SUMITA ARORA Class 11 CS ASSIGNMENT CHAPTER-3 TYPE-B PART-1

Chapter-3 Type-B Solution for Assignment Question No 6-15

6. State the principal of duality in boolean algebra and give the dual of the boolean expression. \((X+Y).(\overline{X}+\overline{Z}).(Y+Z)\)

Sol: According to the principle of duality, a true boolean statement or expression can be converted to its dual form by replacing the “0”s in the statement or expression with “1”s, and vice versa and by replacing the “+”s in the statement or expression with “.”s, and vice versa.

The dual form of \((X+Y).(\overline{X}+\overline{Z}).(Y+Z)\)   is \((X.Y)+(\overline{X}.\overline{Z})+(Y.Z)\)

In the above dual expression all the “+”s were replaced by “.”s, and vice versa.

7. Prove the idempotent law of boolean algebra with the help of a truth table.

Sol:

8. Use the duality theorem to derive another boolean relation from \(A+\overline{A}B = A+B\)

Sol: The duality theorem states that the AND operator is replaced with the OR operator. And OR operator is replaced with AND operator. Binary number 0 is replaced with binary number 1. And Binary number 1 is replaced with binary number 0.

Applying duality principle:

\(A+\overline{A}B = A+B \)
\(A.(\overline{A}+B)    = AB\) (Apply Duality & Distributive law)
\(A\overline{A}+AB    = AB\) (Apply Complement law)
\(0+AB = AB\) (Identity law)
\(AB            = AB\)
Therefore LHS = RHS

9. What would be the complement of the following? (a) \(\overline{A}(B\overline{C}+\overline{B}C)\)    (b) \(xy + \overline{y}z + \overline{z}z\)

Sol (a):  Not the equation

=\(\overline{\overline{A}(B\overline{C}+\overline{B}C)} \) – Apply Demorgan’s first law
=  \(\overline{\overline{A}} ( \overline{(B\overline{C}}) \overline{(\overline{B}C))}\) – Apply Demorgan’s second law (adding + inside and outside( )  )
= \(\overline{\overline{A}}+((\overline{B}+\overline{\overline{C}}) (\overline{\overline{B}}+\overline{C})) \)-  Apply Involution law
= \(A+((\overline{B}+C) (B+\overline{C}))\) – Apply Distribution law
= \(A+( (B+\overline{C})\overline{B}+(B+\overline{C})C )\) – Apply Distribution law
= \(A+ (B\overline{B}+\overline{B}\overline{C}+CB+C\overline{C})\) – Apply Complement law
= \(A+(0+\overline{B}\overline{C}+CB+0)\) – Apply Identity law
= \(A+\overline{B}\overline{C}+CB\)

Sol (b): Not the equation

=  \( \overline{xy + \overline{y}z + \overline{z}z}\) – First law of Demorgan
= \( \overline{(xy)}\)  \( \overline{(\overline{y}z)} \)  \(\overline{(\overline{z}z)} \) –  Second Law of Demorgan
= \( (\overline{x} + \overline{y}) (\overline{\overline{y}}+ \overline{z})( \overline{\overline{z}} + \overline{z}) \)- Apply Involution law
= \( (\overline{x} + \overline{y}) (y + \overline{z})(z + \overline{z}) \) – Apply Complement law
= \( (\overline{x}+\overline{y})(y+\overline{z})1 \) – Apply Identity law
= \( (\overline{x}+\overline{y})(y+\overline{z}) \) – Apply Distribution law
= \(\overline{x}y+\overline{x}\overline{z}+\overline{y}y+\overline{y}\overline{z} \)-  Apply Indempotent law
= \(\overline{x}y+\overline{x}\overline{z}+0+\overline{y}\overline{z} \) – Apply Identity law
= \(\overline{x}y+\overline{x}\overline{z}+\overline{y}\overline{z} \) – Apply Consensus Law
= \(\overline{x}y+\overline{y}\overline{z} \)

10. Find the complement of the following Boolean function: F\(1 = A\overline{B} + \overline{C}\overline{D}\)

Sol: Not the equation:

= \( \overline{A\overline{B} + \overline{C}\overline{D}} \) – Apply Demorgan’s first law
= \( \overline{(A\overline{B})} \) \(\overline{(\overline{C}\overline{D})} \). – Apply DeMorgan’s Second Law
= \( (\overline{A} +\overline{\overline{B}}). (\overline{\overline{C}}+\overline{\overline{D}} ) \) – Apply Involution law
= \( (\overline{A}+ B) (C+D)  \)- Apply Distribution law
= \((C+D)\overline{A}+(C+D)B \) – Apply Distribution law
= \( \overline{A}C+\overline{A}D+BC+BD \)

11. Find the complement of Boolean expression \( (A+\overline{B}+C) (A+\overline{B}C) \)

Sol: Not the equation:

= \( \overline{(A+\overline{B}+C) (A+\overline{B}C) } \) – Apply Demorgan’s second law
= \( \overline{(A+\overline{B}+C)} + \overline{(A+\overline{B}C)} \) – Apply Demorgan’s first law
= \( (\overline{A}.\overline{\overline{B}}.\overline{C}) + ( \overline{A}.(\overline{\overline{B}}+\overline{C})) \) – Apply Involution law
= \( (\overline{A}B\overline{C})+(\overline{A}(B+\overline{C})\) – Apply Distribution law
= \( \overline{A}B\overline{C}+\overline{A}B+\overline{A}\overline{C} \) – Apply Absorption law (A+AB=A)
= \( \overline{A}B+\overline{A}\overline{C} \)

12. Find the complement of Boolean expression \(\overline{A}D+\overline{C}D+\overline{A}B \)

Sol: Not the equation:

=\( \overline{\overline{A}D+\overline{C}D+\overline{A}B } \) – Apply Demorgan’s first law
=\( \overline{(\overline{A}D)}\)    \(\overline{(\overline{C}D)} \)  \(\overline{(\overline{A}B)} \) – Apply Demorgan’s second law
= \( (\overline{\overline{A}}+\overline{D}).(\overline{\overline{C}}+\overline{D}).(\overline{\overline{A}}+\overline{B}) \) – Apply Involution law
= \( (A+\overline{D})(C+\overline{D})(A+\overline{B}) \) – Apply Distribution law
= \( (AC+A\overline{D}+\overline{D}C+\overline{D}\overline{D})(A+\overline{B}) \) – Apply Distribution law
= \( (AC+\overline{D}) (A+\overline{B}) \) – Apply Distribution law
= \( AAC+AC\overline{B}+A\overline{D}+\overline{D}\overline{B} \) – Apply Indempotence law & Absorption law
= \( AC+A\overline{D}+\overline{D}\overline{B} \)