CHAPTER 2 DBM20083 BOOLEAN ALGEBRA

DBM20083
Discrete Mathematic

CHAPTER 2 : BOOLEAN ALGEBRA

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Chapter 2 : Boolean Algebra

 Boolean Function
 Logic Gates
 Minimize of the circuits

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Introduction : Boolean Algebra

 Boolean algebra provides the operations and the
rules for working with the set {0,1}.

 Three operations in Boolean algebra that most
use are
 Complementation
 Boolean sum
 Boolean product

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Introduction : Boolean Algebra

 The complement of an element, denoted with a
bar , is defined by 0ത = 1 1ത = 0.

 The Boolean sum, is denoted by + or by OR,
has the following values:
 1+1 = 1
 1+0 = 1
 0+1 = 1
 0+0 = 0

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Introduction : Boolean Algebra

 The Boolean product, is denoted by . or by
AND, has the following values:
 1.1 = 1
 1.0 = 0
 0.1 = 0
 0.0 = 0

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Introduction : Boolean Algebra

 Example:

1.0 + 0 + 1

 Using the definitions of complementation, the
Boolean Sum and Boolean Product, it follows that

1.0 + 0 + 1 = 0 + 1ത
=0+0=0

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Introduction : Boolean Algebra

 The complement, Boolean sum and Boolean
Product correspond to the logical operations, ~,∨
∧ respectively where 0 corresponds to
F(false) and 1 corresponds to T(True).

 Example: Translate 1.0 + 0 + 1 = 0 into a logical
equivalence. We obtain
 ∧ ∨ ~ ∨ =

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2.1 Boolean Function

2.1.1 Define Boolean Expression and Boolean
function

 A Boolean expression is a logical statement that evaluates
to a Boolean value either TRUE or FALSE. Boolean value
(TRUE or FALSE) are logical connective combine together
to form a Boolean expression. A Boolean function is
represented by a Boolean expression.

 A Boolean value is produced by an Boolean expression.

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2.1 Boolean Function

2.1.1 Define Boolean Expression and Boolean
function

 For example, the equality operator, ==, compares two
values and produces a Boolean value related to whether the
two values are equal to one another.

 Example : print(5==5)

 This statement evaluated to TRUE since the two
operands are equal.

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2.1 Boolean Function

2.1.1 Define Boolean Expression and Boolean
function

 Six common comparison operators:

• == ,
• ! = ,
• < , ℎ
• > , ℎ
• < = , ℎ
• > = , ℎ

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2.1 Boolean Function

2.1.1 Define Boolean Expression and Boolean
function

 Logical operators are:

 ! Logical negation
 && Logical AND
 || Logical OR
 Logical operator operates on Boolean variables or
Boolean expression only.

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2.1 Boolean Function

2.1.1 Define Boolean Expression and Boolean

function

 Example :
• Evaluate the following Boolean expression, given the
following declarations and assignments.

Boolean p=TRUE

Boolean q = FALSE

Boolean r = FALSE
&& ||
&& ||
||


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2.1 Boolean Function

2.1.1 Define Boolean Expression and Boolean

function

 Example :

• Evaluate the following Boolean expression, given the

following declarations and assignments

Boolean p = TRUE

Boolean q = FALSE

Boolean r = FALSE Try this!!!

 ! || &&
 ! ( && )||(! || )

 ||

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2.1 Boolean Function

2.1.1 Define Boolean Expression and Boolean

function

 Example :

• Evaluate the following Boolean expression, given the

following declarations and assignments

Int j=12;

Int k = 15;

Int 4; Try this!!!

 − > ||( == 1&& > 3)
 ! ( − 10 > 4 − )

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2.1 Boolean Function

2.1.1 Define Boolean Expression and Boolean function
 Example : Find the values of the Boolean function
represented by = , , + ҧ

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2.1 Boolean Function

Sum of Products (SOP) & Product of Sums (POS)

 The different forms of canonical expression which includes
the sum of products (SOP) and products of sum (POS)
The canonical expression can be defined as a Boolean
expression which has either min term otherwise max term.
For example, if we have two variables X and Y, then the
canonical expression comprising of min terms will be +
′ ′, whereas the canonical expression comprising the max
terms will be ( + )( ′ + ′)

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2.1 Boolean Function

Sum of Products (SOP) & Product of Sums (POS)

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2.1 Boolean Function

Sum of Products (SOP)

SOP, one kind of Boolean algebra expression.
Different product inputs are added together.
The product of inputs is Boolean logical AND whereas the
sum or addition is Boolean logical OR.
First, what is minterm? When the minimum combinations
of inputs are high then the output will be high. Example,
combinations of AND gate inputs.

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2.1 Boolean Function

Sum of Products (SOP)
The truth table of the min term is shown below:

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2.1 Boolean Function

Types of Sum of Products (SOP)
Three different forms:
 Canonical SOP

 Can be formed with grouping the minterms of the function for
which the output is high or true, and it is also called as the
sum on minterms.

 The expression of the canonical SOP is denoted with sign
summation (∑), and the minterms in the bracket are then wh
en the output is true.

 Non-canonical SOP
 Minimal SOP

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2.1 Boolean Function

Canonical SOP
From the table, the canonical SOP form
can be written as = σ( 1, 2, 3, 5)
By expanding the above summation, we
can get the following function = 1 +
2 + 3 + 5
By substituting the minterms in the above
equation, we get the expression =
′ ′ + ′ ′ + ′ + ′

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2.1 Boolean Function

Non-Canonical SOP
Here, we simplified the product terms.
For example, the expression =
′ ′ + ′ ′ + ′ + ′ =
′ ′ + ′ ′ ′ + +
′ = ′ ′ +
′ + ′
This is still in the form of SOP but it is
non-canonical SOP form.

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2.1 Boolean Function

Product of Sums (POS)
A Boolean expression consisting
purely of maxterm (sum terms) is said
to be a canonical product of sums form.
For example, we have a Boolean
function F defined A and B. So, A and
B are the inputs for F and lets say,
output of F is true i.e F=1 when only
one of the input is true of 1. Now we
draw the truth table for F.

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2.1 Boolean Function

Product of Sums (POS)
Now, we will create a column for the
maxterm using the variables A and B. If
input is 1, we take the complement of
the variable and if input is 0 we take the
variable as is.
To get the desired canonical POS
expression we will multiply the
maxterms (sum terms) for which the
output is 0 (POS form is used when the
output is ‘0’)

= ( + )( ′ + ′)

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2.1 Boolean Function

Product of Sums (POS)
First, we need to denote the
maxterms in shorthand notation.

+ = 0
′ + ′ = 3
 So, = 0 + 3
 This also can be written as =⊓
(0,3)

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2.1 Boolean Function

2.1.2 Use Identities of Boolean

 In mathematics, and identity is a statement true for all
possible values of its variables or variables. The algebraic
identity of x + 0 = x tells us that anything (x) added to zero
equals the original ‘anything” no matter what value that
“anything “ (x) may be . Like ordinary algebra, Boolean
algebra has its own unique identities based on the bivalent
states of Boolean variables.

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2.1 Boolean Function

2.1.2 Use Identities of Boolean

Basic Boolean Algebraic Identities

Additive Multiplicative Inverter
+ 0 = A 0 = 0 0ത = 1
+ 1 = 1 1 = 1ത = 0

+ = = Ӗ =
+ ҧ = 1 ҧ = 0

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2.1 Boolean Function

2.1.2 Use Identities of Boolean

Identity Name AND Form OR Form

Identity Law 1 = 0 + =
Null (Dominance) Law 0 = 0 1 + = 1
Idempotent Law = + =
Inverse Law ҧ = 0 + ҧ = 1
Commutative Law = + = +
Associative Law = ( ) + + = + ( + )
Distributive Law + = ( + )( + ) + = +
Absorption Law + = + =
DeMorgan’s Law = ҧ + ത + = ҧ ത

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2.1 Boolean Function

2.1.2 Use Identities of Boolean

Identity Name

Unit Property + ҧ = 1

Zero property ҧ = 0

Complement Law ҧ =

+ ҧ =
ത =
ത =

Double Complement Law Ӗ =

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2.1 Boolean Function

Example

 Simplify the following expressions by using identities of
Boolean Algebra.

a = ഥ + ഥ ഥ c = ഥ + . ( + ഥ )

b = ( ഥ + )( + ) d = + ഥ .

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2.1 Boolean Function

Solution

a = ഥ + ഥ ഥ
= ഥ + ഥ (Distributive Law)
= ഥ . (Inverse Law)
= ഥ (Identity Law)

b = ( ഥ + )( + )
= ഥ . ( ഥ . ) (D’Morgan Law)
= ഥ . ( ന + ഥ ) (D’Morgan Law)
= ഥ . ( + ഥ ) (Double Complement Law)

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2.1 Boolean Function

Solution

c = ഥ + . ( + ഥ )

= ( ഥ + ) + ( + ) (D’Morgan Law)
= ന . + ( ഥ . ന ) (D’Morgan Law)
= ഥ + ഥ (Double Complement Law)
d = + ഥ .
= ഥ . ((Distributive Law)
= + ഥ + + (Inverse and Idempotent Law)
= ഥ + + (Distributive Law)
= (Dominance Law)
= (Identity Law)

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2.1 Boolean Function

Simplify the following expressions by using
identities of Boolean Algebra
a = ഥ + ഥ
b = ഥ ഥ ഥ + ഥ ഥ ഥ ഥ
c = + . ഥ
d = ഥ + ഥ

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2.1 Boolean Function

Simplify the following expressions by using identities of
Boolean Algebra
a = ഥ + ഥ

b = ഥ ഥ ഥ + ഥ ഥ ഥ ഥ

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2.1 Boolean Function

Simplify the following expressions by using identities of
Boolean Algebra
c = + . ഥ

d = ഥ + ഥ

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2.1 Boolean Function

Exercise 1 :Simplify the following expressions by
using identities of Boolean Algebra

a = + h = ഥ + + ( + )( + ഥ )
b = ( + )( + ) I = + ഥ + ഥ ഥ + ഥ
c = + . ( + ) j = ഥ ഥ ഥ + ഥ ഥ + ഥ ഥ
d = ( ഥ + )( + + ) ഥ k = ഥ + + ഥ ഥ

e = + ഥ + ഥ l = +
f = ( ഥ + )( ഥ + ) m = ( ഥ + )( ഥ + )

g = + + + + n = + ( + )

o = ( ഥ ഥ )( + )

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2.2 Logic Gates

2.1.1 Define Logic Gates

 A logic gate is an elementary building block of a digital circuit.
Most logic gates have two inputs and one output.

 At any given moment, every terminal is in one of the two
binary conditions low(0) or high(1), represented by different
voltage levels. The logic state of a terminal can and
generally does, change often as the circuit processes data.

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2.2 Logic Gates

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2.2 Logic Gates

2.1.2 Construct Combination of Gates

 Any logic circuit no matter how complex can be completely
described using the three basic Boolean operations because
the OR gate, AND gate and NOT circuit are the basic
building blocks of digital systems.

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2.2 Logic Gates

Example

1. Draw the logic circuit to implement the following Boolean

Expressions:

a. = ( + ) c. = ( + )( ത + )

b. = ∙ + d. = + ҧ + ҧ

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2.2 Logic Gates

Example

1. Draw the logic circuit to implement the following Boolean
Expressions:

a. = ( + ) b. = ∙ +

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2.2 Logic Gates

Example

1. Draw the logic circuit to implement the following Boolean
Expressions:
c. = ( + )( ത + ) d. = + ҧ + ҧ

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2.2 Logic Gates

Example

2. Write the Boolean expression for output x for each of the
following circuit diagrams:
a. b.

= ഥ + = +

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2.2 Logic Gates

Example

2. Write the Boolean expression for output x for each of the
following circuit diagrams:
a. b.

= ഥ ( + ) = [ + ( + ) ] ∙

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2.2 Logic Gates

Example

Draw the logic circuit to implement the following Boolean
Expressions:

a. = ത b. = ( + ) ⋅

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2.2 Logic Gates

Example

Draw the logic circuit to implement the following Boolean
Expressions:

a. = ҧ ( + ) b. = [ + ( + ) ] ⋅

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2.2 Logic Gates

Example

Write the Boolean expression for output x for each of the
following circuit diagrams,
a. b.

= ഥ + + + = + ⋅ +

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2.1 Boolean Function

Exercise 2 :Draw the logic circuit to implement
the following Boolean Expressions,

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