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\begin{frame}{Exercise}
\begin{exampleblock}{}
Remove all unit production rules from
\begin{talign}
S \to Aa \mid B &&
A \to a \mid bc \mid B &&
B \to A \mid bb
\end{talign}
\pause
Note that there are no $\lambda$-productions.\\
(So no need to first remove $\lambda$-productions.)
\pause\medskip

We determine all pairs $A \neq B$ with $A \Rightarrow^+ B$:
\begin{talign}
\mpause[1]{S \Rightarrow^+ B} &&
\mpause{A \Rightarrow^+ B} &&
\mpause{B \Rightarrow^+ A} &&
\mpause{S \Rightarrow^+ A}
\end{talign}
\pause\pause\pause\pause\pause
Thus we add the following rules:
\begin{talign}
S &\to Aa \mid B \alert{\mpause[1]{\mid a}\mpause{\mid bc}\mpause{\mid A}\mpause{\mid bb}} \\
A &\to a \mid bc \mid B \alert{\mpause{\mid A}\mpause{\mid bb}} \\
B &\to A \mid bb \alert{\mpause{\mid a}\mpause{\mid bc}\mpause{\mid B}}
\end{talign}
\pause\pause\pause\pause\pause\pause\pause\pause\pause\pause
Removing all unit production rules yields the final result:
\begin{talign}
S &\to a \mid bb \mid bc \mid Aa &&&
A &\to a \mid bb \mid bc &&&
B &\to a \mid bb \mid bc
\end{talign}
\end{exampleblock}
\end{frame}