$¬Q \lor (R \land S) $
$P \lor Q $
$P \land R \land S$
clause 1 and clause 2 can create clause 3?
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3Are you asking if (1 and 2) implies 3? – Aman Kushwaha Jul 27 '21 at 08:53
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$$(¬Q \lor (R \land S)) \land (P \lor Q)$$ $$\equiv ((¬Q \lor (R \land S)) \land P) \lor ((¬Q \lor (R \land S)) \land Q)$$ $$\equiv ((¬Q \lor (R \land S)) \land P) \lor ((\neg Q \land Q) \lor (Q \land R \land S)$$ $$\equiv ((¬Q \lor (R \land S)) \land P) \lor (F \lor (Q \land R \land S)$$ $$\equiv ((¬Q \lor (R \land S)) \land P) \lor (Q \land R \land S) $$ $$\equiv (P \land \neg Q) \lor (P \land R \land S) \lor (Q \land R \land S) $$ Now, $$P \land R \land S \implies (P \land \neg Q) \lor (P \land R \land S) \lor (Q \land R \land S) $$ i.e., $$P \land R \land S \implies (¬Q \lor (R \land S)) \land (P \lor Q)$$ which means clause 3 $\implies$(clause 1 $\land$ clause 2)
Aman Kushwaha
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Here, $F$ is a fallacy (or contradiction). So, $F \lor p \equiv p$ for any statement $p$. – Aman Kushwaha Jul 27 '21 at 09:17