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In modal logic, $\square A$ means "always" and $\lozenge A$ means "sometimes". Does this mean $\square A$ contains $\lozenge A$?

For example, if $\square A$ is true in the Kripke model, is $\lozenge A$ also true? Or should we have exactly "sometimes" for $\lozenge A$?

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  • $\begingroup$ In modal logic $\square$ is usually read as : "It is necessary that ..." and $\lozenge$ as "It is possible that …". Thus, if we equate "It is necessary that ..." with "sometimes", we have that $\lnot \lozenge$ is "not sometimes", i,.e. "never". $\endgroup$ – Mauro ALLEGRANZA Sep 10 '16 at 9:47
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$\Box A$ and $\Diamond A$ can be thought of as analogous to $\forall$ and $\exists$ respectively, only applied to a modal setting. This is because these two quantifiers appear in the semantics of $\Box$ and $\Diamond$:

$w \models \Box A$ iff $\forall u(w \mathcal{R} u \implies u \models A)$

$w \models \Diamond A$ iff $\exists u(w \mathcal{R} u \text{ and } u \models A)$

Thus if there is no $u$ such that $w \mathcal{R} u$, then $w \models \Box A$ vacuously, but $w \not \models \Diamond A$. On the other hand, there may be one or more $u$ such that $w \mathcal{R} u$ and $u \models A$, in which case $w \models \Diamond A$, but there may still be some $u$ with $w \mathcal{R} u$ and $u \not \models A$, in which case $w \not \models \Box A$.

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  • $\begingroup$ so if there exist some u, $w \mathcal{R} u$ which all of them $u \models A$, then both $w \models \Box A$ and $w \models \Diamond A$ are true ? $\endgroup$ – tressa Sep 10 '16 at 9:51
  • $\begingroup$ @tressa yes, that's correct. $\endgroup$ – mrp Sep 10 '16 at 9:55

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