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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.

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    Yes: it is a propositional derivation and in modal logic all propositional rules still hold; see [conjunction introduction](https://en.wikipedia.org/wiki/Conjunction_introduction). – 2017-01-15
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    But this is propositional calculus. I know I can do this in propositional calculus, but ${p, q} \vdash p \land q$ is not the same as ${\lozenge p, \lozenge q} \vdash \lozenge p \land \lozenge q$ or am I missing something? – 2017-01-15
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    The point you're missing is the **definition** of the language of S5, where "$◊p$" is a proposition for any proposition "$p$". – 2017-01-16
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    Thank you! I couldn't find that definition anywhere... Could you also tell me what $\lozenge p \land \lozenge q$ would mean semantically? I understand that under Kripke semantics $\lozenge (p \land q)$ would mean there is some (accessible) world $w$ where $p \land q$. But what would $\lozenge p \land \lozenge q$ mean? There is an accessible world $w_1$ for which $p$ and also another accessible world $w_2$ for which $q$? – 2017-01-17
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    @m-strasser: To notify a user of a message, include `@username` somewhere in the message. I'm surprised that neither Wikipedia nor SEP gives a proper definition of the language of modal logic in their respective articles, but it's very natural, since "$◊p$" is described to denote "possibly p" which is 'clearly' a proposition too. As for your question, indeed your understanding is correct. Note that $◊p \land ◊q$ is true even if $p$ and $q$ are witnessed by different accessible worlds, but $◊(p \land q)$ would only be true if the same accessible world witnesses both $p$ and $q$. – 2017-01-17
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    Actually for Wikipedia, the Modal_logic article links to Modal_operator that gives a roughly correct idea that one uses a modal operator to form new propositions from existing ones, though it's not made formally precise. – 2017-01-17
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    @user21820 thanks, yes if you post this as an answer I would mark it as correct :) – 2017-01-17

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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 $q$.

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$P, Q \vdash P \land Q$ 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.