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?