MathematicsHard115×since 2008Q4385The statement (p∧(∼q))⇒(p⇒(∼q))\left( {p \wedge \left( { \sim q} \right)} \right) \Rightarrow \left( {p \Rightarrow \left( { \sim q} \right)} \right)(p∧(∼q))⇒(p⇒(∼q)) isAa tautologyBequivalent to (∼p)∨(∼q)\left( { \sim p} \right) \vee \left( { \sim q} \right)(∼p)∨(∼q)Ca contradictionDp∨qp \vee qp∨qCheck answerSkip