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I am working through Fitting and Mendelsohn's First Order Modal Logic and have come across the following exercise:

Prove that a frame $\langle \mathcal{G}, \mathcal{R} \rangle$ is transitive if and only if every formula of the form $\Box P \supset \Box \Box P$ is valid in it.

I can prove the left to right direction, but can't even get started on the right to left proof. Any ideas/hints? I know it should be a relatively simple proof but would appreciate any help.

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  • $\begingroup$ For those not familiar with the notation, the superset symbol refers to material implication. $\endgroup$ – nick Aug 4 '15 at 10:43
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Assume that $\mathcal R$ is not transitive, i.e. $x \mathcal R y$ and $y \mathcal R z$ but $\lnot (x \mathcal R z)$.

Consider $x \Vdash p$ and $y \Vdash p$ but $z \nVdash p$. We have $x \Vdash \square p$ and $y \nVdash \square p$.

Thus, $x \nVdash \square \square p$ and this contradicts : $\Vdash \square p \to \square \square p$.

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  • $\begingroup$ Should have known to try a different proof method, thanks! $\endgroup$ – nick Aug 4 '15 at 21:18

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