Consider the following statements regarding database normal forms: 1. Any…

Concept: A relation with only two attributes always satisfies BCNF, because its only possible non-trivial functional dependency must have a superkey as its determinant. Separately, decomposition theory guarantees a lossless-join, dependency-preserving decomposition into 3NF is always achievable (via the synthesis algorithm), whereas a BCNF decomposition is guaranteed lossless-join but NOT guaranteed to preserve every dependency. BCNF is strictly stronger than 3NF because it removes the exception 3NF allows for prime attributes.

1. Statement 1 is true. In a two-attribute relation R(A, B), the only possible non-trivial functional dependencies are A → B or B → A. In either case the determinant is a superkey of R, so the relation trivially satisfies the BCNF condition.

2. Statement 2 is false. Lossless-join decomposition into BCNF is always possible, but preserving every functional dependency at the same time is not always possible. A relation with attributes X, Y, Z and functional dependencies XY → Z and Z → Y cannot be decomposed into BCNF while keeping both dependencies intact — any BCNF-safe split drops one of them.

3. Statement 3 is true. The 3NF synthesis algorithm (one relation per member of a minimal cover of functional dependencies, plus a relation for a candidate key if needed) always produces a decomposition that is both lossless-join and dependency-preserving.

4. Statement 4 is true. 3NF permits a functional dependency X → A to violate the superkey condition provided A is a prime attribute; BCNF removes this exception entirely. Every BCNF relation is therefore also in 3NF, but not every 3NF relation is in BCNF, making BCNF the stricter normal form.

• The combination “1, 2 and 3” wrongly counts statement 2 as true and leaves out statement 4, even though BCNF is demonstrably stricter than 3NF.

• The combination “1, 2 and 4” wrongly counts statement 2 as true and leaves out statement 3, even though the 3NF synthesis algorithm always guarantees a lossless, dependency-preserving decomposition.

• The combination “2, 3 and 4” wrongly counts statement 2 as true and leaves out statement 1, even though a two-attribute relation always satisfies BCNF trivially.

Statements 1, 3 and 4 hold while statement 2 fails, so the correct combination is 1, 3 and 4.