2nd Normal form

Question

Hello, I know this type of question has been asked but I am not getting this. I have a confusion in FD bc->d should it be in 2nd normal form or not.

R={a,b,c,d}

FD={ab->c,bc->d}

Ck={ab}

I want to know whether this is in second normal form or not. Thanks in advance.

Comments

It is in 2nd normal form.As no non-prime attribute is depending on proper subset of candidate key.

Answer 1

The relation is in 2NF as it is in 1NF and there is no partial dependency for any non-key attribute- here $ab$ is the key and there is no FD with only $a$ or $b$ as the determinant and hence no partial dependency is possible.  

$ab \to bc; bc \to d$ where $ab$ is a candidate key and $d$ is a non-prime attribute and $bc$ is not a key. So, this is a transitive dependency and so the Relation is not in 3NF. 

https://gatecse.in/demystifying-database-normalization/