Ā·B̄·C + Ā·B·C + A·B·C — Boolean simplification
coretwo 2-literal terms
Answer
A′ · C + B · C
Why this example is worth doing
Three minterms that share a common literal C, which is the cue to factor before doing anything else: C·(Ā·B̄ + Ā·B + A·B). Inside the bracket the first two terms combine to Ā and the second and third to B, and note that the middle minterm has been used twice. That is legal — idempotence lets you write X = X + X — and it is the algebraic counterpart of overlapping two circles on a K-map. Students who believe each minterm belongs to exactly one group get stuck here.
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′ · C + A′ · B · C + A · B · CA′ · C · (B′ + B) + A · B · C— Factoring (distributive, right to left) — A · B + A · C = A · (B + C)
A′ · C · (B′ + B) + A · B · CA′ · C · 1 + A · B · C— Complement — A + A' = 1
A′ · C · 1 + A · B · CA′ · C + A · B · C— Identity — A · 1 = A
A′ · C + A · B · CA′ · C + A · B · C + B · C— Consensus — A · B + A′ · C + B · C = A · B + A′ · C
A′ · C + A · B · C + B · CA′ · C + B · C— Absorption — A + A · B = A
A′ · B′ · C + A′ · B · C + A · B · C simplifies to A′ · C + B · C in 5 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 |
| # | Cover |
|---|---|
| 1 | A′ · C + B · C |
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 | 0 |
| 7 | 1 | 1 | 1 | 1 |