How can I find candidate key using FD.S?

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VW->XYZ, Z->W, Y->VW

it is Funtional dependency relation S{V,W,X,Y,Z}. candidate key is 'Y'?? or another?? I have a question. Why Z->W break normal form? and what type is broken normal form?

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Renzo On

The relation has three candidate keys:

1. Y
2. VW
3. VZ

and you can verify this by calculating the closure of each of them.

The relation is not in Boyce-Codd Normal Form (BCNF), since in the dependency Z → W the determinant (Z) is not a superkey (while in the other two dependencies the determinants are keys).

Finally, note that the relation is in Third Normal Form (3NF), since W is a prime attribute (that is, is an attribute of a candidate key).