Problem 30 - Discrete Mathematics with Applications

Question

Find the DNFs of the boolean functions $$\begin{array}{|ll||c|} \hline \boldsymbol{x} & \boldsymbol{y} & \boldsymbol{f}(\boldsymbol{x}, \boldsymbol{y}) \\ \hline 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \hline \end{array}$$

Step-by-Step Solution

Answer

The Disjunctive Normal Form of the given boolean function is: \[f(x, y) = \overline{x} \cdot \overline{y} + \overline{x} \cdot y + x \cdot \overline{y}\]

Step 1: Identify the true outputs

First, let's look at the given truth table and identify which rows have a true output (1) for the function \(f(x, y)\). $$\begin{array}{|ll||c|} \hline \boldsymbol{x} & \boldsymbol{y} & \boldsymbol{f}(\boldsymbol{x}, \boldsymbol{y}) \\\ \hline 0 & 0 & 1 \\\ 0 & 1 & 1 \\\ 1 & 0 & 1 \\\ 1 & 1 & 0 \\\ \hline \end{array}$$ In the given table, we can see that the rows 1, 2, and 3 have an output value of 1.

Step 2: Form conjunctions for each true output

Now that we have found the true outputs, we need to form conjunctions (AND) of literals for each row. For the first row, x=0 and y=0. So, the conjunction will be \(\overline{x} \cdot \overline{y}\). For the second row, x=0 and y=1. So, the conjunction will be \(\overline{x} \cdot y\). For the third row, x=1 and y=0. So, the conjunction will be \(x \cdot \overline{y}\).

Step 3: Combine the conjunctions using disjunction

Finally, we need to combine the conjunctions using disjunction (OR) to form the DNF of the function. So, the DNF of the given function is: \(\overline{x} \cdot \overline{y} + \overline{x} \cdot y + x \cdot \overline{y}\). Thus, the Disjunctive Normal Form of the given boolean function is: \[f(x, y) = \overline{x} \cdot \overline{y} + \overline{x} \cdot y + x \cdot \overline{y}\]

Key Concepts

Truth Table

A truth table is a fundamental tool used in logic to represent all possible outcomes of a Boolean function. Each row of the table corresponds to a combination of inputs, and the corresponding output shows the result of the function for that combination. For example, in our exercise, the Boolean function \(f(x, y)\) is depicted using a truth table. It considers only two inputs \(x\) and \(y\), each of which can be either 0 or 1.

The table clearly displays how the function responds to each combination of input values, making it easier to understand the behavior of the logical expression. Truth tables are highly practical for determining the outputs for all variable states. This helps in identifying which rows in the table correspond to a true (1) or false (0) outcome. For our exercise, the truth table helped highlight that the outputs are 1 when the input pairs are \( (0,0), (0,1) \), and \( (1,0) \).

Overall, using truth tables serves as a simple and efficient way to dissect and analyze Boolean functions, providing a clear and organized means to visualize potential combinations and results.

Disjunctive Normal Form (DNF)

The Disjunctive Normal Form (DNF) is an important concept in Boolean algebra where a Boolean function is expressed as an OR (disjunction) of AND (conjunction) clauses. In essence, DNF is a standard way of organizing expressions to simplify logical analysis and computation.

To form a DNF, one must first identify the rows in a truth table where the output is 1. Once these are identified, you create conjunctions for each of these rows using the variables \(x\) and \(y\) and their complements, depending on whether the value in the row is 0 (complement) or 1 (normal). Each conjunction corresponds to a circumstance where the function is true.

For our practical example, the conjunctions based on the true outputs are:

• \(\overline{x} \cdot \overline{y}\) for the inputs \((0,0)\)
• \(\overline{x} \cdot y\) for the inputs \((0,1)\)
• \(x \cdot \overline{y}\) for the inputs \((1,0)\)

These conjunctions are then combined using disjunction or an OR operation to form the entire DNF expression: \( \overline{x} \cdot \overline{y} + \overline{x} \cdot y + x \cdot \overline{y} \). This transformation highlights the instances where the function yields true, providing a systematic approach to Boolean expression simplification.

Conjunction and Disjunction Operations

Conjunction and disjunction are key operations in Boolean algebra that are used to combine logical values and expressions.

A conjunction is an operation commonly referred to as "AND," which results in true if and only if all operand expressions are true. In Boolean functions, conjunctions are represented using a dot "\(\cdot\)" notation or simply by juxtaposition, so \(x \cdot y\) means that both \(x\) and \(y\) must be true (1) for the whole expression to be true.

Conversely, disjunction, denoted by "+", is the "OR" operation. This results in true if any of the operand expressions is true. Thus, in Boolean terms, \(x + y\) is true if either \(x\) or \(y\) or both are true.

In constructing DNFs, conjunctions come into play for creating terms that describe scenarios where the function is true; meanwhile, disjunction is used to combine those terms. Understanding these operations is critical for manipulating and simplifying Boolean expressions.

In our exercise, these operations were employed to consolidate the scenarios from the truth table where the function outputs a true, thereby effectively harnessing these fundamental elements for structuring logical solutions.