• Re: Concise refutation of halting problem proofs V52 [ Ignorant or Dish

    From olcott@21:1/5 to Richard Damon on Sat Jan 22 11:20:56 2022
    XPost: comp.theory, sci.math.symbolic, sci.philosophy

    On 1/22/2022 11:13 AM, Richard Damon wrote:
    On 1/22/22 11:47 AM, olcott wrote:
    On 1/22/2022 10:39 AM, Richard Damon wrote:
    On 1/22/22 11:29 AM, olcott wrote:
    On 1/22/2022 10:23 AM, Richard Damon wrote:

    On 1/22/22 10:48 AM, olcott wrote:
    Halting problem undecidability and infinitely nested simulation (V3) >>>>>
    Take FIFTY TWO, I think you are stuck in an apparent infinite loop.

    You keep on repeating the same basic mistakes.


    We define Linz H to base its halt status decision on the behavior
    of its pure simulation of N steps of its input. N is either the
    number of steps that it takes for its simulated input to reach its >>>>>> final state or the number of steps required for H to match an
    infinite behavior pattern proving that the simulated input would
    never reach its own final state. In this case H aborts the
    simulation of this input and transitions to H.qn.

    Such a pattern NOT existing for the <H^> <H^> pattern, thus your H
    can't correctly abort and becomes non-answering and thus FAILS to
    be a decider.

    The non-existance has been proven and you have ignored that proof,
    showing you have no counter for the proof.

    If you want to claim such a pattern exists, you MUST provide it or
    accept that your logic is just plain flawed as you are claiming the
    existance of something that is impossible.

    In effect, you are saying that if you have a halt decider for you
    halt decider to use, you can write a halt decider.

    FAIL.


    The following simplifies the syntax for the definition of the Linz >>>>>> Turing machine Ĥ, it is now a single machine with a single start
    state. A copy of Linz H is embedded at Ĥ.qx.

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

    Because it is known that the UTM simulation of a machine is
    computationally equivalent to the direct execution of this same
    machine H can always form its halt status decision on the basis of >>>>>> what the behavior of the UTM simulation of its inputs would be.

    When Ĥ applied to ⟨Ĥ⟩ has embedded_H simulate ⟨Ĥ⟩ ⟨Ĥ⟩ these steps
    would keep repeating:
    Ĥ copies its input ⟨Ĥ⟩ to ⟨Ĥ⟩ then embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩...

    But only if H never aborts, if H does abort, then the pattern is
    broken.


    YOU ARE EITHER TOO IGNORANT OR DISHONEST TO ACKNOWLEDGE THE TRUTH OF
    THIS:

    The fact that there are no finite number of steps that the simulated
    input to embedded_H would ever reach its final state conclusively
    proves that embedded_H is correct to abort its simulation of this
    input and transition to Ĥ.qn.


    The problem is that any H that aborts after a finite number of steps,
    gives the wrong answer because it only looked to see if the input
    doesn't halt at some specific finite number.

    OK IGNORANT it is. When there exists no finite (or infinite) number of
    steps such that the simulated input to embedded_H reaches its final
    state then we know that this simulated input does not halt.

    And the only case when that happens is when H does not abort its
    simulation,

    WRONG !!! That happens in every possible case. The simulated input to embedded_H cannot possibly ever reach its final state NO MATTER WHAT !!!



    Halting problem undecidability and infinitely nested simulation (V3)

    https://www.researchgate.net/publication/358009319_Halting_problem_undecidability_and_infinitely_nested_simulation_V3


    --
    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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