Is $\Diamond (p \rightarrow q) \rightarrow (\Diamond p \rightarrow \Diamond q)$ valid in K? The modal logic K is the weakest normal modal system, comprised by classic logic augmented by (K), the necessity distribution axiom schema:
$$\Box (\alpha \rightarrow \beta) \rightarrow (\Box \alpha \rightarrow \Box \beta) \tag{K} $$
We also define the possibility operator as the necessity's dual: 
$$\Box \alpha \equiv\neg \Diamond \neg \alpha $$
Playing with the equivalence for a while, it's easy to prove that, for example, the proposition below is a theorem of K:
$$\vdash_K \Box (p \rightarrow q) \rightarrow (\Diamond p \rightarrow \Diamond q)$$
However, I've been struggling to prove something stronger

$$\Diamond (p \rightarrow q) \rightarrow (\Diamond p \rightarrow \Diamond q)$$

Is it even provable in K? After so many failed attempts, my guess so far is that you need D to derive it. But I'm not sure.
Can you provide a proof/disproof of this proposition? A semantic proof of its vality/invality is also welcome.
Thanks in advance
 A: No, it's reasonably easy to think of a model where it is not true that the possibility of the implication implies the implication of the possibilities.
Take any model where $\Diamond \neg p, \Diamond p, \Box \neg q$ hold.   You can show both that $\Diamond (p\to q)$ and that $\neg (\Diamond p\to \Diamond q)$ are true there in.
Thus $\Diamond(p\to q)\to(\Diamond p\to \Diamond q)$ is not a tautology.
A: No, it isn't valid in $\text K$.
Here's a Kripke semantics countermodel:
$W=\{ w_{0},w_{1} \},\; R=\{(w_{0},w_{0}), (w_{0},w_{1}) \}  ,\; \nu_{w_{0}}(p)=1,\nu_{w_{0}}(q)=0, \; \nu_{w_{1}}(p)=\nu_{w_{1}}(q)=0. $
Will you be fine with checking it yourself?
A: It might be helpful to look at this as (a formula version of) modus ponens under possibility: it would say that if it's possible that $p\to q$ and it's possible that $p$, then it's possible that $q$.  But the two hypothetical possibilities could be incompatible with each other, right?  What we'd need is for it to be possible that ($p\to q$ and $p$), which is not the same.
