Re: ZFC solution to incorrect questions: reject them --Gödel--

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

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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.

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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:
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∀ Barber ∈ People. ¬(∀ Person ∈ People. Shaves(Barber, Person) ⇔ ¬Shaves(Person, Person))
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The following is a true statement:
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¬∃ Barber ∈ People. (∀ Person ∈ People. Shaves(Barber, Person) ⇔ ¬Shaves(Person, Person))
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 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.