MathematicsMedium115×since 2008Q4377The statement (p⇒q)∨(p⇒r)(p \Rightarrow q) \vee(p \Rightarrow r)(p⇒q)∨(p⇒r) is NOT equivalent toA(p∧(∼r))⇒q(p \wedge(\sim r)) \Rightarrow q(p∧(∼r))⇒qB(∼q)⇒((∼r)∨p)(\sim q) \Rightarrow((\sim r) \vee p)(∼q)⇒((∼r)∨p)Cp⇒(q∨r)p \Rightarrow(q \vee r)p⇒(q∨r)D(p∧(∼q))⇒r(p \wedge(\sim q)) \Rightarrow r(p∧(∼q))⇒rCheck answerSkip