Use the table of propositional logical equivalences to show that ¬(p ∨ ¬(p ∧)) is a contradiction.

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$\neg(p\vee \neg(p\wedge q)\\ \iff \neg p \wedge \neg(\neg(p \wedge q) \text{ De Morgan's Law}\\ \iff \neg p \wedge(p \wedge q) \text{ Double Negation Law}\\ \iff (\neg p\wedge p) \wedge q \text{ Associativity Law}\\ \iff F \wedge q \text{ Contradiction}\\ \iff F \text{ Domination Law}\\ \text{Hence the statement is a Contradiction.}$ $\neg (p \lor \neg (p \land q) ⟺ \neg p \land \neg (\neg (p \land q) De Morgan’s Law ⟺ \neg p \land (p \land q) Double Negation Law ⟺ (\neg p \land p) \land q Associativity Law ⟺ F \land q Contradiction ⟺ F Domination Law Hence the statement is a Contradiction.$

Use the table of propositional logical equivalences to show that ¬(p ∨ ¬(p ∧)) is a contradiction.

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Question ID: mtid-5-stid-8-sqid-3155-qpid-1854