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 $|L(B)|&lt;\infty$, 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?
This answers a different version of the question, in which the language in question is $L(A) \setminus L(B)$.
Here is an algorithm for deciding $L$:

Use the product construction to construct a DFA $C$ whose language is $L(A) \setminus L(B)$.
Let $D$ be a DFA whose language is $0^*1^*$.
Use the product construction again to construct a DFA $E$ whose language is $L(C) \Delta L(D)$.
Check (using BFS/DFS) whether some final state of $E$ is reachable from its initial state.
Output &quot;yes&quot; if no final state is reachable from the initial state. Otherwise output &quot;no&quot;.

As you can see, whether the languages encountered along the way are finite or infinite makes absolutely no difference for this algorithm.