
diff -r d9d2da923b7f -r 788f4aa28bc5 etnms/20200205.tex
--- a/etnms/20200205.tex	Thu Feb 06 10:49:23 2020 +0000
+++ b/etnms/20200205.tex	Sat Feb 08 21:34:50 2020 +0000
@@ -1384,100 +1384,31 @@
 Why?
 
 During the first derivative operation, 
-$\rup\backslash a=(    _0[ \ONE\cdot {\bf b}] + _1( _0[ _1\ONE \cdot {\bf a}^*] + [ \ONE \cdot {\bf a}])      )$ 
-is in the form of a sequence regular expression with
-two components, the first
-one $\ONE + \ZERO$ being nullable. 
-Recall again the simplification function definition:
 \begin{center}
-  \begin{tabular}{@{}lcl@{}}
-   
-  $\textit{simp} \; (\textit{SEQ}\;bs\,a_1\,a_2)$ & $\dn$ & $ (\textit{simp} \; a_1, \textit{simp}  \; a_2) \; \textit{match} $ \\
-   &&$\quad\textit{case} \; (\ZERO, \_) \Rightarrow  \ZERO$ \\
-   &&$\quad\textit{case} \; (\_, \ZERO) \Rightarrow  \ZERO$ \\
-   &&$\quad\textit{case} \;  (\ONE, a_2') \Rightarrow  \textit{fuse} \; bs \;  a_2'$ \\
-   &&$\quad\textit{case} \; (a_1', \ONE) \Rightarrow  \textit{fuse} \; bs \;  a_1'$ \\
-   &&$\quad\textit{case} \; (a_1', a_2') \Rightarrow  \textit{SEQ} \; bs \; a_1' \;  a_2'$ \\
-
-  $\textit{simp} \; (\textit{ALTS}\;bs\,as)$ & $\dn$ & $\textit{distinct}( \textit{flatten} ( \textit{map simp as})) \; \textit{match} $ \\
-  &&$\quad\textit{case} \; [] \Rightarrow  \ZERO$ \\
-   &&$\quad\textit{case} \; a :: [] \Rightarrow  \textit{fuse bs a}$ \\
-   &&$\quad\textit{case} \;  as' \Rightarrow  \textit{ALTS}\;bs\;as'$\\ 
-
-   $\textit{simp} \; a$ & $\dn$ & $\textit{a} \qquad \textit{otherwise}$   
-\end{tabular}    
-\end{center}    
+$\rup\backslash a=(    _0[ \ONE\cdot {\bf b}] + _1( _0[ _1\ONE \cdot {\bf a}^*] + [ \ONE \cdot {\bf a}])      )$,
+\end{center}
+\noindent
+ the second derivative gives us
+ \begin{center}
+$\rup\backslash a=(_0( [\ZERO\cdot {\bf b}] + 0) + _1( _0( [\ZERO\cdot {\bf a}^*] + _1[ _1\ONE \cdot {\bf a}^*]) + _1( [\ZERO \cdot {\bf a}] + \ONE)  ))$,
+\end{center}
 
 \noindent
-
-and the difinition of $\flatten$:
- \begin{center}
- \begin{tabular}{c c c}
- $\flatten \; []$ & $\dn$ & $[]$\\
- $\flatten \; \ZERO::rs$ & $\dn$ & $rs$\\
- $\flatten \;(_{\textit{bs}_1}\oplus \textit{rs}_1 ::rs)$ & $\dn$ & $(\map \, (\fuse \, \textit{bs}_1) \,\textit{rs}_1) ::: \flatten(rs)$\\
- $\flatten \; r :: rs$ & $\dn$ & $r::\flatten(rs)$
- \end{tabular}
- \end{center}
- 
- \noindent
-
-If we call $\simp$ on $\rup\backslash a$,
-then we would go throught the third clause of 
-the sequence case:$\quad\textit{case} \;  (\ONE, a_2') \Rightarrow  \textit{fuse} \; bs \;  a_2'$.
-The $\ZERO$ of $(_0\ONE  + \ZERO)$ is thrown away 
-by $\flatten$ and 
-$_0\ONE$ merged into $(_0a  +  _1a^*)$ by simply
-putting its bits($_0$) to the front of the second component:
- ${\bf_0}(_0a  +  _1a^*)$. 
- After a second derivative operation,
- namely, $(_0(_0a  +  _1a^*))\backslash a$, we get 
- $
- _0(_0 \ONE  +  _1(_1\ONE \cdot a^*))
- $, and this simplifies to $_0(_0 \ONE  +  _{11} a^*)$
- by the third clause of the alternative case:
- $\quad\textit{case} \;  as' \Rightarrow  \textit{ALTS}\;bs\;as'$.
-The outmost bit $_0$ stays with 
-the outmost regular expression, rather than being fused to
-its child regular expressions, as what we will later see happens
-to $\simp(\rup\backslash \, s)$.
-If we choose to not simplify in the midst of derivative operations,
-but only do it at the end after the string has been exhausted, 
-namely, $\simp(\rup\backslash \, s)=\simp((\rup\backslash a)\backslash a)$,
-then at the {\bf second} derivative of 
-$(\rup\backslash a)\bf{\backslash a}$
-we will go throught the clause of $\backslash$:
+and this simplifies to
 \begin{center}
-\begin{tabular}{lcl}
-$(\textit{SEQ}\;bs\,a_1\,a_2)\,\backslash c$ & $\dn$ &
-     $(when \; \textit{bnullable}\,a_1)$\\
-					       & &$\textit{ALTS}\,bs\,List(\;\;(\textit{SEQ}\,[]\,(a_1\,\backslash c)\,a_2),$\\
-					       & &$(\textit{fuse}\,(\textit{bmkeps}\,a_1)\,(a_2\,\backslash c))\;\;)$\\
-\end{tabular}
+$ _1(_{011}{\bf a}^* +  _1\ONE)  $ 
 \end{center}
 
-because
-$\rup\backslash a = (_0\ONE  + \ZERO)(_0a  +  _1a^*)$  
-is a sequence
-with the first component being nullable
-(unsimplified, unlike the first round of running$\backslash_{simp}$).
-Therefore $((_0\ONE  + \ZERO)(_0a  +  _1a^*))\backslash a$ splits into
-$\rup\backslash a=(_0( [\ZERO\cdot {\bf b}] + 0) + _1( _0( [\ZERO\cdot a^*] + _1[ _1\ONE \cdot {\bf a}^*]) + _1( [\ZERO \cdot {\bf a}] + \ONE)  ))$ 
+If, after the first derivative we apply simplification we get
+$(_0{\bf b}  + _{101}{\bf  a}^* + _{11}{\bf a}  )$,
+and we do another derivative, getting
+$(\ZERO  + (_{101}(\ONE \cdot _1{\bf a}^*)+_{11}\ONE)$,
+which simplifies to 
+\begin{center}
+$ (_{1011}a^* +  _{11}\ONE)  $ 
+\end{center}
 
-$([(\ZERO + \ZERO)\cdot(_0a  +  _1a^*)] + _0( _0\ONE  + _1[_1\ONE \cdot a^*]))$.
-After these two successive derivatives without simplification,
-we apply $\simp$ to this regular expression, which goes through
-the alternative clause, and each component of 
-$([(\ZERO + \ZERO)\cdot(_0a  +  _1a^*)] + _0( _0\ONE  + _1[_1\ONE \cdot a^*]))$ 
-will be simplified, giving us the list:$[\ZERO, _0(_0\ONE  + _{11}a^*)]$
-This list is then "flattened"--$\ZERO$ will be
-thrown away by $\textit{flatten}$; $ _0(_0\ONE  + _{11}a^*)$
-is opened up to make the list consisting of two separate elements 
-$_{00}\ONE$ and $_{011}a^*$, note that $flatten$ 
-$\fuse$s the bit(s) $_0$ to the front of $_0\ONE $ and $_{11}a^*$.
-The order of simplification, which impacts the order that alternatives
-are  opened up, causes
-the bits to be moved differently.
+
 
 
 




