ECE 331 – Digital System Design
Boolean Algebra
and
Standard Forms of Boolean Expressions
(Lecture #4)
The slides included herein were taken from the materials accompanying
Fundamentals of Logic Design, 6th Edition, by Roth and Kinney,
and were used with permission from Cengage Learning.
Basic Laws and Theorems
Operations with 0 and 1:
1. X + 0 = X 1D. X • 1 = X
2. X + 1 = 1 2D. X • 0 = 0
Idempotent laws:
3. X + X = X 3D. X • X = X
Involution law:
4. (X')' = X
Laws of complementarity:
5. X + X' = 1 5D. X • X' = 0
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Basic Laws and Theorems
Commutative laws:
6. X + Y = Y + X
6D. XY = YX
Associative laws:
7. (X + Y) + Z = X + (Y + Z)
=X+Y+Z
Distributive laws:
8. X(Y + Z) = XY + XZ
Simplification theorems:
9. XY + XY' = X
10. X + XY = X
11. (X + Y')Y = XY
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7D. (XY)Z = X(YZ) = XYZ
8D. X + YZ = (X + Y)(X + Z)
9D. (X + Y)(X + Y') = X
10D. X(X + Y) = X
11D. XY' + Y = X + Y
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Basic Laws and Theorems
DeMorgan's laws:
12. (X + Y + Z +...)' = X'Y'Z'...
12D. (XYZ...)' = X' + Y' + Z' +...
Duality:
13. (X + Y + Z +...)D = XYZ...
13D. (XYZ...)D = X + Y + Z +...
Theorem for multiplying out and factoring:
14. (X + Y)(X' + Z) = XZ + X'Y
14D. XY + X'Z = (X + Z)(X' + Y)
Consensus theorem:
15. XY + YZ + X'Z = XY + X'Z
15D. (X + Y)(Y + Z)(X' + Z) = (X + Y)(X' + Z)
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Duality (13)


The dual of a Boolean expression can be
written by

Replacing AND with OR, and OR with AND

Replacing 0 with 1, and 1 with 0

Leaving literals unchanged
See the Boolean laws and theorems,
previously discussed, for examples of Boolean
expressions and their duals.
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Distributive Law: Example #1
Distributive law (8): X.(Y + Z) = X.Y + X.Z
Use the distributive law to multiply out the
following Boolean expression:
F = (A+B).(C+D).(E+F)
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Distributive Law: Example #2
Distributive law (8D): X + Y.Z = (X+Y).(X+Z)
Use the distributive law to factor the following
Boolean expression:
F = A.B + C.D
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Simplification Theorems: Example #1
Simplification Theorems (9 – 11):
X.Y + X.Y' = X
(X+Y).(X+Y') = X
X + X.Y = X
X.(X+Y) = X
(X+Y').Y = X.Y
X.Y' + Y = X+Y
Use the simplification theorems to simplify the
following Boolean expression:
F = ABC' + AB'C' + A'BC'
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Simplification Theorems: Example #2
Simplification Theorems (9 – 11):
X.Y + X.Y' = X
(X+Y).(X+Y') = X
X + X.Y = X
X.(X+Y) = X
(X+Y').Y = X.Y
X.Y' + Y = X+Y
Use the simplification theorems to simplify the
following Boolean expression:
F = (A'+B'+C').(A+B'+C').(B'+C)
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Simplification Theorems: Example #3
Simplification Theorems (9 – 11):
X.Y + X.Y' = X
(X+Y).(X+Y') = X
X + X.Y = X
X.(X+Y) = X
(X+Y').Y = X.Y
X.Y' + Y = X+Y
Use the simplification theorems to simplify the
following Boolean expression:
F = AB'CD'E + ACD + ACF'GH' +ABCD'E +ACDE' + E'H'
(See Programmed Exercise 3.4 on page 75)
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Consensus Theorem: Example #1
Consensus Theorem:
(15)
X.Y + Y.Z + X'.Z = X.Y + X'.Z
(15D)
(X+Y).(Y+Z).(X'+Z) = (X+Y).(X'+Z)
Use the consensus theorem to simplify the
following Boolean expression:
F = ABC + BCD + A'CD + B'C'D'
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Consensus Theorem: Example #2
Consensus Theorem:
(15)
X.Y + Y.Z + X'.Z = X.Y + X'.Z
(15D)
(X+Y).(Y+Z).(X'+Z) = (X+Y).(X'+Z)
Use the consensus theorem to simplify the
following Boolean expression:
F = (A+C+D')(A+B'+D)(B+C+D)(A+B'+C)
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Consensus Theorem: Example #3
Consensus Theorem:
(15)
X.Y + Y.Z + X'.Z = X.Y + X'.Z
(15D)
(X+Y).(Y+Z).(X'+Z) = (X+Y).(X'+Z)
Use the consensus theorem to simplify the
following Boolean expression:
F = AC' + AB'D + A'B'C + A'CD' + B'C'D'
(See Programmed Exercise 3.5 on page 77)
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DeMorgan's Law
DeMorgan's Law:
(12)
(X + Y + Z + … )' = X'.Y'.Z'...
(12D)
(X.Y.Z… )' = X' +Y' + Z' …
Prove (using a truth table): (X+Y)' = X'.Y'
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X
Y
X+Y
(X + Y)'
X'
Y'
X'.Y'
0
0
0
1
1
1
1
0
1
1
0
1
0
0
1
0
1
0
0
1
0
1
1
1
0
0
0
0
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DeMorgan's Law
Graphical representation of DeMorgan's Law
x
x
y
x
y
y
(X.Y)'
X' + Y'
x
x
y
x
y
y
(X+Y)'
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X'.Y'
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DeMorgan's Law: Example
DeMorgan's Law:
(12)
(X + Y + Z + … )' = X'.Y'.Z'...
(12D)
(X.Y.Z… )' = X' +Y' + Z' …
Find the complement of the following Boolean
expression using DeMorgan's law:
F = (A + (BC)').((AD)' + C.(B' + D))
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Simplifying Boolean Expressions

Boolean algebra can be used in several ways
to simplify a Boolean expression:
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
Combine terms

Eliminate redundant or consensus terms

Eliminate redundant literals

Add redundant terms to be combined with or
allow the elimination of other terms
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Equivalency of Boolean Expressions


Two Boolean expressions are equivalent iff both
expressions evaluate to the same value for all
combinations of the variables in the expressions.
The equivalency can be proven using
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
A Truth table

Boolean algebra theorems to manipulate one
expression until it is identical to the other.

Boolean algebra theorems to reduce both
expressions independently to the same expression.
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Importance of Boolean Algebra



Boolean algebra is used to simplify Boolean
expressions.
Simpler expressions leads to simpler logic circuits.

Reduces cost

Reduces area requirements

Reduces power consumption
The objective of the digital circuit designer is to
design and realize optimal digital circuits.

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Thus, Boolean algebra is an important tool to the
digital circuit designer.
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Problem with Boolean Algebra


In general, there is no easy way to determine when a
Boolean expression has been simplified to a minimum
number of terms or a minimum number of literals.
Karnaugh Maps provide a better mechanism for the
simplification of Boolean expressions.
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Circuit Design: Example
For the following Boolean expression:
F(A,B,C) = A.B.C + A'.B.C + A.B'.C + A.B.C'
1. Draw the circuit diagram
2. Simplify using Boolean algebra
3. Draw the simplified circuit diagram
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Standard Forms of Boolean Expressions
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Standard Forms
There are two standard forms in which all
Boolean expressions can be written:
1. Sum of Products (SOP)
2. Product of Sums (POS)
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Sum of Products (SOP)


Product Term

Logical product = AND operation

A product term is the ANDing of literals

Examples: A.B, A'.B.C, A.C', B.C'.D', A.B.C.D
“Sum of”
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
Logical sum = OR operation

The sum of products is the ORing of product
terms.
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Sum of Products (SOP)



The distributive laws are used to multiply out a
general Boolean expression to obtain the sum
of products (SOP) form.
The distributive laws are also used to convert
a Boolean expression in POS form to one in
SOP form.
A SOP expression is realized using a set of
AND gates (one for each product term) driving
a single OR gate (for the sum).
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Product of Sums (POS)


Sum Term

Logical sum = OR operation

A sum term is the ORing of literals

Examples: A+B, A'+B+C, A+C', B+C'+D'
“Product of”
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
Logical product = AND operation

The product of sums is the ANDing of sum
terms.
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Product of Sums (POS)



The distributive laws are used to factor a
general Boolean expression to obtain the
product of sums (POS) form.
The distributive laws are also used to convert
a Boolean expression in SOP form to one in
POS form.
A POS expression is realized using a set of
OR gates (one for each sum term) driving a
single AND gate (for the product).
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SOP and POS: Examples
For each of the following Boolean expressions,
identify whether it is in SOP or POS form:
1.
2.
3.
4.
5.
6.
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F(A,B,C) = (A+B).(A'+B'+C').(B+C')
F(A,B,C) = A.B.C + B'.C' + A.C' + A'.B.C'
F(A,B,C) = A + B.C + B'.C' + A'.B'.C
F(A,B,C) = (A'+B'+C).(B+C').(A+C').(B')
F(A,B,C) = A.B.C + A'.(B+C) + (A+C').B
F(A,B,C) = A + B + C
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Questions?
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