(A + B)·(A + B̄) — Boolean simplification
corePOS adjacency; dual of example 1
Answer
A
Why this example is worth doing
The product-of-sums mirror of the first example, and a good demonstration that the duality principle is a real labour saver. Most students expand the brackets into A·A + A·B̄ + A·B + B·B̄, then need idempotence and the complement law to clean up the debris. The direct route uses the distributive law in its second, less familiar form — (A + B)·(A + C) = A + B·C — giving A + B·B̄ = A + 0 = A in one step. The page shows both routes side by side.
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 · (A + B′) + B · (A + B′)— Distributive — A · (B + C) = A · B + A · C
A · (A + B′) + B · (A + B′)A · A + A · B′ + B · (A + B′)— Distributive — A · (B + C) = A · B + A · C
A · A + A · B′ + B · (A + B′)A + A · B′ + B · (A + B′)— Idempotent — A · A = A
A + A · B′ + B · (A + B′)A + A · B′ + B · A + B · B′— Distributive — A · (B + C) = A · B + A · C
A + A · B′ + B · A + B · B′A + A · B′ + B · A + 0— Complement — A · A′ = 0
A + A · B′ + B · A + 0A + A · B′ + B · A— Identity — A + 0 = A
A + A · B′ + B · AA + A · B′ + A · B— Commutative — A · B = B · A; A + B = B + A
A + A · B′ + A · BA + A · B— Absorption — A + A · B = A
A + A · BA— Absorption — A + A · B = A
(A + B) · (A + B′) simplifies to A in 9 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 |