TITLE: Are undecidable consequences of Con recursively enumerable?
QUESTION [6 upvotes]: Let $X\subset\Pi_1^0$ be the set of statements which are 
provable in PA$+$Con(PA) but independent of PA. 
Is $X$ recursively enumerable?

REPLY [10 votes]: Here's a proof that doesn't directly diagonalize but instead relies on well-known results, which in turn were proved by diagonalization.  So ultimately, it isn't really easier than Emil's, but it may be easier to find and remember.
I claim first that, if a $\Pi^0_1$ sentence $\phi$ is provable in PA plus $\neg\text{Con}(PA)$, then it is already provable in PA.  This is probably well known, but here's a proof anyway.  The assumption is equivalent to saying that $\text{Con}(PA)$ is provable from PA plus $\neg\phi$.  But since $\neg\phi$ is a $\Sigma^0_1$ sentence, one can also prove from PA plus $\neg\phi$ that PA proves $\neg\phi$.  Combining the preceding two sentences, we get a proof from PA plus $\neg\phi$ that PA plus $\neg\phi$ is consistent.  By Gödel's second incompleteness theorem, it follows that PA plus $\neg\phi$ is inconsistent.  This means that PA proves $\phi$, as claimed.
Now consider the transformation $T$ on $\Pi^0_1$ sentences defined by letting $T(\phi)$ be $\text{Con}(PA)\lor\phi$.  I claim that $T(\phi)$ is in the set $X$ of the question if and only if PA does not prove $\phi$.  To see this, note first that $T(\phi)$ is trivially provable from PA plus $\text{Con}(PA)$.  So $T(\phi)\in X$ if and only if PA doesn't prove $\text{Con}(PA)\lor\phi$.  That's if and only if PA plus $\neg\text{Con}(PA)$ doesn't prove $\phi$.  And, by the claim proved above, that's if and only if PA doesn't prove $\phi$.  
So T is a (trivially computable) many-one (in fact one-one) reduction to $X$ of the set of $\Pi^0_1$ sentences not provable in PA.  The latter set is known not to be recursively enumerable; therefore neither is $X$.