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The problem involves simplifying a Boolean expression using known postulates and theorems.

Given the Boolean function:
\[
F = \overline{A} \cdot B \cdot \overline{C} + \overline{A} \cdot \overline{B} \cdot C + A \cdot \overline{B} \cdot \overline{C}
\]

### Simplification Steps:

1. **Identify common terms**:
   Notice that the expression has three terms, and each term contains two complemented variables.

2. **Group and simplify**:
   Let's look for possible simplifications using Boolean algebra rules:
   \[
   F = \overline{A} \cdot B \cdot \overline{C} + \overline{A} \cdot \overline{B} \cdot C + A \cdot \overline{B} \cdot \overline{C}
   \]

   - The first and second terms have \(\overline{A}\) in common, but they cannot be simplified directly.
   - The second and third terms can be considered for combining \(\overline{B} \cdot C\) and \(\overline{B} \cdot \overline{C}\) with \(A\).

   Let's analyze the terms by applying the Consensus theorem:
   \[
   F = \overline{A} \cdot (B \cdot \overline{C} + \overline{B} \cdot C) + A \cdot \overline{B} \cdot \overline{C}
   \]

3. **Apply the Consensus Theorem**:
   The Consensus Theorem states:
   \[
   AB + \overline{A}C + BC = AB + \overline{A}C
   \]
   But since our terms do not directly fit into the Consensus Theorem, we must simplify the terms individually.

4. **Combine terms directly**:
   Observe that \(\overline{A} \cdot B \cdot \overline{C} + \overline{A} \cdot \overline{B} \cdot C\) is simplified as:
   \[
   \overline{A} \cdot (B \cdot \overline{C} + \overline{B} \cdot C) + A \cdot \overline{B} \cdot \overline{C}
   \]
   Simplifying the expression inside the parentheses:
   \[
   B \cdot \overline{C} + \overline{B} \cdot C
   \]
   Recognizes it as an XOR operation:
   \[
   \overline{A} \cdot (B \oplus C) + A \cdot \overline{B} \cdot \overline{C}
   \]

5. **Final Expression**:
   No further simplifications are apparent using standard Boolean algebra, so the simplified function \(F\) is:
   \[
   F = \overline{A} \cdot (B \oplus C) + A \cdot \overline{B} \cdot \overline{C}
   \]

This is the simplified Boolean expression.

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Would you like further details on the steps, or do you have any questions? Here are some related questions:

1. How does the Consensus theorem help in simplifying Boolean expressions?
2. What is the XOR operation in Boolean algebra, and how does it work?
3. Can De Morgan's laws be applied to simplify other Boolean expressions?
4. What other postulates could be used to simplify different types of Boolean functions?
5. How can Karnaugh maps be used to simplify Boolean expressions?

**Tip:** When simplifying Boolean expressions, always look for common factors and use Boolean theorems like the Consensus theorem and De Morgan’s laws.

Simplify the Boolean function F = A'B'C' + A'BC + AB'C using known postulates and theorems.

Learn how to simplify the Boolean expression F = A'B'C' + A'BC + AB'C using Boolean algebra rules, including the Consensus Theorem and XOR operation. Aimed at high school students in Grades 10-12.

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