A ⊕ B ⊕ C — Boolean simplification
coreirreducible: 4 minterms, no adjacency
Answer
A′ · B′ · C + A′ · B · C′ + A · B′ · C′ + A · B · C
Why this example is worth doing
The important negative result on this page. Three-input XOR is odd-parity, its four minterms sit on diagonally opposite K-map cells, and no two of them differ in only one variable — so nothing combines and the minimal sum-of-products is the canonical sum, four terms of three literals. Two-level minimisation genuinely cannot help. The lesson is that minimality is relative to a target form: as an XOR chain the function costs two gates, which is why parity trees are never built from AND-OR logic.
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 ⊕ CA · B · C + A · B′ · C′ + A′ · B · C′ + A′ · B′ · C— Definition of XOR — a chain of ⊕ is the sum of its odd-parity terms
A ⊕ B ⊕ C simplifies to A · B · C + A · B′ · C′ + A′ · B · C′ + A′ · B′ · C in 1 step. That is a proven minimum (4 terms, 12 literals), not a best effort. It is the only minimum-cost form.
Quine–McCluskey
| # | Term |
|---|---|
| 1 | A′ · B′ · C |
| 2 | A′ · B · C′ |
| 3 | A · B′ · C′ |
| 4 | A · B · C |
| # | Cover |
|---|---|
| 1 | A′ · B′ · C + A′ · B · C′ + A · B′ · C′ + A · B · C |
4 term(s), 12 literal(s); 4 essential prime implicant(s).
Truth table
| # | A | B | C | F |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 1 |
| 2 | 0 | 1 | 0 | 1 |
| 3 | 0 | 1 | 1 | 0 |
| 4 | 1 | 0 | 0 | 1 |
| 5 | 1 | 0 | 1 | 0 |
| 6 | 1 | 1 | 0 | 0 |
| 7 | 1 | 1 | 1 | 1 |