Chapter 3, Problem 26E

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:

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

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

Algebraic expression:

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

The complement of $\text{F'}=\left(\text{x + y'}\right)\cdot \left(\text{x' + y' + z}\right)$

The sum of product for the above expression is as follows:

$\begin{array}{c}\text{F'}=\left(\text{x + y'}\right)\cdot \left(\text{x' + y' + z}\right)\\ =\text{x}\cdot \text{x' + x}\cdot \text{y' + x}\cdot \text{z + x'}\cdot \text{y' + y'}\cdot \text{y' + y'}\cdot \text{z }\left(\text{by Distribution law}\right)\\ =\text{0 + x}\cdot \text{y' + x}\cdot \text{z + x'}\cdot \text{y' + y'}\cdot \text{y' + y'}\cdot \text{z }\left(\text{by Inverse law}\right)\\ =\text{x}\cdot \text{y' + x}\cdot \text{z + x'}\cdot \text{y' + y' + y'}\cdot \text{z }\left(\text{by Idempotent law}\right)\end{array}$

Therefore, the sum of product is “$xy\text{'}+xz+x\text{'}y\text{'}+y\text{'}+y\text{'}z$”.

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:

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

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:

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

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

Proof:

$\begin{array}{c}\text{FF' = 0}\\ \text{F = x'}\cdot \text{y + x}\cdot \text{y}\cdot \text{z'}\\ \text{F'}=\left(\text{x + y'}\right)\cdot \left(\text{x' + y' + z}\right)\\ \text{FF' =}\left(\text{x'}\cdot \text{y + x}\cdot \text{y}\cdot \text{z'}\right)\cdot \left(\left(\text{x + y'}\right)\cdot \left(\text{x' + y' + z}\right)\right)\end{array}$

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

$\begin{array}{c}\text{= }\left(0\right)\cdot \left(0\right)\text{ + }\left(0\right)\cdot \text{y}\cdot \text{z + x'}\cdot \text{x'}\cdot \left(0\right)\text{ + x'}\cdot \left(0\right)\text{ + x'}\cdot \text{z}\cdot \left(0\right)\text{ + x}\cdot \text{x}\cdot \text{z'}\cdot \left(0\right)\\ \text{+ x}\cdot \text{x}\cdot \text{y}\cdot \left(0\right)\text{ + }\left(0\right)\cdot \left(0\right)\cdot \text{z' + x}\cdot \text{z'}\cdot \left(0\right)\text{ + x}\cdot \left(0\right)\cdot \left(0\right)\left(\text{by Inverse law}\right)\\ =0\text{ + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 }\left(\text{by Identity law}\right)\\ =0\end{array}$

Here, LHS and RHS are equal.

Hence it is proved.

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:

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

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:

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

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:

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

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:

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

Explanation of Solution

Proof:

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

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

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

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

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

Here, LHS and RHS are equal.

Hence it is proved.