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\begin{frame}{Example}
\begin{exampleblock}{}
Assume that $L = \{\,a^n b^n c^n \mid n \geq 0\,\}$ was context-free.
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According to the pumping lemma there is $m>0$ such that
\begin{talign}
a^mb^mc^m = uvxyz
\end{talign}
with $|vxy| \leq m$, $|vy| \geq 1$, and $uv^ixy^i z \in L$ for every $i \geq 0$.
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\emph{Note:}
\begin{itemize}\setlength{\itemsep}{0pt}
\item opponent picks $m$,
\item \alert{we pick $w = a^mb^mc^m$},
\item opponent $u,v,x,y,z$.
\end{itemize}
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We do not know $u,v,x,y,z$, but we can reason about them:
\begin{itemize}\setlength{\itemsep}{0pt}
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\item Since $|vxy| \leq m$, it follows that $vy = a^{\,j}b^k$ or $vy = b^{\,j}c^k$.
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\item Since $|vy|\geq 1$ we have $j+k \geq 1$.
\end{itemize}
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Then \alert{$uv^2xy^2z$} does not contain equally many $a$'s, $b$'s and $c$'s.
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\alert{Contradiction}, thus $L$ is not context-free.
\end{exampleblock}
\end{frame}