# PPT - ECE/CS 352 On ```ECE/CS 352: Digital System Fundamentals
Lecture 5 – Basics of
Boolean Algebra
Based on slides by:Charles Kime &amp; Thomas Kaminski
&copy; 2004 Pearson Education, Inc.
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Outline
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Boolean Algebra
Boolean Properties and Identities
Boolean Algebraic Proofs
Useful Theorems
Algebraic Simplification
Complementing Functions
Function Evaluation
Chapter 2
2
Boolean Algebra
 An algebraic structure defined on a set of at least two elements, B,
together with three binary operators (denoted +, &middot; and ) that
satisfies the following basic identities:
1.
3.
5.
7.
9.
X + 0 = X Define existence 2.
X+1 =1
Of 0 and 1
X+X =X
Idempotence
Existence of
X+X =1
Complement
X = X Involution
10. X + Y = Y + X
12. (X + Y) + Z = X + (Y + Z)
14. X(Y + Z) = XY + XZ
16. X + Y = X . Y
4.
6.
8.
X .1 =X
X .0 =0
X .X = X
Dual Functions:
1. Swap +/ &middot;
2. Swap 0/1
X .X = 0
11. XY = YX
Commutative
Associative
13. (XY) Z = X(YZ)
15. X + YZ = (X + Y) (X + Z) Distributive
DeMorgan’s
17. X . Y = X + Y
Chapter 2
3
Some Properties of Identities &amp; the Algebra
 If the meaning is unambiguous, we leave out the symbol “&middot;”
 The identities above are organized into pairs. These pairs
have names as follows:
1-4 Existence of 0 and 1
5-6 Idempotence
7-8 Existence of complement 9 Involution
10-11 Commutative Laws
12-13 Associative Laws
14-15 Distributive Laws
16-17 DeMorgan’s Laws
 The dual of an algebraic expression is obtained by
interchanging + and &middot; and interchanging 0’s and 1’s.
 The identities appear in dual pairs. When there is only
one identity on a line the identity is self-dual, i. e., the
dual expression = the original expression.
Chapter 2
4
Some Properties of Identities &amp; the Algebra (Continued)
 Unless it happens to be self-dual, the dual of an
expression does not equal the expression itself.
 Example: F = (A + C) &middot; B + 0
dual F = (A &middot; C + B) &middot; 1 = A &middot; C + B
 Example: G = X &middot; Y + (W + Z)
dual G =((X+Y) &middot; (W &middot; Z)') = ((X+Y) &middot;(W' + Z')
 Example: H = A &middot; B + A &middot; C + B &middot; C
dual H = (A + B)(A + C)(B + C)
 Are any of these functions self-dual?
(A + BC)(B + C) = AB + AC + BC
Chapter 2
5
Boolean Operator Precedence
 The order of evaluation in a Boolean
expression is:
1. Parentheses
2. NOT
3. AND
4. OR
 Consequence: Parentheses appear
around OR expressions
 Example: F = A(B + C)(C + D)
Chapter 2
6
Example 1: Boolean Algebraic Proof
 A + A&middot;B = A
(Absorption Theorem)
Proof
A +
= A &middot;
= A &middot;
Justification
Steps
A&middot;B
1 + A &middot; B
( 1 + B)
= A &middot; 1
= A
X = X &middot; 1
X &middot; Y + X &middot; Z = X &middot;(Y + Z)
(Distributive Law)
1 + X = 1
X &middot; 1 = X
 Our primary reason for doing proofs is to learn:
•
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Careful and efficient use of the identities and theorems of
Boolean algebra, and
How to choose the appropriate identity or theorem to apply to
make forward progress, irrespective of the application.
Chapter 2
7
Example 2: Boolean Algebraic Proofs
 AB + A’C + BC = AB + A’C (Consensus Theorem)
Proof Steps
Justification
AB + A’C + BC
= AB + A’C + 1 &middot; BC
1 . X = X
= AB + A’C + (A + A’) &middot; BC X + X’ = 1
= AB + A’C + ABC + A’BC
X(Y + Z) = XY + XZ
(Distributive Law)
= AB + ABC + A’C + A’BC
X + Y = Y + X
(Commutative Law)
= AB . 1 + ABC + A’C . 1 + A’C . B
X . 1 = X, X . Y = Y
(Commutative Law)
= AB (1 + C) + A’C (1 + B) X(Y + Z) = XY +XZ
(Distributive Law)
= AB . 1 + A’C . 1 = AB + A’C
X . 1 = X
Chapter 2
.
X
8
Useful Theorems
 x &times;y + x &times;y = y (x + y )(x + y )= y
x  (x + y ) = x
 x + xy = x
 x + x &times;y = x + y x &times;(x + y )= x &times;y
 x &times;y + x &times;z + y &times;z = x &times;y + x &times;z
Minimization
Absorption
Simplification
Consensus
(x + y )&times;(x + z )&times;(y + z ) = (x + y )&times;(x + z )
 x + y = x &times;y
x &times;y = x + y
DeMorgan' s Laws
Chapter 2
9
Expression Simplification
 An application of Boolean algebra
 Simplify to contain the smallest number
of literals (complemented and
uncomplemented variables):
Absorption:
+
A B AXC+D
A BD + AC D + A BCD
XY+
=X
= AB + ABCD + A C D + A C D + A B D
Simplification:
+ X’Y
Y +ABD
= AB + AB(CD) + AXC
(D=+X +D)
= AB + A C + A B D = B(A + AD) +AC
= B (A + D) + A C 5 literals
Chapter 2
10
Complementing Functions
 Use DeMorgan's Theorem to
complement a function:
1. Interchange AND and OR operators
2. Complement each constant value and
literal
 Example: Complement F = xy z + x y z
F = (x + y + z)(x + y + z)
Chapter 2
11
Boolean Function Evaluation
F1 = xy z
F2 = x + yz
F3 = x y z + x y z + x y
F4 = x y + x z
x
0
0
0
0
1
1
1
1
y
0
0
1
1
0
0
1
1
z F1
0 0
1 0
0 0
1 0
0 0
1 0
0 1
1 0
F2
F3
F4
0
1
0
0
1
1
1
1
1
0
0
1
1
1
0
0
0
1
0
1
1
1
0
0
Chapter 2
12
Summary
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Boolean Algebra
Boolean Properties and Identities
Boolean Algebraic Proofs
Useful Theorems
Algebraic Simplification
Complementing Functions
Function Evaluation
Chapter 2
13
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