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  • Re: First step needed to make progress with Olcott

    From Mikko@21:1/5 to olcott on Thu Aug 7 10:39:36 2025
    On 2025-08-06 11:34:45 +0000, olcott said:

    On 8/6/2025 2:18 AM, Mikko wrote:
    On 2025-08-04 14:49:57 +0000, Mr Flibble said:

    First step needed to make progress with Olcott is to get Olcott to admit >>> that he is not working on the Halting Problem or attempting to refute
    Halting Problem proofs but is instead working on the Olcott Problem given >>> his method involves partial rather than total deciders.

    Why should there be any progress with Olcott? The status quo is quite
    good as it is.

    Ah so you are only a troll.

    Are you really that stupid or are you just trolling?
    It is well known that trolling is very different from a desire to
    maintain status quo.

    --
    Mikko

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Fred. Zwarts@21:1/5 to All on Thu Aug 7 11:19:49 2025
    Op 06.aug.2025 om 13:45 schreef olcott:
    On 8/6/2025 4:20 AM, Fred. Zwarts wrote:
    Op 05.aug.2025 om 17:40 schreef olcott:
    On 8/5/2025 3:37 AM, Fred. Zwarts wrote:
    Op 04.aug.2025 om 23:18 schreef olcott:

    The Linz proof specifies that it uses an encoding
    as input yet requires H to report on the behavior
    of Machine M. This is all well and good until one
    realizes that when an input calls its own decider
    that the behavior of the input and the behavior
    of the machine are not the same.

    As usual incorrect claims without evidence.
    *From the bottom of page 319 has been adapted to this*
    https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

    Turing Machine Ĥ applied to its own machine description ⟨Ĥ⟩
        Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
    if Ĥ applied to ⟨Ĥ⟩ halts, and
        Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H  ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
    if Ĥ applied to ⟨Ĥ⟩ does not halt.

    Lines 2 and 4 above do insist that embedded_H
    report on its own behavior.



    Counter factual. embedded_H must report on the behaviour of Ĥ, which
    is not the same thing, even when they have similar behaviour.

    Ĥ.embedded_H is an aspect of Ĥ, thus when it is
    required to report on the behavior of Ĥ applied
    to ⟨Ĥ⟩ it is required to report on its own behavior.
    As usual incorrect claim, based on changing te meaning of the words.
    It must report on the behaviour specified in the input, not on its own behaviour, even when this input specifies behaviour that resembles its
    own behaviour.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to olcott on Thu Aug 7 20:36:41 2025
    On 8/7/25 9:47 AM, olcott wrote:
    On 8/7/2025 4:19 AM, Fred. Zwarts wrote:
    Op 06.aug.2025 om 13:45 schreef olcott:
    On 8/6/2025 4:20 AM, Fred. Zwarts wrote:
    Op 05.aug.2025 om 17:40 schreef olcott:
    On 8/5/2025 3:37 AM, Fred. Zwarts wrote:
    Op 04.aug.2025 om 23:18 schreef olcott:

    The Linz proof specifies that it uses an encoding
    as input yet requires H to report on the behavior
    of Machine M. This is all well and good until one
    realizes that when an input calls its own decider
    that the behavior of the input and the behavior
    of the machine are not the same.

    As usual incorrect claims without evidence.
    *From the bottom of page 319 has been adapted to this*
    https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

    Turing Machine Ĥ applied to its own machine description ⟨Ĥ⟩
        Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞, >>>>> if Ĥ applied to ⟨Ĥ⟩ halts, and
        Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H  ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn >>>>> if Ĥ applied to ⟨Ĥ⟩ does not halt.

    Lines 2 and 4 above do insist that embedded_H
    report on its own behavior.



    Counter factual. embedded_H must report on the behaviour of Ĥ, which
    is not the same thing, even when they have similar behaviour.

    Ĥ.embedded_H is an aspect of Ĥ, thus when it is
    required to report on the behavior of Ĥ applied
    to ⟨Ĥ⟩ it is required to report on its own behavior.
    As usual incorrect claim, based on changing te meaning of the words.

    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
    Ĥ.embedded_H is a named state of machine Ĥ

    It must report on the behaviour specified in the input, not on its own
    behaviour, even when this input specifies behaviour that resembles its
    own behaviour.

    *Repeats until aborted*
    (a) Ĥ copies its input ⟨Ĥ⟩
    (b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
    (c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩


    And, either it *IS* programmed to abort and thus the behavior of the
    machine it is simulating (or its actual correct simulation) WILL reach
    its final state after it stops looking at that behavior, or

    The "decider" just fails to decide and thus isn't a decider.

    Since it must do one or the other, we have shown that it can never
    correctly report that its input is non-halting.

    Remember, every H has been given a different input, as that input is a
    proper representation of the pathological program built from that
    particular H, which are each different, so you have to look at the
    behavior of that specific input, not some "general" input that they all get.

    Unless of course, you admit that you whole process is based on a
    category error that H and the pathological program represented in it
    sinput aren't actually program, and thus you whole work a lie.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Mikko@21:1/5 to olcott on Fri Aug 8 10:14:01 2025
    On 2025-08-07 12:59:33 +0000, olcott said:

    On 8/7/2025 2:39 AM, Mikko wrote:
    On 2025-08-06 11:34:45 +0000, olcott said:

    On 8/6/2025 2:18 AM, Mikko wrote:
    On 2025-08-04 14:49:57 +0000, Mr Flibble said:

    First step needed to make progress with Olcott is to get Olcott to admit >>>>> that he is not working on the Halting Problem or attempting to refute >>>>> Halting Problem proofs but is instead working on the Olcott Problem given >>>>> his method involves partial rather than total deciders.

    Why should there be any progress with Olcott? The status quo is quite
    good as it is.

    Ah so you are only a troll.

    Are you really that stupid or are you just trolling?
    It is well known that trolling is very different from a desire to
    maintain status quo.

    When the status quo of failure to attain mutual agreement
    is "quite good" that indicates that the respondent is not
    interested in an honest dialogue.

    No honest dialogue is possible as long as your contribution is not
    honest. There doesn't seem to be much desire to any dialogue at
    all, whether honest or otherwise. People seem only to want to point
    out errors and dishonesty when they see them.

    --
    Mikko

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Fred. Zwarts@21:1/5 to All on Fri Aug 8 09:54:47 2025
    Op 07.aug.2025 om 15:47 schreef olcott:
    On 8/7/2025 4:19 AM, Fred. Zwarts wrote:
    Op 06.aug.2025 om 13:45 schreef olcott:
    On 8/6/2025 4:20 AM, Fred. Zwarts wrote:
    Op 05.aug.2025 om 17:40 schreef olcott:
    On 8/5/2025 3:37 AM, Fred. Zwarts wrote:
    Op 04.aug.2025 om 23:18 schreef olcott:

    The Linz proof specifies that it uses an encoding
    as input yet requires H to report on the behavior
    of Machine M. This is all well and good until one
    realizes that when an input calls its own decider
    that the behavior of the input and the behavior
    of the machine are not the same.

    As usual incorrect claims without evidence.
    *From the bottom of page 319 has been adapted to this*
    https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

    Turing Machine Ĥ applied to its own machine description ⟨Ĥ⟩
        Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞, >>>>> if Ĥ applied to ⟨Ĥ⟩ halts, and
        Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H  ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn >>>>> if Ĥ applied to ⟨Ĥ⟩ does not halt.

    Lines 2 and 4 above do insist that embedded_H
    report on its own behavior.



    Counter factual. embedded_H must report on the behaviour of Ĥ, which
    is not the same thing, even when they have similar behaviour.

    Ĥ.embedded_H is an aspect of Ĥ, thus when it is
    required to report on the behavior of Ĥ applied
    to ⟨Ĥ⟩ it is required to report on its own behavior.
    As usual incorrect claim, based on changing te meaning of the words.

    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
    Ĥ.embedded_H is a named state of machine Ĥ

    It must report on the behaviour specified in the input, not on its own
    behaviour, even when this input specifies behaviour that resembles its
    own behaviour.

    *Repeats until aborted*
    (a) Ĥ copies its input ⟨Ĥ⟩
    (b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
    (c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩


    and when it aborts, this is a premature abort, because the simulated Ĥ
    would also abort one cycle later. This proves that simulation is not the
    right tool for this analysis, because it must (prematurely) abort in
    order to report, but then it misses the fact that the recursion is only
    finite and there is a final halt state.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to olcott on Fri Aug 8 13:51:17 2025
    On 8/8/25 11:44 AM, olcott wrote:
    On 8/8/2025 2:14 AM, Mikko wrote:
    On 2025-08-07 12:59:33 +0000, olcott said:

    On 8/7/2025 2:39 AM, Mikko wrote:
    On 2025-08-06 11:34:45 +0000, olcott said:

    On 8/6/2025 2:18 AM, Mikko wrote:
    On 2025-08-04 14:49:57 +0000, Mr Flibble said:

    First step needed to make progress with Olcott is to get Olcott
    to admit
    that he is not working on the Halting Problem or attempting to
    refute
    Halting Problem proofs but is instead working on the Olcott
    Problem given
    his method involves partial rather than total deciders.

    Why should there be any progress with Olcott? The status quo is quite >>>>>> good as it is.

    Ah so you are only a troll.

    Are you really that stupid or are you just trolling?
    It is well known that trolling is very different from a desire to
    maintain status quo.

    When the status quo of failure to attain mutual agreement
    is "quite good" that indicates that the respondent is not
    interested in an honest dialogue.

    No honest dialogue is possible as long as your contribution is not
    honest. There doesn't seem to be much desire to any dialogue at
    all, whether honest or otherwise. People seem only to want to point
    out errors and dishonesty when they see them.


    Claude AI proved why HHH(DD)==0 is correct in terms that
    any expert C programmer can understand. https://claude.ai/share/da9e56ba-f4e9-45ee-9f2c-dc5ffe10f00c



    No, it took your incorrect framework and made the same errors you try to do.

    Sorry, AIs can't "prove" something, as they are not programmed to work
    with truth, just to make things that sound good.

    In fact, one of their biggest dangers is they are EXPERT liars, as their training it to make whatever they say SOUND reasonable, even though it
    isn't built on an actual factual chain.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Fred. Zwarts@21:1/5 to All on Sat Aug 9 08:42:07 2025
    Op 08.aug.2025 om 17:44 schreef olcott:
    On 8/8/2025 2:14 AM, Mikko wrote:
    On 2025-08-07 12:59:33 +0000, olcott said:

    On 8/7/2025 2:39 AM, Mikko wrote:
    On 2025-08-06 11:34:45 +0000, olcott said:

    On 8/6/2025 2:18 AM, Mikko wrote:
    On 2025-08-04 14:49:57 +0000, Mr Flibble said:

    First step needed to make progress with Olcott is to get Olcott
    to admit
    that he is not working on the Halting Problem or attempting to
    refute
    Halting Problem proofs but is instead working on the Olcott
    Problem given
    his method involves partial rather than total deciders.

    Why should there be any progress with Olcott? The status quo is quite >>>>>> good as it is.

    Ah so you are only a troll.

    Are you really that stupid or are you just trolling?
    It is well known that trolling is very different from a desire to
    maintain status quo.

    When the status quo of failure to attain mutual agreement
    is "quite good" that indicates that the respondent is not
    interested in an honest dialogue.

    No honest dialogue is possible as long as your contribution is not
    honest. There doesn't seem to be much desire to any dialogue at
    all, whether honest or otherwise. People seem only to want to point
    out errors and dishonesty when they see them.


    Claude AI proved why HHH(DD)==0 is correct in terms that
    any expert C programmer can understand. https://claude.ai/share/da9e56ba-f4e9-45ee-9f2c-dc5ffe10f00c

    A proof based on incorrect assumption is not a proof. It seems your are
    the only one that dos not understand it.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Heathfield@21:1/5 to olcott on Sat Aug 9 15:52:57 2025
    On 09/08/2025 15:40, olcott wrote:
    Assuming that HHH(DD) must report on the
    behavior of DD() is the incorrect assumption of all
    of the proofs.

    No.

    Assuming that HHH(DD) CAN report on the behaviour of DD() is the
    incorrect assumption of all of the proofs.

    1. Assume that a utm halt decider can exist.
    2. Derive contradiction.
    3. Ergo, utm halt decider cannot exist.

    --
    Richard Heathfield
    Email: rjh at cpax dot org dot uk
    "Usenet is a strange place" - dmr 29 July 1999
    Sig line 4 vacant - apply within

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to olcott on Sat Aug 9 17:30:42 2025
    On 8/9/25 10:40 AM, olcott wrote:
    On 8/9/2025 1:42 AM, Fred. Zwarts wrote:
    Op 08.aug.2025 om 17:44 schreef olcott:
    On 8/8/2025 2:14 AM, Mikko wrote:
    On 2025-08-07 12:59:33 +0000, olcott said:

    On 8/7/2025 2:39 AM, Mikko wrote:
    On 2025-08-06 11:34:45 +0000, olcott said:

    On 8/6/2025 2:18 AM, Mikko wrote:
    On 2025-08-04 14:49:57 +0000, Mr Flibble said:

    First step needed to make progress with Olcott is to get Olcott >>>>>>>>> to admit
    that he is not working on the Halting Problem or attempting to >>>>>>>>> refute
    Halting Problem proofs but is instead working on the Olcott
    Problem given
    his method involves partial rather than total deciders.

    Why should there be any progress with Olcott? The status quo is >>>>>>>> quite
    good as it is.

    Ah so you are only a troll.

    Are you really that stupid or are you just trolling?
    It is well known that trolling is very different from a desire to
    maintain status quo.

    When the status quo of failure to attain mutual agreement
    is "quite good" that indicates that the respondent is not
    interested in an honest dialogue.

    No honest dialogue is possible as long as your contribution is not
    honest. There doesn't seem to be much desire to any dialogue at
    all, whether honest or otherwise. People seem only to want to point
    out errors and dishonesty when they see them.


    Claude AI proved why HHH(DD)==0 is correct in terms that
    any expert C programmer can understand.
    https://claude.ai/share/da9e56ba-f4e9-45ee-9f2c-dc5ffe10f00c

    A proof based on incorrect assumption is not a proof. It seems your
    are the only one that dos not understand it.

    Exactly. Assuming that HHH(DD) must report on the
    behavior of DD() is the incorrect assumption of all
    of the proofs.


    Not an "assumption", but a Requrement to be a Halt Decider.

    Sorry, all you are doing is proving that you whole corpus of work is
    based on lies, because that all you have to base your work on.

    To say you are working on the Halting Problem, and then saying your
    criteria isn't that of the Halting Problem is to just admit that you are starting with a lie.

    That you talk about being able to change the definitions, just shows you
    don't understand the rules of LOGIC, and thus you can't even correctly
    claim you are working in that.

    Thus, all you work is proved to be from a worthless base, and thus
    worthless.

    You have been given chances to correct them. but you just doubled down
    on your lies, proving that your work is worthless.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to olcott on Sat Aug 9 17:35:35 2025
    On 8/9/25 11:04 AM, olcott wrote:
    On 8/9/2025 9:52 AM, Richard Heathfield wrote:
    On 09/08/2025 15:40, olcott wrote:
    Assuming that HHH(DD) must report on the
    behavior of DD() is the incorrect assumption of all
    of the proofs.

    No.

    Assuming that HHH(DD) CAN report on the behaviour of DD() is the
    incorrect assumption of all of the proofs.

    1. Assume that a utm halt decider can exist.
    2. Derive contradiction.
    3. Ergo, utm halt decider cannot exist.


    It has always been a false assumption that a halt
    decider must report on the behavior of the direct
    execution of a Turing machine.

    It has always been the DEFINITION, and your rejection just shows that
    you have always been a liar.



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

    Linz requires the second named state of Turing
    machine Ĥ to report on the behavior of itself
    rather than the behavior of its input.

    But they are the same, since the input is the description of the program
    that included itself

    You are just claiming that it is proper logic to just ignore the
    contextual semantic meaning of a word, because you juest don't like it.

    In other words, that LYING is valid logic, and thus you show you don't understand how logic works.



    When the second named state of Turing machine Ĥ
    reports on the behavior of its correct simulation
    of its input

    But it doesn't DO a correct simulaiton of that input, so again, you just
    show that you think lying is acceptable logic.

    That is what has a constant problem with your logic, you confuse what is
    doing what, and call things that are different as if they were the same,
    adn things that are the same, you assume they can be different.

    In other words, you admit to a form of insanity.


    *Repeats until aborted*
    Which happens.> (a) Ĥ copies its input ⟨Ĥ⟩
    (b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
    (c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩


    And the behavior of the input after H stops looking at it is to halt.

    Closing your eyes on the truth doesn't make it go away, it just makes
    you stupid.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Mikko@21:1/5 to olcott on Sun Aug 10 11:52:12 2025
    On 2025-08-08 15:44:39 +0000, olcott said:

    On 8/8/2025 2:14 AM, Mikko wrote:
    On 2025-08-07 12:59:33 +0000, olcott said:

    On 8/7/2025 2:39 AM, Mikko wrote:
    On 2025-08-06 11:34:45 +0000, olcott said:

    On 8/6/2025 2:18 AM, Mikko wrote:
    On 2025-08-04 14:49:57 +0000, Mr Flibble said:

    First step needed to make progress with Olcott is to get Olcott to admit
    that he is not working on the Halting Problem or attempting to refute >>>>>>> Halting Problem proofs but is instead working on the Olcott Problem given
    his method involves partial rather than total deciders.

    Why should there be any progress with Olcott? The status quo is quite >>>>>> good as it is.

    Ah so you are only a troll.

    Are you really that stupid or are you just trolling?
    It is well known that trolling is very different from a desire to
    maintain status quo.

    When the status quo of failure to attain mutual agreement
    is "quite good" that indicates that the respondent is not
    interested in an honest dialogue.

    No honest dialogue is possible as long as your contribution is not
    honest. There doesn't seem to be much desire to any dialogue at
    all, whether honest or otherwise. People seem only to want to point
    out errors and dishonesty when they see them.

    Claude AI proved why HHH(DD)==0 is correct in terms that
    any expert C programmer can understand. https://claude.ai/share/da9e56ba-f4e9-45ee-9f2c-dc5ffe10f00c

    If you can't prove the same then the AI has not proven it in
    a way that you can understand, possibly not at all.

    --
    Mikko

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Mikko@21:1/5 to olcott on Sun Aug 10 11:55:04 2025
    On 2025-08-09 14:40:35 +0000, olcott said:

    On 8/9/2025 1:42 AM, Fred. Zwarts wrote:
    Op 08.aug.2025 om 17:44 schreef olcott:
    On 8/8/2025 2:14 AM, Mikko wrote:
    On 2025-08-07 12:59:33 +0000, olcott said:

    On 8/7/2025 2:39 AM, Mikko wrote:
    On 2025-08-06 11:34:45 +0000, olcott said:

    On 8/6/2025 2:18 AM, Mikko wrote:
    On 2025-08-04 14:49:57 +0000, Mr Flibble said:

    First step needed to make progress with Olcott is to get Olcott to admit
    that he is not working on the Halting Problem or attempting to refute >>>>>>>>> Halting Problem proofs but is instead working on the Olcott Problem given
    his method involves partial rather than total deciders.

    Why should there be any progress with Olcott? The status quo is quite >>>>>>>> good as it is.

    Ah so you are only a troll.

    Are you really that stupid or are you just trolling?
    It is well known that trolling is very different from a desire to
    maintain status quo.

    When the status quo of failure to attain mutual agreement
    is "quite good" that indicates that the respondent is not
    interested in an honest dialogue.

    No honest dialogue is possible as long as your contribution is not
    honest. There doesn't seem to be much desire to any dialogue at
    all, whether honest or otherwise. People seem only to want to point
    out errors and dishonesty when they see them.


    Claude AI proved why HHH(DD)==0 is correct in terms that
    any expert C programmer can understand.
    https://claude.ai/share/da9e56ba-f4e9-45ee-9f2c-dc5ffe10f00c

    A proof based on incorrect assumption is not a proof. It seems your are
    the only one that dos not understand it.

    Exactly. Assuming that HHH(DD) must report on the
    behavior of DD() is the incorrect assumption of all
    of the proofs.

    No, there is no must in the assumptions.

    The thing to be proven is that HHH(DD) does not report on the behavour
    of DD().

    --
    Mikko

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