Is there a dependency preserving, lossless BCNF decomposition for this relational schema?

R(A,B,C) where {AB > C , C > A}. The candidate keys are {A,B} and {C,B}. This is in 3NF but not BCNF because of {C>A} . Now it seems like this can't even be split into a lossless BCNF, let alone a dependency preserving one. Is there some way to prove it? Is there some result that says that if the result of the 3NF algorithm does not give you a BCNF decomposition, then a further BCNF decomposition is not possible?
Edit: I have since figured out that every schema can be split into a lossless BCNF one. It is preserving dependency that is uncertain. Here the lossless decomposition would be R(B,C), R(C,A)

As you have discovered, the decomposition of
R
in the two relationsR1(B, C)
andR2(C, A)
is a lossless decomposition (and both relations are in BCNF). On the other hand, the dependencyAB > C
is not preserved by this decomposition.Note that it is not difficult to convince yourself that, in this particular case, a decomposition of R cannot maintain all the three attributes together, since we have
C > A
, and by joining the decomposed relations the dependencyAB > C
cannot be obtained in any way.This example simply shows that there are cases in which every possible algorithm to decompose a relation in BCNF can produce the loss of one or more dependencies.