Accessibility Relation for Modal Logic I have a simple question. To which accessibility relation does the modal formula $$□(P\lor \lnot P)\iff (□P\lor □\lnot P)$$ correspond? Note that this doesn’t hold for an arbitrary disjunction, but just for instances of the Principle of the Excluded Middle.
 A: I shall present a discussion for exclusive disjunction (xor, denoting it by $\underline{\vee}$). Variations on it can proceed along the same line of thought.
We want to see  what accessibility relation is required for the distributivity property of $\underline{\vee}$ over the necessity operator $\Box$ to hold and vice versa:
$\Box(A\underline{\vee} B)\overset{?}{\leftrightarrow}\Box A\underline{\vee}\Box B\tag{1}$
Let us record first the distributivity laws of modal operators over conjunction and disjunction (in what follows, we shall drop parentheses in accordance with the usual binding power of operators):
$\Box(A\wedge B)\leftrightarrow \Box A\wedge\Box B\tag{2}$
$\Diamond(A\wedge B)\rightarrow\Diamond A\wedge\Diamond B\tag{3}$
$\Box A\vee\Box B\rightarrow \Box(A\vee B)\tag{4}$
$\Diamond(A\vee B)\leftrightarrow\Diamond A\vee \Diamond B\tag{5}$
Notice that 3 and 4 are not reciprocal.  We shall also convert $\Box\phi$ and $\neg\Diamond\neg\phi$ to and fro by definition. Additionally, we shall use the following equivalences (referring to the convenient forms of De Morgan's laws as (7) and (8))
$A\underline{\vee} B\leftrightarrow (A\wedge\neg B)\vee(\neg A\wedge B)\tag{6}$
$A\vee B\leftrightarrow \neg(\neg A\wedge\neg B)\tag{7}$
$A\wedge B\leftrightarrow \neg(\neg A\vee\neg B)\tag{8}$
We begin with the left-to-right direction of (1). Invoking (6), we shall try to derive eventually
$\Box(A\underline{\vee} B)\rightarrow (\Box A\wedge\neg\Box B)\vee(\neg\Box A\wedge\Box B)\tag{$\ast$}$
We rewrite the antecedent of $\ast$:
$\rightarrow\Box((A\wedge\neg B)\vee(\neg A\wedge B))\tag{by 6}$
$\rightarrow\Box\neg(\neg(A\wedge\neg B)\wedge\neg(\neg A\wedge B))\tag{by 7}$
$\rightarrow\neg\Diamond(\neg(A\wedge\neg B)\wedge\neg(\neg A\wedge B))\tag{by definition}$
$\rightarrow\neg(\Diamond\neg(A\wedge\neg B)\wedge\Diamond\neg(\neg A\wedge B))\tag{by 3}$
$\rightarrow\neg(\neg\Box(A\wedge\neg B)\wedge\neg\Box(\neg A\wedge B))\tag{by definition}$
$\rightarrow\neg(\neg(\Box A\wedge\Box\neg B)\wedge\neg(\Box\neg A\wedge\Box B))\tag{by 2}$
$\rightarrow\neg(\neg(\Box A\wedge\neg\Diamond B)\wedge\neg(\neg\Diamond A\wedge\Box B))\tag{by definition}$
$\rightarrow(\Box A\wedge\neg\Diamond B)\vee(\neg\Diamond A\wedge\Box B)\tag{$\ast\ast$}$
Comparing $\ast$ and $\ast\ast$, we see that we need to stipulate the accessibility relation $R$ so that $\Box\phi\leftrightarrow\Diamond\phi$ holds in order to obtain $\ast$. Actually, we need $\Diamond\phi\rightarrow\Box\phi$ at this stage.
By the requirements of possible worlds semantics for $\Box$ and $\Diamond$, this is possible only if  $R$ is a functional (univalent) relation: For each $w$, there is exactly one $w′$ such that $wRw′$. Hence
$$\forall w_{i}\forall w_{j}\forall w_{k}(w_{i}Rw_{j}\wedge w_{i}Rw_{k}\rightarrow w_{j} = w_{k})$$
Notice that a particular instance satisfying this requirement is that $R$ is a reflexive identity relation. So, we turn to right-to-left direction of (1). Invoking (6), we shall try to derive eventually
$\Box A\underline{\vee}\Box B\rightarrow\Box((A\wedge\neg B)\vee(\neg A\wedge B))\tag{$\dagger$}$
We rewrite the antecedent of $\dagger$:
$\rightarrow((\Box A\wedge\neg\Box B)\vee(\neg\Box A\wedge\Box B))\tag{by 6}$
$\rightarrow\neg(\neg\Box A\vee\neg\neg\Box B)\vee\neg(\neg\neg\Box A\vee\neg\Box B))\tag{by 8}$
$\rightarrow\neg(\neg\Box A\vee\Box B)\vee\neg(\Box A\vee\neg\Box B)\tag{double negation elimination}$
$\rightarrow\neg(\Diamond\neg A\vee\Box B)\vee\neg(\Box A\vee\Diamond\neg B))\tag{by definition}$
$\rightarrow\neg(\Diamond\neg A\vee\Diamond B)\vee\neg(\Diamond A\vee\Diamond\neg B)\tag{by $\Box\phi\leftrightarrow\Diamond\phi$}$
$\rightarrow\neg\Diamond(\neg A\vee B)\vee\neg\Diamond(A\vee\neg B))\tag{by 5}$
$\rightarrow\Box\neg(\neg A\vee B)\vee\Box\neg(A\vee\neg B)\tag{by definition}$
$\rightarrow\Box(A\wedge\neg B)\vee\Box(\neg A\wedge B)\tag{by 8}$
$\rightarrow\Box((A\wedge\neg B)\vee (\neg A\wedge B))\tag{by 2}$
We have derived $\dagger$, employing $\Box\phi\leftrightarrow\Diamond\phi$ in between.  It should be remarked that the two stages of (1) are not symmetrical. We need $\Box\phi\rightarrow\Diamond\phi$ at the second stage. Therefore, it is sufficient that $R$ is serial, i.e., $\forall w\exists w'(wRw')$. However, such details seem too much of a digression for one question.
