The modal logic S4.2 with the characteristic axioms

4: $\square \alpha \rightarrow \square \square \alpha$


.2: $\lozenge \square \alpha \rightarrow \square \lozenge \alpha$


T: $\square \alpha \rightarrow \alpha$

is sound and complete for transitive, reflexive and connected frames. Such frames validate the closure principle

CP $\lozenge \square \alpha \wedge \lozenge \square \beta \rightarrow \diamond \square (\alpha \wedge \beta)$

Can someone help me with deriving CP in S4.2?


1 Answer 1


$\let\B\Box\let\D\Diamond$ \begin{align*} \D\B\alpha\land\D\B\beta&\to\D\B\B\alpha\land\D\B\B\beta\\ &\to\B\D\B\alpha\land\D\B\B\beta\\ &\to\D(\D\B\alpha\land\B\B\beta)\\ &\to\D\D(\B\alpha\land\B\beta)\\ &\to\D\D\B(\alpha\land\beta)\\ &\to\D\B(\alpha\land\beta) \end{align*} using the K-provable principle $\B p\land\D q\to\D(p\land q)$ and monotonicity of $\B$ and $\D$. Note that the axiom T is not needed.

  • $\begingroup$ That is a nice proof! I knew that T was not needed, but S4.2 is a more famous logic than K4.2. $\endgroup$ Mar 14, 2014 at 19:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.