The minimum number of gates required to implement the Boolean expression AB +…
The minimum number of gates required to implement the Boolean expression AB + AB' + A'C is
Answer: D. 1 OR gate — Concept: Boolean simplification uses three identities here — the complement law A + A' = 1 (and its dual, A·A' = 0), the factoring/distributive law used as…
- A.
1 AND and 1 OR gate
- B.
2 NAND gates
- C.
3 AND gates and 2 OR gates
- D.
1 OR gate
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Correct answer: D
Concept: Boolean simplification uses three identities here — the complement law A + A' = 1 (and its dual, A·A' = 0), the factoring/distributive law used as A·B + A·B' = A·(B + B'), and the absorption identity A + A'C = A + C (obtained by distributing: A + A'C = (A + A')·(A + C) = 1·(A + C) = A + C).
Application: Reduce AB + AB' + A'C step by step.
Factor the first two terms on A: AB + AB' = A(B + B').
Apply the complement law B + B' = 1, so A(B + B') = A·1 = A. The expression is now A + A'C.
Apply the absorption identity A + A'C = A + C (shown above via the distributive law), leaving the two-literal expression A + C.
A + C uses only the literals A and C in their true (uncomplemented) form combined by a single logical sum, so it can be realized with exactly one 2-input OR gate and nothing else — no AND gate is needed (no product term survives) and no NOT gate is needed (neither surviving literal is complemented).
Cross-check: List all eight input combinations of A, B, C and evaluate the original expression against A + C independently of the algebra above (an exhaustive check, not just a couple of sample rows).
A | B | C | Original expression | A + C |
|---|---|---|---|---|
0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | 0 |
0 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 1 | 1 |
All eight rows match, confirming the simplification exhaustively rather than only algebraically.
Hence the minimum number of gates required is 1 (a single OR gate combining A and C).