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Found in: Page 844

### Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

# Show that each of these identities holds.$${\bf{a)}}$$ $${\bf{x}} \odot {\bf{x = 1}}$$$${\bf{b)}}$$ $${\bf{x}} \odot {\bf{\bar x = 0}}$$$${\bf{c)}}$$ $${\bf{x}} \odot {\bf{y = y}} \odot {\bf{x}}$$

$${\bf{a)}}$$ The given equation $${\bf{x}} \odot {\bf{x = 1}}$$ holds.

$${\bf{b)}}$$ The given equation $${\bf{x}} \odot {\bf{\bar x = 0}}$$ holds.

$${\bf{c)}}$$ The given equation $${\bf{x}} \odot {\bf{y = y}} \odot {\bf{x}}$$ holds.

See the step by step solution

## Step 1: Definition

The complement of an element: $${\bf{\bar 0 = 1}}$$ and $${\bf{\bar 1 = 0}}$$.

The Boolean sum $${\bf{ + }}$$ or $${\bf{OR}}$$ is $${\bf{1}}$$ if either term is $${\bf{1}}$$.

The Boolean product $$\bullet$$ or $${\bf{AND}}$$ is $${\bf{1}}$$ if both terms are $${\bf{1}}$$.

The $${\bf{NOR}}$$ operator $$\downarrow$$ is $${\bf{1}}$$ if both terms are $${\bf{0}}$$.

The $${\bf{XOR}}$$ operator $$\oplus$$ is $${\bf{1}}$$ if one of the terms is $${\bf{1}}$$ (but not both).

The $${\bf{NAND}}$$ operator $$\mid$$ is $${\bf{1}}$$ if either term is $${\bf{0}}$$.

The Boolean operator $$\odot$$ is $${\bf{1}}$$ if both terms have the same value.

## Step 2: Using the Boolean operator

It is given that $${\bf{x}} \odot {\bf{x = 1}}$$.

Now create a table for all possible values of $${\bf{x, y}}$$ and $${\bf{z}}$$.

The Boolean operator $$\odot$$ is $${\bf{1}}$$ if both terms have the same value.

$$\begin{array}{*{20}{r}}{\bf{x}}&{{\bf{ x}}}&{{\bf{ x}} \odot {\bf{x}}}&{\bf{1}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}\\{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{1}}\end{array}$$

Then note that the last two columns of the table are identical.

Hence, it gives $${\bf{x}} \odot {\bf{x = 1}}$$.

## Step 3: Using the Boolean operator

(b)

It is given that $${\bf{x}} \odot {\bf{\bar x = 0}}$$.

Now create a table for all possible values of $${\bf{x, y}}$$ and $${\bf{z}}$$.

The Boolean operator $$\odot$$ is $${\bf{1}}$$ if both terms have the same value.

$$\begin{array}{*{20}{r}}{\bf{x}}&{{\bf{ \bar x}}}&{{\bf{ x}} \odot {\bf{\bar x}}}&{\bf{0}}\\{\bf{0}}&{\bf{1}}&{\bf{0}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\end{array}$$

Then note that the last two columns of the table are identical.

Hence, it gives $${\bf{x}} \odot {\bf{\bar x = 0}}$$.

## Step 4: Using the Boolean operator

It is given that $${\bf{x}} \odot {\bf{y = y}} \odot {\bf{x}}$$.

Now create a table for all possible values of $${\bf{x, y}}$$ and $${\bf{z}}$$.

The Boolean operator $$\odot$$ is $${\bf{1}}$$ if both terms have the same value.

$$\begin{array}{*{20}{r}}{\bf{x}}&{{\bf{ y}}}&{{\bf{ x}} \odot {\bf{y}}}&{{\bf{ y}} \odot {\bf{x}}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}\\{\bf{0}}&{\bf{1}}&{\bf{0}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{1}}\end{array}$$

Then note that the last two columns of the table are identical.

Hence, it gives $${\bf{x}} \odot {\bf{y = y}} \odot {\bf{x}}$$.