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\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}