How do I prove this relation is in the `3FN`?
A relation is in third normal form when for all the FDs X -&gt; Y of a cover F of the dependencies of the relation we have that either X is a superkey, or Y is prime. So we can prove that R is in 3NF if we can show that this condition is true for all the dependencies of F.
The relation has three candidate keys (you can easily see that the closure of each of them contains all the attributes of the relation):
AB
BC
BD

So the only non-prime attribute is E. So, BC-&gt;AE has a determinant which is a superkey, while A-&gt;D and D-&gt;C have a determinate which is a prime attribute.
So all the dependencies satisfy the condition and the relation is in 3NF.