• =?utf-8?Q?Re:_ZFC_solution_to_incorrect_questions:_reject_them_--G=C3=B

    From Mikko@21:1/5 to olcott on Wed Mar 13 12:03:26 2024
    On 2024-03-12 20:38:34 +0000, olcott said:

    On 3/12/2024 3:31 PM, immibis wrote:
    On 12/03/24 20:02, olcott wrote:
    On 3/12/2024 1:31 PM, immibis wrote:
    On 12/03/24 19:12, olcott wrote:
    ∀ H ∈ Turing_Machine_Deciders
    ∃ TMD ∈ Turing_Machine_Descriptions  |
    Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)

    There is some input TMD to every H such that
    Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)

    And it can be a different TMD to each H.

    When we disallow decider/input pairs that are incorrect
    questions where both YES and NO are the wrong answer

    Once we understand that either YES or NO is the right answer, the whole >>>> rebuttal is tossed out as invalid and incorrect.


    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
    BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩


    Once we understand that either YES or NO is the right answer, the whole
    rebuttal is tossed out as invalid and incorrect.

    Does the barber that shaves everyone that does not shave
    themselves shave himself? is rejected as an incorrect question.

    The barber does not exist.

    Russell's paradox did not allow this answer within Naive set theory.

    Naive set theory says that for every predicate P, the set {x | P(x)}
    exists. This axiom was a mistake. This axiom is not in ZFC.

    In Turing machines, for every non-empty finite set of alphabet symbols
    Γ, every b∈Γ, every Σ⊆Γ, every non-empty finite set of states Q, every
    q0∈Q, every F⊆Q, and every δ:(Q∖F)×Γ↛Q×Γ×{L,R}, ⟨Q,Γ,b,Σ,δ,q0,F⟩ is a
    Turing machine. Do you think this is a mistake? Would you remove this
    axiom from your version of Turing machines?

    (Following the definition used on Wikipedia:
    https://en.wikipedia.org/wiki/Turing_machine#Formal_definition)

    The following is true statement:

    ∀ Barber ∈ People. ¬(∀ Person ∈ People. Shaves(Barber, Person) ⇔
    ¬Shaves(Person, Person))

    The following is a true statement:

    ¬∃ Barber ∈ People. (∀ Person ∈ People. Shaves(Barber, Person) ⇔
    ¬Shaves(Person, Person))


    That might be correct I did not check it over and over
    again and again to make sure.

    The same reasoning seems to rebut Gödel Incompleteness:
    ...We are therefore confronted with a proposition which
    asserts its own unprovability. 15 ... (Gödel 1931:43-44)
    ¬∃G ∈ F | G := ~(F ⊢ G)

    Any G in F that asserts its own unprovability in F is
    asserting that there is no sequence of inference steps
    in F that prove that they themselves do not exist in F.

    The barber does not exist and the proposition does not exist.


    When we do this exact same thing that ZFC did for self-referential
    sets then Gödel's self-referential expressions that assert their
    own unprovability in F also cease to exist.

    Although Russel's set cannot be costructed in in ZFC Gödel's set can,
    thus proving that ZFC is incomplete and ZFC augmented with additional
    axioms is either incomplete or inconsistent.

    --
    Mikko

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Mikko@21:1/5 to olcott on Fri Mar 15 13:07:48 2024
    On 2024-03-13 14:25:10 +0000, olcott said:

    On 3/13/2024 5:03 AM, Mikko wrote:
    On 2024-03-12 20:38:34 +0000, olcott said:

    On 3/12/2024 3:31 PM, immibis wrote:
    On 12/03/24 20:02, olcott wrote:
    On 3/12/2024 1:31 PM, immibis wrote:
    On 12/03/24 19:12, olcott wrote:
    ∀ H ∈ Turing_Machine_Deciders
    ∃ TMD ∈ Turing_Machine_Descriptions  |
    Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)

    There is some input TMD to every H such that
    Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)

    And it can be a different TMD to each H.

    When we disallow decider/input pairs that are incorrect
    questions where both YES and NO are the wrong answer

    Once we understand that either YES or NO is the right answer, the whole >>>>>> rebuttal is tossed out as invalid and incorrect.


    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
    BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ >>>>>

    Once we understand that either YES or NO is the right answer, the whole >>>> rebuttal is tossed out as invalid and incorrect.

    Does the barber that shaves everyone that does not shave
    themselves shave himself? is rejected as an incorrect question.

    The barber does not exist.

    Russell's paradox did not allow this answer within Naive set theory.

    Naive set theory says that for every predicate P, the set {x | P(x)}
    exists. This axiom was a mistake. This axiom is not in ZFC.

    In Turing machines, for every non-empty finite set of alphabet symbols >>>> Γ, every b∈Γ, every Σ⊆Γ, every non-empty finite set of states Q, every
    q0∈Q, every F⊆Q, and every δ:(Q∖F)×Γ↛Q×Γ×{L,R}, ⟨Q,Γ,b,Σ,δ,q0,F⟩ is a
    Turing machine. Do you think this is a mistake? Would you remove this
    axiom from your version of Turing machines?

    (Following the definition used on Wikipedia:
    https://en.wikipedia.org/wiki/Turing_machine#Formal_definition)

    The following is true statement:

    ∀ Barber ∈ People. ¬(∀ Person ∈ People. Shaves(Barber, Person) ⇔
    ¬Shaves(Person, Person))

    The following is a true statement:

    ¬∃ Barber ∈ People. (∀ Person ∈ People. Shaves(Barber, Person) ⇔
    ¬Shaves(Person, Person))


    That might be correct I did not check it over and over
    again and again to make sure.

    The same reasoning seems to rebut Gödel Incompleteness:
    ...We are therefore confronted with a proposition which
    asserts its own unprovability. 15 ... (Gödel 1931:43-44)
    ¬∃G ∈ F | G := ~(F ⊢ G)

    Any G in F that asserts its own unprovability in F is
    asserting that there is no sequence of inference steps
    in F that prove that they themselves do not exist in F.

    The barber does not exist and the proposition does not exist.


    When we do this exact same thing that ZFC did for self-referential
    sets then Gödel's self-referential expressions that assert their
    own unprovability in F also cease to exist.

    Although Russel's set cannot be costructed in in ZFC Gödel's set can,
    thus proving that ZFC is incomplete and ZFC augmented with additional
    axioms is either incomplete or inconsistent.


    That is not how it works at all. Russell's paradox pointed out
    incoherence in the notion of a set. ZFC fixed that.

    The inability to show the a self-contradictory sentence is true or false
    is merely the inability to do the logically impossible and places no
    actual limit on anyone or anything.

    If a theory is complete there is a simple computable method to find out
    whether a particular sentence is a theorem or not. That method does not
    work with incomplete theories, and in many cases, including ZF and ZFC,
    no method works.

    --
    Mikko

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Mikko@21:1/5 to olcott on Sun Mar 17 17:44:49 2024
    On 2024-03-15 14:28:49 +0000, olcott said:

    On 3/15/2024 6:07 AM, Mikko wrote:
    On 2024-03-13 14:25:10 +0000, olcott said:

    On 3/13/2024 5:03 AM, Mikko wrote:
    On 2024-03-12 20:38:34 +0000, olcott said:

    On 3/12/2024 3:31 PM, immibis wrote:
    On 12/03/24 20:02, olcott wrote:
    On 3/12/2024 1:31 PM, immibis wrote:
    On 12/03/24 19:12, olcott wrote:
    ∀ H ∈ Turing_Machine_Deciders
    ∃ TMD ∈ Turing_Machine_Descriptions  |
    Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)

    There is some input TMD to every H such that
    Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)

    And it can be a different TMD to each H.

    When we disallow decider/input pairs that are incorrect
    questions where both YES and NO are the wrong answer

    Once we understand that either YES or NO is the right answer, the whole
    rebuttal is tossed out as invalid and incorrect.


    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
    BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ >>>>>>>

    Once we understand that either YES or NO is the right answer, the whole >>>>>> rebuttal is tossed out as invalid and incorrect.

    Does the barber that shaves everyone that does not shave
    themselves shave himself? is rejected as an incorrect question. >>>>>>>>
    The barber does not exist.

    Russell's paradox did not allow this answer within Naive set theory. >>>>>>
    Naive set theory says that for every predicate P, the set {x | P(x)} >>>>>> exists. This axiom was a mistake. This axiom is not in ZFC.

    In Turing machines, for every non-empty finite set of alphabet symbols >>>>>> Γ, every b∈Γ, every Σ⊆Γ, every non-empty finite set of states Q, every
    q0∈Q, every F⊆Q, and every δ:(Q∖F)×Γ↛Q×Γ×{L,R}, ⟨Q,Γ,b,Σ,δ,q0,F⟩ is a
    Turing machine. Do you think this is a mistake? Would you remove this >>>>>> axiom from your version of Turing machines?

    (Following the definition used on Wikipedia:
    https://en.wikipedia.org/wiki/Turing_machine#Formal_definition)

    The following is true statement:

    ∀ Barber ∈ People. ¬(∀ Person ∈ People. Shaves(Barber, Person) ⇔
    ¬Shaves(Person, Person))

    The following is a true statement:

    ¬∃ Barber ∈ People. (∀ Person ∈ People. Shaves(Barber, Person) ⇔
    ¬Shaves(Person, Person))


    That might be correct I did not check it over and over
    again and again to make sure.

    The same reasoning seems to rebut Gödel Incompleteness:
    ...We are therefore confronted with a proposition which
    asserts its own unprovability. 15 ... (Gödel 1931:43-44)
    ¬∃G ∈ F | G := ~(F ⊢ G)

    Any G in F that asserts its own unprovability in F is
    asserting that there is no sequence of inference steps
    in F that prove that they themselves do not exist in F.

    The barber does not exist and the proposition does not exist.


    When we do this exact same thing that ZFC did for self-referential
    sets then Gödel's self-referential expressions that assert their
    own unprovability in F also cease to exist.

    Although Russel's set cannot be costructed in in ZFC Gödel's set can, >>>> thus proving that ZFC is incomplete and ZFC augmented with additional
    axioms is either incomplete or inconsistent.


    That is not how it works at all. Russell's paradox pointed out
    incoherence in the notion of a set. ZFC fixed that.

    The inability to show the a self-contradictory sentence is true or false >>> is merely the inability to do the logically impossible and places no
    actual limit on anyone or anything.

    If a theory is complete there is a simple computable method to find out
    whether a particular sentence is a theorem or not. That method does not
    work with incomplete theories, and in many cases, including ZF and ZFC,
    no method works.


    To say that anything or anyone is in anyway limited or incomplete
    because they lack the ability to do the logically impossible is
    incorrect.

    Human knowledge is not incomplete on the basis of the lack of the
    ability to prove the Liar Paradox is true or false.
    "This sentence is not true." is not true, yet neither true nor false.

    The Liar Paradox is not truth bearer thus has no truth value.
    Tarski concluded that True(L,x) cannot be defined because it
    gets stumped on the Liar Paradox.

    Every sentence in the language of ZFC either is or is not provable.
    Anything else is a logical impossibility.

    There is no complete method to determine whether a sentence in the
    language of ZFC is provable. Existence of such method is a logical impossibility.

    --
    Mikko

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Mikko@21:1/5 to olcott on Mon Mar 18 11:57:23 2024
    On 2024-03-17 17:17:42 +0000, olcott said:

    On 3/17/2024 10:44 AM, Mikko wrote:
    On 2024-03-15 14:28:49 +0000, olcott said:

    On 3/15/2024 6:07 AM, Mikko wrote:
    On 2024-03-13 14:25:10 +0000, olcott said:

    On 3/13/2024 5:03 AM, Mikko wrote:
    On 2024-03-12 20:38:34 +0000, olcott said:

    On 3/12/2024 3:31 PM, immibis wrote:
    On 12/03/24 20:02, olcott wrote:
    On 3/12/2024 1:31 PM, immibis wrote:
    On 12/03/24 19:12, olcott wrote:
    ∀ H ∈ Turing_Machine_Deciders
    ∃ TMD ∈ Turing_Machine_Descriptions  |
    Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)

    There is some input TMD to every H such that
    Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)

    And it can be a different TMD to each H.

    When we disallow decider/input pairs that are incorrect
    questions where both YES and NO are the wrong answer

    Once we understand that either YES or NO is the right answer, the whole
    rebuttal is tossed out as invalid and incorrect.


    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
    BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ >>>>>>>>>

    Once we understand that either YES or NO is the right answer, the whole
    rebuttal is tossed out as invalid and incorrect.

    Does the barber that shaves everyone that does not shave >>>>>>>>>>> themselves shave himself? is rejected as an incorrect question. >>>>>>>>>>
    The barber does not exist.

    Russell's paradox did not allow this answer within Naive set theory. >>>>>>>>
    Naive set theory says that for every predicate P, the set {x | P(x)} >>>>>>>> exists. This axiom was a mistake. This axiom is not in ZFC.

    In Turing machines, for every non-empty finite set of alphabet symbols >>>>>>>> Γ, every b∈Γ, every Σ⊆Γ, every non-empty finite set of states Q, every
    q0∈Q, every F⊆Q, and every δ:(Q∖F)×Γ↛Q×Γ×{L,R}, ⟨Q,Γ,b,Σ,δ,q0,F⟩ is a
    Turing machine. Do you think this is a mistake? Would you remove this >>>>>>>> axiom from your version of Turing machines?

    (Following the definition used on Wikipedia:
    https://en.wikipedia.org/wiki/Turing_machine#Formal_definition) >>>>>>>>
    The following is true statement:

    ∀ Barber ∈ People. ¬(∀ Person ∈ People. Shaves(Barber, Person) ⇔
    ¬Shaves(Person, Person))

    The following is a true statement:

    ¬∃ Barber ∈ People. (∀ Person ∈ People. Shaves(Barber, Person) ⇔
    ¬Shaves(Person, Person))


    That might be correct I did not check it over and over
    again and again to make sure.

    The same reasoning seems to rebut Gödel Incompleteness:
    ...We are therefore confronted with a proposition which
    asserts its own unprovability. 15 ... (Gödel 1931:43-44)
    ¬∃G ∈ F | G := ~(F ⊢ G)

    Any G in F that asserts its own unprovability in F is
    asserting that there is no sequence of inference steps
    in F that prove that they themselves do not exist in F.

    The barber does not exist and the proposition does not exist.


    When we do this exact same thing that ZFC did for self-referential >>>>>>> sets then Gödel's self-referential expressions that assert their >>>>>>> own unprovability in F also cease to exist.

    Although Russel's set cannot be costructed in in ZFC Gödel's set can, >>>>>> thus proving that ZFC is incomplete and ZFC augmented with additional >>>>>> axioms is either incomplete or inconsistent.


    That is not how it works at all. Russell's paradox pointed out
    incoherence in the notion of a set. ZFC fixed that.

    The inability to show the a self-contradictory sentence is true or false >>>>> is merely the inability to do the logically impossible and places no >>>>> actual limit on anyone or anything.

    If a theory is complete there is a simple computable method to find out >>>> whether a particular sentence is a theorem or not. That method does not >>>> work with incomplete theories, and in many cases, including ZF and ZFC, >>>> no method works.


    To say that anything or anyone is in anyway limited or incomplete
    because they lack the ability to do the logically impossible is
    incorrect.

    Human knowledge is not incomplete on the basis of the lack of the
    ability to prove the Liar Paradox is true or false.
    "This sentence is not true." is not true, yet neither true nor false.

    The Liar Paradox is not truth bearer thus has no truth value.
    Tarski concluded that True(L,x) cannot be defined because it
    gets stumped on the Liar Paradox.

    Every sentence in the language of ZFC either is or is not provable.
    Anything else is a logical impossibility.

    There is no complete method to determine whether a sentence in the
    language of ZFC is provable. Existence of such method is a logical
    impossibility.


    ...We are therefore confronted with a proposition which
    asserts its own unprovability. 15 ...(Gödel 1931:43-44)

    ?- G = not(provable(F, G)).
    G = not(provable(F, G)).

    ?- unify_with_occurs_check(G, not(provable(F, G))).
    false.

    Prolog correctly detects a cycle in the evaluation graph of
    the above expression.

    *Is unprovable in F because of its pathological self-reference*
    Not because F is in any way incomplete.

    Per definitium F is incomplete if there is in the language
    of F a sentence that is neither a theorem nor the negation
    or any theorem. ZFC is that way incomplete.

    A theory is called unsovable if there is no complete method
    to determine whether a particular sentence is a theorem.
    In this sense ZFC is unsovable.

    --
    Mikko

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