Proving $(p∨q)∨(p∨¬q)$ is a tautology

Question

I am trying to prove that $(p∨q)∨(p∨¬q)$ is a tautology.

This is a relatively easy problem because it is easy to tell that the statement will be true whenever p or q or ¬q are true. I have never written proofs before, however and I am wondering what the correct way to prove this would be.

I used the distributive law initially so that my problem could be rewritten as p∨p ∨q∨¬q which can then simplify to $p∨q∨¬q.$

What law can I use to say from here that this is a tautology due to the fact that p or q or ¬q will always be true?

Answer 1

$(p∨q)∨(p∨¬q) \equiv p \lor p \lor q \lor \lnot q\quad$ due to associativity and commutativity.

$\equiv p \lor (q \lor \lnot q)\quad$ Simplification, associativity.

$\equiv p \lor (T)\quad$ due to the fact that $q \lor \lnot q$ is always true (the law of the excluded middle).

$\equiv T$ because $p \lor T $ is always true.

Therefore, the initial proposition is a tautology.

Answer 2

Use associativitiy and commutativity: Your statement has $q ∨ ¬q$ which is always true.

Answer 3

An alternative approach using $\lor$ introduction and elimination rules.

\begin{array}{l} q\lor\neg q \\ q\rightarrow(p\lor q)\\ (p\lor q)\rightarrow (p\lor q)\lor(p\lor\lnot q) \\ \lnot q\rightarrow(p\lor\neg q) \\ (p\lor\neg q)\rightarrow(p\lor\neg q)\lor(p\lor q) \\ (q\lor\lnot q)\rightarrow(p\lor\neg q)\lor(p\lor q) \\ (p\lor\neg q)\lor(p\lor q) \end{array}