Is the distribution axiom in the modal system K an equivalence?

Question

It's all in the title, really. Is $(\square p \to \square q) \to \square (p \to q)$ a theorem of system K?

As a follow-up, is it true if we interpret $\square$ as the provability predicate of Peano arithmetic?

Answer 1

No. $\square p\to \square q$ holds whenever $\square p$ is false. This is a very weak statement and by no means implies $\square(p\to q),$ which is very strong. Here is a very simple countermodel: let $a$ be a world accessible to itself where $p$ is false. Then let $b$ be a world accessible to $a$ where $p$ is true and $q$ is false.

For similar reasons this does not hold for provability. Let $p$ be some unprovable statement that is presumably consistent with PA (like $Con(PA),$ for instance). And let $q$ be $0=1.$