Going from (p ∧ ~q) ∨ (~p ∧ q) to (p ∨ q) ∧ (~p ∨~q)

Question

I am confused on how to go from (p ∧ ~q) ∨ (~p ∧ q) to (p ∨ q) ∧ (~p ∨ ~q). I know they are equal because I plugged them into a truth table and all of the rows have the same values. What would be some of the rules I could use to prove these two to be logically equivalent? I think I would need to use DeMorgan's Law at some point but I think there may be a step before that.

Answer 1

You can just expand everything out: \begin{align*} (p \land \neg q) \lor (\neg p \land q) &\equiv (p \lor \neg p) \land (p \lor q) \land (\neg q \lor \neg p) \land (\neg q \lor q) &\text{by Distributivity Law}\\ &\equiv (p \lor \neg p) \land (p \lor q) \land (\neg p \lor \neg q) \land (q \lor \neg q) &\text{by Commutativity Law}\\ &\equiv (\top) \land (p \lor q) \land (\neg p \lor \neg q) \land (\top) &\text{by Inverse Law}\\ &\equiv (p \lor q) \land (\neg p \lor \neg q) &\text{by Identity Law}\\ \end{align*}