Liste des Groupes |
On 3/13/2024 5:03 AM, Mikko wrote:If a theory is complete there is a simple computable method to find outOn 2024-03-12 20:38:34 +0000, olcott said:That is not how it works at all. Russell's paradox pointed out
On 3/12/2024 3:31 PM, immibis wrote:Although Russel's set cannot be costructed in in ZFC Gödel's set can,On 12/03/24 20:02, olcott wrote:When we do this exact same thing that ZFC did for self-referentialOn 3/12/2024 1:31 PM, immibis wrote:Once we understand that either YES or NO is the right answer, the whole rebuttal is tossed out as invalid and incorrect.On 12/03/24 19:12, olcott wrote:Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts∀ H ∈ Turing_Machine_DecidersAnd it can be a different TMD to each H.
∃ 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)
When we disallow decider/input pairs that are incorrectOnce we understand that either YES or NO is the right answer, the whole rebuttal is tossed out as invalid and incorrect.
questions where both YES and NO are the wrong answer
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⟩
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.Russell's paradox did not allow this answer within Naive set theory.Does the barber that shaves everyone that does not shaveThe barber does not exist.
themselves shave himself? is rejected as an incorrect question.
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 barber does not exist and the proposition does not exist.The following is true statement:That might be correct I did not check it over and over
∀ 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))
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.
sets then Gödel's self-referential expressions that assert their
own unprovability in F also cease to exist.
thus proving that ZFC is incomplete and ZFC augmented with additional
axioms is either incomplete or inconsistent.
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.
Les messages affichés proviennent d'usenet.