\begin{frame} \frametitle{Why do we need the conditions?} \begin{example}[$\interpret{{\rm f}}$ not monotone] \smallskip Let $R = \{\; {\rm f}(x) \to {\rm g}({\rm f}(x)) \;\}$ with the $\Sigma$-algebra $(\nat,\interpret{\cdot})$ and \begin{align*} \interpret{{\rm f}}(x) &= x + 1 & \interpret{{\rm g}}(x) &= 0 \end{align*} and $>$ as usual on $\nat$.\\[.5em] \pause Then $R$ is not terminating $${\rm f}(x) \to {\rm g}({\rm f}(x)) \to {\rm g}({\rm g}({\rm f}(x))) \to \ldots$$ \pause but \begin{itemize} \item $>$ is well-founded, and \item $\interpret{{\rm f}(x),\alpha} = \alpha(x)+1 > 0 = \interpret{{\rm g}({\rm f}(x)),\alpha}$. \end{itemize} \smallskip \pause Hence the functions $\interpret{f}$ need to be monotone! \smallskip \end{example} \end{frame}