\begin{frame}{(Strictly) Right Linear Grammars} \begin{block}{Theorem} Let $G$ be a right linear grammar $G$. There exists a \emph{strictly} right linear grammar $H$ such that $L(G) = L(H)$. \end{block} \pause \begin{construction} Let $G=(V,T,S,P)$ be a right linear grammar. \pause\smallskip Assume that we have a production rule $\gamma$ of the form \begin{talign} \alert{A \to u(B)} \end{talign} %with $A,B \in V$ and $u \in T^*$ with $\alert{|u| > 1}$. \pause Then \alert{$u = a w$} for some $a \in T$ and $w \in T^+$. \pause\smallskip Let \alert{$X$} be a \alert{fresh} variable ($X \not\in (V\cup T)$). \pause\smallskip We add $X$ to $V$ and split the rule $\gamma$ into: \begin{talign} \alert{A \to aX} && \alert{X \to w(B)} \end{talign} Then $A \to aX \to aw(B) = u(B)$. \pause It follows $L(G) = L(H)$. \pause\smallskip Repeat splitting until $|u| \le 1$ for all rules. \end{construction} \end{frame}