《离散数学》试题及答案(5)

(4) C→ B (5) B→C (6) C

P Q(4) Q(3)(5) P Q(6)(7) D(1)(8)

(7) C→D (8) D

(9) A→D

所以 { A∨B, C→ B, C→D}蕴涵A→D. 4. 证明：A－(A∩B)

= A∩~(A∩B) ＝A∩(~A∪~B) ＝(A∩~A)∪(A∩~B) ＝ ∪(A∩~B) ＝(A∩~B) ＝A－B 而 (A∪B)－B

= (A∪B)∩~B = (A∩~B)∪(B∩~B) = (A∩~B)∪ = A－B

所以：A－(A∩B) = (A∪B)－B.