MathematicsMedium115×since 2008Q4391Among the two statements (S1):(p⇒q)∧(p∧(∼q))(\mathrm{S} 1):(p \Rightarrow q) \wedge(p \wedge(\sim q))(S1):(p⇒q)∧(p∧(∼q)) is a contradiction and (S2):(p∧q)∨((∼p)∧q)∨(p∧(∼q))∨((∼p)∧(∼q))(\mathrm{S} 2):(p \wedge q) \vee((\sim p) \wedge q) \vee(p \wedge(\sim q)) \vee((\sim p) \wedge(\sim q))(S2):(p∧q)∨((∼p)∧q)∨(p∧(∼q))∨((∼p)∧(∼q)) is a tautologyAboth are false.Bonly (S1) is true.Cboth are true.Donly (S2) is true.Check answerSkip