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Solved Examples on Reduction of Boolean Expression

Examples on Reduction of Boolean Expression: Here, we have set of some of the Solved Examples on Reduction of Boolean Expression.
Submitted by Saurabh Gupta, on November 18, 2019

Example 1: Simplify the given Boolean Expression to minimum no. of variables or literals.

  1. (A+B). (A+B)
  2. ABC + AB + ABC

Answer:

1) (A+B). (A+B)

    = A.A + A. B + B.A + B.B
    = A + A. B + B.A + 0
    = A (1 + B + B) = A

2) ABC + AB + ABC

    = ABC + ABC + AB
    = AB (C + C) +  AB
    = AB +  AB
    = B (A + A) = B. 1 = B

Example 2: Prove that: (A + C) (A.B +C) = 0

Solution:

    LHS     = A.(A + C) (A.B + C)
            = (A. A + AC) (A.B + C)
            = AC (A.B + C)		[Since, A.A = 0]
            = AC. A.B + AC. C
            = 0 = RHS			[Since, A. A = 0 and C. C = 0]
    Hence, our result is proved.

Example 3: Reduce the expression (B +BC) (B + BC) (B + D)

Solution:

    = (B +BC) (B + BC) (B + D)
    = (BB +BBC) (B + D)
    = (B +0) (B +D)		[Since, B.B =0]
    = BB +BD			[Since, B.B =1]
    = B + BD
    = B (1+D) =B 		[Since, 1+D =1].

Example 4: Reduce the expression example

Solution:

example
    = A.BC. (AB +ABC)		        [De-Morgan's Law]
    = A.BC (AB+ABC)
    = A.BC. AB + A.BC. ABC
    = A.A.B.B.C + A.A.B.B.C.C		[Manipulation]
    = 0 + 0 = 0

Example 4: Prove the Associative Law for XOR operation.

Solution:

We know that XOR operation is given as: AB = AB + AB

To prove associativity for XOR operation we are required to prove:

(AB) C =A(BC). Therefore,

example 2

example 3

From equation 1 and 2 we can see LHS = RHS, thus associative law holds true for XOR operation.






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