Does the fact that a modal operator distributive over disjunction imply that a modal operator is distributive over conjunction?

Question

If L is an arbitrary operator on two propositions p and q: Does L(p $\vee$ q) $\Rightarrow$ Lp $\vee$ Lq imply L(p $\land$ q) $\rightarrow$ Lp $\land$ Lq?

Answer 1

No, not necessarily.

When changing the connective, the modality has to interact with the negation ($p \wedge q \equiv \neg (\neg p \vee \neg q)$), so it may behave differently. The negation may change the modality, $L$ may be changed to a dual modality $K$, with each modality having its own syntactic rules.

For example, the exponential in linear logic do not distribute over all connectives.

Answer 2

There is a similar, but more detailed answer on quora https://www.quora.com/Does-the-distributivity-of-a-modal-operator-over-disjunction-imply-distribution-over-conjunction/answer/Sebastian-Liu