Chapter 3, Problem 12E

a.

Explanation of Solution

Given:

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

Truth table for the above expression:

x	y	z	x’	y’	x + y	x + y’	x’ + z	LHS	RHS
0	0	0	1	1	0	1	1	0	0
0	0	1	1	1	0	1	1	0	0
0	1	0	1	0	1	0	1	0	0
0	1	1	1	0	1	0	1	0	0
1	0	0	0	1	1	1	0	0	0
1	0	1	0	1	1	1	1	1	1
1	1	0	0	0	1	1	0	0	0
1	1	1	0	0	1	1	1	1	1

In the above table, the “LHS” and “RHS” columns are equal. Therefore the given expression is valid.

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:

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

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:

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

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

Given:

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

The given expression is validated by the Boolean identities is as follows:

By using distributive law the given expression is modified

$\text{= x}\cdot \text{(x + y')}\cdot \text{(x' + z) + y}\cdot \text{(x + y')}\cdot (\text{x' + z)}$

By using distributive law the above expression is modified

$\text{= x}\cdot \text{(x}\cdot \text{x' + x}\cdot \text{z + x'}\cdot \text{y' + y'}\cdot \text{z) + y}\cdot \text{(x}\cdot \text{x' + x}\cdot \text{z + x'}\cdot \text{y' + y'}\cdot \text{z)}$

By using inverse law the above expression is modified

$\text{= x}\cdot \text{(}0\text{ + x}\cdot \text{z + x'}\cdot \text{y' + y'}\cdot \text{z) + y}\cdot \text{(0 + x}\cdot \text{z + x'}\cdot \text{y' + y'}\cdot \text{z)}$

By using identity law the above expression is modified

$\text{= x}\cdot \text{(x}\cdot \text{z + x'}\cdot \text{y' + y'}\cdot \text{z) + y}\cdot \text{(x}\cdot \text{z + x'}\cdot \text{y' + y'}\cdot \text{z)}$

By using distributive law the above expression is modified

$\text{= x}\cdot \text{x}\cdot \text{z + x}\cdot \text{x'}\cdot \text{y' + x}\cdot \text{y'}\cdot \text{z + y}\cdot \text{x}\cdot \text{z +y}\cdot \text{x'}\cdot \text{y' + y}\cdot \text{y'}\cdot \text{z}$

By using inverse law the above expression is modified

$\text{= x}\cdot \text{x}\cdot \text{z + }(0)\cdot \text{y' + x}\cdot \text{y'}\cdot \text{z + y}\cdot \text{x}\cdot \text{z +}(0)\cdot \text{x + }(0)\cdot \text{z}$

By using null law the above expression is modified

$\text{= x}\cdot \text{x}\cdot \text{z + x}\cdot \text{y'}\cdot \text{z + x}\cdot \text{y}\cdot \text{z}$

By using idempotent law the given expression is modified

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

By using distributive law the above expression is modified

$\text{= x}\cdot \text{z}\cdot \text{(1 + y' + y)}$

By using null law the above expression is modified

$\text{= x}\cdot \text{z}\cdot \text{(1)}$

By using identity law the above expression is modified

$\text{= x}\cdot \text{z}$

Therefore, the given expression is valid.