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\begin{document}
Solitaire Language.
\begin{description}
\item[A] $1$
\item[B] $011$
\item[C] $110$
\item[D] $11(01)^*00(10)^*11$
\item[E] $11(01)^*00(11)^+(10)^*11$
\item[F] $11(01)^*(11)^+00(10)^*11$
\item[G] $11(01)^*(11)^*1011(10)^*11$
\item[H] $11(01)^*1101(11)^*(10)^*11$
\item[I] $11(01)^*(11)^*01$
\item[J] $10(11)^*(10)^*11$
\end{description}
\newpage
Hopping:\\
$1101\to 0011$ and so $11(01)^*\to 11$.\\
Similarly, $(10)^*11\to 11$.

So we can reduce the above cases to
\begin{description}
\item[A] $1$
\item[B] $011$
\item[C] $110$
\item[D] $110011$
\item[E] $1100(11)^+$
\item[F] $(11)^+0011$
\item[G] $(11)^+101111$
\item[H] $111101(11)^*$
\item[I] $110$
\item[J] $011$
\end{description}
Then
\begin{itemize}
\item $B\to A$
\item $C\to A$
\item $D\to I$
\item $E\to J$
\item $F\to I$
\item $G\to F$
\item $H\to E$
\end{itemize}
\newpage
Unhopping\\
$001\to 110$ or $100\to 011$.\\
So,
\begin{itemize}
\item $A\gets B\vee C$
\item $B\gets I\vee J$
\item $C\gets I\vee J$
\item $D\gets D\vee E\vee F$
\item $E\gets E\vee H$
\item $F\gets F\vee G$
\item $G\gets G$
\item $H\gets H$
\item $I\gets I\vee D\vee F$
\item $J\gets J\vee E\vee D$
\end{itemize}

\end{document}
