Basic 3NF doubt .

Question

Suppose we have a relation R(ABCD) where AB is candidate key.

Now if there is a FD B -> AD, then will this FD satisfy the condition of 3NF ? I mean condition of  X->Y where X is super key or Y is prime attribute as here only A is prime attribute on RHS.

I know we can split the above FD as B->A and B->D and here B->A satisfies the condition and B->D doesn't

Any help would be appreciated.

Comments

B $\rightarrow$ D doesn't satisfy even 2NF. So yes B $\rightarrow$ AD doesn't satisfy 3NF

---

When closure of B contains A how can AB be the candidate key?

---

It only assume here.

---

No this Functional dependency doesn’t satisfy the condition of 3NF.

Answer 1

For 3nf : It should have super key on LHS or only prime attributes on the right side.

Not in 3NF, not even in 2NF

Comments

Are sure it's only prime attributes or is it contains a prime attribute?

---

Only.
If you are confused , just decompose the given FD :-)

Answer 2

The condition for 3nf fails when in  x --> y both x and y are non-key attributes. And 2nf fails when x is partial key and y is non-key. Thus in above problem B -> AD, B is partial key and AD is non-key so it fails at 2nf only. To satisfy 3nf it must be in 2nf first.

Answer 3

B -> AD will fail 2NF

B -> A (prime gives prime) and B -> D (prime gives non prime)

To pass 3NF, LHS of FD should be either super key or RHS should be a prime attribute for ALL FD

But for B -> D neither holds so it isn't in 3NF

To be in 2NF, RHS of FD should be given by an any of the complete Candidate Key not a subset of it. (No partial dependency allowed)

B -> D (subset of candidate key gives D )

So it's not even in 2NF.

Here, 

Answer 4

I think everyone is missing something here .

first you said – “AB is candidate key”

and then there is a FD B →  AD

this implies that B – > A

means AB is not a candidate key it is a super key

if B is key and  deriving everything here(Assume) then it is in 3NF even in BCNF