Module 3:Digital Electronics
Boolean Algebra
Quote of the day
“Most people say that it is the intellect
which makes a great scientist. They are
wrong: it is character”.
― Albert Einstein
Postulates that describe Boolean Algebra
P1: The Operations (+) and (·) are closed: i.e. x,y B
a) x + y B
b) x · y B
P2:Identity Law
a) 0 + x = x + 0 = x for every x  B
b) x · 1 = 1 · x = x for every x  B
P3:Commutative Law, for all x, y B
a) x + y = y + x
b) x · y = y · x
Postulates that describe Boolean Algebra
P4: Distributive Law: for all x, y, z B
a) x + (y · z) = (x + y) · (x + z)
b) x · (y + z)= (x · y) + (x · z)
P5:Compliment Law for every element x in B there
exist an element x̄ such that
a) x + x̄ = 1
b) x ·x̄ = 0
P6:There exist at least two elements x, y B such
that x  y.
The algebraic expression written using
variables is referred as Boolean expression.
Principal of Duality
 The dual of an algebraic expression is obtained
by interchanging + and · and interchanging 0’s
and 1’s.
 The Boolean identities appear in dual pairs.
 When there is only one identity on a line the
identity is self-dual, i. e., the dual expression =
the original expression.
 Sometimes, the dot symbol ‘’ (AND operator) is
not written when the meaning is clear.
Summary of Postulates that describe
Boolean Algebra
Postulate
Part (a)
Dual: Part (b)
P2:Identity Law
0+x=x+0=x
x·1=1·x=x
P3:Commutative Law
x+y=y+x
x·y=y·x
P4: Distributive Law
P5:Compliment Law
x + yz = (x + y) (x + z) x (y + z)= xy + xz
x + x̄ = 1
x ·x̄ = 0
Boolean Theorems
• Theorem 1:
a) x + x = x
Proof:
x+x= (x+x)1
=(x+x)·(x+x̄)
=x+x·x̄
=x+0
=x
by P2b
by P5a
by P4a
by P5b
by P2a
Dual of Theorem 1:
b)x · x = x
Proof:
x·x = (x·x)+0
by P2a
=(x·x)+(x·x̄) by P5b
=x·(x+x̄)
by P4b
=x·1
by P5a
=x
by P2b
Boolean Theorems
• Theorem 2:
a) x + 1 = 1
Proof:
x+1 = (x+1) ·1
=(x+1)·(x+x̄)
=x+1·x̄
=x+x̄
=1
Dual of Theorem 2:
b)x · 0 = 0
Proof:
by P2a
by P2b x·0 = (x·0)+0
=(x·0)+(x·x̄) by P5b
by P5a
=x·(0+x)̄
by P4b
by P4a
=x·x̄
by P2a
by P2b
=0
by P5b
by P5a
Boolean Theorems
• Theorem 3:Involution law
Proof: Let x̄ be the compliment of x and (x̄ ) be the
compliment of x̄ .
Then by postulate P5 x + x̄ =1, x· x̄ =0, x̄ +(x̄ )=1
and x̄ ·(x̄ )=0.
(x)  (x)  0
 (x)  x x
By P2a
By
  
 (x)  x  1
By
 ( x)  x  (x)  x By
P5b
P4a
P5a
Boolean Theorems
• Theorem 3:Involution law


Proof contd: (x)  (x)  x  x  x  By P5a

 x  (x)  x
 x0
x

By
P4a
By P5b
By P 2a
Boolean Theorems
• Theorem 4:Associative Law (Proof By Perfect
Induction)
a) x + (y + z) = (x + y) + z
Associative Law
X
Y
Z
X+Y
Y+Z
(X + Y) + Z
X + (Y + Z)
0
0
0
0
0
0
0
0
0
1
0
1
1
1
0
1
0
1
1
1
1
0
1
1
1
1
1
1
1
0
0
1
0
1
1
1
0
1
1
1
1
1
1
1
0
1
1
1
1
1
1
1
1
1
1
1
Boolean Theorems
• Theorem 4:Associative Law (Proof By Perfect
Induction)
b)x · (y · z) = (x · y) · z
Associative Law
X
Y
Z
X·Y
Y·Z
(X · Y) · Z
X· (Y · Z)
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
0
1
1
0
1
0
0
1
0
0
0
0
0
0
1
0
1
0
0
0
0
1
1
0
1
0
0
0
1
1
1
1
1
1
1
Boolean Theorems
• Theorem 5:Demorgans Theorem By Perfect
Induction
a)(x ͞+͞ y) = (x+̄ ȳ)
b) (x ͞·͞ y) = x̄ · ȳ
Demorgan’s Theorem
X
Y
X̄
Ȳ
X+Y
0
X̄·Ȳ
x ͞+͞ y
X̄+Ȳ
x ͞·͞ y
1
X·Y
0
0
0
1
1
1
1
1
0
1
1
0
0
1
0
0
1
1
1
0
0
1
0
1
0
0
1
1
1
1
0
0
1
1
0
0
0
0
Boolean Theorems
• Theorem 6:Absorption
Law
a) x + x · y = x
Proof:
x+(x · y)= (x·1) +(x·y)….
…… by P2b
=x · (1+y) by P4b
=x · 1
by T2a
=x
by P2a
Dual of Theorem 2:
b)x · (x + y) = x
Proof:
x·(x+y)= (x+0) · (x+y)….
…… by P2a
=x+0·y
by P4a
=x+0
by T2b
=x
by P2a
(x)  x
P2:Identity Law
Summery
of the
Boolean
Laws and
theorems
P3:Commutative
Law
P4: Distributive
Law
P5:Compliment
Law
Idempotent law
Theorem 1
Involution Law
Associative Law
Demorgans
Theorem
0+x=x+0=x
x·1=1·x=x
x+y=y+x
x·y=y·x
x + yz = (x + y) (x + z)
x (y + z)= xy + xz
x + x̄ = 1
x ·x̄ = 0
x+x=x
x·x=x
x+1=1
x·0=0
(x)  x
x + (y + z) = (x + y) + z
x · (y · z) = (x · y) · z
(x  y)  x  y
(x  y)  x  y
x+x·y=x
x · (x + y) = x
a) x + x̄ · y = x + y, b) x ·( x̄ + y )= x · y
Absorption law
Simplify Y= AB+ AB̄
•
•
•
•
Y= AB+AB̄ By Distributive Theorem
Y=A(B+B̄)
By Compliment law,
Y=A.1
By Identity law,
Y=A
Simplify Y= (A+B)· (A+B̄ )
Simplify Y= B(DC+DC̄)+AB
Simplify Y= ABC+ĀB+ABC̄
Reduce the following Boolean Expression



f  A B  AC  B C



 

 f  A  AC B B C
  
 f  A B
 f  A B
By Associative and
commutative Law
By Absorption law
f  A B

By Demorgan’s Theorem

 f  A B  A  C  B  C
 


By Demorgan’s Theorem
͞ +AB̄ C(AB+ C)
Simplify Y= AB+ AC
 Y  AB  AC  ABCAB  ABCC
 Y  AB  AC  AC ( B  B)  ABC
0
 Y  AB  A  C  ABC
 Y  A  C  A( B  BC )
 Y  A  C  A( B  C )
 Y  A  C  AB  AC
 Y  A  AB  C  AC
By Distributive Theorem
By Compliment law,
Identity, Demorgan’s
By Distributive Theorem
By Absorption law
By Distributive Theorem
By Associative Theorem
Y  A  B  A  C
By Absorption law
Y  B  C  A  A
By Associative law
 Y  B  C  1 Y  1
By Compliment law,
Identity
Reduce the following Boolean Expression
f  ABC  ABC  ABC  ABC
 f  AC ( B  B)  ABC  ABC
By Distributive Theorem
 f  AC  ABC  ABC
By Compliment law,
Identity
 f  A(C  BC )  ABC
By Distributive Theorem
 f  A(C  B)  ABC
By Absorption law
 f  AC  AB  ABC
By Distributive Theorem
 f  AC  B( A  AC )
By Distributive Theorem
 f  AC  B( A  C)
By Absorption law
 f  AC  AB  BC
By Distributive Theorem
Simplify the following Boolean Expression
ABC  ABC  ABC  ABC  ABC
BC  AC  AB  BCD
CD  A A  CD  AB
Show that AB+ĀC+BC=AB+ĀC
Assignment
• With the help of switching circuit, Input/output
waveforms and truth table explain the operation
of a NOT Gate, AND Gate and OR gate.
• Explain the construction of OR and AND Gates
using Diodes.
• Prove that a) x + x̄ · y = x + y, b) x ·( x̄ + y )= x · y