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Example 3:  Minimise the following expression using laws of Boolean algebra.
              Q.(Q'+P).R.(Q+R)                                                                                [ISC 2012]

              Solution:
              = (Q.Q'+Q.P).R               [Distributive and absorption laws]

              = (0+Q.P).R                  [Complement law]
              = (Q.P).R                    [Properties of 0 and 1]
              = P.Q.R                      [Distributive law]
              Example 4: Prove De Morgan’s law algebraically.

              Solution:
              De Morgan’s Law
              (i) (A+B)' = A'.B'
              (ii) (A.B)' = A'+B'

              To prove (A+B)'=A'.B', we have to apply Complement law X.X'= 0.
              According to De Morgan’s Law, A'.B' is the complement of (A+B).

              Then according to Complement law, (A+B).A'.B' = 0 should be satisfied.
              So, LHS = (A+B).A'.B'

                     = A.A'.B' + B.A'.B'          [Distributive law]
                     = 0+0                        [Complement law]
                     = 0

              To prove (A.B)' = A'+B', we use the other Complement law which states that X+X'=1.
              According to De Morgan’s Law, A'+B' is the complement of (A.B).

              Then according to Complement law, (A.B) + A'+B' = 1 should be satisfied.
              So, LHS = (A.B) + A'+B'
                     = (A.B) + (A'+B')

                     = (A'+B'+A).(A'+B'+B)        [Distributive law]
                     = (1+B').(A'+1)              [Complement law]
                     = 1.1

                     = 1
              Thus (A+B)' = A'.B' and (A.B)' = A'+B' proved.

              Example 5: Minimise the following expression using laws of Boolean algebra.
              (P+Q).(P'+Q).(P'+Q')

              Solution:
              = (P+Q).(P'+Q).(P'+Q')
              = (Q + P.P').(P'+Q')                [Distributive law]
              = Q.(P'+Q')                         [Complement law]

              = P'.Q + Q.Q'                       [Distributive law]




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