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Question

for 2 relations R(A,B) with functional dependency F={A->B} and S(B,C) witht functional dependency S={B->C}  natural join of R and S is in BCNF

Comments

No, it's not in BCNF

Answer 1

R natural join S be Relation G(A,B,C)

With FDs A-->B, B--C

'A' will be primary key for G.

For relation to be in BCNF the determinant(left hand side) of all FDs must be super key.

But in this case for FD B-->C , B is not a super key.

Because of this R natural join S is not in BCNF.