The remainder on dividing 5995^{99}599 by 11 is
(S1)(p⇒q)∨(p∧(∼q))( S 1)(p \Rightarrow q) \vee(p \wedge(\sim q))(S1)(p⇒q)∨(p∧(∼q)) is a tautology(S2) ((∼p)⇒(∼q))∧((∼p)∨q)((\sim p) \Rightarrow(\sim q)) \wedge((\sim p) \vee q)((∼p)⇒(∼q))∧((∼p)∨q) is a contradictionThen