1.The negation of the Boolean expression ((∼q)∧p)⇒((∼p)∨q)((\sim q) \wedge p) \Rightarrow ((\sim p) \vee q)((∼q)∧p)⇒((∼p)∨q) is logically equivalent to:a.p∧qp \wedge qp∧qb.q⇒pq \Rightarrow pq⇒pc.∼(p⇒q)\sim(p \Rightarrow q)∼(p⇒q)d.∼q\sim q∼qLogin to continueOnly logged in users canattempt or see the solution.
