Ā·B̄·C̄ + Ā·B̄·C + Ā·B·C̄ + A·B·C̄ — Boolean simplification
exam standardone 2-literal and one 2-literal term
Answer
A′ · B′ + B · C′
Why this example is worth doing
A four-minterm function where the greedy approach misleads. The obvious first move is to combine the first two into Ā·B̄, and that is correct; the trap is then pairing the third term with the first, which is legal but leaves the fourth stranded and produces a three-term answer. Pairing minterms 2 and 6 into B·C̄ instead covers everything in two terms. This is why the tool computes the minimum with Quine–McCluskey first and then searches for a legal algebraic path to it, rather than rewriting greedily and hoping.
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 · C′ + A · B · C′A′ · B′ · (C′ + C) + A′ · B · C′ + A · B · C′— Factoring (distributive, right to left) — A · B + A · C = A · (B + C)
A′ · B′ · (C′ + C) + A′ · B · C′ + A · B · C′A′ · B′ · 1 + A′ · B · C′ + A · B · C′— Complement — A + A' = 1
A′ · B′ · 1 + A′ · B · C′ + A · B · C′A′ · B′ + A′ · B · C′ + A · B · C′— Identity — A · 1 = A
A′ · B′ + A′ · B · C′ + A · B · C′A′ · B′ + B · C′ · (A′ + A)— Factoring (distributive, right to left) — A · B + A · C = A · (B + C)
A′ · B′ + B · C′ · (A′ + A)A′ · B′ + B · C′ · 1— Complement — A + A' = 1
A′ · B′ + B · C′ · 1A′ · B′ + B · C′— Identity — A · 1 = A
A′ · B′ · C′ + A′ · B′ · C + A′ · B · C′ + A · B · C′ simplifies to A′ · B′ + B · C′ in 6 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′ · B′ |
| 2 | A′ · C′ |
| 3 | B · C′ |
| # | Cover |
|---|---|
| 1 | A′ · B′ + B · C′ |
2 term(s), 4 literal(s); 2 essential prime implicant(s).
Truth table
| # | A | B | C | F |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 1 |
| 1 | 0 | 0 | 1 | 1 |
| 2 | 0 | 1 | 0 | 1 |
| 3 | 0 | 1 | 1 | 0 |
| 4 | 1 | 0 | 0 | 0 |
| 5 | 1 | 0 | 1 | 0 |
| 6 | 1 | 1 | 0 | 1 |
| 7 | 1 | 1 | 1 | 0 |