# Is the $D$ modal logic necessarily serial?

I'm wanting to show in $$D$$ that $$\sim \square (\square p \wedge \square \sim p)$$.

Here's my attempt:

(1) Show $$\sim \square (\square p \wedge \square \sim p)$$

(2) $$\square (\square p \wedge \square \sim p) \quad$$assumption for indirect derivation with double negation

(3)$$\square [\sim (\square p \supset \Diamond p) ]\quad$$ equivalent to line (3)

(4) $$\square p \supset \Diamond p \quad$$ Axiom D

(5) $$\square(\square p \supset \Diamond p )\quad$$ Necessity Inference rule applied to (4)

Now it would be nice if lines 3 and 5 were a contradiction but they aren't quite.

Given a frame $$\langle W,R\rangle$$ we say that it is $$serial$$ iff for every $$w\in W$$ there exists some $$w'\in W$$ such that $$wRw'$$; now it can be shown that the $$D$$ axiom must be valid on every serial frame. It would appear if the converse is true -- namely, that we require every logic system which has $$D$$ as an axiom to be serial -- then in fact the above would give a contradiction because the above proof is given from the perspective of some fixed world $$w$$. If 3 and 5 are true and this frame is serial then there exists $$w'$$ such that $$wRw'$$ with both $$\sim (\square p \supset \Diamond p)$$ and $$(\square p \supset \Diamond p)$$ true at $$w'$$. $$Cleary$$, this would be a contradiction at $$w'$$ and hence for that entire frame. But is it asking to much that every system which adopts $$D$$ is necessarily serial?

• What's particularly unsettling is that if we consider a world $w$ such that there does not exist $w'$ with $wRw'$ then it would be appear that $\square(\square p \wedge \square \sim p)$ is in fact vacuously true (since if it were false we would need to find a world to falsify it but this isn't possible). – Squirtle Nov 13 '17 at 2:44