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

    From olcott@21:1/5 to All on Tue Feb 1 19:33:34 2022
    XPost: comp.theory, sci.logic, sci.math

    On 2/1/2022 6:58 PM, André G. Isaak wrote:
    On 2022-01-30 19:05, olcott wrote:
    On 1/30/2022 7:45 PM, Richard Damon wrote:
    On 1/30/22 7:21 PM, olcott wrote:

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

    These statements need the conditions, that H^ goes to H^.Qy/H^.Qn iff
    H goes to that corresponding state.


    ⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an input to embedded_H in the
    same way that (5,3) is syntactically specified as an input to Sum(5,3)

    Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an input to embedded_H in the >> same way that (1,2) is NOT syntactically specified as an input to
    Sum(5,3)

    I promised myself I wouldn't involve myself in your nonsense any
    further, but here you've made such a terribly inaccurate analogy that I thought I had to comment.

    The inputs to a function such as SUM(X, Y) are two REPRESENTATIONS of integers. If SUM were a Turing Machine, these would be two strings in
    the alphabet of the TM. if this were a C function, X and X would be
    strings of bits which form the twos complement representation of some integer. In neither case would the inputs be actual, mathematical
    integers. C might use the term 'integer' as one of its built in types,
    but C integers are NOT elements of ℤ. They are REPRESENTATIONS of the supported subset of ℤ.

    So ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to embedded_H in the same sense that ⟨5⟩ ⟨3⟩ are
    the inputs to SUM.

    Ĥ ⟨Ĥ⟩ is not the input to embedded_H in the same sense that the actual mathematical integers 3 and 5 are not inputs to SUM.


    We are on the same page so far. (acknowledging when there is agreement
    is an essential part of an honest dialogue).

    If your going to make analogies, at least make ones that are accurate.

    SUM takes REPRESENTATIONS of integers as its inputs, but it answers
    about the ACTUAL integers described by those representations. To talk
    about the sum of two representations is meaningless. Only actual
    integers have sums.

    In exactly the same way, embedded_H takes a REPRESENTATION of some TM ⟨Ĥ⟩ as part of its input but it answers about the ACTUAL TM described by that input, Ĥ.
    To talk about whether a representation of a TM halts is
    meaningless since only actual TMs, not representations of TMs, can halt.
    The conditions which Richard indicates above (following Linz) are
    therefore the correct ones.

    In a previous post which I can't be botherered to find, you claimed that
    when the input to embedded_H is ⟨Ĥ⟩ ⟨Ĥ⟩ that embedded_H can only be expected to answer about its actual inputs and not its 'enclosing TM'.

    Yes, it must answer about its input, but if its input is ⟨Ĥ⟩ ⟨Ĥ⟩, then
    BY THE DEFINITION OF A HALT DECIDER is must determine whether Ĥ applied
    to ⟨Ĥ⟩ halts.

    No you are flat out wrong about this. You are wrong because of your
    ignorance of how deciders work. Deciders compute the mapping from their
    finite string inputs to an accept or reject state on the basis of the
    actual properties of these actual inputs.

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

    An answer of "no" means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ never halts, and nothing
    in the universe can possibly overcome this.

    Because all simulating halt deciders are deciders they are only
    accountable for computing the mapping from their input finite strings to
    an accept or reject state on the basis of whether or not their correct simulation of this input could ever reach its final state.

    embedded_H is only accountable for the behavior of its input ⟨Ĥ⟩ applied to ⟨Ĥ⟩. embedded_H is not accountable for the behavior of the
    computation that it is contained within: Ĥ applied to ⟨Ĥ⟩.



    Halting problem undecidability and infinitely nested simulation (V3)

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


    And that computation happens to be the EXACT SAME
    computation as its 'enclosing TM'. So it is answering about *both*.

    André



    --
    Copyright 2021 Pete Olcott

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