MathematicsMedium115×since 2008Q4381(S1) (p⇒q)∨(p∧(∼q))(\mathrm{S} 1)~(p \Rightarrow q) \vee(p \wedge(\sim q))(S1) (p⇒q)∨(p∧(∼q)) is a tautology (S2) ((∼p)⇒(∼q))∧((∼p)∨q)(\mathrm{S} 2)~((\sim p) \Rightarrow(\sim q)) \wedge((\sim p) \vee q)(S2) ((∼p)⇒(∼q))∧((∼p)∨q) is a contradiction. ThenAonly (S2) is correctBboth (S1) and (S2) are correctConly (S1) is correctDboth (S1) and (S2) are wrongCheck answerSkip