Dual of a Boolean Expression
To get a dual of any Boolean Expression, replace
 OR with AND i.e. + with .
 AND with OR i.e. . with +
 1 with 0
 0 with 1
Example01:
We know, Consensus theorem is
xy + x’z + yz = xy + x’z
The dual of Consensus theorem is
(x + y)(x’ + z)(y + z) = (x + y)(x’ + z)
Example02:
Consider any Boolean Expression
xyz + x’yz’ + y’z = 1
The dual of this Boolean expression is
(x + y + z)(x’ + y + z’)(y’ + z) = 0
Selfdual Functions
When a function is equal to it dual, it is called as a self dual function. 
Example
Consider the function
F (A , B , C) = AB + BC + CA
The dual of this function is
F_{d} (A , B , C)
= (A + B)(B + C)(C + A)
= AB + BC + CA
Since, F (A , B , C) = F_{d} (A , B , C)
∴ F (A , B , C) is a selfdual function.
Conditions for a selfdual function
The necessary and sufficient conditions for a function to be a selfdual function are
 The function must be a Neutral function.
 The function must not contain mutually exclusive terms.
What are mutually exclusive terms?
Consider we have any term X consisting of some variables, then a term obtained by complementing each variable of term X is called as its mutually exclusive term.
Example

Number of Selfdual Functions
where n = number of Boolean variables in the function
Explanation
 We know, for a function to be a selfdual function, the function must be a neutral function.
 For a function to be a neutral function, number of minterms must be equal to number of maxterms. So, we choose half of the terms i.e. 2^{n} / 2 = 2^{n1 }terms.
 Now, for each of these terms, we have two choices whether to include it or not in the selfdual function.
So,
Possible number of selfdual functions
= 2 x 2 x 2 x ……. x 2^{n1}
= 2^{2}^{^(n1)}
Relationship between Neutral Functions and Selfdual Functions
Every selfdual function is surely a neutral function but every neutral function need not be a selfdual function.
Important Property of Selfdual Functions
Selfduality is closed under complementation.
Example
If the function F (A , B , C) = ∑ (0 , 1 , 2 , 4) is a selfdual function, then
Its complement function F’ (A , B , C) = ∑ (3 , 5 , 6 , 7) will also be a selfdual function.
PRACTICE PROBLEM BASED ON SELFDUAL FUNCTIONS
Problem
Consider the following functions
 F (A , B , C) = ∑ (0 , 2 , 3)
 F (A , B , C) = ∑ (0 , 1 , 6 , 7)
 F (A , B , C) = ∑ (0 , 1 , 2 , 4)
 F (A , B , C) = ∑ (3 , 5 , 6 , 7)
Which of the above functions are selfdual functions?
 Only (iii)
 Only (ii)
 Only (iii) and (iv)
 All are selfdual functions
Solution
Condition01:
According to condition01, for a function to be a selfdual function, the function must be a neutral function.
 In all the given options, we have functions of 3 variables A, B and C.
 For a function with 3 Boolean variables, neutral function must contain exactly 2^{n1} = 2^{31} = 4 minterms and 4 maxterms.
 But, Function(i) contains only 3 minterms. So, it is not a neutral function and therefore it can’t be a selfdual function. So, Function(i) gets eliminated.
 We are now left with three other functions which satisfies condition01 and are all neutral functions. We will now use 2^{nd} condition to eliminate the incorrect option(s).
Condition02:
According to condition02, a selfdual function must not contain mutually exclusive terms.
First, let us find which terms are mutually exclusive
A  B  C  Minterms  
0  0  0  0  A’B’C’ 
1  0  0  1  A’B’C 
2  0  1  0  A’BC’ 
3  0  1  1  A’BC 
4  1  0  0  AB’C’ 
5  1  0  1  AB’C 
6  1  1  0  ABC’ 
7  1  1  1  ABC 
From here, pairs of mutually exclusive terms are
(0,7) , (1,6) , (2,5) , (3,4)
 Because mutually exclusive terms are not allowed in selfdual functions, therefore, terms inside the pairs can not appear together.
 But terms 0 and 7 appear together in the function(ii). So, it can not be a selfdual function.
 But functions (iii) and (iv) do not contain any mutually exclusive terms.
∴ Functions (iii) and (iv) are selfdual functions.
Thus,
Option (C) is correct.
Note
Functions (iii) and (iv) are complementary functions. So, if one function is a selfdual function, the other function will also be a selfdual function because selfdual functions are closed under complementation.
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