Is this relation in 3NF as well as in BCNF?

Question

Suppose a relation schema R(A,B,C) and the FDs are

{A -> B, B -> C}

So the superkeys are {A}, {A,B}

Now if we decompose it into 3NF it will be

R1(A,B) with FD {A -> B} and R2(B,C) with FD {B -> C}

Is it in BCNF? I can't determine. Since B was not a superkey in R does {B -> C} in R2 violates BCNF?

Answer 1

{AB} is a superkey, but it's not a candidate key. (It's not a minimal superkey.) The decomposition

• R₁(A B)
• R₂(B C)

is in at least BCNF.

Informally, a relation is in BCNF if every arrow is an arrow out of a candidate key. B is a candidate key in R₂.

The relation R is not in BCNF. The only candidate key in R is A; the FD B->C has an arrow that's not out of a candidate key.

In truth, both R₁ and R₂ are much stronger than BCNF. They're both in 6NF.