# Some questions regarding decidability and semi-decidability of $A/B = \{ w \text{ | }\exists z \in B, wz \in A\}$

I have found two interesting questions regarding the quotient of languages, described as:

$$A/B = \{ w \text{ | }\exists z \in B, wz \in A\}$$

The first one is:

• Let $$A$$ and $$B$$ be regular languages, prove that $$A/B$$ is decidable

By using the proof from this other question, it can be proved that $$A/B$$ is regular if $$A$$ is regular too.

Then, since any regular language is decidable, $$A/B$$ will be regular (and decidable) too.

The second one is:

• Let $$A$$ and $$B$$ be decidable languages, prove that $$A/B$$ is semi-decidable

I have no idea on how to prove this one, I assume that it's somehow related to $$A$$ and $$B$$ being able to stop for any string, whether it's accepted or rejected, but since I don't know whether $$A$$ is regular or not, I don't know what to do.

Let $$\Sigma$$ be your alphabet. You want to design a Turing machine $$T$$ such that, given any word $$w \in \Sigma^*$$, $$T(w)$$ accepts if and only if $$w \in A / B$$. Notice that since you only want to prove that $$A/B$$ is semi-decidable, $$T(w)$$ is not required to reject when $$w \not\in A / B$$.
Since $$A$$ and $$B$$ are decidable, you know that there are two Turing machines $$T_A$$ and $$T_B$$ such that that $$T_A(x)$$ (resp. $$T_B(x)$$) accepts if $$x \in A$$ (resp. $$x \in B$$) and rejects otherwise.
You can then design $$T$$ as follows:
• For each word $$z \in \Sigma^*$$:
• If $$T_B(z)$$ accepts and $$T_A(wz)$$ accepts: