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

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?

• It depends if you mean commutativity with respect to $\lor$ or with respect to $\land$. – Git Gud May 18 '14 at 23:46
• Distribute $\neg$ over $P\land Q$ to find $\top$. – Git Gud May 18 '14 at 23:53
• Do you mean P∧~P instead? P∨~(P∧Q) = P∨~P∨~Q, then replace P∨~P with T? – user3650172 May 18 '14 at 23:58
• No, the only four possible commutations (there are infinite possibilities, but all of them yield one of these four) are $P\lor \neg (P\land Q), \, P\lor \neg (Q\land P), \, \neg (P\land Q)\lor P, \, \neg(Q\land P)\lor P$. The answer to the second question in your comment is yes. – Git Gud May 19 '14 at 0:03
• I see you're a new user. Please read about accepting answers here and here. – Git Gud May 19 '14 at 0:04

## 1 Answer

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. $$= {\large\top}$$

• Does this mean we'll always need to apply DeMorgan's law first then only I can do communtativity and identity law? – user3650172 May 19 '14 at 0:46
• @user3650172 It's associativity; but yes, a negation of something in brackets is a strong indication that DeMorgan's Laws will be needed. – Graham Kemp May 19 '14 at 0:55