Note that it is not immediately obvious that $\llbracket \bderssimp{r}{s} \rrbracket$ (the internal

data structure used in our $\blexer$)

is bounded by a constant $N_r$, where $N$ only depends on the regular expression

$r$, not the string $s$.

When doing a time complexity analysis of any 

lexer/parser based on Brzozowski derivatives, one needs to take into account that

not only the "derivative steps".

%TODO: get a grpah for internal data structure growing arbitrary large

	$\llbracket _{bs}\ONE \rrbracket$ & $\dn$ & $1$\\

	$\llbracket \ZERO \rrbracket$ & $\dn$ & $1$ \\

	$\llbracket _{bs} r_1 \cdot r_2 \rrbracket$ & $\dn$ & $\llbracket r_1 \rrbracket + \llbracket r_2 \rrbracket + 1$\\

$\llbracket _{bs}\mathbf{c} \rrbracket $ & $\dn$ & $1$\\

$\llbracket _{bs}\sum as \rrbracket $ & $\dn$ & $\map \; (\llbracket \_ \rrbracket)\; as   + 1$\\

$\llbracket _{bs} a^* \rrbracket $ & $\dn$ & $\llbracket a \rrbracket + 1$

\end{tabular}

\end{center}

\noindent

Similarly there is a size function for plain regular expressions:

\begin{center}

\begin{tabular}{ccc}

	$\llbracket \ONE \rrbracket_p$ & $\dn$ & $1$\\

	$\llbracket \ZERO \rrbracket_p$ & $\dn$ & $1$ \\

	$\llbracket r_1 \cdot r_2 \rrbracket_p$ & $\dn$ & $\llbracket r_1 \rrbracket_p + \llbracket r_2 \rrbracket_p + 1$\\

$\llbracket \mathbf{c} \rrbracket_p $ & $\dn$ & $1$\\

$\llbracket r_1 \cdot r_2 \rrbracket_p $ & $\dn$ & $\llbracket r_1 \rrbracket_p \; + \llbracket r_2 \rrbracket_p + 1$\\

$\llbracket a^* \rrbracket_p $ & $\dn$ & $\llbracket a \rrbracket_p + 1$

\noindent

The idea of obatining a bound for $\llbracket \bderssimp{a}{s} \rrbracket$

is to get an equivalent form

of something like $\llbracket \bderssimp{a}{s}\rrbracket = f(a, s)$, where $f(a, s)$ 

is easier to estimate than $\llbracket \bderssimp{a}{s}\rrbracket$.

We notice that while it is not so clear how to obtain

a metamorphic representation of $\bderssimp{a}{s}$ (as we argued in chapter \ref{Bitcoded2},

not interleaving the application of the functions $\backslash$ and $\bsimp{\_}$ 

in the order as our lexer will result in the bit-codes dispensed differently),

it is possible to get an slightly different representation of the unlifted versions:

$ (\bderssimp{a}{s})_\downarrow = (\erase \; \bsimp{a \backslash s})_\downarrow$.

This suggest setting the bounding function $f(a, s)$ as 

$\llbracket  (a \backslash s)_\downarrow \rrbracket_p$, the plain size

of the erased annotated regular expression.

This requires the the regular expression accompanied by bitcodes

to have the same size as its plain counterpart after erasure:

	$\asize{a} \stackrel{?}{=} \llbracket \erase(a)\rrbracket_p$. 

\noindent

But there is a minor nuisance: 

the erase function unavoidbly messes with the structure of the regular expression,

due to the discrepancy between annotated regular expression's $\sum$ constructor

and plain regular expression's $+$ constructor having different arity.

$\erase \; _{bs}\sum [] $ & $\dn$ & $\ZERO$\\

$\erase \; _{bs}\sum [a]$ & $\dn$ & $a$\\

$\erase \; _{bs}\sum a :: as$ & $\dn$ & $a + (\erase \; _{[]} \sum as)\quad \text{if $as$ length over 1}$

\noindent

 is turned into a $\ZERO$ during the

Therefore the equality in question does not hold.

$\rerase{\ZERO}$ & $\dn$ & $\RZERO$\\

$\rerase{_{bs}\ONE}$ & $\dn$ & $\RONE$\\

	$\rerase{_{bs}\mathbf{c}}$ & $\dn$ & $\RCHAR{c}$\\

$\rerase{_{bs}r_1\cdot r_2}$ & $\dn$ & $\RSEQ{\rerase{r_1}}{\rerase{r_2}}$\\

$\rerase{_{bs}\sum as}$ & $\dn$ & $\RALTS{\map \; \rerase{\_} \; as}$\\

$\rerase{_{bs} a ^*}$ & $\dn$ & $\rerase{a}^*$

\noindent

$\rrexp$ give the exact correspondence between an annotated regular expression

and its (r-)erased version:

\begin{lemma}

$\rsize{\rerase a} = \asize a$

\end{lemma}

This does not hold for plain $\rexp$s. 

	$(\RALTS{rs})\; \backslash c$ & $\dn$ &  $\RALTS{\map\; (\_ \backslash c) \;rs}$\\

With the new $\rrexp$ datatype in place, one can define its size function,

which precisely mirrors that of the annotated regular expressions:

\end{tabular}

\end{center}

\noindent

\begin{center}

\begin{tabular}{ccc}

	$\llbracket _{bs}\ONE \rrbracket_r$ & $\dn$ & $1$\\

	$\llbracket \ZERO \rrbracket_r$ & $\dn$ & $1$ \\

	$\llbracket _{bs} r_1 \cdot r_2 \rrbracket_r$ & $\dn$ & $\llbracket r_1 \rrbracket_r + \llbracket r_2 \rrbracket_r + 1$\\

$\llbracket _{bs}\mathbf{c} \rrbracket_r $ & $\dn$ & $1$\\

$\llbracket _{bs}\sum as \rrbracket_r $ & $\dn$ & $\map \; (\llbracket \_ \rrbracket_r)\; as   + 1$\\

$\llbracket _{bs} a^* \rrbracket_r $ & $\dn$ & $\llbracket a \rrbracket_r + 1$

\noindent

\subsection{Lexing Related Functions for $\rrexp$}

Everything else for $\rrexp$ will be precisely the same for annotated expressions,

except that they do not involve rectifying and augmenting bit-encoded tokenization information.

As expected, most functions are simpler, such as the derivative:

\begin{center}

  \begin{tabular}{@{}lcl@{}}

  $(\ZERO)\,\backslash_r c$ & $\dn$ & $\ZERO$\\  

  $(\ONE)\,\backslash_r c$ & $\dn$ &

        $\textit{if}\;c=d\; \;\textit{then}\;

         \ONE\;\textit{else}\;\ZERO$\\  

  $(\sum \;\textit{rs})\,\backslash_r c$ & $\dn$ &

  $\sum\;(\textit{map} \; (\_\backslash_r c) \; rs )$\\

  $(r_1\cdot r_2)\,\backslash_r c$ & $\dn$ &

     $\textit{if}\;\textit{rnullable}\,r_1$\\

					       & &$\textit{then}\;\sum\,[(r_1\,\backslash_r c)\cdot\,r_2,$\\

					       & &$\phantom{\textit{then},\;\sum\,}((r_2\,\backslash_r c))]$\\

  & &$\textit{else}\;\,(r_1\,\backslash_r c)\cdot r_2$\\

  $(r^*)\,\backslash_r c$ & $\dn$ &

      $( r\,\backslash_r c)\cdot

       (_{[]}r^*))$

\end{tabular}    

\end{center}  

\noindent

The simplification function is simplified without annotation causing superficial differences.

Duplicate removal without  an equivalence relation:

	$\rdistinct{[]}{rset} $ & $\dn$ & $[]$\\

$\rdistinct{r :: rs}{rset}$ & $\dn$ & $\textit{if}(r \in \textit{rset}) \; \textit{then} \; \rdistinct{rs}{rset}$\\

\noindent

With $\textit{rdistinct}$ one can chain together all the other modules

of $\bsimp{\_}$ (removing the functionalities related to bit-sequences)

and get $\textit{rsimp}$ and $\rderssimp{\_}{\_}$.

We omit these functions, as they are routine. Please refer to the formalisation

(in file BasicIdentities.thy) for the exact definition.

With $\rrexp$ the size caclulation of annotated regular expressions'

simplification and derivatives can be done by the size of their unlifted 

counterpart with the unlifted version of simplification and derivatives applied.

	\begin{itemize}

		\item

$\asize{\bsimp{a}} = \rsize{\rsimp{\rerase{a}}}$

\item

$\asize{\bderssimp{r}{s}} =  \rsize{\rderssimp{\rerase{r}}{s}}$

\end{itemize}

\noindent

In the following content, we will focus on $\rrexp$'s size bound.

We will piece together this bound and show the same bound for annotated regular 

expressions in the end.

 \section{Roadmap to a Finiteness Proof}

 Now that we have defined the $\rrexp$ datatype, and proven that its size changes

 w.r.t derivatives and simplifications mirrors precisely those of annotated regular expressions,

 we aim to bound the size of $r \backslash s$ for any $\rrexp$  $r$.

The way we do it is by two steps:

\begin{itemize}

	\item

		First, we rewrite $r\backslash s$ into something else that is easier

		to bound. This step is especially important for the inductive case 

		$r_1 \cdot r_2$ and $r^*$, where the derivative can grow and bloat in a wild way,

		but after simplification they will always be equal or smaller to a form consisting of an alternative

		list of regular expressions $f \; (g\; (\sum rs))$ with some functions applied to it, where each element will be distinct after the function application.

		The functions $f$ and $g$ are usually from $\flts$ $\distinctBy$ and other appearing in $\bsimp{_}$.

		This way we get a different but equivalent way of expressing : $r\backslash s = f \; (g\; (\sum rs))$, we call the

		right hand side the "closed form" of $r\backslash s$.

	\item

		Then, for such a sum  list of regular expressions $f\; (g\; (\sum rs))$, we can control its size

		by estimation, since $\distinctBy$ and $\flts$ are well-behaved and working together would only 

		reduce the size of a regular expression, not adding to it. The closed form $f\; (g\; (\sum rs)) $ is controlled

		by the size of $rs'$, where every element in $rs'$ is distinct, and each element can be described by som e inductive sub-structures (for example when $r = r_1 \cdot r_2$ then $rs'$ will be solely comprised of $r_1 \backslash s'$ and $r_2 \backslash s''$, $s'$ and $s''$ being sub-strings of $s$), which will have a numeric uppder bound according to inductive hypothesis, which controls $r \backslash s$.

\end{itemize}

\section{Closed Forms}

\subsection{Basic Properties needed for Closed Forms}

\subsubsection{$\textit{rdistinct}$'s Deduplicates Successfully}

The $\textit{rdistinct}$ function, as its name suggests, will

remove duplicates according to the accumulator

and leave only one of each different element in a list:

\begin{lemma}

	The function $\textit{rdistinct}$ satisfies the following

	properties:

	\begin{itemize}

		\item

			If $a \in acc$ then $a \notin (\rdistinct{rs}{acc})$.

		\item

			If list $rs'$ is the result of $\rdistinct{rs}{acc}$,

			All the elements in $rs'$ are unique.

	\end{itemize}

\end{lemma}

\begin{proof}

	The first part is by an induction on $rs$.

	The second part can be proven by using the 

	induction rules of $\rdistinct{\_}{\_}$.

\end{proof}

\noindent

$\rdistinct{\_}{\_}$ will cancel out all regular expression terms

that are in the accumulator, therefore prepending a list $rs_a$ with an arbitrary

list $rs$ whose elements are all from the accumulator, and then call $\rdistinct{\_}{\_}$

on the resulting list, the output will be as if we had called $\rdistinct{\_}{\_}$

without the prepending of $rs$:

\begin{lemma}

	If $rs \subseteq rset$, then 

	$\rdistinct{rs@rsa }{acc} = \rdistinct{rsa }{acc}$.

\end{lemma}

\begin{proof}

	By induction on $rs$.

\end{proof}

\noindent

On the other hand, if an element $r$ does not appear in the input list waiting to be deduplicated,

then expanding the accumulator to include that element will not cause the output list to change:

\begin{lemma}

	The accumulator can be augmented to include elements not appearing in the input list,

	and the output will not change.	

	\begin{itemize}

		\item

		If $r \notin rs$, then $\rdistinct{rs}{acc} = \rdistinct{rs}{\{r\} \cup acc}$.

		\item

			Particularly, when $acc = \varnothing$ and $rs$ de-duplicated, we have\\

			\[ \rdistinct{rs}{\varnothing} = rs \]

	\end{itemize}

\end{lemma}

\begin{proof}

	The first half is by induction on $rs$. The second half is a corollary of the first.

\end{proof}

\noindent

The next property gives the condition for

when $\rdistinct{\_}{\_}$ becomes an identical mapping

for any prefix of an input list, in other words, when can 

we ``push out" the arguments of $\rdistinct{\_}{\_}$:

\begin{lemma}

	If $rs_1$'s elements are all unique, and not appearing in the accumulator $acc$, 

	then it can be taken out and left untouched in the output:

	\[\textit{rdistinct}\;  rs_1 @ rsa\;\, acc

	= rs_1@(\textit{rdistinct} rsa \; (acc \cup rs_1)\]

\end{lemma}

\begin{proof}

By an induction on $rs_1$, where $rsa$ and $acc$ are allowed to be arbitrary.

\end{proof}

\subsubsection{The properties of $\backslash_r$, $\backslash_{rsimp}$, $\textit{Rflts}$ and $\textit{Rsimp}_{ALTS}$} 

We give in this subsection some properties of how $\backslash_r$, $\backslash_{rsimp}$, $\textit{Rflts}$ and $\textit{Rsimp}_{ALTS} $ interact with each other and  with $@$, the concatenation operator.

These will be helpful in later closed form proofs, when

we want to transform the ways in which multiple functions involving

those are composed together

in interleaving derivative and  simplification steps.

When the function $\textit{Rflts}$ 

is applied to the concatenation of two lists, the output can be calculated by first applying the

functions on two lists separately, and then concatenating them together.

\begin{lemma}

	The function $\rflts$ has the below properties:\\

	\begin{itemize}

		\item

	$\rflts \; (rs_1 @ rs_2) = \rflts \; rs_1 @ \rflts \; rs_2$

\item

	If $r \neq \RZERO$ and $\nexists rs_1. r = \RALTS{rs}_1$, then $\rflts \; (r::rs) = r :: \rflts \; rs$

	\end{itemize}

\end{lemma}

\noindent

\begin{proof}

	By induction on $rs_1$.

\end{proof}

\subsection{A Closed Form for the Sequence Regular Expression}

\begin{quote}\it

	Claim: For regular expressions $r_1 \cdot r_2$, we claim that

	\begin{center}

		$ \rderssimp{r_1 \cdot r_2}{s} = 

	\rsimp{(\sum (r_1 \backslash s \cdot r_2 ) \; :: \;(\map \; \rderssimp{r2}{\_} \;(\vsuf{s}{r_1})))}$

	\end{center}

\end{quote}

\noindent