A·B + A·B̄ — Boolean simplification
introadjacency; 1 term, 1 literal
Answer
A
Why this example is worth doing
The smallest simplification in the syllabus and the one that explains every other. B is true in one term and false in the other, and nothing else differs, so whatever B does the value follows A alone. The page names this the adjacency form of the complement law, A·B + A·B̄ = A·(B + B̄) = A·1 = A, and shows the factorisation as a separate step rather than collapsing it — students lose marks for jumping straight to the answer. It is also the algebraic twin of circling two neighbouring 1s on a K-map.
Try your own input in the Boolean algebra simplifier. Simplify any expression to its minimal SOP and POS, one named law at a time.
How the answer is reached
Step by step
A · B + A · B′A · (B + B′)— Factoring (distributive, right to left) — A · B + A · C = A · (B + C)
A · (B + B′)A · 1— Complement — A + A' = 1
A · 1A— Identity — A · 1 = A
A · B + A · B′ simplifies to A in 3 steps. That is a proven minimum (1 term, 1 literal), not a best effort. It is the only minimum-cost form.
Quine–McCluskey
| # | Term |
|---|---|
| 1 | A |
| # | Cover |
|---|---|
| 1 | A |
1 term(s), 1 literal(s); 1 essential prime implicant(s).
Truth table
| # | A | B | F |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 |
| 2 | 1 | 0 | 1 |
| 3 | 1 | 1 | 1 |