Addressed Gerog "can't understand 'erase messes with structure'" comment
% Chapter Template% Main chapter title\chapter{Correctness of Bit-coded Algorithm with Simplification}\label{Bitcoded2} % Change X to a consecutive number; for referencing this chapter elsewhere, use \ref{ChapterX}%Then we illustrate how the algorithm without bitcodes falls short for such aggressive %simplifications and therefore introduce our version of the bitcoded algorithm and %its correctness proof in %Chapter 3\ref{Chapter3}. %\section{Overview}\marginpar{\em Added a completely new \\overview section, \\highlighting\\ contributions.}This chapteris the point from which novel contributions of this PhD project are introducedin detail. The material in thepreviouschapters is necessary for this thesis,because it provides the context for why we need a new framework forthe proof of $\blexersimp$.We will first introduce why aggressive simplifications are needed, after which weprovide our algorithm, contrasting with Sulzmann and Lu's simplifications.We then explain how our simplifications makereusing $\blexer$'s correctness proof impossible.%with some minor modificationsWe discuss possible fixes such as rectification functions and then introduce our proof, which involves a weaker inductiveinvariant than that used in the correctness proof of $\blexer$.\marginpar{Shortened overview.}%material for setting the scene of the formal proof we%are about to describe.\section{Simplifications by Sulzmann and Lu}\marginpar{moved \\simplification \\section to front \\to make coherent\\ sense.}The algorithms $\lexer$ and $\blexer$ work beautifully as functional programs, but not as practical code. One main reason for the slowness is dueto the size of intermediate representations--the derivative regular expressionstend to grow unbounded if the matching involved a large number of possible matches.Consider the derivatives of the following example $(a^*a^*)^*$:%and $(a^* + (aa)^*)^*$:\begin{center} \begin{tabular}{lcl} $(a^*a^*)^*$ & $ \stackrel{\backslash a}{\longrightarrow}$ & $ (a^*a^* + a^*)\cdot(a^*a^*)^*$\\ & $ \stackrel{\backslash a}{\longrightarrow} $ & $((a^*a^* + a^*) + a^*)\cdot(a^*a^*)^* + (a^*a^* + a^*)\cdot(a^*a^*)^*$\\ & $\stackrel{\backslash a}{ \longrightarrow} $ & $\ldots$\\ \end{tabular}\end{center}\noindentFrom the second derivative several duplicate sub-expressions already needs to be eliminated (possiblebitcodes are omitted to make the presentation more concisebecause they are not the key part of the simplifications).A simple-minded simplification function cannot simplifythe third regular expression in the above chain of derivativeregular expressions, namely\begin{center}$((a^*a^* + a^*) + a^*)\cdot(a^*a^*)^* + (a^*a^* + a^*)\cdot(a^*a^*)^*$\end{center}because the duplicates arenot next to each other, and therefore the rule$r+ r \rightarrow r$ from $\textit{simp}$ does not fire.One would expect a better simplification function to work in the following way:\begin{gather*} ((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}.\\ \bigg\downarrow (1) \\ (a^*a^* + a^* \color{gray} + a^* \color{black})\cdot(a^*a^*)^* + \underbrace{(a^*a^* + a^*)\cdot(a^*a^*)^*}_\text{further simp removes this} \\ \bigg\downarrow (2) \\ (a^*a^* + a^* )\cdot(a^*a^*)^* \color{gray} + (a^*a^* + a^*) \cdot(a^*a^*)^*\\ \bigg\downarrow (3) \\ (a^*a^* + a^* )\cdot(a^*a^*)^* \end{gather*}\noindentIn the first step, the nested alternative regular expression$(a^*a^* + a^*) + a^*$ is flattened into $a^*a^* + a^* + a^*$.Now the third term $a^*$ can clearly be identified as a duplicateand therefore removed in the second step. This causes the twotop-level terms to become the same and the second $(a^*a^*+a^*)\cdot(a^*a^*)^*$ removed in the final step.Sulzmann and Lu's simplification function (using our notations) can achieve thissimplification:\begin{center} \begin{tabular}{lcl} $\textit{simp}\_{SL} \; _{bs}(_{bs'}\ONE \cdot r)$ & $\dn$ & $\textit{if} \; (\textit{zeroable} \; r)\; \textit{then} \;\; \ZERO$\\ & &$\textit{else}\;\; \fuse \; (bs@ bs') \; r$\\ $\textit{simp}\_{SL} \;(_{bs}r_1\cdot r_2)$ & $\dn$ & $\textit{if} \; (\textit{zeroable} \; r_1 \; \textit{or} \; \textit{zeroable}\; r_2)\; \textit{then} \;\; \ZERO$\\ & & $\textit{else}\;\;_{bs}((\textit{simp}\_{SL} \;r_1)\cdot (\textit{simp}\_{SL} \; r_2))$\\ $\textit{simp}\_{SL} \; _{bs}\sum []$ & $\dn$ & $\ZERO$\\ $\textit{simp}\_{SL} \; _{bs}\sum ((_{bs'}\sum rs_1) :: rs_2)$ & $\dn$ & $_{bs}\sum ((\map \; (\fuse \; bs')\; rs_1) @ rs_2)$\\ $\textit{simp}\_{SL} \; _{bs}\sum[r]$ & $\dn$ & $\fuse \; bs \; (\textit{simp}\_{SL} \; r)$\\ $\textit{simp}\_{SL} \; _{bs}\sum(r::rs)$ & $\dn$ & $_{bs}\sum (\nub \; (\filter \; (\neg\zeroable)\;((\textit{simp}\_{SL} \; r) :: \map \; \textit{simp}\_{SL} \; rs)))$\\ \end{tabular}\end{center}\noindentThe $\textit{zeroable}$ predicate tests whether the regular expressionis equivalent to $\ZERO$, andcan be defined as:\begin{center} \begin{tabular}{lcl} $\zeroable \; _{bs}\sum (r::rs)$ & $\dn$ & $\zeroable \; r\;\; \land \;\; \zeroable \;_{[]}\sum\;rs $\\ $\zeroable\;_{bs}(r_1 \cdot r_2)$ & $\dn$ & $\zeroable\; r_1 \;\; \lor \;\; \zeroable \; r_2$\\ $\zeroable\;_{bs}r^*$ & $\dn$ & $\textit{false}$ \\ $\zeroable\;_{bs}c$ & $\dn$ & $\textit{false}$\\ $\zeroable\;_{bs}\ONE$ & $\dn$ & $\textit{false}$\\ $\zeroable\;_{bs}\ZERO$ & $\dn$ & $\textit{true}$ \end{tabular}\end{center}\noindentThe \begin{center} \begin{tabular}{lcl} $\textit{simp}\_{SL} \; _{bs}\sum ((_{bs'}\sum rs_1) :: rs_2)$ & $\dn$ & $_{bs}\sum ((\map \; (\fuse \; bs')\; rs_1) @ rs_2)$\\ \end{tabular}\end{center}\noindentclause does flatten the alternative as required in step (1),but $\textit{simp}\_{SL}$ is insufficient if we want to do steps (2) and (3),as these ``identical'' terms have different bit-annotations.They also suggested that the $\textit{simp}\_{SL} $ function should beapplied repeatedly until a fixpoint is reached.We call this construction $\textit{SLSimp}$:\begin{center} \begin{tabular}{lcl} $\textit{SLSimp} \; r$ & $\dn$ & $\textit{while}((\textit{simp}\_{SL} \; r)\; \cancel{=} \; r)$ \\ & & $\quad r := \textit{simp}\_{SL} \; r$\\ & & $\textit{return} \; r$ \end{tabular}\end{center}We call the operation of alternatingly applying derivatives and simplifications(until the string is exhausted) Sulz-simp-derivative,written $\backslash_{SLSimp}$:\begin{center}\begin{tabular}{lcl} $r \backslash_{SLSimp} (c\!::\!s) $ & $\dn$ & $(\textit{SLSimp} \; (r \backslash c)) \backslash_{SLSimp}\, s$ \\$r \backslash_{SLSimp} [\,] $ & $\dn$ & $r$\end{tabular}\end{center}\noindentAfter the derivatives have been taken, the bitcodesare extracted and decoded in the same manneras $\blexer$:\begin{center}\begin{tabular}{lcl} $\textit{blexer\_SLSimp}\;r\,s$ & $\dn$ & $\textit{let}\;a = (r^\uparrow)\backslash_{SLSimp}\, s\;\textit{in}$\\ & & $\;\;\textit{if}\; \textit{bnullable}(a)$\\ & & $\;\;\textit{then}\;\textit{decode}\,(\textit{bmkeps}\,a)\,r$\\ & & $\;\;\textit{else}\;\textit{None}$\end{tabular}\end{center}\noindentWe implemented this lexing algorithm in Scala, and found that the final derivative regular expressionsize still grows exponentially (note the logarithmic scale):\begin{figure}[H] \centering\begin{tikzpicture}\begin{axis}[ xlabel={$n$}, ylabel={size}, ymode = log, legend entries={Final Derivative Size}, legend pos=north west, legend cell align=left]\addplot[red,mark=*, mark options={fill=white}] table {SulzmannLuLexer.data};\end{axis}\end{tikzpicture} \caption{Lexing the regular expression $(a^*a^*)^*$ against strings of the form$\protect\underbrace{aa\ldots a}_\text{n \textit{a}s}$ using Sulzmann and Lu's lexer}\label{SulzmannLuLexer}\end{figure}\noindentAt $n= 20$ we already get an out-of-memory error with Scala's normal JVM heap size settings.In fact their simplification does not improve much overthe simple-minded simplifications we have shown in \ref{fig:BetterWaterloo}.The time required also grows exponentially:\begin{figure}[H] \centering\begin{tikzpicture}\begin{axis}[ xlabel={$n$}, ylabel={time}, %ymode = log, legend entries={time in secs}, legend pos=north west, legend cell align=left]\addplot[red,mark=*, mark options={fill=white}] table {SulzmannLuLexerTime.data};\end{axis}\end{tikzpicture} \caption{Lexing the regular expression $(a^*a^*)^*$ against strings of the form$\protect\underbrace{aa\ldots a}_\text{n \textit{a}s}$ using Sulzmann and Lu's lexer}\label{SulzmannLuLexerTime}\end{figure}\noindentwhich seems like a counterexample for Sulzmann and Lu's linear complexity claimin their paper \cite{Sulzmann2014}:\begin{quote}\it``Linear-Time Complexity Claim \\It is easy to see that each call of one of the functions/operations:simp, fuse, mkEpsBC and isPhi leads to subcalls whose number is bound by the size of the regular expression involved. We claim that thanks to aggressively applying simp this size remains finite. Hence, we can argue that the above mentioned functions/operations have constant time complexity which implies that we can incrementally compute bit-coded parse trees in linear time in the size of the input.'' \end{quote}\noindentThe assumption that the size of the regular expressionsin the algorithmwould stay below a finite constant is not true, at least not in theexamples we considered.The main reason behind this is that (i) Haskell's $\textit{nub}$function requires identical annotations between two annotated regular expressions to qualify as duplicates,and therefore cannot simplify cases like $_{SZZ}a^*+_{SZS}a^*$even if both $a^*$ denote the same language, and(ii) the ``flattening'' only applies to the head of the listin the \begin{center} \begin{tabular}{lcl} $\textit{simp}\_{SL} \; _{bs}\sum ((_{bs'}\sum rs_1) :: rs_2)$ & $\dn$ & $_{bs}\sum ((\map \; (\fuse \; bs')\; rs_1) @ rs_2)$\\ \end{tabular}\end{center}\noindentclause, and therefore is not strong enough to simplify allneeded parts of the regular expression. Moreover,the $\textit{simp}\_{SL}$ function is applied repeatedlyin each derivative step until a fixed point is reached, which makes the algorithm even moreunpredictable and inefficient.%To not get ``caught off guard'' by%these counterexamples,%one needs to be more careful when designing the%simplification function and making claims about them.\section{Our $\textit{Simp}$ Function}We will now introduce our own simplification function.%by making a contrast with $\textit{simp}\_{SL}$.We also describethe ideas behind Sulzmann and Lu's $\textit{simp}\_{SL}$algorithm and why it fails to achieve the desired effect of keeping the sizes of derivatives finitely bounded. In addition, our simplification function will come with a formalcorrectness proof.\subsection{Flattening Nested Alternatives}The idea behind the clause\begin{center} $\textit{simp}\_{SL} \; _{bs}\sum ((_{bs'}\sum rs_1) :: rs_2) \quad \dn \quad _{bs}\sum ((\map \; (\fuse \; bs')\; rs_1) @ rs_2)$\end{center}is that it allowsduplicate removal of regular expressions at different``levels'' of alternatives.For example, this would help with thefollowing simplification:\begin{center}$(a+r)+r \longrightarrow a+r$\end{center}The problem is that only the head elementis ``spilled out''.It is more desirableto flattenan entire list to open up possibilities for further simplificationswith later regular expressions.Not flattening the rest of the elements also means thatthe later de-duplication process does not fully remove further duplicates.For example,using $\textit{simp}\_{SL}$ we cannotsimplify\begin{center} $((a^* a^*)+\underline{(a^* + a^*)})\cdot (a^*a^*)^*+((a^*a^*)+a^*)\cdot (a^*a^*)^*$\end{center}due to the underlined part not being the head of the alternative.We define our flatten operation so that it flattens the entire list: \begin{center} \begin{tabular}{@{}lcl@{}} $\textit{flts} \; (_{bs}\sum \textit{as}) :: \textit{as'}$ & $\dn$ & $(\textit{map} \; (\textit{fuse}\;bs)\; \textit{as}) \; @ \; \textit{flts} \; as' $ \\ $\textit{flts} \; \ZERO :: as'$ & $\dn$ & $ \textit{flts} \; \textit{as'} $ \\ $\textit{flts} \; a :: as'$ & $\dn$ & $a :: \textit{flts} \; \textit{as'}$ \quad(otherwise) \end{tabular} \end{center} \noindentOur $\flts$ operation also throws away $\ZERO$sas they do not contribute to a lexing result.\subsection{Duplicate Removal}After flattening is done, we can deduplicate.The de-duplicate function is called $\distinctBy$,and that is where we make our second improvement overSulzmann and Lu's simplification method.The process goes as follows:\begin{center}$rs \stackrel{\textit{flts}}{\longrightarrow} rs_{flat} \xrightarrow{\distinctBy \; rs_{flat} \; \rerases\; \varnothing} rs_{distinct}$%\stackrel{\distinctBy \; %rs_{flat} \; \erase\; \varnothing}{\longrightarrow} \; rs_{distinct}$\end{center}where the $\distinctBy$ function is defined as:\begin{center} \begin{tabular}{@{}lcl@{}} $\distinctBy \; [] \; f\; acc $ & $ =$ & $ []$\\ $\distinctBy \; (x :: xs) \; f \; acc$ & $=$ & $\quad \textit{if} (f \; x \in acc)\;\; \textit{then} \;\; \distinctBy \; xs \; f \; acc$\\ & & $\quad \textit{else}\;\; x :: (\distinctBy \; xs \; f \; (\{f \; x\} \cup acc))$ \end{tabular}\end{center}\noindentThe reason we define a distinct function under a mapping $f$ is becausewe want to eliminate regular expressions that are syntactically the same,but have different bit-codes.For example, we can remove the second $a^*a^*$ from$_{ZSZ}a^*a^* + _{SZZ}a^*a^*$, because itrepresents a match with shorter initial sub-match (and therefore is definitely not POSIX),and will be discarded by $\bmkeps$ later.\begin{center} $_{ZSZ}\underbrace{a^*}_{ZS:\; match \; 1\; times\quad}\underbrace{a^*}_{Z: \;match\; 1 \;times} + _{SZZ}\underbrace{a^*}_{S: \; match \; 0 \; times\quad}\underbrace{a^*}_{ZZ: \; match \; 2 \; times} $\end{center}%$_{bs1} r_1 + _{bs2} r_2 \text{where} (r_1)_{\downarrow} = (r_2)_{\downarrow}$Due to the way our algorithm works,the matches that conform to the POSIX standard will always be placed further to the left. When we traverse the list from left to right,regular expressions we have already seenwill definitely not contribute to a POSIX value,even if they are attached with different bitcodes.These duplicates therefore need to be removed.To achieve this, we call $\rerases$ as the function $f$ during the distinctionoperation. The function$\rerases$ is very similar to $\erase$, except that it preserves the structurewhen erasing an alternative regular expression.The reason why we use $\rerases$ instead of $\erase$ is thatit keeps the structures of alternative annotated regular expressionswhereas $\erase$ would turn it back into a binary tree structure.Not having to mess with the structure greatly simplifies the finiteness proof in chapter \ref{Finite}.We give the definitions of $\rerases$ here together withthe new datatype used by $\rerases$ (as our plainregular expression datatype does not allow non-binary alternatives).For now we can think of $\rerases$ as the function erase ($(\_)_\downarrow$) defined in chapter \ref{Bitcoded1}and $\rrexp$ as plain regular expressions, but having a general list constructorfor alternatives:\begin{figure}[H]\begin{center} $\rrexp ::= \RZERO \mid \RONE \mid \RCHAR{c} \mid \RSEQ{r_1}{r_2} \mid \RALTS{rs} \mid \RSTAR{r} $\end{center}\caption{$\rrexp$: plain regular expressions, but with $\sum$ alternative constructor}\label{rrexpDef}\end{figure}The function $\rerases$ we define as follows:\begin{center}\begin{tabular}{lcl}$\rerase{\ZERO}$ & $\dn$ & $\RZERO$\\$\rerase{_{bs}\ONE}$ & $\dn$ & $\RONE$\\ $\rerase{_{bs}\mathbf{c}}$ & $\dn$ & $\RCHAR{c}$\\$\rerase{_{bs}r_1\cdot r_2}$ & $\dn$ & $\RSEQ{\rerase{r_1}}{\rerase{r_2}}$\\$\rerase{_{bs}\sum as}$ & $\dn$ & $\RALTS{\map \; \rerase{\_} \; as}$\\$\rerase{_{bs} a ^*}$ & $\dn$ & $\rerase{a}^*$\end{tabular}\end{center}We will provide more details in \ref{whyRerase} for why a newerase function and new datatype is needed. But briefly speakingit is for backward-compatibility with $\blexer$'s correctness proof and the path we (naturally) took during our proof engineering of the finiteness property.\subsection{Putting Things Together}We can now give the definition of our simplification function:%that looks somewhat similar to our Scala code is \begin{center} \begin{tabular}{@{}lcl@{}} $\textit{bsimp} \; (_{bs}a_1\cdot a_2)$ & $\dn$ & $ \textit{bsimp}_{ASEQ} \; bs \;(\textit{bsimp} \; a_1) \; (\textit{bsimp} \; a_2) $ \\ $\textit{bsimp} \; (_{bs}\sum \textit{as})$ & $\dn$ & $\textit{bsimp}_{ALTS} \; \textit{bs} \; (\textit{distinctBy} \; ( \textit{flatten} ( \textit{map} \; bsimp \; as)) \; \rerases \; \varnothing) $ \\ $\textit{bsimp} \; a$ & $\dn$ & $\textit{a} \qquad \textit{otherwise}$ \end{tabular} \end{center} \noindentThe simplification (named $\textit{bsimp}$ for \emph{b}it-coded) does a pattern matching on the regular expression.When it detects that the regular expression is an alternative orsequence, it will try to simplify its children regular expressionsrecursively and then see if one of the children turns into $\ZERO$ or$\ONE$, which might trigger further simplification at the current level.Current level simplifications are handled by the function $\textit{bsimp}_{ASEQ}$,using rules such as $\ZERO \cdot r \rightarrow \ZERO$ and $\ONE \cdot r \rightarrow r$.\begin{center} \begin{tabular}{@{}lcl@{}} $\textit{bsimp}_{ASEQ} \; bs\; a \; b$ & $\dn$ & $ (a,\; b) \textit{match}$\\ &&$\quad\textit{case} \; (\ZERO, \_) \Rightarrow \ZERO$ \\ &&$\quad\textit{case} \; (\_, \ZERO) \Rightarrow \ZERO$ \\ &&$\quad\textit{case} \; (_{bs1}\ONE, a_2') \Rightarrow \textit{fuse} \; (bs@bs_1) \; a_2'$ \\ &&$\quad\textit{case} \; (a_1', a_2') \Rightarrow _{bs}a_1' \cdot a_2'$ \end{tabular}\end{center}\noindentThe most involved part is the $\sum$ clause, where we first call $\flts$ onthe simplified children regular expression list $\textit{map}\; \textit{bsimp}\; \textit{as}$,and then call $\distinctBy$ on that list. The predicate used in $\distinctBy$ for determining whether two elements are the same is $\rerases \; r_1 = \rerases\; r_2$.Finally, depending on whether the regular expression list $as'$ has turned into asingleton or empty list after $\flts$ and $\distinctBy$, $\textit{bsimp}_{ALTS}$decides whether to keep the current level constructor $\sum$ as it is, and removes it when there are fewer than two elements:\begin{center} \begin{tabular}{lcl} $\textit{bsimp}_{ALTS} \; bs \; as'$ & $ \dn$ & $ as' \; \textit{match}$\\ &&$\quad\textit{case} \; [] \Rightarrow \ZERO$ \\ &&$\quad\textit{case} \; a :: [] \Rightarrow \textit{fuse bs a}$ \\ &&$\quad\textit{case} \; as' \Rightarrow _{bs}\sum \textit{as'}$\\ \end{tabular}\end{center}Having defined the $\textit{bsimp}$ function,we add it as a phase after a derivative is taken.\begin{center} \begin{tabular}{lcl} $a \backslash_{bsimp} c$ & $\dn$ & $\textit{bsimp}(a \backslash c)$ \end{tabular}\end{center}%Following previous notations%when extending from derivatives w.r.t.~character to derivative%w.r.t.~string, we define the derivative that nests simplifications %with derivatives:%\comment{simp in the [] case?}We extend this from characters to strings:\begin{center}\begin{tabular}{lcl}$a \backslash_{bsimps} (c\!::\!s) $ & $\dn$ & $(a \backslash_{bsimp}\, c) \backslash_{bsimps}\, s$ \\$a \backslash_{bsimps} [\,] $ & $\dn$ & $a$\end{tabular}\end{center}\noindentThe lexer that extracts bitcodes from the derivatives with simplifications from our $\simp$ functionis called $\blexersimp$:\begin{center}\begin{tabular}{lcl} $\textit{blexer\_simp}\;r\,s$ & $\dn$ & $\textit{let}\;a = (r^\uparrow)\backslash_{bsimp}\, s\;\textit{in}$\\ & & $\;\;\textit{if}\; \textit{bnullable}(a)$\\ & & $\;\;\textit{then}\;\textit{decode}\,(\textit{bmkeps}\,a)\,r$\\ & & $\;\;\textit{else}\;\textit{None}$\end{tabular}\end{center}\noindentThis algorithm keeps the regular expression size small, as we shall demonstrate with some examples in the next section.\subsection{Examples $(a+aa)^*$ and $(a^*\cdot a^*)^*$After Simplification}Recall theprevious $(a^*a^*)^*$ examplewhere $\textit{simp}\_{SL}$ could notprevent the fast growth (over3 million nodes just below $20$ input length)will be reduced to just 15 and stays constant no matter how long theinput string is.This is shown in the graphs below.\begin{figure}[H]\begin{center}\begin{tabular}{ll}\begin{tikzpicture}\begin{axis}[ xlabel={$n$}, ylabel={derivative size}, width=7cm, height=4cm, legend entries={Lexer with $\textit{bsimp}$}, legend pos= south east, legend cell align=left]\addplot[red,mark=*, mark options={fill=white}] table {BitcodedLexer.data};\end{axis}\end{tikzpicture} %\label{fig:BitcodedLexer}&\begin{tikzpicture}\begin{axis}[ xlabel={$n$}, ylabel={derivative size}, width = 7cm, height = 4cm, legend entries={Lexer with $\textit{simp}\_{SL}$}, legend pos= north west, legend cell align=left]\addplot[red,mark=*, mark options={fill=white}] table {BetterWaterloo.data};\end{axis}\end{tikzpicture} \end{tabular}\end{center}\caption{Our Improvement over Sulzmann and Lu's in terms of size of the derivatives.}\end{figure}\noindentGiven the size difference, it is notsurprising that our $\blexersimp$ significantly outperforms$\textit{blexer\_SLSimp}$ by Sulzmann and Lu.Indeed $\blexersimp$ seems to be a correct algorithm that effectivelybounds the size of intermediate representations.\marginpar{\em more connecting material to make narration more coherent.}As promised we will use formal proofs to show that our speculationbased on these experimental results indeed hold.%intuitions indeed hold.In the next section we are going to establish that oursimplification preserves the correctness of the algorithm.\section{Correctness of $\blexersimp$}A natural thought would be to directly re-use the formalproof of $\blexer$'s correctness, with some minor modificationsbut keeping the way the induction is done.However we were not able to find a simple way to re-factorproof of \ref{blexerCorrect} in chapter \ref{Bitcoded1}.\subsection{Why $\textit{Blexer}$'s Proof Does Not Work}The fundamental reason is %we cannot extend the correctness proof of theorem 4because lemma \ref{retrieveStepwise} does not hold anymore when simplifications are involved.\marginpar{\em rephrased things \\so why new \\proof makes sense.}%The proof details are necessary materials for this thesis%because it provides necessary context to explain why we need a%new framework for the proof of $\blexersimp$, which involves%simplifications that cause structural changes to the regular expression.%A new formal proof of the correctness of $\blexersimp$, where the %proof of $\blexer$%is not applicatble in the sense that we cannot straightforwardly extend the%proof of theorem \ref{blexerCorrect} because lemma \ref{retrieveStepwise} does%not hold anymore.%This is because the structural induction on the stepwise correctness%of $\inj$ breaks due to each pair of $r_i$ and $v_i$ described%in chapter \ref{Inj} and \ref{Bitcoded1} no longer correspond to%each other.%In this chapter we introduce simplifications%for annotated regular expressions that can be applied to %each intermediate derivative result. This allows%us to make $\blexer$ much more efficient.%Sulzmann and Lu already introduced some simplifications for bitcoded regular expressions,%but their simplification functions could have been more efficient and in some cases needed fixing.In particular, the correctness theorem of the un-optimised bit-coded lexer $\blexer$ in chapter \ref{Bitcoded1} formalised by Ausaf et al.relies crucially on lemma \ref{retrieveStepwise} that saysany value can be retrieved in a stepwise manner, namely:\begin{equation}\label{eq:stepwise}%eqref: this proposition needs to be referred \vdash v : ((a_\downarrow) \backslash c) \implies \retrieve \; (a \backslash c) \; v= \retrieve \; a \; (\inj \; (a_\downarrow) \; c\; v)\end{equation}%This no longer holds once we introduce simplifications.The regular expressions $a$ and $a\backslash c$ correspond to the intermediateresult before and after the derivative with respect to $c$, and similarly$\inj\; a_\downarrow \; c \; v$ and $v$ correspond to the value before and after the derivative.They go in lockstep pairs\[ (a, \; \inj\; a_\downarrow \; c \; v)\; \text{and} \; (a\backslash c,\; v)\]and the structure of annotated regular expression and value within a pair always align with each other.As $\blexersimp$ integrates$\textit{bsimp}$ by applying it after each call to the derivatives function,%Simplifications are necessary to control the size of derivatives,%but they also destroy the structures of the regular expressions%such that \ref{eq:stepwise} does not hold.\begin{center}\begin{tabular}{lcl} $a \backslash_{bsimps} (c\!::\!s) $ & $\dn$ & $(\textit{bsimp} \; (a \backslash\, c)) \backslash_{bsimps}\, s$ \\%$r \backslash_{bsimps} [\,] $ & $\dn$ & $r$\end{tabular}%\begin{tabular}{lcl}% $\textit{blexer\_simp}\;r\,s$ & $\dn$ &% $\textit{let}\;a = (r^\uparrow)\backslash_{bsimp}\, s\;\textit{in}$\\ % & & $\;\;\textit{if}\; \textit{bnullable}(a)$\\% & & $\;\;\textit{then}\;\textit{decode}\,(\textit{bmkeps}\,a)\,r$\\% & & $\;\;\textit{else}\;\textit{None}$%\end{tabular}\end{center}\noindentit becomes a problem to maintain a similar property as \ref{retrieveStepwise}.Previously without $\textit{bsimp}$ the exact structure of each intermediate regular expression is preserved, %allowing pairs of inhabitation relations %in the form $\vdash v : r \backslash c $ and%$\vdash \inj \; r\; c \; v : r $ to hold in \ref{eq:stepwise}.We can illustrate this using the diagram \ref{fig:inj} in chapter \ref{Inj},by zooming in to the middle bit involving $r_i$, $r_{i+1}$, $v_i$ and $v_{i+1}$,and adding the bottom row to show how bitcodes encoding the lexing informationcan be extracted from every pair $(r_i, \; v_i)$:\begin{center}\label{graph:injZoom} \begin{tikzpicture}[->, >=stealth', shorten >= 1pt, auto, thick] %\node [rectangle ] (1) at (-7, 2) {$\ldots$}; %\node [rectangle, draw] (2) at (-4, 2) {$r_i = _{bs'}(_Za+_Saa)^*$}; %\node [rectangle, draw] (3) at (4, 2) {$r_{i+1} = _{bs'}(_Z(_Z\ONE + _S(\ONE \cdot a)))\cdot(_Za+_Saa)^*$}; %\node [rectangle] (4) at (9, 2) {$\ldots$}; %\node [rectangle] (5) at (-7, -2) {$\ldots$}; %\node [rectangle, draw] (6) at (-4, -2) {$v_i = \Stars \; [\Left (a)]$}; %\node [rectangle, draw] (7) at ( 4, -2) {$v_{i+1} = \Seq (\Alt (\Left \; \Empty)) \; \Stars \, []$}; %\node [rectangle] (8) at ( 9, -2) {$\ldots$}; %\node [rectangle] (9) at (-7, -6) {$\ldots$}; %\node [rectangle, draw] (10) at (-4, -6) {$\textit{bits}_{i} = bs' @ ZZS$}; %\node [rectangle, draw] (11) at (4, -6) {$\textit{bits}_{i+1} = bs'@ ZZS$}; %\node [rectangle] (12) at (9, -6) {$\ldots$}; \node [rectangle ] (1) at (-8, 2) {$\ldots$}; \node [rectangle, draw] (2) at (-5, 2) {$r_i = a_\downarrow$}; \node [rectangle, draw] (3) at (3, 2) {$r_{i+1} = (a\backslash c)_\downarrow$}; \node [rectangle] (4) at (8, 2) {$\ldots$}; \node [rectangle] (5) at (-8, -2) {$\ldots$}; \node [rectangle, draw] (6) at (-5, -2) {$v_i = \inj\; r \; c \; v$}; \node [rectangle, draw] (7) at ( 3, -2) {$v_{i+1} = v$}; \node [rectangle] (8) at ( 8, -2) {$\ldots$}; \node [rectangle] (9) at (-8, -6) {$\ldots$}; \node [rectangle, draw] (10) at (-5, -6) {$\textit{bits}_{i} $}; \node [rectangle, draw] (11) at (3, -6) {$\textit{bits}_{i+1}$}; \node [rectangle] (12) at (8, -6) {$\ldots$}; \path (1) edge [] node {} (2); \path (6) edge [] node {} (5); \path (9) edge [] node {} (10); \path (11) edge [<-] node {} (12); \path (8) edge [] node {} (7); \path (3) edge [] node {} (4); \path (6) edge [dashed,bend right = 30] node {$\retrieve \; a_i \; v_i$} (10); \path (2) edge [dashed,bend left = 48] node {} (10); \path (7) edge [dashed,bend right = 30] node {$\retrieve \; a_{i+1} \; v_{i+1}$} (11); \path (3) edge [dashed,bend left = 45] node {} (11); \path (2) edge [] node {$\backslash c$} (3); \path (2) edge [dashed, <->] node {$\vdash v_i : r_i$} (6); \path (3) edge [dashed, <->] node {$\vdash v_{i+1} : r_{i+1}$} (7); %\path (6) edge [] node {$\vdash v_i : r_i$} (10); %\path (7) edge [dashed, <->] node {$\vdash v_i : r_i$} (11); \path (10) edge [dashed, <->] node {$=$} (11); \path (7) edge [] node {$\inj \; r_{i+1} \; c \; v_i$} (6);% \node [rectangle, draw] (r) at (-6, -1) {$(aa)^*(b+c)$};% \node [rectangle, draw] (a) at (-6, 4) {$(aa)^*(_{Z}b + _{S}c)$};% \path (r)% edge [] node {$\internalise$} (a);% \node [rectangle, draw] (a1) at (-3, 1) {$(_{Z}(\ONE \cdot a) \cdot (aa)^*) (_{Z}b + _Sc)$};% \path (a)% edge [] node {$\backslash a$} (a1);%% \node [rectangle, draw, three sided] (a21) at (-2.5, 4) {$(_{Z}\ONE \cdot (aa)^*)$};% \node [rectangle, draw, three sided1] (a22) at (-0.8, 4) {$(_{Z}b + _{S}c)$};% \path (a1)% edge [] node {$\backslash a$} (a21);% \node [rectangle, draw] (a3) at (0.5, 2) {$_{ZS}(_{Z}\ONE + \ZERO)$};% \path (a22)% edge [] node {$\backslash b$} (a3);% \path (a21)% edge [dashed, bend right] node {} (a3);% \node [rectangle, draw] (bs) at (2, 4) {$ZSZ$};% \path (a3)% edge [below] node {$\bmkeps$} (bs);% \node [rectangle, draw] (v) at (3, 0) {$\Seq \; (\Stars\; [\Seq \; a \; a]) \; (\Left \; b)$};% \path (bs)% edge [] node {$\decode$} (v); \end{tikzpicture} %\caption{$\blexer$ with the regular expression $(aa)^*(b+c)$ and $aab$}\end{center}But $\blexersimp$ introduces simplification after the derivative,making it difficult to align the structures of values and regular expressions.If we change the form of property \ref{eq:stepwise} to adapt to the needs of $\blexersimp$ the precondition of becomes\[ \vdash v : (\textit{bsimp} \; (r\backslash c))\]The inhabitation relation of the other pair no longer holds,because $\inj$ does not work on the simplified value $v$ and the unsimplified regular expression $r$.The retrieve function will not work either.\[ \vdash \inj \; r \; c \; v : r\]It seems unclear what procedures needs to beused to create a new value $v_?$ such that\[ \vdash v_? : r \; \text{and} \; \retrieve \; r \; v_? = \retrieve \; (\textit{bsimp} \; (r\backslash c)) \; v\]hold.%It is clear that once we made %$v$ to align with $\textit{bsimp} \; r_{c}$%in the inhabitation relation, something different than $v_{r}^{c}$ needs to be plugged%in for the above statement to hold.Ausaf et al. \cite{AusafDyckhoffUrban2016}used something they call rectification functions to restore the original value from the simplified value.The idea is that simplification functions not only returns a regular expression,but also a rectification function \[ \textit{simp}^{rect} : Regex \Rightarrow (Value \Rightarrow Value, Regex)%\textit{frect} : Value \Rightarrow Value\]that is recorded recursively,and then applied to the previous value to obtain the correct value for $\inj$ to work on. The recursive case of the lexer is defined as something like\[ \textit{slexer} \; r \; (c\!::\!s) \dn let \;(\textit{frect}, r_c) = \textit{simp}^{rect} \;(r \backslash c) \;\; \textit{in}\;\; \inj \; r \; c \; (\textit{frect} \; (\textit{slexer} \; r_c\; s)) %\textit{match} \; s \; \textit{case} [\]However this approach (including $\textit{slexer}$'s correctness proof) only works without bitcodes, and it limits the kind of simplifications one can introduce.%and they have not yet extended their relatively simple simplifications%to more aggressive ones.See the thesis by Ausaf \cite{Ausaf}for details.%\begin{center}% $\vdash v: (r\backslash c) \implies \retrieve \; (\mathord{?}(\textit{bsimp} \; r_c)) \; v =\retrieve \; r \;(\mathord{?} v_{r}^{c}) $%\end{center}%\noindentWe were not able to use their idea for our very strong simplification rules.Therefore we are taking another route that completelydisposes of lemma \ref{retrieveStepwise},and prove a weakened inductive invariant instead.Let us first explain why lemma \ref{retrieveStepwise}'s requirement is too strong, and suggest a few possible fixes, which leads toour proof which we believe was the most natural and effective method.\subsection{Why Lemma \ref{retrieveStepwise}'s Requirement is too Strong}%From this chapter we start with the main contribution of this thesis, whichThe $\blexer$ proof relies on $r_i, \; v_i$ to match each other in lockstepfor each derivative step $i$, however only $v_0$ is needed and intermediate$v_i$s are purely proof scaffolding.Moreover property \ref{eq:stepwise}is stronger than needed for POSIX lexing: the precondition$\vdash v_{i+1}:r_{i+1}$ is too general in the sense that it allows arbitrary values inhabiting in $r_i$ to retrieve bitcodes.%correspondence between the lexical value and the%regular expression in derivative and injection operations at the same step $i$.%If we revisit the diagram \ref{graph:injZoom} with an exampleConsider a concrete example where $a_i = (_{ZZ}x + _{ZS}y + _{S}x)$ and$a_{i+1} = (_{ZZ}\ONE + \ZERO + _{S}\ONE)$.What is required in the proof of $\blexer$ is that for the POSIX value$v_i = \Left \; (\Left \; Empty)$,the property\[ %\vdash \Left \; (\Left \; Empty) : (\ONE+\ZERO+\ONE) \implies \retrieve \; (_{ZZ}\ONE + \ZERO + _{S}\ONE) \; (\Left \; (\Left \; \Empty) ) = \retrieve \; (_{ZZ}x + _{ZS}y + _{S}x ) \; (\Left \; (\Left \; \Char\; x) )\]holds, and for $\blexersimp$ a property of similar shape to\[ \retrieve \; _{ZZ}\ONE \; \; Empty = \retrieve \; _{ZZ}x \; (\Char\; x)\]needs to hold as well.However for the definitely non-POSIX value $v_i' = \Right \; \Empty$the precondition $\vdash \Right \; \Empty : x+y+x$ holds as well, and thereforethe following equality\[ \retrieve \; (_{ZZ}\ONE + \ZERO + _{S}\ONE) \; (\Right \; \Empty) = \retrieve \; (_{ZZ}x + _{ZS}y + _{S}x ) \; (\Right \; (\Char\; x))\]by lemma \ref{retrieveStepwise} holds for $\blexer$ as well.This cannot hold or be proven anymore with $\blexersimp$ as the correspondingsub-regular expressions and values have been eliminated during the de-duplication procedure of our smplification.We are stuck with a property that holds in $\blexer$ but does not have a counterpartin $\blexersimp$.This needs not hold for the purpose of POSIX lexing though--we know the rightmost subexpression $x$ is not POSIX by the left priority rule.The inductive invariant \ref{eq:stepwise} can be weakened by restricting the precondition$\vdash v_i:r_i$ to $\exists s_i. \; (s_i, r_i) \rightarrow v_i$. We tried this route but it did not work well since we need touse a similar technique as the rectification functions by Ausaf et al, and they can get very complicated with our simplifications.After some trial-and-error we found a property of the form\begin{property} If a POSIX value can be extracted from $a \backslash s$, then it can be extracted from $a \backslash_{bsimps} s$ as well.\end{property}\noindentmost natural to work with, and we defined a binary relation to capturethe connection between $a\backslash s$ and $a \backslash_{bsimps} s$.%and look specifically at%the pairs $v_i, r_i$ and $v_{i+1},\, r_{i+1}$, we get the diagram demonstrating%the invariant that the same bitcodes can be extracted from the pairs:%\tikzset{three sided/.style={% draw=none,% append after command={% [-,shorten <= -0.5\pgflinewidth]% ([shift={(-1.5\pgflinewidth,-0.5\pgflinewidth)}]\tikzlastnode.north east)% edge([shift={( 0.5\pgflinewidth,-0.5\pgflinewidth)}]\tikzlastnode.north west) % ([shift={( 0.5\pgflinewidth,-0.5\pgflinewidth)}]\tikzlastnode.north west)% edge([shift={( 0.5\pgflinewidth,+0.5\pgflinewidth)}]\tikzlastnode.south west) % ([shift={( 0.5\pgflinewidth,+0.5\pgflinewidth)}]\tikzlastnode.south west)% edge([shift={(-1.0\pgflinewidth,+0.5\pgflinewidth)}]\tikzlastnode.south east)% }% }%}%%\tikzset{three sided1/.style={% draw=none,% append after command={% [-,shorten <= -0.5\pgflinewidth]% ([shift={(1.5\pgflinewidth,-0.5\pgflinewidth)}]\tikzlastnode.north west)% edge([shift={(-0.5\pgflinewidth,-0.5\pgflinewidth)}]\tikzlastnode.north east) % ([shift={(-0.5\pgflinewidth,-0.5\pgflinewidth)}]\tikzlastnode.north east)% edge([shift={(-0.5\pgflinewidth,+0.5\pgflinewidth)}]\tikzlastnode.south east) % ([shift={(-0.5\pgflinewidth,+0.5\pgflinewidth)}]\tikzlastnode.south east)% edge([shift={(1.0\pgflinewidth,+0.5\pgflinewidth)}]\tikzlastnode.south west)% }% }%}%%\begin{center}% \begin{tikzpicture}[->, >=stealth', shorten >= 1pt, auto, thick]% %\node [rectangle ] (1) at (-7, 2) {$\ldots$};% %\node [rectangle, draw] (2) at (-4, 2) {$r_i = _{bs'}(_Za+_Saa)^*$};% %\node [rectangle, draw] (3) at (4, 2) {$r_{i+1} = _{bs'}(_Z(_Z\ONE + _S(\ONE \cdot a)))\cdot(_Za+_Saa)^*$};% %\node [rectangle] (4) at (9, 2) {$\ldots$};% %\node [rectangle] (5) at (-7, -2) {$\ldots$};% %\node [rectangle, draw] (6) at (-4, -2) {$v_i = \Stars \; [\Left (a)]$};% %\node [rectangle, draw] (7) at ( 4, -2) {$v_{i+1} = \Seq (\Alt (\Left \; \Empty)) \; \Stars \, []$};% %\node [rectangle] (8) at ( 9, -2) {$\ldots$};% %\node [rectangle] (9) at (-7, -6) {$\ldots$};% %\node [rectangle, draw] (10) at (-4, -6) {$\textit{bits}_{i} = bs' @ ZZS$};% %\node [rectangle, draw] (11) at (4, -6) {$\textit{bits}_{i+1} = bs'@ ZZS$};% %\node [rectangle] (12) at (9, -6) {$\ldots$};% % \node [rectangle ] (1) at (-8, 2) {$\ldots$};% \node [rectangle, draw] (2) at (-5, 2) {$r_i = _{bs'}(_Za+_Saa)^*$};% \node [rectangle, draw] (3) at (3, 2) {$r_{i+1} = _{bs'}(_Z(_Z\ONE + _S(\ONE \cdot a)))\cdot(_Za+_Saa)^*$};% \node [rectangle] (4) at (8, 2) {$\ldots$};% \node [rectangle] (5) at (-8, -2) {$\ldots$};% \node [rectangle, draw] (6) at (-5, -2) {$v_i = \Stars \; [\Left (a)]$};% \node [rectangle, draw] (7) at ( 3, -2) {$v_{i+1} = \Seq (\Alt (\Left \; \Empty)) \; \Stars \, []$};% \node [rectangle] (8) at ( 8, -2) {$\ldots$};% \node [rectangle] (9) at (-8, -6) {$\ldots$};% \node [rectangle, draw] (10) at (-5, -6) {$\textit{bits}_{i} = bs' @ ZZS$};% \node [rectangle, draw] (11) at (3, -6) {$\textit{bits}_{i+1} = bs'@ ZZS$};% \node [rectangle] (12) at (8, -6) {$\ldots$};%%%% \path (1) edge [] node {} (2);% \path (5) edge [] node {} (6);% \path (9) edge [] node {} (10);%% \path (11) edge [] node {} (12);% \path (7) edge [] node {} (8);% \path (3) edge [] node {} (4);%%% \path (6) edge [dashed,bend right = 30] node {$\retrieve \; r_i \; v_i$} (10);% \path (2) edge [dashed,bend left = 48] node {} (10);%% \path (7) edge [dashed,bend right = 30] node {$\retrieve \; r_{i+1} \; v_{i+1}$} (11);% \path (3) edge [dashed,bend left = 45] node {} (11);% % \path (2) edge [] node {$\backslash a$} (3);% \path (2) edge [dashed, <->] node {$\vdash v_i : r_i$} (6);% \path (3) edge [dashed, <->] node {$\vdash v_{i+1} : r_{i+1}$} (7);% %\path (6) edge [] node {$\vdash v_i : r_i$} (10);% %\path (7) edge [dashed, <->] node {$\vdash v_i : r_i$} (11);%% \path (10) edge [dashed, <->] node {$=$} (11);%% \path (7) edge [] node {$\inj \; r_{i+1} \; a \; v_i$} (6);%%% \node [rectangle, draw] (r) at (-6, -1) {$(aa)^*(b+c)$};%% \node [rectangle, draw] (a) at (-6, 4) {$(aa)^*(_{Z}b + _{S}c)$};%% \path (r)%% edge [] node {$\internalise$} (a);%% \node [rectangle, draw] (a1) at (-3, 1) {$(_{Z}(\ONE \cdot a) \cdot (aa)^*) (_{Z}b + _Sc)$};%% \path (a)%% edge [] node {$\backslash a$} (a1);%%%% \node [rectangle, draw, three sided] (a21) at (-2.5, 4) {$(_{Z}\ONE \cdot (aa)^*)$};%% \node [rectangle, draw, three sided1] (a22) at (-0.8, 4) {$(_{Z}b + _{S}c)$};%% \path (a1)%% edge [] node {$\backslash a$} (a21);%% \node [rectangle, draw] (a3) at (0.5, 2) {$_{ZS}(_{Z}\ONE + \ZERO)$};%% \path (a22)%% edge [] node {$\backslash b$} (a3);%% \path (a21)%% edge [dashed, bend right] node {} (a3);%% \node [rectangle, draw] (bs) at (2, 4) {$ZSZ$};%% \path (a3)%% edge [below] node {$\bmkeps$} (bs);%% \node [rectangle, draw] (v) at (3, 0) {$\Seq \; (\Stars\; [\Seq \; a \; a]) \; (\Left \; b)$};%% \path (bs)%% edge [] node {$\decode$} (v);%%% \end{tikzpicture}% %\caption{$\blexer$ with the regular expression $(aa)^*(b+c)$ and $aab$}%\end{center}%When simplifications are added, the inhabitation relation no longer holds,%causing the above diagram to break.%%Ausaf addressed this with an augmented lexer he called $\textit{slexer}$.%%%%we note that the invariant%$\vdash v_{i+1}: r_{i+1} \implies \retrieve \; r_{i+1} \; v_{i+1} $ is too strong%to maintain because the precondition $\vdash v_i : r_i$ is too weak.%It does not require $v_i$ to be a POSIX value %%%which is essential for getting an understanding this thesis%in chapter \ref{Bitcoded1}, which is necessary for understanding why%the proof %%In this chapter,%We contrast our simplification function %with Sulzmann and Lu's, indicating the simplicity of our algorithm.%This is another case for the usefulness %and reliability of formal proofs on algorithms.%These ``aggressive'' simplifications would not be possible in the injection-based %lexing we introduced in chapter \ref{Inj}.%We then prove the correctness with the improved version of %$\blexer$, called $\blexersimp$, by establishing %$\blexer \; r \; s= \blexersimp \; r \; s$ using a term rewriting system.%%----------------------------------------------------------------------------------------% SECTION rewrite relation%----------------------------------------------------------------------------------------In the next section we first introduce the rewriting relation \emph{rrewrite}($\rrewrite$) between two regular expressions,which stands for an atomicsimplification.We then prove properties aboutthis rewriting relation and its reflexive transitive closure.Finally we leverage these properties to showan equivalence between the results generated by$\blexer$ and $\blexersimp$.\subsection{The Rewriting Relation $\rrewrite$($\rightsquigarrow$)}In the $\blexer$'s correctness proof, wedid not directly derive the fact that $\blexer$ generates the POSIX value,but first proved that $\blexer$ generates the same result as $\lexer$.Then we re-usethe correctness of $\lexer$to obtain \begin{center} $(r, s) \rightarrow v \;\; \textit{iff} \;\; \blexer \; r \;s = v$\\ $\nexists v. \; (r, s) \rightarrow v \;\; \textit{iff} \;\; \blexer\; r\;s = \None$.\end{center}%\begin{center}% $(r, s) \rightarrow v \;\; \textit{iff} \;\; \blexer \; r \;s = v$.%\end{center}Here we apply thismodularised technique againby first proving that$\blexersimp \; r \; s $ produces the same output as $\blexer \; r\; s$,and then piecing it together with $\blexer$'s correctness to achieve our maintheorem:\begin{center} $(r, s) \rightarrow v \; \; \textit{iff} \;\; \blexersimp \; r \; s = \Some \;v$ \\ $\nexists v. \; (r, s) \rightarrow v \;\; \textit{iff} \;\; \blexersimp\; r\;s = \None$\end{center}\noindentThe overall idea for the proofof $\blexer \;r \;s = \blexersimp \; r \;s$ is that the transition from $r$ to $\textit{bsimp}\; r$ can bebroken down into smaller rewrite steps of the form:\begin{center} $r \rightsquigarrow^* \textit{bsimp} \; r$\end{center}where each rewrite step, written $\rightsquigarrow$,is an ``atomic'' simplification thatis similar to a small-step reduction in operational semantics (see figure \ref{rrewriteRules} for the rules):\begin{figure}[H]\begin{mathpar} \inferrule * [Right = $S\ZERO_l$]{\vspace{0em}}{_{bs} \ZERO \cdot r_2 \rightsquigarrow \ZERO\\} \inferrule * [Right = $S\ZERO_r$]{\vspace{0em}}{_{bs} r_1 \cdot \ZERO \rightsquigarrow \ZERO\\} \inferrule * [Right = $S_1$]{\vspace{0em}}{_{bs1} ((_{bs2} \ONE) \cdot r) \rightsquigarrow \fuse \; (bs_1 @ bs_2) \; r\\}\\ \inferrule * [Right = $SL$] {\\ r_1 \rightsquigarrow r_2}{_{bs} r_1 \cdot r_3 \rightsquigarrow _{bs} r_2 \cdot r_3\\} \inferrule * [Right = $SR$] {\\ r_3 \rightsquigarrow r_4}{_{bs} r_1 \cdot r_3 \rightsquigarrow _{bs} r_1 \cdot r_4\\}\\ \inferrule * [Right = $A0$] {\vspace{0em}}{ _{bs}\sum [] \rightsquigarrow \ZERO} \inferrule * [Right = $A1$] {\vspace{0em}}{ _{bs}\sum [a] \rightsquigarrow \fuse \; bs \; a} \inferrule * [Right = $AL$] {\\ rs_1 \stackrel{s}{\rightsquigarrow} rs_2}{_{bs}\sum rs_1 \rightsquigarrow rs_2} \inferrule * [Right = $LE$] {\vspace{0em}}{ [] \stackrel{s}{\rightsquigarrow} []} \inferrule * [Right = $LT$] {rs_1 \stackrel{s}{\rightsquigarrow} rs_2}{ r :: rs_1 \stackrel{s}{\rightsquigarrow} r :: rs_2 } \inferrule * [Right = $LH$] {r_1 \rightsquigarrow r_2}{ r_1 :: rs \stackrel{s}{\rightsquigarrow} r_2 :: rs} \inferrule * [Right = $L\ZERO$] {\vspace{0em}}{\ZERO :: rs \stackrel{s}{\rightsquigarrow} rs} \inferrule * [Right = $LS$] {\vspace{0em}}{_{bs} \sum (rs_1 :: rs_b) \stackrel{s}{\rightsquigarrow} ((\map \; (\fuse \; bs_1) \; rs_1) @ rsb) } \inferrule * [Right = $LD$] {\\ \rerase{a_1} = \rerase{a_2}}{rs_a @ [a_1] @ rs_b @ [a_2] @ rs_c \stackrel{s}{\rightsquigarrow} rs_a @ [a_1] @ rs_b @ rs_c}\end{mathpar}\caption{The rewrite rules that generate simplified regular expressions in small steps: $r_1 \rightsquigarrow r_2$ is for bitcoded regular expressions and $rs_1 \stackrel{s}{\rightsquigarrow} rs_2$ for lists of bitcoded regular expressions. Interesting is the LD rule that allows copies of regular expressions to be removed provided a regular expression earlier in the list can match the same strings.}\label{rrewriteRules}\end{figure}\noindentThe rules $LT$ and $LH$ are for rewriting two regular expression listssuch that one regular expressionin the left-hand-side list is rewritable in one stepto the right-hand side's regular expression at the same position.This helps with defining the ``context rule'' $AL$.The reflexive transitive closure of $\rightsquigarrow$ and $\stackrel{s}{\rightsquigarrow}$are defined in the usual way:\begin{figure}[H] \centering\begin{mathpar} \inferrule{\vspace{0em}}{ r \rightsquigarrow^* r \\} \inferrule{\vspace{0em}}{rs \stackrel{s*}{\rightsquigarrow} rs \\} \inferrule{r_1 \rightsquigarrow^* r_2 \land \; r_2 \rightsquigarrow^* r_3}{r_1 \rightsquigarrow^* r_3\\} \inferrule{rs_1 \stackrel{s*}{\rightsquigarrow} rs_2 \land \; rs_2 \stackrel{s*}{\rightsquigarrow} rs_3}{rs_1 \stackrel{s*}{\rightsquigarrow} rs_3}\end{mathpar}\caption{The Reflexive Transitive Closure of $\rightsquigarrow$ and $\stackrel{s}{\rightsquigarrow}$}\label{transClosure}\end{figure}%Two rewritable terms will remain rewritable to each other%even after a derivative is taken:The main point of our rewriting relationis that it is preserved under derivatives,namely\begin{center} $r_1 \rightsquigarrow r_2 \implies (r_1 \backslash c) \rightsquigarrow^* (r_2 \backslash c)$\end{center}And also, if two terms are rewritable to each other,then they produce the same bitcodes:\begin{center} $r \rightsquigarrow^* r' \;\; \textit{then} \; \; \bmkeps \; r = \bmkeps \; r'$\end{center}The decoding phase of both $\blexer$ and $\blexersimp$are the same, which means that if they receive the samebitcodes before the decoding phase,they generate the same value after decoding is done.We will prove the three properties we mentioned above in the next sub-section.\subsection{Important Properties of $\rightsquigarrow$}First we prove some basic facts about $\rightsquigarrow$, $\stackrel{s}{\rightsquigarrow}$, $\rightsquigarrow^*$ and $\stackrel{s*}{\rightsquigarrow}$,which will be needed later.\\The inference rules (\ref{rrewriteRules}) we gave in the previous section have their ``many-steps version'':\begin{lemma}\label{squig1} \hspace{0em} \begin{itemize} \item $rs_1 \stackrel{s*}{\rightsquigarrow} rs_2 \implies _{bs} \sum rs_1 \stackrel{*}{\rightsquigarrow} _{bs} \sum rs_2$ \item $r \rightsquigarrow^* r' \implies _{bs} \sum (r :: rs)\; \rightsquigarrow^*\; _{bs} \sum (r' :: rs)$ \item The rewriting in many steps property is composable in terms of the sequence constructor:\\ $r_1 \rightsquigarrow^* r_2 \implies _{bs} r_1 \cdot r_3 \rightsquigarrow^* \; _{bs} r_2 \cdot r_3 \quad $ and $\quad r_3 \rightsquigarrow^* r_4 \implies _{bs} r_1 \cdot r_3 \rightsquigarrow^* _{bs} \; r_1 \cdot r_4$ \item The rewriting in many steps properties $\stackrel{*}{\rightsquigarrow}$ and $\stackrel{s*}{\rightsquigarrow}$ is preserved under the function $\fuse$:\\ $r_1 \rightsquigarrow^* r_2 \implies \fuse \; bs \; r_1 \rightsquigarrow^* \; \fuse \; bs \; r_2 \quad $ and $rs_1 \stackrel{s}{\rightsquigarrow} rs_2 \implies \map \; (\fuse \; bs) \; rs_1 \stackrel{s*}{\rightsquigarrow} \map \; (\fuse \; bs) \; rs_2$ \end{itemize}\end{lemma}\begin{proof} By an induction on the inductive cases of $\stackrel{s*}{\rightsquigarrow}$ and $\rightsquigarrow^*$ respectively. The third and fourth points are by the properties $r_1 \rightsquigarrow r_2 \implies \fuse \; bs \; r_1 \implies \fuse \; bs \; r_2$ and $rs_2 \stackrel{s}{\rightsquigarrow} rs_3 \implies \map \; (\fuse \; bs) rs_2 \stackrel{s*}{\rightsquigarrow} \map \; (\fuse \; bs)\; rs_3$, which can be inductively proven by the inductive cases of $\rightsquigarrow$ and $\stackrel{s}{\rightsquigarrow}$.\end{proof}\noindentThe inference rules of $\stackrel{s}{\rightsquigarrow}$are defined in terms of the list cons operation, wherewe establish that the $\stackrel{s}{\rightsquigarrow}$ and $\stackrel{s*}{\rightsquigarrow}$ relation is also preserved w.r.t appending and prepending of a list.In addition, wealso prove some relations between $\rightsquigarrow^*$ and $\stackrel{s*}{\rightsquigarrow}$.\begin{lemma}\label{ssgqTossgs} \hspace{0em} \begin{itemize} \item $rs_1 \stackrel{s}{\rightsquigarrow} rs_2 \implies rs @ rs_1 \stackrel{s}{\rightsquigarrow} rs @ rs_2$ \item $rs_1 \stackrel{s*}{\rightsquigarrow} rs_2 \implies rs @ rs_1 \stackrel{s*}{\rightsquigarrow} rs @ rs_2 \; \; \textit{and} \; \; rs_1 @ rs \stackrel{s*}{\rightsquigarrow} rs_2 @ rs$ \item The $\stackrel{s}{\rightsquigarrow} $ relation after appending a list becomes $\stackrel{s*}{\rightsquigarrow}$:\\ $rs_1 \stackrel{s}{\rightsquigarrow} rs_2 \implies rs_1 @ rs \stackrel{s*}{\rightsquigarrow} rs_2 @ rs$ \item $r_1 \rightsquigarrow^* r_2 \implies [r_1] \stackrel{s*}{\rightsquigarrow} [r_2]$ \item $rs_3 \stackrel{s*}{\rightsquigarrow} rs_4 \land r_1 \rightsquigarrow^* r_2 \implies r_2 :: rs_3 \stackrel{s*}{\rightsquigarrow} r_2 :: rs_4$ \item If we can rewrite a regular expression in many steps to $\ZERO$, then we can also rewrite any sequence containing it to $\ZERO$:\\ $r_1 \rightsquigarrow^* \ZERO \implies _{bs}r_1\cdot r_2 \rightsquigarrow^* \ZERO$ \end{itemize}\end{lemma}\begin{proof} The first part is by induction on the list $rs$. The second part is by induction on the inductive cases of $\stackrel{s*}{\rightsquigarrow}$. The third part is by rule induction of $\stackrel{s}{\rightsquigarrow}$. The fourth sub-lemma is by rule induction of $\stackrel{s*}{\rightsquigarrow}$ and using part one to three. The fifth part is a corollary of part four. The last part is proven by rule induction again on $\rightsquigarrow^*$.\end{proof}\noindentNow we are ready to give the proofs of the following properties:\begin{itemize} \item $r \rightsquigarrow^* r'\land \bnullable \; r_1 \implies \bmkeps \; r = \bmkeps \; r'$. \\ \item $r \rightsquigarrow^* \textit{bsimp} \;r$.\\ \item $r \rightsquigarrow r' \implies r \backslash c \rightsquigarrow^* r'\backslash c$.\\\end{itemize}\subsubsection{Property 1: $r \rightsquigarrow^* r'\land \bnullable \; r_1 \implies \bmkeps \; r = \bmkeps \; r'$}Intuitively, this property says we can extract the same bitcodes using $\bmkeps$ from the nullablecomponents of two regular expressions $r$ and $r'$,if we can rewrite from one to the other in finitelymany steps.For convenience, we define a predicate for a list of regular expressionshaving at least one nullable regular expression:\begin{center} $\textit{bnullables} \; rs \quad \dn \quad \exists r \in rs. \;\; \bnullable \; r$\end{center}\noindentThe rewriting relation $\rightsquigarrow$ preserves (b)nullability:\begin{lemma}\label{rewritesBnullable} \hspace{0em} \begin{itemize} \item $\text{If} \; r_1 \rightsquigarrow r_2, \; \text{then} \; \bnullable \; r_1 = \bnullable \; r_2$ \item $\text{If} \; rs_1 \stackrel{s}{\rightsquigarrow} rs_2 \; \text{then} \; \textit{bnullables} \; rs_1 = \textit{bnullables} \; rs_2$ \item $r_1 \rightsquigarrow^* r_2 \implies \bnullable \; r_1 = \bnullable \; r_2$ \end{itemize}\end{lemma}\begin{proof} By rule induction of $\rightsquigarrow$ and $\stackrel{s}{\rightsquigarrow}$. The third point is a result of the second.\end{proof}\noindentFor convenience again,we define $\bmkepss$ on a list $rs$,which extracts the bit-codes on the first $\bnullable$ element in $rs$:\begin{center} \begin{tabular}{lcl} $\bmkepss \; [] $ & $\dn$ & $[]$\\ $\bmkepss \; r :: rs$ & $\dn$ & $\textit{if} \;(\bnullable \; r) \;\; \textit{then} \;\; \bmkeps \; r \; \textit{else} \;\; \bmkepss \; rs$ \end{tabular}\end{center}\noindentIf both regular expressions in a rewriting relation are nullable, then they produce the same bitcodes:\begin{lemma}\label{rewriteBmkepsAux} \hspace{0em} \begin{itemize} \item $r_1 \rightsquigarrow r_2 \implies (\bnullable \; r_1 \land \bnullable \; r_2 \implies \bmkeps \; r_1 = \bmkeps \; r_2)$ \item and $rs_ 1 \stackrel{s}{\rightsquigarrow} rs_2 \implies (\bnullables \; rs_1 \land \bnullables \; rs_2 \implies \bmkepss \; rs_1 = \bmkepss \; rs2)$ \end{itemize}\end{lemma}\begin{proof} By rule induction over the cases that lead to $r_1 \rightsquigarrow r_2$.\end{proof}\noindentWith lemma \ref{rewriteBmkepsAux} in place we are ready to prove itsmany-step version: \begin{lemma} $\text{If} \;\; r \stackrel{*}{\rightsquigarrow} r' \;\; \text{and} \;\; \bnullable \; r, \;\;\; \text{then} \;\; \bmkeps \; r = \bmkeps \; r'$\end{lemma}\begin{proof} By rule induction of $\stackrel{*}{\rightsquigarrow} $. Lemma $\ref{rewritesBnullable}$ gives us both $r$ and $r'$ are nullable. The lemma \ref{rewriteBmkepsAux} solves the inductive case.\end{proof}\subsubsection{Property 2: $r \stackrel{*}{\rightsquigarrow} \textit{bsimp} \; r$}Now we get to the key part of the proof, which says that our simplification's helper functions such as $\distinctBy$ and $\flts$ describereducts of $\stackrel{s*}{\rightsquigarrow}$ and $\rightsquigarrow^* $.The first lemma to prove is a more general version of $rs_ 1 \rightsquigarrow^* \distinctBy \; rs_1 \; \phi$:\begin{lemma} $rs_1 @ rs_2 \stackrel{s*}{\rightsquigarrow} (rs_1 @ (\distinctBy \; rs_2 \; \; \rerases \;\; (\map\;\; \rerases \; \; rs_1)))$\end{lemma}\noindentIt says that for a list made of two parts $rs_1 @ rs_2$, one can throw away the duplicateelements in $rs_2$, as well as those that have appeared in $rs_1$.\begin{proof} By induction on $rs_2$, where $rs_1$ is allowed to be arbitrary.\end{proof}\noindentSetting $rs_2$ to be empty,we get the corollary\begin{corollary}\label{dBPreserves} $rs_1 \stackrel{s*}{\rightsquigarrow} \distinctBy \; rs_1 \; \phi$.\end{corollary}\noindentSimilarly the flatten function $\flts$ describes a reduct of$\stackrel{s*}{\rightsquigarrow}$ as well:\begin{lemma}\label{fltsPreserves} $rs \stackrel{s*}{\rightsquigarrow} \flts \; rs$\end{lemma}\begin{proof} By an induction on $rs$.\end{proof}\noindentThe function $\bsimpalts$ preserves rewritability:\begin{lemma}\label{bsimpaltsPreserves} $_{bs} \sum rs \stackrel{*}{\rightsquigarrow} \bsimpalts \; _{bs} \; rs$\end{lemma}\noindentThe simplification function$\textit{bsimp}$ only transforms the regular expression using steps specified by $\rightsquigarrow^*$ and nothing else:\begin{lemma} $r \stackrel{*}{\rightsquigarrow} \textit{bsimp} \; r$\end{lemma}\begin{proof} By an induction on $r$. The most involved case is the alternative, where we use lemmas \ref{bsimpaltsPreserves}, \ref{fltsPreserves} and \ref{dBPreserves} to do a series of rewriting:\\ \begin{center} \begin{tabular}{lcl} $rs$ & $\stackrel{s*}{\rightsquigarrow}$ & $ \map \; \textit{bsimp} \; rs$\\ & $\stackrel{s*}{\rightsquigarrow}$ & $ \flts \; (\map \; \textit{bsimp} \; rs)$\\ & $\stackrel{s*}{\rightsquigarrow}$ & $ \distinctBy \; (\flts \; (\map \; \textit{bsimp}\; rs)) \; \rerases \; \phi$\\ \end{tabular} \end{center} Using this we can derive the following rewrite sequence:\\ \begin{center} \begin{tabular}{lcl} $r$ & $=$ & $_{bs}\sum rs$\\[1.5ex] & $\rightsquigarrow^*$ & $\bsimpalts \; bs \; rs$ \\[1.5ex] & $\rightsquigarrow^*$ & $\ldots$ \\ [1.5ex] & $\rightsquigarrow^*$ & $\bsimpalts \; bs \; (\distinctBy \; (\flts \; (\map \; \textit{bsimp}\; rs)) \; \rerases \; \phi)$\\[1.5ex] %& $\rightsquigarrow^*$ & $ _{bs} \sum (\distinctBy \; %(\flts \; (\map \; \textit{bsimp}\; rs)) \; \; %\rerases \; \;\phi) $\\[1.5ex] & $\rightsquigarrow^*$ & $\textit{bsimp} \; r$\\[1.5ex] \end{tabular} \end{center} \end{proof}\subsubsection{Property 3: $r_1 \stackrel{*}{\rightsquigarrow} r_2 \implies r_1 \backslash c \stackrel{*}{\rightsquigarrow} r_2 \backslash c$}The rewrite relation $\rightsquigarrow$ changes into $\stackrel{*}{\rightsquigarrow}$after derivatives are taken on both sides:\begin{lemma}\label{rewriteBder} \hspace{0em} \begin{itemize} \item If $r_1 \rightsquigarrow r_2$, then $r_1 \backslash c \rightsquigarrow^* r_2 \backslash c$ \item If $rs_1 \stackrel{s}{\rightsquigarrow} rs_2$, then $ \map \; (\_\backslash c) \; rs_1 \stackrel{s*}{\rightsquigarrow} \map \; (\_ \backslash c) \; rs_2$ \end{itemize}\end{lemma}\begin{proof} By induction on $\rightsquigarrow$ and $\stackrel{s}{\rightsquigarrow}$, using a number of the previous lemmas.\end{proof}\noindentNow we can prove property 3 as an immediate corollary:\begin{corollary}\label{rewritesBder} $r_1 \rightsquigarrow^* r_2 \implies r_1 \backslash c \rightsquigarrow^* r_2 \backslash c$\end{corollary}\begin{proof} By rule induction of $\stackrel{*}{\rightsquigarrow} $ and lemma \ref{rewriteBder}.\end{proof}\noindentThis can be extended and combined with $r \rightsquigarrow^* \textit{bsimp} \; r$to obtain the correspondence between$\blexer$ and $\blexersimp$'s intermediatederivative regular expressions \begin{lemma}\label{bderBderssimp} $a \backslash s \rightsquigarrow^* \bderssimp{a}{s} $\end{lemma}\begin{proof} By an induction on $s$.\end{proof}\subsection{Main Theorem}Now with \ref{bderBderssimp} in place we are ready for the main theorem.\begin{theorem} $\blexer \; r \; s = \blexersimp{r}{s}$\end{theorem}\noindent\begin{proof} We can rewrite in many steps from the original lexer's derivative regular expressions to the lexer with simplification applied (by lemma \ref{bderBderssimp}): \begin{center} $a \backslash s \rightsquigarrow^* \bderssimp{a}{s} $. \end{center} We know that they generate the same bits, if the lexing result is a match: \begin{center} $\bnullable \; (a \backslash s) \implies \bmkeps \; (a \backslash s) = \bmkeps \; (\bderssimp{a}{s})$ \end{center} Now that they generate the same bits, we know they also give the same value after decoding. \begin{center} $\bnullable \; (a \backslash s) \implies \decode \; r \; (\bmkeps \; (a \backslash s)) = \decode \; r \; (\bmkeps \; (\bderssimp{a}{s}))$ \end{center} Which is required by our proof goal: \begin{center} $\blexer \; r \; s = \blexersimp \; r \; s$. \end{center} \end{proof}\noindentAs a corollary,we can link this result with the lemma we proved earlier that \begin{center} $(r, s) \rightarrow v \;\; \textit{iff}\;\; \blexer \; r \; s = \Some \;v$\\ $\nexists v. \; (r, s) \rightarrow v \;\; \textit{iff} \;\; \blexer\; r\;s = \None$.\end{center}and obtain the property that the bit-coded lexer with simplification isindeed correctly generating a POSIX lexing result, if such a result exists.\begin{corollary} $(r, s) \rightarrow v \;\; \textit{iff} \;\; \blexersimp \; r\; s = \Some \; v$\\ $\nexists v. \; (r, s) \rightarrow v \;\; \textit{iff} \;\; \blexersimp\; r\;s = \None$.\end{corollary}\subsection{Comments on the Proof Techniques Used}Straightforward as the proof may seem,the efforts we spent obtaining it were far from trivial.We initially attempted to re-use the argument in \cref{flex_retrieve}. The problem is that both functions $\inj$ and $\retrieve$ require that the annotated regular expressions stay unsimplified, so that one can correctly compare $v_{i+1}$ and $r_i$ and $v_i$ in diagram \ref{graph:inj}.We also tried to prove \begin{center}$\textit{bsimp} \;\; (\bderssimp{a}{s}) = \textit{bsimp} \;\; (a\backslash s)$,\end{center}but this turns out to be not true.A counterexample is\[ a = [(_{Z}1+_{S}c)\cdot [bb \cdot (_{Z}1+_{S}c)]] \;\; \text{and} \;\; s = bb.\]\noindentThen we would have \begin{center} $\textit{bsimp}\;\; ( a \backslash s )$ = $_{[]}(_{ZZ}\ONE + _{ZS}c ) $\end{center}\noindentwhereas \begin{center} $\textit{bsimp} \;\;( \bderssimp{a}{s} )$ = $_{Z}(_{Z} \ONE + _{S} c)$.\end{center}Unfortunately, if we apply $\textit{bsimp}$ differentlywe will always have this discrepancy. This is due to the $\map \; (\fuse\; bs) \; as$ operation happening at different locations in the regular expression.The rewriting relation $\rightsquigarrow^*$ allows us to ignore this discrepancyand view the expressions \begin{center} $_{[]}(_{ZZ}\ONE + _{ZS}c ) $\\ and\\ $_{Z}(_{Z} \ONE + _{S} c)$\end{center}as equal because they were both re-writtenfrom the same expression.The simplification rewriting rulesgiven in \ref{rrewriteRules} are by no meansfinal,one could come up with new rulessuch as $\SEQ r_1 \cdot (\SEQ r_1 \cdot r_3) \rightarrow\SEQs [r_1, r_2, r_3]$.However this does not fit with the proof techniqueof our main theorem, but seem to not violate the POSIXproperty.Having established the correctness of our$\blexersimp$, in the next chapter we shall prove that with our $\simp$ function,for a given $r$, the derivative size is alwaysfinitely bounded by a constant.