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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?

2 Answers2

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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.

Noah Schweber
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    Huh I would not call the $T$-axiom mild. Take into account that in PA assuming $Bew(A)\rightarrow A$ makes the logic inconsistent. – Jsevillamol Jun 06 '17 at 17:17
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    @Jsevillamol That's a very good point - quite silly of me! I almost always work with reflexive frames, where $T$ trivially holds, but of course things get much more interesting if we look at provability-type logics. I've edited my answer to not be nonsense. – Noah Schweber Jun 06 '17 at 17:28
  • @NoahSchweber Are you able to provide a literature reference for such Kripke frames and this class of axioms? – Jim Sep 14 '23 at 18:57
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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).

Confutus
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