1.The proposition p⇒∼(p∧∼q)p \Rightarrow \sim(p \wedge \sim q)p⇒∼(p∧∼q) is equivalent to:a.qqqb.∼p∨q\sim p \vee q∼p∨qc.∼p∧q\sim p \wedge q∼p∧qd.∼p∨∼q\sim p \vee \sim q∼p∨∼qLogin to continueOnly logged in users canattempt or see the solution.
