Does the following set of production rules produce a regular language or not?
$S \to AB \mid b $
$A \to SB$
$B \to AS \mid a$
I have generated following words with above grammar
But I am not able to create a language using it. Is this grammar just context free or is this regular also ? If yes, how to prove that ?
The grammar can be transformed to
$\quad$ $S \to SBB \mid b $
$\quad$ $B \to SBS \mid a$
Let $L_S$ be the language generated by the rules above (with $S$ as the start symbol). $L_S$ is context-free by definition.
We will prove it is not regular. The idea comes from the rule, $B\to SBS$.
Let $L_B$ be the language generated by the two rules above with $B$ as the start symbol.
Claim: We have the following inclusions of languages.
$\quad\{b\}\cup\{b^{i+1}ab^ia\mid i\ge0\}\subseteq L_S$
$\quad\{b^iab^i\mid i\ge0\}\subseteq L_B$
Proof: Use mathematical induction on $n$, the upper bound for $i$.
Claim: We have following inclusions of languages. $\quad L_S\subseteq E_S :=\{ba^{2i}\mid i\ge0\}\cup\{b^{i+1}ab^ia\mid i>0\}\cup\{\text{words that have at least three } a\text{'s}\}$ $\quad L_B\subseteq E_B:=\{a\}\cup\{\text{words that have at least two } a\text{'s}\}$
Proof: It is easy to check that $E_SE_BE_B\subseteq E_S$, $a\in E_S$, $E_BE_SE_B\subseteq E_B$ and $b\in E_B$.
The above two claims implies that $L_S\cap L(b^*ab^*a) = \{b^{i+1}ab^ia\mid i\ge0\}$, which is not a regular language. Since $L(b^*ab^*a)$ is regular, $L_S$ cannot be regular.
Exercise 1. Show that the following grammar generates a regular language.
$\quad$ $S \to SAS \mid b $
$\quad$ $A \to ASA \mid a$
Exercise 2. Show that the following grammar generates a non-regular language.
$\quad$ $S \to AB \mid b $
$\quad$ $B \to SA $
$\quad$ $A \to BS \mid a$
