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From: byron on 11 Nov 2009 03:57
the fault in godels semantic proof is that he constructs a self
referential proposition ie Rq(q) states" I am improvable."

many mathematicians and philosophers ponicare russell and wittgenstein
etc argue self referential propositions only lead to contradiction
and thus must not be used

Wittgenstein shows just that ie godels self referential propisition
leads to contradiction his semantic proof

http://plato.stanford.edu/entries/wittgenstein-mathematics/#WitGodUndMatPro

To refute or undermine this ‘proof,’ Wittgenstein says that if you
have proved ¬P, you have proved that P is provable (i.e., since you
have proved that it is not the case that P is not provable in
Russell's system), and “you will now presumably give up the
interpretation that it is unprovable” (i.e., ‘P is not provable in
Russell's system’), since the contradiction is only proved if we use
or retain this self-referential interpretation (§8).

Thus, Wittgenstein's ‘refutation’ of “Gödel's proof” consists in
showing that no contradiction arises if we do not interpret ‘P’ as ‘P
is not provable in Russell's system’—indeed, without this
interpretation, a proof of P does not yield a proof of ¬P and a proof
of ¬P does not yield a proof of P. In other words, the mistake in the
proof is the mistaken assumption that a mathematical proposition ‘P’
“can be so interpreted that it says: ‘P is not provable in Russell's
system’.” As Wittgenstein says at (§11), “[t]hat is what comes of
making up such sentences.

godels semantic prooof is

http://www.research.ibm.com/people/h/hirzel/papers/canon00-goedel.pdf
We will now prove that the theorem Rq(q) is
undecidable within PM. We can understand this by simply plugging in
the de nitions: Rq(q) , S(q) , q 2 K , :provable(Rq(q)), in other
words, Rq(q) states \I am improvable."
Assuming the theorem Rq(q) were provable, then it would also be true,
i.e. because of (1) :provable(Rq(q)) would be true in contradiction to
the assumption. If on the other
hand :Rq(q) were provable, then we would have q 62 K, i.e. provable(Rq
(q)). That means that both Rq(q) and :Rq(q) would be provable, which
again is impossible.The analogy of this conclusion with the Richard-
antinomy leaps to the eye; there
is also a close kinship with the liar-antinomy, because our
undecidable theorem Rq(q) states that q is in K, i.e. according to (1)
that Rq(q) is not provable. Hence, we have in front of us a theorem
that states its own unprovability.

From the remark that Rq(q) states its own improvability it immediately
follows that Rq(q) is correct, since Rq(q) is in fact unprovable
(because it is undecidable).
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