Chapter 3, Problem 15E

a.

Program Plan Intro

Distributive law:

Consider three variables x, y, and z. The multiplication of variable (x) with the sum of two variables (y and z) is same as the sum of the products (xy and xz).

The representation of distributive law is as follows:

$\begin{array}{l}\text{x + (y}\cdot \text{z) = (x + y) (x + z)}\\ \text{x (y + z) = x}\cdot \text{y + x}\cdot \text{z}\end{array}$

Identity law:

The sum of the variable (x) and the value 0 is “x” and the product of the variable (x) and the value 1 is “x”.

The representation of identity law is as follows:

$\begin{array}{l}\text{0 + x = x}\\ \text{1 }\cdot \text{ x = x}\end{array}$

Inverse law:

The sum of the variable (x) and the complement of the variable (x’) is 1 and the product of the variable (x) and the complement of the variable (x’) is 0.

The representation of inverse law is as follows:

$\begin{array}{l}\text{x + x' = 1}\\ \text{x }\cdot \text{ x = 0}\end{array}$

Idempotent law:

The sum of the variable (x) and the same variable (x) is “x” and the product of the variable (x) and the same variable (x) is “x”.

The representation of idempotent law is as follows:

$\begin{array}{l}\text{x + x = x}\\ \text{x }\cdot \text{ x = x}\end{array}$

Explanation of Solution

Simplification:

$\begin{array}{c}\text{F(x, y, z) = x}\cdot \left(\text{y}\cdot \text{z + y'}\cdot \text{z}\right)\text{+ x}\cdot \text{y + x'}\cdot \text{y + x}\cdot \text{z}\\ \text{= x}\cdot \text{y}\cdot \text{z + x}\cdot \text{y'}\cdot \text{z + x}\cdot \text{y + x'}\cdot \text{y + x}\cdot \text{z }\left(\text{by Distributive law}\right)\\ \text{= x}\cdot \text{z }\left(\text{y + y'}\right)\text{ + x}\cdot \text{y + x'}\cdot \text{y + x}\cdot \text{z }\left(\text{by Distributive law}\right)\\ \text{= x}\cdot \text{z }\left(\text{1}\right)\text{ + x}\cdot \text{y + x'}\cdot \text{y + x}\cdot \text{z }\left(\text{by Inverse law}\right)\end{array}$

$\begin{array}{c}\text{= x}\cdot \text{z + x}\cdot \text{z + x}\cdot \text{y + x'}\cdot \text{y }\left(\text{by Identity law}\right)\\ =\text{ x}\cdot \text{z + x}\cdot \text{y + x'}\cdot \text{y }\left(\text{by Idempotent law}\right)\\ =\text{ x}\cdot \text{z + y }\left(\text{x + x'}\right)\left(\text{by Distributive law}\right)\\ =\text{ x}\cdot \text{z + y }\left(\text{1}\right)\left(\text{by Inverse law}\right)\\ =\text{ x}\cdot \text{z + y }\left(\text{by Identity law}\right)\end{array}$

Therefore, the simplified expression is “$x\cdot z+y$”.

b.

Program Plan Intro

Distributive law:

Consider three variables x, y, and z. The multiplication of variable (x) with the sum of two variables (y and z) is same as the sum of the products (xy and xz).

The representation of distributive law is as follows:

$\begin{array}{l}\text{x + (y}\cdot \text{z) = (x + y) (x + z)}\\ \text{x (y + z) = x}\cdot \text{y + x}\cdot \text{z}\end{array}$

Identity law:

The sum of the variable (x) and the value 0 is “x” and the product of the variable (x) and the value 1 is “x”.

The representation of identity law is as follows:

$\begin{array}{l}\text{0 + x = x}\\ \text{1 }\cdot \text{ x = x}\end{array}$

Inverse law:

The sum of the variable (x) and the complement of the variable (x’) is 1 and the product of the variable (x) and the complement of the variable (x’) is 0.

The representation of inverse law is as follows:

$\begin{array}{l}\text{x + x' = 1}\\ \text{x }\cdot \text{ x = 0}\end{array}$

De Morgan’s Law:

• The complement of the intersection of two sets is the union of their complements.
• The complement of the union of two sets is the intersection of their complements.

The representation of the De Morgan’s law as follows:

$\begin{array}{l}\left(\text{x}\cdot \text{y}\right)\text{' = x' + y'}\\ \left(\text{x + y}\right)\text{' = x' }\cdot \text{ y'}\end{array}$

Double compliment law:

The representation of double compliment law is

$\text{x'' = x}$

Explanation of Solution

Simplification:

$\begin{array}{c}\text{F(x, y, z) = x}\cdot \text{y}\cdot \text{z'' + }\left(\text{y + z}\right)\text{'}\text{ + x'}\cdot \text{y}\cdot \text{z}\\ \text{= x}\cdot \text{y}\cdot \text{z + }\left(\text{y + z}\right)\text{'}\text{ + x'}\cdot \text{y}\cdot \text{z}\text{}\left(\text{by Double Complement law}\right)\\ \text{= x}\cdot \text{y}\cdot \text{z + }\left(\text{y'}\cdot \text{z'}\right)\text{ + x'}\cdot \text{y}\cdot \text{z}\text{}\left(\text{by DeMorgan's law}\right)\\ \text{= y}\cdot \text{z}\left(\text{x + x'}\right)\text{}\text{ + y'}\cdot \text{z' }\left(\text{by Distributive law}\right)\end{array}$

$\begin{array}{c}\text{= y}\cdot \text{z}\left(\text{1}\right)\text{}\text{ + y'}\cdot \text{z' }\left(\text{by Inverse law}\right)\\ \text{= y}\cdot \text{z + y'}\cdot \text{z' }\left(\text{by Identity law}\right)\end{array}$

Therefore, the simplified expression is “$y\cdot z+y\text{'}\cdot z\text{'}$”.

c.

Program Plan Intro

Distributive law:

Consider three variables x, y, and z. The multiplication of variable (x) with the sum of two variables (y and z) is same as the sum of the products (xy and xz).

The representation of distributive law is as follows:

$\begin{array}{l}\text{x + (y}\cdot \text{z) = (x + y) (x + z)}\\ \text{x (y + z) = x}\cdot \text{y + x}\cdot \text{z}\end{array}$

Idempotent law:

The sum of the variable (x) and the same variable (x) is “x” and the product of the variable (x) and the same variable (x) is “x”.

The representation of idempotent law is as follows:

$\begin{array}{l}\text{x + x = x}\\ \text{x }\cdot \text{ x = x}\end{array}$

Null law:

The sum of the variable (x) and the value 1 is “1” and the product of the variable (x) and the value 0 is “0”.

The representation of null law is as follows:

$\begin{array}{l}\text{1 + x = 1}\\ \text{0 }\cdot \text{ x = 0}\end{array}$

Identity law:

The sum of the variable (x) and the value 0 is “x” and the product of the variable (x) and the value 1 is “x”.

The representation of identity law is as follows:

$\begin{array}{l}\text{0 + x = x}\\ \text{1 }\cdot \text{ x = x}\end{array}$

Explanation of Solution

Simplification:

$\begin{array}{c}\text{F(x, y, z) = z}\cdot \left(\text{x}\cdot \text{y' + z}\right)\cdot \left(\text{x + y'}\right)\\ \text{= z}\cdot \left(\left(\text{x}\cdot \text{y' + z}\right)\cdot \text{x +}\left(\text{x}\cdot \text{y' + z}\right)\cdot \text{y'}\text{}\right)\left(\text{by Distributive law}\right)\\ \text{= z}\cdot \left(\left(\text{x}\cdot \text{x}\cdot \text{y' + z}\cdot \text{x}\right)\text{+}\left(\text{x}\cdot \text{y'}\cdot \text{y' + z}\cdot \text{y'}\right)\text{}\right)\text{}\left(\text{by Distributive law}\right)\\ \text{= z}\cdot \left(\text{x}\cdot \text{y' + z}\cdot \text{x + x}\cdot \text{y' + z}\cdot \text{y'}\text{}\right)\text{}\left(\text{by Idempotent law}\right)\end{array}$

$\begin{array}{c}\text{= z}\cdot \left(\text{x}\cdot \text{y' + z}\cdot \text{x + z}\cdot \text{y'}\text{}\right)\text{}\left(\text{by Idempotent law}\right)\\ =\text{z}\cdot \text{x}\cdot \text{y' + z}\cdot \text{z}\cdot \text{x + z}\cdot \text{z}\cdot \text{y' }\left(\text{by Distributive law}\right)\\ =\text{z}\cdot \text{x}\cdot \text{y' + z}\cdot \text{x + z}\cdot \text{y' }\left(\text{by Idempotent law}\right)\\ =\text{z}\cdot \text{y' }\left(\text{x + 1}\right)\text{+ z}\cdot \text{x }\left(\text{by Distributive law}\right)\end{array}$

$\begin{array}{c}=\text{z}\cdot \text{y' }\left(\text{1}\right)\text{+ z}\cdot \text{x }\left(\text{by Null law}\right)\\ =\text{z}\cdot \text{y' + z}\cdot \text{x }\left(\text{by Identity law}\right)\\ =\text{z}\cdot \left(\text{y' + x}\right)\left(\text{by Distributive law}\right)\end{array}$

Therefore, the simplified expression is “$z\cdot \left(y\text{'}+x\right)$”.