MathematicsHard115×since 2008Q4386Let p and q be two statements. Then ∼(p∧(p⇒ ∼q))\sim \left( {p \wedge (p \Rightarrow \, \sim q)} \right)∼(p∧(p⇒∼q)) is equivalent toA(∼p)∨q\left( { \sim p} \right) \vee q(∼p)∨qBp∨(p∧(∼q))p \vee \left( {p \wedge ( \sim q)} \right)p∨(p∧(∼q))Cp∨(p∧q)p \vee \left( {p \wedge q} \right)p∨(p∧q)Dp∨((∼p)∧q)p \vee \left( {\left( { \sim p} \right) \wedge q} \right)p∨((∼p)∧q)Check answerSkip