\begin{frame}[t] \frametitle{Dependency Graphs} \vspace{-1em} \begin{example} \vspace{-1em} \begin{align*} \minus(x,0) &\to x\\ \minus(\s(x),\s(y)) &\to \minus(x,y)\\ \quot(0,\s(y)) &\to 0\\ \quot(\s(x),\s(y)) &\to \s(\quot(\minus(x,y),\s(y))) \end{align*} \vspace{-1em} \pause Dependency graph: analysis which DP-rules may follow each other \vspace{-.5em} \begin{center} \begin{tikzpicture} \begin{scope}[node distance=5mm,very thick] \node (m) {(1) $\minus_\#(\s(x),\s(y)) \to \minus_\#(x,y)$}; \node (qm) [below=of m] {(2) $\quot_\#(\s(x),\s(y)) \to \minus_\#(x,y)$}; \node (qqm) [below=of qm] {(3) $\quot_\#(\s(x),\s(y)) \to \quot_\#(\minus(x,y),\s(y))$}; \draw[->] (qm) -- (m); \draw[->] (qqm) -- (qm); \draw[->] (qqm) to[out=-4,in=4,looseness=2.5] (qqm); \draw[->] (m) to[out=-4,in=4,looseness=2.5] (m); \end{scope} \end{tikzpicture} \end{center} \vspace{-.5em} \pause Idea: consider only strongly connected components $\SN(\{1\}_{\mit{top}}/R)$, $\SN(\{3\}_{\mit{top}}/R)$. \end{example} \end{frame}