Right quotient of $L_1=L(a^*baa^*)$, and $L_2=L(ab^*)$.

Question

Given $L_1=L(a^*baa^*)$, $L_2=L(ab^*)$. The regular expression corresponding to language $L_3=L_1/L_2$ (right quotient) is given by

1. $a^*b$
2. $a^*baa^*$
3. $a^*ba^*$
4. None of the above

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My attempt:

The right quotient (or simply quotient) of a formal language ${\displaystyle L_{1}} $ with a formal language ${\displaystyle L_{2}}$ is the language consisting of strings $w$ such that $wx$ is in ${\displaystyle L_{1}}$ for some string $x$ in ${\displaystyle L_{2}} $.

Given,

$L_1=(a^*baa^*)=\{ba, baa, aba, abaa,\dots\}$

$L_2=(ab^*)=\{a, ab, abb, abbb,\dots\}$

Therefore,

$L_3=L_1/L_2=\{b, ba, ab, aba,\dots\}=(a^*ba^*)$

I have used only string $'a'$ of language $L_2$

Option $(3)$ is true.

But, some where it explained as

$L_1 =\{ba,aba,abaa,baa,\dots\}$

$L_2=\{ab^*\}$

$L_1 / L_2$ operation not successful here

So, Ans $(4)$.

Can you explain it, please?

Answer 1

Suppose that $w\in\{a,b\}^*$, and there is some $x\in L_2$ such that $wx\in L_1$. This $x$ must be $ab^n$ for some $n\ge 0$, so $wab^n\in L_1$. No word in $L_1$ ends in $b$, so $n=0$, and $wa\in L_1$. This is the case if and only if $w$ has the form $a^*ba^*$, so you are correct.

It makes no sense to say that the operation of taking the right quotient is not successful: the right quotient always exists, though it may be empty. Perhaps the person who wrote that answer was under the mistaken impression that $w\in L_1/L_2$ if and only if $wx\in L_1$ for every $x\in L_2$, instead of for at least one $x\in L_2$.