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?
How can I find candidate key using FD.S?
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The relation has three candidate keys:
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 → Wthe 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
Wis a prime attribute (that is, is an attribute of a candidate key).