Given R(A,B,C,D) with FDs F = {AB->CD, C->A, B->D} is decomposed into…

Lossless join check:

• The common attributes of R1(A, B, C) and R2(B, C, D) are B, C.

• (BC)+ = ABCD because C -> A gives A and B -> D gives D.

• Since the common attributes determine the full relation, the decomposition is lossless.

Dependency-preservation check:

• In R1(A, B, C), the dependencies C -> A and AB -> C are preserved.

• In R2(B, C, D), the dependency B -> D is preserved.

• AB -> CD is preserved because AB -> C gives C and B -> D gives D.

Final answer: The decomposition is lossless and dependency preserving.