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 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 Given that a b we know that a a Therefore a is a lower bound of a&b But ab = a^b is glb of a&b. a  a^b But always a^b a Therefore a^b=a Lemma:3 Let (L, , ^) be a lattice, and a,b,c L prove that if avb=a+b c a^b= a b c By the assumption , a c, b c C is an upperbound of a&b We know that a+b is the least upper bound of a&b a+b c

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

State and prove the following laws in a lattice (L, ). 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 = a  (a  b) Similarly, a  (a  b)  a,  a, b  R. Further a  a, a  (a  b) a  a  a  (a  b)  a  a  (a  b)  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 = 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. But a  c is glb of a & c.  y  a  c  a  b  a  c.

Theorem: 4 State and prove modular inequality in a lattice. Proof: Let (L, ) be a lattice For any a, b, c  L, the following inequality holds in L. a  c iff a  (b  c)  (a  b)  c. Case (i) Let a  c. We know that b  a  b c  c,   b, c  L.  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  (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)

In a lattices if a b c, show that (i) ab = 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 = 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  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)  (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  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  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.