SLIDES FOR
CHAPTER 3
BOOLEAN ALGEBRA
This Objectives the book includes:
Study Guide
3.1 Multiplying Out and Factoring Expressions
3.2 Exclusive-OR and Equivalence Operations
3.3 The Consensus Theorem
3.5 Proving the Validity of an Equationng Expressions
Programmed Exercises
Problems
Click the mouse to move to the next page.
Use the ESC key to exit this chapter. Distributive Laws
Given an expression in product-of-sums form, the
corresponding sum-of-products expression can be
obtained by multiplying out, using the two distributive laws:
X(Y + Z) = XY + XZ (3-1)
(X + Y)(X + Z) = X + YZ (3-2)
In addition, the following theorem is very useful for
factoring and multiplying out:
(X + Y)(X′ + Z) = XZ + X′Y (3-3) In the following example, if we were to multiply out by brute
force, we would generate 162 terms, and 158 of these
terms would then have to be eliminated to simplify the
expression. Instead, we will use the distributive laws to
simplify the process.
Example (3-4), p. 63 The same theorems that are useful for multiplying out
expressions are useful for factoring. By repeatedly
applying (3-1), (3-2), and (3-3), any expression can be
converted to a product-of-sums form. Exclusive-OR and
Equivalence Operations
The exclusive-OR operation ( ) is defined as follows:
The equivalence operation ( ) is defined by: We will use the following
symbol for an
exclusive-OR gate:
Section 3.2, p. 64 The following theorems apply to exclusive OR:
Controlled inverter We will use the following
symbol for an
equivalence gate:
Section 3.2, p. 65 The equivalence gate is also called an
exclusive-NOR gate. Example 1:
By (3-6) and (3-17),
F = [(A′B)C + (A′B)′C′ ] + [B′(AC′) + B(AC′)′ ]
= A′BC + (A + B′)C′ + AB′C′ + B(A′ + C)
= B(A′C + A′ + C) + C′(A + B′ + AB′) = B(A′ + C) + C′(A + B′)
Example 2:
= (A′B′ + AB)C′ + (A′B′ + AB) ′C (by (3-6))
= (A′B′ + AB)C′ + (A′B + AB′)C (by (3-19))
= A′B′C′ + ABC′ + A′BC + AB′C
Section 3.2 (p. 66) The Consensus Theorem
The consensus theorem can be stated as follows:
XY + X'Z + YZ = XY + X'Z (3-20)
Dual Form:
(X + Y)(X’ + Z)(Y + Z) = (X + Y)(X’ +(3-21)
Section 3.2 (p. 66-67) Consensus Theorem
Proof
XY + X'Z + Y= (XY + XYZ) + (X'Z + X'YZ)
= XY(1 + Z) + X'Z(1 + Y) = XY + X'Z
Section 3.3 (p. 67) Basic methods for
simplifying functions
1. Combining terms. Use the theorem XY + XY′ = X to combine
two terms. For example,
abc′d′ + abcd′ = abd′ [X = abd′, Y = c] (3-24)
2. Eliminating terms. Use the theorem X + XY = X to eliminate
redundant terms if possible; then try to apply the consensus
theorem (XY + X′Z + YZ = XY + X′Z) to eliminate any consensus
terms. For example,
a′b + a′bc = a′b [X = a′b]
a′bc′ + bcd + a′bd = a′bc′ + bc[X = c, Y = bd, Z = a′b] (3-24)
Section 3.4 (p. 68-69) 3. Eliminating literals. Use the theorem X + X’Y = X + Y to
eliminate redundant literals. Simple factoring may be necessary
before the theorem is applied.
A′B + A′B′C′D′ + ABCD′ = A′(B + B′C′D′) + ABCD′
= A′(B + C′D′) + ABCD′
= B(A′ + ACD′) + A′C′D′
= B(A′ + CD′) + A′C′D′
= A′B + BCD′ + A′C′D′ (3-26) 4. Adding redundant terms. Redundant terms can be
introduced in several ways such as adding xx′, multiplying
by (x + x′), adding yz to xy + x′z, or adding x

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