Three-input AND function check: The statement claims the Boolean expression for a three-input AND gate is Y = A · B + C. Is that correct for a gate whose output is 1 only when A, B, and C are all 1?

Introduction / Context:
Recognizing canonical expressions for basic gates is essential. A three-input AND gate outputs 1 only when all three inputs are 1. The stem proposes Y = A · B + C, which mixes an AND with an OR. We must determine whether this equals the strict three-input AND function or something else. Misidentification leads to incorrect truth tables and wrong hardware implementations.

Given Data / Assumptions:

• Desired function: Y = 1 only when A = B = C = 1.
• Proposed expression: Y = A · B + C.
• Standard operator meanings: · is AND, + is OR.

Concept / Approach:
The true three-input AND is Y = A · B · C. The proposed Y = A · B + C equals 1 when either (A AND B) is 1 or when C is 1. That expression produces many minterms beyond A = B = C = 1, for example A = 0, B = 0, C = 1 gives Y = 1. Therefore, the proposed expression does not implement the strict three-input AND function and the claim is incorrect.

Step-by-Step Solution:

Verification / Alternative check:
Construct K-maps: the three-input AND corresponds to a single cell (A=1,B=1,C=1), whereas A · B + C covers an entire row/column group, clearly not the same region.

Why Other Options Are Wrong:
“Correct” contradicts direct evaluation. Conditional statements about De Morgan or active-low inputs do not transform A · B + C into A · B · C without additional complements and gates.

Common Pitfalls:
Assuming that adding inputs with “+” always tightens a condition—it actually loosens it (more chances to be 1). For strict AND across three signals, use a triple product term.

Final Answer:
Incorrect