Question

Use the results of Exercise $2.16$to give another proof that every regular language is context- free, by showing how to convert a regular expression directly to an equivalent context-free grammar.

Short answer

The results of exercise $2.16$ show that context-free languages are closed under union, concatenation and star operation, thus every regular language is context-free.

Step-by-step solution

Define Regular languages

The context free language is generated by context free grammar. These languages are accepted by Pushdown Automata. These are the superset of regular languages.

Consider context-free languages$L_1$ described as $G_1=(V_1,S,R_1,S_1).$

Consider context-free language $L_2$described as $G_2=(V_2,S,R_2,S_2).$

Explanation

Given a regular language $L$. It has been proven that every regular language has a regular expression that describes it. Let r be a regular expression that describes $L$. That is, let $\left(r\right)$be such that $L$$\left(r\right)=$$L$. Since r is a regular expression it must be of the form:

• a where a$\in \sum ,$

• localid="1662192279709" $\epsilon ,$

• $\Phi$, or be formed recursively from two regular expressions r₁ and r₂ are regular expressions

• $r_{1}U{r}_2,$

• $r_1°r_2,$

$r_1*$

For $r=awherea\in \sum ,$a grammar for,$L\left(a\right)$is $G=(\left\{S\right\},S,\left\{\left(S,a\right)\right\},S)$ For r = ε, a grammar for For a grammar for$L\left(\epsilon \right)isG=(\left\{S\right\},S,\left\{\left(S\to \epsilon \right)\right\},S)$For $r=\Phi ,$a grammar for

$L\left(\epsilon \right)isG=(\left\{S\right\},S,\left\{\right\},S)$

The results of exercise that context-free languages are closed under union, concatenation and star-closure, thus every regular language is context-free.