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computers / comp.ai.philosophy / Re: Concise refutation of halting problem proofs V52 [ ignorance or deception? ]

SubjectAuthor
* Concise refutation of halting problem proofs V52 [ Linz Proof ]olcott
+* Re: Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|`* Re: Concise refutation of halting problem proofs V52 [ Ignorant orolcott
| `* Re: Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|  `* Re: Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|   `- Re: Concise refutation of halting problem proofs V52 [ Ignorant orolcott
`* Re: Concise refutation of halting problem proofs V52 [ Honestolcott
 `* Re: Concise refutation of halting problem proofs V52 [ Honestolcott
  `* Re: Concise refutation of halting problem proofs V52 [ Honestolcott
   `* Re: Concise refutation of halting problem proofs V52 [ Honestolcott
    `* Re: Concise refutation of halting problem proofs V52 [ error orolcott
     `* Re: Concise refutation of halting problem proofs V52 [ error orolcott
      `* Re: Concise refutation of halting problem proofs V52 [ error orolcott
       `* Re: Concise refutation of halting problem proofs V52 [ error orolcott
        +- Re: Concise refutation of halting problem proofs V52 [ error orolcott
        +- Re: Concise refutation of halting problem proofs V52 [ THE KEYolcott
        `* Re: Concise refutation of halting problem proofs V52 [ error orolcott
         `* Re: Concise refutation of halting problem proofs V52 [ error orolcott
          `* Re: Concise refutation of halting problem proofs V52 [ error orolcott
           `* Re: Concise refutation of halting problem proofs V52 [ error orolcott
            +* Re: Concise refutation of halting problem proofs V52 [ error orolcott
            |`* Re: Concise refutation of halting problem proofs V52 [ dishonesty ! ]olcott
            | `* Re: Concise refutation of halting problem proofs V52 [ ignorance? ]olcott
            |  `* Re: Concise refutation of halting problem proofs V52 [ ignorance? ]olcott
            |   `* Re: Concise refutation of halting problem proofs V59 [ key essence ]olcott
            |    +- Re: Concise refutation of halting problem proofs V59 [ key essence ]olcott
            |    +* Re: Concise refutation of halting problem proofs V59 [ key essence ]olcott
            |    |+* Re: Concise refutation of halting problem proofs V59 [ key essence ]olcott
            |    ||+- Re: Concise refutation of halting problem proofs V59 [ ignoranceolcott
            |    ||+- Re: Concise refutation of halting problem proofs V59 [ key essence ]olcott
            |    ||`* Re: Concise refutation of halting problem proofs V59 [ ignoranceolcott
            |    || `* Re: Concise refutation of halting problem proofs V59 [ self-evidentolcott
            |    ||  `* Re: Concise refutation of halting problem proofs V59 [ self-evidentolcott
            |    ||   `* Re: Concise refutation of halting problem proofs V59 [ self-evidentolcott
            |    ||    `- Re: Concise refutation of halting problem proofs V59 [ self-evidentolcott
            |    |`- Re: Concise refutation of halting problem proofs V59 [ key essence ]olcott
            |    `- Re: Concise refutation of halting problem proofs V59 [ key essence ]olcott
            +- Re: Concise refutation of halting problem proofs V52 [ error orolcott
            `* Re: Concise refutation of halting problem proofs V52 [ ignoranceolcott
             `* Re: Concise refutation of halting problem proofs V52 [ dodgy doubleolcott
              `* Re: Concise refutation of halting problem proofs V52 [ dodgy doubleolcott
               +* Re: Concise refutation of halting problem proofs V52 [ dodgy doubleolcott
               |`* Re: Concise refutation of halting problem proofs V52 [ dodgy doubleolcott
               | `* Re: Concise refutation of halting problem proofs V52 [ dodgy doubleolcott
               |  `* Re: Concise refutation of halting problem proofs V52 [ dodgy doubleolcott
               |   `* Re: Concise refutation of halting problem proofs V52 [ dodgy doubleolcott
               |    `* Re: Concise refutation of halting problem proofs V52 [ dodgy doubleolcott
               |     +- Re: Concise refutation of halting problem proofs V52 [ pants on fireolcott
               |     +- Re: Concise refutation of halting problem proofs V52 [ pants on fireolcott
               |     `* Re: Concise refutation of halting problem proofs V52 [ pants on fireolcott
               |      `* Re: Concise refutation of halting problem proofs V52 [ pants on fireolcott
               |       `* Re: Concise refutation of halting problem proofs V52 [ pants on fireolcott
               |        `* Re: Concise refutation of halting problem proofs V52 [ ignorance orolcott
               |         `- Re: Concise refutation of halting problem proofs V52 [ ignorance orolcott
               +* Re: Concise refutation of halting problem proofs V52 [ dodgy doubleolcott
               |`- Re: Concise refutation of halting problem proofs V52 [ dodgy doubleolcott
               `- Re: Concise refutation of halting problem proofs V52 [ dodgy doubleolcott

Pages:123
Subject: Re: Concise refutation of halting problem proofs V52 [ pants on fire ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Fri, 4 Feb 2022 00:14 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
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Subject: Re: Concise refutation of halting problem proofs V52 [ pants on fire
]
Content-Language: en-US
Newsgroups: comp.theory,comp.ai.philosophy,sci.logic,sci.math
References: <ssh8vu$4c0$1@dont-email.me> <st7a2e$oo$1@dont-email.me>
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Followup-To: comp.theory
From: NoO...@NoWhere.com (olcott)
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On 2/3/2022 5:50 AM, Richard Damon wrote:
On 2/3/22 12:24 AM, olcott wrote:
On 2/2/2022 11:09 PM, Richard Damon wrote:
On 2/2/22 11:50 PM, olcott wrote:
On 2/2/2022 10:42 PM, Richard Damon wrote:
On 2/2/22 10:50 PM, olcott wrote:
On 2/2/2022 9:30 PM, Richard Damon wrote:

On 2/2/22 10:12 PM, olcott wrote:
On 2/2/2022 8:57 PM, Richard Damon wrote:
On 2/2/22 9:31 PM, olcott wrote:
On 2/2/2022 8:08 PM, Richard Damon wrote:
On 2/2/22 8:41 PM, olcott wrote:
On 2/2/2022 6:22 PM, Richard Damon wrote:
On 2/2/22 10:20 AM, olcott wrote:
On 2/1/2022 9:58 PM, André G. Isaak wrote:
On 2022-02-01 20:44, olcott wrote:
On 2/1/2022 9:24 PM, André G. Isaak wrote:
On 2022-02-01 19:57, olcott wrote:
On 2/1/2022 8:48 PM, André G. Isaak wrote:
On 2022-02-01 19:37, olcott wrote:
On 2/1/2022 8:08 PM, André G. Isaak wrote:

⟨Ĥ⟩ applied to ⟨Ĥ⟩ is completely meaningless.

Sure and so is the "I am going to go to the" part of
"I am going to go to the store to buy some ice cream."

When you don't cut off what I said in the middle of the sentence then it makes much more sense.

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ?

That's just as meaningless. You can simulate Ĥ applied to ⟨Ĥ⟩ or you can provide ⟨Ĥ⟩ ⟨Ĥ⟩ as the input to a simulator. You cannot simulate ⟨Ĥ⟩ applied to ⟨Ĥ⟩ anymore than you can apply ⟨Ĥ⟩ to ⟨Ĥ⟩.

So you are simply being nit picky about my use of terminology.

Yes, I insist on terminology being used correctly. And any place where you attempt to publish your results will be equally, if not more, nit picky.


It is fine and good that you help correct my terminology.
What is not fine and good is for you to reject the essence of the gist of what I am saying entirely on the basis that I did not say it exactly according to conventions. The is what Ben always did. He never paid any attention to the actual substance of what I was saying.

When ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to simulating halt decider embedded_H and embedded_H correctly determines that its simulated input cannot possibly reach any final state then embedded_H is necessarily correct to transition to Ĥ.qn indicating that its simulated input never halts.

But now you've just hidden your meaningless terminological abuse. "Its simulated input" is only meaningful when it is construed as meaning the simulation of the computation REPRESENTED by the input, i.e. the

Not at all. A simulator simulates a finite string and the actual behavior of this simulated finite string is the ultimate basis of whether or not it specifies a finite sequence of configurations.

No. A simulator simulates a Turing Machine applied to an input. It takes as its input a finite string which represents that Turing Machine/Input pair. It's completely meaningless to talk about simulating a finite string.


It is possible for Turing machines to have blank tapes.

The salient aspect for the Halting problem is whether or not the finite string machine description specifies a finite or infinite sequence of configurations. The ultimate basis for determining this is the actual behavior of the simulated finite string.

Since this equally applies to machines having inputs and machines not having inputs the distinction relative to inputs is moot.

If the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H cannot possibly reach ⟨Ĥ⟩.qn then it is necessarily correct for embedded_H to transition to Ĥ.qn and nothing else in the universe can possibly refute this.

Again, you're falling back on your belief that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ is both meaningful (it isn't) and somehow distinct from H applied to ⟨Ĥ⟩.


The behavior of the simulated input when embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ is the ultimate measure of the halt status of this input.

Which just proves you are not working on the Halting Problem,
No it only proves that you and André don't understand that a halt decider computes the mapping from the inputs to an accept or reject state (here is the part that you two don't understand):

On the basis of the actual behavior specified by the actual input.
On the basis of the actual behavior specified by the actual input.
On the basis of the actual behavior specified by the actual input.


Which is DEFINED by what a the machine the input represents would do,

These words prove themselves true on the basis of their meaning:
The actual behavior of the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H is the ultimate measure of the behavior specified by ⟨Ĥ⟩ ⟨Ĥ⟩.


WRONG, which shows you do not actually know the meaning of the words.
When you disagree that the correct simulation of a machine description of a machine is the ultimate measure of the behavior specified by this machine description it is just like saying that a black cat is not a cat.


The problem is that 'Correct Simulation of a machine description' has an actual meaning, in that the simulation must match the actual behavior of the machine whose description it is simulating, RIGHT?
It must only do exactly what it actually does, if this does not meet expectations then expectations must be incorrect.

Here is what it actually does:
These steps would keep repeating:
   Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
   Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩
   Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates ⟨Ĥ4⟩ ⟨Ĥ5⟩...

And if that is what it actually does, then H NEVER aborts its simulation and thus never give an answer.


When embedded_H correctly matches the above infinite sequence this conclusively proves that its correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly reach ⟨Ĥ⟩.qn. (We don't even need to mention any UTM).



Excepts as previously said, that pattern only exists if H never aborts.

That is not true. The pattern exists for at least any finite number of steps where it can be recognized. The three iterations shown above are plenty enough for it to be recogized.


But if it only exists for a finite number of steps (till it is recognized)

At that point right there embedded_H has complete proof that the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly reach ⟨Ĥ⟩.qn.


--
Copyright 2021 Pete Olcott

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


Subject: Re: Concise refutation of halting problem proofs V52 [ pants on fire ^ 2 ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Organization: A noiseless patient Spider
Date: Fri, 4 Feb 2022 03:10 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
Path: i2pn2.org!i2pn.org!eternal-september.org!reader02.eternal-september.org!.POSTED!not-for-mail
From: polco...@gmail.com (olcott)
Newsgroups: comp.theory,comp.ai.philosophy,sci.logic,sci.math
Subject: Re: Concise refutation of halting problem proofs V52 [ pants on fire
^ 2 ]
Followup-To: comp.theory
Date: Thu, 3 Feb 2022 21:10:38 -0600
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On 2/3/2022 5:50 AM, Richard Damon wrote:
On 2/3/22 12:24 AM, olcott wrote:
On 2/2/2022 11:09 PM, Richard Damon wrote:
On 2/2/22 11:50 PM, olcott wrote:
On 2/2/2022 10:42 PM, Richard Damon wrote:
On 2/2/22 10:50 PM, olcott wrote:


Here is what it actually does:
These steps would keep repeating:
   Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
   Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩
   Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates ⟨Ĥ4⟩ ⟨Ĥ5⟩...

And if that is what it actually does, then H NEVER aborts its simulation and thus never give an answer.


When embedded_H correctly matches the above infinite sequence this conclusively proves that its correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly reach ⟨Ĥ⟩.qn. (We don't even need to mention any UTM).


Excepts as previously said, that pattern only exists if H never aborts.

That is not true. The pattern exists for at least any finite number of steps where it can be recognized. The three iterations shown above are plenty enough for it to be recogized.


But if it only exists for a finite number of steps (till it is recognized)

We are discussing the point in the execution of embedded_H where it has just correctly matched an infinite behavior pattern while it was doing its correct simulation of the first N steps of ⟨Ĥ⟩ ⟨Ĥ⟩

At this point right here embedded_H has complete proof that the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly reach ⟨Ĥ⟩.qn.


--
Copyright 2021 Pete Olcott "Talent hits a target no one else can hit;
Genius hits a target no one else can see." Arthur Schopenhauer


Subject: Re: Concise refutation of halting problem proofs V52 [ pants on fire ^ 2 ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Fri, 4 Feb 2022 03:56 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
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Subject: Re: Concise refutation of halting problem proofs V52 [ pants on fire
^ 2 ]
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Newsgroups: comp.theory,comp.ai.philosophy,sci.logic,sci.math
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Followup-To: comp.theory
From: NoO...@NoWhere.com (olcott)
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On 2/3/2022 9:40 PM, Richard Damon wrote:
On 2/3/22 10:10 PM, olcott wrote:
On 2/3/2022 5:50 AM, Richard Damon wrote:
On 2/3/22 12:24 AM, olcott wrote:
On 2/2/2022 11:09 PM, Richard Damon wrote:
On 2/2/22 11:50 PM, olcott wrote:
On 2/2/2022 10:42 PM, Richard Damon wrote:
On 2/2/22 10:50 PM, olcott wrote:


Here is what it actually does:
These steps would keep repeating:
   Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
   Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩
   Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates ⟨Ĥ4⟩ ⟨Ĥ5⟩...

And if that is what it actually does, then H NEVER aborts its simulation and thus never give an answer.


When embedded_H correctly matches the above infinite sequence this conclusively proves that its correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly reach ⟨Ĥ⟩.qn. (We don't even need to mention any UTM).


Excepts as previously said, that pattern only exists if H never aborts.

That is not true. The pattern exists for at least any finite number of steps where it can be recognized. The three iterations shown above are plenty enough for it to be recogized.


But if it only exists for a finite number of steps (till it is recognized)

We are discussing the point in the execution of embedded_H where it has just correctly matched an infinite behavior pattern while it was doing its correct simulation of the first N steps of ⟨Ĥ⟩ ⟨Ĥ⟩


No, you are CLAIMING (incorrectly) that it has made that determination.

THIS IS WHAT YOU HAVE BEEN DISAGREEING WITH:
As soon as an infinite behavior pattern is correctly recognized in a finite number of steps then it is definitely correct for embedded_H to transition to Ĥ.qn.


--
Copyright 2021 Pete Olcott

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


Subject: Re: Concise refutation of halting problem proofs V52 [ pants on fire ^ 2 ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Fri, 4 Feb 2022 04:45 UTC
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Subject: Re: Concise refutation of halting problem proofs V52 [ pants on fire
^ 2 ]
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On 2/3/2022 10:20 PM, Richard Damon wrote:
On 2/3/22 10:56 PM, olcott wrote:
On 2/3/2022 9:40 PM, Richard Damon wrote:
On 2/3/22 10:10 PM, olcott wrote:
On 2/3/2022 5:50 AM, Richard Damon wrote:
On 2/3/22 12:24 AM, olcott wrote:
On 2/2/2022 11:09 PM, Richard Damon wrote:
On 2/2/22 11:50 PM, olcott wrote:
On 2/2/2022 10:42 PM, Richard Damon wrote:
On 2/2/22 10:50 PM, olcott wrote:


Here is what it actually does:
These steps would keep repeating:
   Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
   Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩
   Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates ⟨Ĥ4⟩ ⟨Ĥ5⟩...

And if that is what it actually does, then H NEVER aborts its simulation and thus never give an answer.


When embedded_H correctly matches the above infinite sequence this conclusively proves that its correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly reach ⟨Ĥ⟩.qn. (We don't even need to mention any UTM).


Excepts as previously said, that pattern only exists if H never aborts.

That is not true. The pattern exists for at least any finite number of steps where it can be recognized. The three iterations shown above are plenty enough for it to be recogized.


But if it only exists for a finite number of steps (till it is recognized)

We are discussing the point in the execution of embedded_H where it has just correctly matched an infinite behavior pattern while it was doing its correct simulation of the first N steps of ⟨Ĥ⟩ ⟨Ĥ⟩


No, you are CLAIMING (incorrectly) that it has made that determination.

THIS IS WHAT YOU HAVE BEEN DISAGREEING WITH:
As soon as an infinite behavior pattern is correctly recognized in a finite number of steps then it is definitely correct for embedded_H to transition to Ĥ.qn.



Except that such a pattern in H^ is a Fairy Dust Powered Unicorn,
So in other words the concept of logical necessity is so far over your head that you cannot begin to fathom it.

When we know that we have a black cat then we know that we have a cat and you dishonestly disagree.

--
Copyright 2021 Pete Olcott

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


Subject: Re: Concise refutation of halting problem proofs V52 [ ignorance or deception? ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.math, sci.logic
Followup: comp.theory
Date: Fri, 4 Feb 2022 05:12 UTC
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Subject: Re: Concise refutation of halting problem proofs V52 [ ignorance or
deception? ]
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From: NoO...@NoWhere.com (olcott)
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On 2/3/2022 11:02 PM, Richard Damon wrote:

On 2/3/22 11:45 PM, olcott wrote:
On 2/3/2022 10:20 PM, Richard Damon wrote:
On 2/3/22 10:56 PM, olcott wrote:
On 2/3/2022 9:40 PM, Richard Damon wrote:
On 2/3/22 10:10 PM, olcott wrote:
On 2/3/2022 5:50 AM, Richard Damon wrote:
On 2/3/22 12:24 AM, olcott wrote:
On 2/2/2022 11:09 PM, Richard Damon wrote:
On 2/2/22 11:50 PM, olcott wrote:
On 2/2/2022 10:42 PM, Richard Damon wrote:
On 2/2/22 10:50 PM, olcott wrote:

That is not true. The pattern exists for at least any finite number of steps where it can be recognized. The three iterations shown above are plenty enough for it to be recogized.


But if it only exists for a finite number of steps (till it is recognized)

We are discussing the point in the execution of embedded_H where it has just correctly matched an infinite behavior pattern while it was doing its correct simulation of the first N steps of ⟨Ĥ⟩ ⟨Ĥ⟩


No, you are CLAIMING (incorrectly) that it has made that determination.

THIS IS WHAT YOU HAVE BEEN DISAGREEING WITH:
As soon as an infinite behavior pattern is correctly recognized in a finite number of steps then it is definitely correct for embedded_H to transition to Ĥ.qn.



Except that such a pattern in H^ is a Fairy Dust Powered Unicorn,
So in other words the concept of logical necessity is so far over your head that you cannot begin to fathom it.

WHY is it logicallyt necessary that the pattern you have presupposed to exist to actually exist?

As I agreed, **IF** H could find such a pattern, it would be correct to abort and go to H.Qn,

Great, yet it took far too long to get agreement on a statement that is true by logical necessity.

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

You already agreed that these steps would repeat if there was a UTM at Ĥ.qx instead of embedded_H:

These steps would keep repeating:
Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩
Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates ⟨Ĥ4⟩ ⟨Ĥ5⟩...



--
Copyright 2021 Pete Olcott

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


Subject: Re: Concise refutation of halting problem proofs V52 [ ignorance or deception? ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Fri, 4 Feb 2022 16:20 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
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Subject: Re: Concise refutation of halting problem proofs V52 [ ignorance or
deception? ]
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References: <ssh8vu$4c0$1@dont-email.me>
<rtKdnX6XWc-RbmT8nZ2dnUU7-T_NnZ2d@giganews.com> <stcthg$3cm$1@dont-email.me>
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<XfqdnbvOjdlgzWb8nZ2dnUU7-X3NnZ2d@giganews.com>
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On 2/3/2022 11:39 PM, Richard Damon wrote:
On 2/4/22 12:12 AM, olcott wrote:
On 2/3/2022 11:02 PM, Richard Damon wrote:

On 2/3/22 11:45 PM, olcott wrote:
On 2/3/2022 10:20 PM, Richard Damon wrote:
On 2/3/22 10:56 PM, olcott wrote:
On 2/3/2022 9:40 PM, Richard Damon wrote:
On 2/3/22 10:10 PM, olcott wrote:
On 2/3/2022 5:50 AM, Richard Damon wrote:
On 2/3/22 12:24 AM, olcott wrote:
On 2/2/2022 11:09 PM, Richard Damon wrote:
On 2/2/22 11:50 PM, olcott wrote:
On 2/2/2022 10:42 PM, Richard Damon wrote:
On 2/2/22 10:50 PM, olcott wrote:

That is not true. The pattern exists for at least any finite number of steps where it can be recognized. The three iterations shown above are plenty enough for it to be recogized.


But if it only exists for a finite number of steps (till it is recognized)

We are discussing the point in the execution of embedded_H where it has just correctly matched an infinite behavior pattern while it was doing its correct simulation of the first N steps of ⟨Ĥ⟩ ⟨Ĥ⟩


No, you are CLAIMING (incorrectly) that it has made that determination.

THIS IS WHAT YOU HAVE BEEN DISAGREEING WITH:
As soon as an infinite behavior pattern is correctly recognized in a finite number of steps then it is definitely correct for embedded_H to transition to Ĥ.qn.



Except that such a pattern in H^ is a Fairy Dust Powered Unicorn,
So in other words the concept of logical necessity is so far over your head that you cannot begin to fathom it.

WHY is it logicallyt necessary that the pattern you have presupposed to exist to actually exist?

As I agreed, **IF** H could find such a pattern, it would be correct to abort and go to H.Qn,

Great, yet it took far too long to get agreement on a statement that is true by logical necessity.

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

You already agreed that these steps would repeat if there was a UTM at Ĥ.qx instead of embedded_H:

These steps would keep repeating:
Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩
Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates ⟨Ĥ4⟩ ⟨Ĥ5⟩...


Right, and if there WAS a UTM at H^.Qx then H is 'just' a UTM, and could never abort its simulation (as if it did, it wouldn't be a UTM) and thus H never answers, and FAILS.

Since you just now agreed that the above is an infinitely repeating pattern if Ĥ.qx was a UTM, (and you only saw three repetitions) then embedded_H could simulate the exact same three repetitions of nested simulations and see the same pattern that you saw.


--
Copyright 2021 Pete Olcott

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


Subject: Re: Concise refutation of halting problem proofs V52 [ dodgy double talk ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Sun, 6 Feb 2022 05:13 UTC
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Subject: Re: Concise refutation of halting problem proofs V52 [ dodgy double
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Followup-To: comp.theory
From: NoO...@NoWhere.com (olcott)
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On 2/1/2022 9:33 PM, André G. Isaak wrote:
On 2022-02-01 20:08, olcott wrote:

embedded_H is exactly a UTM with extra features added.

Apparently you don't know what 'exactly' means. embedded_H is not a UTM *at* *all*.

If embedded_H were a UTM, then

embedded_H would accept ⟨Ĥ⟩ ⟨Ĥ⟩ if Ĥ accepts ⟨Ĥ⟩

If ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by embedded_H cannot possibly ever transition to ⟨Ĥ⟩.qn then it fails to meet the Linz definition:

computation that halts … the Turing machine will halt whenever it enters a final state. (Linz:1990:234) thus the simulated ⟨Ĥ⟩ ⟨Ĥ⟩ does not halt.

This makes it necessarily correct for embedded_H to transition to Ĥ.qn.


--
Copyright 2021 Pete Olcott

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


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