First of all this grammar is right linear. And we know:

Two special types of linear grammars are the following:

- the
**left-linear** or left regular grammars, in which **all nonterminals** in right hand sides are **at the left ends**;
- the
**right-linear** or right regular grammars, in which **all nonterminals** in right hand sides are **at the right ends**.

**Hence the given grammar is regular **and hence the language generated by regular grammar will also be regular. Alternatively we can also write a regular expression for it. Let us see how to do it:

Given grammar:

$G:$ $S\rightarrow aB$

$B \rightarrow bC$

$C\rightarrow xB \mid c$

So, we can reverse substitution from $C$ onwards to $S$ to see what $S$ generates..

Substituting the yield of $C$ in $B$, we have:

$B \rightarrow bxB \mid bc$

which gives $B = (bx)^* bc$

Now, substituting $B$ in $S$ we have:

** ** $\mathbf{S \rightarrow aB}$

** ** $\mathbf{S = a(bx)^*bc}$

**Hence, the corresponding regular expression is : ** $\mathbf{a(bx)^*bc}$