• Re: Concise refutation of halting problem proofs V56 [ key essence ]

    From olcott@21:1/5 to olcott on Sat Jan 29 19:08:07 2022
    XPost: comp.theory, sci.logic, sci.math

    On 1/29/2022 5:51 PM, olcott wrote:
    Correctly answering this single question refutes the Linz proof:
    Which state does Ĥ applied to ⟨Ĥ⟩ transition to correctly ?


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

    Because there are no finite number of steps of the pure simulation of ⟨Ĥ⟩ applied to ⟨Ĥ⟩ by embedded_H such that this simulated input meets
    the Linz definition of halting:

    computation that halts … the Turing machine will halt whenever it enters
    a final state. (Linz:1990:234)

    // Halting defined
    Mathematically this is known as existential quantification:

    ∃n ∈ N number of simulated steps such that:
    embedded_H simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩ ⊢* ⟨Ĥ⟩.qy or embedded_H simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩ ⊢* ⟨Ĥ⟩.qn


    Therefore it is correct to say that the input to embedded_H specifies a sequence of configurations that never halts.



    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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    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From olcott@21:1/5 to All on Sat Jan 29 17:51:38 2022
    XPost: comp.theory, sci.logic, sci.math

    Correctly answering this single question refutes the Linz proof:
    Which state does Ĥ applied to ⟨Ĥ⟩ transition to correctly ?


    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 there are no finite number of steps of the pure simulation of
    ⟨Ĥ⟩ applied to ⟨Ĥ⟩ by embedded_H such that this simulated input meets the Linz definition of halting:

    computation that halts … the Turing machine will halt whenever it enters
    a final state. (Linz:1990:234)

    Therefore it is correct to say that the input to embedded_H specifies a sequence of configurations that never halts.

    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

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