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\noindent
A lemma which might be true, but can also be false, is as follows:

\begin{center}
\begin{tabular}{lll}
If   & (1) & $v_1 \succ_{der\;c\;r} v_2$,\\
     & (2) & $\vdash v_1 : der\;c\;r$, and\\ 
     & (3) & $\vdash v_2 : der\;c\;r$ holds,\\
then &     & $inj\;r\;c\;v_1 \succ_r inj\;r\;c\;v_2$ also holds.  
\end{tabular}
\end{center}

\noindent It essentially states that if one value $v_1$ is 
bigger than $v_2$ then this ordering is preserved under 
injections. This is proved by induction (on the definition of 
$der$\ldots this is very similar to an induction on $r$).
\bigskip

\noindent
The case that is still unproved is the sequence case where we 
assume $r = r_1\cdot r_2$ and also $r_1$ being nullable.
The derivative $der\;c\;r$ is then

\begin{center}
$der\;c\;r = ((der\;c\;r_1) \cdot r_2) + (der\;c\;r_2)$
\end{center}

\noindent 
or without the parentheses

\begin{center}
$der\;c\;r = (der\;c\;r_1) \cdot r_2 + der\;c\;r_2$
\end{center}

\noindent 
In this case the assumptions are

\begin{center}
\begin{tabular}{ll}
(a) & $v_1 \succ_{(der\;c\;r_1) \cdot r_2 + der\;c\;r_2} v_2$\\
(b) & $\vdash v_1 : (der\;c\;r_1) \cdot r_2 + der\;c\;r_2$\\
(c) & $\vdash v_2 : (der\;c\;r_1) \cdot r_2 + der\;c\;r_2$\\
(d) & $nullable(r_1)$
\end{tabular}
\end{center}

\noindent 
The induction hypotheses are

\begin{center}
\begin{tabular}{ll}
(IH1) & $\forall v_1 v_2.\;v_1 \succ_{der\;c\;r_1} v_2
\;\wedge\; \vdash v_1 : der\;c\;r_1 \;\wedge\; 
\vdash v_2 : der\;c\;r_1\qquad$\\
      & $\hfill\longrightarrow
         inj\;r_1\;c\;v_1 \succ{r_1} \;inj\;r_1\;c\;v_2$\smallskip\\
(IH2) & $\forall v_1 v_2.\;v_1 \succ_{der\;c\;r_2} v_2
\;\wedge\; \vdash v_2 : der\;c\;r_2 \;\wedge\; 
\vdash v_2 : der\;c\;r_2\qquad$\\
      & $\hfill\longrightarrow
         inj\;r_2\;c\;v_1 \succ{r_2} \;inj\;r_2\;c\;v_2$\\
\end{tabular}
\end{center}

\noindent 
The goal is

\[
(goal)\qquad
inj\; (r_1 \cdot r_2)\;c\;v_1 \succ_{r_1 \cdot r_2} 
inj\; (r_1 \cdot r_2)\;c\;v_2
\]

\noindent 
If we analyse how (a) could have arisen (that is make a case
distinction), then we will find four cases:

\begin{center}
\begin{tabular}{ll}
LL & $v_1 = Left(w_1)$, $v_2 = Left(w_2)$\\
LR & $v_1 = Left(w_1)$, $v_2 = Right(w_2)$\\
RL & $v_1 = Right(w_1)$, $v_2 = Left(w_2)$\\
RR & $v_1 = Right(w_1)$, $v_2 = Right(w_2)$\\
\end{tabular}
\end{center}


\noindent 
We have to establish our goal in all four cases. 


\subsubsection*{Case LR}

The corresponding rule (instantiated) is:

\begin{center}
\begin{tabular}{c}
$len\,|w_1| \geq len\,|w_2|$\\
\hline
$Left(w_1) \succ_{(der\;c\;r_1) \cdot r_2 + der\;c\;r_2} Right(w_2)$
\end{tabular}
\end{center}

\subsubsection*{Case RL}

The corresponding rule (instantiated) is:

\begin{center}
\begin{tabular}{c}
$len\,|w_1| > len\,|w_2|$\\
\hline
$Right(w_1) \succ_{(der\;c\;r_1) \cdot r_2 + der\;c\;r_2} Left(w_2)$
\end{tabular}
\end{center}


\end{document}  
