\begin{solution}{2}
    \begin{enumerate}
        \item \begin{align*}
                & q_0BabcabB \vdash Bq_1abcabB \\\vdash& Baq_1bcabB \vdash Babq_1cabB \\\vdash& Baq_2bcabB \vdash Bq_2aacabB \\\vdash& q_2BbacabB.
            \end{align*}
        \item \begin{align*}
                & q_0BababB \vdash Bq_1ababB \\\vdash& Baq_1babB \vdash Babq_1abB \\\vdash& Babaq_1bB \vdash Bababq_1B \\\vdash& BababBq_1B \dots
            \end{align*}
        \item See figure \ref{fig:9.2c}.
            
            \begin{figure*}[t]
                \centering
                \begin{tikzpicture}[->, node distance=2cm]
	                \node[state,initial] (q0) {$q_0$};
	                \node[state] (q1) [right=of q0] {$q_1$};
	                \node[state] (q2) [right=of q1] {$q_2$};
                    \draw (q0) -- node[above] {$B/B,R$} ++ (q1);
                    \draw (q1) edge[loop above] node[above,align=left] {$b/b,R$\\$a/a,R$\\$B/B,R$} (q1);
                    \draw (q1) -- node[above] {$c/c,L$} ++ (q2);
                    \draw (q2) edge[loop above] node[above,align=left] {$b/a,L$\\$a/b,L$} (q2);
                \end{tikzpicture}
                \caption{The state machine of exercise 9.2c.}
                \label{fig:9.2c}
            \end{figure*}
        \item Replace all $a$s with $b$s and vice versa on the left of the first $c$, if it exists. If it does not exist, the string is not modified but the machine never terminates.
    \end{enumerate}
\end{solution}