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devel / comp.theory / Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

SubjectAuthor
* How do we know H(P,P)==0 is the correct halt status for the input toolcott
+- How do we know H(P,P)==0 is the correct halt status for the inputolcott
+* How do we know H(P,P)==0 is the correct halt status for the inputwij
|`* How do we know H(P,P)==0 is the correct halt status for the input to H?olcott
| +* How do we know H(P,P)==0 is the correct halt status for the inputwij
| |`* How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | +* How do we know H(P,P)==0 is the correct halt status for the inputwij
| | |`* How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | | `* How do we know H(P,P)==0 is the correct halt status for the inputwij
| | |  `* How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | |   `* How do we know H(P,P)==0 is the correct halt status for the inputwij
| | |    `* How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | |     +- How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| | |     `* How do we know H(P,P)==0 is the correct halt status for the inputwij
| | |      `* How do we know H(P,P)==0 is the correct halt status for the input to H?olcott
| | |       +* How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| | |       |`* How do we know H(P,P)==0 is the correct halt status for the input to H?olcott
| | |       | `* How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| | |       |  `* How do we know H(P,P)==0 is the correct halt status for the input to H?olcott
| | |       |   `* How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| | |       |    `* How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | |       |     `* How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| | |       |      `* How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | |       |       `* How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| | |       |        `* How do we know H(P,P)==0 is the correct halt status for the input to H?olcott
| | |       |         `* How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| | |       |          `* How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | |       |           `* How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| | |       |            `* How do we know H(P,P)==0 is the correct halt status for the input to H?olcott
| | |       |             `* How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| | |       |              `* How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | |       |               `* How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| | |       |                `* How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | |       |                 `* How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| | |       |                  `* How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | |       |                   `* How do we know H(P,P)==0 is the correct halt status for the input to H?Richard Damon
| | |       |                    `* How do we know H(P,P)==0 is the correct halt status for the input to H?olcott
| | |       |                     `* How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| | |       |                      `* How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | |       |                       `* How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| | |       |                        `* How do we know H(P,P)==0 is the correct halt status for the input to H?olcott
| | |       |                         +* How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| | |       |                         |+* How do we know H(P,P)==0 is the correct halt status for the input to H?olcott
| | |       |                         ||`* How do we know H(P,P)==0 is the correct halt status for the input to H?Richard Damon
| | |       |                         || `* How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | |       |                         ||  `- How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| | |       |                         |`* How do we know H(P,P)==0 is the correct halt status for the inputMalcolm McLean
| | |       |                         | +- How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | |       |                         | `* How do we know H(P,P)==0 is the correct halt status for the inputMalcolm McLean
| | |       |                         |  +* How do we know H(P,P)==0 is the correct halt status for the input to H?olcott
| | |       |                         |  |`* How do we know H(P,P)==0 is the correct halt status for the inputAndré G. Isaak
| | |       |                         |  | `* How do we know H(P,P)==0 is the correct halt status for the input to H?olcott
| | |       |                         |  |  `* How do we know H(P,P)==0 is the correct halt status for the inputAndré G. Isaak
| | |       |                         |  |   `* How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | |       |                         |  |    `* How do we know H(P,P)==0 is the correct halt status for the inputAndré G. Isaak
| | |       |                         |  |     `* How do we know H(P,P)==0 is the correct halt status for the input to H?olcott
| | |       |                         |  |      `* How do we know H(P,P)==0 is the correct halt status for the inputAndré G. Isaak
| | |       |                         |  |       `* How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | |       |                         |  |        `* How do we know H(P,P)==0 is the correct halt status for the inputAndré G. Isaak
| | |       |                         |  |         `* How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | |       |                         |  |          `* How do we know H(P,P)==0 is the correct halt status for the inputAndré G. Isaak
| | |       |                         |  |           `* How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | |       |                         |  |            `* How do we know H(P,P)==0 is the correct halt status for the inputAndré G. Isaak
| | |       |                         |  |             `* How do we know H(P,P)==0 is the correct halt status for the input to H?olcott
| | |       |                         |  |              `* How do we know H(P,P)==0 is the correct halt status for the inputAndré G. Isaak
| | |       |                         |  |               `* How do we know H(P,P)==0 is the correct halt status for the input to H?olcott
| | |       |                         |  |                `- How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| | |       |                         |  `* How do we know H(P,P)==0 is the correct halt status for the inputMalcolm McLean
| | |       |                         |   `- How do we know H(P,P)==0 is the correct halt status for the input to H?olcott
| | |       |                         `* How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| | |       |                          +* How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | |       |                          |`* How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| | |       |                          | `* How do we know H(P,P)==0 is the correct halt status for the input to H? [ key axolcott
| | |       |                          |  `- How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| | |       |                          `* How do we know H(P,P)==0 is the correct halt status for the inputwij
| | |       |                           `- How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | |       `* How do we know H(P,P)==0 is the correct halt status for the inputwij
| | |        `* How do we know H(P,P)==0 is the correct halt status for the input to H?olcott
| | |         `* How do we know H(P,P)==0 is the correct halt status for the inputwij
| | |          +* How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | |          |`* How do we know H(P,P)==0 is the correct halt status for the inputwij
| | |          | +- How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | |          | `* How do we know H(P,P)==0 is the correct halt status for the inputdklei...@gmail.com
| | |          |  `* How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | |          |   `* How do we know H(P,P)==0 is the correct halt status for the input to H?Richard Damon
| | |          |    `* How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | |          |     `* How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| | |          |      `* How do we know H(P,P)==0 is the correct halt status for the inputolcott
| | |          |       `- How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| | |          `* How do we know H(P,P)==0 is the correct halt status for the inputChris M. Thomasson
| | |           `* How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| | |            `* How do we know H(P,P)==0 is the correct halt status for the inputChris M. Thomasson
| | |             `* How do we know H(P,P)==0 is the correct halt status for the input to H?olcott
| | |              `* How do we know H(P,P)==0 is the correct halt status for the inputChris M. Thomasson
| | |               `* How do we know H(P,P)==0 is the correct halt status for the input to H?olcott
| | |                `* How do we know H(P,P)==0 is the correct halt status for the inputChris M. Thomasson
| | |                 `* How do we know H(P,P)==0 is the correct halt status for the input to H?olcott
| | |                  `- How do we know H(P,P)==0 is the correct halt status for the input to H?Ben Bacarisse
| | `* How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| |  `* How do we know H(P,P)==0 is the correct halt status for the inputolcott
| |   `- How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
| `* How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
+- How do we know H(P,P)==0 is the correct halt status for the inputRichard Damon
`* How do we know H(P,P)==0 is the correct halt status for the input to H?Ben Bacarisse

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Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

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https://www.novabbs.com/devel/article-flat.php?id=19943&group=comp.theory#19943

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NNTP-Posting-Date: Wed, 18 Aug 2021 19:22:12 -0500
Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input
to H?
Newsgroups: comp.theory
References: <3YOdnecvDsA5Q4r8nZ2dnUU7-TXNnZ2d@giganews.com>
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From: NoO...@NoWhere.com (olcott)
Date: Wed, 18 Aug 2021 19:22:10 -0500
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 by: olcott - Thu, 19 Aug 2021 00:22 UTC

On 8/18/2021 6:51 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/18/2021 2:44 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/17/2021 7:02 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/17/2021 5:14 PM, Ben Bacarisse wrote:
>>>>>
>>>>>>> The facts come from you. They are not a matter of opinion. P(P) halts.
>>>>>>> H(P,P) == 0.
>>>>> ...
>>>>>>> You are happy with that answer, but I can specify a function you can't
>>>>>>> write: B(M, I) such that B(M, I) is non zero if and only if M(I) halts.
>>>>>>> You don't get to choose what the "correct" answer is, so your only
>>>>>>> option if to ignore the challenge.
>>>>>>
>>>>>> At this point I expect and require that those seeking an actual honest
>>>>>> dialogue use the basis that I provided to verify that H does decide
>>>>>> the halt status of its inputs correctly. Everyone else is written off
>>>>>> as dishonest.
>>>>> That's up to you. I can't stop you wasting time on a function no one
>>>>> cares about, but you know you can't write the function I specified, so
>>>>> you are back where you started 17 years ago. There are still
>>>>> undecidable sets.
>>>>
>>>> The function does meet the spec
>>> You don't know that because you have not asked for the proper spec. You
>>> have not even accepted my challenge. Are you accepting the challenge
>>> and proposing H as an implementation without knowing the full spec? I'm
>>> happy for you to say yes if that is what you want to do.
>>
>> I know what the spec is:
>
> No you don't. The spec comes from me, and only if you say you want to
> try the challenge. I am promising you the specification of a function
> you can't write -- a clear demonstration that there are natural
> uncomputable functions.
>

I posted two different correct examples of the spec for the halting
problem if you are talking about something else then this is out-of
scope. You really should never erase the text the you are directly
responding to that seems quite dishonest.

On 8/18/2021 2:54 PM, olcott wrote:
> I know what the spec is:
>
> the Turing machine halting problem. Simply stated, the problem
> is: given the description of a Turing machine M and an input w,
> does M, when started in the initial configuration q0w, perform a
> computation that eventually halts? (Linz:1990:317).
>
> 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
>
> description of a Turing machine M (not a Turing Machine)
> description of an arbitrary computer program (not an executable program)
>
> What is your opinion of what the spec is?

>> What is your opinion of what the spec is?
>
> It's not "the spec", it's my spec. If you take the challenge, I'll
> write it out. I won't bother writing it out properly if you don't want
> to try.
>

--
Copyright 2021 Pete Olcott

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

Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

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Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input
to H?
Newsgroups: comp.theory
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From: NoO...@NoWhere.com (olcott)
Date: Wed, 18 Aug 2021 19:40:10 -0500
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 by: olcott - Thu, 19 Aug 2021 00:40 UTC

On 8/18/2021 7:11 PM, Richard Damon wrote:
> On 8/18/21 7:50 PM, olcott wrote:
>> On 8/18/2021 6:46 PM, Richard Damon wrote:
>>> On 8/18/21 4:45 PM, olcott wrote:
>>>> On 8/18/2021 3:05 PM, dklei...@gmail.com wrote:
>>>>> On Wednesday, August 18, 2021 at 6:50:34 AM UTC-7, olcott wrote:
>>>>>> H does exist. I spent 1.5 years creating the x86utm operating
>>>>>> system. It
>>>>>> took me three full months to get context switching correctly so that
>>>>>> Nested simulation of virtual machines could operate correctly to an
>>>>>> arbitrary recursive depth.
>>>>>>
>>>>>> I had to make sure that I had everything working properly because I do
>>>>>> intend to publish as soon as I have words that are sufficiently
>>>>>> clear to
>>>>>> prove my point. I will have to refactor my code so that it is cleaner.
>>>>>
>>>>> You must be aware that no moderated group will accept your x86
>>>>> equivalence approach as a proof of anything. So your work is
>>>>> quixotic. No amount of refactoring will ever help. No words exist
>>>>> that will justify it. Be prepared to lose.
>>>>>
>>>>
>>>> No I do not accept this because of the Turing equivalence between the
>>>> H(P,P) computation and the simplified Linz Ĥ.qx applied to ⟨Ĥ⟩ ⟨Ĥ⟩
>>>> computation.
>>>>
>>>> People that fail to sufficiently understand Turing equivalence may
>>>> disagree. They are not in my target audience.
>>>>
>>>>
>>>
>>> You actually have just claimed the Turing Equivalence, which you really
>>> need to PROVE to use, and to do that you really need to show H.
>>>
>>> We actually know in fact that it isn't, as your H can not accept any and
>>> all input machine due to it not having its input in an isolated memory
>>> space, thus there are some inputs that the Turing Machine H can accept
>>> that your x86 H can not.
>>>
>>
>> See that is an example of what I mean. I am saying that H(P,P) is
>> equivalent to a simplified Ĥ.qx applied to ⟨Ĥ⟩ ⟨Ĥ⟩ essentially a single
>> freaking computation not an infinite set of computations.
>
> Which just shows that you don't understand the problem at all. A partial
> decider still must be able to be posed to the decider, it just doesn't
> need to be able to get them all right.
As long as it gets the key one of them right all of the proofs have been
refuted.

--
Copyright 2021 Pete Olcott

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

Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

<CThTI.548$vA6.244@fx23.iad>

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Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input
to H?
Newsgroups: comp.theory
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Thu, 19 Aug 2021 00:48 UTC

On 8/18/21 8:40 PM, olcott wrote:
> On 8/18/2021 7:11 PM, Richard Damon wrote:
>> On 8/18/21 7:50 PM, olcott wrote:
>>> On 8/18/2021 6:46 PM, Richard Damon wrote:
>>>> On 8/18/21 4:45 PM, olcott wrote:
>>>>> On 8/18/2021 3:05 PM, dklei...@gmail.com wrote:
>>>>>> On Wednesday, August 18, 2021 at 6:50:34 AM UTC-7, olcott wrote:
>>>>>>> H does exist. I spent 1.5 years creating the x86utm operating
>>>>>>> system. It
>>>>>>> took me three full months to get context switching correctly so that
>>>>>>> Nested simulation of virtual machines could operate correctly to an
>>>>>>> arbitrary recursive depth.
>>>>>>>
>>>>>>> I had to make sure that I had everything working properly because
>>>>>>> I do
>>>>>>> intend to publish as soon as I have words that are sufficiently
>>>>>>> clear to
>>>>>>> prove my point. I will have to refactor my code so that it is
>>>>>>> cleaner.
>>>>>>
>>>>>> You must be aware that no moderated group will accept your x86
>>>>>> equivalence approach as a proof of anything. So your work is
>>>>>> quixotic. No amount of refactoring will ever help. No words exist
>>>>>> that will justify it. Be prepared to lose.
>>>>>>
>>>>>
>>>>> No I do not accept this because of the Turing equivalence between the
>>>>> H(P,P) computation and the simplified Linz Ĥ.qx applied to ⟨Ĥ⟩ ⟨Ĥ⟩
>>>>> computation.
>>>>>
>>>>> People that fail to sufficiently understand Turing equivalence may
>>>>> disagree. They are not in my target audience.
>>>>>
>>>>>
>>>>
>>>> You actually have just claimed the Turing Equivalence, which you really
>>>> need to PROVE to use, and to do that you really need to show H.
>>>>
>>>> We actually know in fact that it isn't, as your H can not accept any
>>>> and
>>>> all input machine due to it not having its input in an isolated memory
>>>> space, thus there are some inputs that the Turing Machine H can accept
>>>> that your x86 H can not.
>>>>
>>>
>>> See that is an example of what I mean. I am saying that H(P,P) is
>>> equivalent to a simplified Ĥ.qx applied to ⟨Ĥ⟩ ⟨Ĥ⟩ essentially a single
>>> freaking computation not an infinite set of computations.
>>
>> Which just shows that you don't understand the problem at all. A partial
>> decider still must be able to be posed to the decider, it just doesn't
>> need to be able to get them all right.
> As long as it gets the key one of them right all of the proofs have been
> refuted.
>

Not by a long shot, since you still are calling a computation that even
you admit halts to be a non-halting computation.

Its only your POOP that you get right, but no one cares about that.

From you current admissions, you aren't even in the same theoretical
basis as the things you want to talk about, so nothing you say actually
matters.

Good luck with your rejection letters.

Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

<87k0kiqlmm.fsf@bsb.me.uk>

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https://www.novabbs.com/devel/article-flat.php?id=19946&group=comp.theory#19946

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input to H?
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 by: Ben Bacarisse - Thu, 19 Aug 2021 01:16 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/18/2021 6:51 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/18/2021 2:44 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/17/2021 7:02 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 8/17/2021 5:14 PM, Ben Bacarisse wrote:
>>>>>>
>>>>>>>> The facts come from you. They are not a matter of opinion. P(P) halts.
>>>>>>>> H(P,P) == 0.
>>>>>> ...
>>>>>>>> You are happy with that answer, but I can specify a function you can't
>>>>>>>> write: B(M, I) such that B(M, I) is non zero if and only if M(I) halts.
>>>>>>>> You don't get to choose what the "correct" answer is, so your only
>>>>>>>> option if to ignore the challenge.
>>>>>>>
>>>>>>> At this point I expect and require that those seeking an actual honest
>>>>>>> dialogue use the basis that I provided to verify that H does decide
>>>>>>> the halt status of its inputs correctly. Everyone else is written off
>>>>>>> as dishonest.
>>>>>> That's up to you. I can't stop you wasting time on a function no one
>>>>>> cares about, but you know you can't write the function I specified, so
>>>>>> you are back where you started 17 years ago. There are still
>>>>>> undecidable sets.
>>>>>
>>>>> The function does meet the spec
>>>> You don't know that because you have not asked for the proper spec. You
>>>> have not even accepted my challenge. Are you accepting the challenge
>>>> and proposing H as an implementation without knowing the full spec? I'm
>>>> happy for you to say yes if that is what you want to do.
>>>
>>> I know what the spec is:
>>
>> No you don't. The spec comes from me, and only if you say you want to
>> try the challenge. I am promising you the specification of a function
>> you can't write -- a clear demonstration that there are natural
>> uncomputable functions.
>
> I posted two different correct examples of the spec for the halting
> problem if you are talking about something else then this is out-of
> scope. You really should never erase the text the you are directly
> responding to that seems quite dishonest.

Yes. I cut them because they are not the specification of the function
that I will challenge you to write. I take it you don't want to know
what the challenge is and you just want to keep waffling about your H?
That's fine. Just say you decline. Don't keep posting specifications
that you hope will be the same as mine. They won't be.

--
Ben.

Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

<87h7fmqllh.fsf@bsb.me.uk>

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input to H?
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 by: Ben Bacarisse - Thu, 19 Aug 2021 01:17 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/18/2021 6:51 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/18/2021 3:03 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> Let's get back to something that you seem to know well:
>>>> Why start again? You just walked away from this discussion before.
>>>> I'll just bring up the errors you made in those threads again...
>>>>
>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>>>> if M applied to wM halts, and
>>>>>
>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>> if M applied to wM does not halt
>>>>>
>>>>> Can you understand that Ĥ applied to ⟨Ĥ⟩ would never stop running if
>>>>> the machine at Ĥ.qx was a UTM?
>>>>
>>>> Sigh. I'm pretty sure I know what you are trying to say (it's "Halts
>>>> with line 15 commented out" again), and I really want to agree with it
>>>> (again) but what you ask is literal nonsense. If the machine at Ĥ.qx is
>>>> a UTM and not H (well, technically it's H' and not H) then asking what Ĥ
>>>> applied to ⟨Ĥ⟩ does is, at best, ambiguous and at worst just daft
>>>> because the name Ĥ is meaningless unless the embedded TM at qx is H
>>>> which is not a UTM. It's what the "hat" means.
>>>>
>>>> If J is a TM like H but without the states and code that makes it stop
>>>> being a UTM (in some cases) then, yes, Ĵ applied to ⟨Ĵ⟩ is a non-halting
>>>> computation. This wording, whilst correct, does not cloud the water
>>>> enough.
>>>
>>> Ah great a breakthrough.
>> You accept my re-wording? You finally understand it? You won't use
>> names to refer to things that they are not? Any of these would be a
>> breakthrough indeed.
>>
>>>> Your use of Ĥ when it's not H at qx is central to you game so
>>>> you won't accept this way of putting it.
>>>
>>> When we adapt Ĥ so that the machine at Ĥ.qx is a UTM and we call this
>>> new machine Ĵ, then Ĵ applied to ⟨Ĵ⟩ never halts.
>> Why did not you not accept this any of the other times I've said it?
>
> I don't know that I didn't.

OK. Odd sort of "breakthrough" if you've known my answer and agreed
with my clearer wording before. Or are you admitting that you don't
read most of the replies so I may have said it before and you just
skimmed over it?

>>>> And have you changed your mind about the computation represented by a
>>>> tape containing ⟨Ĥ⟩ ⟨Ĥ⟩? You falsely asserted, point blank, that that
>>>> tape represents a non-halting computation.
>>>
>>> When we know that Ĵ applied to ⟨Ĵ⟩ never halts then we also know that
>>> Ĥ applied to ⟨Ĥ⟩ never stops running unless the simulating halt
>>> decider at Ĥ.qx aborts the simulation of its input.
>>
>> That's not an answer. Do you still claim that ⟨Ĥ⟩ ⟨Ĥ⟩ (that's your Ĥ,
>> the one you claim to actually have written) represents a non-halting
>> computation?
>
> The above is a simple paraphrase of what you already agreed to:

So what? Why will you not answer? If you think I agreed with you on
this point then please be assured that I was either wrong or you
misunderstood me. My position is clear: based on other statements
you've made about Ĥ, ⟨Ĥ⟩ ⟨Ĥ⟩ represents a halting computation. You have
stated that it does not, but you now appear to be reluctant to say you
stand by that statement. Why are you avoiding this issue? Do you know
it's false and can't bring yourself to say so?

--
Ben.

Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

<GPidnScbL8UfLoD8nZ2dnUU7-ROdnZ2d@giganews.com>

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Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input to H?
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References: <3YOdnecvDsA5Q4r8nZ2dnUU7-TXNnZ2d@giganews.com> <87eeavycxg.fsf@bsb.me.uk> <sf9c3k$7un$1@dont-email.me> <wpidnVhzXtFXtYT8nZ2dnUU78YHNnZ2d@brightview.co.uk> <D6WdnQ0Hy92V0YT8nZ2dnUU7-Q3NnZ2d@giganews.com> <87im06wiup.fsf@bsb.me.uk> <DJGdncQOXNbfHYT8nZ2dnUU7-dPNnZ2d@giganews.com> <874kbqw62q.fsf@bsb.me.uk> <W7udnRlZduvgdof8nZ2dnUU7-IPNnZ2d@giganews.com> <87h7fpuf5v.fsf@bsb.me.uk> <AsSdnUXVrYJ5nYb8nZ2dnUU7-VnNnZ2d@giganews.com> <875yw4v08g.fsf@bsb.me.uk> <oKidneawW_dWu4H8nZ2dnUU7-emdnZ2d@giganews.com> <8735r7u3ab.fsf@bsb.me.uk> <ufKdnZfZ0sUP3YH8nZ2dnUU7-SXNnZ2d@giganews.com> <87wnojsjqd.fsf@bsb.me.uk> <ReKdnb2pB4SVyoH8nZ2dnUU7-SvNnZ2d@giganews.com> <87o89usfll.fsf@bsb.me.uk> <uqadnd39oqwW-ID8nZ2dnUU7-amdnZ2d@giganews.com> <87sfz6qpk9.fsf@bsb.me.uk> <PeqdnehLhMapOYD8nZ2dnUU7-YXNnZ2d@giganews.com> <87k0kiqlmm.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 18 Aug 2021 20:27:27 -0500
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 by: olcott - Thu, 19 Aug 2021 01:27 UTC

On 8/18/2021 8:16 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/18/2021 6:51 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/18/2021 2:44 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/17/2021 7:02 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 8/17/2021 5:14 PM, Ben Bacarisse wrote:
>>>>>>>
>>>>>>>>> The facts come from you. They are not a matter of opinion. P(P) halts.
>>>>>>>>> H(P,P) == 0.
>>>>>>> ...
>>>>>>>>> You are happy with that answer, but I can specify a function you can't
>>>>>>>>> write: B(M, I) such that B(M, I) is non zero if and only if M(I) halts.
>>>>>>>>> You don't get to choose what the "correct" answer is, so your only
>>>>>>>>> option if to ignore the challenge.
>>>>>>>>
>>>>>>>> At this point I expect and require that those seeking an actual honest
>>>>>>>> dialogue use the basis that I provided to verify that H does decide
>>>>>>>> the halt status of its inputs correctly. Everyone else is written off
>>>>>>>> as dishonest.
>>>>>>> That's up to you. I can't stop you wasting time on a function no one
>>>>>>> cares about, but you know you can't write the function I specified, so
>>>>>>> you are back where you started 17 years ago. There are still
>>>>>>> undecidable sets.
>>>>>>
>>>>>> The function does meet the spec
>>>>> You don't know that because you have not asked for the proper spec. You
>>>>> have not even accepted my challenge. Are you accepting the challenge
>>>>> and proposing H as an implementation without knowing the full spec? I'm
>>>>> happy for you to say yes if that is what you want to do.
>>>>
>>>> I know what the spec is:
>>>
>>> No you don't. The spec comes from me, and only if you say you want to
>>> try the challenge. I am promising you the specification of a function
>>> you can't write -- a clear demonstration that there are natural
>>> uncomputable functions.
>>
>> I posted two different correct examples of the spec for the halting
>> problem if you are talking about something else then this is out-of
>> scope. You really should never erase the text the you are directly
>> responding to that seems quite dishonest.
>
> Yes. I cut them because they are not the specification of the function
> that I will challenge you to write.

That is out-of-scope and you know it.
As long as H decides this pattern correctly I have met by burden:

Now we construct a new Turing machine D with H as a subroutine.
This new TM calls H to determine what M does when the input to
M is its own description ⟨M⟩. Once D has determined this information,
it does the opposite. (Sipser:1997:165)

> I take it you don't want to know
> what the challenge is and you just want to keep waffling about your H?
> That's fine. Just say you decline. Don't keep posting specifications
> that you hope will be the same as mine. They won't be.
>

--
Copyright 2021 Pete Olcott

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

Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

<lKidnapLuffcKID8nZ2dnUU7-R-dnZ2d@giganews.com>

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Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input
to H?
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From: NoO...@NoWhere.com (olcott)
Date: Wed, 18 Aug 2021 20:34:54 -0500
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 by: olcott - Thu, 19 Aug 2021 01:34 UTC

On 8/18/2021 8:17 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/18/2021 6:51 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/18/2021 3:03 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> Let's get back to something that you seem to know well:
>>>>> Why start again? You just walked away from this discussion before.
>>>>> I'll just bring up the errors you made in those threads again...
>>>>>
>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>>>>> if M applied to wM halts, and
>>>>>>
>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>>> if M applied to wM does not halt
>>>>>>
>>>>>> Can you understand that Ĥ applied to ⟨Ĥ⟩ would never stop running if
>>>>>> the machine at Ĥ.qx was a UTM?
>>>>>
>>>>> Sigh. I'm pretty sure I know what you are trying to say (it's "Halts
>>>>> with line 15 commented out" again), and I really want to agree with it
>>>>> (again) but what you ask is literal nonsense. If the machine at Ĥ.qx is
>>>>> a UTM and not H (well, technically it's H' and not H) then asking what Ĥ
>>>>> applied to ⟨Ĥ⟩ does is, at best, ambiguous and at worst just daft
>>>>> because the name Ĥ is meaningless unless the embedded TM at qx is H
>>>>> which is not a UTM. It's what the "hat" means.
>>>>>
>>>>> If J is a TM like H but without the states and code that makes it stop
>>>>> being a UTM (in some cases) then, yes, Ĵ applied to ⟨Ĵ⟩ is a non-halting
>>>>> computation. This wording, whilst correct, does not cloud the water
>>>>> enough.
>>>>
>>>> Ah great a breakthrough.
>>> You accept my re-wording? You finally understand it? You won't use
>>> names to refer to things that they are not? Any of these would be a
>>> breakthrough indeed.
>>>
>>>>> Your use of Ĥ when it's not H at qx is central to you game so
>>>>> you won't accept this way of putting it.
>>>>
>>>> When we adapt Ĥ so that the machine at Ĥ.qx is a UTM and we call this
>>>> new machine Ĵ, then Ĵ applied to ⟨Ĵ⟩ never halts.
>>> Why did not you not accept this any of the other times I've said it?
>>
>> I don't know that I didn't.
>
> OK. Odd sort of "breakthrough" if you've known my answer and agreed
> with my clearer wording before. Or are you admitting that you don't
> read most of the replies so I may have said it before and you just
> skimmed over it?
>

It took you six years to finally acknowledge that.

>>>>> And have you changed your mind about the computation represented by a
>>>>> tape containing ⟨Ĥ⟩ ⟨Ĥ⟩? You falsely asserted, point blank, that that
>>>>> tape represents a non-halting computation.
>>>>
>>>> When we know that Ĵ applied to ⟨Ĵ⟩ never halts then we also know that
>>>> Ĥ applied to ⟨Ĥ⟩ never stops running unless the simulating halt
>>>> decider at Ĥ.qx aborts the simulation of its input.
>>>
>>> That's not an answer. Do you still claim that ⟨Ĥ⟩ ⟨Ĥ⟩ (that's your Ĥ,
>>> the one you claim to actually have written) represents a non-halting
>>> computation?
>>
>> The above is a simple paraphrase of what you already agreed to:
>
> So what? Why will you not answer? If you think I agreed with you on

You dishonestly erase the criterion measure that you agreed to showing
that the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ is correctly decided as never halting.

On 5/11/2021 11:10 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> Truism:
>> Every simulation that would never stop unless Halts() stops
>> it at some point specifies infinite execution.
>
> Any algorithm that implements this truism is, of course, a halting
> decider.

> this point then please be assured that I was either wrong or you
> misunderstood me. My position is clear: based on other statements
> you've made about Ĥ, ⟨Ĥ⟩ ⟨Ĥ⟩ represents a halting computation. You have
> stated that it does not, but you now appear to be reluctant to say you
> stand by that statement. Why are you avoiding this issue? Do you know
> it's false and can't bring yourself to say so?
>

--
Copyright 2021 Pete Olcott

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

Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

<875yw2qkfb.fsf@bsb.me.uk>

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https://www.novabbs.com/devel/article-flat.php?id=19950&group=comp.theory#19950

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input to H?
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 by: Ben Bacarisse - Thu, 19 Aug 2021 01:42 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/18/2021 8:16 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/18/2021 6:51 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/18/2021 2:44 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 8/17/2021 7:02 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> On 8/17/2021 5:14 PM, Ben Bacarisse wrote:
>>>>>>>>
>>>>>>>>>> The facts come from you. They are not a matter of opinion. P(P) halts.
>>>>>>>>>> H(P,P) == 0.
>>>>>>>> ...
>>>>>>>>>> You are happy with that answer, but I can specify a function you can't
>>>>>>>>>> write: B(M, I) such that B(M, I) is non zero if and only if M(I) halts.
>>>>>>>>>> You don't get to choose what the "correct" answer is, so your only
>>>>>>>>>> option if to ignore the challenge.
>>>>>>>>>
>>>>>>>>> At this point I expect and require that those seeking an actual honest
>>>>>>>>> dialogue use the basis that I provided to verify that H does decide
>>>>>>>>> the halt status of its inputs correctly. Everyone else is written off
>>>>>>>>> as dishonest.
>>>>>>>> That's up to you. I can't stop you wasting time on a function no one
>>>>>>>> cares about, but you know you can't write the function I specified, so
>>>>>>>> you are back where you started 17 years ago. There are still
>>>>>>>> undecidable sets.
>>>>>>>
>>>>>>> The function does meet the spec
>>>>>> You don't know that because you have not asked for the proper spec. You
>>>>>> have not even accepted my challenge. Are you accepting the challenge
>>>>>> and proposing H as an implementation without knowing the full spec? I'm
>>>>>> happy for you to say yes if that is what you want to do.
>>>>>
>>>>> I know what the spec is:
>>>>
>>>> No you don't. The spec comes from me, and only if you say you want to
>>>> try the challenge. I am promising you the specification of a function
>>>> you can't write -- a clear demonstration that there are natural
>>>> uncomputable functions.
>>>
>>> I posted two different correct examples of the spec for the halting
>>> problem if you are talking about something else then this is out-of
>>> scope. You really should never erase the text the you are directly
>>> responding to that seems quite dishonest.
>>
>> Yes. I cut them because they are not the specification of the function
>> that I will challenge you to write.
>
> That is out-of-scope and you know it.

Not in my opinion, no. A function that that shows there are undecidable
sets should worry you, but for some reason you prefer to stick with
talking about your H that does something entirely unsurprising and
uninteresting.

--
Ben.

Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

<87zgtep4zl.fsf@bsb.me.uk>

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input to H?
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 by: Ben Bacarisse - Thu, 19 Aug 2021 02:01 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/18/2021 8:17 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/18/2021 6:51 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/18/2021 3:03 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> Let's get back to something that you seem to know well:
>>>>>> Why start again? You just walked away from this discussion before.
>>>>>> I'll just bring up the errors you made in those threads again...
>>>>>>
>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>>>>>> if M applied to wM halts, and
>>>>>>>
>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>>>> if M applied to wM does not halt
>>>>>>>
>>>>>>> Can you understand that Ĥ applied to ⟨Ĥ⟩ would never stop running if
>>>>>>> the machine at Ĥ.qx was a UTM?
>>>>>>
>>>>>> Sigh. I'm pretty sure I know what you are trying to say (it's "Halts
>>>>>> with line 15 commented out" again), and I really want to agree with it
>>>>>> (again) but what you ask is literal nonsense. If the machine at Ĥ.qx is
>>>>>> a UTM and not H (well, technically it's H' and not H) then asking what Ĥ
>>>>>> applied to ⟨Ĥ⟩ does is, at best, ambiguous and at worst just daft
>>>>>> because the name Ĥ is meaningless unless the embedded TM at qx is H
>>>>>> which is not a UTM. It's what the "hat" means.
>>>>>>
>>>>>> If J is a TM like H but without the states and code that makes it stop
>>>>>> being a UTM (in some cases) then, yes, Ĵ applied to ⟨Ĵ⟩ is a non-halting
>>>>>> computation. This wording, whilst correct, does not cloud the water
>>>>>> enough.
>>>>>
>>>>> Ah great a breakthrough.
>>>> You accept my re-wording? You finally understand it? You won't use
>>>> names to refer to things that they are not? Any of these would be a
>>>> breakthrough indeed.
>>>>
>>>>>> Your use of Ĥ when it's not H at qx is central to you game so
>>>>>> you won't accept this way of putting it.
>>>>>
>>>>> When we adapt Ĥ so that the machine at Ĥ.qx is a UTM and we call this
>>>>> new machine Ĵ, then Ĵ applied to ⟨Ĵ⟩ never halts.
>>>> Why did not you not accept this any of the other times I've said it?
>>>
>>> I don't know that I didn't.
>>
>> OK. Odd sort of "breakthrough" if you've known my answer and agreed
>> with my clearer wording before. Or are you admitting that you don't
>> read most of the replies so I may have said it before and you just
>> skimmed over it?
>
> It took you six years to finally acknowledge that.

I'd like to see a single citation from any time were I have doubted this
fact. Basically you don't know what I'm saying half the time but you
arrogance prevents you from ever asking for an explanation.

>>>>>> And have you changed your mind about the computation represented by a
>>>>>> tape containing ⟨Ĥ⟩ ⟨Ĥ⟩? You falsely asserted, point blank, that that
>>>>>> tape represents a non-halting computation.
>>>>>
>>>>> When we know that Ĵ applied to ⟨Ĵ⟩ never halts then we also know that
>>>>> Ĥ applied to ⟨Ĥ⟩ never stops running unless the simulating halt
>>>>> decider at Ĥ.qx aborts the simulation of its input.
>>>>
>>>> That's not an answer. Do you still claim that ⟨Ĥ⟩ ⟨Ĥ⟩ (that's your Ĥ,
>>>> the one you claim to actually have written) represents a non-halting
>>>> computation?
>>>
>>> The above is a simple paraphrase of what you already agreed to:
>> So what? Why will you not answer? If you think I agreed with you on
>
> You dishonestly erase the criterion measure that you agreed to showing
> that the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ is correctly decided as never halting.

We won't get anywhere if you keep ducking this question. You made a
false statement in the old thread about this and you won't say if you
stand by that false statement.

You track record in understanding people is poor. But if you think I
agreed, so what? I don't agree now regardless of what you think I've
said in the past. I am allowed to re-state, as clearly as possible, the
facts of the matter and the string ⟨Ĥ⟩ ⟨Ĥ⟩ represents a halting
computation. All I am asking is that you be as clear as I am being. Do
you disagree that the string ⟨Ĥ⟩ ⟨Ĥ⟩ represents a halting computation?

> On 5/11/2021 11:10 AM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> Truism:
>>> Every simulation that would never stop unless Halts() stops
>>> it at some point specifies infinite execution.
>>
>> Any algorithm that implements this truism is, of course, a halting
>> decider.

I don't think you know what I'm saying here. You certainly don't agree
with me as you vehemently deny that you are claiming to have a halt
decider. One day, you might think of asking what I mean *and of
listening to the answer*.

--
Ben.

Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

<2q6dnZ-dGIzeIYD8nZ2dnUU7-ffNnZ2d@giganews.com>

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Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input
to H?
Newsgroups: comp.theory
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<wpidnVhzXtFXtYT8nZ2dnUU78YHNnZ2d@brightview.co.uk>
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From: NoO...@NoWhere.com (olcott)
Date: Wed, 18 Aug 2021 21:04:48 -0500
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 by: olcott - Thu, 19 Aug 2021 02:04 UTC

On 8/18/2021 8:42 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/18/2021 8:16 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/18/2021 6:51 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/18/2021 2:44 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 8/17/2021 7:02 PM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> On 8/17/2021 5:14 PM, Ben Bacarisse wrote:
>>>>>>>>>
>>>>>>>>>>> The facts come from you. They are not a matter of opinion. P(P) halts.
>>>>>>>>>>> H(P,P) == 0.
>>>>>>>>> ...
>>>>>>>>>>> You are happy with that answer, but I can specify a function you can't
>>>>>>>>>>> write: B(M, I) such that B(M, I) is non zero if and only if M(I) halts.
>>>>>>>>>>> You don't get to choose what the "correct" answer is, so your only
>>>>>>>>>>> option if to ignore the challenge.
>>>>>>>>>>
>>>>>>>>>> At this point I expect and require that those seeking an actual honest
>>>>>>>>>> dialogue use the basis that I provided to verify that H does decide
>>>>>>>>>> the halt status of its inputs correctly. Everyone else is written off
>>>>>>>>>> as dishonest.
>>>>>>>>> That's up to you. I can't stop you wasting time on a function no one
>>>>>>>>> cares about, but you know you can't write the function I specified, so
>>>>>>>>> you are back where you started 17 years ago. There are still
>>>>>>>>> undecidable sets.
>>>>>>>>
>>>>>>>> The function does meet the spec
>>>>>>> You don't know that because you have not asked for the proper spec. You
>>>>>>> have not even accepted my challenge. Are you accepting the challenge
>>>>>>> and proposing H as an implementation without knowing the full spec? I'm
>>>>>>> happy for you to say yes if that is what you want to do.
>>>>>>
>>>>>> I know what the spec is:
>>>>>
>>>>> No you don't. The spec comes from me, and only if you say you want to
>>>>> try the challenge. I am promising you the specification of a function
>>>>> you can't write -- a clear demonstration that there are natural
>>>>> uncomputable functions.
>>>>
>>>> I posted two different correct examples of the spec for the halting
>>>> problem if you are talking about something else then this is out-of
>>>> scope. You really should never erase the text the you are directly
>>>> responding to that seems quite dishonest.
>>>
>>> Yes. I cut them because they are not the specification of the function
>>> that I will challenge you to write.
>>
>> That is out-of-scope and you know it.
>
> Not in my opinion, no. A function that that shows there are undecidable
> sets should worry you, but for some reason you prefer to stick with
> talking about your H that does something entirely unsurprising and
> uninteresting.
>

So when I correctly refute the halting problem proofs you say no I did
not refute every proof in the universe and the halting problem proof is
one of these proofs therefore I did not refute the halting problem proof.

THIS IS THE SCOPE OF THE HALTING PROBLEM PROOFS:
Now we construct a new Turing machine D with H as a subroutine.
This new TM calls H to determine what M does when the input to
M is its own description ⟨M⟩. Once D has determined this information,
it does the opposite. (Sipser:1997:165)

--
Copyright 2021 Pete Olcott

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

Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

<E8adnbBrnoijI4D8nZ2dnUU78RXNnZ2d@giganews.com>

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Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input
to H?
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References: <3YOdnecvDsA5Q4r8nZ2dnUU7-TXNnZ2d@giganews.com>
<wpidnVhzXtFXtYT8nZ2dnUU78YHNnZ2d@brightview.co.uk>
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From: NoO...@NoWhere.com (olcott)
Date: Wed, 18 Aug 2021 21:13:14 -0500
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 by: olcott - Thu, 19 Aug 2021 02:13 UTC

On 8/18/2021 9:01 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/18/2021 8:17 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/18/2021 6:51 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/18/2021 3:03 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> Let's get back to something that you seem to know well:
>>>>>>> Why start again? You just walked away from this discussion before.
>>>>>>> I'll just bring up the errors you made in those threads again...
>>>>>>>
>>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>>>>>>> if M applied to wM halts, and
>>>>>>>>
>>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>>>>> if M applied to wM does not halt
>>>>>>>>
>>>>>>>> Can you understand that Ĥ applied to ⟨Ĥ⟩ would never stop running if
>>>>>>>> the machine at Ĥ.qx was a UTM?
>>>>>>>
>>>>>>> Sigh. I'm pretty sure I know what you are trying to say (it's "Halts
>>>>>>> with line 15 commented out" again), and I really want to agree with it
>>>>>>> (again) but what you ask is literal nonsense. If the machine at Ĥ.qx is
>>>>>>> a UTM and not H (well, technically it's H' and not H) then asking what Ĥ
>>>>>>> applied to ⟨Ĥ⟩ does is, at best, ambiguous and at worst just daft
>>>>>>> because the name Ĥ is meaningless unless the embedded TM at qx is H
>>>>>>> which is not a UTM. It's what the "hat" means.
>>>>>>>
>>>>>>> If J is a TM like H but without the states and code that makes it stop
>>>>>>> being a UTM (in some cases) then, yes, Ĵ applied to ⟨Ĵ⟩ is a non-halting
>>>>>>> computation. This wording, whilst correct, does not cloud the water
>>>>>>> enough.
>>>>>>
>>>>>> Ah great a breakthrough.
>>>>> You accept my re-wording? You finally understand it? You won't use
>>>>> names to refer to things that they are not? Any of these would be a
>>>>> breakthrough indeed.
>>>>>
>>>>>>> Your use of Ĥ when it's not H at qx is central to you game so
>>>>>>> you won't accept this way of putting it.
>>>>>>
>>>>>> When we adapt Ĥ so that the machine at Ĥ.qx is a UTM and we call this
>>>>>> new machine Ĵ, then Ĵ applied to ⟨Ĵ⟩ never halts.
>>>>> Why did not you not accept this any of the other times I've said it?
>>>>
>>>> I don't know that I didn't.
>>>
>>> OK. Odd sort of "breakthrough" if you've known my answer and agreed
>>> with my clearer wording before. Or are you admitting that you don't
>>> read most of the replies so I may have said it before and you just
>>> skimmed over it?
>>
>> It took you six years to finally acknowledge that.
>
> I'd like to see a single citation from any time were I have doubted this
> fact. Basically you don't know what I'm saying half the time but you
> arrogance prevents you from ever asking for an explanation.
>
>>>>>>> And have you changed your mind about the computation represented by a
>>>>>>> tape containing ⟨Ĥ⟩ ⟨Ĥ⟩? You falsely asserted, point blank, that that
>>>>>>> tape represents a non-halting computation.
>>>>>>
>>>>>> When we know that Ĵ applied to ⟨Ĵ⟩ never halts then we also know that
>>>>>> Ĥ applied to ⟨Ĥ⟩ never stops running unless the simulating halt
>>>>>> decider at Ĥ.qx aborts the simulation of its input.
>>>>>
>>>>> That's not an answer. Do you still claim that ⟨Ĥ⟩ ⟨Ĥ⟩ (that's your Ĥ,
>>>>> the one you claim to actually have written) represents a non-halting
>>>>> computation?
>>>>
>>>> The above is a simple paraphrase of what you already agreed to:
>>> So what? Why will you not answer? If you think I agreed with you on
>>
>> You dishonestly erase the criterion measure that you agreed to showing
>> that the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ is correctly decided as never halting.
>
> We won't get anywhere if you keep ducking this question. You made a
> false statement in the old thread about this and you won't say if you
> stand by that false statement.
>
> You track record in understanding people is poor. But if you think I
> agreed, so what? I don't agree now regardless of what you think I've
> said in the past. I am allowed to re-state, as clearly as possible, the
> facts of the matter and the string ⟨Ĥ⟩ ⟨Ĥ⟩ represents a halting
> computation. All I am asking is that you be as clear as I am being. Do
> you disagree that the string ⟨Ĥ⟩ ⟨Ĥ⟩ represents a halting computation?

I already totally explained all that.

When-so-ever any input to a simulating halt decider would never halt
unless its simulating halt decider stopped simulating it then this input
is correctly decided as never halting.

The input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts unless its simulating halt decider
stops simulating it.

Can you figure out whether or not this is a yes or no answer to your
question?

I always answer yes or no questions with the reasoning that proves the
yes or no answer. This forces people to actually carefully read all of
my words instead of just glancing at them prior to forming their rebuttal.

>> On 5/11/2021 11:10 AM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> Truism:
>>>> Every simulation that would never stop unless Halts() stops
>>>> it at some point specifies infinite execution.
>>>
>>> Any algorithm that implements this truism is, of course, a halting
>>> decider.
>
> I don't think you know what I'm saying here. You certainly don't agree
> with me as you vehemently deny that you are claiming to have a halt
> decider. One day, you might think of asking what I mean *and of
> listening to the answer*.
>

--
Copyright 2021 Pete Olcott

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

Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

<CljTI.3317$7n1.249@fx27.iad>

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to H?
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Thu, 19 Aug 2021 02:28 UTC

On 8/18/21 10:13 PM, olcott wrote:
> On 8/18/2021 9:01 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/18/2021 8:17 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/18/2021 6:51 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 8/18/2021 3:03 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> Let's get back to something that you seem to know well:
>>>>>>>> Why start again?  You just walked away from this discussion before.
>>>>>>>> I'll just bring up the errors you made in those threads again...
>>>>>>>>
>>>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>>>>>>>> if M applied to wM halts, and
>>>>>>>>>
>>>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>>>>>> if M applied to wM does not halt
>>>>>>>>>
>>>>>>>>> Can you understand that Ĥ applied to ⟨Ĥ⟩ would never stop
>>>>>>>>> running if
>>>>>>>>> the machine at Ĥ.qx was a UTM?
>>>>>>>>
>>>>>>>> Sigh.  I'm pretty sure I know what you are trying to say (it's
>>>>>>>> "Halts
>>>>>>>> with line 15 commented out" again), and I really want to agree
>>>>>>>> with it
>>>>>>>> (again) but what you ask is literal nonsense.  If the machine at
>>>>>>>> Ĥ.qx is
>>>>>>>> a UTM and not H (well, technically it's H' and not H) then
>>>>>>>> asking what Ĥ
>>>>>>>> applied to ⟨Ĥ⟩ does is, at best, ambiguous and at worst just daft
>>>>>>>> because the name Ĥ is meaningless unless the embedded TM at qx is H
>>>>>>>> which is not a UTM.  It's what the "hat" means.
>>>>>>>>
>>>>>>>> If J is a TM like H but without the states and code that makes
>>>>>>>> it stop
>>>>>>>> being a UTM (in some cases) then, yes, Ĵ applied to ⟨Ĵ⟩ is a
>>>>>>>> non-halting
>>>>>>>> computation.  This wording, whilst correct, does not cloud the
>>>>>>>> water
>>>>>>>> enough.
>>>>>>>
>>>>>>> Ah great a breakthrough.
>>>>>> You accept my re-wording?  You finally understand it?  You won't use
>>>>>> names to refer to things that they are not?  Any of these would be a
>>>>>> breakthrough indeed.
>>>>>>
>>>>>>>> Your use of Ĥ when it's not H at qx is central to you game so
>>>>>>>> you won't accept this way of putting it.
>>>>>>>
>>>>>>> When we adapt Ĥ so that the machine at Ĥ.qx is a UTM and we call
>>>>>>> this
>>>>>>> new machine Ĵ, then Ĵ applied to ⟨Ĵ⟩ never halts.
>>>>>> Why did not you not accept this any of the other times I've said it?
>>>>>
>>>>> I don't know that I didn't.
>>>>
>>>> OK.  Odd sort of "breakthrough" if you've known my answer and agreed
>>>> with my clearer wording before.  Or are you admitting that you don't
>>>> read most of the replies so I may have said it before and you just
>>>> skimmed over it?
>>>
>>> It took you six years to finally acknowledge that.
>>
>> I'd like to see a single citation from any time were I have doubted this
>> fact.  Basically you don't know what I'm saying half the time but you
>> arrogance prevents you from ever asking for an explanation.
>>
>>>>>>>> And have you changed your mind about the computation represented
>>>>>>>> by a
>>>>>>>> tape containing ⟨Ĥ⟩ ⟨Ĥ⟩?  You falsely asserted, point blank,
>>>>>>>> that that
>>>>>>>> tape represents a non-halting computation.
>>>>>>>
>>>>>>> When we know that Ĵ applied to ⟨Ĵ⟩ never halts then we also know
>>>>>>> that
>>>>>>> Ĥ applied to ⟨Ĥ⟩ never stops running unless the simulating halt
>>>>>>> decider at Ĥ.qx aborts the simulation of its input.
>>>>>>
>>>>>> That's not an answer.  Do you still claim that ⟨Ĥ⟩ ⟨Ĥ⟩ (that's
>>>>>> your Ĥ,
>>>>>> the one you claim to actually have written) represents a non-halting
>>>>>> computation?
>>>>>
>>>>> The above is a simple paraphrase of what you already agreed to:
>>>> So what?  Why will you not answer?  If you think I agreed with you on
>>>
>>> You dishonestly erase the criterion measure that you agreed to showing
>>> that the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ is correctly decided as never halting.
>>
>> We won't get anywhere if you keep ducking this question.  You made a
>> false statement in the old thread about this and you won't say if you
>> stand by that false statement.
>>
>> You track record in understanding people is poor.  But if you think I
>> agreed, so what?  I don't agree now regardless of what you think I've
>> said in the past.  I am allowed to re-state, as clearly as possible, the
>> facts of the matter and the string ⟨Ĥ⟩ ⟨Ĥ⟩ represents a halting
>> computation.  All I am asking is that you be as clear as I am being.  Do
>> you disagree that the string ⟨Ĥ⟩ ⟨Ĥ⟩ represents a halting computation?
>
>
> I already totally explained all that.
>
> When-so-ever any input to a simulating halt decider would never halt
> unless its simulating halt decider stopped simulating it then this input
> is correctly decided as never halting.
>

WRONG, at least in the real world's definition of the halting problem.
Maybe it is different in your POOP, but in the Real Halting problem,
what matters is what the machine the input is a representation of does,
not that the decider never was able to trace the machine to its final
halting state.

H^(<H^>) Halts, therefore the only right answer to H(<H^>, <H^>) is halting.

H gets it WRONG.

FAIL.

> The input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts unless its simulating halt decider
> stops simulating it.

But it still halts pure based on the code that is H^, thus it is HALTING.

Maybe in POOP that is different, but we don't care about that.

>
> Can you figure out whether or not this is a yes or no answer to your
> question?
>
> I always answer yes or no questions with the reasoning that proves the
> yes or no answer. This forces people to actually carefully read all of
> my words instead of just glancing at them prior to forming their rebuttal.

So, does the machine whose representation was given to H halt or not?

You have previously agreed that the answer is yes.

Does H say that that machine is non-halting?

You have always said that answer is Yes.

Is non-Halting the right answer for a machine that is actually Halting?

You keep on saying yes, but the answer has to be No.

>
>>> On 5/11/2021 11:10 AM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> Truism:
>>>>> Every simulation that would never stop unless Halts() stops
>>>>> it at some point specifies infinite execution.
>>>>
>>>> Any algorithm that implements this truism is, of course, a halting
>>>> decider.
>>
>> I don't think you know what I'm saying here.  You certainly don't agree
>> with me as you vehemently deny that you are claiming to have a halt
>> decider.  One day, you might think of asking what I mean *and of
>> listening to the answer*.
>>
>
>

Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

<cnjTI.3318$7n1.168@fx27.iad>

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https://www.novabbs.com/devel/article-flat.php?id=19955&group=comp.theory#19955

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Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input
to H?
Newsgroups: comp.theory
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Thu, 19 Aug 2021 02:30 UTC

On 8/18/21 10:04 PM, olcott wrote:
> On 8/18/2021 8:42 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/18/2021 8:16 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/18/2021 6:51 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 8/18/2021 2:44 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> On 8/17/2021 7:02 PM, Ben Bacarisse wrote:
>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>
>>>>>>>>>>> On 8/17/2021 5:14 PM, Ben Bacarisse wrote:
>>>>>>>>>>
>>>>>>>>>>>> The facts come from you.  They are not a matter of opinion. 
>>>>>>>>>>>> P(P) halts.
>>>>>>>>>>>> H(P,P) == 0.
>>>>>>>>>> ...
>>>>>>>>>>>> You are happy with that answer, but I can specify a function
>>>>>>>>>>>> you can't
>>>>>>>>>>>> write: B(M, I) such that B(M, I) is non zero if and only if
>>>>>>>>>>>> M(I) halts.
>>>>>>>>>>>> You don't get to choose what the "correct" answer is, so
>>>>>>>>>>>> your only
>>>>>>>>>>>> option if to ignore the challenge.
>>>>>>>>>>>
>>>>>>>>>>> At this point I expect and require that those seeking an
>>>>>>>>>>> actual honest
>>>>>>>>>>> dialogue use the basis that I provided to verify that H does
>>>>>>>>>>> decide
>>>>>>>>>>> the halt status of its inputs correctly. Everyone else is
>>>>>>>>>>> written off
>>>>>>>>>>> as dishonest.
>>>>>>>>>> That's up to you.  I can't stop you wasting time on a function
>>>>>>>>>> no one
>>>>>>>>>> cares about, but you know you can't write the function I
>>>>>>>>>> specified, so
>>>>>>>>>> you are back where you started 17 years ago.  There are still
>>>>>>>>>> undecidable sets.
>>>>>>>>>
>>>>>>>>> The function does meet the spec
>>>>>>>> You don't know that because you have not asked for the proper
>>>>>>>> spec.  You
>>>>>>>> have not even accepted my challenge.  Are you accepting the
>>>>>>>> challenge
>>>>>>>> and proposing H as an implementation without knowing the full
>>>>>>>> spec?  I'm
>>>>>>>> happy for you to say yes if that is what you want to do.
>>>>>>>
>>>>>>> I know what the spec is:
>>>>>>
>>>>>> No you don't.  The spec comes from me, and only if you say you
>>>>>> want to
>>>>>> try the challenge.  I am promising you the specification of a
>>>>>> function
>>>>>> you can't write -- a clear demonstration that there are natural
>>>>>> uncomputable functions.
>>>>>
>>>>> I posted two different correct examples of the spec for the halting
>>>>> problem if you are talking about something else then this is out-of
>>>>> scope. You really should never erase the text the you are directly
>>>>> responding to that seems quite dishonest.
>>>>
>>>> Yes.  I cut them because they are not the specification of the function
>>>> that I will challenge you to write.
>>>
>>> That is out-of-scope and you know it.
>>
>> Not in my opinion, no.  A function that that shows there are undecidable
>> sets should worry you, but for some reason you prefer to stick with
>> talking about your H that does something entirely unsurprising and
>> uninteresting.
>>
>
> So when I correctly refute the halting problem proofs you say no I did
> not refute every proof in the universe and the halting problem proof is
> one of these proofs therefore I did not refute the halting problem proof.
>
> THIS IS THE SCOPE OF THE HALTING PROBLEM PROOFS:
>    Now we construct a new Turing machine D with H as a subroutine.
>    This new TM calls H to determine what M does when the input to
>    M is its own description ⟨M⟩. Once D has determined this information,
>    it does the opposite.  (Sipser:1997:165)
>
>

Right, and in this example, your H says D will be non-halting, when it
turns out that D is a Halting Computation.

Thus H was wrong.

You have NOT refuted that proof.

Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

<7cOdnUWjhL5RVYD8nZ2dnUU7-cXNnZ2d@giganews.com>

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Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input
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From: NoO...@NoWhere.com (olcott)
Date: Wed, 18 Aug 2021 21:58:18 -0500
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 by: olcott - Thu, 19 Aug 2021 02:58 UTC

On 8/18/2021 9:30 PM, Richard Damon wrote:
> On 8/18/21 10:04 PM, olcott wrote:
>> On 8/18/2021 8:42 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/18/2021 8:16 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/18/2021 6:51 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 8/18/2021 2:44 PM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> On 8/17/2021 7:02 PM, Ben Bacarisse wrote:
>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>
>>>>>>>>>>>> On 8/17/2021 5:14 PM, Ben Bacarisse wrote:
>>>>>>>>>>>
>>>>>>>>>>>>> The facts come from you.  They are not a matter of opinion.
>>>>>>>>>>>>> P(P) halts.
>>>>>>>>>>>>> H(P,P) == 0.
>>>>>>>>>>> ...
>>>>>>>>>>>>> You are happy with that answer, but I can specify a function
>>>>>>>>>>>>> you can't
>>>>>>>>>>>>> write: B(M, I) such that B(M, I) is non zero if and only if
>>>>>>>>>>>>> M(I) halts.
>>>>>>>>>>>>> You don't get to choose what the "correct" answer is, so
>>>>>>>>>>>>> your only
>>>>>>>>>>>>> option if to ignore the challenge.
>>>>>>>>>>>>
>>>>>>>>>>>> At this point I expect and require that those seeking an
>>>>>>>>>>>> actual honest
>>>>>>>>>>>> dialogue use the basis that I provided to verify that H does
>>>>>>>>>>>> decide
>>>>>>>>>>>> the halt status of its inputs correctly. Everyone else is
>>>>>>>>>>>> written off
>>>>>>>>>>>> as dishonest.
>>>>>>>>>>> That's up to you.  I can't stop you wasting time on a function
>>>>>>>>>>> no one
>>>>>>>>>>> cares about, but you know you can't write the function I
>>>>>>>>>>> specified, so
>>>>>>>>>>> you are back where you started 17 years ago.  There are still
>>>>>>>>>>> undecidable sets.
>>>>>>>>>>
>>>>>>>>>> The function does meet the spec
>>>>>>>>> You don't know that because you have not asked for the proper
>>>>>>>>> spec.  You
>>>>>>>>> have not even accepted my challenge.  Are you accepting the
>>>>>>>>> challenge
>>>>>>>>> and proposing H as an implementation without knowing the full
>>>>>>>>> spec?  I'm
>>>>>>>>> happy for you to say yes if that is what you want to do.
>>>>>>>>
>>>>>>>> I know what the spec is:
>>>>>>>
>>>>>>> No you don't.  The spec comes from me, and only if you say you
>>>>>>> want to
>>>>>>> try the challenge.  I am promising you the specification of a
>>>>>>> function
>>>>>>> you can't write -- a clear demonstration that there are natural
>>>>>>> uncomputable functions.
>>>>>>
>>>>>> I posted two different correct examples of the spec for the halting
>>>>>> problem if you are talking about something else then this is out-of
>>>>>> scope. You really should never erase the text the you are directly
>>>>>> responding to that seems quite dishonest.
>>>>>
>>>>> Yes.  I cut them because they are not the specification of the function
>>>>> that I will challenge you to write.
>>>>
>>>> That is out-of-scope and you know it.
>>>
>>> Not in my opinion, no.  A function that that shows there are undecidable
>>> sets should worry you, but for some reason you prefer to stick with
>>> talking about your H that does something entirely unsurprising and
>>> uninteresting.
>>>
>>
>> So when I correctly refute the halting problem proofs you say no I did
>> not refute every proof in the universe and the halting problem proof is
>> one of these proofs therefore I did not refute the halting problem proof.
>>
>> THIS IS THE SCOPE OF THE HALTING PROBLEM PROOFS:
>>    Now we construct a new Turing machine D with H as a subroutine.
>>    This new TM calls H to determine what M does when the input to
>>    M is its own description ⟨M⟩. Once D has determined this information,
>>    it does the opposite.  (Sipser:1997:165)
>>
>>
>
> Right, and in this example, your H says D will be non-halting, when it
> turns out that D is a Halting Computation.
>

int main(){ P(P); } is not the same computation as
int main(){ H(P,P); }

The first computation has a dependency on the return value of H(P,P)
that the second computation does not have thus the two computations are
not equivalent.

> Thus H was wrong.
>
> You have NOT refuted that proof.
>

--
Copyright 2021 Pete Olcott

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

Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

<sfklv8$b4o$1@dont-email.me>

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From: jbb...@notatt.com (Jeff Barnett)
Newsgroups: comp.theory
Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input
to H?
Date: Wed, 18 Aug 2021 22:16:36 -0600
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 by: Jeff Barnett - Thu, 19 Aug 2021 04:16 UTC

On 8/18/2021 1:54 PM, olcott wrote:
> On 8/18/2021 2:44 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
<CRAP SNIP>
> I know what the spec is:
>
>      the Turing machine halting problem. Simply stated, the problem
>      is: given the description of a Turing machine M and an input w,
>      does M, when started in the initial configuration q0w, perform a
>      computation that eventually halts? (Linz:1990:317).
>
>      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
>
> description of a Turing machine M (not a Turing Machine)
> description of an arbitrary computer program (not an executable program)
All you have shown is that you know where some version of a spec of
something can be found and copied. There is not a shred of a hint that
you "know" anything. Polly want a cracker?
--
Jeff Barnett

Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

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Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input to H?
Newsgroups: comp.theory
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From: NoO...@NoWhere.com (olcott)
Date: Wed, 18 Aug 2021 23:38:37 -0500
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 by: olcott - Thu, 19 Aug 2021 04:38 UTC

On 8/18/2021 11:16 PM, Jeff Barnett wrote:
> On 8/18/2021 1:54 PM, olcott wrote:
>> On 8/18/2021 2:44 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
> <CRAP SNIP>
>> I know what the spec is:
>>
>>       the Turing machine halting problem. Simply stated, the problem
>>       is: given the description of a Turing machine M and an input w,
>>       does M, when started in the initial configuration q0w, perform a
>>       computation that eventually halts? (Linz:1990:317).
>>
>>       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
>>

>> description of a Turing machine M (not a Turing Machine)
>> description of an arbitrary computer program (not an executable program)

> All you have shown is that you know where some version of a spec of
> something can be found and copied. There is not a shred of a hint that
> you "know" anything. Polly want a cracker?

I have proven that I know more about it than my detractors that continue
to think it is about Turing machines or executing programs instead of
input descriptions of TM's and programs.

--
Copyright 2021 Pete Olcott

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

Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

<sfktoe$fts$1@dont-email.me>

 copy mid

https://www.novabbs.com/devel/article-flat.php?id=19959&group=comp.theory#19959

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Path: i2pn2.org!i2pn.org!eternal-september.org!reader02.eternal-september.org!.POSTED!not-for-mail
From: jbb...@notatt.com (Jeff Barnett)
Newsgroups: comp.theory
Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input
to H?
Date: Thu, 19 Aug 2021 00:29:30 -0600
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 by: Jeff Barnett - Thu, 19 Aug 2021 06:29 UTC

On 8/18/2021 10:38 PM, olcott wrote:
> On 8/18/2021 11:16 PM, Jeff Barnett wrote:
>> On 8/18/2021 1:54 PM, olcott wrote:
>>> On 8/18/2021 2:44 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>> <CRAP SNIP>
>>> I know what the spec is:
>>>
>>>       the Turing machine halting problem. Simply stated, the problem
>>>       is: given the description of a Turing machine M and an input w,
>>>       does M, when started in the initial configuration q0w, perform a
>>>       computation that eventually halts? (Linz:1990:317).
>>>
>>>       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
>>>
>
>>> description of a Turing machine M (not a Turing Machine)
>>> description of an arbitrary computer program (not an executable program)
>
>> All you have shown is that you know where some version of a spec of
>> something can be found and copied. There is not a shred of a hint that
>> you "know" anything. Polly want a cracker?
>
> I have proven that I know more about it than my detractors that continue
> to think it is about Turing machines or executing programs instead of
> input descriptions of TM's and programs.

All you've shown after all these years is that you have memorized the
C-c C-v pair to copy and paste material you do not understand. If your
health problems are as dire as you say, you might switch tacks to
something that will bring you positive feedback. That might easy some of
the anguish I assume you feel. Continuing down the current path will
only maintain the steady distractions that streams of deserved negative
feedback has engendered. Try something that will produce a positive in
your life. The current plan of attack is totally useless. You will only
leave a trace that shows an ignorance of everything you thought was
important. Waste of time and waste of life.

Your use of these newsgroups reminds me of a women who frequented the
online forums that Amazon used to host. They were cutting back the
amount of pictures that people could post, monitoring the forums more
closely for abusive posts until the finally sliced off the forums. Some
one pointed out to this women that she had 10s of thousands of pictures
uploaded. She scanned picture sewing books and uploaded all of it to
accompany sewing patterns and books reviews she wrote. Someone pointed
this out to her and simply asked why Amazon should host all of these
pictures for her? There was enough stuff out there to actually have a
computable cost that added to real money. (Disk have become less
expensive since then ...) The conclusion was that her actions were the
sort that was causing Amazon to want to curtail the forums (plus they
were paying monitors to suppress attacking posts). Her response was that
she had been sick or a child had been very sick (or some such) and this
was her way of releasing horrible moods. At least 20 people then posted
that she should go see a grief councilor and quit screwing up everybody
else's fun. She actually posted back that she thought that Amazon was
obligated to provide her this service; she was entitled!

I have no idea if you feel entitled but that isn't the issue. You have
real problems in your life and many of them are not of recent origin:
You've been at this for decades. Instead of using your time to improve
yourself and your serenity, you insist on surrounding yourself with
distractions. You might as well bury yourself in emotional mud. If you
are so attracted to theoretical computer science, why don't you take a
course in it offered online by a reputable university? Or actually ask
and receive help (from this forum?) as you try to struggle through a
textbook? Linz is about as straightforward as it gets because it tries
to remain as informal as possible without missing the gist of the
subject. If you try hard, you will acquire some beautiful images of
interesting mathematics. That will be a positive distraction for you and
be a good replacement for the crap you continue to invite.

Just a suggestion.
--
Jeff Barnett

Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

<pcqTI.8678$kMKe.1535@fx35.iad>

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https://www.novabbs.com/devel/article-flat.php?id=19960&group=comp.theory#19960

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Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input
to H?
Newsgroups: comp.theory
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Thu, 19 Aug 2021 10:16 UTC

On 8/18/21 10:58 PM, olcott wrote:
> On 8/18/2021 9:30 PM, Richard Damon wrote:
>> On 8/18/21 10:04 PM, olcott wrote:
>>> On 8/18/2021 8:42 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/18/2021 8:16 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 8/18/2021 6:51 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> On 8/18/2021 2:44 PM, Ben Bacarisse wrote:
>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>
>>>>>>>>>>> On 8/17/2021 7:02 PM, Ben Bacarisse wrote:
>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>>
>>>>>>>>>>>>> On 8/17/2021 5:14 PM, Ben Bacarisse wrote:
>>>>>>>>>>>>
>>>>>>>>>>>>>> The facts come from you.  They are not a matter of opinion.
>>>>>>>>>>>>>> P(P) halts.
>>>>>>>>>>>>>> H(P,P) == 0.
>>>>>>>>>>>> ...
>>>>>>>>>>>>>> You are happy with that answer, but I can specify a function
>>>>>>>>>>>>>> you can't
>>>>>>>>>>>>>> write: B(M, I) such that B(M, I) is non zero if and only if
>>>>>>>>>>>>>> M(I) halts.
>>>>>>>>>>>>>> You don't get to choose what the "correct" answer is, so
>>>>>>>>>>>>>> your only
>>>>>>>>>>>>>> option if to ignore the challenge.
>>>>>>>>>>>>>
>>>>>>>>>>>>> At this point I expect and require that those seeking an
>>>>>>>>>>>>> actual honest
>>>>>>>>>>>>> dialogue use the basis that I provided to verify that H does
>>>>>>>>>>>>> decide
>>>>>>>>>>>>> the halt status of its inputs correctly. Everyone else is
>>>>>>>>>>>>> written off
>>>>>>>>>>>>> as dishonest.
>>>>>>>>>>>> That's up to you.  I can't stop you wasting time on a function
>>>>>>>>>>>> no one
>>>>>>>>>>>> cares about, but you know you can't write the function I
>>>>>>>>>>>> specified, so
>>>>>>>>>>>> you are back where you started 17 years ago.  There are still
>>>>>>>>>>>> undecidable sets.
>>>>>>>>>>>
>>>>>>>>>>> The function does meet the spec
>>>>>>>>>> You don't know that because you have not asked for the proper
>>>>>>>>>> spec.  You
>>>>>>>>>> have not even accepted my challenge.  Are you accepting the
>>>>>>>>>> challenge
>>>>>>>>>> and proposing H as an implementation without knowing the full
>>>>>>>>>> spec?  I'm
>>>>>>>>>> happy for you to say yes if that is what you want to do.
>>>>>>>>>
>>>>>>>>> I know what the spec is:
>>>>>>>>
>>>>>>>> No you don't.  The spec comes from me, and only if you say you
>>>>>>>> want to
>>>>>>>> try the challenge.  I am promising you the specification of a
>>>>>>>> function
>>>>>>>> you can't write -- a clear demonstration that there are natural
>>>>>>>> uncomputable functions.
>>>>>>>
>>>>>>> I posted two different correct examples of the spec for the halting
>>>>>>> problem if you are talking about something else then this is out-of
>>>>>>> scope. You really should never erase the text the you are directly
>>>>>>> responding to that seems quite dishonest.
>>>>>>
>>>>>> Yes.  I cut them because they are not the specification of the
>>>>>> function
>>>>>> that I will challenge you to write.
>>>>>
>>>>> That is out-of-scope and you know it.
>>>>
>>>> Not in my opinion, no.  A function that that shows there are
>>>> undecidable
>>>> sets should worry you, but for some reason you prefer to stick with
>>>> talking about your H that does something entirely unsurprising and
>>>> uninteresting.
>>>>
>>>
>>> So when I correctly refute the halting problem proofs you say no I did
>>> not refute every proof in the universe and the halting problem proof is
>>> one of these proofs therefore I did not refute the halting problem
>>> proof.
>>>
>>> THIS IS THE SCOPE OF THE HALTING PROBLEM PROOFS:
>>>     Now we construct a new Turing machine D with H as a subroutine.
>>>     This new TM calls H to determine what M does when the input to
>>>     M is its own description ⟨M⟩. Once D has determined this
>>> information,
>>>     it does the opposite.  (Sipser:1997:165)
>>>
>>>
>>
>> Right, and in this example, your H says D will be non-halting, when it
>> turns out that D is a Halting Computation.
>>
>
> int main(){ P(P); } is not the same computation as
> int main(){ H(P,P); }
>
> The first computation has a dependency on the return value of H(P,P)
> that the second computation does not have thus the two computations are
> not equivalent.

Right, but the first computation is the Computation that the second one
is being asked about.

The DEFINITION of the question in the halting problem, which is what the
second computation is being asked to solve, is what is the behavior of
the first computation.

The Answer given by the second computation does not match the behaviorof
the first, therefore the second computation gave the wrong answer.

PERIOD.

>
>> Thus H was wrong.
>>
>> You have NOT refuted that proof.
>>
>
>

Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

<d16ba639-5c3b-483b-aa99-b62bafbd11a7n@googlegroups.com>

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https://www.novabbs.com/devel/article-flat.php?id=19961&group=comp.theory#19961

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Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input
to H?
From: malcolm....@gmail.com (Malcolm McLean)
Injection-Date: Thu, 19 Aug 2021 10:26:58 +0000
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 by: Malcolm McLean - Thu, 19 Aug 2021 10:26 UTC

On Wednesday, 18 August 2021 at 16:44:44 UTC+1, olcott wrote:
> On 8/18/2021 10:28 AM, Malcolm McLean wrote:
> > On Wednesday, 18 August 2021 at 14:57:10 UTC+1, olcott wrote:
> >>
> >> H has no effect on the machine that it simulates until after its halt
> >> status decision has been made. This conclusively proves that H can
> >> ignore its in execution trace during its halt status analysis.
> >>
> >> Anyone disagreeing with this is either not intelligent or knowledgeable
> >> enough to understand it, or a liar.
> >>
> >> That H does effect the behavior or its input at some other point is
> >> utterly irrelevant to this analysis. We are only answering the single
> >> question: Is it correct for H to ignore its own execution trace during
> >> its halt status analysis?
> >>
> > If H is analysing H, it can't ignore the behaviour of H. That's why your results
> > are wrong despite the execution trace seeming to show a non-halting
> > behaviour.
> >
> Because H only acts as a pure simulator of its input until after
> its halt status decision has been made it has no behavior that
> can possibly effect the behavior of its input. Because of this H
> screens out its own address range in every execution trace that
> it examines. This is why we never see any instructions of H in
> any execution trace after an input calls H.
>
> The above proves itself true entirely on the basis of the meaning
> of its words. There is no possible correct rebuttal there is only
> a failure to comprehend. If you believe that there is a correct
> rebuttal please provide it and I will point out your error.
>
You're wrong here. When H is being called on a program which includes
a call to H, the nested call to H needs to be analysed like any other
call. It can and in fact does affect the halting behaviour of the input.

Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

<87r1eppoe5.fsf@bsb.me.uk>

 copy mid

https://www.novabbs.com/devel/article-flat.php?id=19962&group=comp.theory#19962

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input to H?
Date: Thu, 19 Aug 2021 14:14:42 +0100
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 by: Ben Bacarisse - Thu, 19 Aug 2021 13:14 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/18/2021 8:42 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/18/2021 8:16 PM, Ben Bacarisse wrote:

>>>> Yes. I cut them because they are not the specification of the function
>>>> that I will challenge you to write.
>>>
>>> That is out-of-scope and you know it.
>>
>> Not in my opinion, no. A function that that shows there are undecidable
>> sets should worry you, but for some reason you prefer to stick with
>> talking about your H that does something entirely unsurprising and
>> uninteresting.
>
> So when I correctly refute the halting problem proofs you say no I did
> not refute every proof in the universe and the halting problem proof
> is one of these proofs therefore I did not refute the halting problem
> proof.

No, I'm not saying that.

Anyway, in case you are interested, here is the specification of the
function you can't write:

The computational model is C code with no memory restrictions. Of
course, if you don't actually hit the memory limits of a C
implementation it's just actual C code. I'd be happy to say more about
this model of computation if you think the details will matter to your
solution.

The problem is that of deciding if a function would return if called
from main. A "return decider" (in this model) is a C function returning
_Bool that always returns a value to the expression in which it was
called. A return decider always returns the same value for any
arguments that represnet the same function call expression.

Your mission, should you chose to accept it, is to write a return
decider B with this prototype

typedef uintptr_t data;
typedef void (*function)(data);

extern _Bool B(function, data);

such that B(f, d) returns true if and only if a call of f from main with
argument d returns to main. The two arguments, f and d, are said to
represenet the call expression f(d).

If, rather than just thinking you can do this, you have actual C code,
you should provide either source or a compiled translation unit that can
be linked with this one:

#include <stdint.h>
#include <stdio.h>

typedef uintptr_t data;
typedef void (*function)(data);

extern _Bool B(function, data);

void B_hat(data x) { if (B((function)x, x)) while (1); }

int main(void)
{
printf("%d\n", B(B_hat, (data)B_hat));
fflush(stdout);
B_hat((data)B_hat);
puts("returned");
}

The output should be either

1
returned

or

0

with no further output. Of course you could always just agree that no
such function B can be written.

--
Ben.

Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

<87lf4xpnqa.fsf@bsb.me.uk>

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https://www.novabbs.com/devel/article-flat.php?id=19963&group=comp.theory#19963

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input to H?
Date: Thu, 19 Aug 2021 14:29:01 +0100
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 by: Ben Bacarisse - Thu, 19 Aug 2021 13:29 UTC

olcott <NoOne@NoWhere.com> writes:

> On 8/18/2021 9:01 PM, Ben Bacarisse wrote:

>> We won't get anywhere if you keep ducking this question. You made a
>> false statement in the old thread about this and you won't say if you
>> stand by that false statement.
>>
>> You track record in understanding people is poor. But if you think I
>> agreed, so what? I don't agree now regardless of what you think I've
>> said in the past. I am allowed to re-state, as clearly as possible, the
>> facts of the matter and the string ⟨Ĥ⟩ ⟨Ĥ⟩ represents a halting
>> computation. All I am asking is that you be as clear as I am being. Do
>> you disagree that the string ⟨Ĥ⟩ ⟨Ĥ⟩ represents a halting computation?
>
> I already totally explained all that.

I was not asking for an explanation. I said so explicitly. I was
asking for an answer to a simple question. Your determined avoidance is
very suspicious.

> When-so-ever any input to a simulating halt decider would never halt
> unless its simulating halt decider stopped simulating it then this
> input is correctly decided as never halting.
>
> The input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts unless its simulating halt
> decider stops simulating it.
>
> Can you figure out whether or not this is a yes or no answer to your
> question?

No, I can't. As recently as 12th Aug you incorrectly said

"⟨Ĥ⟩ ⟨Ĥ⟩ is not a string that encodes a halting computation."

but you are suspiciously reluctant to say if you have changed your mind.
Anyway, you are wrong about the string ⟨Ĥ⟩ ⟨Ĥ⟩ though, ironically, I
know you are wrong because of other things you have stated about Ĥ with
even more frequency.

> I always answer yes or no questions with the reasoning that proves the
> yes or no answer.

I see only the waffle about what would happen if the string were not the
string I am asking about.

Since you won't say if you have changed your. I will stick with what
you said before. You are wrong about the string ⟨Ĥ⟩ ⟨Ĥ⟩. It represents
a halting computation. You can't get anywhere until you correct this
obvious error.

--
Ben.

Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

<ALadnUKjVb-x-oP8nZ2dnUU7-S3NnZ2d@giganews.com>

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https://www.novabbs.com/devel/article-flat.php?id=19964&group=comp.theory#19964

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Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input to H?
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
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From: NoO...@NoWhere.com (olcott)
Date: Thu, 19 Aug 2021 09:14:03 -0500
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 by: olcott - Thu, 19 Aug 2021 14:14 UTC

On 8/19/2021 5:26 AM, Malcolm McLean wrote:
> On Wednesday, 18 August 2021 at 16:44:44 UTC+1, olcott wrote:
>> On 8/18/2021 10:28 AM, Malcolm McLean wrote:
>>> On Wednesday, 18 August 2021 at 14:57:10 UTC+1, olcott wrote:
>>>>
>>>> H has no effect on the machine that it simulates until after its halt
>>>> status decision has been made. This conclusively proves that H can
>>>> ignore its in execution trace during its halt status analysis.
>>>>
>>>> Anyone disagreeing with this is either not intelligent or knowledgeable
>>>> enough to understand it, or a liar.
>>>>
>>>> That H does effect the behavior or its input at some other point is
>>>> utterly irrelevant to this analysis. We are only answering the single
>>>> question: Is it correct for H to ignore its own execution trace during
>>>> its halt status analysis?
>>>>
>>> If H is analysing H, it can't ignore the behaviour of H. That's why your results
>>> are wrong despite the execution trace seeming to show a non-halting
>>> behaviour.
>>>
>> Because H only acts as a pure simulator of its input until after
>> its halt status decision has been made it has no behavior that
>> can possibly effect the behavior of its input. Because of this H
>> screens out its own address range in every execution trace that
>> it examines. This is why we never see any instructions of H in
>> any execution trace after an input calls H.
>>
>> The above proves itself true entirely on the basis of the meaning
>> of its words. There is no possible correct rebuttal there is only
>> a failure to comprehend. If you believe that there is a correct
>> rebuttal please provide it and I will point out your error.
>>
> You're wrong here. When H is being called on a program which includes
> a call to H, the nested call to H needs to be analysed like any other
> call. It can and in fact does affect the halting behaviour of the input.

You are doing as bad of a job analyzing this as Robert. You are making
sure to simply ignore key words that I have said.

WHILE H IS MAKING ITS HALT DECISION H IS ACTING AS A PURE SIMULATOR OF
ITS INPUT.

WHILE H IS ACTING AS A PURE SIMULATOR H CANNOT POSSIBLY HAVE ANY EFFECT
ON THE BEHAVIOR OF ITS INPUT.

WHILE H CANNOT POSSIBLY HAVE ANY EFFECT ON THE BEHAVIOR OF ITS INPUT H
NEED NOT EXAMINE ITS OWN EXECUTION TRACE IN ITS HALT STATUS DECISION.

Try and find a specific flaw in that, there are one.

--
Copyright 2021 Pete Olcott

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

Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

<ed49e1ee-ea07-4227-9eaa-d0e22449cb3an@googlegroups.com>

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Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input
to H?
From: malcolm....@gmail.com (Malcolm McLean)
Injection-Date: Thu, 19 Aug 2021 14:39:13 +0000
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 by: Malcolm McLean - Thu, 19 Aug 2021 14:39 UTC

On Thursday, 19 August 2021 at 15:14:11 UTC+1, olcott wrote:
> On 8/19/2021 5:26 AM, Malcolm McLean wrote:
> > On Wednesday, 18 August 2021 at 16:44:44 UTC+1, olcott wrote:
> >> On 8/18/2021 10:28 AM, Malcolm McLean wrote:
> >>> On Wednesday, 18 August 2021 at 14:57:10 UTC+1, olcott wrote:
> >>>>
> >>>> H has no effect on the machine that it simulates until after its halt
> >>>> status decision has been made. This conclusively proves that H can
> >>>> ignore its in execution trace during its halt status analysis.
> >>>>
> >>>> Anyone disagreeing with this is either not intelligent or knowledgeable
> >>>> enough to understand it, or a liar.
> >>>>
> >>>> That H does effect the behavior or its input at some other point is
> >>>> utterly irrelevant to this analysis. We are only answering the single
> >>>> question: Is it correct for H to ignore its own execution trace during
> >>>> its halt status analysis?
> >>>>
> >>> If H is analysing H, it can't ignore the behaviour of H. That's why your results
> >>> are wrong despite the execution trace seeming to show a non-halting
> >>> behaviour.
> >>>
> >> Because H only acts as a pure simulator of its input until after
> >> its halt status decision has been made it has no behavior that
> >> can possibly effect the behavior of its input. Because of this H
> >> screens out its own address range in every execution trace that
> >> it examines. This is why we never see any instructions of H in
> >> any execution trace after an input calls H.
> >>
> >> The above proves itself true entirely on the basis of the meaning
> >> of its words. There is no possible correct rebuttal there is only
> >> a failure to comprehend. If you believe that there is a correct
> >> rebuttal please provide it and I will point out your error.
> >>
> > You're wrong here. When H is being called on a program which includes
> > a call to H, the nested call to H needs to be analysed like any other
> > call. It can and in fact does affect the halting behaviour of the input.
> You are doing as bad of a job analyzing this as Robert. You are making
> sure to simply ignore key words that I have said.
>
> WHILE H IS MAKING ITS HALT DECISION H IS ACTING AS A PURE SIMULATOR OF
> ITS INPUT.
>
> WHILE H IS ACTING AS A PURE SIMULATOR H CANNOT POSSIBLY HAVE ANY EFFECT
> ON THE BEHAVIOR OF ITS INPUT.
>
> WHILE H CANNOT POSSIBLY HAVE ANY EFFECT ON THE BEHAVIOR OF ITS INPUT H
> NEED NOT EXAMINE ITS OWN EXECUTION TRACE IN ITS HALT STATUS DECISION.
>
> Try and find a specific flaw in that, there are one.
>
Let's forget LInz and take this simple function

void loopforever(U32 dummy)
{ while(1);
}

Now loopforever(dummy) doesn't halt, agreed?
this function

int H1(U32 dummy)
{ return H(loopforever, dummy);
}

halts and returns 0 (non-halting). Agreed?

Now consider this function

int H2(u32 dummy)
{ return H(H1, dummy);
}

What does that do?

Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

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https://www.novabbs.com/devel/article-flat.php?id=19966&group=comp.theory#19966

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Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input
to H?
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References: <3YOdnecvDsA5Q4r8nZ2dnUU7-TXNnZ2d@giganews.com>
<87im06wiup.fsf@bsb.me.uk> <DJGdncQOXNbfHYT8nZ2dnUU7-dPNnZ2d@giganews.com>
<874kbqw62q.fsf@bsb.me.uk> <W7udnRlZduvgdof8nZ2dnUU7-IPNnZ2d@giganews.com>
<87h7fpuf5v.fsf@bsb.me.uk> <AsSdnUXVrYJ5nYb8nZ2dnUU7-VnNnZ2d@giganews.com>
<875yw4v08g.fsf@bsb.me.uk> <oKidneawW_dWu4H8nZ2dnUU7-emdnZ2d@giganews.com>
<8735r7u3ab.fsf@bsb.me.uk> <ufKdnZfZ0sUP3YH8nZ2dnUU7-SXNnZ2d@giganews.com>
<87wnojsjqd.fsf@bsb.me.uk> <ReKdnb2pB4SVyoH8nZ2dnUU7-SvNnZ2d@giganews.com>
<87o89usfll.fsf@bsb.me.uk> <uqadnd39oqwW-ID8nZ2dnUU7-amdnZ2d@giganews.com>
<87sfz6qpk9.fsf@bsb.me.uk> <PeqdnehLhMapOYD8nZ2dnUU7-YXNnZ2d@giganews.com>
<87k0kiqlmm.fsf@bsb.me.uk> <GPidnScbL8UfLoD8nZ2dnUU7-ROdnZ2d@giganews.com>
<875yw2qkfb.fsf@bsb.me.uk> <2q6dnZ-dGIzeIYD8nZ2dnUU7-ffNnZ2d@giganews.com>
<87r1eppoe5.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Thu, 19 Aug 2021 10:14:09 -0500
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 by: olcott - Thu, 19 Aug 2021 15:14 UTC

On 8/19/2021 8:14 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/18/2021 8:42 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/18/2021 8:16 PM, Ben Bacarisse wrote:
>
>>>>> Yes. I cut them because they are not the specification of the function
>>>>> that I will challenge you to write.
>>>>
>>>> That is out-of-scope and you know it.
>>>
>>> Not in my opinion, no. A function that that shows there are undecidable
>>> sets should worry you, but for some reason you prefer to stick with
>>> talking about your H that does something entirely unsurprising and
>>> uninteresting.
>>
>> So when I correctly refute the halting problem proofs you say no I did
>> not refute every proof in the universe and the halting problem proof
>> is one of these proofs therefore I did not refute the halting problem
>> proof.
>
> No, I'm not saying that.
>
> Anyway, in case you are interested, here is the specification of the
> function you can't write:
>
> The computational model is C code with no memory restrictions. Of
> course, if you don't actually hit the memory limits of a C
> implementation it's just actual C code. I'd be happy to say more about
> this model of computation if you think the details will matter to your
> solution.
>
> The problem is that of deciding if a function would return if called
> from main. A "return decider" (in this model) is a C function returning
> _Bool that always returns a value to the expression in which it was
> called. A return decider always returns the same value for any
> arguments that represnet the same function call expression.
>
> Your mission, should you chose to accept it, is to write a return
> decider B with this prototype
>
> typedef uintptr_t data;
> typedef void (*function)(data);
>
> extern _Bool B(function, data);
>
> such that B(f, d) returns true if and only if a call of f from main with
> argument d returns to main. The two arguments, f and d, are said to
> represenet the call expression f(d).
>
> If, rather than just thinking you can do this, you have actual C code,
> you should provide either source or a compiled translation unit that can
> be linked with this one:
>
> #include <stdint.h>
> #include <stdio.h>
>
> typedef uintptr_t data;
> typedef void (*function)(data);
>
> extern _Bool B(function, data);
>
> void B_hat(data x) { if (B((function)x, x)) while (1); }
>
> int main(void)
> {
> printf("%d\n", B(B_hat, (data)B_hat));
> fflush(stdout);
> B_hat((data)B_hat);
> puts("returned");
> }
>
> The output should be either
>
> 1
> returned
>
> or
>
> 0
>
> with no further output. Of course you could always just agree that no
> such function B can be written.
>

The x86utm operating system cannot call any C functions.
Good job on the use of C. My own use of unsigned integers as function
pointers is unconventional and not as portable. I did this on purpose so
that x86utm would have a single standard tape element type.

#include <stdint.h>
#include <stdio.h>

typedef uintptr_t data;
typedef void (*function)(data);

extern _Bool B(function, data);

void B_hat(data x) { if (H((u32)x, (u32)x)) while (1); }

int main2()
{ OutputHex(H((u32)B_hat, (u32)B_hat));
B_hat((u32)B_hat);
OutputString("returned");
}

_B_hat()
[00000efc](01) 55 push ebp
[00000efd](02) 8bec mov ebp,esp
[00000eff](03) 8b4508 mov eax,[ebp+08]
[00000f02](01) 50 push eax
[00000f03](03) 8b4d08 mov ecx,[ebp+08]
[00000f06](01) 51 push ecx
[00000f07](05) e850feffff call 00000d5c
[00000f0c](03) 83c408 add esp,+08
[00000f0f](02) 85c0 test eax,eax
[00000f11](02) 740b jz 00000f1e
[00000f13](05) ba01000000 mov edx,00000001
[00000f18](02) 85d2 test edx,edx
[00000f1a](02) 7402 jz 00000f1e
[00000f1c](02) ebf5 jmp 00000f13
[00000f1e](01) 5d pop ebp
[00000f1f](01) c3 ret
Size in bytes:(0036) [00000f1f]

_main2()
[00000f2c](01) 55 push ebp
[00000f2d](02) 8bec mov ebp,esp
[00000f2f](05) 68fc0e0000 push 00000efc
[00000f34](05) 68fc0e0000 push 00000efc
[00000f39](05) e81efeffff call 00000d5c
[00000f3e](03) 83c408 add esp,+08
[00000f41](01) 50 push eax
[00000f42](05) e8e5f3ffff call 0000032c
[00000f47](03) 83c404 add esp,+04
[00000f4a](05) 68fc0e0000 push 00000efc
[00000f4f](05) e8a8ffffff call 00000efc
[00000f54](03) 83c404 add esp,+04
[00000f57](05) 6823030000 push 00000323
[00000f5c](05) e8dbf3ffff call 0000033c
[00000f61](03) 83c404 add esp,+04
[00000f64](01) 5d pop ebp
[00000f65](01) c3 ret
Size in bytes:(0058) [00000f65]

_main()
[00000f6c](01) 55 push ebp
[00000f6d](02) 8bec mov ebp,esp
[00000f6f](05) e8b8ffffff call 00000f2c
[00000f74](02) 33c0 xor eax,eax
[00000f76](01) 5d pop ebp
[00000f77](01) c3 ret
Size in bytes:(0012) [00000f77]

machine stack stack machine assembly
address address data code language
======== ======== ======== ========= =============
....[00000f6c][00101c0a][00000000] 55 push ebp
....[00000f6d][00101c0a][00000000] 8bec mov ebp,esp
....[00000f6f][00101c06][00000f74] e8b8ffffff call 00000f2c
....[00000f2c][00101c02][00101c0a] 55 push ebp
....[00000f2d][00101c02][00101c0a] 8bec mov ebp,esp
....[00000f2f][00101bfe][00000efc] 68fc0e0000 push 00000efc
....[00000f34][00101bfa][00000efc] 68fc0e0000 push 00000efc
....[00000f39][00101bf6][00000f3e] e81efeffff call 00000d5c

Begin Local Halt Decider Simulation at Machine Address:efc
....[00000efc][00211caa][00211cae] 55 push ebp
....[00000efd][00211caa][00211cae] 8bec mov ebp,esp
....[00000eff][00211caa][00211cae] 8b4508 mov eax,[ebp+08]
....[00000f02][00211ca6][00000efc] 50 push eax
....[00000f03][00211ca6][00000efc] 8b4d08 mov ecx,[ebp+08]
....[00000f06][00211ca2][00000efc] 51 push ecx
....[00000f07][00211c9e][00000f0c] e850feffff call 00000d5c
....[00000efc][0025c6d2][0025c6d6] 55 push ebp
....[00000efd][0025c6d2][0025c6d6] 8bec mov ebp,esp
....[00000eff][0025c6d2][0025c6d6] 8b4508 mov eax,[ebp+08]
....[00000f02][0025c6ce][00000efc] 50 push eax
....[00000f03][0025c6ce][00000efc] 8b4d08 mov ecx,[ebp+08]
....[00000f06][0025c6ca][00000efc] 51 push ecx
....[00000f07][0025c6c6][00000f0c] e850feffff call 00000d5c
Local Halt Decider: Infinite Recursion Detected Simulation Stopped
....[00000f3e][00101c02][00101c0a] 83c408 add esp,+08
....[00000f41][00101bfe][00000000] 50 push eax
---[00000f42][00101bfe][00000000] e8e5f3ffff call 0000032c
0 ....[00000f47][00101c02][00101c0a] 83c404 add esp,+04
....[00000f4a][00101bfe][00000efc] 68fc0e0000 push 00000efc
....[00000f4f][00101bfa][00000f54] e8a8ffffff call 00000efc
....[00000efc][00101bf6][00101c02] 55 push ebp
....[00000efd][00101bf6][00101c02] 8bec mov ebp,esp
....[00000eff][00101bf6][00101c02] 8b4508 mov eax,[ebp+08]
....[00000f02][00101bf2][00000efc] 50 push eax
....[00000f03][00101bf2][00000efc] 8b4d08 mov ecx,[ebp+08]
....[00000f06][00101bee][00000efc] 51 push ecx
....[00000f07][00101bea][00000f0c] e850feffff call 00000d5c

Begin Local Halt Decider Simulation at Machine Address:efc
....[00000efc][0026c772][0026c776] 55 push ebp
....[00000efd][0026c772][0026c776] 8bec mov ebp,esp
....[00000eff][0026c772][0026c776] 8b4508 mov eax,[ebp+08]
....[00000f02][0026c76e][00000efc] 50 push eax
....[00000f03][0026c76e][00000efc] 8b4d08 mov ecx,[ebp+08]
....[00000f06][0026c76a][00000efc] 51 push ecx
....[00000f07][0026c766][00000f0c] e850feffff call 00000d5c
....[00000efc][002b719a][002b719e] 55 push ebp
....[00000efd][002b719a][002b719e] 8bec mov ebp,esp
....[00000eff][002b719a][002b719e] 8b4508 mov eax,[ebp+08]
....[00000f02][002b7196][00000efc] 50 push eax
....[00000f03][002b7196][00000efc] 8b4d08 mov ecx,[ebp+08]
....[00000f06][002b7192][00000efc] 51 push ecx
....[00000f07][002b718e][00000f0c] e850feffff call 00000d5c
Local Halt Decider: Infinite Recursion Detected Simulation Stopped
....[00000f0c][00101bf6][00101c02] 83c408 add esp,+08
....[00000f0f][00101bf6][00101c02] 85c0 test eax,eax
....[00000f11][00101bf6][00101c02] 740b jz 00000f1e
....[00000f1e][00101bfa][00000f54] 5d pop ebp
....[00000f1f][00101bfe][00000efc] c3 ret
....[00000f54][00101c02][00101c0a] 83c404 add esp,+04
....[00000f57][00101bfe][00000323] 6823030000 push 00000323
---[00000f5c][00101bfe][00000323] e8dbf3ffff call 0000033c
returned
....[00000f61][00101c02][00101c0a] 83c404 add esp,+04
....[00000f64][00101c06][00000f74] 5d pop ebp
....[00000f65][00101c0a][00000000] c3 ret
....[00000f74][00101c0a][00000000] 33c0 xor eax,eax
....[00000f76][00101c0e][00100000] 5d pop ebp
....[00000f77][00101c12][00000008] c3 ret
Number_of_User_Instructions(2)
Number of Instructions Executed(47451)


Click here to read the complete article
Re: How do we know H(P,P)==0 is the correct halt status for the input to H?

<CbSdna_sjJJB6IP8nZ2dnUU7-KednZ2d@giganews.com>

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https://www.novabbs.com/devel/article-flat.php?id=19967&group=comp.theory#19967

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Subject: Re: How do we know H(P,P)==0 is the correct halt status for the input to H?
Newsgroups: comp.theory
References: <3YOdnecvDsA5Q4r8nZ2dnUU7-TXNnZ2d@giganews.com> <87im06wiup.fsf@bsb.me.uk> <DJGdncQOXNbfHYT8nZ2dnUU7-dPNnZ2d@giganews.com> <874kbqw62q.fsf@bsb.me.uk> <W7udnRlZduvgdof8nZ2dnUU7-IPNnZ2d@giganews.com> <87h7fpuf5v.fsf@bsb.me.uk> <AsSdnUXVrYJ5nYb8nZ2dnUU7-VnNnZ2d@giganews.com> <875yw4v08g.fsf@bsb.me.uk> <oKidneawW_dWu4H8nZ2dnUU7-emdnZ2d@giganews.com> <8735r7u3ab.fsf@bsb.me.uk> <ufKdnZfZ0sUP3YH8nZ2dnUU7-SXNnZ2d@giganews.com> <87wnojsjqd.fsf@bsb.me.uk> <ReKdnb2pB4SVyoH8nZ2dnUU7-SvNnZ2d@giganews.com> <87im02sepy.fsf@bsb.me.uk> <U7KdnRnxgLHX7YD8nZ2dnUU7-QvNnZ2d@giganews.com> <87v942qpkj.fsf@bsb.me.uk> <W4CdnecJQbhMP4D8nZ2dnUU7-WXNnZ2d@giganews.com> <87h7fmqllh.fsf@bsb.me.uk> <lKidnapLuffcKID8nZ2dnUU7-R-dnZ2d@giganews.com> <87zgtep4zl.fsf@bsb.me.uk> <E8adnbBrnoijI4D8nZ2dnUU78RXNnZ2d@giganews.com> <87lf4xpnqa.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Thu, 19 Aug 2021 10:16:43 -0500
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 by: olcott - Thu, 19 Aug 2021 15:16 UTC

On 8/19/2021 8:29 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/18/2021 9:01 PM, Ben Bacarisse wrote:
>
>>> We won't get anywhere if you keep ducking this question. You made a
>>> false statement in the old thread about this and you won't say if you
>>> stand by that false statement.
>>>
>>> You track record in understanding people is poor. But if you think I
>>> agreed, so what? I don't agree now regardless of what you think I've
>>> said in the past. I am allowed to re-state, as clearly as possible, the
>>> facts of the matter and the string ⟨Ĥ⟩ ⟨Ĥ⟩ represents a halting
>>> computation. All I am asking is that you be as clear as I am being. Do
>>> you disagree that the string ⟨Ĥ⟩ ⟨Ĥ⟩ represents a halting computation?
>>
>> I already totally explained all that.
>
> I was not asking for an explanation. I said so explicitly. I was
> asking for an answer to a simple question. Your determined avoidance is
> very suspicious.
>
>> When-so-ever any input to a simulating halt decider would never halt
>> unless its simulating halt decider stopped simulating it then this
>> input is correctly decided as never halting.
>>
>> The input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts unless its simulating halt
>> decider stops simulating it.
>>
>> Can you figure out whether or not this is a yes or no answer to your
>> question?
>
> No, I can't. As recently as 12th Aug you incorrectly said
>
> "⟨Ĥ⟩ ⟨Ĥ⟩ is not a string that encodes a halting computation."
>

The input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts unless its simulating halt decider
stops simulating it, thus conclusively proving that the input to Ĥ.qx
⟨Ĥ⟩ ⟨Ĥ⟩ never halts.

> but you are suspiciously reluctant to say if you have changed your mind.
> Anyway, you are wrong about the string ⟨Ĥ⟩ ⟨Ĥ⟩ though, ironically, I
> know you are wrong because of other things you have stated about Ĥ with
> even more frequency.
>
>> I always answer yes or no questions with the reasoning that proves the
>> yes or no answer.
>
> I see only the waffle about what would happen if the string were not the
> string I am asking about.
>
> Since you won't say if you have changed your. I will stick with what
> you said before. You are wrong about the string ⟨Ĥ⟩ ⟨Ĥ⟩. It represents
> a halting computation. You can't get anywhere until you correct this
> obvious error.
>

--
Copyright 2021 Pete Olcott

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

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