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

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.

• I can’t help with the specific frame condition, but the formula is valid in frames that are both transitive and symmetric. Commented Aug 14 at 18:33

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!

• Thanks! I am now away from the computer and I will check the proof more thoroughly when I have the opportunity, but it seems persuasive. Brief motivation: you’re right about epistemic logic and defining belief within it; as most systems (Stalnaker’s and Lenzen’s) use S4.2 for knowledge to do it (and define a KD45 belief), I was wondering what is in fact the weakest epistemic logic that encode some belief operator as $\Diamond \Box$. D is obviously equivalent to .2 on this reading, which corresponds to convergence, and necessitation is secured by reflexivity, but I was troubled by K. Commented Aug 16 at 4:19
• Now when I’m looking at it, it seems to me that if $R$ is reflexive, then every „super-directed” frame would be also convergent, which would be quite cool :) Commented Aug 16 at 4:43
• I agree, reflexivity also implies that "super-directed" equals convergence, so this truly is a condition that lives away from the "usual" frame conditions. Also some things I realized in the mean time, in case you are interested in metalogical properties: let SD be the system K+"Axiom above". By some heavy duty theorems (namely the Fine-Van Benthem theorem), if this logic is complete, it must be canonical. So the only way to prove this logic complete will be through a canonical model argument. I have briefly sketched one, and it seems to me that this should in fact be the case. Commented Aug 16 at 7:20

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.

• Thanks, that is helpful! If I understand correctly, the argument proves that the axiom is derivable in KB4(=KB5) and suggests that it is derivable in K5 - which is in fact the case, as I checked in umsu proof generator. Though, as the axiom is derivable in S4.2, being euclidean in not necessary, but it certainly is sufficient. Commented Aug 15 at 11:03