BOOLEAN ALGEBRA.ppt

Boolean Algebra
 The basis for Boolean algebra:
 Boolean algebra is the algebra of binary values.
 we refer to these values as 0 and 1
 although we could also refer to them as true and false
 the basic Boolean algebra operations are
 AND, OR, NOT.

Basic Boolean Identities
 As with algebra, there will be Boolean
operations that we will want to simplify
 We apply the following Boolean identities to help
 For instance, in algebra, x = y * (z + 0) + (z * 0) can be
simplified to just x = y * z

Terminology
 Element 0 is called “FALSE”.
 Element 1 is called “TRUE”.
 ‘+’ operation “OR”,‘*’ operation “AND” and ’ operation “NOT”.
 Juxtaposition implies * operation: ab = a * b
 Operator order of precedence is: (), ’, *, +.
a+bc = a+(b*c) ≠ (a+b)*c
ab’ = a(b’) ≠ (a*b)’
 Single Bit Boolean Algebra(1’ = 0 and 0’ = 1)
+ 0 1
0 0 1
1 1 1
* 0 1
0 0 0
1 0 1

Truth Tables
xy = x AND y = x * y x + y = x OR y x bar = NOT x
AND is true only if OR is true if either NOT inverts the bit
both inputs are true inputs are true We will denote x bar as ~X

Boolean Functions and
Expressions
Definition: Let B = {0, 1}. The variable x is called
a Boolean variable if it assumes values only from
B.
A function from Bn, the set {(x1, x2, …, xn) |xiB,
1  i  n}, to B is called a Boolean function of
degree n.
Boolean functions can be represented using
expressions made up from the variables and
Boolean operations.

Expressions
The Boolean expressions in the variables x1,
x2, …, xn are defined recursively as follows:
 0, 1, x1, x2, …, xn are Boolean expressions.
 If E1 and E2 are Boolean expressions, then (-
E1),
(E1E2), and (E1 + E2) are Boolean expressions.
Each Boolean expression represents a Boolean
function. The values of this function are obtained
by substituting 0 and 1 for the variables in the
expression.

Boolean Expressions
 We form Boolean expressions out of Boolean
operations on Boolean variables or Boolean values
 So, like algebraic expressions, we can create more
complex Boolean expressions as we might need
 Consider the expression:
 F = X + ~Y*Z
 What is it’s truth table?
Notice that it is easier to derive the
truth table for the entire expression
by breaking it into subexpressions
So first we determine ~Y
next, ~Y * Z
finally, X + ~Y*Z

Expressions
Example: Give a Boolean expression for the
Boolean function F(x, y) as defined by the
following table:
x y F(x, y)
0 0 0
0 1 1
1 0 0
1 1 0
Possible solution: F(x, y) = (-x)y

Expressions
Another Example: Possible solution I:
F(x, y, z) = -(xz + y)
0
0
1
1
F(x, y, z)
1
0
1
0
z
0
0
1
0
1
0
0
0
y
x
0
0
0
1
1
0
1
0
1
1
1
1
0
1
0
1
Possible solution II:
F(x, y, z) = (-(xz))(-y)

Duality
The dual of a Boolean expression is obtained by
interchanging Boolean sums and Boolean
products and interchanging 0s and 1s.
Examples:
The dual of x(y + z) is x +( yz)
The dual of -x1 + (-y + z) is (-x + 0)((-y)z).
The dual of a Boolean function F represented by a Boolean
expression is the function represented by the dual of this expression.

The dual of a statement S is obtained by interchanging * and +; 0
and 1.
Dual of (a*1)*(0+a’) = 0 is (a+0)+(1*a’) = 1
Dual of any theorem in a Boolean Algebra is also a theorem.
This is called the Principle of Duality.
Principle of Duality

Theorem 1
The complement of x is unique
Proof :
Assume x1' and x2' are both complements of x.
x + x1' = 1, x • x1' = 0, x + x2' = 1, x • x2' = 0
x1' = x1' • 1 1 is the identity for AND
= x1' • (x + x2') substitution, x + x2' = 1
= (x1' • x) + (x1' • x2') AND distributes over OR
= (x • x1') + (x1' • x2') AND is commutative
= 0 + (x1' • x2') substitution, x • x1' = 0
= (x • x2') + (x1' • x2') substitution, x • x2' = 0
= (x2' • x) + (x2' • x1') AND is commutative, twice
= x2' • (x + x1') AND distributes over OR
= x2' • 1 substitution, x + x1' = 1
= x2' 1 is the identity for AND
Thus, any two elements that are the complement of x are equal.
This implies that x' is unique

Theorem 2
x + 1 = 1
Proof:
x + 1 = 1 • (x + 1) 1 is the identity for AND
= (x + x') • (x + 1) Complement, x + x' = 1
= x + (x' • 1) OR distributes over AND
= x + x' 1 is the identity for AND
= 1 Complement, x + x' = 1
Theorem 3
AND's identity is the complement of OR's identity
0' = 1
Proof:
0' = 0 + 0' 0 is the identity for OR
= 1 Complement, x + x' = 1

Theorem 4
Idempotent
x + x = x
Proof:
x + x = (x + x) • 1 1 is the identity for AND
= (x + x) • (x + x') Complement, x + x' = 1
= x + (x • x') OR distributes over AND
= x + 0 Complement, x • x' = 0
= x 0 is the identity for OR

Theorem 5
Involution
(x')' = x
Proof:
Let x' be the complement of x and (x')' be the complement of x'.
x + x' = 1, xx' = 0, x' + (x')' = 1, and x'(x')' = 0
(x')' = (x')' + 0 0 is the identity for OR
= (x')' + xx' Substitution, xx' = 0
= [(x')' + x][(x')' + x'] OR distributes over AND
= [x + (x')'][x' + (x')'] OR is commutative (P3), twice
= [x + (x')'] • 1 Substitution, x' + (x')' = 1
= [x + (x')'][x + x'] Substitution, x + x' = 1
= x + [(x')' • x'] OR distributes over AND
= x + [x' • (x')'] AND is commutative
= x + 0 Substitution, x'(x')' = 0
= x 0 is the identity for OR

Theorem 6
Absorption
x + xy = x
Proof:
x + xy = (x • 1) + xy 1 is the identity for AND
= x(1 + y) AND distributes over OR
= x(y + 1) OR is commutative
= x • 1 x + 1 = 1 (Thm 2-1)
= x 1 is the identity for AND
Theorem 7
x + x'y = x + y
Proof:
x + x'y = (x + x') (x + y) OR distributes over AND
= 1 • (x + y) Complement x + x' = 1
= x + y 1 is the identity for AND

Theorem 8
DeMorgan's Law 1
(x + y)' = x' y'
Proof:
By Theorem 1 (complements are unique) and Postulate P9 (complement), for every x in a Boolean
algebra there is a unique x' such that
x + x' = 1 and x • x' = 0
So it is sufficient to show that x'y' is the complement of x + y. We'll do this by showing that (x + y)
+
(x'y') = 1 and (x + y) • (x'y') = 0
(x + y) + (x'y') = [(x + y) + x'] [(x + y) + y'] OR distributes over AND
= [(y + x) + x'] [(x + y) + y'] OR is commutative
= [y + (x + x')] [x + (y + y')] OR is associative,twice
= (y + 1)(x + 1) Complement, x + x' = 1, twice
= 1 • 1 x + 1 = 1 (Thm 2-1), twice
= 1 Idempotent, x • x = x (Thm 4-2)
Also,
(x + y)(x'y') = (x'y')(x + y) AND is commutative
= [(x'y')x] + [(x'y')y] AND distributes over OR
= [(y'x')x] + [(x'y')y] AND is commutative
= [y'(x'x)] + [x'(y'y)] AND is associative (Thm 8-2),twice
= [y'(xx')] + [x'(yy')] AND is commutative, twice
= [y' • 0] + [x' • 0] Complement, x • x' = 0, twice
= 0 + 0 x • 0 = 0 (Thm 2-2), twice
= 0 Idempotent, x + x = x (Thm 4-1)

Prove x + yz = (x + y)(x + z)
x + yz = (x + y)(x + z)
= xx + xz + xy + yz
= x + xz + xy + yz
= x (1 + z + y) + yz
= x (1) + yz
= x + yz
Prove (x + y) · (x + y') = x
x = (x + y) · (x + y')
= xx + xy + xy' + yy'
= x + xy + xy' + yy'
= x + xy + xy' + 0
= x (1 + y + y')
= x (1)
= x

BOOLEAN ALGEBRA.ppt

More Related Content

What's hot (20)

Similar to BOOLEAN ALGEBRA.ppt (20)

Recently uploaded (20)

BOOLEAN ALGEBRA.ppt