% Chapter Template

\chapter{Regular Expressions and POSIX Lexing-Preliminaries} % Main chapter title

\label{Inj} % In chapter 2 \ref{Chapter2} we will introduce the concepts
%and notations we 
% used for describing the lexing algorithm by Sulzmann and Lu,
%and then give the algorithm and its variant and discuss
%why more aggressive simplifications are needed. 


\marginpar{Gerog comment: addressed attributing work correctly problem,
showed clearly these definitions come from Ausaf et al.}

This is a preliminary chapter which describes the results of
Sulzmann and Lu \cite{Sulzmann2014} and part of the formalisations by
Ausaf et al \ref{AusafDyckhoffUrban2016} (mainly the definitions).
These results are not part of this PhD work,
but the details are included to provide necessary context
for our work on the correctness proof of $\blexersimp$,
as we show in chapter \ref{Bitcoded2} how the proofs break down when
simplifications are applied.

In the coming section, the definitions of basic notions 
for regular languages and regular expressions are given.
This is essentially a description in ``English''
the functions and datatypes used by Ausaf et al. 
\cite{AusafDyckhoffUrban2016} \cite{Ausaf}
in their formalisation in Isabelle/HOL.
We include them as we build on their formalisation,
and therefore inherently use these definitions.


% In the next section, we will briefly
% introduce Brzozowski derivatives and Sulzmann
% and Lu's algorithm, which the main point of this thesis builds on.
%We give a taste of what they 
%are like and why they are suitable for regular expression
%matching and lexing.
\section{Formal Specification of POSIX Matching 
and Brzozowski Derivatives}
%Now we start with the central topic of the thesis: Brzozowski derivatives.
Brzozowski \cite{Brzozowski1964} first introduced the 
concept of a \emph{derivative} of regular expression in 1964.
The derivative of a regular expression $r$
with respect to a character $c$, is written as $r \backslash c$.
This operation tells us what $r$ transforms into
if we ``chop'' off a particular character $c$ 
from all strings in the language of $r$ (defined
later as $L \; r$).
%To give a flavour of Brzozowski derivatives, we present
%two straightforward clauses from it:
%\begin{center}
%	\begin{tabular}{lcl}
%		$d \backslash c$     & $\dn$ & 
%		$\mathit{if} \;c = d\;\mathit{then}\;\ONE\;\mathit{else}\;\ZERO$\\
%$(r_1 + r_2)\backslash c$     & $\dn$ & $r_1 \backslash c \,+\, r_2 \backslash c$\\
%	\end{tabular}
%\end{center}
%\noindent
%The first clause says that for the regular expression
%denoting a singleton set consisting of a single-character string $\{ d \}$,
%we check the derivative character $c$ against $d$,
%returning a set containing only the empty string $\{ [] \}$
%if $c$ and $d$ are equal, and the empty set $\varnothing$ otherwise.
%The second clause states that to obtain the regular expression
%representing all strings' head character $c$ being chopped off
%from $r_1 + r_2$, one simply needs to recursively take derivative
%of $r_1$ and $r_2$ and then put them together.
Derivatives have the property
that $s \in L \; (r\backslash c)$ if and only if 
$c::s \in L \; r$ where $::$ stands for list prepending.
%This property can be used on regular expressions
%matching and lexing--to test whether a string $s$ is in $L \; r$,
%one simply takes derivatives of $r$ successively with
%respect to the characters (in the correct order) in $s$,
%and then test whether the empty string is in the last regular expression.
With this property, derivatives can give a simple solution
to the problem of matching a string $s$ with a regular
expression $r$: if the derivative of $r$ w.r.t.\ (in
succession) all the characters of the string matches the empty string,
then $r$ matches $s$ (and {\em vice versa}).  
%This makes formally reasoning about these properties such
%as correctness and complexity smooth and intuitive.
There are several mechanised proofs of this property in various theorem
provers,
for example one by Owens and Slind \cite{Owens2008} in HOL4,
another one by Krauss and Nipkow \cite{Nipkow98} in Isabelle/HOL, and
yet another in Coq by Coquand and Siles \cite{Coquand2012}.

In addition, one can extend derivatives to bounded repetitions
relatively straightforwardly. For example, the derivative for 
this can be defined as:
\begin{center}
	\begin{tabular}{lcl}
		$r^{\{n\}} \backslash c$     & $\dn$ & $r \backslash c \cdot
		r^{\{n-1\}} (\text{when} n > 0)$\\
	\end{tabular}
\end{center}
\noindent
Experimental results suggest that  unlike DFA-based solutions
for bounded regular expressions,
derivatives can cope
large counters
quite well.

There have also been 
extensions of derivatives to other regex constructs.
For example, Owens et al include the derivatives
for the \emph{NOT} regular expression, which is
able to concisely express C-style comments of the form
$/* \ldots */$ (see \cite{Owens2008}).
Another extension for derivatives is
regular expressions with look-aheads, done
by Miyazaki and Minamide
\cite{Takayuki2019}.
%We therefore use Brzozowski derivatives on regular expressions 
%lexing 



Given the above definitions and properties of
Brzozowski derivatives, one quickly realises their potential
in generating a formally verified algorithm for lexing: the clauses and property
can be easily expressed in a functional programming language 
or converted to theorem prover
code, with great ease.
Perhaps this is the reason why derivatives have sparked quite a bit of interest
in the functional programming and theorem prover communities in the last
fifteen or so years (
\cite{Almeidaetal10}, \cite{Berglund14}, \cite{Berglund18},
\cite{Chen12} and \cite{Coquand2012}
to name a few), despite being buried in the ``sands of time'' \cite{Owens2008}
after they were first published by Brzozowski.


However, there are two difficulties with derivative-based matchers:
First, Brzozowski's original matcher only generates a yes/no answer
for whether a regular expression matches a string or not.  This is too
little information in the context of lexing where separate tokens must
be identified and also classified (for example as keywords
or identifiers). 
Second, derivative-based matchers need to be more efficient in terms
of the sizes of derivatives.
Elegant and beautiful
as many implementations are,
they can be still quite slow. 
For example, Sulzmann and Lu
claim a linear running time of their proposed algorithm,
but that was falsified by our experiments. The running time 
is actually $\Omega(2^n)$ in the worst case.
A similar claim about a theoretical runtime of $O(n^2)$ 
is made for the Verbatim \cite{Verbatim}
%TODO: give references
lexer, which calculates POSIX matches and is based on derivatives.
They formalized the correctness of the lexer, but not their complexity result.
In the performance evaluation section, they analyzed the run time
of matching $a$ with the string 
\begin{center}
	$\underbrace{a \ldots a}_{\text{n a's}}$.
\end{center}
\noindent
They concluded that the algorithm is quadratic in terms of 
the length of the input string.
When we tried out their extracted OCaml code with the example $(a+aa)^*$,
the time it took to match a string of 40 $a$'s was approximately 5 minutes.


\subsection{Sulzmann and Lu's Algorithm}
Sulzmann and Lu~\cite{Sulzmann2014} overcame the first 
problem with the yes/no answer 
by cleverly extending Brzozowski's matching
algorithm. Their extended version generates additional information on
\emph{how} a regular expression matches a string following the POSIX
rules for regular expression matching. They achieve this by adding a
second ``phase'' to Brzozowski's algorithm involving an injection
function. This injection function in a sense undoes the ``damage''
of the derivatives chopping off characters.
In earlier work, Ausaf et al provided the formal
specification of what POSIX matching means and proved in Isabelle/HOL
the correctness
of this extended algorithm accordingly
\cite{AusafDyckhoffUrban2016}.

The version of the algorithm proven correct
suffers however heavily from a 
second difficulty, where derivatives can
grow to arbitrarily big sizes. 
For example if we start with the
regular expression $(a+aa)^*$ and take
successive derivatives according to the character $a$, we end up with
a sequence of ever-growing derivatives like 

\def\ll{\stackrel{\_\backslash{} a}{\longrightarrow}}
\begin{center}
\begin{tabular}{rll}
$(a + aa)^*$ & $\ll$ & $(\ONE + \ONE{}a) \cdot (a + aa)^*$\\
& $\ll$ & $(\ZERO + \ZERO{}a + \ONE) \cdot (a + aa)^* \;+\; (\ONE + \ONE{}a) \cdot (a + aa)^*$\\
& $\ll$ & $(\ZERO + \ZERO{}a + \ZERO) \cdot (a + aa)^* + (\ONE + \ONE{}a) \cdot (a + aa)^* \;+\; $\\
& & $\qquad(\ZERO + \ZERO{}a + \ONE) \cdot (a + aa)^* + (\ONE + \ONE{}a) \cdot (a + aa)^*$\\
& $\ll$ & \ldots \hspace{15mm}(regular expressions of sizes 98, 169, 283, 468, 767, \ldots)
\end{tabular}
\end{center}
 
\noindent where after around 35 steps we usually run out of memory on a
typical computer.  Clearly, the
notation involving $\ZERO$s and $\ONE$s already suggests
simplification rules that can be applied to regular regular
expressions, for example $\ZERO{}\,r \Rightarrow \ZERO$, $\ONE{}\,r
\Rightarrow r$, $\ZERO{} + r \Rightarrow r$ and $r + r \Rightarrow
r$. While such simple-minded simplifications have been proved in 
the work by Ausaf et al. to preserve the correctness of Sulzmann and Lu's
algorithm \cite{AusafDyckhoffUrban2016}, they unfortunately do
\emph{not} help with limiting the growth of the derivatives shown
above: the growth is slowed, but the derivatives can still grow rather
quickly beyond any finite bound.

Therefore we want to look in this thesis at a second
algorithm by Sulzmann and Lu where they
overcame this ``growth problem'' \cite{Sulzmann2014}.
In this version, POSIX values are 
represented as bit sequences and such sequences are incrementally generated
when derivatives are calculated. The compact representation
of bit sequences and regular expressions allows them to define a more
``aggressive'' simplification method that keeps the size of the
derivatives finite no matter what the length of the string is.
They make some informal claims about the correctness and linear behaviour
of this version, but do not provide any supporting proof arguments, not
even ``pencil-and-paper'' arguments. They write about their bit-coded
\emph{incremental parsing method} (that is the algorithm to be formalised
in this dissertation)


  
  \begin{quote}\it
  ``Correctness Claim: We further claim that the incremental parsing
  method [..] in combination with the simplification steps [..]
  yields POSIX parse trees. We have tested this claim
  extensively [..] but yet
  have to work out all proof details.'' \cite[Page 14]{Sulzmann2014}
\end{quote}  
Ausaf and Urban made some initial progress towards the 
full correctness proof but still had to leave out the optimisation
Sulzmann and Lu proposed.
Ausaf  wrote \cite{Ausaf},
  \begin{quote}\it
``The next step would be to implement a more aggressive simplification procedure on annotated regular expressions and then prove the corresponding algorithm generates the same values as blexer. Alas due to time constraints we are unable to do so here.''
\end{quote}  
This thesis implements the aggressive simplifications envisioned
by Ausaf and Urban,
together with a formal proof of the correctness of those simplifications.


One of the most recent work in the context of lexing
%with this issue
is the Verbatim lexer by Egolf, Lasser and Fisher~\cite{Verbatim}.
This is relevant work for us and we will compare it later with 
our derivative-based matcher we are going to present.
There is also some newer work called
Verbatim++~\cite{Verbatimpp}, which does not use derivatives, 
but deterministic finite automaton instead.
We will also study this work in a later section.
%An example that gives problem to automaton approaches would be
%the regular expression $(a|b)^*a(a|b)^{\{n\}}$.
%It requires at least $2^{n+1}$ states to represent
%as a DFA.


\section{Basic Concepts}
Formal language theory usually starts with an alphabet 
denoting a set of characters.
Here we use the datatype of characters from Isabelle,
which roughly corresponds to the ASCII characters.
In what follows, we shall leave the information about the alphabet
implicit.
Then using the usual bracket notation for lists,
we can define strings made up of characters as: 
\begin{center}
\begin{tabular}{lcl}
$\textit{s}$ & $\dn$ & $[] \; |\; c  :: s$
\end{tabular}
\end{center}
where $c$ is a variable ranging over characters.
The $::$ stands for list cons and $[]$ for the empty
list.
For brevity, a singleton list is sometimes written as $[c]$.
Strings can be concatenated to form longer strings in the same
way we concatenate two lists, which we shall write as $s_1 @ s_2$.
We omit the precise 
recursive definition here.
We overload this concatenation operator for two sets of strings:
\begin{center}
\begin{tabular}{lcl}
$A @ B $ & $\dn$ & $\{s_A @ s_B \mid s_A \in A \land s_B \in B \}$\\
\end{tabular}
\end{center}
We also call the above \emph{language concatenation}.
The power of a language is defined recursively, using the 
concatenation operator $@$:
\begin{center}
\begin{tabular}{lcl}
$A^0 $ & $\dn$ & $\{ [] \}$\\
$A^{n+1}$ & $\dn$ & $A @ A^n$
\end{tabular}
\end{center}
The union of all powers of a language   
can be used to define the Kleene star operator:
\begin{center}
\begin{tabular}{lcl}
 $A*$ & $\dn$ & $\bigcup_{i \geq 0} A^i$ \\
\end{tabular}
\end{center}

\noindent
However, to obtain a more convenient induction principle 
in Isabelle/HOL, 
we instead define the Kleene star
as an inductive set: 

\begin{center}
\begin{mathpar}
	\inferrule{\mbox{}}{[] \in A*\\}

\inferrule{s_1 \in A \;\; s_2 \in A*}{s_1 @ s_2 \in A*}
\end{mathpar}
\end{center}
\noindent
We also define an operation of "chopping off" a character from
a language, which we call $\Der$, meaning \emph{Derivative} (for a language):
\begin{center}
\begin{tabular}{lcl}
$\textit{Der} \;c \;A$ & $\dn$ & $\{ s \mid c :: s \in A \}$\\
\end{tabular}
\end{center}
\noindent
This can be generalised to ``chopping off'' a string 
from all strings within a set $A$, 
namely:
\begin{center}
\begin{tabular}{lcl}
$\textit{Ders} \;s \;A$ & $\dn$ & $\{ s' \mid s@s' \in A \}$\\
\end{tabular}
\end{center}
\noindent
which is essentially the left quotient $A \backslash L$ of $A$ against 
the singleton language with $L = \{s\}$.
However, for our purposes here, the $\textit{Ders}$ definition with 
a single string is sufficient.

The reason for defining derivatives
is that they provide another approach
to test membership of a string in 
a set of strings. 
For example, to test whether the string
$bar$ is contained in the set $\{foo, bar, brak\}$, one can take derivative of the set with
respect to the string $bar$:
\begin{center}
\begin{tabular}{lll}
	$S = \{foo, bar, brak\}$ & $ \stackrel{\backslash b}{\rightarrow }$ & 
	$\{ar, rak\}$ \\
				 & $\stackrel{\backslash a}{\rightarrow}$ & $\{r \}$\\
				 & $\stackrel{\backslash r}{\rightarrow}$ & $\{[]\}$\\
				 %& $\stackrel{[] \in S \backslash bar}{\longrightarrow}$ & $bar \in S$\\
\end{tabular}	
\end{center}
\noindent
and in the end, test whether the set
contains the empty string.\footnote{We use the infix notation $A\backslash c$
	instead of $\Der \; c \; A$ for brevity, as it will always be
	clear from the context that we are operating
on languages rather than regular expressions.}

In general, if we have a language $S$,
then we can test whether $s$ is in $S$
by testing whether $[] \in S \backslash s$.
With the sequencing, Kleene star, and $\textit{Der}$ operator on languages,
we have a  few properties of how the language derivative can be defined using 
sub-languages.
For example, for the sequence operator, we have
something similar to a ``chain rule'':
\begin{lemma}
\[
	\Der \; c \; (A @ B) =
	\begin{cases}
	((\Der \; c \; A) \, @ \, B ) \cup (\Der \; c\; B) , &  \text{if} \;  [] \in A  \\
	 (\Der \; c \; A) \,  @ \, B, & \text{otherwise}
	 \end{cases}	
\]
\end{lemma}
\noindent
This lemma states that if $A$ contains the empty string, $\Der$ can "pierce" through it
and get to $B$.
The language derivative for $A*$ can be described using the language derivative
of $A$:
\begin{lemma}
$\textit{Der} \;c \;(A*) = (\textit{Der}\; c A) @ (A*)$\\
\end{lemma}
\begin{proof}
There are two inclusions to prove:
\begin{itemize}
\item{$\subseteq$}:\\
The set 
\[ S_1 = \{s \mid c :: s \in A*\} \]
is enclosed in the set
\[ S_2 = \{s_1 @ s_2 \mid s_1 \, s_2.\;  s_1 \in \{s \mid c :: s \in A\} \land s_2 \in A* \}. \]
This is because for any string $c::s$ satisfying $c::s \in A*$,
%whenever you have a string starting with a character 
%in the language of a Kleene star $A*$, 
%then that
the character $c$, together with a prefix of $s$
%immediately after $c$ 
forms the first iteration of $A*$, 
and the rest of the $s$ is also $A*$.
This coincides with the definition of $S_2$.
\item{$\supseteq$}:\\
Note that
\[ \Der \; c \; (A*) = \Der \; c \;  (\{ [] \} \cup (A @ A*) ) \]
holds.
Also the following holds:
\[ \Der \; c \;  (\{ [] \} \cup (A @ A*) ) = \Der\; c \; (A @ A*) \]
where the $\textit{RHS}$ can be rewritten
as \[ (\Der \; c\; A) @ A* \cup (\Der \; c \; (A*)) \]
which of course contains $\Der \; c \; A @ A*$.
\end{itemize}
\end{proof}

\noindent
The clever idea of Brzozowski was to find counterparts of $\Der$ and $\Ders$
for regular expressions.
To introduce them, we need to first give definitions for regular expressions,
which we shall do next.

\subsection{Regular Expressions and Their Meaning}
The \emph{basic regular expressions} are defined inductively
 by the following grammar:
\[			r ::=   \ZERO \mid  \ONE
			 \mid  c  
			 \mid  r_1 \cdot r_2
			 \mid  r_1 + r_2   
			 \mid r^*         
\]
\noindent
We call them basic because we will introduce
additional constructors in later chapters, such as negation
and bounded repetitions.
We use $\ZERO$ for the regular expression that
matches no string, and $\ONE$ for the regular
expression that matches only the empty string.\footnote{
Some authors
also use $\phi$ and $\epsilon$ for $\ZERO$ and $\ONE$
but we prefer this notation.} 
The sequence regular expression is written $r_1\cdot r_2$
and sometimes we omit the dot if it is clear which
regular expression is meant; the alternative
is written $r_1 + r_2$.
The \emph{language} or meaning of 
a regular expression is defined recursively as
a set of strings:
%TODO: FILL in the other defs
\begin{center}
\begin{tabular}{lcl}
$L \; \ZERO$ & $\dn$ & $\varnothing$\\
$L \; \ONE$ & $\dn$ & $\{[]\}$\\
$L \; c$ & $\dn$ & $\{[c]\}$\\
$L \; (r_1 + r_2)$ & $\dn$ & $ L \; r_1 \cup L \; r_2$\\
$L \; (r_1 \cdot r_2)$ & $\dn$ & $ L \; r_1 @ L \; r_2$\\
$L \; (r^*)$ & $\dn$ & $ (L\;r)*$
\end{tabular}
\end{center}
\noindent
%Now with language derivatives of a language and regular expressions and
%their language interpretations in place, we are ready to define derivatives on regular expressions.
With $L$, we are ready to introduce Brzozowski derivatives on regular expressions.
We do so by first introducing what properties they should satisfy.

\subsection{Brzozowski Derivatives and a Regular Expression Matcher}
%Recall, the language derivative acts on a set of strings
%and essentially chops off a particular character from
%all strings in that set, Brzozowski defined a derivative operation on regular expressions
%so that after derivative $L(r\backslash c)$ 
%will look as if it was obtained by doing a language derivative on $L(r)$:
%Recall that the language derivative acts on a 
%language (set of strings).
%One can decide whether a string $s$ belongs
%to a language $S$ by taking derivative with respect to
%that string and then checking whether the empty 
%string is in the derivative:
%\begin{center}
%\parskip \baselineskip
%\def\myupbracefill#1{\rotatebox{90}{\stretchto{\{}{#1}}}
%\def\rlwd{.5pt}
%\newcommand\notate[3]{%
%  \unskip\def\useanchorwidth{T}%
%  \setbox0=\hbox{#1}%
%  \def\stackalignment{c}\stackunder[-6pt]{%
%    \def\stackalignment{c}\stackunder[-1.5pt]{%
%      \stackunder[-2pt]{\strut #1}{\myupbracefill{\wd0}}}{%
%    \rule{\rlwd}{#2\baselineskip}}}{%
%  \strut\kern7pt$\hookrightarrow$\rlap{ \footnotesize#3}}\ignorespaces%
%}
%\Longstack{
%\notate{$\{ \ldots ,\;$ 
%	\notate{s}{1}{$(c_1 :: s_1)$} 
%	$, \; \ldots \}$ 
%}{1}{$S_{start}$} 
%}
%\Longstack{
%	$\stackrel{\backslash c_1}{\longrightarrow}$
%}
%\Longstack{
%	$\{ \ldots,\;$  \notate{$s_1$}{1}{$(c_2::s_2)$} 
%	$,\; \ldots \}$
%}
%\Longstack{
%	$\stackrel{\backslash c_2}{\longrightarrow}$ 
%}
%\Longstack{
%	$\{ \ldots,\;  s_2
%	,\; \ldots \}$
%}
%\Longstack{
%	$ \xdashrightarrow{\backslash c_3\ldots\ldots} $	
%}
%\Longstack{
%	\notate{$\{\ldots, [], \ldots\}$}{1}{$S_{end} = 
%	S_{start}\backslash s$}
%}
%\end{center}
%\begin{center}
%	$s \in S_{start} \iff [] \in S_{end}$
%\end{center}
%\noindent
Brzozowski noticed that $\Der$
can be ``mirrored'' on regular expressions which
he calls the derivative of a regular expression $r$
with respect to a character $c$, written
$r \backslash c$. This infix operator
takes regular expression $r$ as input
and a character as a right operand.
The derivative operation on regular expression
is defined such that the language of the derivative result 
coincides with the language of the original 
regular expression being taken 
derivative with respect to the same characters, namely
\begin{property}

\[
	L \; (r \backslash c) = \Der \; c \; (L \; r)
\]
\end{property}
\noindent
Next, we give the recursive definition of derivative on
regular expressions so that it satisfies the properties above.
%The derivative function, written $r\backslash c$, 
%takes a regular expression $r$ and character $c$, and
%returns a new regular expression representing
%the original regular expression's language $L \; r$
%being taken the language derivative with respect to $c$.
\begin{table}
	\begin{center}
\begin{tabular}{lcl}
		$\ZERO \backslash c$ & $\dn$ & $\ZERO$\\  
		$\ONE \backslash c$  & $\dn$ & $\ZERO$\\
		$d \backslash c$     & $\dn$ & 
		$\mathit{if} \;c = d\;\mathit{then}\;\ONE\;\mathit{else}\;\ZERO$\\
$(r_1 + r_2)\backslash c$     & $\dn$ & $r_1 \backslash c \,+\, r_2 \backslash c$\\
$(r_1 \cdot r_2)\backslash c$ & $\dn$ & $\mathit{if} \, [] \in L(r_1)$\\
	&   & $\mathit{then}\;(r_1\backslash c) \cdot r_2 \,+\, r_2\backslash c$\\
	&   & $\mathit{else}\;(r_1\backslash c) \cdot r_2$\\
	$(r^*)\backslash c$           & $\dn$ & $(r\backslash c) \cdot r^*$\\
\end{tabular}
	\end{center}
\caption{Derivative on Regular Expressions}
\label{table:der}
\end{table}
\noindent
The most involved cases are the sequence case
and the star case.
The sequence case says that if the first regular expression
contains an empty string, then the second component of the sequence
needs to be considered, as its derivative will contribute to the
result of this derivative:
\begin{center}
	\begin{tabular}{lcl}
		$(r_1 \cdot r_2 ) \backslash c$ & $\dn$ & 
		$\textit{if}\;\,([] \in L(r_1))\; 
		\textit{then} \; (r_1 \backslash c) \cdot r_2 + r_2 \backslash c$ \\
		& & $\textit{else} \; (r_1 \backslash c) \cdot r_2$
	\end{tabular}
\end{center}
\noindent
Notice how this closely resembles
the language derivative operation $\Der$:
\begin{center}
	\begin{tabular}{lcl}
		$\Der \; c \; (A @ B)$ & $\dn$ & 
		$ \textit{if} \;\,  [] \in A \; 
		\textit{then} \;\, ((\Der \; c \; A) @ B ) \cup 
		\Der \; c\; B$\\
		& & $\textit{else}\; (\Der \; c \; A) @ B$\\
	\end{tabular}
\end{center}
\noindent
The derivative of the star regular expression $r^*$ 
unwraps one iteration of $r$, turns it into $r\backslash c$,
and attaches the original $r^*$
after $r\backslash c$, so that 
we can further unfold it as many times as needed:
\[
	(r^*) \backslash c \dn (r \backslash c)\cdot r^*.
\]
Again,
the structure is the same as the language derivative of the Kleene star: 
\[
	\textit{Der} \;c \;(A*) \dn (\textit{Der}\; c A) @ (A*)
\]
In the above definition of $(r_1\cdot r_2) \backslash c$,
the $\textit{if}$ clause's
boolean condition 
$[] \in L(r_1)$ needs to be 
somehow recursively computed.
We call such a function that checks
whether the empty string $[]$ is 
in the language of a regular expression $\nullable$:
\begin{center}
		\begin{tabular}{lcl}
			$\nullable(\ZERO)$     & $\dn$ & $\mathit{false}$ \\  
			$\nullable(\ONE)$      & $\dn$ & $\mathit{true}$ \\
			$\nullable(c)$ 	       & $\dn$ & $\mathit{false}$ \\
			$\nullable(r_1 + r_2)$ & $\dn$ & $\nullable(r_1) \vee \nullable(r_2)$ \\
			$\nullable(r_1\cdot r_2)$  & $\dn$ & $\nullable(r_1) \wedge \nullable(r_2)$ \\
			$\nullable(r^*)$       & $\dn$ & $\mathit{true}$ \\
		\end{tabular}
\end{center}
\noindent
The $\ZERO$ regular expression
does not contain any string and
therefore is not \emph{nullable}.
$\ONE$ is \emph{nullable} 
by definition. 
The character regular expression $c$
corresponds to the singleton set $\{c\}$, 
and therefore does not contain the empty string.
The alternative regular expression is nullable
if at least one of its children is nullable.
The sequence regular expression
would require both children to have the empty string
to compose an empty string, and the Kleene star
is always nullable because it naturally
contains the empty string.  
\noindent
We have the following two correspondences between 
derivatives on regular expressions and
derivatives on a set of strings:
\begin{lemma}\label{derDer}
	\mbox{}
	\begin{itemize}
		\item
$\textit{Der} \; c \; L(r) = L (r\backslash c)$
\item
	$c\!::\!s \in L(r)$ \textit{iff} $s \in L(r\backslash c)$.
	\end{itemize}
\end{lemma}
\begin{proof}
	By induction on $r$.
\end{proof}
\noindent
which are the main properties of derivatives
that enables us later to reason about the correctness of
derivative-based matching.
We can generalise the derivative operation shown above for single characters
to strings as follows:

\begin{center}
\begin{tabular}{lcl}
$r \backslash_s (c\!::\!s) $ & $\dn$ & $(r \backslash c) \backslash_s s$ \\
$r \backslash_s [\,] $ & $\dn$ & $r$
\end{tabular}
\end{center}

\noindent
When there is no ambiguity, we will 
omit the subscript and use $\backslash$ instead
of $\backslash_s$ to denote
string derivatives for brevity.
Brzozowski's derivative-based
regular-expression matching algorithm can then be described as:

\begin{definition}
$\textit{match}\;s\;r \;\dn\; \nullable \; (r\backslash s)$
\end{definition}

\noindent
Assuming the string is given as a sequence of characters, say $c_0c_1 \ldots c_n$, 
this algorithm, presented graphically, is as follows:

\begin{equation}\label{matcher}
\begin{tikzcd}
r_0 \arrow[r, "\backslash c_0"]  & r_1 \arrow[r, "\backslash c_1"] & 
r_2 \arrow[r, dashed]  & r_n  \arrow[r,"\textit{nullable}?"] & 
\;\textrm{true}/\textrm{false}
\end{tikzcd}
\end{equation}

\noindent
 It is relatively
easy to show that this matcher is correct, namely
\begin{lemma}
	$\textit{match} \; s\; r  = \textit{true} \; \textit{iff} \; s \in L(r)$
\end{lemma}
\begin{proof}
	By induction on $s$ using the property of derivatives:
	lemma \ref{derDer}.
\end{proof}
\begin{figure}
\begin{center}
\begin{tikzpicture}
\begin{axis}[
    xlabel={$n$},
    ylabel={time in secs},
    %ymode = log,
    legend entries={Naive Matcher},  
    legend pos=north west,
    legend cell align=left]
\addplot[red,mark=*, mark options={fill=white}] table {NaiveMatcher.data};
\end{axis}
\end{tikzpicture} 
\caption{Matching the regular expression $(a^*)^*b$ against strings of the form
$\protect\underbrace{aa\ldots a}_\text{n \textit{a}s}
$ using Brzozowski's original algorithm}\label{NaiveMatcher}
\end{center}
\end{figure}
\noindent
If we implement the above algorithm naively, however,
the algorithm can be as slow as backtracking lexers, as shown in 
\ref{NaiveMatcher}.
Note that both axes are in logarithmic scale.
Around two dozen characters
this algorithm already ``explodes'' with the regular expression 
$(a^*)^*b$.
To improve this situation, we need to introduce simplification
rules for the intermediate results,
such as $r + r \rightarrow r$ or $\ONE \cdot r \rightarrow r$,
and make sure those rules do not change the 
language of the regular expression.
One simple-minded simplification function
that achieves these requirements 
is given below (see Ausaf et al. \cite{AusafDyckhoffUrban2016}):
\begin{center}
	\begin{tabular}{lcl}
		$\simp \; r_1 \cdot r_2 $ & $ \dn$ & 
		$(\simp \; r_1,  \simp \; r_2) \; \textit{match}$\\
					  & & $\quad \case \; (\ZERO, \_) \Rightarrow \ZERO$\\
					  & & $\quad \case \; (\_, \ZERO) \Rightarrow \ZERO$\\
					  & & $\quad \case \; (\ONE, r_2') \Rightarrow r_2'$\\
					  & & $\quad \case \; (r_1', \ONE) \Rightarrow r_1'$\\
					  & & $\quad \case \; (r_1', r_2') \Rightarrow r_1'\cdot r_2'$\\
		$\simp \; r_1 + r_2$ & $\dn$ & $(\simp \; r_1, \simp \; r_2) \textit{match}$\\
				     & & $\quad \; \case \; (\ZERO, r_2') \Rightarrow r_2'$\\
				     & & $\quad \; \case \; (r_1', \ZERO) \Rightarrow r_1'$\\
				     & & $\quad \; \case \; (r_1', r_2') \Rightarrow r_1' + r_2'$\\
		$\simp \; r$ & $\dn$ & $r\quad\quad (otherwise)$
				   
	\end{tabular}
\end{center}
If we repeatedly apply this simplification  
function during the matching algorithm, 
we have a matcher with simplification:
\begin{center}
	\begin{tabular}{lcl}
		$\derssimp \; [] \; r$ & $\dn$ & $r$\\
		$\derssimp \; c :: cs \; r$ & $\dn$ & $\derssimp \; cs \; (\simp \; (r \backslash c))$\\
		$\textit{matcher}_{simp}\; s \; r $ & $\dn$ & $\nullable \; (\derssimp \; s\;r)$
	\end{tabular}
\end{center}
\begin{figure}
\begin{tikzpicture}
\begin{axis}[
    xlabel={$n$},
    ylabel={time in secs},
    %ymode = log,
    %xmode = log,
    grid = both,
    legend entries={Matcher With Simp},  
    legend pos=north west,
    legend cell align=left]
\addplot[red,mark=*, mark options={fill=white}] table {BetterMatcher.data};
\end{axis}
\end{tikzpicture} 
\caption{$(a^*)^*b$ 
against 
$\protect\underbrace{aa\ldots a}_\text{n \textit{a}s}$ Using $\textit{matcher}_{simp}$}\label{BetterMatcher}
\end{figure}
\noindent
The running time of $\textit{ders}\_\textit{simp}$
on the same example of Figure \ref{NaiveMatcher}
is now ``tame''  in terms of the length of inputs,
as shown in Figure \ref{BetterMatcher}.

So far, the story is use Brzozowski derivatives and
simplify as much as possible, and at the end test
whether the empty string is recognised 
by the final derivative.
But what if we want to 
do lexing instead of just getting a true/false answer?
Sulzmann and Lu \cite{Sulzmann2014} first came up with a nice and 
elegant (arguably as beautiful as the definition of the 
Brzozowski derivative) solution for this.
\marginpar{explicitly attribute the work}
For the same reason described
at the beginning of this chapter, 
we introduce the formal
semantics of $\POSIX$ lexing by 
Ausaf et al.\cite{AusafDyckhoffUrban2016}, 
followed by the first lexing algorithm by 
Sulzmanna and Lu \cite{Sulzmann2014} 
that produces the output conforming
to the $\POSIX$ standard. 
%TODO: Actually show this in chapter 4.
In what follows 
we choose to use the Isabelle-style notation
for function and datatype constructor applications, where
the parameters of a function are not enclosed
inside a pair of parentheses (e.g. $f \;x \;y$
instead of $f(x,\;y)$). This is mainly
to make the text visually more concise.

\section{Values and the Lexing Algorithm by Sulzmann and Lu}
In this section, we present a two-phase regular expression lexing 
algorithm.
The first phase takes successive derivatives with 
respect to the input string,
and the second phase does the reverse, \emph{injecting} back
characters, in the meantime constructing a lexing result.
We will introduce the injection phase in detail slightly
later, but as a preliminary we have to first define 
the datatype for lexing results, 
called \emph{value} or
sometimes also \emph{lexical value}.  
Values and regular
expressions correspond to each other 
as illustrated in the following
table:

\begin{center}
	\begin{tabular}{c@{\hspace{20mm}}c}
		\begin{tabular}{@{}rrl@{}}
			\multicolumn{3}{@{}l}{\textbf{Regular Expressions}}\medskip\\
			$r$ & $::=$  & $\ZERO$\\
			& $\mid$ & $\ONE$   \\
			& $\mid$ & $c$          \\
			& $\mid$ & $r_1 \cdot r_2$\\
			& $\mid$ & $r_1 + r_2$   \\
			\\
			& $\mid$ & $r^*$         \\
		\end{tabular}
		&
		\begin{tabular}{@{\hspace{0mm}}rrl@{}}
			\multicolumn{3}{@{}l}{\textbf{Values}}\medskip\\
			$v$ & $::=$  & \\
			&        & $\Empty$   \\
			& $\mid$ & $\Char \; c$          \\
			& $\mid$ & $\Seq\,v_1\, v_2$\\
			& $\mid$ & $\Left \;v$   \\
			& $\mid$ & $\Right\;v$  \\
			& $\mid$ & $\Stars\,[v_1,\ldots\,v_n]$ \\
		\end{tabular}
	\end{tabular}
\end{center}
\noindent
A value has an underlying string, which 
can be calculated by the ``flatten" function $|\_|$:
\begin{center}
	\begin{tabular}{lcl}
		$|\Empty|$ & $\dn$ &  $[]$\\
		$|\Char \; c|$ & $ \dn$ & $ [c]$\\
		$|\Seq \; v_1, \;v_2|$ & $ \dn$ & $ v_1| @ |v_2|$\\
		$|\Left \; v|$ & $ \dn$ & $ |v|$\\
		$|\Right \; v|$ & $ \dn$ & $ |v|$\\
		$|\Stars \; []|$ & $\dn$ & $[]$\\
		$|\Stars \; v::vs|$ &  $\dn$ & $ |v| @ |\Stars(vs)|$
	\end{tabular}
\end{center}
Sulzmann and Lu used a binary predicate, written $\vdash v:r $,
to indicate that a value $v$ could be generated from a lexing algorithm
with input $r$. They call it the value inhabitation relation,
defined by the rules.
\begin{figure}[H]
\begin{mathpar}
	\inferrule{\mbox{}}{\vdash \Char \; c : \mathbf{c}} \hspace{2em}

	\inferrule{\mbox{}}{\vdash \Empty :  \ONE} \hspace{2em}

\inferrule{\vdash v_1 : r_1 \;\; \vdash v_2 : r_2 }{\vdash \Seq \; v_1,\; v_2 : (r_1 \cdot r_2)}

\inferrule{\vdash v_1 : r_1}{\vdash \Left \; v_1 : r_1+r_2}

\inferrule{\vdash v_2 : r_2}{\vdash \Right \; v_2:r_1 + r_2}

\inferrule{\forall v \in vs. \vdash v:r \land  |v| \neq []}{\vdash \Stars \; vs : r^*}
\end{mathpar}
\caption{The inhabitation relation for values and regular expressions}\label{fig:inhab}
\end{figure}
\noindent
The condition $|v| \neq []$ in the premise of star's rule
is to make sure that for a given pair of regular 
expression $r$ and string $s$, the number of values 
satisfying $|v| = s$ and $\vdash v:r$ is finite.
This additional condition was
imposed by Ausaf and Urban to make their proofs easier.
Given a string and a regular expression, there can be
multiple values for it. For example, both
$\vdash \Seq(\Left \; ab)(\Right \; c):(ab+a)(bc+c)$ and
$\vdash \Seq(\Right\; a)(\Left \; bc ):(ab+a)(bc+c)$ hold
and the values both flatten to $abc$.
Lexers therefore have to disambiguate and choose only
one of the values to be generated. $\POSIX$ is one of the
disambiguation strategies that is widely adopted.

Ausaf et al. \cite{AusafDyckhoffUrban2016} 
formalised the property 
as a ternary relation.
The $\POSIX$ value $v$ for a regular expression
$r$ and string $s$, denoted as $(s, r) \rightarrow v$, can be specified 
in the following rules\footnote{The names of the rules are used
as they were originally given in \cite{AusafDyckhoffUrban2016}.}:
\begin{figure}[p]
\begin{mathpar}
	\inferrule[P1]{\mbox{}}{([], \ONE) \rightarrow \Empty}
		
	\inferrule[PC]{\mbox{}}{([c], c) \rightarrow \Char \; c}

	\inferrule[P+L]{(s,r_1)\rightarrow v_1}{(s, r_1+r_2)\rightarrow \Left \; v_1}

	\inferrule[P+R]{(s,r_2)\rightarrow v_2\\ s \notin L \; r_1}{(s, r_1+r_2)\rightarrow \Right \; v_2}

	\inferrule[PS]{(s_1, v_1) \rightarrow r_1 \\ (s_2, v_2)\rightarrow r_2\\
		\nexists s_3 \; s_4. s_3 \neq [] \land s_3 @ s_4 = s_2 \land 
		s_1@ s_3 \in L \; r_1 \land s_4 \in L \; r_2}{(s_1 @ s_2, r_1\cdot r_2) \rightarrow
	\Seq \; v_1 \; v_2}

	\inferrule[P{[]}]{\mbox{}}{([], r^*) \rightarrow \Stars([])}

	\inferrule[P*]{(s_1, v) \rightarrow v \\ (s_2, r^*) \rightarrow \Stars \; vs \\
		|v| \neq []\\ \nexists s_3 \; s_4. s_3 \neq [] \land s_3@s_4 = s_2 \land
		s_1@s_3 \in L \; r \land s_4 \in L \; r^*}{(s_1@s_2, r^*)\rightarrow \Stars \;
	(v::vs)}
\end{mathpar}
\caption{The inductive POSIX rules given by Ausaf et al.
	\cite{AusafDyckhoffUrban2016}.
This ternary relation, written $(s, r) \rightarrow v$, 
formalises the POSIX constraints on the
value $v$ given a string $s$ and 
regular expression $r$.
}
\label{fig:POSIXDef}
\end{figure}\afterpage{\clearpage}
\noindent

%\begin{figure}
%\begin{tikzpicture}[]
%    \node [minimum width = 6cm, rectangle split, rectangle split horizontal, 
%	    rectangle split parts=2, rectangle split part fill={red!30,blue!20}, style={draw, rounded corners, inner sep=10pt}]
%	    (node1)
%	    {$r_{token1}$
%	    \nodepart{two}  $\;\;\; \quad r_{token2}\;\;\;\quad$ };
%	    %\node [left = 6.0cm of node1] (start1) {hi};
%	    \node [left = 0.2cm of node1] (middle) {$v.s.$};
%    \node [minimum width = 6cm, left = 0.2cm of middle, rectangle split, rectangle split horizontal, 
%	    rectangle split parts=2, rectangle split part fill={red!30,blue!20}, style={draw, rounded corners, inner sep=10pt}]
%	    (node2)
%	    {$\quad\;\;\;r_{token1}\quad\;\;\;$
%	    \nodepart{two}  $r_{token2}$ };
%	    \node [below = 0.1cm of node2] (text1) {\checkmark preferred by POSIX};
%		\node [above = 1.5cm of middle, minimum width = 6cm, 
%			rectangle, style={draw, rounded corners, inner sep=10pt}] 
%			(topNode) {$s$};
%	    \path[->,draw]
%	        (topNode) edge node {split $A$} (node2)
%	        (topNode) edge node {split $B$} (node1)
%		;
%			
%
%\end{tikzpicture}
%\caption{Maximum munch example: $s$ matches $r_{token1} \cdot r_{token2}$}\label{munch}
%\end{figure}
The above $\POSIX$ rules follow the intuition described below: 
\begin{itemize}
	\item (Left Priority)\\
		Match the leftmost regular expression when multiple options for matching
		are available. See P+L and P+R where in P+R $s$ cannot
		be in the language of $L \; r_1$.
	\item (Maximum munch)\\
		Always match a subpart as much as possible before proceeding
		to the next part of the string.
		For example, when the string $s$ matches 
		$r_{part1}\cdot r_{part2}$, and we have two ways $s$ can be split:
		Then the split that matches a longer string for the first part
		$r_{part1}$ is preferred by this maximum munch rule.
		The side-condition 
		\begin{center}
		$\nexists s_3 \; s_4. s_3 \neq [] \land s_3 @ s_4 = s_2 \land 
		s_1@ s_3 \in L \; r_1 \land s_4 \in L \; r_2$
		\end{center}
		in PS causes this.
		%(See
		%\ref{munch} for an illustration).
\end{itemize}
\noindent
These disambiguation strategies can be 
quite practical.
For instance, when lexing a code snippet 
\[ 
	\textit{iffoo} = 3
\]
using a regular expression 
for keywords and 
identifiers:
%(for example, keyword is a nonempty string starting with letters 
%followed by alphanumeric characters or underscores):
\[
	r_{keyword} + r_{identifier}.
\]
If we want $\textit{iffoo}$ to be recognized
as an identifier
where identifiers are defined as usual (letters
followed by letters, numbers or underscores),
then a match with a keyword (if)
followed by
an identifier (foo) would be incorrect.
POSIX lexing generates what is included by lexing.

\noindent
We know that a POSIX 
value for regular expression $r$ is inhabited by $r$.
\begin{lemma}
$(r, s) \rightarrow v \implies \vdash v: r$
\end{lemma}
\noindent
The main property about a $\POSIX$ value is that 
given the same regular expression $r$ and string $s$,
one can always uniquely determine the $\POSIX$ value for it:
\begin{lemma}
$\textit{if} \,(s, r) \rightarrow v_1 \land (s, r) \rightarrow v_2\quad  \textit{then} \; v_1 = v_2$
\end{lemma}
\begin{proof}
By induction on $s$, $r$ and $v_1$. The inductive cases
are all the POSIX rules. 
Probably the most cumbersome cases are 
the sequence and star with non-empty iterations.
We shall give the details for proving the sequence case here.

When we have 
\[
	(s_1, r_1) \rightarrow v_1 \;\, and \;\, 
	(s_2, r_2) \rightarrow v_2  \;\, and \;\, 
	\nexists s_3 \; s_4. s_3 \neq [] \land s_3 @ s_4 = s_2 \land 
		s_1@ s_3 \in L \; r_1 \land s_4 \in L \; r_2
\]
we know that the last condition 
excludes the possibility of a 
string $s_1'$ longer than $s_1$ such that 
\[
(s_1', r_1) \rightarrow v_1'   \;\; 
and\;\; (s_2', r_2) \rightarrow v_2'\;\; and \;\;s_1' @s_2' = s 
\]
hold.
A shorter string $s_1''$ with $s_2''$ satisfying
\[
(s_1'', r_1) \rightarrow v_1''
\;\;and\;\; (s_2'', r_2) \rightarrow v_2'' \;\;and \;\;s_1'' @s_2'' = s 
\]
cannot possibly form a $\POSIX$ value either, because
by definition, there is a candidate
with a longer initial string
$s_1$. Therefore, we know that the POSIX
value $\Seq \; a \; b$ for $r_1 \cdot r_2$ matching
$s$ must have the 
property that 
\[
	|a| = s_1 \;\; and \;\; |b| = s_2.
\]
The goal is to prove that $a = v_1 $ and $b = v_2$.
If we have some other POSIX values $v_{10}$ and $v_{20}$ such that 
$(s_1, r_1) \rightarrow v_{10}$ and $(s_2, r_2) \rightarrow v_{20}$ hold,
then by induction hypothesis $v_{10} = v_1$ and $v_{20}= v_2$, 
which means this "other" $\POSIX$ value $\Seq(v_{10}, v_{20})$
is the same as $\Seq(v_1, v_2)$. 
\end{proof}
\noindent
We have now defined what a POSIX value is and shown that it is unique.
The problem is to generate
such a value in a lexing algorithm using derivatives.

\subsection{Sulzmann and Lu's Injection-based Lexing Algorithm}

Sulzmann and Lu extended Brzozowski's 
derivative-based matching
to a lexing algorithm by a second phase 
after the initial phase of successive derivatives.
This second phase generates a POSIX value 
if the regular expression matches the string.
The algorithm uses two functions called $\inj$ and $\mkeps$.
The function $\mkeps$ constructs a POSIX value from the last
derivative $r_n$:
\begin{ceqn}
\begin{equation}\label{graph:mkeps}
\begin{tikzcd}
r_0 \arrow[r, "\backslash c_0"]  & r_1 \arrow[r, "\backslash c_1"] & r_2 \arrow[r, dashed, "\ldots"] & r_n \arrow[d, "mkeps" description] \\
	        & 	              & 	            & v_n       
\end{tikzcd}
\end{equation}
\end{ceqn}
\noindent
In the above diagram, again we assume that
the input string $s$ is made of $n$ characters
$c_0c_1 \ldots c_{n-1}$ 
The last derivative operation 
$\backslash c_{n-1}$ generates the derivative $r_n$, for which
$\mkeps$ produces the value $v_n$. This value
tells us how the empty string is matched by the (nullable)
regular expression $r_n$, in a POSIX way.
The definition of $\mkeps$ is
	\begin{center}
		\begin{tabular}{lcl}
			$\mkeps \; \ONE$ 		& $\dn$ & $\Empty$ \\
			$\mkeps \; (r_{1}+r_{2})$	& $\dn$ 
						& $\textit{if}\; (\nullable \; r_{1}) \;\,
							\textit{then}\;\, \Left \; (\mkeps \; r_{1})$\\ 
						& & $\phantom{if}\; \textit{else}\;\, \Right \;(\mkeps \; r_{2})$\\
			$\mkeps \; (r_1 \cdot r_2)$ 	& $\dn$ & $\Seq\;(\mkeps\;r_1)\;(\mkeps \; r_2)$\\
			$\mkeps \; r^* $	        & $\dn$ & $\Stars\;[]$
		\end{tabular}
	\end{center}


\noindent 
The function prefers the left child $r_1$ of $r_1 + r_2$ 
to match an empty string if there is a choice.
When there is a star to match the empty string,
we give the $\Stars$ constructor an empty list, meaning
no iteration is taken.
The result of $\mkeps$ on a $\nullable$ $r$ 
is a POSIX value for $r$ and the empty string:
\begin{lemma}\label{mePosix}
$\nullable\; r \implies (r, []) \rightarrow (\mkeps\; r)$
\end{lemma}
\begin{proof}
	By induction on $r$.
\end{proof}
\noindent
After the $\mkeps$-call, Sulzmann and Lu inject back the characters one by one
in reverse order as they were chopped off in the derivative phase.
The function for this is called $\inj$. This function 
operates on values, unlike $\backslash$ which operates on regular expressions.
In the diagram below, $v_i$ stands for the (POSIX) value 
for how the regular expression 
$r_i$ matches the string $s_i$ consisting of the last $n-i$ characters
of $s$ (i.e. $s_i = c_i \ldots c_{n-1}$ ) from the previous lexical value $v_{i+1}$.
After injecting back $n$ characters, we get the lexical value for how $r_0$
matches $s$. 
\begin{figure}[H]
\begin{center}	
\begin{ceqn}
\begin{tikzcd}
r_0 \arrow[r, dashed] \arrow[d]& r_i \arrow[r, "\backslash c_i"]  \arrow[d]  & r_{i+1}  \arrow[r, dashed] \arrow[d]        & r_n \arrow[d, "mkeps" description] \\
v_0           \arrow[u]                 & v_i  \arrow[l, dashed]                              & v_{i+1} \arrow[l,"inj_{r_i} c_i"]                 & v_n \arrow[l, dashed]         
\end{tikzcd}
\end{ceqn}
\end{center}
\caption{The two-phase lexing algorithm by Sulzmann and Lu \cite{AusafDyckhoffUrban2016},
	matching the regular expression $r_0$ and string of the form $[c_0, c_1, \ldots, c_{n-1}]$.
	The first phase involves taking successive derivatives w.r.t the characters $c_0$,
	$c_1$, and so on. These are the same operations as they have appeared in the matcher
	\ref{matcher}. When the final derivative regular expression is nullable (contains the empty string),
	then the second phase starts. First, $\mkeps$ generates a POSIX value which tells us how $r_n$ matches
	the empty string, by always selecting the leftmost 
	nullable regular expression. After that, $\inj$ ``injects'' back the character in reverse order as they 
	appeared in the string, always preserving POSIXness.}\label{graph:inj}
\end{figure}
\noindent
The function $\textit{inj}$ as defined by Sulzmann and Lu
takes three arguments: a regular
expression ${r_{i}}$, before the character is chopped off, 
a character ${c_{i}}$ (the character we want to inject back) and 
the third argument $v_{i+1}$ the value we want to inject into. 
The result of an application 
$\inj \; r_i \; c_i \; v_{i+1}$ is a new value $v_i$ such that
\[
	(s_i, r_i) \rightarrow v_i
\]
holds.
The definition of $\textit{inj}$ is as follows: 
\begin{center}
\begin{tabular}{l@{\hspace{1mm}}c@{\hspace{5mm}}l}
  $\textit{inj}\;(c)\;c\,Empty$            & $\dn$ & $\Char\,c$\\
  $\textit{inj}\;(r_1 + r_2)\;c\; (\Left\; v)$ & $\dn$ & $\Left  \; (\textit{inj}\; r_1 \; c\,v)$\\
  $\textit{inj}\;(r_1 + r_2)\,c\; (\Right\;v)$ & $\dn$ & $\Right \; (\textit{inj}\;r_2\;c  \; v)$\\
  $\textit{inj}\;(r_1 \cdot r_2)\; c\;(\Seq \; v_1 \; v_2)$ & $\dn$  & 
  $\Seq \; (\textit{inj}\;r_1\;c\;v_1) \; v_2$\\
  $\textit{inj}\;(r_1 \cdot r_2)\; c\;(\Left \; (\Seq \; v_1\;v_2) )$ & 
  $\dn$  & $\Seq \; (\textit{inj}\,r_1\,c\,v_1)\; v_2$\\
  $\textit{inj}\;(r_1 \cdot r_2)\; c\; (\Right\; v)$ & $\dn$  & $\Seq\; (\textit{mkeps}\; r_1) \; (\textit{inj} \; r_2\;c\;v)$\\
  $\textit{inj}\;(r^*)\; c \; (\Seq \; v\; (\Stars\;vs))$         & $\dn$  & $\Stars\;\,((\textit{inj}\;r\;c\;v)\,::\,vs)$\\
\end{tabular}
\end{center}
\noindent 
The function recurses on 
the shape of regular
expressions and values.
Intuitively, each clause analyses 
how $r_i$ could have transformed when being 
derived by $c$, identifying which subpart
of $v_{i+1}$ has the ``hole'' 
to inject the character back into.
Once the character is
injected back to that sub-value; 
$\inj$ assembles all parts
to form a new value.

For instance, the last clause is an
injection into a sequence value $v_{i+1}$
whose second child
value is a star and the shape of the 
regular expression $r_i$ before injection 
is a star.
We therefore know 
the derivative 
starts on a star and ends as a sequence:
\[
	(r^*) \backslash c \longrightarrow r\backslash c \cdot r^*
\]
during which an iteration of the star
had just been unfolded, giving the below
value inhabitation relation:
\[
	\vdash \Seq \; v \; (\Stars \; vs) : (r\backslash c) \cdot r^*.
\]
The value list $vs$ corresponds to
matched star iterations,
and the ``hole'' lies in $v$ because
\[
	\vdash v: r\backslash c.
\]
Finally, 
$\inj \; r \;c \; v$ is prepended
to the previous list of iterations and then
wrapped under the $\Stars$
constructor, giving us $\Stars \; ((\inj \; r \; c \; v) ::vs)$.

Putting all together, Sulzmann and Lu obtained the following algorithm
outlined in the injection-based lexing diagram \ref{graph:inj}:
\begin{center}
\begin{tabular}{lcl}
	$\lexer \; r \; [] $ & $=$ & $\textit{if} \; (\nullable \; r)\; \textit{then}\;  \Some(\mkeps \; r) \; \textit{else} \; \None$\\
	$\lexer \; r \;c::s$ & $=$ & $\textit{case}\; (\lexer \; (r\backslash c) \; s) \;\textit{of}\; $\\
	& & $\quad \phantom{\mid}\; \None \implies \None$\\
	& & $\quad \mid           \Some(v) \implies \Some(\inj \; r\; c\; v)$
\end{tabular}
\end{center}
\noindent

\subsection{Examples on How Injection and Lexer Works}
We will provide a few examples on how $\inj$ and $\lexer$
works by showing their values in each recursive call on some 
concrete examples.
We start with the call $\lexer \; (a+aa)^*\cdot c\; aac$
on the lexer, note that the value's character constructor
$\Char \;c$ is abbreviated as $c$ for readability.
Similarly the last derivative's sub-expression
is abbreviated as $r_{=0}$\footnote{which is equal to
$((\ZERO + (\ZERO a+ \ZERO))\cdot(a+aa)^* + (\ZERO + \ZERO a)\cdot (a+aa)^*)+
((\ZERO+\ZERO a)\cdot(a+aa)^* + (\ZERO+\ZERO a)\cdot(a+aa)^*),
$.}
whose language interpretation is equivalent to that of
$\ZERO$ and therefore not crucial to be displayed fully expanded,
as they will not be injected into.%our example run.
\begin{center}
	\begin{tabular}{lcl}
		$(a+\textcolor{magenta}{a}\textcolor{blue}{a})^* \cdot \textcolor{red}{c}$ & $\stackrel{\backslash \textcolor{magenta}{a}}{\rightarrow}$ & $((\ONE + \textcolor{magenta}{\ONE} \textcolor{blue}{a})\cdot (a+aa)^*)\cdot \textcolor{red}{c}+\ZERO$\\
				   %& $\stackrel{\rightarrow}{\backslash a}$ & $((\ONE + \ONE a)\cdot (a+aa)^*)\cdot c + \ZERO$\\
				    & $\stackrel{\backslash \textcolor{blue}{a}}{\rightarrow}$  &
				    $(((\ZERO+(\ZERO a+ \textcolor{blue}{\ONE}))\cdot (a+aa)^* + (\ONE+\ONE a)\cdot (a+aa)^* )\cdot \textcolor{red}{\mathbf{c}} + \ZERO)+\ZERO$\\
				    & $\stackrel{\backslash \textcolor{red}{c}}{\rightarrow}$  &
				    $((r_{=0}\cdot c + \textcolor{red}{\ONE})+\ZERO)+\ZERO$\\
				    & $\stackrel{\mkeps}{\rightarrow}$ & $\Left (\Left \; (\Right \; \textcolor{red}{\Empty}))$ \\
				   & $\stackrel{\inj \;\textcolor{red}{c} }{\rightarrow}$ & 
				   $\Left \; (\Left \; (\Seq \;(\Left \; (\Seq \; (\Right \; (\Right\; \textcolor{blue}{\Empty})) \; \Stars \, [])) \; \textcolor{red}{c}))$\\
				   & $\stackrel{\inj \;\textcolor{blue}{a}}{\rightarrow}$ & 
				   $\Left\; (\Seq \; (\Seq\; (\Right \; (\Seq \; \textcolor{magenta}{\Empty }\; \textcolor{blue}{ \mathbf{a}}) )
				   \;\Stars \,[]) \; \textcolor{red}{c})$\\
				   & $\stackrel{\inj \;\textcolor{magenta}{a}}{\rightarrow}$ & $\Seq \; (\Stars \; [\Right \; (\Seq \; \mathbf{\textcolor{magenta}{a}} \; \textcolor{blue}{a})]) \; \textcolor{red}{ c}$

				   %$\inj \; r \; c\A$
		%$\inj \; (a\cdot b)\cdot c \; \Seq \; \ONE \; b$ & $=$ & $(a+e)$\\
	\end{tabular}
\end{center}
\noindent
We have assigned different colours for each character,
as well as their corresponding locations in values and regular expressions.
The most recently injected character is marked with a bold font.
%Their corresponding derivative steps have been marked with the same
%colour. We also mark the specific location where the coloured character
%was injected into with the same colour, before and after the injection occurred.
%TODO: can be also used to motivate injection->blexer in Bitcoded1
To show the details of how $\inj$ works,
we zoom in to the second injection above, 
illustrating the recursive calls involved: 
\begin{center}
	\begin{tabular}{lcl}
$\inj \;\quad ((\ONE + {\ONE} 
	\textcolor{blue}{a})\cdot (a+aa)^*)\cdot 
	c + \ZERO \; \quad \textcolor{blue}{a} \; $ & &\\
$\quad \Left \; 
(\Left \; (\Seq \;(\Left \; (\Seq \; (\Right \; (\Right\; 
	\textcolor{blue}{\Empty})) \; \Stars \, [])) \; c))$ & \\$=$\\
	$\Left \; (\inj \; ((\ONE + \ONE 
	\textcolor{blue}{a})\cdot (a+aa)^*)\cdot 
	c \quad  \textcolor{blue}{a} \quad (\Left \; (\Seq\; ( \Left \; (\Seq \; (\Right \; (\Right\; 
	\textcolor{blue}{\Empty})) \; \Stars \, []) ) \; c ) )\;\;)$ &\\ $=$\\
	$\Left \; (\Seq \; (\inj \; (\ONE + \ONE \textcolor{blue}{a})\cdot(a+aa)^* \quad \textcolor{blue}{a} \quad
	(\Left \; (\Seq \; (\Right \; (\Right\;\textcolor{blue}{\Empty})))) ) \; c ) $ & \\$=$\\
	$\Left \; (\Seq \; (\Seq \; (\inj \quad (\ONE + \ONE \textcolor{blue}{a}) \quad \textcolor{blue}{a}\quad \Right \;(\Right \; \textcolor{blue}{\Empty}) ) \; \Stars \,[])\; c)$ &\\ $=$ \\
	$\Left \; (\Seq \; (\Seq \; (\Right \; (\inj \; \ONE \textcolor{blue}{a} \quad \textcolor{blue}{a}\quad (\Right \;\textcolor{blue}{\Empty})))) \; \Stars \, [])$
																		   & \\ $=$\\
																		   $\Left \; (\Seq \; (\Seq \; (\Right \; (\Seq \; (\mkeps \; \ONE)\;(\inj \;\textcolor{blue}{a} \; \textcolor{blue}{a} \; \textcolor{blue}{\Empty})) ) ) \; \Stars \, [] )$
																		   & \\ $=$\\
																		   $\Left \; (\Seq \; (\Seq \; (\Right \; (\Seq \; \Empty \; \textcolor{blue}{a}) ) ) \; \Stars \, [] )$
 	\end{tabular}
\end{center}
\noindent
We will now introduce the proofs related to $\inj$ and
$\lexer$. These proofs have originally been found by Ausaf et al.
in their 2016 work \cite{AusafDyckhoffUrban2016}.
Proofs to some of these lemmas have also been discussed in more detail in 
Ausaf's thesis \cite{Ausaf}.
Nevertheless, we still introduce these proofs, to make this thesis
self-contained as we show inductive variants used in these proofs break
after more simplifications are introduced.
%Also note that lemmas like \ref{}described in this thesis is a more faithful
%representation of the actual accompanying Isabelle formalisation, and 
%therefore 



Recall that lemma 
\ref{mePosix} tells us that
$\mkeps$ always generates the POSIX value.
The function $\inj$ preserves the POSIXness, provided
the value before injection is POSIX, namely
\begin{lemma}\label{injPosix}
	If$(r \backslash c, s) \rightarrow v $,
	then $(r, c :: s) \rightarrow (\inj r \; c\; v)$.
\end{lemma}
\begin{proof}
	By induction on $r$.
	The non-trivial cases are sequence and star.
	When $r = a \cdot b$, there can be
	three cases for the value $v$ satisfying $\vdash v:a\backslash c$.
	We give the reasoning why $\inj \; r \; c \; v$ is POSIX in each
	case.
	\begin{itemize}
		\item
			$v = \Seq \; v_a \; v_b$.\\
			The ``not nullable'' clause of the $\inj$ function is taken:
			\begin{center}
				\begin{tabular}{lcl}
					$\inj \; r \; c \; v$ &   $=$ & $ \inj \;\; (a \cdot b) \;\; c \;\; (\Seq \; v_a \; v_b) $\\
					& $=$ & $\Seq \; (\inj \;a \; c \; v_a) \; v_b$
				\end{tabular}
			\end{center}
			We know that there exists a unique pair of
			$s_a$ and $s_b$ satisfying	
				$(a \backslash c, s_a) \rightarrow v_a$,
				$(b , s_b) \rightarrow v_b$, and
				$\nexists s_3 \; s_4. s_3 \neq [] \land s_a @ s_3 \in 
				L \; (a\backslash c) \land
				s_4 \in L \; b$.
			The last condition gives us
			$\nexists s_3 \; s_4. s_3 \neq [] \land (c :: s_a )@ s_3 \in 
				L \; a \land
				s_4 \in L \; b$.
			By induction hypothesis, $(a, c::s_a) \rightarrow \inj \; a \; c \; v_a $ holds,
			and this gives us
			\[
				(a\cdot b, (c::s_a)@s_b) \rightarrow \Seq \; (\inj \; a\;c \;v_a) \; v_b.
			\]
		\item
			$v = \Left \; (\Seq \; v_a \; v_b)$\\
			The argument is almost identical to the above case,	
			except that a different clause of $\inj$ is taken:
			\begin{center}
				\begin{tabular}{lcl}
					$\inj \; r \; c \; v$ &   $=$ & $ \inj \;\; (a \cdot b) \;\; c \;\; (\Left \; (\Seq \; v_a \; v_b)) $\\
					& $=$ & $\Seq \; (\inj \;a \; c \; v_a) \; v_b$
				\end{tabular}
			\end{center}
			With similar reasoning, 

			\[
				(a\cdot b, (c::s_a)@s_b) \rightarrow \Seq \; (\inj \; a\;c \;v_a) \; v_b.
			\]
			again holds.
		\item 
			$v = \Right \; v_b$\\
			Again the injection result would be 
			\begin{center}
				\begin{tabular}{lcl}
					$\inj \; r \; c \; v$ &   $=$ & $ \inj \;\; (a \cdot b) \;\; c \;\; \Right \; (v_b) $\\
					& $=$ & $\Seq \; (\mkeps \; a) \; (\inj \;b \; c\; v_b)$
				\end{tabular}
			\end{center}
			We know that $a$ must be nullable,
			allowing us to call $\mkeps$ and get
			\[
				(a, []) \rightarrow \mkeps \; a.
			\]
			Also, by inductive hypothesis
			\[
				(b, c::s) \rightarrow \inj\; b \; c \; v_b
			\]
			holds.
			In addition, as
			$\Right \;v_b$  instead of $\Left \ldots$ is 
			the POSIX value for $v$, it must be the case
			that $s \notin L \;( (a\backslash c)\cdot b)$.
			This tells us that 
			\[
				\nexists s_3 \; s_4.
				s_3 @s_4 = s  \land s_3 \in L \; (a\backslash c) 
				\land s_4 \in L \; b
			\]
			which translates to
			\[
				\nexists s_3 \; s_4. \; s_3 \neq [] \land
				s_3 @s_4 = c::s  \land s_3 \in L \; a 
				\land s_4 \in L \; b.
			\]
			(Which says there cannot be a longer 
			initial split for $s$ other than the empty string.)
			Therefore we have $\Seq \; (\mkeps \; a) \;(\inj \;b \; c\; v_b)$
			as the POSIX value for $a\cdot b$.
	\end{itemize}
	The star case can be proven similarly.
\end{proof}
\noindent







%TODO: Cut from previous lexer def, need to make coherent
The central property of the $\lexer$ is that it gives the correct result
according to
POSIX rules. 
\begin{theorem}\label{lexerCorrectness}
	The $\lexer$ based on derivatives and injections is correct: 
	\begin{center}
		\begin{tabular}{lcl}
			$\lexer \; r \; s = \Some(v)$ & $ \Longleftrightarrow$ & $ (r, \; s) \rightarrow v$\\
			$\lexer \;r \; s = \None $ & $\Longleftrightarrow$ & $ \neg(\exists v. (r, s) \rightarrow v)$
		\end{tabular}
	\end{center}
\end{theorem} 
\begin{proof}
By induction on $s$. $r$ generalising over an arbitrary regular expression.
The $[]$ case is proven by an application of lemma \ref{mePosix}, and the inductive case
by lemma \ref{injPosix}.
\end{proof}
\noindent
As we did earlier in this chapter with the matcher, one can 
introduce simplification on the regular expression in each derivative step.
However, due to lexing, one needs to do a backward phase (w.r.t the forward derivative phase)
and ensure that
the values align with the regular expression at each step.
Therefore one has to
be careful not to break the correctness, as the injection 
function heavily relies on the structure of 
the regular expressions and values being aligned.
This can be achieved by recording some extra rectification functions
during the derivatives step and applying these rectifications in 
each run during the injection phase.
With extra care
one can show that POSIXness will not be affected
by the simplifications listed here \cite{AusafDyckhoffUrban2016}. 
\begin{center}
	\begin{tabular}{lcl}
		$\simp \; r_1 \cdot r_2 $ & $ \dn$ & 
		$(\simp \; r_1,  \simp \; r_2) \; \textit{match}$\\
					  & & $\quad \case \; (\ZERO, \_) \Rightarrow \ZERO$\\
					  & & $\quad \case \; (\_, \ZERO) \Rightarrow \ZERO$\\
					  & & $\quad \case \; (\ONE, r_2') \Rightarrow r_2'$\\
					  & & $\quad \case \; (r_1', \ONE) \Rightarrow r_1'$\\
					  & & $\quad \case \; (r_1', r_2') \Rightarrow r_1'\cdot r_2'$\\
		$\simp \; r_1 + r_2$ & $\dn$ & $(\simp \; r_1, \simp \; r_2) \textit{match}$\\
				     & & $\quad \; \case \; (\ZERO, r_2') \Rightarrow r_2'$\\
				     & & $\quad \; \case \; (r_1', \ZERO) \Rightarrow r_1'$\\
				     & & $\quad \; \case \; (r_1', r_2') \Rightarrow r_1' + r_2'$\\
		$\simp \; r$ & $\dn$ & $r\quad\quad (otherwise)$
				   
	\end{tabular}
\end{center}

However, one can still end up
with exploding derivatives,
even with the simple-minded simplification rules allowed
in an injection-based lexer.
\section{A Case Requiring More Aggressive Simplifications}
For example, when starting with the regular
expression $(a^* \cdot a^*)^*$ and building just over
a dozen successive derivatives 
w.r.t.~the character $a$, one obtains a derivative regular expression
with millions of nodes (when viewed as a tree)
even with the mentioned simplifications.
\begin{figure}[H]
\begin{center}
\begin{tikzpicture}
\begin{axis}[
    xlabel={$n$},
    ylabel={size},
    legend entries={Simple-Minded Simp, Naive Matcher},  
    legend pos=north west,
    legend cell align=left]
\addplot[red,mark=*, mark options={fill=white}] table {BetterWaterloo.data};
\addplot[blue,mark=*, mark options={fill=white}] table {BetterWaterloo1.data};
\end{axis}
\end{tikzpicture} 
\end{center}
\caption{Size of $(a^*\cdot a^*)^*$ against $\protect\underbrace{aa\ldots a}_\text{n \textit{a}s}$}
\end{figure}\label{fig:BetterWaterloo}
   
That is because Sulzmann and Lu's 
injection-based lexing algorithm keeps a lot of 
"useless" values that will not be used. 
These different ways of matching will grow exponentially with the string length.
Consider the case
\[
	r= (a^*\cdot a^*)^* \quad and \quad
	s=\underbrace{aa\ldots a}_\text{n \textit{a}s}
\]
as an example.
This is a highly ambiguous regular expression, with
many ways to split up the string into multiple segments for
different star iterations,
and for each segment 
multiple ways of splitting between 
the two $a^*$ sub-expressions.
When $n$ is equal to $1$, there are two lexical values for
the match:
\[
	\Stars \; [\Seq \; (\Stars \; [\Char \; a])\; (\Stars \; [])] \quad (v1)
\]
and
\[
	\Stars \; [\Seq \; (\Stars \; [])\; (\Stars \; [\Char \; a])] \quad (v2)
\]
The derivative of $\derssimp \;s \; r$ is
\[
	(a^*a^* + a^*)\cdot(a^*a^*)^*.
\]
The $a^*a^*$ and $a^*$ in the first child of the above sequence
correspond to value 1 and value 2, respectively.
When $n=2$, the number goes up to 7:
\[
	\Stars \; [\Seq \; (\Stars \; [\Char \; a, \Char \; a])\; (\Stars \; [])]
\]

\[
	\Stars \; [\Seq \; (\Stars \; [\Char \; a])\; (\Stars \; [\Char \; a])]
\]

\[
	\Stars \; [\Seq \; (\Stars \; [])\; (\Stars \; [\Char \; a, \Char \; a])]
\]

\[
	\Stars \; [\Seq \; (\Stars \; [\Char \; a])\; (\Stars \; []), \Seq \; (\Stars \; [\Char\;a])\; (\Stars\; []) ]
\]

\[
	\Stars \; [\Seq \; (\Stars \; [\Char \; a])\; (\Stars \; []), 
	 	   \Seq \; (\Stars \; [])\; (\Stars \; [\Char \; a])
		  ] 
\]

\[
	\Stars \; [
	 	   \Seq \; (\Stars \; [])\; (\Stars \; [\Char \; a]),
	 	   \Seq \; (\Stars \; [])\; (\Stars \; [\Char \; a])
		  ] 
\]
and
\[
	\Stars \; [
	 	   \Seq \; (\Stars \; [])\; (\Stars \; [\Char \; a]),
		   \Seq \; (\Stars \; [\Char \; a])\; (\Stars \; [])
		  ] 
\]
And $\derssimp \; aa \; (a^*a^*)^*$ is
\[
	((a^*a^* + a^*)+a^*)\cdot(a^*a^*)^* + 
	(a^*a^* + a^*)\cdot(a^*a^*)^*.
\]
which removes two out of the seven terms corresponding to the
seven distinct lexical values.

It is not surprising that there are exponentially many 
distinct lexical values that cannot be eliminated by 
the simple-minded simplification of $\derssimp$. 
A lexer without a good enough strategy to 
deduplicate will naturally
have an exponential runtime on highly
ambiguous regular expressions because there
are exponentially many matches.
For this particular example, it seems
that the number of distinct matches growth
speed is proportional to $(2n)!/(n!(n+1)!)$ ($n$ being the input length).

On the other hand, the
 $\POSIX$ value for $r= (a^*\cdot a^*)^*$  and 
$s=\underbrace{aa\ldots a}_\text{n \textit{a}s}$ is  
\[
	\Stars\,
	[\Seq \; (\Stars\,[\underbrace{\Char(a),\ldots,\Char(a)}_\text{n iterations}]), \Stars\,[]].
\]
At any moment, the  subterms in a regular expression
that will potentially result in a POSIX value is only
a minority among the many other terms,
and one can remove the ones that are not possible to 
be POSIX.
In the above example,
\begin{equation}\label{eqn:growth2}
	((a^*a^* + \underbrace{a^*}_\text{A})+\underbrace{a^*}_\text{duplicate of A})\cdot(a^*a^*)^* + 
	\underbrace{(a^*a^* + a^*)\cdot(a^*a^*)^*}_\text{further simp removes this}.
\end{equation}
can be further simplified by 
removing the underlined term first,
which would open up possibilities
of further simplification that removes the
underbraced part.
The result would be 
\[
	(\underbrace{a^*a^*}_\text{term 1} + \underbrace{a^*}_\text{term 2})\cdot(a^*a^*)^*.
\]
with corresponding values
\begin{center}
	\begin{tabular}{lr}
		$\Stars \; [\Seq \; (\Stars \; [\Char \; a, \Char \; a])\; (\Stars \; [])]$  & $(\text{term 1})$\\
		$\Stars \; [\Seq \; (\Stars \; [\Char \; a])\; (\Stars \; [\Char \; a])]  $ &  $(\text{term 2})$
	\end{tabular}
\end{center}
Other terms with an underlying value, such as
\[
	\Stars \; [\Seq \; (\Stars \; [])\; (\Stars \; [\Char \; a, \Char \; a])]
\]
do not to contribute a POSIX lexical value,
and therefore can be thrown away.

Ausaf et al. \cite{AusafDyckhoffUrban2016} 
have come up with some simplification steps, 
and incorporated the simplification into $\lexer$.
They call this lexer $\textit{lexer}_{simp}$ and proved that
\[
	\lexer \; r\; s = \textit{lexer}_{simp} \; r \; s
\]
The function $\textit{lexer}_{simp}$
involves some fiddly manipulation of value rectification,
which we omit here.
however those steps
are not yet sufficiently strong, to achieve the above effects.
And even with these relatively mild simplifications, the proof
is already quite a bit more complicated than the theorem \ref{lexerCorrectness}.
One would need to prove something like this: 
\[
	\textit{If}\; (\textit{snd} \; (\textit{simp} \; r\backslash c), s) \rightarrow  v  \;\;
	\textit{then}\;\; (r, c::s) \rightarrow 
	\inj\;\, r\,  \;c \;\, ((\textit{fst} \; (\textit{simp} \; r \backslash c))\; v).
\]
instead of the simple lemma \ref{injPosix}, where now $\textit{simp}$
not only has to return a simplified regular expression,
but also what specific simplifications 
have been done as a function on values
showing how one can transform the value
underlying the simplified regular expression
to the unsimplified one.

We therefore choose a slightly different approach
also described by Sulzmann and Lu to
get better simplifications, which uses
some augmented data structures compared to 
plain regular expressions.
We call them \emph{annotated}
regular expressions.
With annotated regular expressions,
we can avoid creating the intermediate values $v_1,\ldots v_n$ and a 
second phase altogether.
We introduce this new datatype and the 
corresponding algorithm in the next chapter.




