Page 31 - Computer science 868 Class 12
P. 31

Thus, the Principle of duality states that each Boolean expression has its dual expression which can be obtained by
                 replacing:
                 •  AND (.) with OR (+) or OR (+) with AND (.)
                 •  0 with 1 or 1 with 0
                 •  Complements remain unchanged

                 Let us see some examples.
                 Example 1: Find the dual of (A.B') + 1 + (A.B.C)

                 Solution: The dual is (A+B').0.(A+B+C)
                 Example 2: Find the dual of X.(Y+1).(Z+0)
                 Solution: The dual is X + Y.0 + Z.1

                 Example 3: Prove that the dual of the XOR gate is equal to its complement.
                 Solution:
                 The expression of a two-variable XOR gate is A'.B + A.B'

                 Its dual = (A'+B).(A+B')
                 Complement = (A'.B + A.B')'
                             = (A'.B)'.(A.B')'
                             = (A"+B').(A'+B")
                             = (A+B').(A'+B)

                             = (A'+B).(A+B') which is equal to the dual expression.

                 Simplification of a Boolean expression using laws
                 Boolean laws can be applied for proving or simplifying a Boolean expression. Simplifying involves eliminating terms or
                 removing variables, thus deriving a reduced expression that is easier to design using the logic circuit.
                 Example 1: Simplify (A+C).(A+A.D) + AC + C                                                     [ISC 2015]

                 Solution:
                 = (A+C).(A+A.D) + C + C.A
                 = (A+C).A+C                         [Absorption law]

                 = A+C                               [Absorption law]
                 Example 2: Prove (x'+z) + [(y'+z).(x'+y)]' = 1                                                 [ISC 2017]

                 Solution:
                 = (x'+z) + (y'+z)' + (x'+y)'        [De Morgan’s law]
                 = (x'+z) + y".z' + x".y'
                 = (x'+z) + y.z' + x.y'              [Involution law]

                 = (x'+x.y') + (z+y.z')
                 = (x+x').(x'+y') + (z+z').(z+y)              [Distributive law]
                 = 1.(x'+y') + 1.(z+y)
                 = x'+y'+y+z                         [Property of 1]

                 = x'+1+z                            [Property of 1]
                 = 1                                 [Property of 1]


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