I have a question on the definition of 3NF given by Chris Date in his book "Database Design and Relational Theory", page 78.
The definition given in the book is: "A relvar R is in 3NF iff for every non-trivial FD X -> Y, either X is a superkey, or Y is a subkey."
(For Date "Y is a subkey" means that Y is contained in a candidate key, and no assumption is made on the cardinality of the set Y in the Date definition.)
It seems to me, however, that this definition is not equivalent to the usual definition (that can be found in other references) saying that "R is in 3NF if for every FD X -> Y, either the FD is trivial, or X is a superkey, or every element in Y\X is contained in a candidate key".
Consider now the relvar with 5 attributes R(A,B,C,D,E) with the following FD cover:
{A,B} -> C,
{C,D} -> E,
E -> B
These imply {A,E} -> {B,C}. The candidate keys of R are K1 = {A,B,D}, K2 = {A,C,D} and K3 = {A,E,D} and so the FD {A,E} -> {B,C} shows that R is not in 3NF if we use Date's definition.
However, it is in 3NF if we use the "usual" definition (since every attribute is actually contained in a candidate key).
Is there something I do not understand? Or is Date really using another (stronger than the usual one) definition of 3NF?