The given Boolean function is:

$f(x, y, z) = (x + y) \overline{z}$

To find the Disjunctive Normal Form (DNF) and Conjunctive Normal Form (CNF), let's break it down step by step.

1. Truth Table Construction

We need to construct a truth table for the function to determine both forms. Let's first determine the values of $(x + y)$ and then apply the complement of $z$ to get $(x + y) \overline{z}$.

2. Disjunctive Normal Form (DNF)

DNF is obtained by taking the OR (disjunction) of all the minterms where the function value is 1.

From the truth table, $f(x, y, z) = 1$ for the following rows:

• Row 3: $x = 0, y = 1, z = 0$
• Row 5: $x = 1, y = 0, z = 0$
• Row 7: $x = 1, y = 1, z = 0$

Thus, the DNF is:

$DNF = (\overline{x} \cdot y \cdot \overline{z}) + (x \cdot \overline{y} \cdot \overline{z}) + (x \cdot y \cdot \overline{z})$

3. Conjunctive Normal Form (CNF)

CNF is obtained by taking the AND (conjunction) of all the maxterms where the function value is 0.

From the truth table, $f(x, y, z) = 0$ for the following rows:

• Row 1: $x = 0, y = 0, z = 0$
• Row 2: $x = 0, y = 0, z = 1$
• Row 4: $x = 0, y = 1, z = 1$
• Row 6: $x = 1, y = 0, z = 1$
• Row 8: $x = 1, y = 1, z = 1$

Thus, the CNF is:

$CNF = (x + y + \overline{z}) \cdot (x + y + z) \cdot (x + \overline{y} + z) \cdot (\overline{x} + y + z) \cdot (\overline{x} + \overline{y} + z)$

Final Answer:

• DNF: $(\overline{x} \cdot y \cdot \overline{z}) + (x \cdot \overline{y} \cdot \overline{z}) + (x \cdot y \cdot \overline{z})$
• CNF: $(x + y + \overline{z}) \cdot (x + y + z) \cdot (x + \overline{y} + z) \cdot (\overline{x} + y + z) \cdot (\overline{x} + \overline{y} + z)$

Would you like more details on these forms or have any specific questions?

Related Questions:

1. How do you simplify Boolean expressions to their minimal forms?
2. What is the significance of DNF and CNF in Boolean algebra?
3. How can Karnaugh maps be used to simplify Boolean functions?
4. What is the difference between minterms and maxterms?
5. Can Boolean functions be represented graphically using logic gates?

Tip:

When dealing with Boolean expressions, visualizing the function with logic gates can often make the problem clearer and help in understanding its behavior.