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Expert-verified Found in: Page 818 ### Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095 # Find the duals of these Boolean expressions.$$\begin{array}{l}{\bf{a) x + y}}\\{\bf{b) \bar x\bar y}}\\{\bf{c) xyz + \bar x\bar y\bar z}}\\{\bf{d) x\bar z + x \times 0 + \bar x \times 1}}\end{array}$$

a) The dual of the given expression $$x + y$$ is $$xy$$

b) The dual of the given expression $$\bar x\bar y$$ is $$\bar x + \bar y$$

c) The dual of the given expression$$xyz + \bar x\bar y\bar z$$ is $$\left( {x + y + z} \right)\left( {\bar x + \bar y + \bar z} \right)$$

d) The dual of the given expression$$x\bar z + x \times 0 + \bar x \times 1$$ is $$\left( {x + \bar z} \right)\left( {x + 1} \right)\left( {\bar x + 0} \right)$$

See the step by step solution

## Step 1: Definition

The complement of an element: $$\bar 0 = 1$$ and $$\bar 1 = 0$$

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

The Boolean product or $$A{\bf{ }}N{\bf{ }}D$$ is $$1$$ if both terms are $$1$$ .

The dual of a Boolean expression interchanges the Boolean sum and the Boolean product, and interchanges $$0$$ and $$1$$ .

## Step 2:(a) Using the dual of a Boolean expression

$$x + y$$

The dual of $$x + y$$ replaces the Boolean sum $$+$$ by the Boolean product.

Therefore, the dual $$= x \times y = xy$$

## Step 3:(b)Using the dual of a Boolean expression

$$\bar x\bar y$$

Note: $$\bar x\bar y$$ implies the Boolean product $$\bar x \times \bar y$$

The dual of $$\bar x \times \bar y$$ replaces the Boolean product $$\times$$ by the Boolean sum $$+$$

Therefore, the dual $$= \bar x + \bar y$$

## Step 4:(c)Using the dual of a Boolean expression

$$xyz + \bar x\bar y\bar z$$

Note: $$xyz + \bar x\bar y\bar z$$ implies $$\left( {x \times y \times z} \right) + \left( {\bar x \times \bar y \times \bar z} \right)$$

The dual of $$x \times y \times z + \bar x \times \bar y \times \bar z$$ replaces the Boolean product $$\times$$ by the Boolean sum $$+$$ and replaces the Boolean sum $$+$$ by the Boolean product.

Therefore, the dual $$= \left( {x + y + z} \right) \times \left( {\bar x + \bar y + \bar z} \right) = \left( {x + y + z} \right)\left( {\bar x + \bar y + \bar z} \right)$$

## Step 5:(d)Using the dual of a Boolean expression

$$x\bar z + x \times 0 + \bar x \times 1$$

Note: $$x\bar z + x \times 0 + \bar x \times 1$$ implies $$\left( {x \times \bar z} \right) + \left( {x \times 0} \right) + \left( {\bar x \times 1} \right)$$

The dual of $$\left( {x \times \bar z} \right) + \left( {x \times 0} \right) + \left( {\bar x \times 1} \right)$$ replaces the Boolean product $$\times$$by the Boolean sum $$+$$, replaces the Boolean sum $$+$$ by the Boolean product ; replaces $$0$$ by $$1$$ and replaces $$1$$ by $$0$$ .

Therefore, the dual $$= \left( {x + \bar z} \right) \times \left( {x + 1} \right) \times \left( {\bar x + 0} \right) = \left( {x + \bar z} \right)\left( {x + 1} \right)\left( {\bar x + 0} \right)$$ ### Want to see more solutions like these? 