Use the table of propositional logical equivalences to show that ¬(p ∨
¬(p ∧)) is a contradiction.
¬(p∨¬(p∧q) ⟺ ¬p∧¬(¬(p∧q) De Morgan’s Law ⟺ ¬p∧(p∧q) Double Negation Law ⟺ (¬p∧p)∧q Associativity Law ⟺ F∧q Contradiction ⟺ F Domination LawHence the statement is a Contradiction.\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.}¬(p∨¬(p∧q)⟺¬p∧¬(¬(p∧q) De Morgan’s Law⟺¬p∧(p∧q) Double Negation Law⟺(¬p∧p)∧q Associativity Law⟺F∧q Contradiction⟺F Domination LawHence the statement is a Contradiction.
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