Page 61 - Computer science 868 Class 12
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37.  Using the truth table, state whether the following proposition is a tautology, contingency or contradiction:
                       ∼(A ∧ B) ∨ (∼A → B)                                                                       [ISC 2018]
                   Ans.    A        B     ∼(A ∧ B)  (∼A → B) ∼(A ∧ B) ∨ (∼A → B)
                           0        0        1        0            1
                           0        1        1        1            1
                           1        0        1        1            1
                           1        1        0        1            1
                         The final column has all 1's. Hence, it is a tautology.
                   38.  (a)  State the Commutative law and prove it with the help of a truth table.              [ISC 2018]
                   Ans.  Commutative law states that the interchanging of the order of operands in a Boolean equation does not change its result.
                        Using OR operator → A + B = B + A
                        Using AND operator → A * B = B * A
                                                         Truth Table
                           A        B       A+B      B+A
                           0        0        0        0
                           0        1        1        1
                           1        0        1        1
                           1        1        1        1
                        (b)   Convert the following expression into its canonical POS form:
                           F (X, Y, Z) = (X+Y').(Y'+Z)                                                           [ISC 2018]
                   Ans.  F (X, Y, Z) = (X+Y').(Y'+Z)
                        = (X+Y'+0).(Y'+Z+0)
                        = (X+Y'+ (Z.Z')).(Y'+Z+(X.X'))
                        = (X+Y'+Z).(X+Y'+Z').(X+Y'+Z).(X'+Y'+Z)
                        = (X+Y'+Z).(X+Y'+Z').(X'+Y'+Z)
                        (c)    Find the dual of
                          (A' + B).(1 + B') = A'+B                                                               [ISC 2018]
                   Ans.  (A' + B).(1 + B') = A'+B
                        Dual = A'.B + 0.B' = A'.B
                        (d)   Verify the following proposition with the help of a truth table:
                          (P ∧ Q) ∨ (P ∧ ∼ Q) = P                                                                [ISC 2018]
                   Ans.  Verify: (P∧Q)∨(P∧∼Q) = P
                        (e)   If F (A, B, C) = A' (BC' + B'C), then find F'                                      [ISC 2018]
                   Ans.  F (A, B, C) = A'(BC'+B'C)
                        F' = A" + (BC' + B'C)'
                        = A+ (BC')'.(B'C)'
                        = A + (B' + C).(B + C')
                    39.  Given the Boolean function F(A, B, C, D) = Σ (2, 4, 8, 9, 10, 12, 13).                  [ISC 2018]
                        Reduce the above expression by using 4-variable Karnaugh map, showing the various groups (i.e. octal, quads and pairs).
                   Ans.  F(A, B, C, D) = Σ (0, 2, 4, 8, 9, 10, 12, 13)
                             C'D'   C'D    CD    CD'

                       A'B'  0   1  1   0  3  0  2   1

                       A'B  4   1  5   0  7   0  6   0
                       AB   12   1  13   1  15   0  14   0

                       AB'  8   1  9   1  11   0  10   1

                          = There are three quads:
                        Quad 1: (m +m +m +m ) = C'D'
                                        12
                                            8
                                 0
                                    4
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                                                                                                      Boolean Algebra   59
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