Answer to Question #249491 in Discrete Mathematics for ffh

Question

Show following equivalence without considering the truth table.

(𝑝̅ ∧( 𝑞̅∧𝑟)) ∨(𝑞 ∧𝑟) ∨(𝑝 ∧𝑟)↔𝑟

Answer 1

$(\lnot p \land(\neg q \land r))\lor(q\land r) \lor(p\land r)\leftrightarrow(r \land(\neg q \land \neg p))\lor(r\land (q\lor p)) \leftrightarrow$

$\leftrightarrow r\land ((\lnot p \land \lnot q) \lor (q \lor p)) \leftrightarrow r\land (\lnot(p\lor q) \lor (p\lor q)) \leftrightarrow r\land 1 \leftrightarrow r$