A·B + Ā·C + B·C — Boolean simplification
coreconsensus; the B·C term is redundant
Answer
A′ · C + A · B
Why this example is worth doing
The consensus theorem. B·C is the consensus term of A·B and Ā·C — it contains the literals left over when A and Ā cancel — and it is covered by the other two, so it can be deleted without changing a single row. This matters commercially, not just algebraically: deleting it removes an AND gate and two wires from the synthesised circuit. It is also the classic hazard-removal term, so the page notes that in a real gate-level design you sometimes deliberately keep the redundant term back.
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 · CA · B + A′ · C— Consensus — A · B + A′ · C + B · C = A · B + A′ · C
A · B + A′ · C + B · C simplifies to A · B + A′ · C in 1 step. 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 | F |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 1 |
| 2 | 0 | 1 | 0 | 0 |
| 3 | 0 | 1 | 1 | 1 |
| 4 | 1 | 0 | 0 | 0 |
| 5 | 1 | 0 | 1 | 0 |
| 6 | 1 | 1 | 0 | 1 |
| 7 | 1 | 1 | 1 | 1 |