https://r2.erweima.ai/mathbot/image/1726015872628mecmztii.jpg

The given function in the image is:

\[
F = \overline{A}BC + A\overline{B}\overline{C} + AB\overline{C}
\]

We will simplify this Boolean expression using Boolean algebra rules.

### Step 1: Group similar terms
Let's group the second and third terms, as they both contain \( \overline{C} \):

\[
F = \overline{A}BC + (\overline{B}\overline{C}A + AB\overline{C})
\]

### Step 2: Apply Distributive Property
We can factor out \( A\overline{C} \) from the grouped terms:

\[
F = \overline{A}BC + A\overline{C}(\overline{B} + B)
\]

Since \( \overline{B} + B = 1 \) (Complement Law):

\[
F = \overline{A}BC + A\overline{C}(1)
\]

This simplifies to:

\[
F = \overline{A}BC + A\overline{C}
\]

### Step 3: Final Simplified Expression
The expression \( \overline{A}BC + A\overline{C} \) is the simplified Boolean function.

Thus, the simplified result is:

\[
F = \overline{A}BC + A\overline{C}
\]

---

Would you like more details or have any questions about the steps?

### Related questions:
1. What is the Complement Law in Boolean algebra?
2. How can the Distributive Property be used in Boolean algebra?
3. Can you simplify a Boolean function further using Karnaugh maps?
4. How would you simplify Boolean expressions using truth tables?
5. What are the most common Boolean identities used in simplification?

**Tip:** When simplifying Boolean expressions, always look for terms that can be factored or grouped for easier simplification.

Simplifique a função F = A̅BC + AB̅C̅ + ABC̅ usando postulados e teoremas conhecidos.

Learn how to simplify the Boolean function F = A̅BC + AB̅C̅ + ABC̅ using fundamental Boolean postulates and theorems, including the Complement Law and Distributive Property. Suitable for students studying digital logic design.

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