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On 3/12/24 8:14 PM, olcott wrote:It this was true then it would be a little nuts for anyoneOn 3/12/2024 9:51 PM, Richard Damon wrote:CAN'T EXIST.On 3/12/24 4:14 PM, olcott wrote:>On 3/12/2024 6:00 PM, Richard Damon wrote:>On 3/12/24 2:44 PM, olcott wrote:>On 3/12/2024 4:31 PM, Richard Damon wrote:>On 3/12/24 1:38 PM, olcott wrote:>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)
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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.
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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⟩
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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.
>>>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.
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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?
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(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.
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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)
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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.
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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.
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And you end up with a very weak logic system that can't even have the full properties of the Natuarl Numbers.
Natural numbers never really did have the property of provability.
This was something artificially contrived that never really belonged
to them.
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No, Godel showed (or maybe used a previous proof) that you can use the Mathematics of Natural Numbers to test if a proof is valid.
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You just don't understand it. It really is very related to how Turing Machines work, which can be converted to a mathematical model.
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There is a field that looks at the comparability of Computations to Logic, so they are all really quite related.
This is refuted.
...We are therefore confronted with a proposition which
asserts its own unprovability. 15 ...(Gödel 1931:43-44)
Right, seen in the meta-Theory from F.
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based on immblis Russell's Paradox reply
¬∃G ∈ F | G ↔ ~(F ⊢ G) // is simply false
Then a proof must exist in F that G is True, So G can't be false.
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Note, The statement G is NOT a statement about itself being provable, that is only a semantic revealed in the RIGHT meta-theory.
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*Not in my actual example where F has its own provability operator*
Provability in uncomputable.
This requires that:>It doesn't.No one can prove that they themselves do not exist.>>
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.
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No FINITE sequence of inference steps.
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Thus G cannot possibly derive a sequence of inference
steps that prove that they themselves do not exist.
It shows that no proof exists of its truth.There cannot possibly be any set of inference steps that prove
It also shows that it can't be false, as if it were, then it would be proven to be true, and you can't prove a false statement true in a consistant system.
I am talking about this.>Then you aren't talking about Godel, who you mentioned just before.Note, the key is that G doesn't assert that, G is a statement about math, that only when interpreted in a meta-theory sees that.I am not referring to that one.
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So, you LIE.*That lie might possibly cost you your soul*
Great finally some more mutual agreement.>Right, You can't have, IN THE SYSTEM, a statement that something must be true and unprovable, as that is, itself, a contradiction.G is actually a statment about the existance of a number that matches a complected and carefully constructed relationship, that is fully computable. The Existance or non-existance of such a number is a pure binary thing, either it WILL or it WON'T.No formal system can possibly have any sequence of inference
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The complicated relationship deals with a way to encode as a number ANY analyitic statement in F (since all statements are just strings, and strings can be encoded into a number), and a calculation to see if that statement actually is a proof starting with the enumerated truth makers of F, through the valid and allowed logical operatons to the statement of G. A number that satisfies this relation, encodes a proof of G, and any such proof, can always be encoded in to a number.
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step that prove that they themselves do not exist because
this is self-contradictory, not because the system is incomplete.
In my G this is obvious.
When true means provable from axioms then unprovable means>No, you have proved that a direct statement in F, can't assert that a statement is True but Unprovable.
Whether or not Gödel's G is isomorphic to mine I have proved
this this is false by providing the counter-example of my G.
But Godel G, doesn't directly assert it, but creates mathematics based on a meta-theory that establish it.*This directly asserts it*
That is the power of Mathematics.That hardly anyone knows that epistemological antinomies
>Which you don't understand, and is refering to logic in Meta-F, which you don't handle at all.
...14 Every epistemological antinomy can likewise be used for a similar undecidability proof...(Gödel 1931:43-44)
...We are therefore confronted with a proposition which>It shows that F defines an epistemological statement to be true, which is equivent to F being inconsistant.G is the statement that no such number exist. So, if G is false, then such a number exists, and then in the Meta-Theory, we can decode that number into a proof in F that G must be true.>
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If this is the case, then F must be inconsistant, and the proof starts with the presumption that we are dealing with a consistant logic system.
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Or it could be the epistemological antinomy of trying to find a
sequence of inference steps in F that prove that they themselves
do not exist.
As you point out, you can't make the statement in F, you need a level of indirection with a meta-system to make it work.--
>You are repeating yourself, showing you self to be just the CHILDISH pathetic ignorant hypocritical pathological lying idiot.But, if G is true, then there can be no such number, and thus from our knowledge in the mete-theory, we know that no proof CAN exist for that fact.Or it could be the epistemological antinomy of trying to find a
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Thus, the only possiblity for this statement, is to be True, but unprovable.
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sequence of inference steps in F that prove that they themselves
do not exist.
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Meta-F can see this and F is not incomplete if it cannot accomplish
an epistemological antinomy.
>The big part of the theory is showing that such a relationship can be made, and that shows that proof checkers are a computable function.>
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Given a system, and an enumeration of its basic truths, we can check if a proof present actually proves the result it claims to.
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