Trying to get BCNF out of this relation
Yes, it is.

Moreover, a correct table composed by two attributes is always in BCNF.

In fact, if we have R(A,B):

Proof


If A->B but B doesn't imply A, A is key => R(A,B) is in BCNF
If B->A but A doesn't imply B, B is key => R(A,B) is in BCNF
If B->A and A->B, keys: {A}, {B}; => R(A,B) is in BCNF