\documentclass{article}

\input{exer-defs}

%\renewcommand{\thepage}{}
\newcommand{\app}{\mathbin{\mathtt{@}}}
\newcommand{\rev}{\mathtt{rev}}
\newcommand{\hd}{\mathtt{hd}}
\newcommand{\length}{\mathtt{length}}
\newcommand{\cons}{\mathbin{\mathtt{::}}}
\newcommand{\xs}{\mathit{xs}}
\newcommand{\ys}{\mathit{ys}}
\newcommand{\ws}{\mathit{ws}}
\newcommand{\zs}{\mathit{zs}}
\newcommand{\us}{\mathit{us}}
\newcommand{\vs}{\mathit{vs}}
\newcommand{\ns}{\mathit{ns}}
\newcommand{\ms}{\mathit{ms}}
\newcommand{\ls}{\mathit{ls}}
\newcommand{\zss}{\mathit{zss}}
\newcommand{\myinsert}{\mathtt{insert}}
\newcommand{\scanZeroSum}{\mathtt{scanZeroSum}}
\newcommand{\extend}{\mathtt{extend}}
\newcommand{\bal}{\mathtt{bal}}
\newcommand{\balanced}{\mathtt{balanced}}
\newcommand{\splitRev}{\mathtt{splitRev}}
\newcommand{\None}{\mathtt{None}}
\newcommand{\Some}{\mathtt{Some}}

\begin{document}
\begin{center}\large\bf
CIS 705 --- Programming Languages --- Spring 2009
\end{center}
\begin{center}\Large\bf
Assignment 3
\end{center}
\begin{center}\large\bf
Model Answers to Exercises 1 and 3
\end{center}

\section*{Exercise 1}
\renewcommand{\thesection}{1}
\theoremnumber{1}

\subsection*{Inefficient Version (\texttt{balanced-inefficient.ml})}

The ``inefficient'' version of the \texttt{balanced} function uses the
function \texttt{sublists} to generate all the sublists of its input,
$\mathit{ns}$, listed in strictly ascending order.  It then
filters-out those sublists that don't satisfy the predicate
\texttt{isBalanced}.  When called with a list $\mathit{xs}$,
\texttt{isBalanced} returns \texttt{false}, if its input is empty or
doesn't sum to $0$.  Otherwise, it uses \texttt{sublists} to generate
all the sublists of $\mathit{xs}$.  It then checks that each of these
sublists is either empty, has the same length as $\mathit{xs}$ (and so
is $\mathit{xs}$), or has a non-$0$ sum.  As soon as a nonempty,
proper sublist with a sum of $0$ is found, it returns \texttt{false}.

The inefficiency of this version of \texttt{balanced} is mainly due to
its reliance on generating all the sublists of lists.  For example,
the list consisting of the first $200$ natural numbers has 20,101
sublists, the list consisting of the first $300$ natural numbers
has 45,151, and my attempt at running \texttt{sublists} with
the first $1000$ natural numbers resulted in a stack overflow.

As a first example, let's consider how \texttt{balanced} works with
input $\mathit{ns}=[1;2;3;4;5;-12;3;4;5;-15]$.  There are $50$
sublists of this list:
\begin{quotation}
\raggedright
\noindent
$[\,]$, $[-15]$, $[-12]$, $[-12; 3]$, $[-12; 3;
4]$, $[-12; 3; 4; 5]$, $[-12; 3; 4; 5; -15]$, $[1]$, $[1; 2]$, $[1; 2;
3]$, $[1; 2; 3; 4]$, $[1; 2; 3; 4; 5]$, $[1; 2; 3; 4; 5; -12]$, $[1;
2; 3; 4; 5; -12; 3]$, $[1; 2; 3; 4; 5; -12; 3; 4]$,\\
$[1; 2; 3; 4; 5; -12; 3; 4; 5]$, $[1; 2; 3; 4; 5; -12; 3; 4; 5; -15]$,
$[2]$, $[2; 3]$, $[2; 3; 4]$, $[2; 3; 4; 5]$, $[2; 3; 4; 5; -12]$,
$[2; 3; 4; 5; -12; 3]$, $[2; 3; 4; 5; -12; 3; 4]$, $[2; 3; 4; 5; -12;
3; 4; 5]$, \\
$[2; 3; 4; 5; -12; 3; 4; 5; -15]$, $[3]$, $[3; 4]$, $[3; 4; 5]$, $[3;
4; 5; -15]$, $[3; 4; 5; -12]$, $[3; 4; 5; -12; 3]$, $[3; 4; 5; -12; 3;
4]$, $[3; 4; 5; -12; 3; 4; 5]$, $[3; 4; 5; -12; 3; 4; 5; -15]$, $[4]$,
$[4; 5]$, $[4; 5; -15]$, $[4; 5; -12]$, $[4; 5; -12; 3]$, $[4; 5; -12;
3; 4]$, $[4; 5; -12; 3; 4; 5]$, $[4; 5; -12; 3; 4; 5; -15]$, $[5]$,
$[5; -15]$, $[5; -12]$, $[5; -12; 3]$, $[5; -12; 3; 4]$, $[5; -12; 3;
4; 5]$, $[5; -12; 3; 4; 5; -15]$.
\end{quotation}
When the predicate \texttt{isBalanced} is called on these $50$ lists,
all but the following $5$ are rejected for either being empty or not
summing to $0$:
\begin{quotation}
\noindent
$[-12; 3; 4; 5]$, $[1; 2; 3; 4; 5; -12; 3; 4; 5; -15]$, $[3; 4; 5; -12]$,
$[4; 5; -12; 3]$, $[5; -12; 3; 4]$.
\end{quotation}
For each element $\mathit{xs}$ of this list, \texttt{isBalanced} is
forced to generate all the sublists of $\mathit{xs}$,
verifying that each such sublist
$\mathit{ys}$ is either empty, $\mathit{xs}$ or
has a non-$0$ sum.  All but the second of the five candidates check out,
and, of course, the second candidate is $\mathit{ns}$ itself,
which has all of the other four sublists as nonempty, proper sublists
with sum $0$.
Note that the list of all sublists of $\mathit{ns}$ was computed twice:
once by \texttt{balanced}, and once by \texttt{isBalanced}.

As a second example, let's consider how \texttt{balanced}
works with $\mathit{ns}=[1; 2; 3; 0; -3; -2; -1]$.
There are $29$ sublists of
this list:
\begin{quotation}
\raggedright
\noindent
$[\,]$, $[-3]$, $[-3; -2]$, $[-3; -2; -1]$, $[-2]$, $[-2; -1]$,
$[-1]$, $[0]$, $[0; -3]$, $[0; -3; -2]$, $[0; -3; -2; -1]$, $[1]$,
$[1; 2]$, $[1; 2; 3]$, $[1; 2; 3; 0]$, $[1; 2; 3; 0; -3]$, $[1; 2; 3;
0; -3; -2]$, $[1; 2; 3; 0; -3; -2; -1]$, $[2]$, $[2; 3]$, $[2; 3; 0]$,
$[2; 3; 0; -3]$, $[2; 3; 0; -3; -2]$, $[2; 3; 0; -3; -2; -1]$, $[3]$,
$[3; 0]$, $[3; 0; -3]$, $[3; 0; -3; -2]$, $[3; 0; -3; -2; -1]$.
\end{quotation}
When the predicate \texttt{isBalanced} is called on these $29$ lists,
all but the following $4$ are rejected for either being empty or not
summing to $0$:
\begin{quotation}
\noindent
$[0]$, $[1; 2; 3; 0; -3; -2; -1]$, $[2; 3; 0; -3; -2]$, $[3; 0; -3]$.
\end{quotation}
For each element $\mathit{xs}$ of this list, \texttt{isBalanced} is
forced to generate all the sublists of $\mathit{xs}$, verifying that
each such sublist $\mathit{ys}$ is either empty, $\mathit{xs}$ or has
a non-$0$ sum.  This time, only the first candidate checks out.  The
sublists of $[0]$---i.e., $[\,]$ and $[0]$---are either empty or $[0]$
itself.  All of the other candidates have $[0]$ as a nonempty, proper
sublist.

\subsection*{Efficient Version (\texttt{balanced.ml})}

The ``efficient'' version of the \texttt{balanced} function uses
the tail-recursive auxiliary function \texttt{bal} to do its work.
\texttt{bal} has three arguments:
\begin{itemize}
\item $\mathit{ms}$, which is the suffix of the overall input,
  $\mathit{ns}$, that remains to be processed;

\item $\mathit{xs}$, which is the reversal of the longest nice (having
  no nonempty sublists summing to $0$) suffix of the part of
  $\mathit{ns}$ that's already been processed; and

\item $\mathit{zss}$, which is all the balanced sublists of the
  part of $\mathit{ns}$ that's already been processed, listed in
  strictly ascending order.
\end{itemize}
To begin with, $\mathit{ms}=\mathit{ns}$, $\mathit{xs}=[\,]$ and
$\mathit{zss}=[\,]$, and when $\mathit{ms}$ becomes empty,
$\mathit{zss}$ is \texttt{balanced}'s answer.

When $\mathit{ms}$ is nonempty, the \texttt{extend} function is used
to process its next element, $m$.  The function \texttt{scanZeroSum}
is first used to look for a prefix of $\mathit{xs}$ such that
the sum of $m$ and the prefix's sum is $0$.
\begin{itemize}
\item If there is no such prefix, then \texttt{extend} communicates
  back to \texttt{bal} that no new balanced sublists of $\mathit{ns}$
  have been found, and that the new version of $\mathit{xs}$ will be
  $m\cons\mathit{xs}$, and then \texttt{bal} iterates, replacing
  $\mathit{ms}$ with its tail, $\mathit{xs}$ with $m\cons\mathit{xs}$,
  and leaving $\mathit{zss}$ unchanged.  Only one list cell needed to
  be created in this case, but length of $\mathit{xs}$
  additions were needed.  All of the processing is done
  tail-recursively.  In the worst case, $\mathit{xs}$ can grow to be
  as long as $\mathit{ns}$.

\item On the other hand, if there is a prefix $\mathit{ys}$ of $xs$
  such that $m$ plus the sum of $\mathit{ys}$ is $0$, then
  \texttt{extend} communicates back to \texttt{bal} that the reversal of
  $m\cons\mathit{ys}$ is a newly found balanced sublist of $\mathit{ns}$
  and that all but the last element of $m\cons\mathit{ys}$ should be
  the new version of $\mathit{xs}$, and then
  \texttt{bal} iterates, replacing $\mathit{ms}$ with its tail,
  $\mathit{xs}$ with all but the last element of $m\cons\mathit{ys}$, and
  adding the reversal of $m\cons\mathit{ys}$ to $\mathit{zss}$.
  It takes two times the length of $\mathit{ys}$ plus $2$ list
  cell allocations to build the new version of $\mathit{xs}$
  and find the new element of $\mathit{zss}$.  And
  all of this processing is done tail-recursively, except for
  the insertion of the new answer into $\mathit{zss}$.
  If $\mathit{ys}$ is short (when $m=0$, it will be empty), then
  the new version of $\mathit{xs}$ will also be short, making subsequent
  steps more efficient.
\end{itemize}

As a first example (also considered with the inefficient version),
let's consider how \texttt{balanced} works with input
$\mathit{ns}=[1;2;3;4;5;-12;3;4;5;-15]$.  We start out with
\begin{displaymath}
\mathit{ms}=[1; 2; 3; 4; 5; -12; 3; 4; 5; -15],\qquad
\mathit{xs}=[\,],\qquad
\mathit{zss}=[\,].
\end{displaymath}
The first five steps simply add elements to the front of
$\mathit{xs}$.  First, $1$ is added, because there is
no prefix of $\mathit{xs}$ whose sum plus $1$ is $0$,
\begin{displaymath}
\mathit{ms}=[2; 3; 4; 5; -12; 3; 4; 5; -15],\qquad
\mathit{xs}=[1],\qquad
\mathit{zss}=[\,],
\end{displaymath}
Next $2$ is added,
\begin{displaymath}
\mathit{ms}=[3; 4; 5; -12; 3; 4; 5; -15],\qquad
\mathit{xs}=[2; 1],\qquad
\mathit{zss}=[\,].
\end{displaymath}
Then $3$ is added,
\begin{displaymath}
\mathit{ms}=[4; 5; -12; 3; 4; 5; -15],\qquad
\mathit{xs}=[3; 2; 1],\qquad
\mathit{zss}=[\,].
\end{displaymath}
Then $4$ is added,
\begin{displaymath}
\mathit{ms}=[5; -12; 3; 4; 5; -15],\qquad
\mathit{xs}=[4; 3; 2; 1],\qquad
\mathit{zss}=[\,].
\end{displaymath}
And finally $5$ is added,
\begin{displaymath}
\mathit{ms}=[-12; 3; 4; 5; -15],\qquad
\mathit{xs}=[5; 4; 3; 2; 1],\qquad
\mathit{zss}=[\,].
\end{displaymath}
Because the next element of $\mathit{ms}$ is now $-12$, and
$[5;4;3]$ is a prefix of $\mathit{xs}$ whose sum plus
$-12$ is $0$, in the next step, $\mathit{xs}$ becomes all
but the last element of $[-12;5;4;3]$, and the reversal of
$[-12;5;4;3]$ is inserted into $\mathit{zss}$,
\begin{displaymath}
\mathit{ms}=[3; 4; 5; -15],\qquad
\mathit{xs}=[-12; 5; 4],\qquad
\mathit{zss}=[[3; 4; 5; -12]].
\end{displaymath}
Because the next element of $\mathit{ms}$ is now $3$, and
$[-12;5;4]$ is a prefix of $\mathit{xs}$ whose sum plus
$3$ is $0$, in the next step, $\mathit{xs}$ becomes all
but the last element of $[3;-12;5;4]$, and the reversal of
$[3;-12;5;4]$ is inserted into $\mathit{zss}$,
\begin{displaymath}
\mathit{ms}=[4; 5; -15],\qquad
\mathit{xs}=[3; -12; 5],\qquad
\mathit{zss}=[[3; 4; 5; -12]; [4; 5; -12; 3]].
\end{displaymath}
Because the next element of $\mathit{ms}$ is now $4$, and
$[3;-12;5]$ is a prefix of $\mathit{xs}$ whose sum plus
$4$ is $0$, in the next step, $\mathit{xs}$ becomes all
but the last element of $[4;3;-12;5]$, and the reversal of
$[4;3;-12;5]$ is inserted into $\mathit{zss}$,
\begin{displaymath}
\mathit{ms}=[5; -15],\qquad
\mathit{xs}=[4; 3; -12],\qquad
\mathit{zss}=[[3; 4; 5; -12]; [4; 5; -12; 3]; [5; -12; 3; 4]].
\end{displaymath}
Because the next element of $\mathit{ms}$ is now $5$, and
$[4;3;-12]$ is a prefix of $\mathit{xs}$ whose sum plus
$5$ is $0$, in the next step, $\mathit{xs}$ becomes all
but the last element of $[5;4;3;-12]$, and the reversal of
$[5;4;3;-12]$ is inserted into $\mathit{zss}$,
\begin{displaymath}
\mathit{ms}=[-15],\qquad
\mathit{xs}=[5; 4; 3],\qquad
\mathit{zss}=[[-12; 3; 4; 5]; [3; 4; 5; -12]; [4; 5; -12; 3]; [5; -12; 3; 4]].
\end{displaymath}
Because the next element of $\mathit{ms}$ is now $-15$, and
there is no prefix of $\mathit{xs}$ whose sum plus $-15$ is
$0$, in the next step, $-15$ is simply added to the
front of $\mathit{xs}$,
\begin{displaymath}
\mathit{ms}=[\,],\qquad
\mathit{xs}=[-15; 5; 4; 3],\qquad
\mathit{zss}=[[-12; 3; 4; 5]; [3; 4; 5; -12]; [4; 5; -12; 3]; [5; -12; 3; 4]].
\end{displaymath}
Finally, $\mathit{ms}$ is empty, so the result is $\mathit{zss}$.

As a second example (also considered with the inefficient version),
let's consider how \texttt{balanced} works with input
$\mathit{ns}=[1; 2; 3; 0; -3; -2; -1]$.  We start out with
\begin{displaymath}
\mathit{ms}=[1; 2; 3; 0; -3; -2; -1],\qquad
\mathit{xs}=[],\qquad
\mathit{zss}=[].
\end{displaymath}
First, we add $1$ to $\mathit{xs}$,
\begin{displaymath}
\mathit{ms}=[2; 3; 0; -3; -2; -1],\qquad
\mathit{xs}=[1],\qquad
\mathit{zss}=[].
\end{displaymath}
Next, we add $2$ to $\mathit{xs}$,
\begin{displaymath}
\mathit{ms}=[3; 0; -3; -2; -1],\qquad
\mathit{xs}=[2; 1],\qquad
\mathit{zss}=[].
\end{displaymath}
Next, we add $3$ to $\mathit{xs}$,
\begin{displaymath}
\mathit{ms}=[0; -3; -2; -1],\qquad
\mathit{xs}=[3; 2; 1],\qquad
\mathit{zss}=[].
\end{displaymath}
Because the next element of $\mathit{ms}$ is now $0$, and
$[\,]$ is a prefix of $\mathit{xs}$ whose sum plus
$0$ is $0$, in the next step, $\mathit{xs}$ becomes all
but the last element of $[0]$, and the reversal of
$[0]$ is inserted into $\mathit{zss}$,
\begin{displaymath}
\mathit{ms}=[-3; -2; -1],\qquad
\mathit{xs}=[],\qquad
\mathit{zss}=[[0]].
\end{displaymath}
The rest of the steps add $-3$, $-2$ and $-1$ to $\mathit{xs}$, without
adding anything to $\mathit{zss}$.

\section*{Exercise 3}
\renewcommand{\thesection}{3}
\theoremnumber{1}

\subsection*{Preliminaries}

In what follows, we'll abbreviate \texttt{List.rev}, \texttt{List.hd}
and \texttt{List.length} to \texttt{rev}, \texttt{hd} and
\texttt{length}, respectively.
$\length$ is defined so that, for all lists $\xs$ and $\ys$ of values
of the same type, $\length(\xs\app\ys)=\length\,\xs + \length\,\ys$.
And $\rev$ is defined so that, for all lists $\xs$ and $\ys$ of values
of the same type, $\rev(\xs\app\ys) = \rev\,\ys \app \rev\,\xs$.
We'll use the following lemma about $\rev$ repeatedly and without
citation:

\begin{lemma}
\label{RevLem}
\begin{enumerate}[\quad(1)]
\item For all lists $\xs$, $\rev(\rev\,\xs)=\xs$.

\item For all lists $\xs$ and $\ys$ of values of the same type,
  $\xs=\ys$ iff $\rev\,\xs = \rev\,\ys$.

\item For all lists $\xs$ and $\ys$ of values of the same type,
  $\xs = \rev\,ys$ iff $\rev\,\xs = \ys$.

\item For all lists $\xs$, $\length\,\xs=\length(\rev\,\xs)$.
\end{enumerate}
\end{lemma}

\begin{proof}
(1) can be proved by induction on $\xs$. (2) and (3) follow
immediately from (1). And (4) follows by induction on $\xs$.
\end{proof}

\begin{lemma}
\label{SuffixesOfNiceLem}
For all nice lists $\xs$ and suffixes $\us$ and $\vs$ of $\xs$, if
$\us$ and $\vs$ have the same sum, then $\us=\vs$.
\end{lemma}

\begin{proof}
Suppose $\xs$ is a nice list and $\us$ and $\vs$ are suffixes of
$\xs$ with the same sum.  We must show that $\us=\vs$.  We'll
consider the case where $\us$ is no longer than $\vs$, the other
case being symmetric.  Thus $\vs=\ws\app\us$ for some $\ws$.  Since
$\ws$ is a sublist of $\vs$, which is a sublist of $\xs$, we have
that $\ws$ is a sublist of $\xs$.  Since $\us$ and $\vs$ have the
same sum, and the sum of $\vs$ is the sum of $\ws$ plus the sum of
$\us$, it follows that $\ws$ has a sum of $0$.  Thus, because $\xs$
is nice and $\ws$ is a sublist of $\xs$, we have that $\ws=[\,]$, so
that $\us = [\,]\app\us = \ws\app\us = \vs$.
\end{proof}

\begin{lemma}
\label{BalancedSuffixLem}
There is at most one balanced suffix of a list of integers $\ns$.
\end{lemma}

\begin{proof}
  Suppose $\us$ and $\vs$ are balanced suffixes of a list of integers
  $\ns$.  We must show that they are equal.  We'll consider the case
  where $\us$ is no longer than $\vs$, the other case being symmetric.
  Thus $\vs=\ws\app\us$ for some $\ws$.  Because $\us$ and $\vs$ both
  sum to $0$, the sum of $\ws$ must also be $0$.  Thus, since
  $\ws\app\us=\vs$ is balanced, we have that either $\ws=[\,]$ or
  $\us=[\,]$, as otherwise both of $\ws$ and $\us$ would be proper,
  nonempty sublists of $\vs$ with sum $0$.  In the first case, we have
  that $\us = [\,]\app\us = \ws\app\us = \vs$.  And, in the second
  case, we have that $[\,]=\us$ is balanced---contradiction.  Thus
  $\us=\vs$.
\end{proof}

\begin{lemma}
\label{NicePlusNonZeroImpliesBalancedLem}
If $\xs$ is a nice list, $n$ is a nonzero integer, and the sum of $\xs$ plus
$n$ is $0$, then $\xs\app[n]$ is balanced.
\end{lemma}

\begin{proof}
Suppose $\xs$ is a nice list, $n$ is a nonzero integer, and the sum
of $\xs$ plus $n$ is $0$.  We must show that $\xs\app[n]$ is balanced.
Because it clearly has a sum of $0$ and is nonempty, it remains to
show that it has no proper, nonempty sublist with sum $0$.  Suppose,
toward a contradiction, that $\us$ is a proper, nonempty sublist of
$\xs\app[n]$ with sum $0$.  There are two cases to consider.
\begin{itemize}
\item Suppose $\us$ is a sublist of $\xs$.  Because $\us$ is nonempty
  and sums to $0$, and $\xs$ is nice, we have a contradiction.

\item Suppose $\us$ is a suffix of $\xs\app[n]$.  Because $\us$ is
  nonempty, there is a suffix $\vs$ of $\xs$ such that
  $\us=\vs\app[n]$.  Because $\us$ sums to $0$, it follows that $\vs$
  sums to $-n$.  But the sum of $\xs$ plus $n$ is $0$, and thus $\xs$
  also sums to $-n$.  Because $\xs$ and $\vs$ are suffixes of the same
  nice list ($\xs$) and have the same sum,
  Lemma~\ref{SuffixesOfNiceLem} tells us that $\xs=\vs$.  But then
  $\us=\vs\app[n]=\xs\app[n]$, contradicting that $\us$ is a proper
  sublist of $\xs\app[n]$.
\end{itemize}
\end{proof}

\begin{lemma}
\label{NiceCharactLem}
Suppose $\ns$ and $\xs$ are lists of integers.  The following statements
are equivalent:
\begin{enumerate}[\quad(1)]
\item the reversals of the prefixes of $\xs$ are the nice suffixes of
  $\ns$;

\item $\rev\,\xs$ is the longest nice suffix of $\ns$.
\end{enumerate}
\end{lemma}

\begin{proof}
Suppose that $\ns$ and $\xs$ are lists of integers.
\begin{itemize}
\item ((1) implies (2))\quad Suppose the reversals of the prefixes of
  $\xs$ are the nice suffixes of $\ns$.  We must show that $\rev\,\xs$ is
  the longest nice suffix of $\ns$.  Because $\xs$ is
  a prefix of itself, $\rev\,\xs$ is a nice suffix of $\ns$.  Suppose,
  toward a contradiction, that there is nice suffix $\ms$ of $\ns$
  that is longer than $\rev\,\xs$.  Then $\ms=\rev\,\ys$ for some
  prefix $\ys$ of $\xs$.  Because $\rev\,\ys=\ms$ is longer than
  $\rev\,\xs$, we have that $\ys=\rev(\rev\,\ys)$ is longer than
  $\xs=\rev(\rev\,\xs)$---contradicting the fact that $\ys$ is a
  prefix of $\xs$.  Thus $\rev\,\xs$ is the longest nice suffix of
  $\ns$.

\item ((2) implies (1))\quad Suppose $\rev\,\xs$ is the longest nice
  suffix of $\ns$.  We must show that the reversals of the prefixes of
  $\xs$ are the nice suffixes of $\ns$.

  First, suppose that $\ys$ is a prefix of $\xs$.  Then
  $\xs=\ys\app\us$ for some $\us$, so that
  $\rev\,\us\app\rev\,\ys=\rev(\ys\app\us)=\rev\,\xs$.
  Because $\rev\,\ys$ is a suffix of $\rev\,\xs$, which is in
  turn a suffix of $\ns$, we have that $\rev\,\ys$ is a suffix
  of $\ns$.  And since $\rev\,\ys$ is a sublist of the nice
  list $\rev\,\xs$, it follows that $\rev\,\ys$ is nice, as any nonempty
  sublist of $\rev\,\ys$ with a sum of $0$ would also be a nonempty
  sublist of $\rev\,\xs$.  Thus $\rev\,\ys$ is a nice suffix of $\ns$.

  Second, suppose that $\ms$ is a nice suffix of $\ns$.  Because
  $\rev\,\xs$ is the longest nice suffix of $\ns$, it follows that
  $\ms$ is a suffix of $\rev\,\xs$.  Thus there is a $\us$ such that
  $\rev\,\xs = \us\app\ms$, so that $\xs = \rev(\rev\,\xs) =
  \rev(\us\app\ms) = \rev\,\ms\app\rev\,\us$.  Hence $\rev\,\ms$ is
  a prefix of $\xs$ whose reversal is $\ms$.
\end{itemize}
\end{proof}

\subsection*{Correctness Proof for \texttt{extend}}

Suppose $\ns$ is a list of integers, $n$ is an integer, and the
reversals of the prefixes of $\xs$ are the nice suffixes of $\ns$.  By
Lemma~\ref{NiceCharactLem}, we have that $\rev\,\xs$ is the longest
nice suffix of $\ns$.  There are two cases to consider.
\begin{itemize}
\item Suppose there is no prefix of $\xs$ whose sum, when added to
  $n$, yields $0$.  Then $\scanZeroSum\,n\,\xs$ returns $\None$, so
  that $\extend\,n\,\xs$ returns $(\None, n\cons\xs)$.  Thus we must
  show that:
  \begin{enumerate}[\quad(1)]
  \item there is no balanced suffix of $\ns\app[n]$, and

  \item the reversals of the prefixes of $n\cons\xs$ are the nice
    suffixes of $\ns\app[n]$, which, by Lemma~\ref{NiceCharactLem}, is
    equivalent to showing that $\rev\,xs\app[n]=\rev(n\cons\xs)$ is
    the longest nice suffix of $\ns\app[n]$.
  \end{enumerate}
  We have that $n\neq 0$, since otherwise $[\,]$ would be a prefix of
  $\xs$ whose sum, when added to $n$, yields $0$.

  Suppose, toward a contradiction, that there is a balanced suffix of
  $\ns\app[n]$.  Because balanced lists are nonempty, it follows that
  there is a suffix $\ms$ of $\ns$ such that $\ms\app[n]$ is balanced.
  Because $\ms$ is a proper sublist of $\ms\app[n]$, it follows that
  $\ms$ is nice (a nonempty sublist of $\ms$ with sum $0$ would
  be a proper, nonempty sublist of $\ms\app[n]$).
  Thus, since $\ms$ is a nice suffix of $\ns$, we have
  that $\rev\,\ms$ is a prefix of $\xs$.  Because $\ms\app[n]$ is
  balanced, we have that the sum of $\ms$, when added to $n$, yields
  $0$.  But then the sum of $\rev\,\ms$, when added to $n$, also
  yields $0$, so that there is prefix of $\xs$ whose sum, when added
  to $n$, yields $0$---contradiction.  Thus there is no balanced
  suffix of $\ns\app[n]$, i.e., (1) holds.

  It remains to show (2).  Suppose, toward a contradiction, that
  $\rev\,\xs\app[n]$ is not nice.  Because $n\neq 0$ and $\rev\,\xs$
  is nice, it follows that there is a nonempty suffix $\us$ of
  $\rev\,\xs$ such that the sum of $\us$, when added to $n$, yields
  $0$.  Because $\us$ is a sublist of $\rev\,\xs$, and $\rev\,\xs$ is
  nice, we have that $\us$ is nice.  Thus
  Lemma~\ref{NicePlusNonZeroImpliesBalancedLem} tells us that
  $\us\app[n]$ is balanced. Since $\us$ is a suffix of $\rev\,\xs$,
  and $\rev\,xs$ is a suffix of $\ns$, we have that $\us$ is a suffix
  of $\ns$, so that $\us\app[n]$ is a balanced suffix of
  $\ns\app[n]$---contradiction.  Thus we have that $\rev\,\xs\app[n]$
  is a nice suffix of $\ns\app[n]$.

  Finally, suppose, toward a contradiction, that there is a nice
  suffix $\us$ of $\ns\app[n]$ that is longer than $\rev\,\xs\app[n]$.
  Then $\us=\ms\app[n]$, where $\ms$ is a longer suffix of $\ns$ than
  $\rev\,xs$.  Because $\us$ is nice and $\ms$ is a sublist of $\us$,
  we have that $\ms$ is nice, so that $\ms$ is a nice suffix of $\ns$
  that is longer than $\rev\,xs$---contradiction.  This concludes the
  proof of (2).

\item Suppose there is a prefix of $\xs$ whose sum, when added to $n$,
  yields $0$.  Let $\ys$ be the shortest such prefix.
  Then $\scanZeroSum\,n\,\xs$ returns
  $\Some\,i$, where $i$ is the length of $\ys$.  Since $0\leq i$ and
  $i$ is less-than-or-equal-to the length of $n\cons\xs$,
  $\splitRev\,i\,(n\cons\xs)$ returns $(\us,\vs)$, where
  $\us$ is the reversal of the first $i$ elements of $n\cons\xs$,
  and $\vs$ is the remaining elements of $n\cons\xs$.
  Thus $\extend\,n\,\xs$ returns $(\Some(\hd\,\vs\cons\us), \rev\,\us)$.

  Because $i$ is the length of $\ys$, and $\ys$ is a prefix of $\xs$,
  it follows that $\rev\,\us$ is all but the last element of
  $n\cons\ys$, and that $\hd\,\vs$ is the last element of $n\cons\ys$.
  Thus $n\cons\ys = \rev\,\us\app[\hd\,\vs]$.  Hence $\rev(n\cons\ys)
  = \rev(\rev\,\us\app[\hd\,\vs]) = \rev[\hd\,\vs]\app\rev(\rev\,\us)
  = [\hd\,\vs]\app\us = \hd\,\vs\cons\us$.  Thus $\extend\,n\,\xs$
  returns $\Some(\rev(n\cons\ys))$ paired with all but the last
  element of $n\cons\ys$.

  Hence we must show that:
  \begin{enumerate}[\quad(1)]
  \item $\rev(n\cons\ys)$ is the unique balanced suffix of
    $\ns\app[n]$, which, by Lemma~\ref{BalancedSuffixLem}, is
    equivalent to showing that $\rev\,\ys\app[n]=\rev(n\cons\ys)$ is
    a balanced suffix of $\ns\app[n]$, and

  \item the reversals of the prefixes of all but the last element of
    $n\cons\ys$ are the nice suffixes of $\ns\app[n]$, which, by
    Lemma~\ref{NiceCharactLem}, is equivalent to showing that the reversal of
    all but the last element of $n\cons\ys$ is the longest nice suffix
    of $\ns\app[n]$.
  \end{enumerate}

  We know that $n$ plus the sum of $\ys$ is $0$, and thus that
  $\rev\,\ys\app[n]$ has a sum of $0$.  And, clearly
  $\rev\,\ys\app[n]$ is nonempty.  So for (1) it remains to show that
  $\rev\,\ys\app[n]$ has no proper, nonempty sublist with a sum of
  $0$.  To this end, suppose, toward a contradiction, that $\zs$ is a
  proper, nonempty sublist of $\rev\,\ys\app[n]$ with a sum of $0$.
  Because $\ys$ is a prefix of $\xs$, $\rev\,\ys$ is a nice suffix of
  $\ns$.  Thus $\zs$ is not a sublist of $\rev\,\ys$, so that
  $\zs=\ws\app[n]$ for some suffix $\ws$ of $\rev\,\ys$.  Since the
  sum of $\zs$ is $0$, we have that the sum of $\ws$ is $-n$.  But,
  since the sum of $\rev\,\ys\app[n]$ is $0$, we also have that the
  sum of $\rev\,\ys$ is $-n$.  Because $\rev\,\ys$ is nice,
  $\rev\,\ys$ and $\ws$ are suffixes of $\rev\,\ys$, and $\rev\,\ys$
  and $\ws$ have identical sums, Lemma~\ref{SuffixesOfNiceLem} tells
  us that $\rev\,\ys=\ws$.  But this means that
  $\zs=\ws\app[n]=\rev\,\ys\app[n]$, showing that $\zs$ is not a
  proper sublist of $\rev\,\ys\app[n]$---contradiction.  This completes the
  proof of (1).

  For (2), there are two cases to consider.
  \begin{itemize}
  \item Suppose $n=0$.  Then $\ys=[\,]$, by its definition.  Hence all
    but the last element of $n\cons\ys$ is $[\,]$.  So we must show
    that $[\,] = \rev[\,]$ is the longest nice suffix of $\ns\app[n]$.
    $[\,]$ is nice and is a suffix of any list.  Furthermore, any
    longer suffix of $\ns\app[n]$ would have $[0]$ as a sublist, and
    so wouldn't be nice.  So $[\,]$ is the longest nice suffix of
    $\ns\app[n]$.

  \item Suppose $n\neq 0$.  Then the sum of $\ys$ is $-n$, so that
    $\ys=\us\app[m]$ for some $\us$ and $m$.  We must show that
    $\rev\,\us\app[n]=\rev(n\cons\us)$ is the longest nice suffix of
    $\ns\app[n]$.  We have that $\rev\,\ys\app[n] =
    \rev(\us\app[m])\app[n] = m\cons\rev\,\us\app[n]$.  From (1), we
    know that $\rev\,\ys\app[n]$ is a balanced suffix of $\ns\app[n]$.
    Since $\rev\,\us\app[n]$ is a suffix of $\rev\,\ys\app[n]$, we
    have that $\rev\,\us\app[n]$ is a suffix of $\ns\app[n]$.
    Furthermore, because $\rev\,\us\app[n]$ is a proper sublist of the
    balanced list $\rev\,\ys\app[n]$, it follows that
    $\rev\,\us\app[n]$ is nice (if $\rev\,\us\app[n]$ had a nonempty
    sublist with sum $0$, then $\rev\,\ys\app[n]$ would have a proper,
    nonempty sublist with sum $0$).  Furthermore, any longer suffix of
    $\ns\app[n]$ would contain $m\cons\rev\,\us\app[n] =
    \rev\,\ys\app[n]$ (which has a sum of $0$) as a sublist, and so
    wouldn't be nice.  Thus $\rev\,\us\app[n]$ is the longest nice suffix of
    $\ns\app[n]$.
  \end{itemize}
\end{itemize}

\subsection*{Correctness Proof for \texttt{balanced}}

Suppose $\ns$ is a list of integers.  First, we'll show that,
under the assumption that the auxiliary function $\bal$ is correct,
$\balanced\;\ns$ returns the balanced sublists of $\ns$, listed in
strictly ascending order.  Then, we'll prove that $\bal$ is correct.

We have that $\ns$ is a suffix of itself, and the prefix of $\ns$ of
length $\length\,\ns - \length\,\ns$ is $[\,]$.  Since the reversals
of the prefixes of $[\,]$ (there is only one, $[\,]$) are the nice
suffixes of $[\,]$ (there is only one, $[\,]$), we have that the
reversals of the prefixes of $[\,]$ are the nice suffixes of the
prefix of $\ns$ of length $\length\,\ns - \length\,\ns$.  Since there
are no balanced sublists of $[\,]$, we have that $[\,]$ is the balanced
sublists of $[\,]$, listed in strictly ascending order, i.e., is the
balanced sublists of the prefix of $\ns$ of length $\length\,\ns -
\length\,\ns$, listed in strictly ascending order.  Thus,
by $\bal$'s specification, $\bal\,\ns\,[\,]\,[\,]$ returns
the balanced sublists of $\ns$, listed in strictly ascending order.
And this, in turn, is what $\balanced\,\ns$ returns.

To prove that $\bal$ is correct, suppose $\ms$ is a suffix of $\ns$,
and the reversals of the prefixes of $\xs$ are the nice suffixes of
the prefix of $\ns$ of length $\length\,\ns-\length\,\ms$, and $\zss$ is
the balanced sublists of the prefix of $\ns$ of length
$\length\,\ns-\length\,\ms$, listed in strictly ascending order.  We
must prove that $\bal\,\ms\,\xs\,\zss$ returns all the balanced
sublists of $\ns$, listed in strictly ascending order.
There are two cases to consider.
\begin{itemize}
\item Suppose $\ms=[\,]$.  Then the prefix of $\ns$ 
  of length $\length\,\ns-\length\,\ms$ is $\ns$, and so
  $\zss$ is the balanced sublists of $\ns$, listed
  in strictly ascending order.  But this is what $\bal$ returns.

\item Suppose $\ms=m\cons\ms'$ for some integer $m$ and list of
  integers $\ms'$.  Because $\ms$ is a suffix of $\ns$, we have that
  $\ms'$ is a suffix of $\ns$.  Let $\ls$ be the prefix of $\ns$ of
  length $\length\,\ns-\length\,\ms$.  Then $\ls\app[m]$ is the prefix
  of $\ns$ of length $\length\,\ns-\length\,\ms'$.  There are
  two subcases to consider.
  \begin{itemize}
  \item Suppose there is no balanced suffix of $\ls\app[m]$.  Because
    the reversals of the prefixes of $\xs$ are the nice suffixes of
    $\ls$, we have that $\extend\,m\,\xs$ returns $(\None,\ys)$, where
    the reversals of the prefixes of $\ys$ are the nice suffixes of
    $\ls\app[m]$.  Because $\zss$ is the balanced sublists of $\ls$,
    listed in strictly ascending order, and there is no balanced
    suffix of $\ls\app[m]$, we have that $\zss$ is the balanced
    sublists of $\ls\app[m]$, listed in strictly ascending order.
    Thus, by the specification of $\bal$, we have that
    $\bal\,\ms'\,\ys\,\zss$ returns the balanced sublists of $\ns$,
    listed in strictly ascending order.  But this is what $\bal$ returns.

  \item Suppose there is a balanced suffix of $\ls\app[m]$.
    Because the reversals of the prefixes of $\xs$ are the nice
    suffixes of $\ls$, we have that $\extend\,m\,\xs$ returns
    $(\Some\,zs,\ys)$, where $\zs$ is the unique balanced suffix of
    $\ls\app[m]$, and the reversals of the prefixes of $\ys$ are the
    nice suffixes of $\ls\app[m]$.
    Since the elements of $\zss$ are sorted in strictly ascending order,
    $\myinsert\,\zs\,\zss$ consists of $\zs$ plus all
    of the elements in $\zss$, listed in strictly ascending order.
    Because $\zss$ is the balanced sublists of $\ls$,
    and $\zs$ is the unique balanced suffix
    of $\ls\app[m]$, it follows that $\myinsert\,\zs\,\zss$ is the balanced
    sublists of $\ls\app[m]$, listed in strictly ascending order.
    Thus, by the specification of $\bal$,
    we have that $\bal\,\ms'\,\ys\,(\myinsert\,\zs\,\zss)$ returns the
    balanced sublists of $\ns$, listed in strictly ascending order.
    But this is what $\bal$ returns.
  \end{itemize}
\end{itemize}
\end{document}