A·(Ā + B) — Boolean simplification
coredistribution then null law
Answer
A · B
Why this example is worth doing
Distributing produces A·Ā + A·B, and the first product is identically 0 by the complement law, so the whole term vanishes and A·B survives. The teaching point is that a term going to 0 disappears from a sum while a term going to 1 swallows it — the two null laws behave oppositely and are routinely swapped. This example is also the standard proof of the absorption variant A·(Ā + B) = A·B, which is worth memorising in its own right because it appears constantly inside larger expressions.
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 · (A′ + B)A · A′ + A · B— Distributive — A · (B + C) = A · B + A · C
A · A′ + A · B0 + A · B— Complement — A · A′ = 0
0 + A · BA · B— Identity — A + 0 = A
A · (A′ + B) simplifies to A · B in 3 steps. That is a proven minimum (1 term, 2 literals), not a best effort. It is the only minimum-cost form.
Quine–McCluskey
| # | Term |
|---|---|
| 1 | A · B |
| # | Cover |
|---|---|
| 1 | A · B |
1 term(s), 2 literal(s); 1 essential prime implicant(s).
Truth table
| # | A | B | F |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 |
| 2 | 1 | 0 | 0 |
| 3 | 1 | 1 | 1 |