Canonical Forms:

There are two types of canonical forms:

  1. Disjunctive Normal Forms or Sum of Products or (SOP).
  2. Conjunctive Normal Forms or Products of Sums or (POS).

Disjunctive Normal Forms or Sum of Products or (SOP):

A Boolean expression over ({0, 1}, ∨,∧,') is said to be in disjunctive normal form if it is a join of minterms

Example: (x1'∧x2'∧x3')∨( x1'∧x2∧x3' )∨(x1∧x2∧x3) is a Boolean expression in disjunctive normal form.

Since there are three min-terms x1'∧x2'∧x3',x1'∧x2∧x3 and x1∧x2∧x3.

Max-term: A Boolean Expression of n variables x1,x2,....xnis said to be a max-term if it is of the form x1∨x2∨..........∨xn

where xi is used to denote xi or xi'.

Conjunctive Normal Forms or Products of Sums or (POS):

A Boolean expression over ({0, 1}, ∨,∧,') is said to be in a disjunctive normal form if it is a meet of max-terms

Example:

        (x1∨x2∨x3)∧( x1∨x2∨x3 )∧(x1∨x2∨x3 )∧(x1'∨x2∨x3' )∧(x1'∧x2'∧x3)

is a Boolean expression in conjunctive normal form consisting of five max-terms.

Obtaining A Disjunctive Normal Form:

Consider a function from {0, 1}n to {0, 1}. A Boolean expression can be obtained in disjunctive normal forms corresponding to this function by having a min-term corresponding to each ordered n-tuples of 0's and 1's for which the value of the function is 1.

Obtaining A Conjunctive Normal Form:

Consider a function from {0, 1}n to {0, 1}. A Boolean expression can be obtained in conjunctive normal forms corresponding to this function by having a max-term corresponding to each ordered n-tuples of 0's and 1's for which the value of function is0.

Example: Express the following function in

  1. Disjunctive Normal Form
  2. Conjunctive Normal Form
ff
(0, 0, 0)1(1, 0, 0)0
(0, 0, 1)0(1, 0, 1)1
(0, 1, 0)1(1, 1, 0)0
(0, 1, 1)0(1, 1, 1)1

Solution:

  1. (x1'∧ x2' ∧ x3') ∨(x1'∧x2∧ x3' )∨(x1∧x2'∧x3 )∨(x1∧x2∧x3)
  2. (x1'∨x2'∨x3')∧( x1'∨x2∨x3 )∧(x1∨x2'∨x3' )∧(x1∨x2∨x3')

Principle of Duality:

The dual of any expression E is obtained by interchanging the operation + and * and also interchanging the corresponding identity elements 0 and 1, in original expression E.

Example: Write the dual of following Boolean expressions:

1. (x1*x2) + (x1*x3')         2. (1+x2)*( x1+1)
3. (a ∧(b∧c))

Solution:

1. ( x1+x2)*( x1+x3')         2. (0*x2)+( x1*0)
3. (a ∨(b∧c))

Note: The dual of any theorem in a Boolean algebra is also a theorem.