1. 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

2. 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

3. 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.

4. 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.

5. 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.

6. 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)

7. 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

8. 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).

9. 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)

10. 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

11. 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.