Simplify the function y(A,B,C) = Σ (0,1,3,6,7) using Boolean algebra

We are tasked with simplifying the Boolean function \( y(A, B, C) = \Sigma(0, 1, 3, 6, 7) \), which means that the function is expressed in terms of a sum of minterms for the given indices.

### Step 1: Write the Minterms in Binary

First, represent the numbers (0, 1, 3, 6, 7) in binary based on the three variables \( A \), \( B \), and \( C \).

\[
\begin{aligned}
0 & : (A'B'C') \\
1 & : (A'B'C) \\
3 & : (A'BC) \\
6 & : (AB'C) \\
7 & : (ABC)
\end{aligned}
\]

Thus, the function can be written as:

\[
y(A, B, C) = A'B'C' + A'B'C + A'BC + AB'C + ABC
\]

### Step 2: Simplify Using Boolean Algebra

1. **Group common terms**:
   We can group minterms where possible:
   \[
   y = A'B'C' + A'B'C + A'BC + AB'C + ABC
   \]
   
   Notice that the first three terms have \( A' \) in common:
   \[
   y = A'(B'C' + B'C + BC) + AB'C + ABC
   \]

2. **Simplify the grouped terms**:
   We can simplify the expression inside the parentheses:
   \[
   B'C' + B'C + BC = B'(C' + C) + BC
   \]
   Since \( C' + C = 1 \) (a basic Boolean identity), we get:
   \[
   B' + BC
   \]
   Therefore, the simplified expression becomes:
   \[
   y = A'(B' + BC) + AB'C + ABC
   \]

3. **Further simplifications**:
   Now distribute the terms:
   \[
   y = A'B' + A'BC + AB'C + ABC
   \]
   
   Group the terms \( A'BC \) and \( ABC \):
   \[
   y = A'B' + AB'C + (A'BC + ABC)
   \]
   
   Factor out \( C \) from the last group:
   \[
   y = A'B' + AB'C + C(A'B + AB)
   \]
   
   Simplify \( A'B + AB \) (this is an XOR pattern, which simplifies to \( B \)):
   \[
   y = A'B' + AB'C + BC
   \]

### Final Simplified Expression

The simplified Boolean expression is:
\[
y(A, B, C) = A'B' + AB'C + BC
\]

This is the minimal form of the function.

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Would you like further details or any clarifications?

Here are 5 related questions for practice:
1. How would you represent the function if the given minterms were \( \Sigma(1, 2, 5, 6) \)?
2. What is the simplified Boolean expression for \( y(A,B,C) = \Sigma(2,3,6,7) \)?
3. How can you simplify the Boolean expression \( AB + A'B' + AB' \) using Boolean algebra?
4. What are the key Boolean identities used to simplify logic expressions?
5. How do Karnaugh maps (K-maps) assist in simplifying Boolean functions?

**Tip**: Always try to group common factors when simplifying Boolean expressions to reduce the number of terms.

Learn how to simplify the Boolean function y(A,B,C) = Σ(0,1,3,6,7) using Boolean algebra. This solution covers key concepts like minterms, distributive law, and XOR pattern simplification. Ideal for high school students in Grades 10-12.

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