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