--- a/proof.tex Fri Oct 05 00:40:52 2012 +0100
+++ b/proof.tex Fri Oct 05 16:26:04 2012 +0100
@@ -127,7 +127,47 @@
holds for every set of strings $A$ and $B$. That means the right-hand side of (d) is also $Der\,c\,(L(r_1)) \cup Der\,c\,(L(r_2))$,
because $L(r_1 + r_2) = L(r_1) \cup L(r_2)$. And we are done with the fourth case.
-\item Fifth Case:
+\item Fifth Case: $P(r_1 \cdot r_2)$ is $L(der\,c\,(r_1 \cdot r_2)) = Der\,c\,(L(r_1 \cdot r_2))$ (e). We can assume already:
+
+\begin{center}
+\begin{tabular}{ll}
+$P(r_1)$: & $L(der\,c\,r_1) = Der\,c\,(L(r_1))$ (I)\\
+$P(r_2)$: & $L(der\,c\,r_2) = Der\,c\,(L(r_2))$ (II)
+\end{tabular}
+\end{center}
+
+Let us first consider the case where $nullable(r_1)$ holds. Then
+
+\[
+der\,c\,(r_1 \cdot r_2) = ((der\,c\,r_1) \cdot r_2) + (der\,c\,r_2).
+\]
+
+The corresponding language of the right-hand side is
+
+\[
+(L(der\,c\,r_1) \,@\, L(r_2)) \cup L(der\,c\,r_2).
+\]
+
+By the induction hypotheses (I) and (II), this is equal to
+
+\[
+(Der\,c\,(L(r_1)) \,@\, L(r_2)) \cup (Der\,c\,(L(r_2)).\;\;(**)
+\]
+
+We also know that $L(r_1 \cdot r_2) = L(r_1) \,@\,L(r_2)$. We have to know what
+$Der\,c\,(L(r_1) \,@\,L(r_2))$ is.
+
+Let us analyse what
+$Der\,c\,(A \,@\, B)$ is for arbitrary sets of strings $A$ and $B$. If $A$ does \emph{not}
+contain the empty string, then every string in $A\,@\,B$ is of the form $s_1 \,@\, s_2$ where
+$s_1 \in A$ and $s_2 \in B$. So if $s_1$ starts with $c$ then we just have to remove it. Consequently,
+$Der\,c\,(A \,@\, B) = (Der\,c\,(A)) \,@\, B$. This case does not apply here though, because we already
+proved that if $r_1$ is nullable, then $L(r_1)$ contains the empty string. In this case, every string
+in $A\,@\,B$ is either of the form $s_1 \,@\, s_2$, with $s_1 \in A$ and $s_2 \in B$, or
+$s_3$ with $s_3 \in B$. This means $Der\,c\,(A \,@\, B) = ((Der\,c\,(A)) \,@\, B) \cup Der\,c\,B$.
+But this proves that (**) is $Der\,c\,(L(r_1) \,@\, L(r_2))$.
+
+Similarly in the case where $r_1$ is \emph{not} nullable.
\item Sixth Case:
\end{itemize}