Is $(\Diamond p \rightarrow \Diamond q)\rightarrow \Diamond (p \rightarrow q)$ invalid in K? Is it true that the formula bellow is invalid in K: $(\Diamond p \rightarrow \Diamond q)\rightarrow \Diamond (p \rightarrow q)$ Because we could construct a counter model where $(\Diamond p \rightarrow \Diamond q)$ is true by having $\Diamond p$ false that would make $(\Diamond p \rightarrow \Diamond q$) hold. Such a model would be for example a model with a single world with no successors with no valuation. Then in this model $\Diamond (p \rightarrow q)$ does not hold, because there does not exist a successor of x such that xRy and that in y $p \rightarrow q$ holds Is this a correct counterexample?

If you regard a model with just **one possible world** and your accesibility relation $R$ is **reflexive** ($wRw$ for all possible worlds), then your counter example works:

Say $\Diamond p$ **false**

$\Rightarrow$ $p$ **false** in $w$ (where $w$ is the only / static world) ..since $R$ is reflexive

$\Rightarrow$ $p \rightarrow q$ **true** in $w$

$\Rightarrow$ $\Diamond (p \rightarrow q)$ **true** in $w$ ..since $R$ is reflexive

$\Rightarrow$ $(\Diamond p \rightarrow \Diamond q) \rightarrow \Diamond (p \rightarrow q)$ **true** in $w$ ..since $(\Diamond p \rightarrow \Diamond q)$ and $\Diamond (p \rightarrow q)$ are **true** in $w$