# Question regarding 2NF partial key dependency

If the functional depencies are
AB-->CD
BC-->D
IS the relation still in 2NF?,I mean since AB is the key and the 2nd BC,out of which B is part of the key,is the relation still in 2NF??

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Can't BC->D be shortened to C->D since AB->C? – Petter Brodin Jan 19 '13 at 11:05

Assuming your relation is `ABCD` and assuming you meant the following functional dependencies:

• AB-->C
• BC-->D

Then you need to decompose this relation to get it to third normal form (3NF). This is because D is not determined by the key of `ABCD`. To bring this to 3NF you need to have two relations: `ABC` and `BCD`. Now each non-key attribute is fully dependent on the key of its relation (and nothing else).

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Shouldn't you explain why/how you removed the `AB->D` dependency? – ypercubeᵀᴹ Jan 19 '13 at 13:27
@ypercube, I thought about that and thought that it was probably a typo based on short-cutting the listing of the original relation (`ABCD`). It seemed pretty unlikely that D is determined by both AB and BC. Why not write ABC-->D? I couldn't think of a practical case that met the dependencies as stated. – Joel Brown Jan 19 '13 at 13:35
Oh, I thought you meant that we can derive `AB->D` from the other 2 dependencies. – ypercubeᵀᴹ Jan 19 '13 at 13:36

From wikipedia, 2nd Normal Form:

Second normal form (2NF) is a normal form used in database normalization. 2NF was originally defined by E.F. Codd in 1971. A table that is in first normal form (1NF) must meet additional criteria if it is to qualify for second normal form. Specifically: a table is in 2NF if and only if it is in 1NF and no non-prime attribute is dependent on any proper subset of any candidate key of the table. A non-prime attribute of a table is an attribute that is not a part of any candidate key of the table.

There is only one candidate key, the `AB`.

About the other two non-prime attributes, `C` and `D`:

• `C` depends only on the candidate key `AB`.

• `D` depends on on the candidate key `AB` and on `BC`. `BC` is neither a candidate key nor a proper subset of (the only one) candidate key. (It doesn't matter that `B` is a subset of the candidate key, because `D` does not depend on just `B`.)

So, the relation is in 2NF.

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