MathematicsMedium115×since 2008Q4367Consider the following two propositions: P1:∼(p→∼q)P1: \sim (p \to \sim q)P1:∼(p→∼q) P2:(p∧∼q)∧((∼p)∨q)P2:(p \wedge \sim q) \wedge (( \sim p) \vee q)P2:(p∧∼q)∧((∼p)∨q) If the proposition p→((∼p)∨q)p \to (( \sim p) \vee q)p→((∼p)∨q) is evaluated as FALSE, then :AP1 is TRUE and P2 is FALSEBP1 is FALSE and P2 is TRUECBoth P1 and P2 are FALSEDBoth P1 and P2 are TRUECheck answerSkip