
diff -r 2072a8d54e3e -r ae6010c14e49 ChengsongTanPhdThesis/Chapters/Cubic.tex
--- a/ChengsongTanPhdThesis/Chapters/Cubic.tex	Sun Nov 06 23:06:10 2022 +0000
+++ b/ChengsongTanPhdThesis/Chapters/Cubic.tex	Mon Nov 07 21:31:07 2022 +0000
@@ -42,10 +42,10 @@
 include this new construct. With bounded repetitions
 we are able to out-compete other verified lexers such as
 Verbatim++ on regular expressions which involve a lot of
-repetitions. We also present the idempotency property proof
-of $\bsimp$, which leverages the idempotency proof of $\rsimp$.
-This reinforces our claim that the fixpoint construction
-originally required by Sulzmann and Lu can be removed in $\blexersimp$.
+repetitions. %We also present the idempotency property proof
+%of $\bsimp$, which leverages the idempotency proof of $\rsimp$.
+%This reinforces our claim that the fixpoint construction
+%originally required by Sulzmann and Lu can be removed in $\blexersimp$.
 \\
 Last but not least, we present our efforts and challenges we met
 in further improving the algorithm by data
@@ -541,8 +541,674 @@
 
 
 \subsubsection{Finiteness Proof Augmentation}
-The bounded repetitions can be handled
-using techniques similar to stars.
+The bounded repetitions are
+very similar to stars, and therefore the treatment
+is similar, with minor changes to handle some slight complications
+when the counter reaches 0.
+The exponential growth is similar:
+\begin{center}
+	\begin{tabular}{ll}
+		$r^{\{n\}} $ & $\longrightarrow_{\backslash c}$\\
+		$(r\backslash c)  \cdot  
+		r^{\{n - 1\}}*$ & $\longrightarrow_{\backslash c'}$\\
+		\\
+		$r \backslash cc'  \cdot r^{\{n - 2\}}* + 
+		r \backslash c' \cdot r^{\{n - 1\}}*$ &
+		$\longrightarrow_{\backslash c''}$\\
+		\\
+		$(r_1 \backslash cc'c'' \cdot r^{\{n-3\}}* + 
+		r \backslash c''\cdot r^{\{n-1\}}) + 
+		(r \backslash c'c'' \cdot r^{\{n-2\}}* + 
+		r \backslash c'' \cdot r^{\{n-1\}}*)$ & 
+		$\longrightarrow_{\backslash c'''}$ \\
+		\\
+		$\ldots$\\
+	\end{tabular}
+\end{center}
+Again, we assume that $r\backslash c$, $r \backslash cc'$ and so on
+are all nullable.
+The flattened list of terms for $r^{\{n\}} \backslash_{rs} s$
+\begin{center}
+	$[r_1 \backslash cc'c'' \cdot r^{\{n-3\}}*,\;
+	r \backslash c''\cdot r^{\{n-1\}}, \; 
+	r \backslash c'c'' \cdot r^{\{n-2\}}*, \;
+	r \backslash c'' \cdot r^{\{n-1\}}*,\; \ldots ]$  
+\end{center}
+that comes from 
+\begin{center}
+		$(r_1 \backslash cc'c'' \cdot r^{\{n-3\}}* + 
+		r \backslash c''\cdot r^{\{n-1\}}) + 
+		(r \backslash c'c'' \cdot r^{\{n-2\}}* + 
+		r \backslash c'' \cdot r^{\{n-1\}}*)+ \ldots$ 
+\end{center}
+are made of sequences with different tails, where the counters
+might differ.
+The observation for maintaining the bound is that
+these counters never exceed $n$, the original
+counter. With the number of counters staying finite,
+$\rDistinct$ will deduplicate and keep the list finite.
+We introduce this idea as a lemma once we describe all
+the necessary helper functions.
+
+Similar to the star case, we want
+\begin{center}
+	$\rderssimp{r^{\{n\}}}{s} = \rsimp{\sum rs}$.
+\end{center}
+where $rs$
+shall be in the form of 
+$\map \; f \; Ss$, where $f$ is a function and
+$Ss$ a list of objects to act on.
+For star, the object's datatype is string.
+The list of strings $Ss$
+is generated using functions 
+$\starupdate$ and $\starupdates$.
+The function that takes a string and returns a regular expression
+is the anonymous function $
+(\lambda s'. \; r\backslash s' \cdot r^{\{m\}})$.
+In the NTIMES setting,
+the $\starupdate$ and $\starupdates$ functions are replaced by 
+$\textit{nupdate}$ and $\textit{nupdates}$:
+\begin{center}
+	\begin{tabular}{lcl}
+		$\nupdate \; c \; r \; [] $ & $\dn$ & $[]$\\
+		$\nupdate \; c \; r \; 
+		(\Some \; (s, \; n + 1) \; :: \; Ss)$ & $\dn$ & %\\
+						     $\textit{if} \; 
+						     (\rnullable \; (r \backslash_{rs} s))$ \\
+						     & & $\;\;\textit{then} 
+						     \;\; \Some \; (s @ [c], n + 1) :: \Some \; ([c], n) :: (
+						     \nupdate \; c \; r \; Ss)$ \\
+						     & & $\textit{else} \;\; \Some \; (s @ [c], n+1) :: (
+						     \nupdate \; c \; r \; Ss)$\\
+		$\nupdate \; c \; r \; (\textit{None} :: Ss)$ & $\dn$ & 
+		$(\None :: \nupdate  \; c \; r \; Ss)$\\
+							      & & \\
+	%\end{tabular}
+%\end{center}
+%\begin{center}
+	%\begin{tabular}{lcl}
+		$\nupdates \; [] \; r \; Ss$ & $\dn$ & $Ss$\\
+		$\nupdates \; (c :: cs) \; r \; Ss$ &  $\dn$ &  $\nupdates \; cs \; r \; (
+		\nupdate \; c \; r \; Ss)$
+	\end{tabular}
+\end{center}
+\noindent
+which take into account when a subterm
+\begin{center}
+	$r \backslash_s s \cdot r^{\{n\}}$
+\end{center}
+counter $n$
+is 0, and therefore expands to 
+\begin{center}
+$r \backslash_s (s@[c]) \cdot r^{\{n\}} \;+
+\; \ZERO$ 
+\end{center}
+after taking a derivative.
+The object now has type 
+\begin{center}
+$\textit{option} \;(\textit{string}, \textit{nat})$
+\end{center}
+and therefore the function for converting such an option into
+a regular expression term is called $\opterm$:
+
+\begin{center}
+	\begin{tabular}{lcl}
+	$\opterm \; r \; SN$ & $\dn$ & $\textit{case} \; SN\; of$\\
+				 & & $\;\;\Some \; (s, n) \Rightarrow 
+				 (r\backslash_{rs} s)\cdot r^{\{n\}}$\\
+				 & & $\;\;\None  \Rightarrow 
+				 \ZERO$\\
+	\end{tabular}
+\end{center}
+\noindent
+Put together, the list $\map \; f \; Ss$ is instantiated as
+\begin{center}
+	$\map \; (\opterm \; r) \; (\nupdates \; s \; r \;
+	[\Some \; ([c], n)])$.
+\end{center}
+For the closed form to be bounded, we would like
+simplification to be applied to each term in the list.
+Therefore we introduce some variants of $\opterm$,
+which help conveniently express the rewriting steps 
+needed in the closed form proof.
+\begin{center}
+	\begin{tabular}{lcl}
+	$\optermOsimp \; r \; SN$ & $\dn$ & $\textit{case} \; SN\; of$\\
+				 & & $\;\;\Some \; (s, n) \Rightarrow 
+				 \textit{rsimp} \; ((r\backslash_{rs} s)\cdot r^{\{n\}})$\\
+				 & & $\;\;\None  \Rightarrow 
+				 \ZERO$\\
+				 \\
+	$\optermosimp \; r \; SN$ & $\dn$ & $\textit{case} \; SN\; of$\\
+				 & & $\;\;\Some \; (s, n) \Rightarrow 
+				 (\textit{rsimp} \; (r\backslash_{rs} s)) 
+				 \cdot r^{\{n\}}$\\
+				 & & $\;\;\None  \Rightarrow 
+				 \ZERO$\\
+				 \\
+	$\optermsimp \; r \; SN$ & $\dn$ & $\textit{case} \; SN\; of$\\
+				 & & $\;\;\Some \; (s, n) \Rightarrow 
+				 (r\backslash_{rsimps} s)\cdot r^{\{n\}}$\\
+				 & & $\;\;\None  \Rightarrow 
+				 \ZERO$\\
+	\end{tabular}
+\end{center}
+
+
+For a list of 
+$\textit{option} \;(\textit{string}, \textit{nat})$ elements,
+we define the highest power for it recursively:
+\begin{center}
+	\begin{tabular}{lcl}
+		$\hpa \; [] \; n $ & $\dn$ & $n$\\
+		$\hpa \; (\None :: os) \; n $ &  $\dn$ &  $\hpa \; os \; n$\\
+		$\hpa \; (\Some \; (s, n) :: os) \; m$ & $\dn$ & 
+		$\hpa \;os \; (\textit{max} \; n\; m)$\\
+		\\
+		$\hpower \; rs $ & $\dn$ & $\hpa \; rs \; 0$\\
+	\end{tabular}
+\end{center}
+
+Now the intuition that an NTIMES regular expression's power
+does not increase can be easily expressed as
+\begin{lemma}\label{nupdatesMono2}
+	$\hpower \; (\nupdates \;s \; r \; [\Some \; ([c], n)]) \leq n$
+\end{lemma}
+\begin{proof}
+	Note that the power is non-increasing after a $\nupdate$ application:
+	\begin{center}
+		$\hpa \;\; (\nupdate \; c \; r \; Ss)\;\; m \leq 
+		 \hpa\; \; Ss \; m$.
+	 \end{center}
+	 This is also the case for $\nupdates$:
+	\begin{center}
+		$\hpa \;\; (\nupdates \; s \; r \; Ss)\;\; m \leq 
+		 \hpa\; \; Ss \; m$.
+	 \end{center}
+	 Therefore we have that
+	 \begin{center}
+		 $\hpower \;\; (\nupdates \; s \; r \; Ss) \leq
+		  \hpower \;\; Ss$
+	 \end{center}
+	 which leads to the lemma being proven.
+
+ \end{proof}
+
+
+We also define the inductive rules for
+the shape of derivatives of the NTIMES regular expressions:\\[-3em]
+\begin{center}
+	\begin{mathpar}
+		\inferrule{\mbox{}}{\cbn \;\ZERO}
+
+		\inferrule{\mbox{}}{\cbn \; \; r_a \cdot (r^{\{n\}})}
+
+		\inferrule{\cbn \; r_1 \;\; \; \cbn \; r_2}{\cbn \; r_1 + r_2}
+
+		\inferrule{\cbn \; r}{\cbn \; r + \ZERO}
+	\end{mathpar}
+\end{center}
+\noindent
+A derivative of NTIMES fits into the shape described by $\cbn$:
+\begin{lemma}\label{ntimesDersCbn}
+	$\cbn \; ((r' \cdot r^{\{n\}}) \backslash_{rs} s)$ holds.
+\end{lemma}
+\begin{proof}
+	By a reverse induction on $s$.
+	For the inductive case, note that if $\cbn \; r$ holds,
+	then $\cbn \; (r\backslash_r c)$ holds.
+\end{proof}
+In addition, for $\cbn$-shaped regular expressioins, one can flatten
+them:
+\begin{lemma}\label{ntimesHfauPushin}
+	If $\cbn \; r$ holds, then $\hflataux{r \backslash_r c} = 
+	\textit{concat} \; (\map \; \hflataux{\map \; (\_\backslash_r c) \;
+	(\hflataux{r})})$
+\end{lemma}
+\begin{proof}
+	By an induction on the inductive cases of $\cbn$.
+\end{proof}
+		
+
+
+This time we do not need to define the flattening functions for NTIMES only,
+because $\hflat{\_}$ and $\hflataux{\_}$ work on NTIMES already:
+\begin{lemma}\label{ntimesHfauInduct}
+$\hflataux{( (r\backslash_r c) \cdot r^{\{n\}}) \backslash_{rsimps} s} = 
+ \map \; (\opterm \; r) \; (\nupdates \; s \; r \; [\Some \; ([c], n)])$
+\end{lemma}
+\begin{proof}
+	By a reverse induction on $s$.
+	The lemma \ref{ntimesDersCbn} is used.
+\end{proof}
+\noindent
+We have a recursive property for NTIMES with $\nupdate$ 
+similar to that for STAR,
+and one for $\nupdates $ as well:
+\begin{lemma}\label{nupdateInduct1}
+	(i) $\textit{concat} \; (\map \; (\hflataux{\_} \circ (
+	\opterm \; r)) \; Ss) = \map \; (\opterm \; r) \; (\nupdate \;
+	c \; r \; Ss)$\\
+	(ii) $\textit{concat} \; (\map \; \hflataux{\_}\; 
+	\map \; (\_\backslash_r x) \;
+		(\map \; (\opterm \; r) \; (\nupdates \; xs \; r \; Ss))) =
+	\map \; (\opterm \; r) \; (\nupdates \;(xs@[x]) \; r\;Ss)$ holds.
+\end{lemma}
+\begin{proof}
+	(i) is by an induction on $Ss$.
+	(ii) is by an induction on $xs$.
+\end{proof}
+
+The $\nString$ predicate is defined for conveniently
+expressing that there are no empty strings in the
+$\Some \;(s, n)$ elements generated by $\nupdate$:
+\begin{center}
+	\begin{tabular}{lcl}
+		$\nString \; \None$  & $\dn$ & $ \textit{true}$\\
+		$\nString \; (\Some \; ([], n))$ & $\dn$ & $ \textit{false}$\\
+		$\nString \; (\Some \; (c::s, n))$  & $\dn$ & $ \textit{true}$\\
+	\end{tabular}
+\end{center}
+\begin{lemma}\label{nupdatesNonempty}
+	If for all elements $opt$ in $\textit{set} \; Ss$,
+	$\nString \; opt$ holds, the we have that
+	for all elements $opt$ in $\textit{set} \; (\nupdates \; s \; r \; Ss)$,
+	$\nString \; opt$ holds.
+\end{lemma}
+\begin{proof}
+	By an induction on $s$, where $Ss$ is set to vary over all possible values.
+\end{proof}
+
+\noindent
+
+\begin{lemma}\label{ntimesClosedFormsSteps}
+	The following list of equalities or rewriting relations hold:\\
+	(i) $r^{\{n+1\}} \backslash_{rsimps} (c::s) = 
+	\textit{rsimp} \; (\sum (\map \; (\opterm \;r \;\_) \; (\nupdates \;
+	s \; r \; [\Some \; ([c], n)])))$\\
+	(ii)
+	\begin{center}
+	$\sum (\map \; (\opterm \; r) \; (\nupdates \; s \; r \; [
+	\Some \; ([c], n)]))$ \\ $ \sequal$\\
+	 $\sum (\map \; (\textit{rsimp} \circ (\opterm \; r))\; (\nupdates \;
+	 s\;r \; [\Some \; ([c], n)]))$\\
+ 	\end{center}
+	(iii)
+	\begin{center}
+	$\sum \;(\map \; (\optermosimp \; r) \; (\nupdates \; s \; r\; [\Some \;
+	([c], n)]))$\\ 
+	$\sequal$\\
+	 $\sum \;(\map \; (\optermsimp r) \; (\nupdates \; s \; r \; [\Some \;
+	([c], n)])) $\\
+	\end{center}
+	(iv)
+	\begin{center}
+	$\sum \;(\map \; (\optermosimp \; r) \; (\nupdates \; s \; r\; [\Some \;
+	([c], n)])) $ \\ $\sequal$\\
+	 $\sum \;(\map \; (\optermOsimp r) \; (\nupdates \; s \; r \; [\Some \;
+	([c], n)])) $\\
+	\end{center}
+	(v)
+	\begin{center}
+	 $\sum \;(\map \; (\optermOsimp r) \; (\nupdates \; s \; r \; [\Some \;
+	 ([c], n)])) $ \\ $\sequal$\\
+	  $\sum \; (\map \; (\textit{rsimp} \circ (\opterm \; r)) \;
+	  (\nupdates \; s \; r \; [\Some \; ([c], n)]))$
+  	\end{center}
+\end{lemma}
+\begin{proof}
+	Routine.
+	(iii) and (iv) make use of the fact that all the strings $s$
+	inside $\Some \; (s, m)$ which are elements of the list
+	$\nupdates \; s\;r\;[\Some\; ([c], n)]$ are non-empty,
+	which is from lemma \ref{nupdatesNonempty}.
+	Once the string in $o = \Some \; (s, n)$ is 
+	nonempty, $\optermsimp \; r \;o$,
+	$\optermosimp \; r \; o$ and $\optermosimp \; \; o$ are guaranteed
+	to be equal.
+\end{proof}
+
+\begin{theorem}\label{ntimesClosedForm}
+	The derivative of $r^{\{n+1\}}$ can be described as an alternative
+	containing a list
+	of terms:\\
+	$r^{\{n+1\}} \backslash_{rsimps} (c::s) = \textit{rsimp} \; (
+	\sum (\map \; (\optermsimp \; r) \; (\nupdates \; s \; r \;
+	[\Some \; ([c], n)])))$
+\end{theorem}
+\begin{proof}
+	By the rewriting steps described in lemma \ref{ntimesClosedFormsSteps}.
+\end{proof}
+
+\begin{lemma}\label{nupdatesNLeqN}
+	For an element $o$ in $\textit{set} \; (\nupdates \; s \; r \;
+	[\Some \; ([c], n)])$, either $o = \None$, or $o = \Some
+	\; (s', m)$ for some string $s'$ and number $m \leq n$.
+\end{lemma}
+
+
+\begin{lemma}\label{ntimesClosedFormListElemShape}
+	For any element $r'$ in $\textit{set} \; (\map \; (\optermsimp \; r) \; (
+	\nupdates \; s \; r \; [\Some \; ([c], n)]))$,
+	we have that $r'$ is either $\ZERO$ or $r \backslash_{rsimps} s' \cdot
+	r^{\{m\}}$ for some string $s'$ and number $m \leq n$.
+\end{lemma}
+\begin{proof}
+	Using lemma \ref{nupdatesNLeqN}.
+\end{proof}
+
+\begin{theorem}\label{ntimesClosedFormBounded}
+	Assuming that for any string $s$, $\llbracket r \backslash_{rsimps} s
+	\rrbracket_r \leq N$ holds, then we have that\\
+	$\llbracket r^{\{n+1\}} \backslash_{rsimps} s \rrbracket_r \leq
+	\textit{max} \; (c_N+1)* (N + \llbracket r^{\{n\}} \rrbracket+1)$,
+	where $c_N = \textit{card} \; (\textit{sizeNregex} \; (
+	N + \llbracket r^{\{n\}} \rrbracket_r+1))$.
+\end{theorem}
+\begin{proof}
+We have that for all regular expressions $r'$ in $\textit{set} \; (\map \; (\optermsimp \; r) \; (
+	\nupdates \; s \; r \; [\Some \; ([c], n)]))$,
+	$r'$'s size is less than or equal to $N + \llbracket r^{\{n\}} 
+	\rrbracket_r + 1$
+because $r'$ can only be either a $\ZERO$ or $r \backslash_{rsimps} s' \cdot
+r^{\{m\}}$ for some string $s'$ and number 
+$m \leq n$ (lemma \ref{ntimesClosedFormListElemShape}).
+In addition, we know that the list 
+$\map \; (\optermsimp \; r) \; (
+\nupdates \; s \; r \; [\Some \; ([c], n)])$'s size is at most
+$c_N = \textit{card} \; 
+(\sizeNregex \; ((N + \llbracket r^{\{n\}} \rrbracket) + 1))$.
+This gives us $\llbracket \llbracket r \backslash_{rsimps} \;s \llbracket_r
+\leq N * c_N$.
+\end{proof}
+
+
+Assuming we have that
+\begin{center}
+	$\rderssimp{r^*}{s} = \rsimp{(\sum \map \; (\lambda s'. r\backslash s' \cdot r^*) \; Ss)}$.
+\end{center}
+holds.
+The idea of $\starupdate$ and $\starupdates$
+is to update $Ss$ when another
+derivative is taken on $\rderssimp{r^*}{s}$
+w.r.t a character $c$ and a string $s'$
+respectively.
+Both $\starupdate$ and $\starupdates$ take three arguments as input:
+the new character $c$ or string $s$ to take derivative with, 
+the regular expression
+$r$ under the star $r^*$, and the
+list of strings $Ss$ for the derivative $r^* \backslash s$ 
+up until this point  
+such that 
+\begin{center}
+$(r^*) \backslash s = \sum_{s' \in sSet} (r\backslash s') \cdot r^*$ 
+\end{center}
+is satisfied.
+
+Functions $\starupdate$ and $\starupdates$ characterise what the 
+star derivatives will look like once ``straightened out'' into lists.
+The helper functions for such operations will be similar to
+$\sflat{\_}$, $\sflataux{\_}$ and $\sflataux{\_}$, which we defined for sequence.
+We use similar symbols to denote them, with a $*$ subscript to mark the difference.
+\begin{center}
+	\begin{tabular}{lcl}
+		$\hflataux{r_1 + r_2}$ & $\dn$ & $\hflataux{r_1} @ \hflataux{r_2}$\\
+		$\hflataux{r}$ & $\dn$ & $[r]$
+	\end{tabular}
+\end{center}
+
+\begin{center}
+	\begin{tabular}{lcl}
+		$\hflat{r_1 + r_2}$ & $\dn$ & $\sum (\hflataux {r_1} @ \hflataux {r_2}) $\\
+		$\hflat{r}$ & $\dn$ & $r$
+	\end{tabular}
+\end{center}
+\noindent
+These definitions are tailor-made for dealing with alternatives that have
+originated from a star's derivatives.
+A typical star derivative always has the structure of a balanced binary tree:
+\begin{center}
+	$(r_1 \backslash cc'c'' \cdot r^* + r \backslash c'') + 
+	(r \backslash c'c'' \cdot r^* + r \backslash c'' \cdot r^*) $ 
+\end{center}
+All of the nested structures of alternatives
+generated from derivatives are binary, and therefore
+$\hflat{\_}$ and $\hflataux{\_}$ only deal with binary alternatives.
+$\hflat{\_}$ ``untangles'' like the following:
+\[
+	\sum ((r_1 + r_2) + (r_3 + r_4))  + \ldots \;
+	\stackrel{\hflat{\_}}{\longrightarrow} \;
+	\RALTS{[r_1, r_2, \ldots, r_n]} 
+\]
+Here is a lemma stating the recursive property of $\starupdate$ and $\starupdates$,
+with the helpers $\hflat{\_}$ and $\hflataux{\_}$\footnote{The function $\textit{concat}$ takes a list of lists 
+			and merges each of the element lists to form a flattened list.}:
+\begin{lemma}\label{stupdateInduct1}
+	\mbox
+	For a list of strings $Ss$, the following hold.
+	\begin{itemize}
+		\item
+			If we do a derivative on the terms 
+			$r\backslash_r s \cdot r^*$ (where $s$ is taken from the list $Ss$),
+			the result will be the same as if we apply $\starupdate$ to $Ss$.
+			\begin{center}
+				\begin{tabular}{c}
+			$\textit{concat} \; (\map \; (\hflataux{\_} \circ ( (\_\backslash_r x)
+			\circ (\lambda s.\;\; (r \backslash_r s) \cdot r^*)))\; Ss )\;
+			$\\
+			\\
+			$=$ \\
+			\\
+			$\map \; (\lambda s. (r \backslash_r s) \cdot (r^*)) \; 
+			(\starupdate \; x \; r \; Ss)$
+				\end{tabular}
+			\end{center}
+		\item
+			$\starupdates$ is ``composable'' w.r.t a derivative.
+			It piggybacks the character $x$ to the tail of the string
+			$xs$.
+			\begin{center}
+				\begin{tabular}{c}
+					$\textit{concat} \; (\map \; \hflataux{\_} \; 
+					(\map \; (\_\backslash_r x) \; 
+					(\map \; (\lambda s.\;\; (r \backslash_r s) \cdot 
+					(r^*) ) \; (\starupdates \; xs \; r \; Ss))))$\\
+					\\
+					$=$\\
+					\\
+					$\map \; (\lambda s.\;\; (r\backslash_r s) \cdot (r^*)) \;
+					(\starupdates \; (xs @ [x]) \; r \; Ss)$
+				\end{tabular}
+			\end{center}
+	\end{itemize}
+\end{lemma}
+			
+\begin{proof}
+	Part 1 is by induction on $Ss$.
+	Part 2 is by induction on $xs$, where $Ss$ is left to take arbitrary values.
+\end{proof}
+			
+
+Like $\textit{createdBySequence}$, we need
+a predicate for ``star-created'' regular expressions:
+\begin{center}
+	\begin{mathpar}
+		\inferrule{\mbox{}}{ \textit{createdByStar}\; \RSEQ{ra}{\RSTAR{rb}} }
+
+		\inferrule{  \textit{createdByStar} \; r_1\; \land  \; \textit{createdByStar} \; r_2 }{\textit{createdByStar} \; (r_1 + r_2) } 
+	\end{mathpar}
+\end{center}
+\noindent
+All regular expressions created by taking derivatives of
+$r_1 \cdot (r_2)^*$ satisfy the $\textit{createdByStar}$ predicate:
+\begin{lemma}\label{starDersCbs}
+	$\textit{createdByStar} \; ((r_1 \cdot r_2^*) \backslash_r s) $ holds.
+\end{lemma}
+\begin{proof}
+	By a reverse induction on $s$.
+\end{proof}
+If a regular expression conforms to the shape of a star's derivative,
+then we can push an application of $\hflataux{\_}$ inside a derivative of it:
+\begin{lemma}\label{hfauPushin}
+	If $\textit{createdByStar} \; r$ holds, then
+	\begin{center}
+		$\hflataux{r \backslash_r c} = \textit{concat} \; (
+		\map \; \hflataux{\_} (\map \; (\_\backslash_r c) \;(\hflataux{r})))$
+	\end{center}
+	holds.
+\end{lemma}
+\begin{proof}
+	By an induction on the inductivev cases of $\textit{createdByStar}$.
+\end{proof}
+%This is not entirely true for annotated regular expressions: 
+%%TODO: bsimp bders \neq bderssimp
+%\begin{center}
+%	$(1+ (c\cdot \ASEQ{bs}{c^*}{c} ))$
+%\end{center}
+%For bit-codes, the order in which simplification is applied
+%might cause a difference in the location they are placed.
+%If we want something like
+%\begin{center}
+%	$\bderssimp{r}{s} \myequiv \bsimp{\bders{r}{s}}$
+%\end{center}
+%Some "canonicalization" procedure is required,
+%which either pushes all the common bitcodes to nodes
+%as senior as possible:
+%\begin{center}
+%	$_{bs}(_{bs_1 @ bs'}r_1 + _{bs_1 @ bs''}r_2) \rightarrow _{bs @ bs_1}(_{bs'}r_1 + _{bs''}r_2) $
+%\end{center}
+%or does the reverse. However bitcodes are not of interest if we are talking about
+%the $\llbracket r \rrbracket$ size of a regex.
+%Therefore for the ease and simplicity of producing a
+%proof for a size bound, we are happy to restrict ourselves to 
+%unannotated regular expressions, and obtain such equalities as
+%TODO: rsimp sflat
+% The simplification of a flattened out regular expression, provided it comes
+%from the derivative of a star, is the same as the one nested.
+
+
+
+Now we introduce an inductive property
+for $\starupdate$ and $\hflataux{\_}$.
+\begin{lemma}\label{starHfauInduct}
+	If we do derivatives of $r^*$
+	with a string that starts with $c$,
+	then flatten it out,
+	we obtain a list
+	of the shape $\sum_{s' \in sS} (r\backslash_r s') \cdot r^*$,
+	where $sS = \starupdates \; s \; r \; [[c]]$. Namely,
+	\begin{center}
+		$\hflataux{(\rders{( (\rder{c}{r_0})\cdot(r_0^*))}{s})} = 
+		\map \; (\lambda s_1. (r_0 \backslash_r s_1) \cdot (r_0^*)) \; 
+		(\starupdates \; s \; r_0 \; [[c]])$
+	\end{center}
+holds.
+\end{lemma}
+\begin{proof}
+	By an induction on $s$, the inductive cases
+	being $[]$ and $s@[c]$. The lemmas \ref{hfauPushin} and \ref{starDersCbs} are used.
+\end{proof}
+\noindent
+
+$\hflataux{\_}$ has a similar effect as $\textit{flatten}$:
+\begin{lemma}\label{hflatauxGrewrites}
+	$a :: rs \grewrites \hflataux{a} @ rs$
+\end{lemma}
+\begin{proof}
+	By induction on $a$. $rs$ is set to take arbitrary values.
+\end{proof}
+It is also not surprising that $\textit{rsimp}$ will cover
+the effect of $\hflataux{\_}$:
+\begin{lemma}\label{cbsHfauRsimpeq1}
+	$\rsimp{(r_1 + r_2)} = \rsimp{(\RALTS{\hflataux{r_1} @ \hflataux{r_2}})}$
+\end{lemma}
+
+\begin{proof}
+	By using the rewriting relation $\rightsquigarrow$
+\end{proof}
+And from this we obtain a proof that a star's derivative will be the same
+as if it had all its nested alternatives created during deriving being flattened out:
+For example,
+\begin{lemma}\label{hfauRsimpeq2}
+	$\createdByStar{r} \implies \rsimp{r} = \rsimp{\RALTS{\hflataux{r}}}$
+\end{lemma}
+\begin{proof}
+	By structural induction on $r$, where the induction rules 
+	are these of $\createdByStar{\_}$.
+	Lemma \ref{cbsHfauRsimpeq1} is used in the inductive case.
+\end{proof}
+
+
+%Here is a corollary that states the lemma in
+%a more intuitive way:
+%\begin{corollary}
+%	$\hflataux{r^* \backslash_r (c::xs)} = \map \; (\lambda s. (r \backslash_r s) \cdot
+%	(r^*))\; (\starupdates \; c\; r\; [[c]])$
+%\end{corollary}
+%\noindent
+%Note that this is also agnostic of the simplification
+%function we defined, and is therefore of more general interest.
+
+Together with the rewriting relation
+\begin{lemma}\label{starClosedForm6Hrewrites}
+	We have the following set of rewriting relations or equalities:
+	\begin{itemize}
+		\item
+			$\textit{rsimp} \; (r^* \backslash_r (c::s)) 
+			\sequal
+			\sum \; ( ( \sum (\lambda s. (r\backslash_r s) \cdot r^*) \; (
+			\starupdates \; s \; r \; [ c::[]] ) ) )$
+		\item
+			$r \backslash_{rsimp} (c::s) = \textit{rsimp} \; ( (
+			\sum ( (\map \; (\lambda s_1. (r\backslash s_1) \; r^*) \;
+			(\starupdates \;s \; r \; [ c::[] ])))))$
+		\item
+			$\sum ( (\map \; (\lambda s. (r\backslash s) \; r^*) \; Ss))
+			\sequal
+			 \sum ( (\map \; (\lambda s. \textit{rsimp} \; (r\backslash s) \;
+			 r^*) \; Ss) )$
+		\item
+			$\map \; (\lambda s. (\rsimp{r \backslash_r s}) \cdot (r^*)) \; Ss
+			\scfrewrites
+			\map \; (\lambda s. (\rsimp{r \backslash_r s}) \cdot (r^*)) \; Ss$
+		\item
+			$( ( \sum ( ( \map \ (\lambda s. \;\;
+			(\textit{rsimp} \; (r \backslash_r s)) \cdot r^*) \; (\starupdates \;
+			s \; r \; [ c::[] ])))))$\\
+			$\sequal$\\
+			$( ( \sum ( ( \map \ (\lambda s. \;\;
+			( r \backslash_{rsimp} s)) \cdot r^*) \; (\starupdates \;
+			s \; r \; [ c::[] ]))))$\\
+	\end{itemize}
+\end{lemma}
+\begin{proof}
+	Part 1 leads to part 2.
+	The rest of them are routine.
+\end{proof}
+\noindent
+Next the closed form for star regular expressions can be derived:
+\begin{theorem}\label{starClosedForm}
+	$\rderssimp{r^*}{c::s} = 
+	\rsimp{
+		(\sum (\map \; (\lambda s. (\rderssimp{r}{s})\cdot r^*) \; 
+		(\starupdates \; s\; r \; [[c]])
+		)
+		)
+	}
+	$
+\end{theorem}
+\begin{proof}
+	By an induction on $s$.
+	The lemmas \ref{rsimpIdem}, \ref{starHfauInduct}, \ref{starClosedForm6Hrewrites} 
+	and \ref{hfauRsimpeq2}
+	are used.	
+	In \ref{starClosedForm6Hrewrites}, the equalities are
+	used to link the LHS and RHS.
+\end{proof}
+
+
+
+
 The closed form for them looks like:
 %\begin{center}
 %	\begin{tabular}{llrclll}
@@ -598,6 +1264,9 @@
 \end{center}
 \noindent
 
+
+
+
 %----------------------------------------------------------------------------------------
 %	SECTION 1
 %----------------------------------------------------------------------------------------
@@ -663,3 +1332,11 @@
 % \]
 % 
 %
+%\begin{center}
+%	\begin{tabular}{lcl}
+%		$\ntset \; r \; (n+1) \; c::cs $ & $\dn$ & $\nupdates \;
+%		cs \; r \; [\Some \; ([c], n)]$\\
+%		$\ntset \; r\; 0 \; \_$ &  $\dn$ &  $\None$\\
+%		$\ntset \; r \; \_ \; [] $ & $ \dn$ & $[]$\\
+%	\end{tabular}
+%\end{center}




