Incompleteness: ~G, the proof of G, and the numeral of G's proof
- From: "Scott H" <nospam>
- Date: Sat, 26 Apr 2008 20:32:52 -0400
I have another question.
Let T be a consistent theory supporting arithmetic, T' be its reflection in
T, and T'' its reflection in T'. Let G be Goedel's undecidable statement in
T, [~Pr Aq[~Pr Aq x]].
By the incompleteness theorem, T + ~G is consistent. However, adding ~G to T
does not amount to adding ~G' to T' or ~G'' to T''. T' and T'' remain the
same, the latter a reflection of the former.
Since T + ~G doesn't specify a proof of G' in T' yet, let's add a constant
symbol c to T along with the axiom, [Pr G' => Pr(c, G')].
If T'' is still a reflection of T', based on the same axioms, then how can
G' be provable in T' without G'' being provable in T'' (contradicting G')? A
proof of G' in T', c, would not have a numeral in T'' that proves G'' --
assuming we can speak of c's numeral. c's status in T'' suddenly changes.
As was explained to me earlier, the definition of omega-consistency accounts
only for numerals and not other constant symbols. Therefore, T is still not
omega-consistent after the addition of c and its axiom because c isn't a
numeral in T. However, since c inherits all mathematically inductive
properties in T -- supposedly including "primitive recursive
definability" -- wouldn't c have a numeral in T'? How can c be proof of G'
in T' but not have a numeral in T'' that would prove G''?
.
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