Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|cite|improve this question

1 Answer 1

up vote 5 down vote accepted

$\let\B\Box\let\D\Diamond$ \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.

share|cite|improve this answer
That is a nice proof! I knew that T was not needed, but S4.2 is a more famous logic than K4.2. – Frode Bjørdal Mar 14 '14 at 19:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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