# Prove or refute transitivity property for $\square \neg A \lor \lozenge \lozenge A$

I don't know how should I solve this problem in modal logic:

Does $$\square \neg A \lor \lozenge \lozenge A$$ satisfy transitivity ?

First, I assume that you are talking about normal modal logics.

Transitivity is not enough. Your model needs to be reflexive (try to prove your formula in S4 using analytical tables or some calculus).

As for a counter model for your formula in K4 (minimal transitive normal logic), here it is: $w_0\longrightarrow w_1$. Set $w_0\nvDash A$ and $w_1\vDash A$ and see that your formula doesn't hold in $w_0$.

Now, a converse question. Can $\Box\neg A\vee\diamond\diamond A$ hold in a non-transitive model (where $\neg\forall w_1,w_2,w_3(R(w_1,w_2)\wedge R(w_2,w_3)\Rightarrow R(w_1,w_3))$)?

Sure, it can, because it holds in any model, where $\Box\neg A$ or $\diamond\diamond A$ holds and they can of course be non-transitive.

This axiom, rewritten equivalently as $\Diamond A\to\Diamond\Diamond A$, expresses the well-known property that the frame is $dense$, i.e., it satisfies the property: "between any two points $x$ and $z$ such that $xRz$, there exists a point $y$ such that $xRy$ and $yRz$."

Density has no relation to transitivity: you can easily build a transitive but not dense frame; a dense but not transitive frame; a transitive dense frame, e.g., $(\mathbb{Q},<)$; a frame that is neither transitive nor dense. So the answer to your question is: yes, since a dense frame does not have to be transitive.