Question #179054

~(~p^q) ^(pvq) =p


1
Expert's answer
2021-04-15T07:28:03-0400

¬(¬p∧q)∧(p∨q)=(p∨¬q)∧(p∨q){DeMorgan′s}=p∧(¬q∨q){Distributive}=p∧1{Identity}=p\neg(\neg p \wedge q)\wedge (p \vee q)=\\ (p \vee \neg q) \wedge (p \vee q)\{De Morgan's\}=\\ p \wedge (\neg q \vee q)\{Distributive\}=\\ p \wedge 1 \{ Identity\}=\\ p


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