The internal structure of an equivalence class of sets of rankings

Up to this point, we still have only a trivial characterization of what it means for two sets of rankings to be equivalent: they are when they are true in exactly the same tableaux.

Despite the fact that we have already learned a great deal about equivalence classes of sets of rankings — for instance, we now know they are in one-one correspondence with equivalence classes of tableaux — we still cannot tell, given an arbitrary pair of sets of

rankings, whether they are in the same equivalence class or not without comparing directly the sets of tableaux they are true in. And given that such sets are infinite, this brute force method of checking does not look too attractive. But the following fact will help us find the means to test for equivalency effectively.

(86) For each equivalence classCof sets of rankings, there exists a unique minimal proper extension S^P+ set of rankings within that class.

Proof of 86. Each equivalence class of sets of rankings has its minimal proper representative S^{M in}. AsS^{M in}is proper by 84, it is equal to its own minimal proper extension by 76. Hence there always exists oneS^P+ per class.

For uniqueness, suppose there are two distinct minimal proper extensions in some C. By 76, both must be proper. Then by 83, both are equal to theS^{M in} representative of class C. But then they are equal, contrary to assumption. ⊣

The last fact we have to prove before we can test for equivalency easily is the following one:

(87) For any set of rankings S and its minimal proper extension S^P+, S and S^P+ are equivalent sets of rankings.

Proof of 87. Suppose towards contradiction that were not so. Fix someS andS^P+ which are not equivalent.

By construction ofS^P+, any φ∈S must be a refinement of a ranking fromS^P+. Therefore the set of total refinements ofSis a subset of the set of total refinements ofS^P+. From that, it follows thatS^P+⊧S.

Thus ifS /⊧S^P+, it must be witnessed by some tableauM whereS is true, butS^P+ is not. We fix a particular rowr ofM whereS, but not S^P+, is true. W.l.o.g., we assumerhas a single L in Ci.

rcannot be true in S by virtue of being accounted for byS^Co, forS^P+ has the same core.

Supposer is accounted for by the non-core part ofS. Then in each φ∈S, there is an atomic ranking accounting for the L in Ciin r. Each rankingψ∈ S^P+ contains the part φ^Ci_j from some φj∈S. Therefore eachψhas to account forrjust as well as anyφdoes.

Thus it is impossible to find any row at which S is true, but S^P+ is not. Hence the two are compatible with the same set of tableaux, and thusS andS^P+ are equivalent. ⊣

These results allow us to derive the following simple corollary:

(88) Sets of rankings S₁ and S₂ are equivalent iffS₁^P+=S₂^P+.

As we can (for finiteCon) effectively compute the minimal proper extension of any set of rankings, and as those are also finite and thus easy to compare, we have an effective test for equivalency of two arbitrary sets of rankings.

Moreover, from 87 we learn exactly which sets are in a given equivalence class: those whose proper minimal extension is its S^{M in} representative. This, in turn, gives us a way

to generate the members of the equivalence class smaller than S^{M in}: we can subtract rankings from S^{M in} without offending OT-compatibility as long as the result still has the same minimal proper extension.

In order to preserve what the minimal proper extension is, we need to make sure that the full range of Ci-alternatives present is still in the set. It is not necessary to have the same alternative present in S^{M in} more than once, so as long as each has at least one instance, the set remains equivalent to S^{M in}. This can be illustrated with the following example:

(89) ₍C1≫C3) ∧ (C1≫C4)

iiiiiiiiiiiiiiii

UU UU UU UU UU UU UU UU

(C1≫C3) ∧ (C2≫C4) (C2≫C3) ∧ (C1≫C4) (C2≫C3) ∧ (C2≫C4)

UUUUUUUUUUUUUUUU iiiiiiiiiiiiiiii

(90)

C1 C2 C3 C4

W W L e

W W e L

The set of rankings consisting of the four partial rankings in 89 is proper. It corresponds to the tableau in 90, and can be decomposed into S_C3 and S_C4, both of which contain 2 alternatives:

(91) SC3= {(C1≫C3), (C2≫C3)}

SC4= {(C1≫C4), (C2≫C4)}

In the picture 89, each pair of rankings connected by a line can be transformed into each other by exchanging a particularCi-alternative in one of them for the other possible Ci-alternative.

It is easy to see that the pairs of rankings opposite from each other — those pairs not connected by a line, and maximally different from each other — are each sufficient to recover the sets in 91. All four rankings are included in the S^{M in} representative of the equivalence class in question, but as long as the opposite rankings in 89 are kept, we can subtract one or two of the rankings maximal in the corresponding tableau 90 preserving OT-equivalency of the set.

Thus the set S^{M in}, which records all possible combinations of Ci-alternatives, and not just a subset of those sufficient for the recovery of S_Ci sets, is not truly a minimal representative. Yet the same example of the set in 89 shows that though there may be smaller sets T ⊂S^{M in} still in the same equivalence class, those T will not in the general case be unique. Therefore even though S^{M in} is bulkier than needed, it can serve as the unique representative of the class, but smaller equivalent T may not. Still, in practice

it is of course better to have smaller representations. But even more efficient than any equivalent T ⊂S^{M in} for recording the equivalence class would be direct writing down of the core rankingS^Co and non-empty setsS_Ciinto whichS^{M in}of the equivalence class may be decomposed.