Answer
contradiction of the fact that $R_n$ = $R_{n+1}$.
Work Step by Step
We know from Exercise 58d that the equivalence classes of Rk are a refinement of the equivalence classes of $R_{k−1}$ for each positive integer k.
The equivalence classes are finite sets, and finite sets cannot be refined indefinitely (the most refined they can be is for each equivalence
class to contain just one state).
Therefore this sequence of refinements must remain unchanged from some point onward.
It remains to show that as soon as we have Rn = Rn+1, then
$R_n$ = $R_m$ for all m > n, from which it follows that Rn = R∗,
and so the equivalence classes for these two relations will be
the same.
By induction, it suffices to show that if Rn = Rn+1, then Rn+1 = Rn+2. Suppose that $R_{n+1}$ = $R_{n+2}$.
This means that there are states s and t that are (n+1)-equivalent but not
(n + 2)-equivalent.
Thus there is a string x of length n + 2 such that, say, f (s, x) is final but f (t, x) is nonfinal.
Write x = aw, where a ∈ I . Then f (s, a) and f (t, a) are not (n + 1)-equivalent, because w drives the first to a final state
and the second to a nonfinal state.
But f (s, a) and f (t, a) are n-equivalent, because s and t are (n+1)-equivalent.
This contradicts the fact that $R_n$ = $R_{n+1}$.
