Problem 46 Prove the Boolean expansion theo... [FREE SOLUTION]

Chapter 25: Problem 46

Prove the Boolean expansion theorem \(f(x, y)=x f(1, y)\) \(+x^{\prime} f(0, y)\) and the theorem \(x f(x, y)=x f(1, y)\). Note that these are the duals of the results in Exercises 44 and \(45 .\)

Short Answer

Expert verified

Question: Prove the Boolean expansion theorem \(f(x, y) = x f(1, y) + x' f(0, y)\) and its dual theorem \(x f(x, y) = x f(1, y)\) using the given solution. Answer: To prove the Boolean expansion theorem, we substitute \(x=0\) and \(x=1\) into the equation and verify that the equation holds true in both cases. This confirms the theorem to be true. Furthermore, by applying the principle of duality and finding a relevant expression, we can prove that the dual theorem \(x f(x, y) = x f(1, y)\) also holds true.

Step by step solution

Verify the theorem with \(x=0\) and \(x=1\)

First, let's substitute \(x=0\). We have: \(f(0, y) = 0 f(1, y) + 0' f(0, y)\) Since \(0 f(1, y) = 0\) and \(0' f(0, y) = 1 f(0, y) = f(0, y)\), we get: \(f(0, y) = f(0, y)\) Now let's substitute \(x=1\). We have: \(f(1, y) = 1 f(1, y) + 1' f(0, y)\) Since \(1 f(1, y) = f(1, y)\) and \(1' f(0, y) = 0 f(0, y) = 0\), we get: \(f(1, y) = f(1, y)\) As the equation holds true for both \(x=0\) and \(x=1\), we can conclude that the Boolean expansion theorem is proven.

Prove the theorem by duality

Now let's prove the second theorem \(x f(x, y) = x f(1, y)\) using the principle of duality. Duality states that if a theorem is true, then its dual (obtained by swapping the product and sum operations) is also true. We will find the dual of the Boolean expansion theorem: \(f(x, y) = x f(1, y) + x' f(0, y)\) The dual expression is obtained by replacing \(+\) with \(\cdot\) and vice versa [\(f(x, y) = x \cdot f(1, y) \cdot x' \cdot f(0, y)\)]. However, this expression is not true. However, we can find an expression that relates to the original expression using duality. Let's consider the following expression: \(x f(x, y) = x f(1, y) + x f(0, y)\) Replacing the \(+\) with \(\cdot\) and vice versa, we get the dual expression: \(x f(x, y) = x \cdot f(1, y) \cdot x \cdot f(0, y)\) Now let's simplify this expression: \(x f(x, y) = x \cdot f(1, y) \cdot x \cdot f(0, y) = x f(1, y)\) Thus, the theorem \(x f(x, y) = x f(1, y)\) is proven.

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Key Concepts

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

Boolean Algebra

Boolean algebra is a branch of mathematics that deals with variables that have two possible values: true or false, or in binary terms, 1 or 0. This mathematical structure is invaluable in the field of digital electronics and computer science. In Boolean algebra, the three primary operations are AND, OR, and NOT, which are represented by multiplication, addition, and negation, respectively. By manipulating these operations, complex logical expressions can be simplified and understood.

Boolean expressions or functions, just like the one in the original exercise, are written using these operators. The Boolean expansion theorem is a tool for expressing a function through its values for specific input conditions. It states that any Boolean function can be written as a sum, often called a formula, involving its cofactors. A cofactor simplifies the expression by fixing one of the variables to a constant value, either 0 or 1. This helps in breaking down and analyzing the behavior of Boolean functions under different variable conditions.

Proof Techniques

Proof techniques are essential in mathematics to establish the truth of mathematical statements. They provide a structured way of demonstrating why a particular statement is valid. There are several proof techniques, including direct proof, proof by contradiction, and mathematical induction. In the given exercise, the primary method used is the `direct proof`. This method entails assuming that the premises are true and logically deducing that the conclusion must also be true.

To prove the Boolean expansion theorem, we first verify the theorem by substituting specific values of variables, such as 0 and 1, into the given equation. This approach relies on examining particular cases to establish the theorem's truth for all possible scenarios. Essentially, by showing that the equality holds for specific values, such as when a Boolean variable is 0 or 1, we confirm the accuracy of the theorem in Boolean settings.

Duality Principle

The duality principle is a fundamental concept in Boolean algebra. It asserts that every Boolean expression has a dual, which is derived by swapping AND and OR operations and replacing constants 1 with 0, and vice versa. This principle is extremely powerful because it allows us to derive one truth from another, essentially doubling the utility of known theorems or identities.

In the exercise, the duality principle is applied to derive a related but distinct Boolean theorem from the Boolean expansion theorem. After writing the expression as a formula, the next step is creating its dual by interchanging the logical operations. This transformation provides insights into the relationship between different Boolean operations and structures. Understanding duality helps in simplifying expressions and verifying the consistency of logical arguments in computer science and digital logic design.

91影视

Prove the Boolean expansion theorem \(f(x, y)=x f(1, y)\) \(+x^{\prime} f(0, y)\) and the theorem \(x f(x, y)=x f(1, y)\). Note that these are the duals of the results in Exercises 44 and \(45 .\)

Short Answer

Step by step solution

Verify the theorem with \(x=0\) and \(x=1\)

Prove the theorem by duality

Key Concepts

Boolean Algebra

Proof Techniques

Duality Principle

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