# Prove this logical equivalence with laws

Prove without using truth tables: $$(((p \vee r) ∧ q) \vee (p \vee r)) ∧ (\neg p \vee r) ⇔ r$$ I tried but I always get stuck when applying like 4 laws, and i don't even know if i using them correctly, i think is the ¬p that is given me problems here, please help

This its what i have so far

((q ∧ p) v (q ∧ r) v (p v r)) ∧ (¬p v r)

((q ∧ p) v ((q ∧ r) v r) v p) ∧ (¬p v r)

((q ∧ p) v r v p)) ∧ (¬p v r)

((q ∧ p) v p v r)) ∧ (¬p v r)

(p v r) ∧ (¬p v r)

• Show us what you got so far, so we can build up on it. Commented Oct 2, 2020 at 12:35
• You're there. You just have to use distributivity of ∧ on v to simplify the last expression. And you should use ⇔ to show that each of the expressions you have written are equivalent to each other.
– MasB
Commented Oct 2, 2020 at 13:04
• ooh, now i see it, i was so focused on the ¬p that i didn't notice that, thanks Commented Oct 2, 2020 at 13:13

You were just missing the last step.

((q ∧ p) v (q ∧ r) v (p v r)) ∧ (¬p v r)

((q ∧ p) v ((q ∧ r) v r) v p) ∧ (¬p v r)

((q ∧ p) v r v p)) ∧ (¬p v r)

((q ∧ p) v p v r)) ∧ (¬p v r)

(p v r) ∧ (¬p v r)

(p v ¬p) ∧ r

r