Question

The minimum number of NOR gates required to implement A(A+B)(A+B+C) is equal to

• A. zero
• B. 3
• C. 4
• D. 7

Hint:

Always simplify Boolean expressions completely using absorption laws before calculating gate counts.
If the final simplified expression reduces to a single literal, the gate requirement is always 0.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This question asks for the minimum number of universal NOR gates required to implement a given Boolean expression.
To find the minimum number of gates, we first need to simplify the Boolean expression to its simplest form.

Step 2: Key Formula or Approach:
We will use the laws of Boolean algebra, specifically the Absorption Law:
\[ X(X+Y) = X \]
And the distributive and idempotent properties:
\[ X \cdot X = X \]
\[ X + XY = X(1+Y) = X \]

Step 3: Detailed Explanation:

• Let the Boolean expression be $F$:
\[ F = A(A+B)(A+B+C) \]

• Apply the Absorption Law to the first two terms $A(A+B)$:
\[ A(A+B) = A \cdot A + A \cdot B = A + AB = A(1+B) = A \]

• Now, substitute this back into the original expression:
\[ F = A(A+B+C) \]

• Expand the expression:
\[ F = A \cdot A + A \cdot B + A \cdot C \]
\[ F = A + AB + AC \]

• Apply the absorption property again:
\[ F = A(1 + B + C) \]

• Since $1 + B + C = 1$ in Boolean logic:
\[ F = A \cdot 1 = A \]

• The final simplified expression is simply $F = A$.

• Since the output is directly equal to the input $A$, no logic gates are required to implement this function.

Step 4: Final Answer:
The minimum number of NOR gates required to implement the function is zero, which corresponds to Option (A).