Proof that $p \vee [(\neg p \vee \neg q) \vee (p \vee q)] \iff \top.$

Question

I just began logic and proofs there is a proof in the midterm of $2019$ that I could not do.

Here is the statement :

$p \vee [(\neg p \vee \neg q) \vee (p \vee q)] \iff \top.$

I started using Distributive Laws and got a really long mess. I thought maybe if I developed maybe I'd use De Morgan's law or Absorption Laws but none really came up.

Any HELP would be a lot appreciated ( I'm not looking for a full answer but maybe going step by step )

Answer 1

Let $p$ and $q$ be propositions, and consider the expression $$p \vee [(\neg p \vee \neg q) \vee (p \vee q)].$$

Applying the Associative Law, we have that

$$p \vee [(\neg p \vee \neg q) \vee (p \vee q)] \iff [p \vee (\neg p \vee \neg q)] \vee (p \vee q).$$

Applying once more the same rule yields

$$[p \vee (\neg p \vee \neg q)] \vee (p \vee q) \iff [(p \vee \neg p) \vee \neg q] \vee (p \vee q).$$

Note that $p \vee \neg p$ is a tautology. So $p \vee \neg p \iff \top.$ So

$$[(p \vee \neg p) \vee \neg q] \vee (p \vee q) \iff [\top \vee \neg q] \vee (p \vee q).$$

Since $\top$ is a tautology, the disjunction on any proposition with $\top$ will be always true, and hence a tautology. So

$$[\top \vee \neg q] \vee (p \vee q) \iff \top \vee (p \vee q) \iff \top.$$

$\square$