By induction to n one shows that if there is a string of length n that
is accepted by q but not by r, then q and r can be shown
non-equivalent using the rules above.
<P>
Induction base (n=0): if the empty string is accepted by q but not by
r, then q is an accept state and r isn't, and the first rule settles that.
<P>
Induction step: Suppose the statement is established for n.
Let aw be a word of length n+1 that is accepted by q but not by r.
Then w must be a word of length n that is accepted by delta(q,a) but
not by delta(r,a). By induction, the rules above show that delta(q,a)
and delta(r,a) are nonequivalent. Now the second rule shows that q and
r are non-equivalent.