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Boolean Algebra 2/23/01
Assume that a Boolean Algebra B has elements that take one of two
values 0 or 1.
Axioms - Rules of Boolean Algebra.
1a 0 • 0 = 0
1b 1 + 1 = 1
2a 1 • 1 = 1
2b 0 + 0 = 0
3a 0 • 1 = 1 • 0 = 0
3b 1 + 0 = 0 + 1 = 1
4a If x=0, then x'=1
4b If x=1, then x'=0
Single variable theorems from axioms:
5a x • 0 = 0 Prove by induction substituting x=0 and x=1
5b x + 1 = 1 and use the axioms.
6a x • 1 = x
6b x + 0 = x
7a x • x = x
7b x + x = x
8a x • x' = 0
8b x + x' = 1
9 x'' = x
1

Duality
Given a logic expression, its dual is obtained by replacing all
+ operations with • operations and vice versa, and by replacing all
0s with 1s and vice versa.
The dual of any true statement is also a true statement.
Pairs of single variable theorems are duals.
Two and three variable properties:
10a x • y = y • x Commutative
10b x + y = y + x (10a's dual)
11a x • (y • z) = (x • y) • z Associative
11b x + (y + z) = (x + y) + z
12a x • (y + z) = x • y + x • z Distributive
12b x + y • z = (x + y) • (x + z)
13a x + x • y = x Absorption
13b x • (x + y) = x
14a x • y + x • y' = x
14b (x + y) • (x + y') = x
15a (x • y)' = x' + y' DeMorgans Theorem
15b (x + y)' = x' • y'
16a x + x' • y = x + y Can prove by induction or by
16b x • (x' + y) = x + y algebraic manipulation.
2

Proof of DeMorgans Theorem 15b: (x + y) = x' • y'
x y
0 0
0 1
1 0
1 1
LHS
x+y (x+y)'
0 1
1 0
1 0
1 0
RHS
x' y' x'•y'
1 1 1
1 0 0
0 1 0
0 0 0
Problem 2.1 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
Problem 2.2 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
Truth tables or algebraic manipulation can be used to prove two
expressions are equivalent. Algebraic manipulation can also be
used to reduce an expression (or circuit).
3

Simplifican of Boolean Functions
x y f
0 0
0 1
1 0
1 1
1
1
1
0
x
y
f = x'•y' + x'•y + x•y'
= x'(y' + y) + xy'
= x'(1) + xy'
= x' + xy'
x
y
f
4
f

F = yz + xyz' + xyz + xy'z' + xy'z
y
z
x
y
z'
x
y
z
x
y'
z'
x
y'
z
F = yz + xyz' + xyz + xy'z' + xy'z
= yz + x(yz' + yz + y'z' + y'z)
= yz + x(y(z + z') + y'(z' + z))
= yz + x(y(1) + y'(1))
= yz + x(y + y')
= yz + x(1)
= yz + x
y
z
x
F
F
5

F(x,y,z) = ∑(1, 2, 3, 6)
= x'y'z + x'yz' + x'yz + xyz'
= x'y'z + x'yz + yz'(x'+x)
= x'y'z + x'yz + yz'
= x'z(y'+y) + yz'
= x'z(1) + yz'
= x'z + yz'
F(x,y,z) = ∑(1, 5, 6)
= x'y'z + xy'z + xyz'
= y'z(x'+x) + xyz'
= y'z + xyz'
6

Venn Diagrams:
x
x
y
x
x + y
x y
x y
x
x y
z
y
x y z
(x + y)'
x
z
y
y
x y + x z
7