Applying Commutivity Law to a Tautology $P\lor \neg (P \land Q) $

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How do I apply Commutivity law to a tautology: $P\lor \neg(P \land Q)$?

I understand the it is $A\lor B = B\lor A$, but how can this apply to the above tautology?

Do I assume $P$ as $A$, and $\neg (P\land Q)$ as $B$?


I just checked the answer, the answer is: $\neg Q \lor \top$. Where did the $\top$ come from?

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You don't simplify the statement using commutativity.

Take the statement: $$P \lor \neg (P \land Q)$$ Apply DeMorgan's Laws: The negation of a conjunction is the disjunction of the negations. $$= P\lor (\neg P \lor \neg Q)$$ Disjunctions are associative. $$= (P\lor \neg P)\lor \neg Q$$ The disjunction of a preposition and its negation is a tautology. $$= {\large\top}\lor \neg Q$$ The disjunction of a tautology and a preposition is a tautology. $$= {\large\top}$$