What would be the 3NF of this relation?

Question

R(ABCDE)

AB->CDE
A->C
D->E

Solve:

1NF We presume that this is already in 1NF.

2NF AB is the candidate key here. A->C is the violation.

So, we decompose them like the following:

R1 = (AC) + (AB) = (ABC)

R2 = R - (AC) + (AB) = (BDE) + (AB) = (ABDE)

3NF

???

Answer 1

The third normal form of your relational schema is the following:

R1 (A B D)    
R2 (A C)    
R3 (D E)

You can verify it by finding a canonical cover of the set of dependencies, which is:

A B → D
A → C
D → E