boolenalgebralecturenotes-220915100409-6e6a97bd (1).pptx

BOOLEAN ALGEBRA
Let X be a nonempty set with two binary operations + and ∗, a unary
operation ‘, and two distinct elements 0 and 1. Then X is called a Boolean
algebra if the following axioms hold where A, B, C are any elements in X:
Commutative Laws
A + B = B
+ A A ∗ B =
B ∗ A
Distributive Laws
A ∗ B +
C = (A
∗ B) + (A ∗ C) A + B ∗
C =
A +
B ∗ (A
+ C)
Complement Laws

BOOLEAN ALGEBRA
The operations +, ∗, and ‘ are called sum, product, and
complement, respectively.
Boolean Algebra is the mathematics of digital systems.
Definition of terms
 Variable- symbol used to represent a logical quantity. Any variable can
have a value of 1 or 0.
 Complement- inverse of a variable. A or A
 Literal is a variable/ complement of a variable.
Boolean addition is equivalent to the OR operation.
Basic rules for Boolean addition:
0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 =
1
In Boolean algebra, a sum term is a sum of
literals. e.g. A + B

BOOLEAN ALGEBRA
Basic rules useful in manipulating and simplifying Boolean expressions:
Idempotent laws:
𝐀 ∗ 𝐀
= 𝐀
𝐀 + 𝐀 = 𝐀
Boundedness laws:
𝐀 + 1 = 𝟏 𝐀 ∗ 𝟎 = 𝟎
Absorption laws:
𝐀 + 𝐀 ∗ 𝐁
= 𝐀
𝐀 ∗ (𝐀 + 𝐁) = 𝐀
Involution law: 𝐀 = 𝐀
Boolean multiplication is equivalent to the AND operation.
Basic rules for Boolean multiplication:
0 ∗ 0 = 0 0 ∗ 1 = 1 1 ∗ 0 = 1 1 ∗ 1 = 1
In Boolean algebra, a product term is a product of
literals. e.g A ∗ B

BOOLEAN ALGEBRA
De Morgan Laws
The complement of a product of variables is equal to the sum of the
complements of the variables
XY = X + Y
The complement of a sum of variables is equal to the product of
the complements of the variables
X + Y = X. Y

BOOLEAN ALGEBRA
Simplification of Boolean expressions using Boolean Algebra
 Its reducing a particular expression to its simplest form or change its form
to a more convenient one.
 Use basic laws, rules and theorems of Boolean Algebra
 Simplified Boolean expression uses the fewest gates possible to
implement
a given expression.
Example:
Simplify the following Boolean Expression:
𝐀𝐁 + 𝐀𝐂 + 𝐀𝐁𝐂
= 𝐀𝐁 𝐀𝐂 + 𝐀𝐁𝐂
= ( 𝐀 + 𝐁 )(𝐀 + 𝐂) +
𝐀𝐁𝐂
apply De Morgan law
apply De Morgan law
apply the rule 𝐀 + 𝐁𝐀 + 𝐂 = 𝐀
+ 𝐁𝐂
= (𝐀 + 𝐁𝐂) +
𝐀𝐁𝐂
= 𝐀 𝟏
+ 𝐁𝐂
+
𝐁𝐂
apply the rule 𝟏 + 𝐀 =
𝟏
= 𝐀
+ 𝐁𝐂

BOOLEAN ALGEBRA
Truth Table
 It is a table showing the inputs and the corresponding outputs of a logic
expression or circuit.
Steps taken when constructing a truth table
 List the input variables combinations of 0s and 1s in a binary
sequence ( 2n combinations for n inputs).
 Place a 1 or 0 in the output column for each combination of input
variables that was determined in the evaluation.
Example
Construct a truth table and put the results of logic circuit A(B +
CD)
 4 variables means 24 combinations for n inputs.

A B C D A(B + CD)
0 0 0 0 0
0 0 0 1 0
0 0 1 0 0
0 0 1 1 0
0 1 0 0 0
0 1 0 1 0
0 1 1 0 0
0 1 1 1 0
1 0 0 0 0
1 0 0 1 0
1 0 1 0 0
1 0 1 1 1
1 1 0 0 1
1 1 0 1 1
1 1 1 0 1
1 1 1 1 1

BOOLEAN ALGEBRA
Logic Gates
 A logic gate is an circuit that perform basic logical operation.
 In reality, gates consist of one to six transistors, but digital designers
think of them as a single unit.
 Integrated circuits contain collections of gates suited to a particular
purpose.
 The three simplest gates:
 AND
 OR
 NOT

BOOLEAN ALGEBRA
Distinctive Shape
Symbol
Truth Table

BOOLEAN ALGEBRA
Not gate/ Inverter
 performs inversion/ complementation operation- changes one logic level
to the opposite level
 The negation indicator is a bubble.

BOOLEAN ALGEBRA
And Gate
 Its composed of 2/ more inputs and a single output.
 the total number of possible combinations of binary inputs is
given by N = 2n where N is the input combinations and n is the
number of input variables.
 performs logical multiplication
 produces a HIGH output when all of the inputs are HIGH

BOOLEAN ALGEBRA
Or Gate
 Its composed of 2/ more inputs and a single output.
 performs logical addition.
 produces a HIGH output when any of the inputs is HIGH.

BOOLEAN ALGEBRA
NAND gate
 Contraction of NOT-AND i.e. AND function with a complemented output.
 produces a LOW output only when all of the inputs are HIGH; the output
will be HIGH when any of the inputs is LOW.
 DeMorgan’s Theorem: AB = A + B
NOR gate
 Contraction of NOT-OR i.e. OR function with a complemented
output.
 produces a LOW output when any of the inputs are HIGH; the output will
be HIGH when all of the inputs are LOW.
 DeMorgan’s Theorem: A + B = AB

BOOLEAN ALGEBRA
Truth Table Distinctive Shape
Symbol

BOOLEAN ALGEBRA
 Boolean algebra provides a concise way to express the operation of a logic
circuit formed by a combination of logic gates so that the output can be
determined for various combinations of input values.
 To derive a Boolean expression for a given logic circuit, begin at the left-
most inputs and work towards the final output, writing the expression for
each gate.
Example:
𝐀
𝐀 + 𝐁
𝐀 𝐀 + 𝐁 =
𝐀𝐁
𝐀
𝐁
𝐀 + 𝐁 +
𝐀𝐁

BOOLEAN ALGEBRA
Standard Forms of Boolean Expressions
 Standardization makes the evaluation, simplification and implementation of
Boolean expression much more systematic and easier.
 Domain of Boolean expression is the set of variables contained in the
expression in complemented or un complemented form.
 The domain of the expression 𝐀 + 𝐀𝐁𝐂 + 𝐁𝐂𝐃 is the set of
variables
A, B, C ,D.
 All Boolean expressions, regardless of their form can be converted into 2
standard forms:
 Sum-of-Product
 Product-of-Sum

BOOLEAN ALGEBRA
Sum of Products
 Its formed when 2 or more product terms that are summed by
Boolean addition.
 It can contain a single-variable term as in 𝐀 + 𝐀𝐁𝐂 + 𝐁𝐂𝐃
 In SOP expression, a single overbar cannot extend over more than one
variable.
 SOP expression can have 𝐀𝐁𝐂 but not 𝐀𝐁𝐂
 Implementing a SOP expression simply ORing the outputs of 2 or more
AND
gates( AND-OR implementation of SOP expression).
 SOP can be implemented by NAND-NOR.
 Conversion of a general expression to SOP form is done using Boolean
algebra techniques.
 A standard SOP expression is one in which all the variables in the domain
appear in each product term in the expression
NB : Important in constructing truth table and in the K-map simplification

BOOLEAN ALGEBRA
 To converting product terms to standard SOP
 Multiply each nonstandard product term by A + A = 1 where A is
the missing variable.
 Repeat the step above until all resulting product terms contain all
variables in the domain.
 A standard product term = 1 for only one combination of variable
values.
 A product term is implemented with an AND gate whose output is 1 only if
each of its inputs is 1.
 SOP expression is =1 only if one or more product terms in the expression
=1.
ABCD = 1010 = 1

BOOLEAN ALGEBRA
Example:
Convert ABC + AB + ABCD into standard SOP form.
Standardize the first product term
ABC(D + D) = ABCD + ABCD
Standardize the second product term
AB (C + C) = ABC
+ ABC ABC(D + D) =
ABCD + ABCD ABC (D +
D) = ABCD + ABCD
Standard SOP
ABCD+ ABCD + ABCD+ ABCD +
ABCD+ ABCD + ABCD

BOOLEAN ALGEBRA
Product of Sums
 Formed when 2 or more sum terms multiplied.
 A POS expression can contain a single-variable term.
 In a POS expression, a single overbar cannot extend over more than
one variable.
 Implementing a POS expression requires ANDing the outputs of 2 or
more OR gates.
 A standard POS expression is one in which all variables in the domain
is in each sum term.
To convert sum terms to standard POS
 Add to each non-standard product term AA = 0
where A is the
missing variable.
 Apply rule A + BC = A + BA + C
 Repeat step 1, until all resulting sum terms contain all variables in

BOOLEAN ALGEBRA
 A standard sum term =0 for only one combination of variable values and is 1
for all other combinations of values for variables.
A + B + C + D = 0 + 1 + 0 + 1 = 0
 A sum term is implemented with OR gate whose output is 0 only if each of
its inputs =0.
 A POS expression = 0 only if one or more of the sum in the expression =0.

BOOLEAN ALGEBRA
Standard SOP to Standard POS Conversion
 Binary values of the product terms in a given standard SOP expression are
present in the equivalent standard POS expression.
 To converting standard SOP to standard POS
 Evaluate each product term in the SOP. Determine the numbers that
represent the product term.
REMEMBER: Standard product term =1
 Determine all binary not included in the evaluation in 1
 Write the equivalent sum term for each binary number from step 2 and
expression in POS form.

BOOLEAN ALGEBRA
Example
Convert the SOP expression to an equivalent POS expression:
ABC + ABC + ABC + ABC + ABC
Binary values are 000, 010, 011, 101,111
There are 3 variables in the domain of this expression and 23
possible combinations.
SOP expression contains 5 of these combinations so the POS must
contain the
other which are 001, 100 and 110.
A + B + C A + B + C
(A + B + C)
 To convert a standard POS to a standard SOP
 Evaluate each sum term in the POS. Determine the numbers
that
represent the sum term.
REMEMBER: Standard sum term =0

BOOLEAN ALGEBRA
 Steps considered when entering a SOP expression into the TRUTH
Table
 Convert the SOP expression standard form
 List all possible combinations of binary values of the variables in the
variables in the expressions.
 Place 1 in the output column for each binary value that makes the
standard SOP =1.
 Place 0 for all the remaining binary values.
Example
Develop a truth table for the standard SOP expression ABC + ABC +
ABC.
ABC: 001 ABC: 100 ABC: 111

BOOLEAN ALGEBRA
Inputs Output
A B C X Product Term
0 0 0 0
0 0 1 1 ABC
0 1 0 0
0 1 1 0
1 0 0 1 ABC
1 0 1 0
1 1 0 0
1 1 1 1 ABC

BOOLEAN ALGEBRA
 Steps considered when entering a POS expression into
the TRUTH Table
 Convert the POS expression standard form
 List all possible combinations of binary values of the variables in the
variables in the expressions.
 Place 0 in the output column for each binary value that makes the
standard POS=0.
 Place 1 for all the remaining binary values.
Example
Develop a truth table for the standard POS expression 𝐀𝐁𝐂 +
𝐀𝐁𝐂 +
𝐀𝐁𝐂.
𝐀𝐁𝐂: 001 𝐀𝐁𝐂: 𝟏𝟎𝟎
𝐀𝐁𝐂: 111

BOOLEAN ALGEBRA
Karnaugh Map
 K map provides a systematic method for simplifying Boolean expression
and if properly used will introduce the simplest SOP/ POS expression
possible(minimum expression).
 Its an array of cells in which each cell represents a binary value of the input
variable.
 Cells are arranged in a way so that simplification of a given expression is a
matter of grouping the cells.
 K-map can be used for expressions with 2,3,4 and 5 variables.
 The Quine McClusky method can be used for higher numbers of variables.
 The number of cells in a K-map = total number of possible input variable
combinations.

BOOLEAN ALGEBRA
3 variable K-map array of 8 cells
 Cells in a K-maps are arranged so that there is a single-variable change
between adjacent cells( adjacency is defined by a single-variable change).
 Physically, each cell is adjacent to the cells that are immediately next to it on
any of its 4 sides( a cell is not adjacent to the cells to any of its corner –cells
that diagonally touch each other)
C
AB 0 1
00
01
11
10
ABC ABC
ABC ABC
ABC ABC
ABC ABC

BOOLEAN ALGEBRA
 Top row cells are adjacent to the corresponding bottom row cells.
 Cells in the outer left column are adjacent to the corresponding cells in the
outer right column( wrap- around adjacency).
K-map SOP Minimization
 For an SOP expression in standard form, a 1 is placed on the K-map for each
product term in the expression.
 The cells that do not have 1 are the cells for which the expression is 0.
 When working with SOP expressions the 0s are left off the map.

BOOLEAN ALGEBRA
C
AB 0 1
00
01
11
10
1
1
1 1
Example
Map the standard SOP expression on the K-map ABC + ABC +
ABC + ABC
ABC
ABC
ABC
ABC
 For a nonstandard SOP expression , convert it to standard form before
you use a K-map.

BOOLEAN ALGEBRA
 Simplification of SOP expressions using K-maps is the process
of getting the fewest possible terms with the fewest possible
variables(minimum expression)
 The minimum expression is obtained by grouping 1s i.e.
enclosing
those adjacent cells containing 1s.
RULES
 Groups must contain either 1,2,4,8 or 16 cells(2n cells).
 Each cell in the group must be adjacent to 1 or more cells in that same
group.
 Include the largest possible number of 1s in a group.
 Each 1 on the map must be included in at least 1 group. The 1s
already in a group can be included in another group.
 Groups may overlap.
 Maximize the size the groups and minimize the number of groups.

BOOLEAN ALGEBRA
Determination of SOP expressions from K-map
 Each group of cells containing 1s creates one product term composed of
variables that stay the same within a group . i.e. variables that do not change
from in complemented to uncomplemented or vice versa.
 Determine the minimum product term for each group.
3 variable map
1 cell group yields a 3 variable term.
8 cell group yields a value of 1 for
expression.
4 variable map

BOOLEAN ALGEBRA
Example
Minimize 𝐁𝐂𝐃 + 𝐀𝐁𝐂𝐃 + 𝐀𝐁𝐂𝐃 + 𝐀𝐁𝐂𝐃 + 𝐀𝐁𝐂𝐃
+ 𝐀𝐁𝐂𝐃 +
𝐀𝐁𝐂𝐃 + 𝐀𝐁𝐂𝐃 + 𝐀𝐁C𝐃 using a K-map.
The first product term is not in standard form. Converting it into
standard form it becomes A𝐁𝐂 𝐃 + 𝐀𝐁𝐂 𝐃
A group of 8 cells
formed by adjacent
outer column produces
𝐃
A group of 4 cells
formed by adjacent
outer column produces
𝐁𝐂
CD
AB 00 01 11 10
00
01
11
10
1 1 1
1 1
1 1
1 1 1

BOOLEAN ALGEBRA
 With POS expressions in standard form, 0s representing the standard terms
are on the K-map.
Example
Map the standard POS expression on the K-map
(𝐀 + 𝐁 + 𝐂 + 𝐃)(𝐀 + 𝐁 + 𝐂 + 𝐃)(𝐀 + 𝐁 + 𝐂 + 𝐃)(𝐀
+ 𝐁 + 𝐂 + 𝐃)
 For a nonstandard POS expression , convert it to standard form before you
use a K-map.
K-maps POS Minimization
1.
2.
Determine the binary value of each sum term in the standard POS
expression with binary value =0.
Group the 0s to minimum sum terms
Example
Map the standard POS expression on the K-map
(𝐁 + 𝐂 + 𝐃)(𝐀 + 𝐁 + 𝐂 + 𝐃)(𝐀 + 𝐁 + 𝐂 + 𝐃)(𝐀 + 𝐁 + 𝐂
+ 𝐃)(𝐀 + 𝐁 +
𝐂 + 𝐃)

BOOLEAN ALGEBRA
Conversion between POS and SOP using the K-map
 With POS expressions, all cells that do not contain 0s contain 1s from which
SOP expression is derived.
 Likewise for an SOP expression, all the cells that do not contain 1s contain
0s, from which the POS expression is derived.
 This provides a good way to compare both minimum forms of an expression
to determine if one of them can be implemented with fewer gates than the
other.
Example
Convert the following standard POS expression into a minimum POS
expression, a standard SOP expression and a minimum SOP expression
(𝐀 + 𝐁 + 𝐂 + 𝐃)(𝐀 + 𝐁 + 𝐂 + 𝐃)(𝐀 + 𝐁 + 𝐂 + 𝐃)(𝐀
+ 𝐁 + 𝐂 + 𝐃)(𝐀 +
𝐁 + 𝐂 + 𝐃) (𝐀 + 𝐁 + 𝐂 + 𝐃)(𝐀 + 𝐁 + +𝐂 + 𝐃)

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