(2) M↑A↓ (cid:3) P1↓

Theorem 50.  (1) M↑PS↓ (cid:3) AC↓
Proof. (1) Use the same counterexample as in the proof of Theorem 48(2) and note that τ is also PS↓. However, c1(1, 3) = 1
but c2( f (1), f (3)) = 2, so τ is not AC↓.
(2) Let  S1 = S2 = {1, 2},  E1 = ∅ and  E2 = {(cid:3)1, 2, a(cid:4)}. Also let  f be the identity function,  R = ∅ and  τ = (cid:3) f , R, d1, d2(cid:4). 
Then τ is A↓ but not P1↓ since E1 = ∅. (cid:2)

We can also prove similar results for the hierarchy of reﬁnement properties.

Theorem 51. 
(1) M↑P1↓A↓ (cid:3) PL↓
(3) M↑PW↓A↓ (cid:3) P↓
Proof. (1-4) Follows from Theorem 18 since Examples 17(1, 3, 4) are A↓ for τ = (cid:3) f , R, d1, d2(cid:4) and Example 17(2) is A↓ for 
τ = (cid:3) f , R, w1, d2(cid:4), where w1 = d1 except that w1(1, 4) = 2. (cid:2)

(2) M↑PL↓A↓ (cid:3) PW↓
(4) M↑P↓A↓ (cid:3) PS↓

The major reason for these results is that reﬁnement is deﬁned without any metrics; even if we have a guarantee that
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