7.31

Consider the schema \(R = (A,B,C,D,E,G)\) and the set \(F\) of functional dependencies: \[ AB \rightarrow CD \\ B \rightarrow D \\ DE \rightarrow B \\ DEG \rightarrow AB \\ AC \rightarrow DE \\ \]

\(R\) is not in BCNF for many reasons, one of which arises from the functional dependency \(AB \rightarrow CD\). Explain why \(AB \rightarrow CD\) shows that \(R\) is not in BCNF and then use the BCNF decomposition algorithm starting with \(AB \rightarrow CD\) to generate a BCNF decomposition of \(R\). Once that is done, determine whether your result is or is not dependency preserving, and explain your reasoning.

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\(AB \rightarrow CD\) is NOT a trivial functional dependency.

Also, \((AB)^+ = \{A, B, C, D, E\}\). That is, \(AB\) is NOT a superkey.

Thus the relation \(R\) is NOT in BCNF.

BCNF decomposition of \(R\):-

\[ \{ \{A,B,G\}, \{A,B,E\}, \{A,B,C\}, \{B,D\} \} \]

Clearly the above decomposition is NOT dependency preserving since the functional dependency \(DE \rightarrow B\) cannot be tested by using only one relation.