Consider the relation R = (A,B,C,D,E, F) with the following functional…

Answer: The following decompositions are lossless and dependency-preserving: (A, B, C, D), (A, B, D, F), (E, F) and (A, B, C, D, F), (E, F)

Reasoning for the decomposition (A, B, C, D), (A, B, D, F), (E, F):

• Check losslessness for the join of (A,B,C,D) and (A,B,D,F): their intersection is {A, B, D}. Using A → C we have {A, B, D} → C, so the intersection functionally determines all attributes of (A,B,C,D). Therefore the join of these two relations is lossless.

• Now join that result with (E,F): the common attribute is F, and F → E, so F functionally determines all attributes of (E,F). Hence this final join is also lossless. Overall the decomposition is lossless.

• Dependency preservation: A → C and BC → D are both contained in the relation (A,B,C,D), and F → E is contained in (E,F). Thus all original FDs are preserved by the projections.

Reasoning for the decomposition (A, B, C, D, F) and (E, F):

• The intersection is {F}. Since F → E, the intersection determines the attributes of (E,F), so the join is lossless.

• Dependency preservation: A → C and BC → D are present in the relation (A,B,C,D,F), and F → E is present in (E,F). Thus all FDs are preserved.

Why the decomposition (A, B, C) and (C, D, E, F) fails:

• Losslessness: the intersection is {C}, but C does not functionally determine all attributes of either relation (there is no FD with C on the left), so the join can be lossy.

• Dependency preservation: BC → D is not preserved because B is in the first relation while D is only in the second; there is no single relation containing B, C and D to enforce BC → D.

Conclusion: The two decompositions listed at the start are both lossless and dependency-preserving; the decomposition into (A,B,C) and (C,D,E,F) is not.