For example, consider the following relation:

R(A, B, C, D, E) 

With the functional dependencies (FDs) shown below:

{A→B,  BC→E,  D→C}

Here candidate key is {AD} and the relation is not in second normal form (2NF) because of the partial dependencies A→B, D→C.

Now, if I decompose R as follows:

R1(AB) ,  R2(CD),  R3(ADE)

...these relations are now in 2NF, but the FD BC→E is lost.

How can I decompose R to comply with 2NF and preserve the FDs?


1 Answer 1


One decomposition in 3NF (and so also in 2NF) is:


This decomposition can be obtained with the so-called synthesis algorithm for 3NF, it is a lossless decomposition and preserves the Functional Dependencies.

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