Theory $T$ that is consistent, such that $ T + \mathop{Con}(T)$ is inconsistent

Solution 1:

Yes. If $T$ is any consistent theory to which Goedel's theorems apply (e.g. $PA$, $ZFC$, etc.), then the theory $T'=T+\neg Con(T)$ is also consistent. But $T'$ proves "$T'$ is inconsistent," since any contradiction in $T$ is also a contradiction in $T'$ since $T'\supseteq T$. (See also the comment threads here.)

Note that just because a theory $S$ proves "$S$ is inconsistent," does not mean that $S$ is inconsistent! This is a common stumbling point. The point of the above paragraph is exactly that any "reasonable" theory gives rise to a consistent theory proving its own inconsistency.

Why is $T'$ consistent? Well, suppose it weren't. Then by the Soundness Theorem that would mean that every model of $T$ satisfied "$Con(T)$". Then by the Completeness Theorem (no, that's not a typo :P), we would have $T$ proves $Con(T)$ - but this contradicts the Incompleteness Theorem.