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

Is it possible for a theory $T$ to be consistent, but for $T$ + $\mathop{Con(T)}$ to be inconsistent?

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.
• @Demi Note that "$T+Con(T)$ is inconsistent" means exactly "$T$ proves $\neg Con(T)$" (by the Completeness Theorem). So if you want a natural example, you need a natural theory $T$ which proves its own inconsistency. Such a theory would not be very interesting from a standard mathematics perspective, since it is clearly incorrect (either $T$ is consistent, in which case it's wrong about its own inconsistency; or $T$ is not consistent, in which case $T$ proves lots of incorrect things). So I'd say that you aren't going to get natural examples. (cont'd) – Noah Schweber Oct 1 '16 at 23:52
• That said, note that we can't easily prove that e.g. $ZFC$ doesn't prove $\neg Con(ZFC)$ - note that this would imply that $ZFC$ is consistent! So we'd need a theory capable of proving $Con(ZFC)$ already, in order to do this. – Noah Schweber Oct 1 '16 at 23:53
• @user1952009 Every axiom of $T$ is also an axiom of $T'$, so any contradiction in $T$ is also a contradiction in $T'$. Moreover, $T'$ can prove the previous sentence (assuming $T$ is strong enough - say, $T$ is at least as strong as $I\Sigma_1$, a tiny subtheory of $PA$, although even this is almost certainly overkill). So since $T'$ proves "$T$ is inconsistent," $T'$ also proves "$T'$ is inconsistent." – Noah Schweber Oct 1 '16 at 23:54