Impossibility of theories proving consistency of each other?

Question

By Godel's second incompleteness theorem, a consistent theory (to which the theorem applies) cannot prove itself consistent. I learned that it's also impossible to have a pair of consistent theories each proving the consistency of the other. But I can't see how this follows from the second theorem. Or is there something more involved?

Answer 1

Assume that $A$ proves $B$ is consistent and $B$ proves $A$ is consistent. We want to prove that this means that $A$ proves $A$ is consistent.

We can take "A is consistent" to means "$0=1$ is not provable in $A$". To prove this by contradiction in $A$, we work in $A$ and assume "$0=1$ is provable in $A$". Then $B$ would prove "$0=1$ is provable in $A$", because $B$ is a sufficiently strong theory of arithmetic. More importantly, because $A$ is sufficiently strong, $A$ will prove that "$0=1$ is provable in $A$" implies "$B$ proves that '$0=1$ is provable in $A$'". Notice there are two levels of quoting there.

But we assumed that $B$ proves $A$ is consistent, which means that $B$ proves "$0=1$ is not provable in $A$". Moreover, because $A$ is sufficiently strong, $A$ proves that "$B$ proves that '$0=1$ is not provable in $A$'".

Now we have a contradiction: from the assumption "$0=1$ is provable in $A$", we have that $A$ proves both "$B$ proves that '$0=1$ is not provable in $A$'" and "$B$ proves that '$0=1$ is provable in $A$'". So, under that extra assumption, $A$ proves that "$B$ is inconsistent". But we also assumed $A$ proves "$B$ is consistent". So $A$ proves that the extra assumption is false, which means that $A$ proves "$0=1$ is not provable in $A$",

So, overall, under the usual assumptions that $A$ and $B$ are sufficiently strong, if $A$ proves $B$ is consistent and $B$ proves $A$ is consistent, then $A$ proves $A$ is consistent. That is impossible by the incompleteness theorem.