Given a relation R(A, B, C) with the set of functional dependencies F = {A ->…

Concept: A lossless-join decomposition of a relation R is a split into fragments whose natural join reconstructs R exactly; a dependency-preserving decomposition is one whose fragments' projected functional dependencies, taken together, imply every dependency in the original set F. For a binary split R1, R2 (with R1 union R2 = R), the standard lossless-join test says the split is lossless exactly when the shared attributes R1 intersect R2 functionally determine R1 or determine R2. Two general theorems govern existence: a lossless-join decomposition of R always exists for any relation schema and any FD set (the BCNF decomposition algorithm is guaranteed, by construction, to always split off a lossless fragment at each step); a dependency-preserving decomposition of R also always exists for any relation schema and FD set (the 3NF synthesis algorithm constructs one) -- even though a decomposition that is simultaneously lossless, dependency-preserving, and in BCNF is not always achievable together.

Application:

1. Compute the closure of A under F = {A -> B, B -> C, C -> B}: starting from {A}, A -> B adds B giving {A, B}; then B -> C adds C, giving A+ = {A, B, C} = R. So A is a (minimal) candidate key of R.

2. Split R into R1(A, B) and R2(B, C). Check the lossless-join test on the shared attributes: R1 ∩ R2 = {B}, and B -> C holds, so B functionally determines {B, C} = R2. The test (R1 ∩ R2) -> R2 is satisfied, so this split is lossless.

3. Project F onto each fragment: R1(A, B) retains A -> B; R2(B, C) retains both B -> C and C -> B. Together these projected dependencies reconstruct all three original dependencies, so this same split is also dependency-preserving.

4. So a single split, R1(A, B) and R2(B, C), is simultaneously lossless and dependency-preserving -- demonstrating that a decomposition with each property exists.

Cross-check: Try a different split, R1(A, B) and R2(A, C). Here R1 ∩ R2 = {A}; since A is a key, A functionally determines all of R (hence both R1 and R2), so this split is also lossless by the same test. But projecting F onto it keeps only A -> B and A -> C; the dependencies B -> C and C -> B are lost, so this split is lossless but not dependency-preserving. A third split, R1(A, B) and R2(C) (no shared attribute at all), fails the lossless test outright: the empty intersection determines nothing, so joining the fragments back can produce spurious rows. None of this contradicts either statement: the statements ask whether a decomposition with each property EXISTS, not whether every decomposition has it -- the R1(A, B)/R2(B, C) split above already exhibits one with both properties at once.

Result: Since a decomposition of R exists that is both lossless and dependency-preserving, both (i) and (ii) are true.