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?

  • $\begingroup$ My question is rather the opposite of this implication. I am assuming the frame is dense and want to show the implication must hold $\endgroup$
    – A_alk
    Apr 18 at 12:37
  • $\begingroup$ 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 $\endgroup$ Apr 18 at 13:15
  • 1
    $\begingroup$ this question shouldn't be marked as a duplicate, because the other question was about proving the opposite direction of this implication $\endgroup$ Apr 18 at 20:26

1 Answer 1


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, $\langle F,V\rangle, w \vDash \square A$.


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