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  • Re: Concise refutation of halting problem proofs V52 [ ignorance or dec

    From olcott@21:1/5 to Richard Damon on Thu Feb 3 23:12:41 2022
    XPost: comp.theory, sci.math, sci.logic

    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

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From olcott@21:1/5 to Richard Damon on Fri Feb 4 10:20:47 2022
    XPost: comp.theory, sci.logic, sci.math

    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

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)