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= π(110,111,001,101)                       [Writing 1 in complemented literal and 0 in normal literal]
                 = π(6, 7, 1, 5)

                 Alternatively, using the truth table method

                                     X          Y          Z         (X'+Y')       (Y+Z')    (X'+Y').(Y+Z')
                                     0          0          0           1            1             1
                                     0          0          1           1            0             0

                                     0          1          0           1            1             1
                                     0          1          1           1            1             1
                                     1          0          0           1            1             1
                                     1          0          1           1            0             0
                                     1          1          0           0            1             0

                                     1          1          1           0            1             0
                 Taking the maxterms with 0’s in the final column, we get

                 (X+Y+Z').(X'+Y+Z').(X'+Y'+Z).(X'+Y'+Z') as the POS
                   OR
                 = π(1, 5, 6, 7)

                 Example 2: Convert the following SOP to canonical SOP
                 XY + Z'

                 Solution:
                                                                          st
                                                                                                             nd
                 = X.Y.1 + Z'.1.1                           [Z is missing in 1  term and X and Y are missing in the 2  term]
                 = X.Y.(Z+Z') + Z'.(X+X').(Y+Y')            [Replacing 1 with the missing literal]
                 = X.Y.Z + X.Y.Z' + (X.Z'+X'.Z').(Y+Y')      [Distributive law]
                 = X.Y.Z + X.Y.Z' + X.Y.Z' + X'.Y.Z' + X.Y'.Z' + X'.Y'.Z'
                 = X.Y.Z + X.Y.Z' + X'.Y.Z' + X.Y'.Z' + X'.Y'.Z'      [Eliminating duplicates]
                 = Σ(111, 110, 010, 100, 000)               [Writing 0 for complemented and 1 for normal literal]

                   OR
                 = Σ(7, 6, 2, 4, 0)
                 Alternatively, using the truth table method

                                       X         Y          Z           X.Y         Z'         X.Y+Z'

                                       0         0          0           0           1            1
                                       0         0          1           0           0            0
                                       0         1          0           0           1            1
                                       0         1          1           0           0            0
                                       1         0          0           0           1            1
                                       1         0          1           0           0            0
                                       1         1          0           1           1            1
                                       1         1          1           1           0            1




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