Show following equivalence without considering the truth table.
(𝑝̅ ∧( 𝑞̅∧𝑟)) ∨(𝑞 ∧𝑟) ∨(𝑝 ∧𝑟)↔𝑟
(¬p∧(¬q∧r))∨(q∧r)∨(p∧r)↔(r∧(¬q∧¬p))∨(r∧(q∨p))↔(\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(¬p∧(¬q∧r))∨(q∧r)∨(p∧r)↔(r∧(¬q∧¬p))∨(r∧(q∨p))↔
↔r∧((¬p∧¬q)∨(q∨p))↔r∧(¬(p∨q)∨(p∨q))↔r∧1↔r\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↔r∧((¬p∧¬q)∨(q∨p))↔r∧(¬(p∨q)∨(p∨q))↔r∧1↔r
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