离散A(答案).docVIP

Method 1: (p ( (q ( r) ( (p ( ((q ( r) since s → t ≡?s∨t (((( p) ( ((q ( r) since s → t ≡?s∨t ( p ( ((q ( r) from the double negation law ( (q ( (p ( r) by the associative and commutative laws for disjunction ( q ( (p ( r) since s → t ≡?s∨t Consequently (p ( (q ( r) and q ( (p ( r) are logically equivalent. Method 2 : We construct the truth table for these propositions in the table. p q r (p ( (q ( r) q ( (p ( r) T T T T T T T F T T T F T T T T F F T T F T T T T F T F F F F F T T T F F F T T Since the truth values of (p ( (q ( r) and q ( (p ( r) agree, these propositions are logically equivalent. ∵a,b,c and m are integers such that m ( 2,c 0,and a ( b (mod m) ∴ a = km + b , where k is an integer. ∴ ac = (km + b) ( c = kmc + bc = k (mc) + bc ∴ ac ( bc (mod mc) Hence , if a,b,c,and m are integers such that m ( 2,c 0,and a ( b (mod m),then ac ( bc (mod mc) We find that A[2] = A ⊙ A = B[2] = B ⊙ B = So A[2] ( B[2] = ( = 4 The Hasse Diagram of ({2,4,6,9,12,18,27,36,48,60,72},|) is shown in Figure 1. . ( FIGURE 1 ) Since 27 and 48 have no upper bounds in ({2,4,6,9,12,18,27,36,48,60,72},|) , they certainly do not have a least upper bound. Hence , this poset is not a lattice. 5. Let R1 = {(a , c),(b , d),(c , a),(d , b),(e , d)}, then the reflexive closure of R 1is R1 ( ( = {(a , c),(b , d),(c , a),(d , b),(e , d)} ( {(a , a),(b , b),(c ,c),(d , d),(e , e)} = {(a , c),(b , d),(c , a),(d , b),(e , d), (a , a),(b , b),(c ,c),(d , d),(e , e)} Let R2 = {(a , c),(b , d),(c , a),(d , b),(e , d), (a , a),(b , b),(c ,c),(d , d),(e , e)}, then the symmetric closure of R2 is R2 ( R2 -1 = {(a , c),(b , d),(c , a),(d , b),(e , d), (a , a),(b , b),(c ,c),(d , d),(e , e)} ( {(c , a),(d , b),(a , c),(b , d),(d , e), (a , a),(b , b),(c ,c),(d , d),(e , e)} = {(a , c),(b , d),(c , a),(d , b),