Question

What is the simplified form of the Boolean expression (A + B)(A + C)?

• A. A + B + C
• B. AB + C
• C. AC + BC
• D. A + BC

Hint:

Memorizing the second distributive law, $(X+Y)(X+Z) = X+YZ$, is very useful as it provides a direct shortcut for simplifying expressions of this common form.

Answer 1

The Correct Option is D

We can simplify the given Boolean expression using the distributive law of Boolean algebra, which states that $X(Y+Z) = XY + XZ$ and also $(X+Y)(X+Z) = X + YZ$.
Method 1: Using the second distributive law directly.
Let $X = A$, $Y = B$, and $Z = C$.
The expression is $(A+B)(A+C)$.
According to the law $(X+Y)(X+Z) = X + YZ$, the simplified form is $A + BC$.
Method 2: Expanding using the first distributive law.
$(A+B)(A+C) = A(A+C) + B(A+C)$
$= A \cdot A + A \cdot C + B \cdot A + B \cdot C$
Using the idempotent law ($A \cdot A = A$):
$= A + AC + AB + BC$
Factor out A from the first three terms:
$= A(1 + C + B) + BC$
Using the identity law ($1 + X = 1$):
$= A(1) + BC$
$= A + BC$.
Both methods yield the same simplified form.