Let

$A/B=\left\{w\right|wx\in aAforsomex\in B\}$Show that if$A$is context free and$B$is regular, then$A/B$is context free

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 $G_2=(V_2,S,R_2,S_2).$

A language is regular if it can be expressed in terms of regular expression.A regular expression can also be described as a sequence of pattern that defines a string. Regular expressions are used to match character combinations in strings.

Yes, if $A$is context free and $B$is regular, then $A/B$is context freez.

Let $M=\left(Q,\Sigma ,\delta ,q_0,F\right)$be a deterministic finite automata for $A$.

And consider $M$be a deterministic finite automata for $B$and $M$ ^I be a pushdown deterministic automata for $A$. As given in the question $A$is context free and $B$is regular, then $A/B$is context free.

Here construct a pushdown deterministic automata for $A/B$by modifying $M$’ so that it can run simultaneously (using the Cartesian product) $M$’s finite state control in addition to its own finite state control. Initially, only the original logic of $M$’ is run.

The modified machine, using nondeterminism, will start up $M$after it will read. From this point onward, the machine guesses a string $x,$and processes $x,$by both $M$and $M$’.The machine will accept only if both machines simultaneously accept.

Yes. Let $M=\left(Q,\Sigma ,\delta ,q_0,F\right)$be a deterministic finite automata for $A$.

Here a state $q\in Q,$

And define , $Lq=\left\{x\right|\delta \left(q,x\right)\in F\}.$

Then $M_0=\left(Q,\Sigma ,\delta ,q_0,F_0\right)$is a deterministic finite automata for $A/B$where

$F_0=\left\{q\right|L_0\cup B\ne E\}.$

Hence, if$A$ is context free and $B$is regular, then $A/B$is context free is proved.