# Validity of $(\lozenge\phi\wedge\lozenge\psi) \rightarrow (\lozenge(\phi\wedge\lozenge\psi)\vee\lozenge(\psi\wedge\lozenge\phi))$ in modal logic

I'm trying to answer a question, one part of which asks me to provide either an informal semantic argument or a counterexample to determine whether $$(\lozenge\phi\wedge\lozenge\psi) \rightarrow (\lozenge(\phi\wedge\lozenge\psi)\vee\lozenge(\psi\wedge\lozenge\phi))$$ is a valid formula in system K.

It then asks me, for those parts of the question I've found to be false, to determine what conditions I can impose on the accessibility relation in order to make it valid.

I haven't been able to come up with either a proof or a counterexample, so I'd really appreciate any help you could offer. If it turns out that it isn't valid, then some sort of hint for what conditions are necessary would be really helpful too. Thanks!

Consider a frame with three nodes $$\{a,b,c\}.$$ Node $$a$$ has $$b$$ and $$c$$ accessible from it (and $$a$$ is not accessible from $$a).$$ No nodes are accessible from $$b$$ and $$c.$$ At node $$b,$$ $$\phi$$ is true and $$\psi$$ is false, whereas at node $$c,$$ $$\phi$$ is false and $$\psi$$ is true. The statement in question is not true at node $$a,$$ so it is not valid.
One well-known (but strong) frame condition that guarantees the statement is valid is that the accessibility relation is an equivalence relation, i.e. system S5. In this case, if $$\lozenge\phi$$ and $$\lozenge \psi$$ hold at some node $$a$$, then the accessible worlds $$b$$ and $$c$$ at which $$\phi$$ and $$\psi$$ hold respectively are accessible to each other, so $$\lozenge\phi$$ and $$\lozenge \psi$$ hold at both $$b$$ and $$c.$$ Thus, even $$\lozenge(\phi\land \lozenge \psi)\land \lozenge(\psi\land \lozenge\phi)$$ (with a $$\land,$$ not a $$\lor$$) holds at $$a.$$
A weaker condition that will get us only what we need is that if $$aRb$$ and $$aRc$$ then $$bRc$$ or $$cRb.$$ (On Wikipedia, they call this $$H$$ and note that it is given by the axiom $$\square (\square A\to B)\lor \square(\square B\to A).$$)