New answers tagged modal-logic
For : (1) implies (2) : 1) $(p \land q) \rightarrow p$ --- tautology 2) $\square ( (p \land q) \rightarrow p)$ --- from 1) by Rule of Necessitation 3) $\square (p \land q) \rightarrow \square p$ --- from 2) by Ax K and modus ponens 4) $(p \land q) \rightarrow q$ --- tautology 5) $\square (p \land q) \rightarrow \square q$ --- from 4) as in 2)-3) by Ax K ...
Ok I got it. If $p$ is false at all worlds, then at a "bottommost" world $n$, $n\models\neg p$ and $n\models\square p$, and $n\models\neg(p\leftrightarrow\square p)$ and therefore (from transitivity and converse wellfoundedness) $0 \not\models\square(p\leftrightarrow\square p)$ (where $0$ is the "first" world) whence $0\models \square(p\leftrightarrow\square ...
The complexity theoretical aspect of some such logics is investigated in: Vardi: The Complexity of Epistemic Reasoning (1989) The paper also includes a reference to Vardi's '86 paper which apparently contains more on such logics (I have not read that one, though).
If by "describes" you mean "modally defines" (i.e., there is a modal formula $\varphi$ such that for every frame: the frame satisfies the property iff the $\varphi$ is valid in the frame), then the answer seems to be no: By the Goldblatt-Thomason Theorem (see e.g. Blackburn, de Rijke, Venema, Modal Logic (2002) p.142) modally definable classes of frames need ...
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