of simulating halt decider H that
>> the input to H(P,P) never reaches its final state.
>
> H(P,P)==0 is wrong because P(P) halts. You keep trying to explain some
> other decision problem that you think H is getting right. For the
> halting problem -- the one you've been "studying" for more than 14 years
> -- H(P,P)==0 is only correct if P(P) does not halt and you've told us
> that is does.
>
>> It is like I am telling there are no integers between 1 and 2 and you
>> just don't believe it.
>
> No, its like you are tell me that H(P,P)==false is the right answer from
> a halt decider when P(P) is a halting computation. In fact it's very
> much like that. Almost exactly like that in fact.
>
>> It seems to be intuitively true that H(P,P) should report that its
>> input halts because P(P) halts.
>
> No. I have no intuition about what you even mean because inputs don't
> do anything. What is true by definition (no intuition required) is that
> H(P,P) should be false only if P(P) does not halt.
>
>> This intuition
>
> I don't have that intuition. What "the input" does is meaningless.

Every halt decider is ONLY tasked with determining the behavior of its
input. That you say otherwise is very stupid.

--
Copyright 2021 Pete Olcott "Great spirits have always encountered
violent opposition from mediocre minds." Einstein

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versus logic ]

<sls5k8$reo$1@dont-email.me>

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Subject: Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [
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 by: André G. Isaak - Tue, 2 Nov 2021 20:04 UTC

On 2021-11-02 11:32, olcott wrote:
> On 11/2/2021 12:25 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 11/2/2021 11:15 AM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>
>>>>> As long as H(P,P)==0 is correct none of my other "errors" are of any
>>>>> consequence what-so-ever.
>>>>
>>>> That's why I said one error really count: H(P,P)==0 is not correct
>>>> because P(P) halts.  How is it that you can keep ignoring this?
>>>
>>> It is a verified fact that for every possible (abort / do not abort)
>>> behavior of every possible encoding of simulating halt decider H that
>>> the input to H(P,P) never reaches its final state.
>>
>> H(P,P)==0 is wrong because P(P) halts.  You keep trying to explain some
>> other decision problem that you think H is getting right.  For the
>> halting problem -- the one you've been "studying" for more than 14 years
>> -- H(P,P)==0 is only correct if P(P) does not halt and you've told us
>> that is does.
>>
>>> It is like I am telling there are no integers between 1 and 2 and you
>>> just don't believe it.
>>
>> No, its like you are tell me that H(P,P)==false is the right answer from
>> a halt decider when P(P) is a halting computation.  In fact it's very
>> much like that.  Almost exactly like that in fact.
>>
>>> It seems to be intuitively true that H(P,P) should report that its
>>> input halts because P(P) halts.
>>
>> No.  I have no intuition about what you even mean because inputs don't
>> do anything.  What is true by definition (no intuition required) is that
>> H(P,P) should be false only if P(P) does not halt.
>>
>>> This intuition
>>
>> I don't have that intuition.  What "the input" does is meaningless.
>
> Every halt decider is ONLY tasked with determining the behavior of its
> input. That you say otherwise is very stupid.

Why don't you go back and reread definition of Halt Decider given in
12.1 in Linz.

Let w_M be a string that describes a Turing machine M, and let w be a
string in M’s alphabet. A solution of the halting problem is a Turing
machine H, which for any w_M and w performs the computation:

q_0 w_M w ⊢* x_1 q_y x_2 if M applied to w halts, and
q_0 w_M w ⊢* x_1 q_n x_2 if M applied to w does not halt

The input to this machine is w_M plus the input string w. The conditions
specified to the two final states make no mention whatsoever of w_M.
They talk about *M* applied to w.

That is, a Halt decider is tasked with determining the behaviour of the
Turing Machine M described by the input, not the behaviour of the input
w_M itself (which is meaningless anyways).

After all this time you still don't even understand what a Halt Decider
is, so how can you possibly expect to construct one?

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

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versus logic ]

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 by: olcott - Tue, 2 Nov 2021 20:24 UTC

On 11/2/2021 3:04 PM, André G. Isaak wrote:
> On 2021-11-02 11:32, olcott wrote:
>> On 11/2/2021 12:25 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 11/2/2021 11:15 AM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>>>> As long as H(P,P)==0 is correct none of my other "errors" are of any
>>>>>> consequence what-so-ever.
>>>>>
>>>>> That's why I said one error really count: H(P,P)==0 is not correct
>>>>> because P(P) halts.  How is it that you can keep ignoring this?
>>>>
>>>> It is a verified fact that for every possible (abort / do not abort)
>>>> behavior of every possible encoding of simulating halt decider H that
>>>> the input to H(P,P) never reaches its final state.
>>>
>>> H(P,P)==0 is wrong because P(P) halts.  You keep trying to explain some
>>> other decision problem that you think H is getting right.  For the
>>> halting problem -- the one you've been "studying" for more than 14 years
>>> -- H(P,P)==0 is only correct if P(P) does not halt and you've told us
>>> that is does.
>>>
>>>> It is like I am telling there are no integers between 1 and 2 and you
>>>> just don't believe it.
>>>
>>> No, its like you are tell me that H(P,P)==false is the right answer from
>>> a halt decider when P(P) is a halting computation.  In fact it's very
>>> much like that.  Almost exactly like that in fact.
>>>
>>>> It seems to be intuitively true that H(P,P) should report that its
>>>> input halts because P(P) halts.
>>>
>>> No.  I have no intuition about what you even mean because inputs don't
>>> do anything.  What is true by definition (no intuition required) is that
>>> H(P,P) should be false only if P(P) does not halt.
>>>
>>>> This intuition
>>>
>>> I don't have that intuition.  What "the input" does is meaningless.
>>
>> Every halt decider is ONLY tasked with determining the behavior of its
>> input. That you say otherwise is very stupid.
>
> Why don't you go back and reread definition of Halt Decider given in
> 12.1 in Linz.
>
> Let w_M be a string that describes a Turing machine M, and let w be a
> string in M’s alphabet. A solution of the halting problem is a Turing
> machine H, which for any w_M and w performs the computation:
>
> q_0 w_M w ⊢* x_1 q_y x_2 if M applied to w halts, and
> q_0 w_M w ⊢* x_1 q_n x_2 if M applied to w does not halt
>
> The input to this machine is w_M plus the input string w. The conditions
> specified to the two final states make no mention whatsoever of w_M.
> They talk about *M* applied to w.
>

By saying it that way people can be confused (the same way that Linz
himself was confused) into believing that the question is about whether
or not the halt decider itself at Ĥ.qx halts.

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

The behavior of the input to the halt decider after Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩
has nothing to do with whether or not the halt decider itself halts.

The fact that the input to H(P,P) never reaches its final state under
both aborting and not aborting behaviors for every possible encoding of
simulating halt decider H conclusively proves that the input to H(P,P)
never halts and H(P,P)==0 is correct.

Shown on pages 4 and 5 of this paper:
Halting problem undecidability and infinitely nested simulation
May 2021 PL Olcott

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

The key part to pay attention to it the execution trace of the first
seven lines of the x86 code for P:

Begin Local Halt Decider Simulation at Machine Address:c36
machine stack stack machine assembly
address address data code language
======== ======== ======== ========= =============
[00000c36][002117ca][002117ce] 55 push ebp
[00000c37][002117ca][002117ce] 8bec mov ebp,esp
[00000c39][002117ca][002117ce] 8b4508 mov eax,[ebp+08]
[00000c3c][002117c6][00000c36] 50 push eax // push P
[00000c3d][002117c6][00000c36] 8b4d08 mov ecx,[ebp+08]
[00000c40][002117c2][00000c36] 51 push ecx // push P
[00000c41][002117be][00000c46] e820fdffff call 00000966 // call H(P,P)

--
Copyright 2021 Pete Olcott

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

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versus logic ]

<sls7ku$aom$1@dont-email.me>

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From: agis...@gm.invalid (André G. Isaak)
Newsgroups: comp.theory,comp.ai.philosophy,sci.logic,sci.math
Subject: Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [
intuition versus logic ]
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 by: André G. Isaak - Tue, 2 Nov 2021 20:38 UTC

On 2021-11-02 14:24, olcott wrote:
> On 11/2/2021 3:04 PM, André G. Isaak wrote:
>> On 2021-11-02 11:32, olcott wrote:
>>> On 11/2/2021 12:25 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 11/2/2021 11:15 AM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>>>> As long as H(P,P)==0 is correct none of my other "errors" are of any
>>>>>>> consequence what-so-ever.
>>>>>>
>>>>>> That's why I said one error really count: H(P,P)==0 is not correct
>>>>>> because P(P) halts.  How is it that you can keep ignoring this?
>>>>>
>>>>> It is a verified fact that for every possible (abort / do not abort)
>>>>> behavior of every possible encoding of simulating halt decider H that
>>>>> the input to H(P,P) never reaches its final state.
>>>>
>>>> H(P,P)==0 is wrong because P(P) halts.  You keep trying to explain some
>>>> other decision problem that you think H is getting right.  For the
>>>> halting problem -- the one you've been "studying" for more than 14
>>>> years
>>>> -- H(P,P)==0 is only correct if P(P) does not halt and you've told us
>>>> that is does.
>>>>
>>>>> It is like I am telling there are no integers between 1 and 2 and you
>>>>> just don't believe it.
>>>>
>>>> No, its like you are tell me that H(P,P)==false is the right answer
>>>> from
>>>> a halt decider when P(P) is a halting computation.  In fact it's very
>>>> much like that.  Almost exactly like that in fact.
>>>>
>>>>> It seems to be intuitively true that H(P,P) should report that its
>>>>> input halts because P(P) halts.
>>>>
>>>> No.  I have no intuition about what you even mean because inputs don't
>>>> do anything.  What is true by definition (no intuition required) is
>>>> that
>>>> H(P,P) should be false only if P(P) does not halt.
>>>>
>>>>> This intuition
>>>>
>>>> I don't have that intuition.  What "the input" does is meaningless.
>>>
>>> Every halt decider is ONLY tasked with determining the behavior of
>>> its input. That you say otherwise is very stupid.
>>
>> Why don't you go back and reread definition of Halt Decider given in
>> 12.1 in Linz.
>>
>> Let w_M be a string that describes a Turing machine M, and let w be a
>> string in M’s alphabet. A solution of the halting problem is a Turing
>> machine H, which for any w_M and w performs the computation:
>>
>> q_0 w_M w ⊢* x_1 q_y x_2 if M applied to w halts, and
>> q_0 w_M w ⊢* x_1 q_n x_2 if M applied to w does not halt
>>
>> The input to this machine is w_M plus the input string w. The
>> conditions specified to the two final states make no mention
>> whatsoever of w_M. They talk about *M* applied to w.
>>
>
> By saying it that way people can be confused (the same way that Linz

Saying it that way is how halt decider is *defined*. Not just by Linz,
but by *everyone* who has ever discussed this problem.

If you want to discuss the halting problem, this is the definition you
must use.

> himself was confused) into believing that the question is about whether
> or not the halt decider itself at Ĥ.qx halts.

(A) There is no halt decider at Ĥ.qx
(B) There is nothing to get confused about. The input to the submachine
at Ĥ.qx is the string ⟨Ĥ⟩ ⟨Ĥ⟩. Any decision made by Ĥ.qx (which is not a
halting decision) makes a decision about Ĥ applied to ⟨Ĥ⟩.

Ĥ, Ĥ.qx, and w_Ĥ (or ⟨Ĥ⟩ to use your notion) are three different things.
Why would anyone confuse w_Ĥ with Ĥ.qx?

> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
> The behavior of the input to the halt decider after Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩
> has nothing to do with whether or not the halt decider itself halts.

Nor does the definition given above claim that it does. Any comfusion
here is purely confusion on your part.

André

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

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versus logic ]

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 by: olcott - Tue, 2 Nov 2021 20:43 UTC

On 11/2/2021 3:38 PM, André G. Isaak wrote:
> On 2021-11-02 14:24, olcott wrote:
>> On 11/2/2021 3:04 PM, André G. Isaak wrote:
>>> On 2021-11-02 11:32, olcott wrote:
>>>> On 11/2/2021 12:25 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 11/2/2021 11:15 AM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>>>> As long as H(P,P)==0 is correct none of my other "errors" are of
>>>>>>>> any
>>>>>>>> consequence what-so-ever.
>>>>>>>
>>>>>>> That's why I said one error really count: H(P,P)==0 is not correct
>>>>>>> because P(P) halts.  How is it that you can keep ignoring this?
>>>>>>
>>>>>> It is a verified fact that for every possible (abort / do not abort)
>>>>>> behavior of every possible encoding of simulating halt decider H that
>>>>>> the input to H(P,P) never reaches its final state.
>>>>>
>>>>> H(P,P)==0 is wrong because P(P) halts.  You keep trying to explain
>>>>> some
>>>>> other decision problem that you think H is getting right.  For the
>>>>> halting problem -- the one you've been "studying" for more than 14
>>>>> years
>>>>> -- H(P,P)==0 is only correct if P(P) does not halt and you've told us
>>>>> that is does.
>>>>>
>>>>>> It is like I am telling there are no integers between 1 and 2 and you
>>>>>> just don't believe it.
>>>>>
>>>>> No, its like you are tell me that H(P,P)==false is the right answer
>>>>> from
>>>>> a halt decider when P(P) is a halting computation.  In fact it's very
>>>>> much like that.  Almost exactly like that in fact.
>>>>>
>>>>>> It seems to be intuitively true that H(P,P) should report that its
>>>>>> input halts because P(P) halts.
>>>>>
>>>>> No.  I have no intuition about what you even mean because inputs don't
>>>>> do anything.  What is true by definition (no intuition required) is
>>>>> that
>>>>> H(P,P) should be false only if P(P) does not halt.
>>>>>
>>>>>> This intuition
>>>>>
>>>>> I don't have that intuition.  What "the input" does is meaningless.
>>>>
>>>> Every halt decider is ONLY tasked with determining the behavior of
>>>> its input. That you say otherwise is very stupid.
>>>
>>> Why don't you go back and reread definition of Halt Decider given in
>>> 12.1 in Linz.
>>>
>>> Let w_M be a string that describes a Turing machine M, and let w be a
>>> string in M’s alphabet. A solution of the halting problem is a Turing
>>> machine H, which for any w_M and w performs the computation:
>>>
>>> q_0 w_M w ⊢* x_1 q_y x_2 if M applied to w halts, and
>>> q_0 w_M w ⊢* x_1 q_n x_2 if M applied to w does not halt
>>>
>>> The input to this machine is w_M plus the input string w. The
>>> conditions specified to the two final states make no mention
>>> whatsoever of w_M. They talk about *M* applied to w.
>>>
>>
>> By saying it that way people can be confused (the same way that Linz
>
> Saying it that way is how halt decider is *defined*. Not just by Linz,
> but by *everyone* who has ever discussed this problem.
>
> If you want to discuss the halting problem, this is the definition you
> must use.
>

If everyone "defines" that the halt decider is wrong when it correctly
reports what the actual behavior of its actual input would be then
everyone (besides me) is wrong.

--
Copyright 2021 Pete Olcott

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

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versus logic ]

<sls97c$mj5$1@dont-email.me>

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From: agis...@gm.invalid (André G. Isaak)
Newsgroups: comp.theory
Subject: Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [
intuition versus logic ]
Date: Tue, 2 Nov 2021 15:05:47 -0600
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 by: André G. Isaak - Tue, 2 Nov 2021 21:05 UTC

On 2021-11-02 14:43, olcott wrote:

> If everyone "defines" that the halt decider is wrong when it correctly
> reports what the actual behavior of its actual input would be then
> everyone (besides me) is wrong.

The definition tells you what a halt decider *is*. It doesn't define it
as 'wrong'. It defines what question it is supposed to answer.

The input to a halt decider is a string. Strings don't *have* halting
behaviour so your position above is entirely incoherent.

The string w_Ĥ w_Ĥ describes the computation where the Turing Machine Ĥ
is applied to the string w_Ĥ. That computation halts. And it is the
behaviour of that computation that a halting decider must report when
given w_Ĥ w_Ĥ as an input BECAUSE THAT'S WHAT A HALT DECIDER IS DEFINED
TO DO.

Your claim that the "input" doesn't halt even though the actual machine
does simply illustrates that you are clueless about what the actual
halting problem is.

André

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

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versus logic ]

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intuition versus logic ]
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 by: olcott - Tue, 2 Nov 2021 21:41 UTC

On 11/2/2021 4:05 PM, André G. Isaak wrote:
> On 2021-11-02 14:43, olcott wrote:
>
>> If everyone "defines" that the halt decider is wrong when it correctly
>> reports what the actual behavior of its actual input would be then
>> everyone (besides me) is wrong.
>
> The definition tells you what a halt decider *is*. It doesn't define it
> as 'wrong'. It defines what question it is supposed to answer.
>
> The input to a halt decider is a string. Strings don't *have* halting
> behaviour so your position above is entirely incoherent.
>

In computability theory, the halting problem is the problem of
determining, from a description of an arbitrary computer program and an
input, whether the program will finish running, or continue to run
forever. https://en.wikipedia.org/wiki/Halting_problem

A halt decider only need answer whether or not the correct simulation of
its input would ever reach a final state of this input by a simulating
halt decider.

In every case where the answer is "no" every encoding of any simulating
halt decider would correctly decide the halt status of this input as
"not halting".

> The string w_Ĥ w_Ĥ describes the computation where the Turing Machine Ĥ
> is applied to the string w_Ĥ. That computation halts. And it is the
> behaviour of that computation that a halting decider must report when
> given w_Ĥ w_Ĥ as an input BECAUSE THAT'S WHAT A HALT DECIDER IS DEFINED
> TO DO.
>
> Your claim that the "input" doesn't halt even though the actual machine
> does simply illustrates that you are clueless about what the actual
> halting problem is.
>
> André
>

--
Copyright 2021 Pete Olcott

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

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [ intuition versus logic ]

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From: agis...@gm.invalid (André G. Isaak)
Newsgroups: comp.theory
Subject: Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [
intuition versus logic ]
Date: Tue, 2 Nov 2021 15:49:10 -0600
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 by: André G. Isaak - Tue, 2 Nov 2021 21:49 UTC

On 2021-11-02 15:41, olcott wrote:
> On 11/2/2021 4:05 PM, André G. Isaak wrote:
>> On 2021-11-02 14:43, olcott wrote:
>>
>>> If everyone "defines" that the halt decider is wrong when it
>>> correctly reports what the actual behavior of its actual input would
>>> be then everyone (besides me) is wrong.
>>
>> The definition tells you what a halt decider *is*. It doesn't define
>> it as 'wrong'. It defines what question it is supposed to answer.
>>
>> The input to a halt decider is a string. Strings don't *have* halting
>> behaviour so your position above is entirely incoherent.
>>
>
> In computability theory, the halting problem is the problem of
> determining, from a description of an arbitrary computer program and an
> input, whether the program will finish running, or continue to run
> forever. https://en.wikipedia.org/wiki/Halting_problem

The definition of 'halting problem' is what it is.

Note that the above definition doesn't make any mentions of
'simulations' just as the more formal definition used by Linz's does not.

> A halt decider only need answer whether or not the correct simulation of
> its input would ever reach a final state of this input by a simulating
> halt decider.

A 'correct simulation', presumably, would be one that acts identically
to the actual TM being simulated. That means that if the actual TM halts
the simulation also must halt. Which means that your simulation is not a
'correct simulation'.

If a simulation can have a different behaviour then the actual program,
then that simulation can't be used to answer the question which a halt
decider is supposed to answer.

You really ought to consider a different line of 'work'. Have you
thought of maybe inventing a perpetual motion machine or of squaring the
circle? Maybe you'd have more luck with those.

André

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

Re: H(P,P)==0 is correct for every simulating halt decider H --- V2 [ impossibly incorrect ]

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 by: olcott - Tue, 2 Nov 2021 22:51 UTC

On 11/2/2021 4:49 PM, André G. Isaak wrote:
> On 2021-11-02 15:41, olcott wrote:
>> On 11/2/2021 4:05 PM, André G. Isaak wrote:
>>> On 2021-11-02 14:43, olcott wrote:
>>>
>>>> If everyone "defines" that the halt decider is wrong when it
>>>> correctly reports what the actual behavior of its actual input would
>>>> be then everyone (besides me) is wrong.
>>>
>>> The definition tells you what a halt decider *is*. It doesn't define
>>> it as 'wrong'. It defines what question it is supposed to answer.
>>>
>>> The input to a halt decider is a string. Strings don't *have* halting
>>> behaviour so your position above is entirely incoherent.
>>>
>>
>> In computability theory, the halting problem is the problem of
>> determining, from a description of an arbitrary computer program and
>> an input, whether the program will finish running, or continue to run
>> forever. https://en.wikipedia.org/wiki/Halting_problem
>
> The definition of 'halting problem' is what it is.
>
> Note that the above definition doesn't make any mentions of
> 'simulations' just as the more formal definition used by Linz's does not.
>
>> A halt decider only need answer whether or not the correct simulation
>> of its input would ever reach a final state of this input by a
>> simulating halt decider.
>
> A 'correct simulation', presumably, would be one that acts identically
> to the actual TM being simulated. That means that if the actual TM halts
> the simulation also must halt. Which means that your simulation is not a
> 'correct simulation'.
>

There are no freaking presumptions about it. As long as the simulation
of P(P) matches its x86 source code then the simulation is correct.

If you don't agree with this then you are either:
(a) A liar
(b) Stupid
(c) Clueless

_P()
[00000c36](01) 55 push ebp
[00000c37](02) 8bec mov ebp,esp
[00000c39](03) 8b4508 mov eax,[ebp+08] // 2nd Param
[00000c3c](01) 50 push eax
[00000c3d](03) 8b4d08 mov ecx,[ebp+08] // 1st Param
[00000c40](01) 51 push ecx
[00000c41](05) e820fdffff call 00000966 // call H
[00000c46](03) 83c408 add esp,+08
[00000c49](02) 85c0 test eax,eax
[00000c4b](02) 7402 jz 00000c4f
[00000c4d](02) ebfe jmp 00000c4d
[00000c4f](01) 5d pop ebp
[00000c50](01) c3 ret
Size in bytes:(0027) [00000c50]

A correct simulation of P on input P can be empirically validated by the
simulation of the first seven x86 instruction of P.

Begin Local Halt Decider Simulation at Machine Address:c36
machine stack stack machine assembly
address address data code language
======== ======== ======== ========= =============
[00000c36][002117ca][002117ce] 55 push ebp
[00000c37][002117ca][002117ce] 8bec mov ebp,esp
[00000c39][002117ca][002117ce] 8b4508 mov eax,[ebp+08]
[00000c3c][002117c6][00000c36] 50 push eax // push P
[00000c3d][002117c6][00000c36] 8b4d08 mov ecx,[ebp+08]
[00000c40][002117c2][00000c36] 51 push ecx // push P
[00000c41][002117be][00000c46] e820fdffff call 00000966 // call H(P,P)

Because the simulation of P(P) matches its x86 source-code it is
necessarily correct and impossibly incorrect.

The only possibility for every encoding of simulating halt decider H is:

(a) H fails to abort its simulation (and thus fails to be a decider) and
the first seven instructions of P infinitely repeat in which case P
never reaches its final state of 0xc50.

(b) H aborts its simulation of P at some address of P between 0xc36 and
0xc41 in which case the input P fails to every reach its final state of
0xc50.

One the basis of the correct simulation of the x86 source-code of P(P)
we can see that it is utterly impossible for P to ever reach its final
state of 0xc50.

> If a simulation can have a different behaviour then the actual program,
> then that simulation can't be used to answer the question which a halt
> decider is supposed to answer.
>
> You really ought to consider a different line of 'work'. Have you
> thought of maybe inventing a perpetual motion machine or of squaring the
> circle? Maybe you'd have more luck with those.
>
> André
>

--
Copyright 2021 Pete Olcott

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

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