Consider two finite languages, $L_A$ and $L_B$, potentially over different alphabets. Now since these languages are finite, there exist minimal acyclic deterministic finite-state automata to decide them. We are given that $L_A$ is bigger than $L_B$: $|L_A| > |L_B|$.

We are given a function $f$ which maps elements of $L_A$ to $L_B$: $\ f{:}L_A \rightarrow L_B$. We also have the constraint that acceptance is preserved under $f$: $\ \ x \in L_A$ iff $f(x) \in L_B$, modeled below by formula $\eqref{eq1}$.

Here's the question: if two strings $u$ and $v$ map to the same Myhill-Nerode equivalence class in $B$ under $f$ [modeled by formula $\eqref{eq2}$], do they map to the same equivalence class in $A$ [modeled by formula $\eqref{eq3}$]? That is, if we are given formulas $\eqref{eq1}$ and $\eqref{eq2}$, can we conclude formula $\eqref{eq3}$?

$\forall x((x \in L_A) \leftrightarrow (f(x) \in L_B)) \tag{1} \label{eq1}$

$\forall z((f(uz) \in L_B) \leftrightarrow (f(vz) \in L_B)) \tag{2} \label{eq2}$

$\forall z((uz \in L_A) \leftrightarrow (vz \in L_A)) \tag{3} \label{eq3}$

Is there some existing result which subsumes what I'm trying to do here? It feels very similar to the closure property under homomorphisms. Or do we need to prove the result, for instance by two applications of formula $\eqref{eq1}$ to formula $\eqref{eq2}$ to yield theorem $\eqref{eq3}$?


1 Answer 1


Yes, (1) and (2) imply formula (3).

$\forall z$, $(uz \in L_A) \leftrightarrow (f(uz)\in L_B) \leftrightarrow (f(vz)\in L_B) \leftrightarrow (vz\in L_A)$.

  • 1
    $\begingroup$ It does not matter whether $L_A$ is finite and whether $L_B$ is finite. It does not matter whether $L_A$ is regular and whether $L_B$ is regular. $\endgroup$
    – John L.
    Jul 6, 2022 at 20:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.