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Proof.
We have shown that K*N(E) is a fixed point of fE, so we only need to show that K*N(E) is the greatest fixed point. Let B be a fixed point of fB. We want to show that B ⊆ KkN(E) for each value k≥1. We will proceed by induction on k. By hypothesis,
B = fE(B) = K1N(E∩B) ⊆ K1N(E)by monotonicity, so we have the k=1 case. Now suppose that for k=m, B ⊆ KmN(E). Then by monotonicity,
We also have:
(i) K1N(B) ⊆ K1N KmN(E) = Km+1N(E)
by monotonicity, so combining (i) and (ii) we have:
(ii) B = K1N(E∩B) ⊆ K1N(B)
B ⊆ K1N(B) ⊆ Km+1N(E)completing the induction.
Peter Vanderschraaf peterv@cyrus.andrew.cmu.edu | Giacomo Sillari Carnegie Mellon University gsillari@andrew.cmu.edu |