Given two regular expressions $R$ and $S$ on an alphabet $\Sigma$ it is possible to decide their equivalence as follows:

- build two finite automata $M_R$ and $M_S$ such that $L(R) = L(M_R)$ and $L(S) = L(M_S)$
- build an automaton $M$ such that $L(M) = (L(M_R) - L(M_S)) \cup (L(M_S) - L(M_R))$
- test emptyness of $L(M)$ using a reachability algorithm on $M$

I was wondering if there is another way to decide equivalence. Suppose $M_R$ and $M_S$ are the minimal DFA (without epsilon-moves) such that $L(R) = L(M_R)$ and $L(S) = L(M_S)$. If they have a different number of states, then $R$ and $S$ are not equivalent. Otherwise let $m$ be the number of states of the two automata. Is it true that $L(M_R) = L(M_S)$ iff ${x \in L(M_R) : |x| \leq m +1 } = {x \in L(M_S) : |x| \leq m +1 }$? How to prove that with the Myhill-Nerode theorem?

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