Unit-5

Lattice & Boolean algebra

Q1) Define lattices.

A1)

Lattices-

A lattice is a partially ordered set in which every pair of elements a,b has a greatest lower bound and a least upper bound

For example-

Suppose S be a non-empty set and L = P(S); that means ( is partially ordered.

If X and Y are two elements of L, then-

Hence ( is a Lattice.

Q2) Let (L, ≤) be a lattice in which ‘⋅’ And ‘+’ denote the operations of meet and join respectively.

Then

a≤b ⇔a ⋅b = a ⇔a + b = b ∀a, b, c ∈L

A2)

Suppose a ≤b

As we know that a ≤a so that a ≤a.b but according to definition a.b≤a

So that- a ≤b ⇒a ⋅b = a

Let a ⋅b = a

But this is possible only if a ≤b which means a ⋅b = a ⇒a ≤b

∴a≤b⇒a ⋅b = a and a ⋅b = a ⇒a ≤b

Combine these two, we get

a≤b ⇔a ⋅b = a

Now let,

a⋅b = a,

Then we have

b + (a ⋅b) = b + a + a + b

But,

b + (a ⋅b) = b

Hence a + b = b

Similarly by assuming a + b = b

We can show that a ⋅b = a

Hence a ≤b ⇔a ⋅b = a ⇔a + b = b = a

Q3) Define isomorphic lattice.

A3)

Isomorphic Lattice-

Two lattices L and L’are said to be isomorphic if there is a one-to-one correspondence

f: L →L’

Such that

f (a ∧b) = f (a) ∧f (b) and f (a ∨b) = f (a) ∨f (b)

For any elements a, b in L.

Q4) What is direct product?

A4)

Let (L1, *, +) and (L2,∧, ∨) be two lattices. The algebraic system (L1 × L2,...,+)

In which the binary operation + and ‘’ ⋅are on L1 × L2 defined as

(a1, b1) ⋅(a2 ,b2 ) = (a1 ∗a2 , b1 ∧b2 )

(a1, b1 )+ (a2 ,b2 ) = (a1 ⊕a2 , b1 ∨b2 )

For all (a1, b1) and (a2, b2) (a2 ,b2 ) ∈L1 × L2

Is called the direct product of the lattices L1 and L2.

Q5) Define distributive lattice.

A5)

A lattice (L, ⋅, +) is called a distributive lattice if for any a, b, c ∈L

a⋅ (b + c) = (a ⋅b) + (a ⋅c) and

a + (b ⋅c) = (a + b) ⋅ (a + c)

Note- The power set of a non-empty set A is a lattice under the operation ∩and ∪is a distributive lattice.

Q6) Define Boolean algebra.

A6)

A Boolean algebra is a distributive complemented lattice having atleast two elements as well as 0 and 1.

A Boolean algebra is generally denoted by a 6-tuple, (B, +, ⋅, 1, 0, 1) where (B, +, ⋅) is a lattice with two binary operations + and ⋅, called the join and meet respectively is a unary operation in B. The elements 0 and 1 are the least and greatest elements of the lattice (B, +, ⋅). The following axioms are satisfied:

1. There exist at least two elements a, b in B and that a ≠b.

2. ∀a, b ∈B

(i) a+ b ∈B

(ii) a⋅b ∈B

3. for all a, b ∈B

(i) a+ b = b + a

(ii) a⋅b = b ⋅a

Commutative laws

4. Associative laws: for all a, b, c ∈B

(i) a+ (b + c) = (a + b) + c

(ii) a⋅(b ⋅c) = (a ⋅b) ⋅c

5. Distributive laws: for all a, b, c ∈B

(i) a+ (b ⋅c) = (a + b) ⋅(a + c)

(ii) a⋅(b + c) = (a ⋅b) + (a ⋅c)

6. (i) Existence of zero: There exists of B such that

a + 0 = a ∀a ∈B

The element 0 is called the zero element

(ii) Existence of unit: There exists 1 ∈B sum that

a⋅1 = a ∀a ∈B

The element 1 is called the unit element.

7. Existence of complement: ∀a ∈B there exists an element a′∈B such that

(i) a+ a′ = 1 and (ii) a ⋅a′ = 0

Q7) Define equivalent Boolean expression.

A7)

Two Boolean expression α() and β() are said to be equivalent if one can be obtained from the other by a finite number of application of the identities of Boolean algebra.

Q8) Simplify z (y+z) ((x + y + z).

A8)

z(y + z) (x + y + z)

= (z y + z z) (x + y + z)

= (z y + z) (x + y + z)

= z (y + 1) (x + y + z)

= z (x + y + z)

= z x + z y + z z

= z x + z y + z

= z (x + y + 1)

= z (x + 1)

= z

Q9) Simplify Y = (P + Q) (P + Q′) (P′+ Q)

A9)

Y = (P + Q) (P + Q′) (P′+ Q)

= (P P+ P Q′+ P Q + Q Q′) (P′+ Q)

= (P + P Q′+ P Q + 0) (P′+ Q)

= (P + P Q′+ P Q) (P′+ Q)

= P P1 + P Q + P Q′P′+ P Q′Q + P Q P′+ P Q Q

= 0 + P Q + 0 + 0 + 0 + P Q

= P Q + P Q

= P Q

Q10) Minimize the expression ++ A B

A10)

+ + A B

= + + + A B

= + + + A B

= + + A B

= + A B +

Hence