Re: Incompleteness: Will there always be unprovable statements?

Scott H says...

Many of us are in the habit of saying that any system strong enough to
support arithmetic must be incomplete. But what about the theorem that any
consistent first-order theory has a consistent and complete extension?

If you look at Godel's proof, you see that there is a technical requirement
that theoremhood be definable. That is, there has to be a formula Provable(x)
with the interpretation: x is a code for a theorem. If the theory is a
recursively axiomatizable extension of Peano Arithmetic, then there will
be such a formula, but if the theory is not recursively axiomatizable, then
there may be no such formula. For example, let T = the set of all true
statements of arithmetic. Then there is no formula Provable(x) in the
language of arithmetic formalizing "x is a code for a true statement
of arithmetic". That's a consequence of Tarski's undefinability theorem,
http://en.wikipedia.org/wiki/Tarski's_indefinability_theorem.

You *can* define arithmetical truth in ZF, however. But then you can't
define truth in ZF inside ZF.

--
Daryl McCullough
Ithaca, NY

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