modal logic proof question

Question

If we have $$ \vdash A \rightarrow B $$then we can derive $$ \vdash \Box A \rightarrow \Box B $$ By necessity

But we should then also be able to derive $$ \vdash \diamond A \rightarrow \diamond B $$

Intuitivly, it makes sense, but I am not sure how to prove it. Any help?

Answer 1

1) $\vdash A \to B$

2) $\vdash \lnot B \to \lnot A$

3) $\vdash \Box \lnot B \to \Box \lnot A$

4) $\vdash \lnot \Box \lnot A \to \lnot \Box \lnot B$.