MathematicsMedium115×since 2008Q4398Among the statements (S1) : (p⇒q)∨((∼p)∧q)(p \Rightarrow q) \vee((\sim p) \wedge q)(p⇒q)∨((∼p)∧q) is a tautology (S2) : (q⇒p)⇒((∼p)∧q)(q \Rightarrow p) \Rightarrow((\sim p) \wedge q)(q⇒p)⇒((∼p)∧q) is a contradictionAneither (S1) and (S2) is TrueBonly (S2) is TrueCboth (S1)(\mathrm{S} 1)(S1) and (S2)(\mathrm{S} 2)(S2) are TrueDonly (S1) is TrueCheck answerSkip