Boolean algebra simplification: Reduce the expression (A + B)(A + C) to its simplest sum-of-terms form.

Introduction / Context:
Simplifying Boolean expressions is central to minimizing logic circuits, reducing gate count, power, and delay. The expression (A + B)(A + C) appears frequently in control logic and can be simplified by applying distributive and absorption laws.

Given Data / Assumptions:

• Variables A, B, C are Boolean (0 or 1).
• Operators: + is OR, juxtaposition (e.g., AB) is AND.
• Target: a minimal, logically equivalent expression.

Concept / Approach:
Use distributive and absorption properties: X + XY = X and X(X + Y) = X. By distributing A across the product and removing redundant terms, we reach a compact form with fewer gates when implemented in hardware.

Step-by-Step Solution:
Start: (A + B)(A + C).Distribute: = A(A + C) + B(A + C).Simplify first term: A(A + C) = A.Second term: B(A + C) = AB + BC.Combine: A + AB + BC.Apply absorption: A + AB = A.Result: A + BC.

Verification / Alternative check:
Construct a Karnaugh map for A, B, C and group minterms; the minimal cover yields A + BC. Alternatively, truth-table evaluation confirms equivalence across all 8 input combinations.

Why Other Options Are Wrong:
AB + AC is the result of partial distribution without final simplification. A + B + C overstates the function, turning don’t-care situations into always-on conditions. AC + B does not match the original for all inputs.

Common Pitfalls:
Stopping after distribution; forgetting the absorption rule; mixing operator precedence. Always look for X + XY = X patterns to reduce expressions.

Final Answer:
A + BC