Question #249491

Show following equivalence without considering the truth table.

(š‘Ģ… ∧( š‘žĢ…āˆ§š‘Ÿ)) ∨(š‘ž āˆ§š‘Ÿ) ∨(š‘ āˆ§š‘Ÿ)ā†”š‘Ÿ


Expert's answer

(¬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

↔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



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