Page 63 - Computer science 868 Class 12
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SOP Expression for X (A, F, C, I) : X (A ,F, C, I) = Σ (1, 2, 3, 5, 6, 7, 8, 9, 10, 11)
A'F'C'I + A'F'CI' + A'F'CI + A'FC'I + A'FCI' + A'FCI + AF'C'I' + AF'C'I +AF'CI' + AF'CI
(b) Using the truth table, state whether the following proposition is a tautology, contingency or a contradiction:
∼(A ∧ B) ∨ ( ∼A => B) [ISC 2018]
Ans. ∼(A∧B) ∨ (∼A=>B) is a Tautology, Contradiction or Contingency
A B ∼A ∼B ∼(A ∧ B) (∼A => B) ∼(A ∧ B) ∨ (∼A => B)
0 0 1 1 1 0 1
0 1 1 0 1 1 1
1 0 0 1 1 1 1
1 1 0 0 0 1 1
Hence, it is a Tautology.
(c) Simplify the following expression, using Boolean laws: [ISC 2018]
A • (A' + B) • C • (A + B)
Ans. A • (A' + B) • C • (A + B)
= (AA' + AB).(AC + BC)
= (0 + AB).(AC + BC)
= ABC + ABC
= ABC
42. State the law represented by the following proposition and prove it with the help of a truth table: P ∨ P = P [ISC 2017]
Ans. The law represented by the given proposition is Idempotent law.
Truth Table
P P P∨P P∨P⇔P
T T T T
F F F F
43. State the Principle of Duality. [ISC 2017]
Ans. Principal of duality states that each Boolean expression has its dual expression which can be obtained by replacing:
• AND (.) with OR (+) or OR (+) with AND (.)
• 0 with 1 or 1 with 0
• Complements remain unchanged
44. Find the complement of the following Boolean expression using De Morgan’s law: [ISC 2017]
F(a, b, c) = (b' + c) + a
Ans. ((b' + c) + a)'
= (b'+c)'. a'
= ((b')' . c') . a'
= bc'a'
45. If (∼P => Q) then write its: [ISC 2017]
(i) Inverse (ii) Converse
Ans. (i) Inverse of the given statement is P => ∼Q
(ii) Converse of the given statement is Q >= ∼P
46. Given the Boolean function F(A, B, C, D) = Σ (2, 3, 4, 5, 6, 7, 8, 10, 11).
Reduce the above expression by using 4-variable Karnaugh map, showing the various groups (i.e. octal, quads and pairs). [ISC 2017]
Ans. F(A, B, C, D) = Σ (2, 3, 4, 5, 6, 7, 8, 10, 11)
CD
AB C'D' C'D CD CD'
A'B' 0 1 1 3 1 2
A'B 1 4 1 5 1 7 1 6
AB
12 13 15 14
AB' 1 1 1
8 9 11 10
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Boolean Algebra 61

