Can I derive (pos p and pos q) from pos p, pos q in S5?

Question

Is there a theorem of S5 that allows me to derive $\lozenge p \land \lozenge q$ from $\lozenge p$ and $\lozenge q$? I know I can derive $\lozenge p \land \lozenge q$ if I have simply $p$ and $q$, but my question is if it is possible if the propositions are prefixed with the possibility operator.

Answer 1

In most modal logics, the language is the same as for propositional logic except that adding a modal operator in front of any proposition yields yet another proposition. In particular if $p$ is a proposition then $◊p$ also is a proposition. Since we retain classical logic including the inference rules, we can from $◊p$ and $◊q$ deduce $( ◊p \land ◊q )$, since the latter is merely the conjunction of 2 propositions that we are already given. Note that $( ◊p \land ◊q )$ is weaker (holds in more situations) than $◊( p \land q )$, because $( ◊p \land ◊q )$ is true as long as some accessible world witnesses $p$ and some (not necessarily the same) accessible world witnesses $q$, whereas $◊( p \land q )$ is true only when some single accessible world witnesses both $p$ and $q$.

Answer 2

$P, Q \vdash P \land Q$ is derivable in propositional logic no matter what $P$ and $Q$ are. In particular, we can take $P = \lozenge p$ and $Q= \lozenge q$. As Mauro pointed out, since S5 includes propositional logic, this derivation works in S5.