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

