130 \item Fifth Case:

   130 \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:

   131 

   132 \begin{center}

   133 \begin{tabular}{ll}

   134 $P(r_1)$: & $L(der\,c\,r_1) = Der\,c\,(L(r_1))$ (I)\\

   135 $P(r_2)$: & $L(der\,c\,r_2) = Der\,c\,(L(r_2))$ (II)

   136 \end{tabular}

   137 \end{center}

   138 

   139 Let us first consider the case where $nullable(r_1)$ holds. Then 

   140 

   141 \[

   142 der\,c\,(r_1 \cdot r_2) = ((der\,c\,r_1) \cdot r_2) + (der\,c\,r_2).

   143 \]

   144 

   145 The corresponding language of the right-hand side is 

   146 

   147 \[

   148 (L(der\,c\,r_1) \,@\, L(r_2)) \cup L(der\,c\,r_2).

   149 \]

   150 

   151 By the induction hypotheses (I) and (II), this is equal to

   152 

   153 \[

   154 (Der\,c\,(L(r_1)) \,@\, L(r_2)) \cup (Der\,c\,(L(r_2)).\;\;(**)

   155 \]

   156 

   157 We also know that $L(r_1 \cdot r_2) = L(r_1) \,@\,L(r_2)$.  We have to know what

   158 $Der\,c\,(L(r_1) \,@\,L(r_2))$ is.

   159 

   160 Let us analyse what

   161 $Der\,c\,(A \,@\, B)$ is for arbitrary sets of strings $A$ and $B$. If $A$ does \emph{not}

   162 contain the empty string, then every string in $A\,@\,B$ is of the form $s_1 \,@\, s_2$ where

   163 $s_1 \in A$ and $s_2 \in B$. So if $s_1$ starts with $c$ then we just have to remove it. Consequently,

   164 $Der\,c\,(A \,@\, B) = (Der\,c\,(A)) \,@\, B$. This case does not apply here though, because we already 

   165 proved that if $r_1$ is nullable, then $L(r_1)$ contains the empty string. In this case, every string

   166 in  $A\,@\,B$ is either of the form $s_1 \,@\, s_2$, with $s_1 \in A$ and $s_2 \in B$, or

   167 $s_3$ with $s_3 \in B$. This means $Der\,c\,(A \,@\, B) = ((Der\,c\,(A)) \,@\, B) \cup Der\,c\,B$.

   168 But this proves that (**) is $Der\,c\,(L(r_1) \,@\, L(r_2))$.

   169 

   170 Similarly in the case where $r_1$ is \emph{not} nullable.