BOOLEAN ALGEBRA

TRUTH TABLE

Truth table is a table which represents all the possible values of logical variables /

statements along with all the possible results of the given combinations of values.

Eg:

X Y R

1 1 1

1 0 0

0 1 0

0 0 0

1 represents TRUE and 0 represents FALSE.

TAUTOLOGY

If result of any logical statement or expression is always TRUE or 1, it is called

Tautology.

FALLACY

If result of any logical statement or expression is always FALSE or 0, it is called

Fallacy.

NOT OPERATOR

This Operator operates on single variables.

Truth Table

X Result

0 0

1 1

OR OPERATOR

Operator denotes operation called logical addition.

1

X Y X+ Y

0 0 0

0 1 1

1 0 1

1 1 1

AND OPERATOR

AND Operator denotes operation called logical Multiplication.

X Y X . Y

0 0 0

0 1 0

1 0 0

1 1 1

Evaluation of Boolean Expressions Using Truth Table.

1. X + (Y.Z)’

X Y Z Y.Z (Y.Z)’ X+(Y.Z)’

0 0 0 0 1 1

0 0 1 0 1 1

0 1 0 0 1 1

0 1 1 1 0 0

1 0 0 0 1 1

1 0 1 0 1 1

1 1 0 0 1 1

1 1 1 1 0 1

BASIC LOGIC GATES

Logic Gate

2

A Gate is simply an electronic circuit which operates on one or more signals to

produce an output signal.

There are three types of logic gates:-

1. Inverter (NOT Gate)

2. OR gate

3. AND gate

Inverter (NOT Gate)

An inverter (NOT gate) is a gate with only one input signal and one output signal.

The output signal is always the opposite of the input state.

Truth Table of NOT Gate

X X’

0 0

1 1

NOT Gate Symbol

OR Gate

The OR Gate has two or more input signals but only one output signal. If anyone

input signal is high (1) then output signal will be high (1).

X Y X+ Y

0 0 0

0 1 1

1 0 1

1 1 1

OR Gate Symbol

3

AND Gate

The AND Gate has two or more input signals but only one output signal. If

anyone input signal is low (0) then output signal will be low otherwise high

X Y X. Y

0 0 0

0 1 0

1 0 0

1 1 1

AND Gate Symbol

BASIC POSTULATES OF BOOLEAN ALGEBRA

I. If x != 0 then x=1 and if x!=1 then x=0

II. OR Relations (logical Addition)

0 + 0 = 0

0 + 1 = 1

1 + 0 = 1

1 + 1 = 1

4

III. AND Relations (Logical Multiplication)

0 . 0 = 0

0 . 1 = 1

1 . 0 = 1

1 . 1 = 1

IV. Complement rules

0’ = 1

1’ = 0

PRINCIPLE OF DUALITY

This states that starting with a Boolean relation another Boolean relation can be

derived by

1. Changing each OR sign to an AND sign.

2. Changing each AND sign to OR sign.

3. Replacing each 0 by 1 and each 1 by 0.

The derived relation using duality principal is called dual of original

expression.

BASIC THEOREMS OF BOOLEAN ALGEBRA

1. Properties of 0 and 1

a. 0 + X = X

b. 1 + X = 1

c. 0 . X = 0

d. 1. X = X

2. Indempotence Law

(a) X + X = X

(b) X . X = X

Proof

(a) X + X = X

X X R

5

0 0 0

1 1 1

Proof

(b) X . X = X

X X R

0 0 0

1 1 1

3. Involution

X’’ = X

Proof

X X’ X’’

0 1 0

1 0 1

(9) Complementarity Law

(a) X + X’ = 1

(b) X . X’ = 0

Proof

X X’ X+X’

0 1 1

1 0 1

X X’ X.X’

0 1 0

1 0 0

(10) Commutative Law

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i. X + Y= Y + X

X Y X + Y Y + X

0 0 0 0

0 1 1 1

1 0 1 1

1 1 1 1

ii.X.Y = Y. X

X Y X.Y Y.X

0 0 0 0

0 1 0 0

1 0 0 0

1 1 1 1

(11) The Associative Law

(i) X + (Y + Z) = (X + Y) + Z (ii) X(YZ) =(XY)Z

Truth table for X+(Y+Z) = (X+Y)+Z is given below :

Input Output

X Y Z Y+Z X+Y X+(Y+Z) (X+Y)+Z

7

0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1

0

1

1

1

0

1

1

1

0

0

1

1

1

1

1

1

0

1

1

1

1

1

1

1

0

1

1

1

1

1

1

1

Comparing the columns X+(Y+Z) and (X+Y)+Z, we see both of these are

identical, Hence proved. Since (i) is proved, (ii) is dual of rule (ii), hence it is also

proved.

(12) The Distributive Law

(i) X(Y+Z) = XY+XZ (ii) X+YZ =(X+Y)(X+Z)

Truth table for X(Y+Z) = XY+XZ are given below:

Input Output

X Y Z Y+Z XY XZ X(Y+Z) XY+XZ

8

0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1

0

1

1

1

0

1

1

1

0

0

0

0

0

0

1

1

0

0

0

0

0

1

0

1

0

0

0

0

0

1

1

1

0

0

0

0

0

1

1

1

Both the columns X(Y+Z) and XY+YZ are identical, hence proved.

The algebraic proof of law X+YZ=(X+Y)(X+Z)

RHS = (X+Y)(X+Z)

=XX+XZ+XY+YZ

=X+XZ+XY+YZ

=X+XY+XZ+YZ

=X(1+Y)+Z(X+Y)

=X.1+Z(X+Y)

=X+XZ+YZ=X(1+Z)+YZ

=X +YZ

= LHS, Hence proved.

Eg: State Distributive law and verify the same using truth table.

Ans. If X, Y, Z are Boolean Variables then

X.(Y + Z) = X.Y + X.Z or X+Y.Z = (X+Y).(X+Z)

X Y Z Y+Z X.(Y+Z) X.Y X.Z X.Y+X.Z

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0 0 0 0 0 0 0 0

0 0 1 1 0 0 0 0

0 1 0 1 0 0 0 0

0 1 1 1 0 0 0 0

1 0 0 0 0 0 0 0

1 0 1 1 1 0 1 1

1 1 0 1 1 1 0 1

1 1 1 1 1 1 1 1

(13) Absorption law

(i) X+XY = X (ii) X(X+Y) =X

(i) Truth table for X+XY = X is given below :

Input Output

X Y XY X+XY

0 0

0 1

1 0

1 1

0

0

0

1

0

0

1

1

Both the columns X+XY and X are identical, hence proved.

(ii) Truth table for X.(X+Y) = X is given below :

Input Output

10

X Y X+Y X(X+Y)

0 0

0 1

1 0

1 1

0

1

1

1

0

0

1

1

Column X and X(X+Y) are identical, hence proved.

DEMORGAN’S THEOREMS

1. Demorgan’s First Theorem

(X+Y)’ = X’Y’

Proof

(X+Y)+(X’Y’) = 1

(X+Y)+X’Y’=((X+Y)+X’).((X+Y)+Y’)

= (X+X’+Y).(X+Y+Y’)

= (1+Y).(X+1)

= 1.1

= 1

(X+Y).(X’Y’) = 0

(X+Y).(X’Y’) = X’Y’.(X+Y)

= X’Y’X+X’Y’Y

= 0.Y+X’.0

= 0

Hence proved

2. Demorgan’s Second Theorem

(X.Y)’ = X’+Y’

Proof

XY + (X’+Y’) = 1

=(X’+Y’)+XY

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=(X’+Y’+X).(X’+Y’+Y)

=(X+X’+Y’).(X’+Y+Y’)

=(1+Y’).(X’+1)

=1.1

=1

XY. (X’+Y’) = 0

= XY.( X’+Y’) ->X(Y+Z)=XY+XZ

= XYX’+XYY’

= 0.Y+X.0

= 0+0

= 0

Note:- Find the dual of the following expression.

(X+Y)+Z=X+(Y+Z)

The dual is (X.Y).Z=X.(Y.Z)

Derivation of Boolean expression

1. Minterms

Minterms is a product of all the literals (with or without the bar) within the

logic system.

Example: Convert X+Y to minterm

X+Y

= X.1+Y.1 (Because X.1 = X)

= X.(Y+Y’)+Y.(X+X’) (Because X+X’ = 1)

= XY+XY’+XY+X’Y By Distributive law

= XY+XY’+X’Y (Because X+X = X)

2. Maxterms

Maxterm is a sum of all the literals (with or without the bar) within the logic

system.

Example:

If the value of the variables are X=0, Y=1 and Z=1 then the Maxterm will be

X+Y’+Z’

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3. Canonical Expression

Boolean expression composed entirely either of minterms or Maxterm is referred

to as canonical expression. Canonical expression can be represented in two forms:

a. Sum – of – Products (SOP) form

ii) Product – of – Sum (POS) form

Canonical Sum of Products form

When a Boolean expression is represented purely as sum of minterms

(Product terms), it is said to be canonical Sum of Product form.

Truth table for Product Terms (3 input)

X Y Z ProductTerms/

Minterms

0 0 0 X’Y’Z’

0 0 1 X’Y’Z

0 1 0 X’YZ’

0 1 1 X’YZ

1 0 0 XY’Z’

1 0 1 XY’Z

1 1 0 XYZ’

1 1 1 XYZ

Canonical sum of products expression can also be represented by the following Short

hand notation.

Example: - F=∑(1,4,5,6,7)

= m1 + m4 + m5 + m6 + m7

= 001 + 100 + 101 + 110 + 111 (Boolean values for decimal

numbers)

= X’Y’Z + XY’Z’+ XY’Z+XYZ’+XYZ

Canonical Product of sum form

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When a Boolean expression is represented purely as product of maxterms, it is

said to be in canonical Product of Sum form of expression.

Truth table for Sum Terms (3 input)

X Y Z SumTerms/Maxterms

0 0 0 X+Y+Z

0 0 1 X+Y+Z’

0 1 0 X+Y’+Z

0 1 1 X+Y’+Z’

1 0 0 X’+Y+Z

1 0 1 X’+Y+Z’

1 1 0 X’+Y’+Z

1 1 1 X’+Y’+Z’

Canonical product of sum expression can also be represented by the following Short

hand notation.

Example(1): - F=∏ (1,4,5,6,7)

=M1 . M4. M5. M6.M7

= 001 + 100 + 101 + 110 + 111 (Boolean values for decimal

numbers)

= (X+Y+Z’)( X’+Y+Z)( X’+Y+Z’)( X’+Y’+Z)( X’+Y’+Z’)

Minimization of Boolean expression

1. Algebraic method

Example : - Reduce X’Y’Z’ + X’YZ’ + XY’Z’ + XYZ’

= X’(Y’Z’ + YZ’) + X(Y’Z’+YZ’)

=X’(Z’(Y’+Y)) + X(Z’(Y’+Y))

=X’(Z’.1)+X(Z’.1)

=X’Z’+XZ’

=Z’(X’+X)

=Z’.1

=Z’

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2. Simplification using karnaugh maps

Karnaugh map

Karnaugh map or K – map is a graphical display of the fundamental products in a

truth table. K – Map is nothing but a rectangle made up of certain number of squares

represents a Maxterm or minterm

SOP reduction using K-maps

For a function of n variables, there would be a map of 2n squares each represents

a minterm. Following are two, three, four variables k-maps for SOP reduction.

Two variables

Three variables

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Four variable:

1) Reduce the following Boolean expression using K-Map

F(U,V,W,Z) = ∑ (0, 1, 2, 3, 4, 10, 11)

Ans.

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1. Obtain the simplified form of a boolean expression using Karnaugh map.

F(U,V,W,X) = ∑ (0,3, 4, 5, 7, 11, 13, 15)

[00]W’Z’ [01] W’Z [11]WZ [10]WZ’

[00] U’V’

[01] U’V

[11] UV

[00] UV’

1

1

1 1 1

1 1

1

2 quads, 1 pair.

Quad 1(m3+m7+m11+m15) reduces to WZ

Quad 2(m5+m7+m13+m15) reduces to VZ

Pair 1(m0,m4) reduces to U’W’Z’

Therefore F=WZ + VZ + U’W’Z’

POS reduction using K-maps : -

For a function of n variables, there would be a map of 2n squares each represents

a maxterm. Following are two, three, four variables k-maps for POS reduction.

Two variables

Y Y’

X+Y

0

X+Y’

1

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X’+Y

2

X’+Y’

3

X

X’

Three variables Y+Z Y+Z’ Y’+Z’ Y’+ZX+Y+Z 0

X+Y+Z’ 1

X+Y’+Z’ 3

X+Y’+Z 2

X’+Y+Z

4

X’+Y+Z’

5

X’+Y’+Z’

7

X’+Y’+Z

6

X

X’

Four variables

Y+Z Y+Z’ Y’+Z’ Y’+ZW+X+Y+Z 0

W+X+Y+Z’ 1

W+X+Y’+Z’ 3

W+X+Y’+Z 2

W+X’+Y+Z 4

W+X’+Y+Z’ 5

W+X’+Y’+Z’ 7

W+X’+Y’+Z 6

W’+X’+Y+Z 12

W’+X’+Y+Z’ 13

W’+X’+Y’+Z’ 15

W’+X’+Y’+Z 14

W’+X+Y+Z 8

W’+X+Y+Z’ 9

W’+X+Y’+Z’ 11

W’+X+Y’+Z 10

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W+X

W+X’

W’+X’

W’+X

More about Logic Gates

NAND and NOR Gates

● NAND and NOR gates can greatly simplify circuit diagrams. As we will see you

can use these gates wherever you could use AND, OR, and NOT.

NAND Gate:-

The NAND gate has two or more input signals but only one output signal. If all of

the inputs are (high), then the output produced is 0 (low).

01110111010

19

0A B

BA

NAND

NOR Gates:-

The NOR gates has two or more input signals but only one output signal. If all

inputs are 0 (i.e., low), then the output signal is 1 (high).

011001010100

A BBA

20

NOR

XOR and XNOR Gates

XOR is used to choose between two mutually exclusive inputs. Unlike OR, XOR is true

only when one input or the other is true, not both

A B AÅB

0 0 0

0 1 1

1 0 1

1 1 0

XOR Gate (Exclusive OR gate):-

The XOR gate can also have two or more inputs but produces one output signal.

Exclusive-OR gate different from OR gate. OR gate produces output 1 for any

input combination having one or more 1’s, but XOR gate produces output 1 for

only those input combinations that have odd number 1’s.

XOR

21

XNOR Gate (Exclusive NOR gate)

The XNOR Gate is logically equivalent to an inverted XOR i.e., XOR gate

followed by a NOT gate (inventor). Thus XNOR produces 1 (high) output when

the input combination has even number of 1’s.

A B A B

0 0 1

0 1 0

1 0 0

1 1 1

XNOR

SOLVED PROBLEMS

1 mark Questions

1) Why are AND and NOR gates called Universal gates? (1)

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Ans. NAND and NOR gates are less expensive and easier to design. Also, other

switching functions (AND, OR) can easily be implemented using NAND/NOR

gates. Thus, these (NAND, NOR) gates are also reffered to as Universal Gates.

(2) Write the Sum of Products form of the function G(U,V,W). Truth table

representation of G is as follows: (1)

U V W G

0 0 0 0

0 0 1 0

0 1 0 1

0 1 1 0

1 0 0 1

1 0 1 0

1 1 0 1

1 1 1 1

Ans. To get the product of sums form, we need to add maxterms for all those input

combinations that produce output as 0. Thus ,

G(U,V,W) = (U + V + W) (U + V + W’) (U + V’ + W’) (U’ + V + W’)

(3) Convert P+Q’R to product of sum

Ans:

P+Q’R=(P+Q’) (P+R’) [ by the rule A+BC= (A+B)(A+C)

=(P+Q’+RR’) ( P+QQ’+R) [ by the rule AA’=0

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= (P+Q’+R) (P+Q’+R’) (P+Q+R) (P+Q’+R) [ by the rule A+BC=

(A+B)(A+C)

(4) Concert X+YZ into sum of product

Ans:

X+YZ=X.1+YZ.1 [by the rule A.1=A

= X(Y+Y’)+YZ(X+X’) [ by the rule A+A’=1

=XY+XY’+YZX+YZX’

=XY.1+XY’.1+XYZ+YZX’ [by the rule A.1=A

=XY(Z+Z’)+XY’(Z+Z’)+XYZ+X’YZ [ by the rule A+A’=1

=XYZ+XYZ’+XY’Z+XY’Z’+XYZ+X’YZ

2 marks Questions

(1) Write the equivalent Boolean Expression for the following Logic circuit. (2)

1. Prove that X.(X+Y)=X by truth table method. (2)

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Ans.

X Y X+Y X.(X+Y)

0 0 0 0

1.1 1 0

2.0 1 1

1 1 1 1

From the above table it is obvious that X.(X+Y) = X because both the columns

are identical.

2. Prove that X.(X+Y)=X by algebric method (2)

Ans. LHS = X.(X+Y)

= X.X+X.Y

=X+X.Y

=X.(1+Y)

=X.1 = X = RHS

3. State the distributive laws of Boolean algebra. How are they different from

distributive laws of ordinary algebra. (2)

Ans. Distributive laws of Boolean algebra state that

i. X(Y+Z) = XY+XZ

ii.X+YZ =(X+Y)(X+Z)

Ist law X(Y+Z) = XY+XZ holds good for all values of X, Y and Z in ordinary

algebra whereas X+YZ =(X+Y)(X+Z) holds good only for two values (0,1) of X,

Y and Z.

4. In Boolean algebra, verify using truth table that (X + Y)’ = X’ Y’ for each X,

Y in (0, 1). (2)

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Ans. As it is a 2-variable expression, truth table will be as follows :

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

5. State Demorgan’s laws. Verify one of the Demorgan’s laws using truth

tables. (2)

Ans. De Morgan’s first theorem. It states that X + Y = X . Y

De Morgan’s second theorem. It states that X . Y = X + Y

Truth table for second theorem

X Y X.Y X.Y X Y X+Y

0 0 0 1 1 1 1

0

1

1

0

0

0

1

1

1

0

0

1

1

1

1 1 1 0 0 0 0

X.Y and X+Y are identical.

6. Draw the logic circuit diagram for the following expression : (2)

Y = a b + b c + c a

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7. Prepare a truth table for X’ Y Z’ + X Y’ (2)

Ans. Truth table for is given below :

Input O

u

t

p

u

t

X Y Z X’ Y’ Z’ X’YZ’ XY’ X’YZ’ + XY’

0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1

1

1

1

1

0

0

0

0

1

1

0

0

1

1

0

0

1

0

1

0

1

0

1

0

0

0

1

0

0

0

0

0

0

0

0

0

1

1

0

0

0

0

1

0

1

1

0

0

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(9) Write the equivalent expression for the following logic circuit : (2)

Ans. F=(AC)’ +(BA)’+(BC)’

(10) Write the equivalent expression for the following logic circuit: (2)

Ans. (X + Y’)(X’ + Y)(X’ + Y’)

4 marks Questions

(1) Obtain the simplified form of a boolean expression using Karnaugh map. (4)

F(u,v,w,x) = ∑ (0, 3, 4, 5, 7, 11, 13, 15)

[00]W Z [01] W Z [11]W Z [10]W Z

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[00] U V

[01] U V

[11] U V

[00] U V

1

1

1 1 1

1 1

1

2 quads, 1 pair.

Quad 1(m3+m7+m11+m15) reduces to WZ

Quad 2(m5+m7+m13+m15) reduces to VZ

Pair 1(m0,m4) reduces to UWZ

Therefore F=WZ + VZ + UWZ8. By means of truth table, demonstrate the validity of the following Postulates /

Laws of Boolean algebra: (4)

a. Commulative law

b. Idempotent law

Ans.

(a) The commulative law states that

(i) X+Y = Y+X (ii) X.Y =Y.X

(i) Truth table for X+Y = Y+X is given below :

Input Output

X Y X+Y Y+X

0 0 0 0

0 1 1 1

1 0 1 1

29

1 1 1 1

Comparing the columns X+Y and Y+X, we see both of these are identical. Hence proved.

(ii) Truth table for X.Y = Y.X is given below:

Input Output

X Y X.Y Y.X

0 0 0 0

0 1 0 0

1 0 0 0

1 1 1 1

Comparing the columns X.Y and Y.X, we see both of these are identical. Hence proved.

(b) The Idempotent law states that

(i) X+X = X ii) X.X =X

(i) Truth table for X+X = X is given below :

Input Output

X X X+X

0 0

1 1

0

1

(ii) Truth table for X.X = X is given below :

Input Output

X X X.X

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0 0

1 1

0

1

POINTS TO REMEMBER

● Binary Decision

● Logical Statements

● TRUTH TABLE :-Truth table is a table which represents all the possible values

of logical variables /statements along with all the possible results of the given

combinations of values

● TAUTOLOGY:-If result of any logical statement or expression is always TRUE

or 1, it is called Tautology.

● FALLACY : - If result of any logical statement or expression is always FALSE

or 0, it is called Fallacy.

● PRINCIPLE OF DUALITY

This states that starting with a Boolean relation another Boolean relation can be

derived by

1. Changing each OR sign to an AND sign.

2. Changing each AND sign to OR sign.

3. Replacing each 0 by 1 and each 1 by 0.

The derived relation using duality principal is called dual of original

expression.

● BASIC THEOREMS OF BOOLEAN ALGEBRA

4. Properties of 0 and 1

5. 0+x=x

6. 1+x=1

7. 0.x=0

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8. 1.x=x

9. Indempotence law

(a) x+x=x

(b) x.x=x

● Involution

x’’=x

● Complementarity law

(a) x+x’=1

(b) x.x’= 0

● Cummtative law

x + y= y+x

● The associative law

(i) X+(Y+Z) = (X+Y)+Z (ii) X(YZ) =(XY)Z

● The distributive law

(i) X(Y+Z) = XY+XZ (ii) X+YZ =(X+Y)(X+Z)

● Absorption law

(i) X+XY = X (ii) X(X+Y) =X

● DEMORGAN’S THEOREMS

(X+Y)’=X’Y’

(X.Y)’=X’+Y’

● Minterms: - Minterms is a product of all the literals (with or without the bar)

within the logic system.

● Maxterms: - Maxterm is a sum of all the literals (with or without the bar) within

the logic system

● Canonical sum of product form:- When a Boolean expression is represented

purely as sum of minterms or product terms, it is said to be canonical sum of

product form.

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● Canonical Product of sum form: - When a Boolean expression is represented

purely as product of maxterms, it is said to be in canonical product of sum form of

expression.

● NOR Gates:-

The NOR gates has two or more input signals but only one output signal. If all

inputs are 0 (i.e., low), then the output signal is 1 (high).

● NAND Gate:-

The NAND gate has two or more input signals but only one output signal. If all of

the inputs are (high), then the output produced is 0 (low).

● XOR Gate (Exclusive OR gate):-

The XOR gate can also have two or more inputs but produces one output signal.

Exclusive-OR gate different from OR gate. OR gate produces output 1 for any

input combination having one or more 1’s, but XOR gate produces output 1 for

only those input combinations that have odd number 1’s.

● XNOR Gate (Exclusive NOR gate)

The XNOR Gate is logically equivalent to an inverted XOR i.e., XOR gate

followed by a NOT gate (inventor). Thus XNOR produces 1 (high) output when

the input combination has even number of 1’s.

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