Conjunction of Disjunctions Consequence

Do we have this somewhere already?

Theorem
Suppose that $(p \lor q) \land (r \lor s)$.

Then $p \lor r \lor (q \land s)$

Proof
By distribution, $(p \land (r \lor s)) \lor (q \land r) \lor (q \land s)$.

$p \land (r \lor s) \implies p$ by simplification.

$q \land r \implies r$ by simplification.

$q \land s \implies q \land s$ by identity.

Thus the result holds by constructive dilemma and modus ponens.