Is this table in third normal form? I have a table L(A, B, C). Functional dependencies are: A->B, A->C, B->A, B->C. Is this table in 3nf? 
My thoughts: A and B are keys of this table. Table is in 3nf if there are no transitive dependencies between keys and non-prime attributes. There is a transitive dependence A->B->C, so it's not in 3nf? Or it does not work in this way, because B is a key? Hopefully I didn't confuse you too much. Thanks.
Here's one attempt (this is not a day to day activity for me, so I may be doing some weird errors below):

A and B are clearly candidate keys of R. Therefore C is the only non-prime attribute of R

R is in 3NF iff:

a) R is in 2NF

b) Every non-prime attribute of R (C) is non-transitively dependent on every superkey of R.

The superkeys of R is (A, B), (A, C), (A, B, C)

By reflexivity (A, C) -> C AND (A, B, C) -> C
By composition of A->C and B->C we get (A, B) -> (C, C), i.e. (A, B) -> C

C is therefore non-transitively dependent on every superkey of R, and hence R is in 3NF.