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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)

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    $\begingroup$ Show us what you got so far, so we can build up on it. $\endgroup$
    – Natalie Clarius
    Commented Oct 2, 2020 at 12:35
  • $\begingroup$ 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. $\endgroup$
    – MasB
    Commented Oct 2, 2020 at 13:04
  • $\begingroup$ ooh, now i see it, i was so focused on the ¬p that i didn't notice that, thanks $\endgroup$
    – semqdr
    Commented Oct 2, 2020 at 13:13

1 Answer 1

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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

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