Simplify a logic expression

Question

I'm studying to my exam and I have some doubts.

The expression: ¬(P ∨ Q) ∨ (¬P ∨ Q)

The result: ¬P ∨ Q

The objective is to simplify.

I'm stuck at (¬P ∧ ¬Q) ∨ ¬P ∨ Q

I could make the distributive, but then what?

Answer 1

I think this proof should help, in case you need any step explained:

By DeMorgan's Law: $$\lnot(P\lor Q)\lor(\lnot P \lor Q)\iff(\lnot P \land \lnot Q) \lor (\lnot P \lor Q )$$

By the commutative property: $$ \iff (\lnot P\lor Q)\lor(\lnot P\land\lnot Q)$$

By the distribution property: $$ \iff ((\lnot P\lor Q)\lor\lnot P)\land((\lnot P\lor Q)\lor\lnot Q)$$

By the commutative property: $$ \iff ((Q\lor \lnot P)\lor\lnot P)\land((\lnot P\lor Q)\lor\lnot Q)$$

By the associative property: $$ \iff (Q\lor (\lnot P\lor\lnot P))\land((\lnot P\lor Q)\lor\lnot Q)$$

By the tautology: $$ \iff (Q\lor \lnot P)\land((\lnot P\lor Q)\lor\lnot Q)$$

By the commutative property: $$ \iff (\lnot P \lor Q )\land((\lnot P\lor Q)\lor\lnot Q)$$

By the absorption property: $$ \iff (\lnot P \lor Q )$$

Done.

Answer 2

Note that $(A\land B)\lor B=B$, and then we have $$(\lnot P \land \lnot Q) \lor \lnot P \lor Q = \lnot P \lor Q.$$