A·B + Ā·C + B·C + A·B·D — Boolean simplification
edge caseconsensus and absorption together; 2 terms
Answer
A′ · C + A · B
Why this example is worth doing
A four-variable expression built so that two different redundancies hide in it at once. B·C is the consensus of A·B and Ā·C and can be deleted; A·B·D is absorbed by A·B, since any row making A·B·D true already makes A·B true. Neither removal is visible from the term count alone, and D disappears from the answer entirely — a result students often assume must be a mistake. The derivation labels the two steps with different law names so it is clear they are different phenomena.
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′ · C + B · C + A · B · DA · B + A′ · C + B · C— Absorption — A + A · B = A
A · B + A′ · C + B · CA · B + A′ · C— Consensus — A · B + A′ · C + B · C = A · B + A′ · C
A · B + A′ · C + B · C + A · B · D simplifies to A · B + A′ · C in 2 steps. That is a proven minimum (2 terms, 4 literals), not a best effort. It is the only minimum-cost form.
Quine–McCluskey
| # | Term |
|---|---|
| 1 | A′ · C |
| 2 | B · C |
| 3 | A · B |
| # | Cover |
|---|---|
| 1 | A′ · C + A · B |
2 term(s), 4 literal(s); 2 essential prime implicant(s).
Truth table
| # | A | B | C | D | F |
|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 1 | 0 |
| 2 | 0 | 0 | 1 | 0 | 1 |
| 3 | 0 | 0 | 1 | 1 | 1 |
| 4 | 0 | 1 | 0 | 0 | 0 |
| 5 | 0 | 1 | 0 | 1 | 0 |
| 6 | 0 | 1 | 1 | 0 | 1 |
| 7 | 0 | 1 | 1 | 1 | 1 |
| 8 | 1 | 0 | 0 | 0 | 0 |
| 9 | 1 | 0 | 0 | 1 | 0 |
| 10 | 1 | 0 | 1 | 0 | 0 |
| 11 | 1 | 0 | 1 | 1 | 0 |
| 12 | 1 | 1 | 0 | 0 | 1 |
| 13 | 1 | 1 | 0 | 1 | 1 |
| 14 | 1 | 1 | 1 | 0 | 1 |
| 15 | 1 | 1 | 1 | 1 | 1 |