Disjunction and conjunction in modal logic Is there a system of modal logic in which
$\Box (A\vee B)\leftrightarrow\Box A\vee\Box B$,
$\Box (A\wedge B)\leftrightarrow\Box A\wedge\Box B$,
$\Diamond (A\vee B)\leftrightarrow\Diamond A\vee\Diamond B$,
$\Diamond (A\wedge B)\leftrightarrow\Diamond A\wedge\Diamond B$
hold?
Alternatively, is there a modal system in which
$\Box (A\vee B)\leftrightarrow\Box A\vee\Box B$
and
$\Diamond (A\wedge B)\leftrightarrow\Diamond A\wedge\Diamond B$?
are theorems?
 A: A quick comment - such systems correspond to a very narrow class of Kripke frames.
Think about the formula "$\Box(A\vee \neg A)$." This is a tautology in most modal systems, and so by the rule "$\Box(A\vee B)\iff\Box A\vee\Box B$", a system with your property would satisfy "$\Box A\vee\Box \neg A$" for all sentences $A$.
From this, it's easy to that the Kripke frames satisfying this rule are exactly those in which every world sees exactly one world (possibly itself). In particular, assuming rule "$T$" ($\Box A\implies A$), we have $\Box A\iff A$ and so the modalities trivialize.
A: In a system of modal logic based on Lukasiewicz 3-valued logic, these are all theorems.
Lp corresponds to $\Box$p
Mp corresponds to $\Diamond$p
In this system, $\Box$ (p $\lor$~p) is not a theorem. It is a statement of the law of bivalence, which does not generally hold in L3. It can be regarded as a contingent statement, applicable to some but not all statements. This version of modal logic also requires that one swallow the existence of propositions for which $\Diamond$(p $\land$ ~p).
