Question #249491

Show following equivalence without considering the truth table.

(𝑝̅ ∧( 𝑞̅∧𝑟)) ∨(𝑞 ∧𝑟) ∨(𝑝 ∧𝑟)↔𝑟


1
Expert's answer
2021-10-11T15:40:41-0400

(¬p(¬qr))(qr)(pr)(r(¬q¬p))(r(qp))(\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

r((¬p¬q)(qp))r(¬(pq)(pq))r1r\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



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