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The modal logic S4.2 with the characteristic axioms

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

and

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

and

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?

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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.

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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

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