\begin{minipage}[t]{8in}
\begin{tabular}{ll}
void  & insert\index{Set!insert}(Item e);
\end{tabular}\\
 
 \parbox[b]{2in} \ \ 
\begin{minipage}[t]{4in}If the reference count of \verb+store+ is greater
			   than 1, then other collection objects are also 
			   using
			   it as their implementation.  Rather than changing
			   its contents, a copy will be made ($O(n)$), 
			   and the reference
			   counts will be updated.  Finally, check to 
			   ensure that $e$ is not currently a member, then
			   call \rar{store}{insert(e)}. to insert $e$.\end{minipage}
\end{minipage}
\vspace{0.2in}


\begin{minipage}[t]{8in}
\begin{tabular}{ll}
void  & append\index{Set!append}(Item e);
\end{tabular}\\
 
 \parbox[b]{2in} \ \ 
\begin{minipage}[t]{4in}After considering reference counts as above, 
			   check to ensure that $e$ is not a member, then
			   call \rar{store}{append(e)}. \end{minipage}
\end{minipage}
\vspace{0.2in}


\begin{minipage}[t]{8in}
\begin{tabular}{ll}
void  & remove\index{Set!remove}(Item e);
\end{tabular}\\
 
 \parbox[b]{2in} \ \ 
\begin{minipage}[t]{4in}After considering reference counts as above, 
			   remove $e$ from \verb+store+.\end{minipage}
\end{minipage}
\vspace{0.2in}


\begin{minipage}[t]{8in}
\begin{tabular}{ll}
int  & occurrences\index{Set!occurrences}(Item);
\end{tabular}\\
 
			
 \parbox[b]{2in} \ \ 
\begin{minipage}[t]{4in} Return 1 or 0.  $O(1)$.  \end{minipage}
\end{minipage}
\vspace{0.2in}
 

\begin{minipage}[t]{8in}
\begin{tabular}{ll}
int  & permutationQ\index{Set!permutationQ}();
\end{tabular}\\
 
			
 \parbox[b]{2in} \ \ 
\begin{minipage}[t]{4in} Always returns TRUE. $O(1)$.\end{minipage}
\end{minipage}
\vspace{0.2in}

