264 Logic as a Tool (e) Since ¬C ∧ B is true then B is true and ¬C is true, hence C is false Since ¬(A ∨ C ) → C is true and C is false, then ¬(A ∨ C ) is false, that is, A ∨ C is true Now since C is false, A must be true 1.1.7 (b) Suppose R is false Since P ∨ R is true, then P is true For (P ∨ Q) → R to be true, P ∨ Q must be false Hence P and Q is false Contradiction Hence R is true (d) Suppose Q is true Since ¬P → ¬Q must be true, then ¬P must be false and thus P is true For P ∧ ¬R to be false, ¬R must be false But this is a contradiction with ¬R true Hence Q is false (e) Suppose P is true Since P → Q is true then ¬Q is false Hence S ∧ ¬Q is false But R ∨ (S ∧ ¬Q) is true so R must be true This contradicts ¬R is true, which was given, hence P must be false (i) Since P → R is false then P is true and R is false But with P → Q true, this means Q is true Since Q → (R ∨ S ) is true then R ∨ S must be true Since R is false we have that S is true 1.1.8 (g) Tautology: p q (p ∨ ¬q ) → ¬(q ∧ ¬p) F F T T F T F T F F T T T F T T T F T F T T T T ↑ T F T T F T F F T T F F F T F T (i) Neither a tautology nor a contradictory formula: p F F F F T T T T q F F T T F F T T r F T F T F T F T ¬(¬p TT TT FT FT FF FF TF TF ↔ F F T T T T F F q) F F T T F F T T ∧ T T F F F F F T ↑ (r F T F T F T F T ∨ T T F T T T F T ¬q ) T T F F T T F F (j) Contradictory formula: p F F F F T T T T q F F T T F F T T r F T F T F T F T ¬ F F F F T F F F ((p ∧ ¬q ) → r) FFTTF FFTTT FFFTF FFFTT TTTFF TTTTT TFFTF TFFTT ↔(¬(q ∨ r) FTFFF FFFTT FFTTF FFTTT FTFFF FFFTT FFTTF FFFTT ↑ → T T T T F T T T ¬p) T T T T F F F F