〉) = 〈R(x),R(y)〉. Rigid Pairs entails that when R is a counterpairing and S is any plurality of ordered pairs, an ordered pair is one of R(S) iff it is 〈R(x),R(y)〉 for some 〈x, y〉 that is one of S. That is, R(S)(z1z2) ↔ ∃x∃y(Rxz2 ∧ Sxy ∧ Ryz2): in other words, R(S) = R ◦ S ◦ R−1. Thus if we were to require not only counterpairings but the converses of counterpairings to be universally defined functions, we would be guaranteed that R(S) is a universally defined function whenever R and S are counterpairings. And it is hard to think of any other plausible way to guarantee this. R ◦ S ◦ R−1 will certainly not have a universal domain unless R−1 does; and unless S is very special, it will not be functional unless R−1 is. So if ordered pairs are rigid, the desire to vindicate (17a) and (17b) pushes us towards the view that every counterpairing is a global permutation-some ordered pairs such that everything is the first element of exactly one of them, and everything is the second element of exactly one of them. The challenge posed by (17c) is to find a meaning for 'S is accessible from R' that makes it true. We can find one by substituting Analysis F* into our analysis 'at R '. In the simple case where the counterpart relation is qualitative, this is easily done: at R^φ(t1, . . . , tn) ↔ at R∃Q(CQ ∧ φ(Q(t1), . . . ,Q(tn))) ↔ ∃Q(CQ ∧ φ(Q(R(t1)), . . . ,Q(R(tn)))) ↔ ∃Q(CQ ∧ at(Q ◦ R)φ(t1, . . . , tn)) Thus we can accept (17c) provided we understand 'S is accessible from R' to mean 'there is a counterpairing Q such that S = Q ◦ R'. If counterpairings are one-to-one, this is equivalent to 'S ◦ R−1 is a counterpairing'.40 40What about non-qualitative counterpart relations? Suppose C(S) is analysed as ψ(a1, . . . , a j,S), where ψ is qualitative and a1 . . . a j are some singular or plural terms. Then, in the third and fourth lines of the above derivation, instead of 'C(Q)' we should have 'ψ(R(a1), . . . ,R(a j),Q)'; so our analysis of 'S is accessible fr