Godel DISPROOF
From: |-|erc (gotcha_at_beauty.com)
Date: 06/08/04
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Date: Tue, 08 Jun 2004 07:29:24 GMT
"The existence of G is equivalent to inconsistency of T, where T is
a Factual_Consistent_Model"
"Daryl McCullough" <daryl@atc-nycorp.com> wrote
> |-|erc says...
>
> >> >Is G true?
> >> >No, it has no proof therefore it can't be declared true.
>
> [stuff deleted]
>
> >Truth is what's proven.
>
> Can you prove that?
yes
truth
absolute truth
substantiated truth
derived truth
proven truth
proof
which line is this sequence of equivalent statements is erronous?
>
> >We assume Proof(x) is a standard formula. Godel *then* derived it
> >is impossible.
>
> Both of those statements are false. Godel did not *assume*
> that there is a formula in PA defining provability, he *derived*
> such a formula. And he didn't prove that such a formula is
no, he made minimal assumptions about the theory, he had NO
internal workings of Proof() bar its numerical indexing.
> impossible---he proved that such a formula is incomplete---there
> are some statements that are not provable, and are also not
> disprovable.
meaning the *total* function Proof() is impossible.
>
> >If you are examining the proof again you can't assume
> >Proof(x) is impossible.
>
> If you are talking about the fact that if G says that it
> is not provable, then G must be true, that doesn't count
> as a proof of G, because it has an additional assumption:
> that Peano Arithmetic is consistent. What you can prove
> is that
>
> If PA is consistent, then G
>
> But you can't prove G outright.
Then split the models in half.
CASE 1/ PA is consistent
Godel nomenclature without regard to PAs consistency
CASE 2/ PA is not consistent
other Godel nomenclature
BECAUSE ALL YOUR STATEMENTS ARE RELIANT ON THAT CLAUSE!
>
> >anything that is proven is then TRUE,
>
> Yes, PA was constructed so that it can only prove true
> statements about arithmetic. But the question is whether
> this fact about PA can be proved *using* PA (and nothing more).
> It can't.
cyclic internal catch 22 errors abound in computer theory
>
> >"If G has no proof then it is true"
> << this is a proof, a self defeating proof.
> >
> >Required to prove G.
> >CASE 1
> >G has no proof
> >if G has no proof then it is true
> >G has no proof -> G
> >Modus Ponens
> >G
>
> That's correct. If G has no proof, then G is true.
>
> >CASE 2
> >G has a proof
> >AX, X has a proof -> X
>
> That statement is not provable in Peano Arithmetic.
>
what on earth are you going on about?
you don't even have a defn of proof, how cares about peano arithmetic
its the domain of predicate calculus.
its the most trivial implication in maths, and is this where your argument
holds or is this another rabbit run into obscurity to spend 100s of posts
defining the domain of PROOF(x) -> x then its not even part of your
argument anyway?
Proof has to do with axioms and rules of inference.
Truth has to do with interpretations of sentences.
You can read a sentence and know its true, it doesn't require a proof?
Whats a formula parser for then? when do you know the truth from an axiom,
a proof of a sentence?
This is the garbage BELIEVING "this statement G has no proof" and
DISBELIEVING Proof(x)<->x will lead you to conclude.
"Aatu Koskensilta" <aatu.koskensilta@xortec.fi> wrote in
> |-|erc wrote:
>
> > (1) T |- A <=> ~A
> >
> > If T |- A, then by (1) T |- ~A. But since ~ represents
> > falsity in T, and since A is Pi_1 and T can only prove true
> > Pi_1 statements, we would have T |/- A, which contradicts the assumption
> > that T |/- A. Since (1) asserts that A is equivalent to its own
> > falsity, we see that A is actually true
>
> I showed - in outline - how to construct a sentence G, s.t.
>
> T |- G <=> ~\phi("G")
>
> Please outline a construction for an A, s.t.
>
> T |- A <=> ~A
>
> This would be an interesting feat, since the existence of such an A is
> equivalent to to inconsistency of T, which contradicts the assumption
> that T is consistent.
I'll do the opposite on route to what I'm asserting,
I'll show why G is in the same class as A.
T |- C <=> \phi("X") <-> X
T |- G <=> ~\phi("G")
> >>(1) T |- G <=> ~\phi("G")
> >>
> >>If T |- G, then by (1) T |- ~\phi("G"). But since \phi represents
> >>provability in T, and since \phi("G") is Pi_1 and T can only prove true
> >>Pi_1 statements, we would have T |/- G, which contradicts the assumption
> >>that T |/- G. Since (1) asserts that G is equivalent to its own
> >>unprovability, we see that G is actually true.
G
By T |- C, G -> \phi("G"), since \phi represents provability in T we would have
by T |- G <=> ~\phi("G"), then \phi("G") -> ~G we have G is false.
~G
Therefore the existence of G is equivalent to inconsistency of T, where T is
a Factual_Consistent_Model where \phi(formula)<->formula, containing only proven facts.
Herc
-- unless being willing to be crucified again. Of course Jesus can come back, but it would be a terrible abuse, people would attack, ridicule, lie, compaign about the person, and they would call for psychs. And they would come to get Jesus committed, drugged, e-schocked before the comeback would be known to all of the people. Barbara Schwarz
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