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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