LATTICES AND BOOLEAN ALGEBRA Lemma 1 Show that

LATTICES AND BOOLEAN ALGEBRA Lemma: 1 Show that a b a, b a+b for any two elements a&b in a lattice( L, ) Proof: Let a, b L we know that a b is glb of a & b a^b= a b is a lower bound of a and b a b a, a b b Also a+b=a b is a lub of a&b a+b is an upper bound of a & b a a+b; b a+b Lemma: 2 Let (L, ) be a lattice, and a, b L (i)a b =b (ii) a^b=a If a b, then show that Proof: (i) Given that a b we know that b b b is upper bond of a&b But a b a+b is lub of a& b but always b (a b ) a b b but always b (a b ) § a b=b

LATTICES AND BOOLEAN ALGEBRA § § § Theorem : 1 Let (L, , ) be a lattice, and a, b, c, d, L. (i) a b c d (ii) a b c d § Proof: (i) Let x = c d, x = is lub of c & d x is also an upper bound of c & d. c x, d x. Now a c, c x a x; b d, d x b x. x is also an upper bound of a & b. & a b is lub of a & b Thus a b x = c d. (ii) Let y = a b y is glb of a & b. y is lower bound of a & b. y a; y b. Now y a, a c y c. y b, b d y d. y is a lower bound of c & d. c d is glb of c & d. Thus y c d a b c d

§ Theorem : 2 § State and prove the following laws in a lattice (L, ). § § (i) absorption laws (ii) idempotent laws (iii) Commutative laws. Proof: Case (i) The absorption laws are a a ( ( b) = a; a (a b) = a, § We know that a a (a b), a, b L, Also a b a, a a a (a b) a a = a (a b) Similarly, a (a b) a, b R. Further a a, a (a b) a a a (a b) = a Case (ii) The idempotent laws are a a ; a a = a. We have a a a, a a a. But a a is lub of a & a. a a is glb of a a & a. Here a a a, a a a = a; a a = a Case (iii) The commutative laws are a b = b a ; a b = b a. For this, a b is lub of a & b. b a is lub of b & a. We know that the lub of a & b = lub of b & a, a b = b a. Similarly a b = b a.

§ § § Theorem : 3 State and prove Isotonicity property in a lattice Proof: Let (L, ) be a lattice. For a, b, c, L, the following properties called isotonicity laws. b c a * b a * c ; a b a c. (ie) b c a b a c ; a b a c. Let us assume that b c. Claim : a b a c Let x = a c. Then x is lub of a & c. x is an upper bound of a & c. a x, c x But b c, c x b x Also a x. x is upper bound of a & b But a b is lub of a & b. a b x = a c. Claim : a b a c. Let y = a b y is glb of a & b y is a lower bound of a & b y a, y b. Using b c, y c. y is a lower bound of a & c.

§ § § Theorem: 4 State and prove modular inequality in a lattice. Proof: Let (L, ) be a lattice § Case (i) Let a c. We know that b a b c c, b, c L. For any a, b, c L, the following inequality holds in L. a c iff a (b c) (a b) c. c is an upper bound of a & b c. But x = a (b c) is lub of a a & b c. x = a (b c) c Now we know that a a, b c b By isotonicity property, x=a (b c) a b By (1) and (2) x x (a b) c x=a (b c) (a b) c Case(ii): Conversly, let a (b c) (a b) c we have a a y , V y L a (a b) c (by assumption) c a c hence the proof.

§ Theorem: 5 § Let (L, ) be a lattice. For any a, b, c L § Prove the following inequailities hold in L. § § (i) a (b c) (a b) (a c) (ii) a (b c) > (a b) (a c) § Proof: Let a, b, c L. § (i) a a b; a a c. a = a a (a b) (a c) (by a result)………. . (1) Also b a b ; c a c b c (a b) (a c) (by a result)……………. . (2) By (1) & (2), a (b c) (a b) (a c) (ii) We know that a b a, a c a § § § = (a b) (a c) a a=a (by a result)………. . (3) Also a b b, a c c = (a b) ( a c) c………………. (4) By (3) & (4) (a b) (a c) a (b c)

§ Theorem: 6 § In a lattices if a b c, show that § (i) a b = b c § (ii) (a b) (b c) = (a b) ( a c) § Proof: § a b a b=b, a b =a § b c b c=c, b c =b § a c a c=c, a c =a § a b = b c ……………. . . (i) follows § Now (a b) ( b c) = a b =b § (a b) (a c) = b c =b ………………. . (ii) follows

§ § § § Theorem: 7 In a lattice (L, ), show that (i) (a b) (c d) (a c) (b d) (ii) (a b) (b c ) (c a) (a b) (b c) (c a), a, b, c L. Proof: Let a, b, c L. Then a b a (or) b a b a c a a b b b c Using (1), (2) & (3), we get a b (a b) (b c) (c a) Similarly b c (a b) (b c) (c a) c a (a b) (b c) (c a) This proves (ii) We have a a c, b b d (a b) (b d) We know that c a c d b d c d (a b) (b d) By (4) & (5), (a b) (c d) (a b) (b d). This proves (i).

§ § § § Theorem: 8 In a lattice (L, ), Prove that for a, b, c L, (i) (a b) (a c) a (b (a c)) (ii) (a b) (a c) ≥ a (b (a c)) Proof: We know that a b a, a c a (a b) (a c) a a = a Also a b b, a c. (a b) (a c) b (a c) From (1) & (2), (a b) (a c) a (b (a c)) This proves (i) We know that a a b ; a a c a = a a (a b) (a c) Further b a b ; a c b (a c) (a b) (a c) By (3) & (4), a (b (a c)) (a b) (a c). This proves (ii)

§ § Definition: (Distributive lattice): Let (L, ) be a lattice under (ie) (L, ) is a lattice in which both lub and glb of any two elts exist in L). Then (K ) is called distributive lattice iff § § a (b c) = (a b) (a c) a (b c) = (a b) (a c) , a, b, c L

§ § Theorem: 9 Show that every chain is a distributive lattice. Proof: Let (l, ) be a chain. Let a, b, c L. Then there are the following possible cases. Case (i): a b c Then a (b c) = a c =a (a b) (a c) = a a = a a (b c) = a b = b ( b c = c) (a b) (a c) = b c = b ( a b = b; a c = c) Both distributive laws hold. Case (ii); Let a ≥ b ≥ c. The a c = b; a b = a, a c = c; a c = a Now a (b c) = a b = b (a b) (a c) = b c = b Also a (b c) = a b = a; ( b c a), (a b) (a c) = a a = a Both distributive laws hold.