$\sf K$-proof of $(\Box p \land \Diamond q) \rightarrow \Diamond (p\land q)$

Question

$\sf K$-proof of $(\Box p \land \Diamond q) \rightarrow \Diamond (p\land q)$

I am stuck on the proof of this using the axiom system $\sf K$.

This is the exercise 1.6.1 in Blackburn's modal logic. Any help, please?

Thank you.

Answer 1

Note that we want to prove (from an appropriate hypothesis) a statement of the form "$\neg\Box\neg\varphi$" - this means that we're going to have to use the contrapositive form of rule K. (Basically, because that's all we've got in system K.)

Putting it in "$\Box$-only" form, our hypothesis is $$(*)\quad\Box p\wedge\neg\Box \neg q.$$ Now this has the form $$A\wedge\neg B,$$ which is to say it has the form $$\neg (A\implies B).$$ Specifically, our hypothesis is equivalent to $$(**)\quad \neg(\Box p\implies \Box\neg q).$$ Now think about what the contrapositive of rule $K$ lets us do here ...

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Incidentally, this is a good example of a situation where semantic reasoning is much more efficient than syntactic reasoning. It's easy to show that "$(\Box p\wedge\Diamond q)\implies\Diamond (p\wedge q)$" is valid in every Kripke frame, hence by the usual completeness theorem for K is a K-tautology.