ĐẠI HỌC

CÔNG NGHỆ

VIETNAM NATIONAL UNIVERSITY, HANOI

University of Engineering & Technology

Digital Design

Lecture 3: Boolean and Switching Algebra

Xuan-Tu Tran, PhD

Faculty of Electronics and Telecommunication

Smart Integrated Systems (SIS) Laboratory

Email: tutx@vnu.edu.vn

www.uet.vnu.edu.vn/~tutx

ĐẠI HỌC

CÔNG NGHỆ

History

Developed by George Boole

in his book (a treatise): “An Investigation of the Laws of Thought”

no application was made of Boolean Algebra until the late 1930s

Nakashima in Japan (1937) and Shannon at MIT (1938), each

independently applied the algebra of Boole to the analysis of networks of

relays (in telephone systems).

10/8/2010

Xuan-Tu Tran

2

ĐẠI HỌC

CÔNG NGHỆ

The Huntington postulates

In 1904, Huntington found that

all of the results and implications of the algebra described by Boole could

be derived from only six basic postulates.

Huntington postulates: The set {B, +, ·, ¯ }

B is the set of elements or constants of the algebra

the symbols + and · are two binary operators(*)

the overbar ¯ is a unary operator(*)

is a Boolean algebra if the following hold true:

(*)

10/8/2010

The terms binary operator and unary operator refer to the number of arguments

involved in the operation: two or one, respectively.

Xuan-Tu Tran

3

ĐẠI HỌC

CÔNG NGHỆ

1.

The Huntington postulates

Closure (khép kín)

For all elements a and b in the set B,

a+b∈B

a·b∈B

2.

Existing 0 and 1 elements

There exists a 0 element in B such that

for every element a in B, 0 + a = a + 0 = a

There exists a 1 element in B such that

for every element a in B, a = a · 1 = a

10/8/2010

Xuan-Tu Tran

4

ĐẠI HỌC

CÔNG NGHỆ

3.

The Huntington postulates

Commutativity (giao hoán)

a+b=b+a

a·b=b·a

4.

Distributivity (phân phối)

a · (b + c) = a · b + a · c

a + (b · c) = (a + b) · (a + c)

10/8/2010

Xuan-Tu Tran

5

The Huntington postulates

ĐẠI HỌC

CÔNG NGHỆ

5.

∀a ∈ B, there exists an element a-bar in the set B such that

a + a =1

a⋅a = 0

6.

There exists at least two distinct elements in B

Switching algebra is a Boolean algebra in which the number of

elements in the set B is precisely 2.

10/8/2010

Xuan-Tu Tran

6

ĐẠI HỌC

CÔNG NGHỆ

Switching Algebra

is a Boolean algebra in which the number of elements in the set B is

precisely 2

The two binary operators, represented by the signs + and ·, are called

the OR and the AND, respectively.

The unary operator, represented by the overbar ¯ , is called the NOT or

the complement operator.

10/8/2010

Xuan-Tu Tran

7

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications

Theorem 1 – idempotence

(i)

a+a = a

(ii)

a⋅a = a

(Định lý hấp thụ)

Proof

a + a = (a + a ) ⋅1

= (a + a ) ⋅ (a + a )

= a + a⋅a

= a+0

=a

10/8/2010

P-2(ii)

P-5(i)

P-4(ii)

P-5(ii)

P-2(i)

Xuan-Tu Tran

a ⋅ a= a ⋅ a + 0

= a⋅a + a⋅a

= a ⋅ (a + a )

= a ⋅1

=a

P-2(i)

P-5(ii)

P-4(i)

P-5(i)

P-2(ii)

8

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications

Theorem 2

(i)

(ii)

Proof

a ⋅0 = 0 + a⋅0

= a⋅a + a ⋅0

= a ⋅ ( a + 0)

= a ⋅ (a )

=0

a ⋅0 = 0⋅a = 0

a +1 = 1+ a = 1

a +1

P-2(i)

P-5(ii)

Principle

of duality

P-4(i)

P-2(ii)

P-5(i)

P-4(ii)

P-2(i)

P-2(ii)

P-5(ii)

P-5(i)

Exercise for students

10/8/2010

Xuan-Tu Tran

9

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications

Theorem 3

Let a be an element of B. Then a-bar is unique.

Proof

Assume that a has 2 distinct

complements (not equal), a-bar and b.

Then by P-5, we must have that:

a +b =1

and

a + a =1

and

a⋅a = 0

and

a ⋅b = 0

10/8/2010

a = a ⋅1

b = b ⋅1

= a ⋅ ( a + b ) = b( a + a )

= a ⋅a + a ⋅b = b⋅a + b⋅a

= 0 + a ⋅b

= 0+b⋅a

= a ⋅b

= a ⋅b

Xuan-Tu Tran

a =b

P-2(ii)

P-4(i)

P-2(i)

10

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications – Basic logic operations

From this point, we will restrict our attention to switching algebras

only.

Switching algebra is basically a two-element Boolean algebra which,

obviously, has the two elements 0 and 1.

AND operation

x

y

z=x·y

0

0

0

Theorem 1(ii)

0

1

0

Theorem 2(i)

1

0

0

Postulate 3(ii)

1

10/8/2010

Symbols

1

1

Theorem 1(ii)

Xuan-Tu Tran

x

y

x

y

z

&

&

z

11

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications – Basic logic operations

Symbols

OR operation

x

y

z=x+y

0

0

0

Derived from which

0

1

1

theorem, postulate?

1

0

1

Exercise for Students

1

1

1

x

y

x

y

≥1

≥1

z

Symbols

NOT operation

10/8/2010

z

x

z = x-bar

Derived from which

0

1

theorem, postulate?

1

0

Exercise for Students

Xuan-Tu Tran

x

x

z

11

z

12

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications – Basic logic operations

NAND operation

Symbols

x

y

z=x·y

0

0

1

0

1

1

1

0

1

1

1

0

x

y

z=x+y

0

0

1

0

1

0

1

0

0

1

1

0

x

y

z

x

y

&

&

z

NOR operation

10/8/2010

Symbols

x

y

x

y

Xuan-Tu Tran

z

≥1

≥1

z

13

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications – Basic logic operations

XOR operation

x

y

z=x⊕y

0

0

0

0

1

1

1

0

1

1

1

Symbols

x

y

x

y

0

z

=1

=1

z

z = x ⊕ y = xy + x y

- Construct XOR gate from NOT, AND, and OR ?

10/8/2010

Xuan-Tu Tran

14

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications – Basic logic operations

NXOR operation

x

y

z=x⊕y

0

0

1

0

1

0

1

0

0

1

1

Symbols

x

y

x

y

1

z

=1

=1

z

z = x ⊕ y = x y + xy

- Construct NXOR gate from NOT, AND, and OR ?

10/8/2010

Xuan-Tu Tran

15

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications

Theorem 4 – Involution

Định lý phủ định của phủ định

(x ) = x

Let x be a switching variable. Then

Proof

10/8/2010

x

x

(x )

0

1

0

1

0

1

Since the left column is identical to the

right column and we have listed all

possibilities, we have proved the result.

Xuan-Tu Tran

16

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications

Theorem 5

(i)

Let x and y be two switching variables. Then

(ii)

x+ x⋅ y = x

x ⋅ (x + y ) = x

Proof

10/8/2010

x

y

x⋅ y

x

+

x⋅ y

0

0

0

0

+

0

0

0

1

0

0

+

0

0

1

0

0

1

+

0

1

1

1

1

1

+

1

1

Xuan-Tu Tran

17

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications

Theorem 6

Let x, y, and z be switching variables. Then

1.

(i)

Associativity (định lý kết hợp)

(ii)

2.

(i)

(ii)

3.

x+x⋅y = x+ y

x ⋅ (x + y ) = x ⋅ y

Consensus (đồng tâm)

(i)

(ii)

10/8/2010

x ⋅ ( y ⋅ z ) = (x ⋅ y ) ⋅ z

x + ( y + z ) = (x + y ) + z

x⋅ y + x ⋅z + y⋅z = x⋅ y + x ⋅z

(x + y ) ⋅ (x + z ) ⋅ ( y + z ) = (x + y ) ⋅ (x + z )

Xuan-Tu Tran

18

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications – De Morgan’s Theorem

De Morgan’s Theorem

Let x and y be two switching variables. Then

(x + y ) = x ⋅ y

(x ⋅ y ) = x + y

Proof

This theorem may easily verified by complete enumeration.

Exercise for Students

10/8/2010

Xuan-Tu Tran

19

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications – De Morgan’s Theorem

Used for the evaluation of complements of switching expressions.

For example,

(x + y ⋅ (z + w ))

= ( x ) ⋅ ( y ⋅ ( z + w ))

(

)

= ( x ) ⋅ (y + ( z ) ⋅ w )

= (x ) ⋅ y + (z + w )

= x ⋅ (y + z ⋅ w )

= x⋅ y + x⋅z⋅w

10/8/2010

Xuan-Tu Tran

20

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications – De Morgan’s Theorem

Corollary of De Morgan’s Theorem

(x1 + x 2 + L + x n ) = x1 ⋅ x 2 ⋅ L ⋅ x n

(x1 ⋅ x 2 ⋅ L ⋅ x n ) = x1 + x 2 + L + x n

Expand the following function:

((x ⋅ ( y + z ))⋅ ( y + w ⋅ z )⋅ (x + z ))

10/8/2010

Xuan-Tu Tran

21

ĐẠI HỌC

CÔNG NGHỆ

Truth tables

Truth table

a simple way of representing a switching function is to make a list of the

possible variable assignments and note the value the function takes on for

each assignment. This list is called Truth Table.

How to create the Truth Table of a switching function?

For example,

10/8/2010

F ( x , y , z ) = x + yz

Xuan-Tu Tran

22

ĐẠI HỌC

CÔNG NGHỆ

Truth tables

F ( x , y , z ) = x + yz

x

y

z

F (x, y, z)

0

0

0

1

0

0

1

1

0

1

0

1

0

1

1

1

2n assignment

1

0

0

0

possibilities

1

0

1

0

1

1

0

1

1

1

1

0

- Write the Truth Table for the following switching function: F (x, y, z) =

10/8/2010

Xuan-Tu Tran

x⋅ y⋅z + x⋅ y

23

ĐẠI HỌC

CÔNG NGHỆ

Truth tables

Find a switching function from a truth table

10/8/2010

x

y

F (x, y)

0

0

0

0

1

0

1

0

0

1

1

1

Xuan-Tu Tran

F (x, y) ???

24

ĐẠI HỌC

CÔNG NGHỆ

Canonical forms: Minterm - Maxterm

Alternatives to the tabular representation

To list only the assignments for which a function is 1 or,

To list those for which the function is 0

Canonical representation

F = 1 whenever x = 1 and y = 1

F ( x , y ) = xy

10/8/2010

Xuan-Tu Tran

x

y

F (x, y)

0

0

0

0

1

0

1

0

0

1

1

1

25

CÔNG NGHỆ

VIETNAM NATIONAL UNIVERSITY, HANOI

University of Engineering & Technology

Digital Design

Lecture 3: Boolean and Switching Algebra

Xuan-Tu Tran, PhD

Faculty of Electronics and Telecommunication

Smart Integrated Systems (SIS) Laboratory

Email: tutx@vnu.edu.vn

www.uet.vnu.edu.vn/~tutx

ĐẠI HỌC

CÔNG NGHỆ

History

Developed by George Boole

in his book (a treatise): “An Investigation of the Laws of Thought”

no application was made of Boolean Algebra until the late 1930s

Nakashima in Japan (1937) and Shannon at MIT (1938), each

independently applied the algebra of Boole to the analysis of networks of

relays (in telephone systems).

10/8/2010

Xuan-Tu Tran

2

ĐẠI HỌC

CÔNG NGHỆ

The Huntington postulates

In 1904, Huntington found that

all of the results and implications of the algebra described by Boole could

be derived from only six basic postulates.

Huntington postulates: The set {B, +, ·, ¯ }

B is the set of elements or constants of the algebra

the symbols + and · are two binary operators(*)

the overbar ¯ is a unary operator(*)

is a Boolean algebra if the following hold true:

(*)

10/8/2010

The terms binary operator and unary operator refer to the number of arguments

involved in the operation: two or one, respectively.

Xuan-Tu Tran

3

ĐẠI HỌC

CÔNG NGHỆ

1.

The Huntington postulates

Closure (khép kín)

For all elements a and b in the set B,

a+b∈B

a·b∈B

2.

Existing 0 and 1 elements

There exists a 0 element in B such that

for every element a in B, 0 + a = a + 0 = a

There exists a 1 element in B such that

for every element a in B, a = a · 1 = a

10/8/2010

Xuan-Tu Tran

4

ĐẠI HỌC

CÔNG NGHỆ

3.

The Huntington postulates

Commutativity (giao hoán)

a+b=b+a

a·b=b·a

4.

Distributivity (phân phối)

a · (b + c) = a · b + a · c

a + (b · c) = (a + b) · (a + c)

10/8/2010

Xuan-Tu Tran

5

The Huntington postulates

ĐẠI HỌC

CÔNG NGHỆ

5.

∀a ∈ B, there exists an element a-bar in the set B such that

a + a =1

a⋅a = 0

6.

There exists at least two distinct elements in B

Switching algebra is a Boolean algebra in which the number of

elements in the set B is precisely 2.

10/8/2010

Xuan-Tu Tran

6

ĐẠI HỌC

CÔNG NGHỆ

Switching Algebra

is a Boolean algebra in which the number of elements in the set B is

precisely 2

The two binary operators, represented by the signs + and ·, are called

the OR and the AND, respectively.

The unary operator, represented by the overbar ¯ , is called the NOT or

the complement operator.

10/8/2010

Xuan-Tu Tran

7

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications

Theorem 1 – idempotence

(i)

a+a = a

(ii)

a⋅a = a

(Định lý hấp thụ)

Proof

a + a = (a + a ) ⋅1

= (a + a ) ⋅ (a + a )

= a + a⋅a

= a+0

=a

10/8/2010

P-2(ii)

P-5(i)

P-4(ii)

P-5(ii)

P-2(i)

Xuan-Tu Tran

a ⋅ a= a ⋅ a + 0

= a⋅a + a⋅a

= a ⋅ (a + a )

= a ⋅1

=a

P-2(i)

P-5(ii)

P-4(i)

P-5(i)

P-2(ii)

8

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications

Theorem 2

(i)

(ii)

Proof

a ⋅0 = 0 + a⋅0

= a⋅a + a ⋅0

= a ⋅ ( a + 0)

= a ⋅ (a )

=0

a ⋅0 = 0⋅a = 0

a +1 = 1+ a = 1

a +1

P-2(i)

P-5(ii)

Principle

of duality

P-4(i)

P-2(ii)

P-5(i)

P-4(ii)

P-2(i)

P-2(ii)

P-5(ii)

P-5(i)

Exercise for students

10/8/2010

Xuan-Tu Tran

9

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications

Theorem 3

Let a be an element of B. Then a-bar is unique.

Proof

Assume that a has 2 distinct

complements (not equal), a-bar and b.

Then by P-5, we must have that:

a +b =1

and

a + a =1

and

a⋅a = 0

and

a ⋅b = 0

10/8/2010

a = a ⋅1

b = b ⋅1

= a ⋅ ( a + b ) = b( a + a )

= a ⋅a + a ⋅b = b⋅a + b⋅a

= 0 + a ⋅b

= 0+b⋅a

= a ⋅b

= a ⋅b

Xuan-Tu Tran

a =b

P-2(ii)

P-4(i)

P-2(i)

10

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications – Basic logic operations

From this point, we will restrict our attention to switching algebras

only.

Switching algebra is basically a two-element Boolean algebra which,

obviously, has the two elements 0 and 1.

AND operation

x

y

z=x·y

0

0

0

Theorem 1(ii)

0

1

0

Theorem 2(i)

1

0

0

Postulate 3(ii)

1

10/8/2010

Symbols

1

1

Theorem 1(ii)

Xuan-Tu Tran

x

y

x

y

z

&

&

z

11

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications – Basic logic operations

Symbols

OR operation

x

y

z=x+y

0

0

0

Derived from which

0

1

1

theorem, postulate?

1

0

1

Exercise for Students

1

1

1

x

y

x

y

≥1

≥1

z

Symbols

NOT operation

10/8/2010

z

x

z = x-bar

Derived from which

0

1

theorem, postulate?

1

0

Exercise for Students

Xuan-Tu Tran

x

x

z

11

z

12

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications – Basic logic operations

NAND operation

Symbols

x

y

z=x·y

0

0

1

0

1

1

1

0

1

1

1

0

x

y

z=x+y

0

0

1

0

1

0

1

0

0

1

1

0

x

y

z

x

y

&

&

z

NOR operation

10/8/2010

Symbols

x

y

x

y

Xuan-Tu Tran

z

≥1

≥1

z

13

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications – Basic logic operations

XOR operation

x

y

z=x⊕y

0

0

0

0

1

1

1

0

1

1

1

Symbols

x

y

x

y

0

z

=1

=1

z

z = x ⊕ y = xy + x y

- Construct XOR gate from NOT, AND, and OR ?

10/8/2010

Xuan-Tu Tran

14

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications – Basic logic operations

NXOR operation

x

y

z=x⊕y

0

0

1

0

1

0

1

0

0

1

1

Symbols

x

y

x

y

1

z

=1

=1

z

z = x ⊕ y = x y + xy

- Construct NXOR gate from NOT, AND, and OR ?

10/8/2010

Xuan-Tu Tran

15

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications

Theorem 4 – Involution

Định lý phủ định của phủ định

(x ) = x

Let x be a switching variable. Then

Proof

10/8/2010

x

x

(x )

0

1

0

1

0

1

Since the left column is identical to the

right column and we have listed all

possibilities, we have proved the result.

Xuan-Tu Tran

16

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications

Theorem 5

(i)

Let x and y be two switching variables. Then

(ii)

x+ x⋅ y = x

x ⋅ (x + y ) = x

Proof

10/8/2010

x

y

x⋅ y

x

+

x⋅ y

0

0

0

0

+

0

0

0

1

0

0

+

0

0

1

0

0

1

+

0

1

1

1

1

1

+

1

1

Xuan-Tu Tran

17

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications

Theorem 6

Let x, y, and z be switching variables. Then

1.

(i)

Associativity (định lý kết hợp)

(ii)

2.

(i)

(ii)

3.

x+x⋅y = x+ y

x ⋅ (x + y ) = x ⋅ y

Consensus (đồng tâm)

(i)

(ii)

10/8/2010

x ⋅ ( y ⋅ z ) = (x ⋅ y ) ⋅ z

x + ( y + z ) = (x + y ) + z

x⋅ y + x ⋅z + y⋅z = x⋅ y + x ⋅z

(x + y ) ⋅ (x + z ) ⋅ ( y + z ) = (x + y ) ⋅ (x + z )

Xuan-Tu Tran

18

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications – De Morgan’s Theorem

De Morgan’s Theorem

Let x and y be two switching variables. Then

(x + y ) = x ⋅ y

(x ⋅ y ) = x + y

Proof

This theorem may easily verified by complete enumeration.

Exercise for Students

10/8/2010

Xuan-Tu Tran

19

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications – De Morgan’s Theorem

Used for the evaluation of complements of switching expressions.

For example,

(x + y ⋅ (z + w ))

= ( x ) ⋅ ( y ⋅ ( z + w ))

(

)

= ( x ) ⋅ (y + ( z ) ⋅ w )

= (x ) ⋅ y + (z + w )

= x ⋅ (y + z ⋅ w )

= x⋅ y + x⋅z⋅w

10/8/2010

Xuan-Tu Tran

20

ĐẠI HỌC

CÔNG NGHỆ

Algebra Implications – De Morgan’s Theorem

Corollary of De Morgan’s Theorem

(x1 + x 2 + L + x n ) = x1 ⋅ x 2 ⋅ L ⋅ x n

(x1 ⋅ x 2 ⋅ L ⋅ x n ) = x1 + x 2 + L + x n

Expand the following function:

((x ⋅ ( y + z ))⋅ ( y + w ⋅ z )⋅ (x + z ))

10/8/2010

Xuan-Tu Tran

21

ĐẠI HỌC

CÔNG NGHỆ

Truth tables

Truth table

a simple way of representing a switching function is to make a list of the

possible variable assignments and note the value the function takes on for

each assignment. This list is called Truth Table.

How to create the Truth Table of a switching function?

For example,

10/8/2010

F ( x , y , z ) = x + yz

Xuan-Tu Tran

22

ĐẠI HỌC

CÔNG NGHỆ

Truth tables

F ( x , y , z ) = x + yz

x

y

z

F (x, y, z)

0

0

0

1

0

0

1

1

0

1

0

1

0

1

1

1

2n assignment

1

0

0

0

possibilities

1

0

1

0

1

1

0

1

1

1

1

0

- Write the Truth Table for the following switching function: F (x, y, z) =

10/8/2010

Xuan-Tu Tran

x⋅ y⋅z + x⋅ y

23

ĐẠI HỌC

CÔNG NGHỆ

Truth tables

Find a switching function from a truth table

10/8/2010

x

y

F (x, y)

0

0

0

0

1

0

1

0

0

1

1

1

Xuan-Tu Tran

F (x, y) ???

24

ĐẠI HỌC

CÔNG NGHỆ

Canonical forms: Minterm - Maxterm

Alternatives to the tabular representation

To list only the assignments for which a function is 1 or,

To list those for which the function is 0

Canonical representation

F = 1 whenever x = 1 and y = 1

F ( x , y ) = xy

10/8/2010

Xuan-Tu Tran

x

y

F (x, y)

0

0

0

0

1

0

1

0

0

1

1

1

25

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