• Re: Concise refutation of halting problem proofs V47, [ H is an objecti

    From olcott@21:1/5 to Richard Damon on Sun Jan 9 10:40:09 2022
    XPost: comp.theory, sci.logic, sci.math

    On 1/8/2022 9:59 PM, Richard Damon wrote:
    On 1/8/22 10:20 PM, olcott wrote:
    On 1/8/2022 8:55 PM, Richard Damon wrote:
    On 1/8/22 8:41 PM, olcott wrote:
    // Simplified Linz(1990) Ĥ
    // and Strachey(1965) P
    void P(ptr x)
    {
       if (H(x, y))
         HERE: goto HERE;
    }

    H and P are defined according to the standard HP counter-example
    template shown above.

    H bases its halt status decision on the behavior of the simulation
    of its input.

    Then P demonstrates an infinitely repeating pattern that cannot
    possibly ever reach its final state.

    This conclusively proves that the input to H meets the Linz
    definition of non-halting:

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

    thus the sufficiency condition for H to report that its input
    specifies a non-halting computation.

    Halting problem undecidability and infinitely nested simulation V2
    https://www.researchgate.net/publication/356105750_Halting_problem_undecidability_and_infinitely_nested_simulation_V2




    Full Proof with Request for Rebuttal
    We have gone around the circle of this MANY times, and you keep just
    rearranging things and not every answering the refutation.

    The problem is that you are simply too stupid to ever understand that
    P specifies a sequence of configurations that never reach its final
    state and thus is correctly determined to be a non-halting computation
    according to Linz.

    And you are too stupid to see that it doesn't if H(P,P) returns 0, as
    this just proved.


    It is always correct for H to report on what the behavior of its input
    would be if H did not interfere with the behavior of this input.
    H is an objective observer.

    It is never correct for H to report on what the behavior of its input
    would be if H did interfere with the behavior of this input.
    H is not an objective observer.


    Your failure to point out an error will be taken as an admission that
    you accept that your logic is incorrect.

    FAIR WARNING.


    Malcolm, Kaz and Flibble are not too stupid to understand this.

    Ben, André and Mike are not interested in understanding what I say
    they are only interested in finding some basis for rebuttal. If there
    is at least one minor point that I have not proven completely they
    count everything that I say as incorrect on the basis of this minor
    point.




    --
    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 Sun Jan 9 13:11:40 2022
    XPost: comp.theory, sci.logic, sci.math

    On 1/9/2022 12:06 PM, Richard Damon wrote:
    On 1/9/22 12:57 PM, olcott wrote:
    On 1/9/2022 11:43 AM, Richard Damon wrote:
    On 1/9/22 11:40 AM, olcott wrote:
    On 1/8/2022 9:59 PM, Richard Damon wrote:
    On 1/8/22 10:20 PM, olcott wrote:
    On 1/8/2022 8:55 PM, Richard Damon wrote:
    On 1/8/22 8:41 PM, olcott wrote:
    // Simplified Linz(1990) Ĥ
    // and Strachey(1965) P
    void P(ptr x)
    {
       if (H(x, y))
         HERE: goto HERE;
    }

    H and P are defined according to the standard HP counter-example >>>>>>>> template shown above.

    H bases its halt status decision on the behavior of the
    simulation of its input.

    Then P demonstrates an infinitely repeating pattern that cannot >>>>>>>> possibly ever reach its final state.

    This conclusively proves that the input to H meets the Linz
    definition of non-halting:

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

    thus the sufficiency condition for H to report that its input
    specifies a non-halting computation.

    Halting problem undecidability and infinitely nested simulation V2 >>>>>>>> https://www.researchgate.net/publication/356105750_Halting_problem_undecidability_and_infinitely_nested_simulation_V2




    Full Proof with Request for Rebuttal
    We have gone around the circle of this MANY times, and you keep
    just rearranging things and not every answering the refutation.

    The problem is that you are simply too stupid to ever understand
    that P specifies a sequence of configurations that never reach its >>>>>> final state and thus is correctly determined to be a non-halting
    computation according to Linz.

    And you are too stupid to see that it doesn't if H(P,P) returns 0,
    as this just proved.


    It is always correct for H to report on what the behavior of its
    input would be if H did not interfere with the behavior of this input. >>>> H is an objective observer.

    It is never correct for H to report on what the behavior of its
    input would be if H did interfere with the behavior of this input.
    H is not an objective observer.

    IMPROPERLY PHRASED, H must report on what the machine that its input
    represents will do, even if that includes a copy of itself. That is
    not H 'interfering' with the behavior of that machine.

    It is Impossible for the copy of a decider doing the deciding to
    'interfere' with the behavior of a machine, as that behavior is
    defined independent of the decider.

    Yes, the aborting of a simulation by the copy of the decider doing
    the deciding doesn't affect the behavior of the machine it is
    deciding on, a copy of it IN the machine it is trying to decide on,
    DOES, as it IS part of the machine it is deciding on.

    FAIL.


    The fact that you can't keep the different copies of H separate shows
    your lack of reasoning ability.


    When H reports on what the behavior of its simulated input would be if
    H did not interfere, it is the same for P, infinite loops, or infinite
    recursion, H must only reject its input as non-halting.


    Except it isn't 'interference' for the copy of H in the input to do what
    it is programmed to do.


    It is the job of H to determine what the behavior of the input would be
    if H did not interfere with this behavior.

    Alternatively H could correctly recognize inputs that would never stop
    running if H did not interfere and then report that every input does
    halt when H does interfere.

    Such an H could simply accept every input in that some of its inputs
    halt on their own and the other inputs must be aborted. In this case It
    would be impossible to create an input that H would not correctly decide.

    In fact, it is interference for the H that is deciding to NOT let the
    copy of H inside P to do that that H is programmed to do.

    DEFINITIONS, you know, The correct answer for H(<X>, y) is basd on what
    X(y) would do when run.

    The behavior of 'the input' is what that program would do when run
    'without outside interference', that means that copy of H inside P does
    what it will do.

    Since H(P,P) returns 0, ALL Copies of H(P,P) return 0, so the copy
    inside P does thins, and P(P) will Halt.

    FAIL.

    This has been PROVEN and no actual rebuttal provided, so you have
    conceded the point.



    Your failure to point out an error will be taken as an admission
    that you accept that your logic is incorrect.

    FAIR WARNING.


    Malcolm, Kaz and Flibble are not too stupid to understand this.

    Ben, André and Mike are not interested in understanding what I say >>>>>> they are only interested in finding some basis for rebuttal. If
    there is at least one minor point that I have not proven
    completely they count everything that I say as incorrect on the
    basis of this minor point.










    --
    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)