A + A·B — Boolean simplification
introabsorption; 1 term, 1 literal
Answer
A
Why this example is worth doing
Absorption is the law students distrust most, because the second term looks like it must contribute something. It cannot: every row where A·B is 1 already has A equal to 1, so the OR gate's second input never changes an output that was 0. The page shows the two-row truth-table proof next to the derivation A + A·B = A·(1 + B) = A·1 = A, which routes through the annulment law 1 + B = 1 — the step people skip and then cannot reproduce in an exam.
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 · BA— Absorption — A + A · B = A
A + A · B simplifies to A in 1 step. 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 |