\begin{frame}
  \frametitle{BCNF Synthesis Algorithm: Example}

  \begin{exampleblock}{}
    Consider $R = (A,B,C,D,E)$ with the canonical set of FDs
    \begin{talign}
      A \to D &&
      B \to C &&
      B \to D &&
      D \to E
    \end{talign}
    Here $\{\,A,B\,\}$ is the only minimal key. \pause Is $R$ in BCNF? \pause No.
    \begin{enumerate}
    \pause
      \item Maximise the right-hand sides of the FDs:
    \pause
        \begin{talign}
          A \to D,E &&
          B \to C,D,E &&
          D \to E
        \end{talign}
      \item \pause Split off violating FD's one by one:
        \begin{itemize}
        \pause
          \item $\mathcal{S} = \{\, R_0(\ul{A},\ul{B},C,D,E) \,\}$
        \pause
          \item $A \to D,E$ violates BCNF of $R_0$
        \pause
          \item $\mathcal{S} = \{\, R_0(\ul{A},\ul{B},C),\; R_1(\ul{A},D,E) \,\}$
        \pause
          \item $B \to C,D,E$ violates BCNF of $R_0$
        \pause
          \item $\mathcal{S} = \{\, R_0(\ul{A},\ul{B}),\; R_1(\ul{A},D,E),\; R_2(\ul{B},C) \,\}$
        \pause
          \item $D \to E$ violates BCNF of $R_1$ 
        \pause
          \item $\mathcal{S} = \{\, R_0(\ul{A},\ul{B}),\; R_1(\ul{A},D),\; R_2(\ul{B},C),\; R_3(\ul{D},E) \,\}$\pause\ - done!
        \end{itemize}
    \end{enumerate}
  \end{exampleblock}
  Note that we lost the dependency $B \to D$!  
\end{frame}