KnowledgeBoat Logo

Computer Science

Reduce the following Boolean expression into their simplest forms:

  1. {(CD)' + A} + A + C.D + A.B
  2. A.{B + C (A.B + A.C)'}

Boolean Algebra

10 Likes

Answer

{(CD)' + A} + A + C.D + A.B


  {(CD)' + A} + A + C.D + A.B
= C' + D' + A + A + C.D + A.B           [De-Morgan's Law]
= C' + D' + A + C.D + A.B               [∵ A+A=A]
= A(1 + B) + C' + D' + C.D  
= A + C' + D' + C.D                     [∵ 1+B=1]
= (A + C' + D' + C).(A + C' + D' + D)   [Distributive Law]
= (A + D' + 1).(A + C' + 1)             [∵ C'+C=1, D'+D=1]
= 1.1                                   [∵ A+D'+1=1, A+C'+1=1]
= 1

A.{B + C (A.B + A.C)'}


A.{B + C(A.B + A.C)'}
A.{B + C[(A.B)'.(A.C)']}                [De-Morgan's Law]
A.{B + C[(A' + B').(A' + C')]}          [De-Morgan's Law]
A.{B + C[A'A' + A'C' + A'B' + B'C']}    [Distributive Law]
A.{B + C[A' + A'C' + A'B' + B'C']}      [∵ A'A'=A']
A.{B + C[A'(1 + C' + B') + B'C']}
A.{B + C[A'.1 + B'C']}                  [∵ 1 + C' + B' = 1]
A.{B + A'C + B'CC'}                     [Distributive Law]
A.{B + A'C + 0}                         [∵ CC'=0]
AB + AA'C                               [Distributive Law]
AB                                      [∵ AA'C=0]

Answered By

5 Likes


Related Questions