Modal logic: condition corresponding to $\Diamond \Box (A \Rightarrow B)\Rightarrow (\Diamond \Box A \Rightarrow \Diamond \Box B)$?

Question

In normal modal logic, what would be the condition on the accessibility relation corresponding to the following axiom (the analogue of the distribution axiom for $\Diamond \Box$ instead of $\Box$):

$\Diamond \Box (A \Rightarrow B)\Rightarrow (\Diamond \Box A \Rightarrow \Diamond \Box B)$?

I know that this schema is derivable in S5 and S4.2, but not in S4; what I'm trying to do is to find out is what is possibly the weakest logic that validates it. I did some digging, but I haven't found any useful procedure for coming up with semantic analogues (first- or second-order) to axioms which cannot be put in the general form required by Lemmon-Scott theorem. I'd deeply appreciate any help with this question, even just pointing the relevant literature that might be of help, as I am not a mathematician but philosopher myself.

Answer 1

If I have not made any mistakes (I invite you to check thouroughly, correspondence proofs can often have mysterious holes), I think this axiom defines some strange variant of directedness: if $xRy$ and $xRz$ then there is some $w$ such that $xRw$ and whenever $wRk$ then $yRk$ and $zRk$. Call this a super-directed frame, and a $w$ in these conditions a "witness" for super-directedness.

Let me provide a correspondence proof. This assumes familiarity with Kripke semantics, but I hope that will not be a problem. First, assume that $(X,R)$ is a super-directed frame. Suppose that $x\Vdash \Diamond \Box (a\rightarrow b)$ and $x\Vdash \Diamond \Box a$. Let $y$ and $z$ respectively be witnesses. Since $xRy$ and $xRz$, by super-directedness, there is some $w$ such that $xRw$ in the right conditions. Now if $wRk$ then because $yRk$, $k\Vdash a$; and because $zRk$, $k\Vdash a\rightarrow b$, so $k\Vdash b$. So $w\Vdash \Box b$, as desired.

Now assume that $(X,R)$ is not super-directed. Let $xRy$ and $xRz$ with no $w$ witnessing super-directedness. Let $V$ be a valuation such that $V(a)=\{k : yRk\}$ and $V(b)=\{k : yRk \wedge zRk\}$. Then $y\Vdash \Box a$ and note that $z\Vdash \Box (a\rightarrow b)$: if $zRm$, and $m\Vdash a$, then by definition, $yRm$, so $m\in V(b)$, i.e., $m\Vdash b$. Now assume that $x\Vdash \Diamond \Box b$. Then by definition, $xRw$, and $w\Vdash \Box b$. But then whenever $wRk$, then $yRk$ and $zRk$ -- i.e., $w$ is a super-directed witness.

Note that if you assume transitivity, then super-directedness coincides with directedness: if $xRy$ and $xRz$, any successor of $y$ and $z$, say $w$, will be a super-directed witness in the above sense.

Some further notes: the principle you are considering is indeed the Kripke axiom for the composed modality $\Diamond \Box$. This is equivalent to adding the two normality conditions for the I know that in epistemic logic this modality is sometimes considered, since it can be useful to axiomatise forms of belief. I have searched in the literature for whether anything is known about adding these normality rules for composed modalities, but have not found anything. I would be interested to know what is your motivation for studying this principle!

Answer 2

Not the sought answer precisely, but an answer, anyway, hoping to be of some help:

In what follows, I shall employ the derived rule transitivity of implication (TI) of propositional calculus:

$$p\rightarrow q, q\rightarrow r\vdash p\rightarrow r$$

Also, for my argument, my reference will be the section 4 of An Introduction to Modal Logic by E. J. Lemmon (in collaboration with D. Scott, edited by K. Segerberg, 1977).

So, note that

$$\Box(p\rightarrow q)\rightarrow(\Diamond p\rightarrow\Diamond q)$$

which can be straightforwardly derived from axiom K, the definition $\Box\phi\leftrightarrow\neg\Diamond\neg\phi$, and propositional calculus, without calling in axiom D.

By taking $p$ and $q$ as $\Box A$ and $\Box B$, respectively, we can write

(1) $\Box(\Box A\rightarrow\Box B)\rightarrow(\Diamond\Box A\rightarrow\Diamond\Box B)$

By axiom K, we have

(2) $\Box\Box(A\rightarrow B)\rightarrow\Box(\Box A\rightarrow\Box B)$

From (1) and (2), by TI

(3) $\Box\Box(A\rightarrow B)\rightarrow(\Diamond\Box A\rightarrow\Diamond\Box B)$

By axiom B

(4) $\Diamond\Box\Box(A\rightarrow B)\rightarrow\Box(A\rightarrow B)$

By axiom 4

(5) $\Box(A\rightarrow B)\rightarrow\Box\Box(A\rightarrow B)$

From (4) and (5), by TI

(6) $\Diamond\Box(A\rightarrow B)\rightarrow\Box\Box(A\rightarrow B)$

From (3) and (6), by TI

(7) $\Diamond\Box(A\rightarrow B)\rightarrow(\Diamond\Box A\rightarrow\Diamond\Box B)$

Therefore, we need the symmetric relation by B and transitive relation by 4. Since a symmetric relation is also euclidean, when it is transitive, we may also say that the formula holds on euclidean frames.