Assume that $L = \{\,a^n b^n c^n \mid n \geq 0\,\}$ was context-free.
    According to the pumping lemma there is $m>0$ such that
      \mpause[1]{a^mb^mc^m} = uvxyz
    with $|vxy| \leq m$, $|vy| \geq 1$, and $uv^ixy^i z \in L$ for every $i \geq 0$.

    Since $|vxy| \leq m$, \alert{$vy = a^{\,j}b^k$} or \alert{$vy = b^{\,j}c^k$} for some $j,k \ge 0$.
    Since $|vy|\geq 1$ we have \alert{$j+k \geq 1$}.

    Then \alert{$uv^2xy^2z$} does not contain equally many $a$'s, $b$'s and $c$'s.

    \alert{Contradiction}, thus $L$ is not context-free.
    \item opponent picks $m$,
    \item we pick $w = a^mb^mc^m$,
    \item opponent $u,v,x,y,z$