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A B (A+B)' A'.B' (A.B)' A'+B'
0 0 1 1 1 1
0 1 0 0 1 1
1 0 0 0 1 1
1 1 0 0 0 0
Algebraically, they can be represented as:
(A+B)' = A'.B'
(A.B)' = A'+B'
The proof using the truth table is given here.
From the above law, if we break the whole complement, the individual complements of the variables are obtained
along with a change in sign, i.e., “break the line and change the sign”. This process is called demorganisation and is
used to find the complement of a Boolean expression.
For example, A+B = A . B
Some examples based on the complement are given below.
Example 1: Find the complement of the following Boolean expression
F(a, b, c) = (b+c') + a [ISC 2017]
Solution:
((b+c') + a)'
= (b+c')'.a'
= b'.c".a' [c"=c]
= b'.c.a'
Example 2: Find the complement of the following Boolean expression
A.B'+ A'+ B.C [ISC 2016]
Solution:
(A.B'+A'+B.C)'
= (A.B')'.A".(B.C)' [A"=A]
= (A'+B").A.(B'+C')
= (A'+B).A.(B'+C')
Principle of Duality
You must have noted that all the Boolean laws mentioned above are written in pairs. In these pairs, the second law
is obtained by replacing OR (0) with AND (1) and vice-versa in the first law. This process of conversion is called the
Principle of Duality and the expressions are called dual expressions.
Consider the expression, A+0 = A.
If we replace + with . and 0 with 1, we get A.1=A.
Hence, the two expressions are dual to each other.
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