I just stumbled across this. In classic logic, the negation of a valid formula is unsatisfiable and vice versa. Given the usual Kripke semantics definitions of modal logic K (see below), this law seems only to hold in the former direction. But I failed to construct a counterexample for the opposite direction. So do you know a modal formula that is unsatisfiable in K and whose negation is not valid?

Some definitions: A K-model is a structure $M = (W,R,I)$ consisting of a non-empty graph of "worlds" $(W,R)$ and a family $I$ of valuations, providing a pc-valuation for each world $w \in W$. For a modal formula $H$, let $(M,w) \models H $ be defined as usual in K and let $M \models H$ iff $(M,w)\models H$ holds for all worlds $w\in W$. Based on this I call $H$ valid iff $M \models H$ holds for each K-structure $M$ and I call $H$ unsatisfiable iff $M\not\models H$ holds for each K-structure.


1 Answer 1


How about $H = a \land \Diamond \neg a$, where $a$ is some atomic proposition? Consider a model $M = (W, R, I)$. If $M$ has a world $w$ such that $M, w \models H$, then there must exist some $w' \in W$ (with $wRw'$) such that $M, w' \models \neg a$, meaning that $M, w' \not\models H$. Hence $H$ is unsatisfiable.

On the other hand, $\neg H = \neg a \lor \Box a$ is clearly not valid, since we can create a model that does not satisfy it.

  • $\begingroup$ You spelled out the conditions for the satisfiability of $H$ as opposed to its unsatisfiability. $\endgroup$
    – PW_246
    Sep 9, 2023 at 15:38

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