Simplify the given function F(A,B,C)= ∑(0,1,5).

To simplify the Boolean function \( F(A, B, C) = \sum(0, 1, 5) \), follow these steps:

### Step 1: Express the function in minterm form
The sum of minterms notation \( \sum(0, 1, 5) \) refers to the Boolean function being true for the binary representations of the decimal numbers 0, 1, and 5.

- Decimal 0 = \( A'B'C' \) (000)
- Decimal 1 = \( A'B'C \)  (001)
- Decimal 5 = \( AB'C \)   (101)

Thus, the function in its expanded minterm form is:

\[
F(A, B, C) = A'B'C' + A'B'C + AB'C
\]

### Step 2: Simplify the expression
Look for common terms or factors to simplify the Boolean expression.

- Factor out \( A'B' \) from the first two terms:

\[
F(A, B, C) = A'B'(C' + C) + AB'C
\]

- Apply the complement law \( C' + C = 1 \):

\[
F(A, B, C) = A'B' + AB'C
\]

### Step 3: Check for further simplifications
There are no more common factors or simplifications, so the simplified Boolean function is:

\[
F(A, B, C) = A'B' + AB'C
\]

This is the simplest form of the function.

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Would you like more details or have any questions? Here are 5 related questions to deepen your understanding:

1. How would the function change if the minterms were different?
2. What is the significance of using the complement law in Boolean algebra?
3. Can you derive the Boolean function from the truth table for \( F(A, B, C) \)?
4. How can you implement the simplified function using logic gates?
5. How does Karnaugh mapping help in simplifying Boolean expressions?

**Tip:** Factorization is a powerful technique to simplify Boolean functions by grouping terms and applying Boolean laws like complement or identity.

Learn how to simplify the Boolean function F(A,B,C)= ∑(0,1,5) using Boolean algebra techniques. This problem includes the use of minterms and the complement law, suitable for high school students in Grades 9-12.

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