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.


1 Answer 1


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:
      • Accept.
  • $\begingroup$ Oh! I see. That was simple...thank you. $\endgroup$
    – Lightsong
    Jan 5, 2021 at 14:57

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