# Computing automaton for $L(A) / L(B)$ gives ones for $A,B$

I’m trying to figure out whether infinite language change the answer.

Show that the following language is decidable: $$L=\{\langle A,B \rangle : \text{ A,B are DFAs, L(B) is finite, and L(A)/ L(B)=L(0^*1^*) }\}.$$

(I am talking about right division.)

We know how to check whether the language of a DFA is finite, and given two DFAs, we know how to check whether their languages are equal. The algorithms I know to the above problems uses the DFA’s, so it is necessary having the DFA’s in order to decide those problems.

I’m trying to figure out whether $$|L(B)|=\infty$$ changes the answer. To the best of my understanding, because $$|L(B)|<\infty$$, we can explicitly construct a DFA that accepts $$L(A)/ L(B)$$, whereas if $$L (B)=\infty$$ all we know is about the existence of $$DFA$$ that accepts $$L(A)/ L(B)$$.

However, even if $$L(B)$$ is an infinite language, since there is a finite number of DFAs, one of which accepts $$L(A) / L(B)$$, I can certainly know that there is a Turing machine that decides the language $$L$$. Right?