How to show $\vdash_{\mathbf{S5}} \Diamond \Box p\to \Box p$?

Question

How to show $\mathbf{S5}$-proof of $\Diamond \Box p\to \Box p$? It is a part of exercise 1.6.3 in Blackburn's Modal Logic.

Answer 1

I assume your definition of $\mathbf{S5}$ is the normal modal logic K together with \begin{align*} (T)&\colon \square P\rightarrow P\\ (4)&\colon \square P\rightarrow \square \square P\\ (B)&\colon P\rightarrow \square \lozenge P \end{align*}

Claim: $\lozenge \square p\rightarrow \square p$.

Proof sketch: Assume $\lozenge \square p$. By $(4)$, $\lozenge \square \square p$. By the contrapositive of $(B)$, $\square p$.

In a bit more detail: As an instance of $(4)$, $\square p\rightarrow \square \square p$ is a theorem. By necessitation, $(K)$, and contrapositive, we have $\lozenge\square p\rightarrow \lozenge\square \square p$. As an instance of $(B)$, $\lozenge p \rightarrow \square \lozenge\lozenge p$ is a theorem. The contrapositive of this is $\lozenge \square \square p\rightarrow \square p$. Composing these implications, we have $\lozenge \square p\rightarrow \square p$, as desired.