DNF conversion in Propositional Logic: Commutative Property [closed]

I want to deduce the below function to a Distributive normal formula (DNF) and Conjunctive normal formula (CNF)

$$(P \land \neg Q) \lor (\neg Q \lor R)$$ Next I assume Commutative holds. (the question; does it ?) $$(\neg Q \lor R) \lor (P \land \neg Q)$$ Using Distributive property: $$((\neg Q \lor R)\lor P) \land ((\neg Q\lor R)\lor \neg Q)$$ again Distributive law: $$(\neg Q \lor P \lor R \lor P) \land (\neg Q \lor \neg Q \lor R \lor \neg Q)$$

OR Can I take the bracket off without applying Distributive law : $$(\neg Q \lor R\lor P) \land (\neg Q\lor R\lor \neg Q)$$

Further, simplify: $$(\neg Q \lor P \lor R) \land (R \lor \neg Q)$$

Am I doing this correct ?

New contributor
OmegaD is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.