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Chapter 12: Problem 38

Find the dual of each boolean property. $$(x+y) z=x z+y z$$

Short Answer

Expert verified
The dual of the given Boolean property \((x+y)z = xz+yz\) is: \((x \cdot y) + z = (x + z) \cdot (y + z)\).

Step by step solution

01

The original Boolean property is: \((x+y)z = xz+yz\). #Step 2: Apply the duality rules to the property#

Apply the duality rules as mentioned in the analysis, we will get the dual property as: 1. Replace \((+) \rightarrow(\cdot)\) and \((\cdot) \rightarrow (+)\) 2. Since no 0's and 1's are present in the given property, we don't need to replace them. #Step 3: Substitute and simplify#
02

By applying the duality rules, we get the dual of the given property: \((x \cdot y) + z = (x + z) \cdot (y + z)\) #Step 4: Present the result#

The dual of the given Boolean property \((x+y)z = xz+yz\) is: \((x \cdot y) + z = (x + z) \cdot (y + z)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Boolean Properties
Boolean properties are foundational rules in Boolean algebra that describe how variables interact using typical operations like AND, OR, and NOT. Understanding these properties is crucial as they enable us to simplify Boolean expressions while designing digital circuits.

Some key Boolean properties include:
  • Identity Law: Describes the impact of identity elements: 0 and 1. For example, any Boolean variable added to 0 remains unchanged, i.e., \( x + 0 = x \), and any Boolean variable ANDed with 1 is unchanged, i.e., \( x \cdot 1 = x \).
  • Domination Law: Involves dominant elements where \( x + 1 = 1 \) and \( x \cdot 0 = 0 \).
  • Idempotent Law: This law states that a variable ORed or ANDed with itself does not change its value: \( x + x = x \) and \( x \cdot x = x \).
  • Complement Law: Points out that a variable and its complement sum to 1 or nullify each other: \( x \cdot \overline{x} = 0 \) and \( x + \overline{x} = 1 \).
These properties form the bedrock of simplifying expressions in Boolean algebra.
Duality Principle
The duality principle is a powerful concept in Boolean algebra that highlights symmetrical dual behavior in Boolean expressions. The principle states that every Boolean expression remains valid if we swap AND operations with OR operations and replace 0's with 1's, and vice versa.

Applying the duality principle helps in deriving the dual expression of a given Boolean property. For example, consider the expression \((x+y)z = xz + yz\). Applying duality involves:
  • Swapping the + (OR) with \(\cdot\) (AND), and vice versa.
  • Since our expression doesn't contain 0's or 1's, those don't need swapping in this case.
After swapping, we get the dual expression \((x \cdot y) + z = (x + z) \cdot (y + z)\). This dual reflects a complementary perspective of the initial expression and maintains logical consistency across dual operations.
Boolean Expressions
Boolean expressions are the language of Boolean algebra. They are composed of variables and logical operators and are fundamental in designing logic circuits and solving logical equations.

These expressions use:
  • Variables: Can hold binary values (0 or 1) representing true or false states.
  • Operators: Like AND (\(\cdot\)), OR (+), and NOT (\(\overline{x}\)), which define the relationships between variables.
For instance, the expression \((x + y)z\) combines variables and operators to yield a result based on the input values of \(x\) and \(y\). Simplifying Boolean expressions enables clearer circuit designs. Tools like Boolean properties and the duality principle streamline this simplification process, ensuring efficient logic described in a minimal form.

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Most popular questions from this chapter

Using a Karnaugh map, simplify each boolean expression. $$x y z+x y z^{\prime}+x^{\prime} y^{\prime} z^{\prime}+x^{\prime} y^{\prime} z$$

Use the following definition of the binary operator \(\mathrm{XOR}\) , denoted by \(\oplus,\) for Exercises \(69-81 .\) $$ x \oplus y=\left\\{\begin{array}{ll}{1} & {\text { if exactly one of the bits } x \text { and } y \text { is } 1} \\ {0} & {\text { otherwise }}\end{array}\right. $$ Is \(\\{\oplus\\}\) functionally complete?

Construct a logic table for each boolean function defined by each boolean expression. $$\left(x+y^{\prime}+z\right)\left(x y^{\prime} z\right)$$

Construct a logic table for each boolean expression. $$(x \downarrow x) \downarrow(y \downarrow y)$$

Represent each sum of minterms in a Karnaugh map. $$w x y z+w x y^{\prime} z+w^{\prime} x y z+w^{\prime} x y^{\prime} z$$
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