# 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.

• Not a theorem, as answered below. Also not a theorem of $\mathbf D$. But if you are interested, you can try to prove $\Box (p \rightarrow q) \rightarrow (\Diamond p \rightarrow \Diamond q)$, which is a theorem of $\mathbf K$. Commented Dec 13, 2016 at 17:43

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.

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?

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.