1.The compound statement (p∨q)∧(∼p)⇒q(p \vee q) \wedge (\sim p) \Rightarrow q(p∨q)∧(∼p)⇒q is equivalent to:a.p∨qp \vee qp∨qb.p∧∼qp \wedge \sim qp∧∼qc.∼(p∧q)\sim(p \wedge q)∼(p∧q)d.∼(p∧q)∨(p∧∼q)\sim(p \wedge q) \vee (p \wedge \sim q)∼(p∧q)∨(p∧∼q)Login to continueOnly logged in users canattempt or see the solution.
