7.14

Show that there can be more than one canonical cover for a given set of functional dependencies, using the following set of dependencies: \[ X \rightarrow YZ, Y \rightarrow XZ, \text{ and } Z \rightarrow XY \]

---

Consider the first functional dependency. We can verify that \(Z\) is extraneous in \(X \rightarrow YZ\) and delete it. Subsequently, we can similarly check that \(X\) is extraneous in \(Y \rightarrow XZ\) and delete it, and that \(Y\) is extraneous in \(Z \rightarrow XY\) and delete it, resulting in a canonical cover \(X \rightarrow Y, Y \rightarrow Z, Z \rightarrow X\).

However, we can also verify that \(Y\) is extraneous in \(X \rightarrow YZ\) and delete it. Subsequently, we can similarly check that \(Z\) is extraneous in \(Y \rightarrow XZ\) and delete it, and that \(X\) is extraneous in \(Z \rightarrow XY\) and delete it, resulting in a canonical cover \(X \rightarrow Z, Y \rightarrow X, Z \rightarrow Y\).