# Prove that if a frame is dense then $\Box \Box A \to \Box A$ is valid [duplicate]

I am having trouble with the following question on modal logic.

A modal logic frame $$M =\langle W, R\rangle$$ is dense whenever $$\forall x\forall y(Rxy\rightarrow\exists z(Rxz \land Rzy))$$

Prove that the formula $$\Box\Box A\to\Box A$$ is valid in every dense frame.

We don't make any assumptions about the frames. How can this be done?

• My question is rather the opposite of this implication. I am assuming the frame is dense and want to show the implication must hold Apr 18 at 12:37
• You don’t need to assume the relation is transitive. Once you assume the frame is dense, and you assume that box box A is true at some world x (with arbitrary valuation function), then consider any world that x sees Apr 18 at 13:15
• this question shouldn't be marked as a duplicate, because the other question was about proving the opposite direction of this implication Apr 18 at 20:26

Let $$F=\langle W,R\rangle$$ be a dense frame. Take some arbitrary valuation function $$V$$, and some arbitrary world $$w \in W$$. Suppose $$\langle F,V\rangle, w \vDash \square\square A$$. Take any $$u$$ s.t. $$Rwu$$. Then there exists some $$z$$ s.t. $$Rwz$$ and $$Rzu$$. So $$\langle F,V\rangle, z \vDash \square A$$, and $$\langle F,V\rangle, u \vDash A$$. Since $$u$$ was arbitrary, $$\langle F,V\rangle, w \vDash \square A$$.