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\begin{frame}{\alt<1-11>{Termination}{\alert{Termination is undecidable!}\onslide<12>{}}}
\vspace{-.5ex}
\begin{goal}{}
Assume there would be a \emph{program $\terminatoronitself$}\; with the behaviour: \vspace{-.5ex}
\begin{itemize}\setlength{\itemsep}{0pt}
\item input: a program \forestgreen{$P$}
\item output: \emph{yes} if \forestgreen{$P$} terminates on input \forestgreen{$P$},
\emph{no} otherwise
\end{itemize}
\end{goal}
\pause

\begin{center}\vspace{-.25ex}
\begin{tikzpicture}[p/.style={rectangle,minimum width=37mm,fill=yellow!50!orange!20,draw=black,dashed,rounded corners=2mm,inner sep=2mm,align=left}]
\node (p) [p] {p = read input;\\[-.5ex]\hspace{1.1cm}\scalebox{.8}{\vdots}\\[-.5ex]print result;};
\draw [decorate,decoration={brace,amplitude=10pt},xshift=-4pt,yshift=0pt]
($(p.south west) + (-2mm,0)$) to node [left,xshift=-3mm] {program $\terminatoronitself$} ($(p.north west) + (-2mm,0)$);

\mpause[1]{
\node (p')  at (p.south) [p,anchor=north] {\dm{if} result = yes\\\dm{then}\;\, loop forever\\\dm{else}\;\; terminate};

\draw [decorate,decoration={brace,amplitude=10pt},xshift=-4pt,yshift=0pt]
($(p.north east) + (2mm,0)$) to node [right,xshift=3mm] {program $\diagonalisator$} ($(p'.south east) + (2mm,0)$);
}
\end{tikzpicture}
\end{center}\vspace{-1.2ex}
\pause\pause

\begin{goal}{}
What happens if we run $\diagonalisator$ with input $\diagonalisator$?\vspace{-.5ex}
\begin{itemize}\setlength{\itemsep}{0pt}
\pause
\item initial part $\terminatoronitself$\, decides whether $\diagonalisator$ terminates on input $\diagonalisator$
\pause
\item if the result is \alert{yes}, then $\diagonalisator$ runs forever\pause{} $\,$ \alert{$\xmark$}
\pause
\item if the result is \alert{no}, then $\diagonalisator$ terminates\pause{} $\,$ \alert{$\xmark$}
\end{itemize}
\end{goal}
\pause
\vspace{-.5ex}
Thus $\terminatoronitself$ has made a mistake!\pause{} And thus also $\terminator\,$! \pause