\documentclass[11pt]{article}

\begin{document}

\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}

\noindent 

This means we can also assume in this case

\[

(e)\quad len\,|w_1| \geq len\,|w_2|

\] 

\noindent 

which is the premise of the rule above.

Instantiating $v_1$ and $v_2$ in the assumptions (b) and (c)

gives us

\begin{center}

\begin{tabular}{ll}

(b*) & $\vdash Left(w_1) : (der\;c\;r_1) \cdot r_2 + der\;c\;r_2$\\

(c*) & $\vdash Right(w_2) : (der\;c\;r_1) \cdot r_2 + der\;c\;r_2$\\

\end{tabular}

\end{center}

\noindent Since these are assumptions, we can further analyse

how they could have arisen according to the rules of $\vdash

\_ : \_\,$. This gives us two new assumptions

\begin{center}

\begin{tabular}{ll}

(b**) & $\vdash w_1 : (der\;c\;r_1) \cdot r_2$\\

(c**) & $\vdash w_2 : der\;c\;r_2$\\

\end{tabular}

\end{center}

\noindent 

Looking at (b**) we can further analyse how this

judgement could have arisen. This tells us that $w_1$

must have been a sequence, say $u_1\cdot u_2$, with

\begin{center}

\begin{tabular}{ll}

(b***) & $\vdash u_1 : der\;c\;r_1$\\

       & $\vdash u_2 : r_2$\\

\end{tabular}

\end{center}

\noindent 

Instantiating the goal means we need to prove

\[

inj\; (r_1 \cdot r_2)\;c\;(Left(u_1\cdot u_2)) \succ_{r_1 \cdot r_2} 

inj\; (r_1 \cdot r_2)\;c\;(Right(w_2))

\]

\noindent 

We can simplify this according to the rules of $inj$:

\[

(inj\; r_1\;c\;u_1)\cdot u_2 \succ_{r_1 \cdot r_2} 

(mkeps\;r_1) \cdot (inj\; r_2\;c\;w_2)

\]

\noindent

This is what we need to prove. There are only two rules that

can be used to prove this judgement:

\begin{center}

\begin{tabular}{cc}

\begin{tabular}{c}

$v_1 = v_1'$\qquad $v_2 \succ_{r_2} v_2'$\\

\hline

$v_1\cdot v_2 \succ_{r_1\cdot r_2} v_1'\cdot v_2'$

\end{tabular} &

\begin{tabular}{c}

$v_1 \succ_{r_1} v_1'$\\

\hline

$v_1\cdot v_2 \succ_{r_1\cdot r_2} v_1'\cdot v_2'$

\end{tabular}

\end{tabular}

\end{center}

\noindent 

Using the left rule would mean we need to show that

\[

inj\; r_1\;c\;u_1 = mkeps\;r_1

\]

\noindent 

but this can never be the case.\footnote{Actually Isabelle

found this out after analysing its argument. ;o)} Lets assume

it would be true, then also if we flat each side, it must hold

that

\[

|inj\; r_1\;c\;u_1| = |mkeps\;r_1|

\]

\noindent 

But this leads to a contradiction, because the right-hand side

will be equal to the empty list, or empty string. This is 

because we assumed $nullable(r_1)$ and there is a lemma

called \texttt{mkeps\_flat} which shows this. On the other

side we know by assumption (b***) and lemma \texttt{v4} that 

the other side needs to be a string starting with $c$ (since

we inject $c$ into $u_1$). The empty string can never be equal 

to something starting with $c$\ldots therefore there is a 

contradiction.

That means we can only use the rule on the right-hand side to 

prove our goal. This implies we need to prove

\[

inj\; r_1\;c\;u_1 \succ_{r_1} mkeps\;r_1

\]

\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}