Boolean Algebra
Dr Amber Habib
Mathematical Sciences Foundation
St. Stephen’s College
Delhi 110007
Algebra of Sets 1
Fix a set X and consider the behaviour of its
subsets relative to the operations of union
(∪), intersection (∩). For every A, B, C ⊂ X,
we have:
• Idempotent Law: A ∪A = A ∩A = A.
• Commutative Law:
A ∪B = B ∪A, A ∩B = B ∩A.
• Associative Law:
A ∪ (B ∪ C) = (A ∪B) ∪ C
A ∩ (B ∩ C) = (A ∩B) ∩ C
• Distributive Law:
A ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C)
A ∪ (B ∩ C) = (A ∪B) ∩ (A ∪ C)
1
Algebra of Sets 2
Moreover, there are two special subsets: ∅and X. Relative to these, the operations
obey:
• A ∪ ∅ = A, A ∩ ∅ = ∅.
• A ∪X = X, A ∩X = A.
In particular ∅ serves as identity for ∪, while
X does the same for ∩.
Finally, we have the operation of complemen-
tation: Every A ⊂ X has a unique comple-
ment A′ (with respect to X), and
• A ∪A′ = X, A ∩A′ = ∅.
2
Algebra of Logic 1
Now we consider propositional calculus. We
let ∨ stand for ‘or’ and ∧ stand for ‘and’.
Then for all statements A, B, C, we have:
• Idempotent Law: A ∨A = A ∧A = A.
• Commutative Law:
A ∨B = B ∨A, A ∧B = B ∧A.
• Associative Law:
A ∨ (B ∨ C) = (A ∨B) ∨ C
A ∧ (B ∧ C) = (A ∧B) ∧ C
• Distributive Law:
A ∧ (B ∨ C) = (A ∧B) ∨ (A ∧ C)
A ∨ (B ∧ C) = (A ∨B) ∧ (A ∨ C)
3
Algebra of Logic 2
We write T for the statement which is neces-
sarily true and F for the one which is neces-
sarily false. Relative to these, the operations
obey:
• A ∨ F = A, A ∧ F = F .
• A ∨ T = T , A ∧ T = A.
In particular F serves as identity for ∨, while
T does the same for X.
Finally, we have the operation of negation:
Every statement A has a unique negation ¬A,
and
• A ∨ ¬A = T , A ∧ ¬A = F .
4
Boolean Algebras
An abstract Boolean Algebra is a set B with
• Two binary operations ∨ (“join”)and ∧(“meet”),
• Two special elements denoted 0 (“zero”)
and 1 (“unity”),
• An operation ′ (“complement”),
such that for all a, b, c ∈ B we have:
5
Laws of Boolean Algebra
• Idempotent Law: a ∨ a = a ∧ a = a.
• Commutative Law:
a ∨ b = b ∨ a, a ∧ b = b ∧ a.
• Associative Law:
a ∨ (b ∨ c) = (a ∨ b) ∨ c
a ∧ (b ∧ c) = (a ∧ b) ∧ c
• Distributive Law:
a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c)
a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c)
• a ∨ 0 = a, a ∧ 0 = 0.
• a ∨ 1 = 1, a ∧ 1 = a.
• a ∨ a′ = 1, a ∧ a′ = 0.
6
The Simplest Example
Let B = {0,1}. Define
• 0 ∧ 0 = 0 ∨ 0 =
1 ∧ 1 = 1 ∨ 1 =
• 0 ∧ 1 =
0 ∨ 1 =
• 0′ =
1′ =
7
The Simplest Example
Let B = {0,1}. Define
• 0 ∧ 0 = 0 ∨ 0 = 0,
1 ∧ 1 = 1 ∨ 1 = 1.
• 0 ∧ 1 =
0 ∨ 1 =
• 0′ =
1′ =
8
The Simplest Example
Let B = {0,1}. Define
• 0 ∧ 0 = 0 ∨ 0 = 0,
1 ∧ 1 = 1 ∨ 1 = 1.
• 0 ∧ 1 = 0,
0 ∨ 1 = 1.
• 0′ =
1′ =
9
The Simplest Example
Let B = {0,1}. Define
• 0 ∧ 0 = 0 ∨ 0 = 0,
1 ∧ 1 = 1 ∨ 1 = 1.
• 0 ∧ 1 = 0,
0 ∨ 1 = 1.
• 0′ = 1,
1′ = 0.
10
The Simplest Example 2
The previous example could also be obtained
as follows: Let B consist of the subsets of a
singleton set X = {x}. Let ∨ = ∪, ∧ = ∩,
and ′ stand for complement. Define
• 0 = ∅
• 1 = X
Then the rules for combining 0,1 are exactly
as on the previous slide.
11
Absorption Law
Let B be a Boolean algebra. If a, b ∈ B, then
a ∧ (a ∨ b) = a ∨ (a ∧ b) = a.
12
Absorption Law
Let B be a Boolean algebra. If a, b ∈ B, then
a ∧ (a ∨ b) = a ∨ (a ∧ b) = a.
Proof. First, by Distributive Law,
a ∧ (a ∨ b) = (a ∧ a) ∨ (a ∧ b) = a ∨ (a ∧ b).
13
Absorption Law
Let B be a Boolean algebra. If a, b ∈ B, then
a ∧ (a ∨ b) = a ∨ (a ∧ b) = a.
Proof. First, by Distributive Law,
a ∧ (a ∨ b) = (a ∧ a) ∨ (a ∧ b) = a ∨ (a ∧ b).
Then,
a ∧ (a ∨ b) = [a ∧ (a ∨ b)] ∨ [b ∧ b′]
14
Absorption Law
Let B be a Boolean algebra. If a, b ∈ B, then
a ∧ (a ∨ b) = a ∨ (a ∧ b) = a.
Proof. First, by Distributive Law,
a ∧ (a ∨ b) = (a ∧ a) ∨ (a ∧ b) = a ∨ (a ∧ b).
Then,
a ∧ (a ∨ b) = [a ∧ (a ∨ b)] ∨ [b ∧ b′]
= (a ∨ b) ∧ (a ∨ b′) ∧ (a ∨ b ∨ b)
∧(a ∨ b ∨ b′)
15
Absorption Law
Let B be a Boolean algebra. If a, b ∈ B, then
a ∧ (a ∨ b) = a ∨ (a ∧ b) = a.
Proof. First, by Distributive Law,
a ∧ (a ∨ b) = (a ∧ a) ∨ (a ∧ b) = a ∨ (a ∧ b).
Then,
a ∧ (a ∨ b) = [a ∧ (a ∨ b)] ∨ [b ∧ b′]
= (a ∨ b) ∧ (a ∨ b′) ∧ (a ∨ b ∨ b)
∧(a ∨ b ∨ b′)
= (a ∨ b) ∧ (a ∨ b′) ∧ 1
= (a ∨ b) ∧ (a ∨ b′)
= a ∨ (b ∧ b′)
= a ∨ 1 = a.
16
Cancellation Law
Let B be a Boolean algebra. Suppose there
are a, b, c ∈ B such that
a ∨ b = a ∨ c
a ∧ b = a ∧ c.
Then b = c.
17
Cancellation Law
Let B be a Boolean algebra. Suppose there
are a, b, c ∈ B such that
a ∨ b = a ∨ c
a ∧ b = a ∧ c.
Then b = c.
Proof. We repeatedly use the Absorption
Law:
b = b ∧ (a ∨ b)
18
Cancellation Law
Let B be a Boolean algebra. Suppose there
are a, b, c ∈ B such that
a ∨ b = a ∨ c
a ∧ b = a ∧ c.
Then b = c.
Proof. We repeatedly use the Absorption
Law:
b = b ∧ (a ∨ b)
= b ∧ (a ∨ c)
= (a ∧ b) ∨ (b ∧ c)
= (a ∧ c) ∨ (b ∧ c)
= c ∧ (a ∨ b)
= c ∧ (a ∨ c) = c.
19
Uniqueness of Complement
Let B be a Boolean algebra. Suppose a, b ∈ Bsuch that
a ∨ b = 1
a ∧ b = 0.
Then b = a′.
20
Uniqueness of Complement
Let B be a Boolean algebra. Suppose a, b ∈ Bsuch that
a ∨ b = 1
a ∧ b = 0.
Then b = a′.
Proof. We have
a ∨ b = a ∨ a′ = 1
a ∧ b = a ∧ a′ = 0.
Hence, by the Cancellation Law, b = a′.
Corollary: a′′ = a.
Corollary: 0′ = 1, 1′ = 0.
21
De Morgan’s Laws
Let B be a Boolean algebra and a, b ∈ B.
Then
• (a ∨ b)′ = a′ ∧ b′.
• (a ∧ b)′ = a′ ∨ b′.
22
De Morgan’s Laws
Let B be a Boolean algebra and a, b ∈ B.
Then
• (a ∨ b)′ = a′ ∧ b′.
• (a ∧ b)′ = a′ ∨ b′.
Proof. Let c = (a ∨ b)′. Then
(a ∨ b) ∧ (a′ ∧ b′) = (a ∧ a′ ∧ b′) ∨ (b ∧ a′ ∧ b′)
= 0 ∨ 0 = 0,
(a ∨ b) ∨ (a′ ∧ b′) = (a ∨ b ∨ a′) ∧ (a ∨ b ∨ b′)
= 1 ∧ 1 = 1.
23
Switching Circuits 1
Our aim is to study circuits such as the fol-lowing:
|b′
| |a b
|a
• Each mark, such as |a
represents aswitch (in this case, named a).
• Each switch has two states, “on” and“off” (or 1 and 0).
• If two switches always have the same state,we consider them to be the same. In par-ticular, they have the same name.
• If two switches always have opposite states,we call one the complement of the other,and denote it by a ′ (e.g. b and b′).
24
Switching Circuits 2
|b′
| |a b
|a
The specific question is: How do the states
of the individual switches affect the state of
the entire circuit? Or, which combinations
of states of individual switches lead to the
entire circuit being “on” (current can pass
from one end to the other) or “off” (current
cannot pass).
In the above example, the circuit is on exactly
when a is on, and b is off. Hence we could
just as well use the simpler circuit
| |a b′
25
Series Connection
| |a b
a ∧ b
Parallel Connection
|a
|b
a ∨ b
26
Series Connection
| |a b
a ∧ b
Parallel Connection
|a
|b
a ∨ b
A Series-Parallel Circuit
|b′
| |a b
|a
27
Series Connection
| |a b
a ∧ b
Parallel Connection
|a
|b
a ∨ b
A Series-Parallel Circuit
|b′
| |a b
|a
((a ∧ b) ∨ a) ∧ b′
28
Series-Parallel Circuits
Not only switches, but circuits can be placed
in series or parallel connections. Suppose we
have circuits A and B, which we denote by:
A B
We can connect them in series:
A B A ∧B
Or in parallel:
A
B
A ∨B
29
The Algebra of Switching Circuits 1
Consider two circuits A and B made fromswitches a, b, c, . . . . We consider them equalif for any choice of states of a, b, c, . . . , A andB have the same state.
We have defined two operations ∧ and ∨ onthe set of switching circuits. We have thefollowing identities for these operations:
• Idempotent Law: A ∨A = A ∧A = A.
A A =A
A
= A
• Commutative Law:
A ∨B = B ∨A, A ∧B = B ∧A.
• Associative Law:
A ∨ (B ∨ C) = (A ∨B) ∨ C
A ∧ (B ∧ C) = (A ∧B) ∧ C
30
The Algebra of Switching Circuits 2
Distributive Law
• A ∧ (B ∨ C) = (A ∧B) ∨ (A ∧ C)
|A
|B
|C
=|A
|B
|A
|C
• A ∨ (B ∧ C) = (A ∨B) ∧ (A ∨ C)
|A
|B
|C
=|A
|B
|A
|C
31
The Algebra of Switching Circuits 3Zero, Unity, Complements
We introduce two special switches:
• The switch named 1 is always on.
• The switch named 0 is always off.
It is easy to see that for any circuit A,
• A ∨ 0 = A, A ∧ 0 = 0.
• A ∨ 1 = 1, A ∧ 1 = A.
Two circuits are termed complementary ifthey are always in opposite states, and wethen name them A and A′. We have
• A ∨A′ = 1
• A ∧A′ = 0
32
The Boolean Algebra of Switching Circuits
The previous few slides show that switching
circuits follow the rules of Boolean algebra.
This enables us to develop systematic meth-
ods for analyzing them, especially for replac-
ing circuits by smaller ones with the same
behaviour.
For instance, consider the equality
|b′
| |a b
|a
= | |a b′
This can be derived by algebra:
((a ∧ b) ∨ a) ∧ b′ = a ∧ b′,
by the Absorption Law.
33
Switching Circuits and Truth-Tables
Recall that truth-tables are a useful way of
exploring the structure of logic. Since switch-
ing circuits have the same algebraic struc-
ture, we can expect a similar role for truth-
tables in this context. For instance, the cir-
cuit given by the expression a ∧ b′ has the
table:
a�b 0 10 0 01 1 0
or
a b a ∧ b′
0 0 00 1 01 1 01 0 1
Similarly, the circuit (a ∧ b) ∨ (a′ ∧ b′) has the
table:
a�b 0 10 1 01 0 1
or
a b a ∧ b′
0 0 10 1 01 1 11 0 0
34
Karnaugh Maps
Given a switching circuit, we can generate
its table in a mechanical way. Conversely,
given its table, we can generate the Boolean
expression corresponding to the circuit.
Karnaugh maps are a way of using the table
to generate a simple or efficient expression
for the circuit.
35
Karnaugh Maps For Two Switches
We start by looking at some possibilities for
a circuit involving two switches a and b:
a′ ∧ b′
a�b 0 10 1 01 0 0
a ∧ b′
a�b 0 10 0 01 1 0
aa�b 0 10 0 01 1 1
b′
a�b 0 10 1 01 1 0
(a′ ∧ b′) ∨ (a ∧ b)a�b 0 10 1 01 0 1
a′ ∨ b′
a�b 0 10 1 11 1 0
1a�b 0 10 1 11 1 1
0a�b 0 10 0 01 0 0
36
Karnaugh Maps For Two Switches 2
From the tables on the last slide, we canconclude the following:
• Whenever 1’s occur in vertical or hori-zontal pairs, the pair of boxes representsa single switch (or its complement – soone switch drops out).
• Whenever 1’s occur in a 2 × 2, square,the square represents the constant 1 (twoswitches drop out).
We illustrate the use of these observationsby an example:
a�b 0 10 1 11 1 0
1 1 ⇒ a′
11
⇒ b′
Hence the circuit is represented by a′ ∨ b′.37
Karnaugh Maps For Three Switches
We use tables of the following format:
a�bc 00 01 11 1001
Example 1:
a�bc 00 01 11 100 1 1 11 1
1 1 ⇒ a′ ∧ c11
⇒ b′ ∧ c′
So the circuit is represented by
(a′ ∧ c) ∨ (b′ ∧ c′).
38
Top Related