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We have frame $\mathcal{F} = (W,R)$ that $R$ is reflexive and transitive and $\forall x,y,z (xRy \wedge xRz \wedge y\ne z \rightarrow yRz \vee zRy)$ ,prove $\mathcal{F} \models \square(\square p \rightarrow q) \vee \square(\square q \rightarrow p )$.

I assume that if $\mathcal{F} \nvDash \square(\square p \rightarrow q)$ then I must show $\mathcal{F} \models \square(\square q \rightarrow p) $.

so because $\mathcal{x} \nvDash \square(\square p \rightarrow q)$ I have $\exists y \;xRy((\,\forall z \; yRz \; z\Vdash q) \rightarrow y \Vdash p) $ but I don't have any idea how to get $\mathcal{x} \models \square(\square q \rightarrow p)$.

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Try proving the contrapositive, i.e.: If for a valuation $\sigma$ we have $\mathcal{F},\sigma \not\models \Box(\Box p \to q) \lor \Box(\Box q \to p)$, then there are worlds $x,y,z \in W$ such that $xRy, xRz, y\neq z$ and neither $yRz$ nor $zRy$ holds. From the first statement you get good candidates for $x,y,z$.

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  • $\begingroup$ Thank you so much,you are right ,now I can prove it. $\endgroup$ – haleh Jan 10 '17 at 17:17

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