• Re: Concise refutation of halting problem proofs V42 [compute the mappi

    From olcott@21:1/5 to Ben Bacarisse on Wed Dec 29 22:02:44 2021
    XPost: comp.theory, sci.logic, sci.math

    On 12/29/2021 1:16 PM, Ben Bacarisse wrote:
    olcott <NoOne@NoWhere.com> writes:

    *My criterion measure with Ben's notational conventions*
    H.q0 wM w ⊢* H.qy iff UTM(wM, w) halts
    H.q0 wM w ⊢* H.qn iff UTM(wM, w) does not halt

    It's not your criterion, it's mine. It's also Linz's. If yours is not
    the same as his (and mine), it should be different in some way. (And we
    can safely ignore it, since no one cares about your alternative
    criteria.)


    It has always been my criterion measure that halt deciders base their
    halt status decision on the pure simulation of their inputs.

    That your notational conventions very cleanly express this idea as
    applied to the Linz H is a good breakthrough.

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

    As soon as people understand that this same criterion measure applies recursively to every embedded_H at Ĥ.qx, then they will see how this
    criterion measure provides the means for embedded_H to correctly compute
    the mapping from ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn.



    --
    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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  • From olcott@21:1/5 to All on Thu Dec 30 09:15:05 2021
    XPost: comp.theory, sci.logic, sci.math

    On 12/29/2021 11:18 PM, André G. Isaak wrote:
    On 2021-12-29 21:51, olcott wrote:
    On 12/29/2021 10:44 PM, Richard Damon wrote:
    On 12/29/21 11:02 PM, olcott wrote:
    On 12/29/2021 1:16 PM, Ben Bacarisse wrote:
    olcott <NoOne@NoWhere.com> writes:

    *My criterion measure with Ben's notational conventions*
    H.q0 wM w ⊢* H.qy  iff UTM(wM, w) halts
    H.q0 wM w ⊢* H.qn  iff UTM(wM, w) does not halt

    It's not your criterion, it's mine.  It's also Linz's.  If yours is >>>>> not
    the same as his (and mine), it should be different in some way.
    (And we
    can safely ignore it, since no one cares about your alternative
    criteria.)


    It has always been my criterion measure that halt deciders base
    their halt status decision on the pure simulation of their inputs.

    That your notational conventions very cleanly express this idea as
    applied to the Linz H is a good breakthrough.

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

    As soon as people understand that this same criterion measure
    applies recursively to every embedded_H at Ĥ.qx, then they will see
    how this criterion measure provides the means for embedded_H to
    correctly compute the mapping from ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn.


    Yes each level can use the test to determine if the halt decider at
    that level got the right answer or not.

    The issue is that when we are doing this check, you check JUST that
    level, and not lower levels.

    No you are wrong.

    No, he isn't.

    The specification you give above (where YET AGAIN you omit the
    conditions)

    The conditions are provided by H, because H is copied to Ĥ.qx we need
    not state them again.

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

    To make things simple we name the copy of H at Ĥ.qx embedded_H.

    Because the criterion measure is applied as if the embedded_H at Ĥ.qx
    was replaced by a UTM and the TM description of Ĥ would include a TM description of this UTM the execution of Ĥ applied to ⟨Ĥ⟩ would

    endlessly repeat:
    copy its input ⟨Ĥ⟩ to ⟨Ĥ⟩ and then simulate this input with the UTM...

    In order for embedded_H to see what this machine would do it must
    actually perform a pure simulation of N steps of its input ⟨Ĥ⟩ ⟨Ĥ⟩. This
    may or may not include simulating embedded_H.

    As soon as embedded_H sees that these steps would otherwise endlessly
    repeat it aborts its simulation and transitions to Ĥ.qn.

    Because it is applying a criterion measure that we know is correct:
    A halt decider is always correct when it bases its halt status decision
    on the behavior of the UTM simulation of its inputs.
    We know that this transition to Ĥ.qn is correct.

    is for a case where the input to Ĥ is a description of
    itself, but the actual specification of the machine can't make that assumption. So really we should be using:


    The criterion measure is merely provided to humans so that humans can understand that the basis of the halt status decision of embedded_H is
    sound.

    Because it is obvious that the basis is sound when applied to H then
    this same sound basis carries over to embedded_H.

    Ĥ.q0 ⟨wM⟩ ⊢* Ĥ.qx wM wM⟩ ⊢* Ĥ.qy ∞ iff UTM(wM, wM) halts
    Ĥ.q0 ⟨wM⟩ ⊢* Ĥ.qx wM wM⟩ ⊢* Ĥ.qn   iff UTM(wM, wM) does not halt

    Ĥ has absolutely no knowledge of what it will be given as an input. In

    For purposes of discussion we are simply ignoring every other input and
    only focusing on the one input.

    It is simpler for the human mind to see all the relevant details in one
    place and not have to imagine all of the changes to see the case-at-hand.

    The way that you specify the criterion measure is vague. When I define
    the criterion measure for H this applies the criterion measure at the
    point where H is copied Ĥ.qx.

    the case we considered above wM was ⟨Ĥ⟩, but wM could have been a description of any machine. Ĥ has no way of knowing if wM is a
    description of itself, or a description of some TM which simply prints
    its input string to the tape in reverse order (which wouldn't involve simulating its input at all).

    H thus embedded_H only makes its halt status decision on the basis of
    the simulation of its input. H can only see what its input does by
    simulating its input.

    So the conditions given above cannot
    possibly apply to any 'lower level' since Ĥ knows nothing about any of
    those 'lower levels'.

    André


    It only applies to lower levels when there is actual pathological self-reference(Olcott 2004) as there is with the definition of Ĥ. Ĥ is
    asking itself whether or not it stops running and then looping if it
    says yes.


    --
    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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  • From olcott@21:1/5 to Richard Damon on Thu Dec 30 10:11:06 2021
    XPost: comp.theory, sci.logic, sci.math

    On 12/30/2021 9:57 AM, Richard Damon wrote:
    On 12/30/21 10:20 AM, olcott wrote:
    On 12/30/2021 8:19 AM, Richard Damon wrote:
    On 12/29/21 11:51 PM, olcott wrote:
    On 12/29/2021 10:44 PM, Richard Damon wrote:
    On 12/29/21 11:02 PM, olcott wrote:
    On 12/29/2021 1:16 PM, Ben Bacarisse wrote:
    olcott <NoOne@NoWhere.com> writes:

    *My criterion measure with Ben's notational conventions*
    H.q0 wM w ⊢* H.qy  iff UTM(wM, w) halts
    H.q0 wM w ⊢* H.qn  iff UTM(wM, w) does not halt

    It's not your criterion, it's mine.  It's also Linz's.  If yours >>>>>>> is not
    the same as his (and mine), it should be different in some way.
    (And we
    can safely ignore it, since no one cares about your alternative
    criteria.)


    It has always been my criterion measure that halt deciders base
    their halt status decision on the pure simulation of their inputs. >>>>>>
    That your notational conventions very cleanly express this idea as >>>>>> applied to the Linz H is a good breakthrough.

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

    As soon as people understand that this same criterion measure
    applies recursively to every embedded_H at Ĥ.qx, then they will
    see how this criterion measure provides the means for embedded_H
    to correctly compute the mapping from ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn.


    Yes each level can use the test to determine if the halt decider at
    that level got the right answer or not.

    The issue is that when we are doing this check, you check JUST that
    level, and not lower levels.

    No you are wrong.



    Source? (Some REAL reference)


    H.q0 wM w ⊢* H.qy  iff UTM(wM, w) halts
    H.q0 wM w ⊢* H.qn  iff UTM(wM, w) does not halt

    The source is the correct reasoning of how the above criterion measure
    would be applied when H is copied to Ĥ.qx.

    We copy the ALGORITHM of H, the detailed step by step instructions



    You keep thinking that a subsequent level UTM magically turns into
    embedded_H and aborts its simulation.

    There IS no UTM in the machine. PERIOD. (unless H IS a UTM)

    H always bases its halt status decision on the behavior of the UTM
    simulation of its input. Thus embedded_H bases its halt status decision
    on the behavior of the UTM simulation of its input. This means that H
    and embedded_H continue a pure simulation of N steps of their input
    until their input halts on its own or this input demonstrates an
    infinite behavior pattern.

    Since it is obvious to humans that replacing embedded_H at Ĥ.qx with a
    UTM would cause Ĥ applied to ⟨Ĥ⟩ to never stop running humans can see that the pure simulation of N steps of the input to embedded_H would
    never halt on their own for every N from 0 to ∞.

    Thus correctly applying the criterion measure of H to the input to
    embedded_H results in the "does not halt" criterion measure being met.


    --
    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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