Theory of sets of OT rankings

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Theory of sets of OT rankings

Igor Yanovich MIT

yanovich@mit.edu August 31, 2012

Abstract

Many OT tableaux are compatible with more than one total ranking. This paper develops the formal theory of sets of OT rankings compatible with a given tableau, including methods for efficiently working with them employing equivalent sets of partial rankings. The resulting theory allows for lossless and monotonic OT learning.

1 Introduction

Consider an Optimality-Theoretic language learner which just received input equivalent to the following comparative tableau:¹

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C1 C2 C3 C4 C5

W W e e L

W e W e L

e e W W L

There are 60 total rankings of the 5-constraint set {C1, C2, C3, C4, C5} that are compatible with 1. They are given in 2. Since classical OT only views total constraint rankings as possible grammars, the learner has to choose one of those 60, even though the information it actually has does not favor one of them over the others. The learner thus have to add further assumptions regarding how to make that choice: e.g., just pick a ranking at random, or perhaps try placing markedness over faithfulness first ([Smolensky, 1996], a.m.o.), or use the size of different sets like 2 as a clue ([Riggle, 2008]), etc.

1All tableaux in the paper are given in the comparative format introduced by [Prince, 2000]. The ordering of the constraints in the tableau doesnot represent the ranking.

I use namesC1,C2, ... for constraints instead of meaningful names likeDeporAlignin order to stress that the results in the paper are general for any sets of OT constraints.

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⎪⎪⎪⎨⎪⎪⎪

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C1≫C3≫C2≫C4≫C5 C3≫C1≫C2≫C4≫C5 C1≫C2≫C3≫C4≫C5 C3≫C2≫C1≫C4≫C5 C1≫C2≫C4≫C3≫C5 C3≫C2≫C4≫C1≫C5 C2≫C1≫C4≫C3≫C5 C2≫C3≫C4≫C1≫C5 C2≫C4≫C1≫C3≫C5 C2≫C4≫C3≫C1≫C5 C2≫C1≫C3≫C4≫C5 C2≫C3≫C1≫C4≫C5 C1≫C3≫C4≫C2≫C5 C3≫C1≫C4≫C2≫C5 C1≫C4≫C3≫C2≫C5 C3≫C4≫C1≫C2≫C5 C1≫C4≫C2≫C3≫C5 C3≫C4≫C2≫C1≫C5 C4≫C1≫C2≫C3≫C5 C4≫C3≫C2≫C1≫C5 C4≫C2≫C1≫C3≫C5 C4≫C2≫C3≫C1≫C5 C4≫C1≫C3≫C2≫C5 C4≫C3≫C1≫C2≫C5 C1≫C3≫C2≫C5≫C4 C3≫C1≫C2≫C5≫C4 C1≫C2≫C3≫C5≫C4 C3≫C2≫C1≫C5≫C4 C2≫C1≫C3≫C5≫C4 C2≫C3≫C1≫C5≫C4 C1≫C3≫C4≫C5≫C2 C3≫C1≫C4≫C5≫C2 C1≫C4≫C3≫C5≫C2 C3≫C4≫C1≫C5≫C2 C4≫C1≫C3≫C5≫C2 C4≫C3≫C1≫C5≫C2 C1≫C4≫C2≫C5≫C3 C4≫C1≫C2≫C5≫C3 C1≫C2≫C4≫C5≫C3 C4≫C2≫C1≫C5≫C3 C2≫C1≫C4≫C5≫C3 C2≫C4≫C1≫C5≫C3 C2≫C3≫C4≫C5≫C1 C3≫C2≫C4≫C5≫C1 C2≫C4≫C3≫C5≫C1 C3≫C4≫C2≫C5≫C1 C4≫C2≫C3≫C5≫C1 C4≫C3≫C2≫C5≫C1 C1≫C3≫C5≫C2≫C4 C3≫C1≫C5≫C2≫C4 C1≫C3≫C5≫C4≫C2 C3≫C1≫C5≫C4≫C2 C1≫C4≫C5≫C2≫C3 C4≫C1≫C5≫C2≫C3 C1≫C4≫C5≫C3≫C2 C4≫C1≫C5≫C3≫C2 C2≫C3≫C5≫C1≫C4 C3≫C2≫C5≫C1≫C4 C2≫C3≫C5≫C4≫C1 C3≫C2≫C5≫C4≫C1

⎫⎪⎪⎪⎪⎪⎪⎪⎪

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However, it is well known that such selection of a total ranking based on incomplete data leads to genuine overcommitment, and the learner often has to start from scratch after it receives a new piece of information, as in Multi-Recursive Constraint Demotion (MRCD) of [Tesar, 1997]. While the amount of available data increases monotonically during the learning, the output hypothesis of the learner does not monotonically improve in the general case. Thus some OT learners, such as MRCD, just reject monotonicity of the output; others, like GLA of [Boersma, 1997], simulate gradual learning by only allowing minimal adjustments, but do not even guarantee convergence (see [Pater, 2008]).

The purpose of this paper is to provide the formal theory which allows us to reject the dogma that OT grammars have to be single total rankings. There are both substantial and practical reasons for declaring only total rankings to be proper OT grammars. Substantial reasons essentially have the following form: “Natural language speakers behave as if they have total-ranking grammars”. This is an empirical claim, and I will have nothing to say about it in this paper. But practical reasons, however, are just that it is not easy to work with sets like the one in 2. An average human analyst would have to spend a considerable

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amount of time to process a representation like that and extract the useful information in it. What I do in this paper is build the technical tools that allow one to work with sets of OT rankings nearly with the same ease as with individual rankings. For instance, I show that in an important sense, the set of partial rankings in 3 is a strict equivalent of 2.

(3) ⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

(C1≫C5) ∧ (C3≫C5) (C1≫C5) ∧ (C4≫C5) (C2≫C5) ∧ (C3≫C5)

⎫⎪⎪⎪⎪⎬

⎪⎪⎪⎪⎭

The developed theory of sets of OT rankings has a number of applications. One of them is a solution for the general form of the OT Ranking problem. In the literature, only the narrow form of the Ranking problem is usually addressed: the problem of finding some ranking compatible with a given tableau. The general form of the Ranking problem is to findall rankings compatible with a given tableau. Our theory of sets of OT rankings provides a solution for it.² Among other results, our theory introduces manageable methods for working with large sets of rankings, and a test for whether a set of rankings captures all the information in some tableau, and thus is a legitimate OT grammar hypothesis (we will show that not all sets are).

The form of the paper will be somewhat unusual for a phonological paper: the nar- rative will not revolve around any particular empirical phenomenon, but will develop a comprehensive formal analysis of the space of sets of OT rankings instead. The down side of not concentrating on a single empirical phenomenon is obvious; however, during the previous two decades of OT development, many fundamental questions have been left unanswered by investigations focussing on specific immediate applications: e.g., the first solution to the general form of the OT Ranking problem has only been reported in [Brasoveanu and Prince, 2005], and even that only as a side result; the failure of the GLA algorithm, described by [Boersma, 1997], to converge in the general case has only been proved a decade later in [Pater, 2008] with the help of a specific counterexample;

etc. On the other hand, a formal analysis of the theory of OT rows in [Prince, 2002]

allowed for many subsequent practical advances, including the Fusional Reduction of [Brasoveanu and Prince, 2011]. The current paper attempts to do for sets of OT rankings what Prince did for rows and tableaux.

I will describe the theory of sets of OT rankings mostly as a finished product. The actual proofs often get long and technical, and are provided in the online appendices. This way they can be safely skipped if one simply wishes to apply the results, but all the heavy machinery used is provided for reuse as well.

2The first solution to the general form of the Ranking problem was reported in [Brasoveanu and Prince, 2005], but it was procedural, obtained as a side-product of a certain algorithm. As a result, Brasoveanu and Prince’s solution is not directly scalable: for instance, if a new datum is added to the tableau, solving the Ranking problem for the updated tableau would require running their algorithm from scratch. In contrast, our investigation of the theory of sets of rankings reveals how the solutions to the Ranking problem for related tableaux are related to each other.

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Section 2 gives a short summary of the normal form results for OT tableaux and equivalence-preserving tableau transformations. Section 3 develops a theory of partial OT rankings, conservatively interpreted as equivalent to certain sets of total OT rankings rather than objects in their own right (unlike, e.g., in the work by Arto Antilla and coauthors).

Section 4 introduces sets of partial rankings, and, using those, develops the general theory of sets of OT rankings, both partial and total. Section 5 concludes the paper.

2 Preliminaries

Throughout the paper we will often use the notions of a “possible row” and a “possible tableau”. A possible row for the purposes of this paper is simply any tuple of W-s, L-s and e-s of the appropriate length, and similarly for tableaux. Of course, many, if not most, actual choices of Con and Gen would make some of our “possible rows” impossible: for instance, if whenever someC1 is violated,C2 must be violated as well, then we will never actually have a row with an L inC1, but a W inC2. Here we will work with the space of all logically possible, all conceivable tableaux rather than with all tableaux possible given a fixed choice of Con and Gen. Our analysis creates a baseline with which the way a particular pair of Conand Gen restrict the range of possible tableaux can be compared.

We will frequently use “inline notation” for rows, where, e.g., (W, L, L, L) is a 4- constraint comparative row. By default, we take the constraint names for such an inline row to beC1,C2, and so forth. For a rowr, we denote byL(r)the set of constraints where rhas an L;W(r)is the set of constraints with a W in them. Thus we can define a new row r1 as follows: W(r1) ∶= {C1, C3}, L(r1) ∶= {C4}. For a constraint set {C1, C2, C3, C4}, that defines the rowr1=(W, e, W, L).

Since we will be studying rankings and sets of rankings in this paper, we will not be interested in the distinctions between tableaux which are OT-equivalent — that is, such tableaux with which exactly the same rankings are compatible. Rather than dealing with individual tableaux, we will be mostly dealing withequivalence classes of tableaux. It is trivial that there exist OT tableaux which are compatible with exactly the same sets of total rankings, and equivalence-preserving operations on tableaux have been studied, a.o., in [Hayes, 1997], [Prince, 2002], [Prince, 2006]. Building on those results, [Author, 2012] defines a normal form for OT tableaux and proves the following theorems:

(4) Normal form existence theorem

An arbitrary (finite) tableauT can be transformed into an equivalent normal form tableau by a (finite) sequence of equivalence-preserving transformations of the following types: row splittings and mergers; inference eliminations and introductions;

Generalized W Removal-s and additions; contradictory jumps; and row swaps.

(5) Normal form uniqueness theorem

In each equivalence class of OT tableaux, there is at most one normal form tableau.

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The operations mentioned in 4 are either trivial or familiar from the earlier literature.

Row splitting is dividing a row with several L-s into several single-L rows, and row mergers is its inverse. Inference elimination is deleting a row entailed (in [Prince, 2002]-s sense by the rest of the tableau, and inference introduction is its inverse. Generalized W Removal was described by [Prince, 2006], and it involves replacing a W with an e if that W did not actually do any useful work. Again, false W additions is just the inverse. Contradictory jump is adding or subtracting any row from a tableau not compatible with any rankings whatsoever as long as the result is also a contradictory tableau. Finally, row swaps are just swaps in the linear order of rows.

What the theorems 4 and 5 establish is that each equivalence class of OT tableaux has a unique representative: its normal form tableau. Moreover, any finite tableau can be computably transformed into the normal form. We can thus work only normal form tableaux without loss of generality, which makes proving results about compatibility with rankings a lot easier. As it is thus harmless to sloppily talk of tableaux meaning equivalence classes of tableaux, we will often do so.

The normal form of [Author, 2012], used in the above theorems, is as follows:

(6) Normal form for OT tableaux:

1. The only contradictory tableau in the normal form is the one-row tableau with a single L in the first constraint.

2. Each row has at most a single L.

3. There are no rows which can be inference-eliminated.

4. In multiple-W rows, there are no false W-s.

5. The rows are are ordered according to some strict total order of the set of all possible rows.

Here is an example of a normal form tableau:

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C1 C2 C3 C4 C5

W e L e e

W e e L e

W e e e L

e W e L e

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3 Partial rankings

3.1 Introducing partial rankings

We can view rankings as formulas in a certain logical language, and OT tableaux as models in which those formulas may be true or false.³ A ranking istrue in a tableau simply when it is OT-compatible with it (that is, when it selects the designated winner). A ranking φ viewed as a formula says for each pair of constraints Ci and Cj whether Ci≫Cj, or Cj ≫ Ci, or neither. We call an elementary building block of a ranking, Ci ≫ Cj, an atomic ranking. Full definitions behind the logical interpretation of rankings are given in the Appendix A.

In classical OT [Prince and Smolensky, 1993], the grammar is a total ordering of a fixed set of constraints Con, so all pairs of constraints are ranked. But the notion of OT-compatibility can be easily and conservatively extended to partial rankings:

(8) A ranking φis OT-compatible witha row r iff for every L in r in cellCi, there is a constraintCj dominating Ciinφsuch thatCj has a W in r.

The only new thing in 8, compared to the classical version, is that it does not guarantee that there will be a single W covering all the L-s in a row, even though every L still has to be covered by some W.

A special case of a partial ranking familiar from the literature is astratified hierarchy, [Tesar and Smolensky, 1996]. A stratified hierarchy is essentially a total order on a partition ofCon, with the cells of the partition called strata. Thus a constraint in a stratumA is ranked above all the constraints from lower strata, below all the constraints from higher strata, and is non-ranked with the other constraints fromA. There are many partial rankings which are not stratified hierarchies. E.g., a ranking which ranksC1 overC2, but does not rankC3 with respect to either is not a stratified hierarchy (if we were to treat C3 as a part of some stratum A, than the stratum A would not be ranked with respect to the strata containing C1 andC2.)

Our treatment of partial rankings is different from two treatments suggested in the literature, that of violation cancellation (“crucial tie”), and that of optionality.

According to our definition, if C1 andC2 are not ranked with respect to each other, it simply means that a W in either cannot cover an L in the other.

On the mark cancellation view, if C1 and C2 are not ranked, they are in a crucial tie, and basically behave as a single super-constraint. This view only makes sense for stratified hierarchies, but not for partial rankings in general. For consider a partial ranking like this:

(9) C1≫C2≫C3 C4≫C5

3When studying relations between different tableaux, [Prince, 2002] used a similar logical interpretation of OT: he viewedtableaux as formulas, and rankings as models. This immediately allows one to use the logical apparatus, so we apply the same strategy “from the other end”, so to speak.

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C4 in this ranking is not ranked with respect toC1,C2 and C3. If we are to interpret mutual non-ranking as imposing a crucial tie, then we have to say for the ranking in 9 there are crucial ties in all three pairsC1-C4,C2-C4, andC3-C4. Presuming that crucial ties are transitive, we derive a contradiction with the fact thatC1≫C2≫C3 according to the ranking.

The optionality interpretation of partial ranking (cf. [Anttila, 1997], [Anttila and Cho, 1998], a.o.) says that a grammar with an unranked pair of constraints ranks them one way half the time, and the other way another half.⁴ As the result, different output forms may be generated. This view does not derive a contradiction for a ranking like 9. But its treatment of non-ranking is different from ours.

SupposeC1 have a W in a given row,C2 an L. Let rankingφ not rankC1 and C2, and also put no other W on top ofC2’s L. On the crucial tie interpretation of non-ranking (if the number of violations ofC1 andC2 was the same), the winner is decided by other constraints ranked lower;

on Anttila’s view, the designated winner of the row will win some of the time; and on our view, it will always lose, as the L inC2 is not covered by any W according to rankingφ.

Our treatment is conservative because an uncovered L in our interpretation of partiality always leads to failure. The crucial tie and the optionality treatments of mutual unranking say that under certain circumstances, uncovered L-s are fine.

We can think of partial rankings as true possible grammars, or alternatively we can view them simply as convenient abbreviations for certain sets of total rankings, as we will see below. No “ontological commitment” is thus required, though given that sets of rankings should at least be allowed the status of properhypotheses about the grammar in OT learning if we want to allow for lossless learning at all, thinking of partial rankings as possible grammars is quite reasonable.⁵

A rankingφentails a ranking ψiff at every possible comparative row/tableau where φ is true,ψ is true as well. Ifφentails ψ, we writeφ⊧ψ.

(10) For rankings φ,ψ, φ⊧ψiff ψis true in (compatible with) every tableau thatφis true in (compatible with).

The fewer tableaux a ranking is compatible with, the more rankings it entails. In the limit, a ranking not compatible with any tableau by definition entails all other rankings.

If we have an arbitraryφand someφ^′which is a refinement ofφ(that is, which has all atomic rankings thatφhas, and then some more, see 11), then φnecessarily entailsφ^′.

4When there are more than two unranked constraints, the arithmetics gets more complicated. Each total refinement of the underspecified grammar creates a tableau for the input. The frequency of a given output is the ratio of the number of tableaux with that output to the number of total refinements of the total rankings (and thus different tableaux for the same input).

5[Tesar and Smolensky, 1996, pp. 28-29] argue that unless the data are generated by a total ranking, their RCD learning algorithm would not converge, and thus one should assume that grammars have to be total rankings. This argument does not apply in our case because Tesar and Smolensky assume the crucial-tie interpretation of mutual un-ranking, which is different from ours. RCD converges just as well on tableaux generated by partial ranking grammars under our interpretation as on tableaux generated by total ranking grammars.

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(11) φ^′ is arefinement of φ iff ∀Ci, Cj∶ (Ci≫Cj) ∈φ→ (Ci≫Cj) ∈φ^′. If φ^′ is a refinement ofφand φ≠φ^′, thenφ^′ is aproper refinementof φ.

(12) If φ^′ is a refinement ofφ, thenφ⊧φ^′.

So for any φ and its refinement φ^′, we have that φ ⊧ φ^′. What about φ and some ψ which is not φ’s refinement? It turns out neither can asymmetrically entail the other, unless a particular choice of Conand Gen forces that.

(13) If φand ψare not refinements of each other, then φ/⊧ψand ψ /⊧φ.

The proofs of 12 and 13 are provided in Appendix B.

12 and 13 essentially say that the relations “being a refinement of” and “being asymmetrically entailed by” coincide. .

The following important fact holds, in two formulations:

(14) a. If every proper refinement ofφis true in a tableauM, thenφis also true inM. b. If every proper refinement adding only one atomic ranking to φ is true in M,

thenφis true inM.

It is easier to see why the second claim is true. If the refinements ofφby bothCi≫Cj and Cj ≫ Ci are true in tableau M, that means pairwise ranking of Ci and Cj is not crucial for being compatible with M as long as all atomic rankings from φ are present.

The first claim easily follows from the second by induction.

14 means that the truth of a partial ranking may be viewed asdependent, or parasitic, on the truth of its proper refinements. If all of them are true in some tableau, then the original partial ranking has to be true there. If at least one proper refinement is false, than the ancestor partial ranking is false, by 12.

The following is an important special case of 14 allowing us to identify partial rankings with sets of their total refinements:

(15) Partial ranking φis true in tableau M iff every total refinement ofφis true inM. This means that a partial ranking and the set of its total refinements can be used interchangeably. They are equivalent objects, so everything we say about partial rankings can also be said about corresponding sets of total rankings. It is just that partial rankings are much easier to work with than sets of their total refinements.

Note that while every partial ranking corresponds to the unique set of its total refinements, not every set of total rankings corresponds to a partial ranking: e.g., the set consisting ofC1≫C2≫C3 andC3≫C2≫C1 does not correspond to any.

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3.2 Correspondence between partial rankings and tableaux

In this section, we will show that for every partial ranking, there exists a special tableau which contains exactly the same amount of information. Thus there is duality between partial rankings and such tableaux: they are essentially two different ways to represent the same object. In itself, this result is of limited usefulness, because the duality is restricted to a very special class of tableaux. But it will serve as a model for the similar general result in Section 4, which puts an arbitrary tableau into a correspondence with the set of all rankings true in it.

A ranking φ is maximal with respect to a tableau M, or M-maximal, iff φ is true in M, but none of φ’s ancestors is. (The mnemonic is “a ranking which is maximally underspecified, but still true in tableau M”.)

Sometimes, there will be only one maximal ranking for a given tableau, but not always:

(16) C1 C2 C3

W L e C1≫C2 is maximal

(17) C1 C2 C3

W L W C1≫C2 is maximal, C3≫C2 is maximal

17 makes it clear that in the general case, one tableau may have more than one maximal ranking. But for every partial rankingφ, there exists a special tableau for whichφ is the only maximal ranking. In particular, for the rankingC1≫C2, 16 is such a tableau.

It turns out that we can compute such a special tableau from any partial ranking, and vice versa, given such a tableau, we can recover the only partial ranking maximal for it.

For rankingφ, we will call such a tableau φ’srepresentative tableau, orMφ.

The construction procedure ofM_φfor an arbitraryφis simple. For each atomic ranking Ci≫Cj inφ that is meaningful (that is, not entailed by transitivity), build a row r_Ci,Cj such that its Cicell has a W, its Cj cell has an L, and all other cells have e-s. r_Ci,Cj is not compatible with any ranking which does not include the atomic rankingCi≫Cj: the L in theCj cell must be dominated, and there is only one W to do that inr_Ci,Cj, the one inCi, so no other atomic ranking but Ci≫Cj can make r_Ci,Cj happy.

Combine all such rows into a single tableau. Now if we subtract any meaningful atomic ranking fromφ, the result will not be compatible with the constructed tableau because the row corresponding to the subtracted ranking will become false. Of course, it is harmless to add more atomic rankings, constructing a refinement of φ, but we cannot take things away without making the resulting ranking false.

(18) Representative tableau M_φ for a partial ranking φ:

M_φ∶= {r ∣ (∣W(r)∣ = ∣L(r)∣ =1) ∧ [∃(Ci≫Cj) ∈φ∶ Ci∈W(r) ∧ Cj∈L(r)]}

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The inverse of the procedure is easy to define, too: given a tableau with onlyrCi,Cjrows, we can recover for each such row the corresponding atomic rankingCi≫Cj. Combining all such atomic rankings, we get the originalφ.

What is the class of tableaux which are representative for a partial ranking? First, by construction all of them are normal form tableaux: since partial rankings cannot contain contradictory atomic rankings, the resulting tableaux are not contradictory; since we only used meaningful atomic rankings, no rows in the resulting tableau are superfluous; all rows have just one L; finally, there can be no false W-s, for each row only has one W.

Moreover, all normal form tableaux that only have one-W-one-L rows are representative for some partial ranking: they only contain rCi,Cj rows, and the procedure above is guaranteed to result in a proper partial ranking.

There is thus a one-one correspondence between partial rankings and one-W-one-L normal form tableaux: the ranking construction procedure above outputs a ranking for any such tableau, and it is easy to see that no two rankings are mapped to the same representative tableau by 18.

What is the significance of that correspondence? Assuming the OT learning perspective, suppose we have a set of data which is equivalent to a normal form tableauMwith only one- W-one-L rows. The correspondence immediately provides us with a grammar hypothesis that exactly matches our data: the partial rankingφfor whichM is representative. Indeed, asφ is the only maximal ranking forM,φ’s total refinements are the only total rankings true in M. Thus choosing φ as a hypothesis does not rule out any rankings compatible with the data, and does not erroneously rule in any which are not.

The correspondence also provides a criterion for when exactly the faithful grammar hypothesis corresponding to a set of data cannot be a single partial ranking, and has to be a set of rankings: it is precisely when the normal form of the set of data has rows which are not one-W-one-L rows. It is not surprising that the criterion has to do with whether there are rows with more than one W: it has long been known that disjunctivity arises from such rows as (W, W, L). But not every tableau with multiple-W rows implies true disjunctivity — only if a normal form contains such rows can we say that the equivalence class of tableaux is genuinely disjunctive. In what follows, we call such special normal form tableaux with only one-W-one-L rowsnon-disjunctive tableaux, and the domain of all non-disjunctive tableaux, we callM⁻.

To sum up, there is duality between partial rankings and non-disjunctive tableaux:

the corresponding ranking and tableau contain exactly the same information, and may be transformed into each other by a computable procedure. We can thus use the two objects interchangeably, viewing them as different representations of the same entity.

3.3 Duality between partial rankings and sets of tableaux

Another important duality is between partial rankings and sets of all tableaux compatible with them. For each ranking φ, its representative set of tableaux σ_φ contains all

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tableaux which φcan account for. In this section, we will study such σφ-s, and learn how they correspond to representative tableauxM_φ.

(19) For any distinct rankings φand ψ, we have σ_φ≠σ_ψ.

Obviously, some sets of tableaux which are not σφ for any φ. How can we tell if an arbitrary set of tableaux is theσ-set for some ranking? It turns out anyσ_φcan be generated using theentailment closure of representative tableauM_φ. This provides a testing criterion for which sets of tableaux areσ-sets.

Take an arbitrary non-disjunctive tableau N with a row r not entailed by M_φ. By construction ofM_φ, rankingφcan only account for rows inM_φor rows entailed by it (note that this is entailment between tableaux, not between rankings), so φ is not compatible with r, and with N as a whole. Without loss of generality, suppose N is in the normal form. Thenφis not compatible with any equivalence class of tableaux whose normal form N has a row not entailed byM_φ.

On the other hand, any tableau N consisting only of rows from M_φand rows entailed by it is bound to makeφtrue. Thus we can defineσ_φin terms ofM_φthrough its entailment closureM_φ^En:

(20) The entailment closure tableau M^En is such that r∈M^En iffM entails r.

(21) σM ∶= {N ∣N^{N o} ⊆M^En}, whereN^{N o} is the normal form ofN.

For σMφ we write σφ, since such a set characterizes not only the representative tableau M_φ, but also the rankingφ.

As an illustration, for the tableauM_φ in 22, which is the representative tableau of the rankingφ∶= (C1≫C2) ∧ (C2≫C3), the correspondingM_φ^En is given in 23.

(22) M_φ=

C1 C2 C3

W L e

e W L

(23) M_φ^En=

C1 C2 C3

W L e

e W L

W e L

W L L

W L W

W W L

As σ_φ sets contain an infinite number of individual tableaux, they cannot be fully computed in a finite amount of time. However, they contain only a relatively small number

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of equivalence classes of tableaux. It is thus more convenient to think ofσ-sets as of sets of normal form tableaux, which would be finite for a finiteCon.

ComputingM_φfromσ_φviewed as a set of normal form tableaux is then simple: we just combine all tableaux inσφ into a single big tableau, and eliminate all superfluous rows.⁶

WhileM_φis a non-disjunctive tableau, the normal form members ofσ_φcan be disjunctive. E.g., for the non-disjunctive tableau in 22, a genuinely disjunctive tableau with a single row (W, W, L) is within the correspondingσ set.

Later in Section 4.3, the criterion allowing us to test an arbitrary set of tableaux for whether it is a σ-set will become important. In case of σ-sets corresponding to partial rankings, the criterion is as follows:

(24) Set of tableaux σ is in Σ⁻ iff there is a tableauM^En such that:

• M^En belongs to a non-disjunctive equivalence class (that is, its normal form is in M⁻), and

• for every row r entailed byM^En,r∈M^En,⁷ and

• for every normal form tableau N ∈σ, every row q∈N is also inM^En, and

• for every normal form tableau P ⊆M^En,P ∈σ, and

• σ is closed under OT-equivalence-preserving operations.

Let’s work through the clauses of the definition in 24. The first clause ensures thatMÊn is generated by some non-disjunctive normal form tableauM_φ. The second clause says that MÊn is closed under row entailment. The third clause says that all rows we can find in the normal form part of σ are included in MÊn. The fourth clause establishes the other direction: every normal form subset ofMÊnis required to be in σ. Finally, the last clause guarantees that not only normal form tableaux get into σ, but their whole equivalence classes, and at the same time that there are no “stray” non-normal form tableaux that crawled intoσ without their normal form being there.

If we simply remove the first clause of 24 (the clause requiring non-disjunctivity), we get a definition of a bigger domain Σ which we will discuss in Section 4.3.

Thus there exist duality between three kinds of objects: partial rankings, non-disjunctive tableaux, and sets of tableaux in the domain Σ⁻ characterized in 24. We can work with whichever of the three is most convenient, and then easily transfer the results into the two other domains.

The duality results for individual partial rankings that we obtained in the last two sections will serve as a model for similar duality results for sets of partial rankings in Section 4.

6Simply combining all the rows of an arbitrary set of tableaux may result in a contradictory tableau.

But all tableaux in σφ are entailed by Mφ. As such, they must be consistent with each other, and no contradiction may arise.

7Note that all tableaux in the equivalence class entail the same set of rows.

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3.4 The domain of partial rankings as an algebraic structure

What are the relations between different partial rankings that have relevance for OT? First, it is entailment between rankings. Here is the algebraic structure that entailment imposes on the domain of partial rankings (let’s call that domain Φ):

(25) _Λ

C1≫C2

:

:: :: :: :: :: :: :: ::

(C1≫C2) ∧ (C1≫C3)

yyrrrrrrrrrrrrrrrrrrrrrrrrr

:

:: :: :: :: :: :: ::

:: (C1≫C2) ∧ (C3≫C2)

%%L LL LL LL LL LL LL LL LL LL LL LL LL

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

This picture shows only the small part of the actual structure even for the very small Con ∶= {C1, C2, C3}. Λ is the empty ranking not ordering any pair of constraints, and it entails all other rankings. The partial ranking (C1 ≫ C2) entails every other non- trivial ranking in the picture: they are its refinements. (C1 ≫ C2) ∧ (C1 ≫ C3) and (C1 ≫ C2) ∧ (C3 ≫ C2) are still underspecified; they do not decide, respectively, the ranking between C2 andC3, and between C1 andC3. The three rankings at the bottom of the picture are total rankings: they are fully specified in our smallCon.

If there is a direct arrow from φ to ψ in the picture, it means that we can get ψ from φ by adding one atomic ranking, and transitively closing the result. So each arrow connects two rankings which are at minimal distance from each other in the structure. The reason why a partial ranking can have immediate descendants at different levels is transitive closure of rankings. For instance, when we add toC1≫C2 an atomic rankingC2≫C3, we effectively add also the third atomic ranking C1 ≫ C3 as well, by transitivity. Had partial rankings not been necessarily transitively closed, the structure would have looked differently. Note that there is no direct arrow from (C1 ≫ C2) to (C1 ≫C3) ∧ (C3 ≫ C2) ∧ (C1 ≫ C2): there is no single atomic ranking that we can add to the former to immediately get the latter.

The entailment relation obeys reflexivity (every ranking entails itself), antisymmetry (if ranking φentails ψ and they are not equal, then ψ does not entail φ) and transitivity (if φ entails ψ, and ψ entails ξ, then φ also entails ξ), and thus is a partial order. The structure ⟨Φ,Λ,⊧⟩ (the set Φ of partial rankings, with a distinguished element Λ, plus a relation of entailment between rankings) is a poset (partially ordered set) with a bound

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on one side. We can thus say that more underspecified rankings are greater than their refinements, treating⊧ as≥. The empty ranking Λ is the maximal element of the order.

The refinements of a ranking can be called itsdaughters ordescendants, bringing in the tree terminology, as our poset structure can be viewed as a tree with the empty ranking Λ as the root. Then we have a natural name for the inverse of the refinement relation: a rankingais aancestor for a rankingb iffb is a refinement ofa.

Recall that an M-maximal ranking is such a ranking that it is true in M, but none of its immediate ancestors is. All descendants of an M-maximal ranking in the structure

⟨Φ,Λ,⊧⟩ have to be true in M, so the maximal ranking defines a “triangle” of rankings compatible with M, and is the greatest element in this “triangle”.

For σ-sets, the notion of entailment imported from the domain of partial rankings Φ says thatσ_φ⊧σ_ψ when σ_φ⊆σ_ψ: ifφ⊧ψ, then ψ is compatible with all the tableauxφis compatible with, and then possibly with some more, which exactly describes the situation whenσ_φ⊆σ_ψ.

For representative tableaux, entailment between equivalence classes M_φ and M_ψ is most easily characterized using the entailment closures M_φÊn and M_ψÊn (see 20 above for the definition). φ⊧ψ exactly when M_φÊn⊆M_ψÊn. This is, of course, not surprising, given the relation betweenσ_φ and M_φÊn, 21.

In addition to entailment, there are two natural operations on partial rankings: ranking- intersection∩^r and ranking-union∪^r . Take some arbitrary φand ψ. The set of all atomic rankings that they share is the ranking-intersection φ∩^rψ. The transitive closure of the union of all atomic rankings from either ranking is the ranking-union φ∪rψ.

(26) For rankings φand ψ,φ∩^rψ is the smallest⁸ ranking ξ s.t.

∀Ci, Cj∶ [(Ci≫Cj) ∈φ ∧ (Ci≫Cj) ∈ψ] → (Ci≫Cj) ∈ξ.

(27) For rankingsφandψ,φ∪rψis defined iff¬∃Ci, Cj∶ (Ci≫Cj) ∈φ∧ (Cj≫Ci) ∈ψ.

If φ∪^rψis defined, then it is the smallest rankingξ s.t.

∀Ci, Cj∶ [(Ci≫Cj) ∈φ ∨ (Ci≫Cj) ∈ψ] → (Ci≫Cj) ∈ξ.

Note that the ranking-intersection ∩^r behaves simply as set intersection for sets of atomic rankings contained in the argument partial rankings, and is always defined.but ranking-unionφ∪rψ isnot a simple set union of the sets of atomic rankings in φand ψ.

First, we need to apply transitive closure to that set union to get a proper OT ranking.

Second, ifφand ψcontradict each other, then φ∪^rψ is undefined.

What is the OT sense of those two operations? Imagine a very sophisticated OT learner that selects several overcommitted hypothesis according to a complex system of biases. Suppose that learner thinks, after receiving some data, that the following two hypotheses are highly probable: φ∶=Onset >> NoCoda >>Parse and ψ ∶=Onset >>

Parse>>NoCoda. Now the ranking-intersectionφ∩^rφ=(Onset >>Parse) ∧(Onset

8Where “smallest” means “containing the least number of meaningful atomic rankings”.

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>>NoCoda) contains the atomic rankings which, according to our learner, would be even more highly probable than grammar hypothesesφand ψ.

On the other hand, imagine another sophisticated OT learner which processes several different sets of data in parallel, and has found that one set of data can be accounted for by simplyOnset>>NoCoda, and that another set can be accounted for by NoCoda>>

Parse. The ranking-union∪r of those two partial rankingsOnset>>NoCoda>>Parse is the smallest partial ranking which accounts for both sets of data at the same time.

Of course, these examples are only meant as illustrative sketches: one would hardly need to use very sophisticated learners to learn simple syllable theories. But it is easy to apply the same idea in a more realistic setting.

∩^r and ∪^r are straightforwardly connected to entailment:

(28) φ∩rψ ⊧ φ

φ ⊧ φ∪rψ

Thusφ∩^rψis an ancestor for bothφand ψ, andφ∪^rψ (if it exists) is a descendant for both of them. Moreover, the partial rankingφ∩^rψis the most specified common ancestor, andφ∪rψ is the least specified common descendant ofφandψ. Entailment among partial rankings is definable in terms of ∩^r: φ⊧ψ iff(φ∩^rψ) =φ.

Both ranking-union and ranking-intersection are associative, commutative, distribu- tive, and obey absorption (when all unions involved are defined). As ranking-unions are not always defined, the structure ⟨Φ,Λ,⊧,∩^r,∪^r⟩ is not an algebraic substructure of the powerset algebra generated by the set of atomic rankings with set-theoretic operations on it. Optimality Theory carves out a non-trivial part of that powerset algebra, excluding contradictory “pseudo-rankings”.

3.5 Summary of Section 3

In this section, we have defined the notion of OT-compatibility for partial rankings under which such rankings are compatible with a tableauT iff all their total refinements are. Thus partial rankings may be viewed as notational shortcuts for certain sets of total rankings.

In the next section, we will generalize that result, studying sets of partial rankings that can be such shortcuts for arbitrary sets of total OT rankings which are compatible with one and the same tableau.

Furthermore, we have established that for any partial ranking φ, we can compute its representative tableau M_φ such that φis the only M_φ-maximal ranking, and that we can also do the inverse computation ofφfromM_φ. Similarly, we have described duality between rankings φ and sets σφ of all tableaux compatible with them, and provided computable procedures for going back and forth between φ and σ_φ. That means that φ, M_φ and σ_φ are essentially different forms in which the very same information about grammar is represented, and thus they can be used interchangeably. Again, in the following section we will generalize this to arbitrary sets of OT rankings.

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4 Sets of partial rankings

If we have a non-disjunctive tableau, then there exists a unique partial ranking maximal in it, and that ranking represents all total rankings compatible with the tableau. In other words, the correspondent partial ranking is the solution of the OT Ranking problem for the case of non-disjunctive tableaux. In this section, we study sets of rankings corresponding in the same way to an arbitrary tableau, and thus derive the full solution of the general form of the OT Ranking problem which allows us to perform lossless OT grammar learning.

The structure of this section is as follows. Section 4.1 sets the stage by defining crucial basic notions. Section 4.2 solves the general form of the OT Ranking problem and provides the characterization of proper sets of rankings (in other words, of proper OT grammar hypotheses.) Section 4.3 develops an analysis of the relations between proper sets and all other sets of rankings, which shows that each proper set of rankings is the minimal proper representative S^{M in} of its equivalence class, and provides a method for finding an equivalent proper set from an arbitrary set of rankings. Finally, Section 4.4 discusses the algebraic structure of the domain of proper sets of rankings, which is shown to be generated by⋓-closure of a small number of very simple sets of rankings used as atoms.

4.1 OT-compatibility for sets of partial rankings

A partial ranking φis true in a tableau M whenever every ranking from the set of all its total refinements is true in M. So when φ is the only M-maximal ranking, it is exactly the faithful grammar hypothesis given the data inM. But for a disjunctive tableau such as 29, there is more than one maximal ranking. A grammar hypothesis not losing any information from the tableau in such case must be something other than a single partial ranking. In this section, we will define OT-compatibility for objects which play that role:

for sets of partial rankings.

(29) C1 C2 C3

W W L C1≫C2 is maximal, C3≫C2 is maximal

We the usual set notation for sets of (total or partial) rankings: {φ, ψ}, for rankingsφ, ψ. Uppercase letters such asS,T, ... are used as variables over such sets.

A set S of rankings is compatible with a row r iff all of the rankings in the set are true there. Compatibility/truth in a tableau, entailment and equivalence result straightforwardly from that:

(30) A set of rankings S= {φ1, ..., φn}is true in a row riff for all φi∈S, φi is true inr.

(31) S= {φ₁, ..., φ_n} is true in a tableau M iff for allφ_i∈S, φ_i is true inM. (32) For sets of rankingsS,T, S⊧T iffT is true in every tableauS is true in.

(33) For sets of rankingsS,T, S and T are equivalent iff S⊧T and T ⊧S.

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There are two distinguished sets of rankings: ∅and{Λ}. ∅is the empty set of rankings, and it is not compatible with any row whatsoever, including even L-less rows: there is simply no ranking in ∅ to satisfy the definition of truth above. {Λ}, on the other hand, does contain one ranking, though it is the empty one. Unlike ∅, the set {Λ} is true in L-less rows. We call all other sets of rankings but those twonon-trivial.

For singleton sets of rankings, the notions of truth and entailment coincide with the corresponding notions for single partial rankings. Thus we identify singleton sets with individual rankings (i.e., we do not distinguish betweenφand {φ}).

By the definition of truth for sets of rankings, if φ ⊧ ψ, then the set {..., φ, ψ, ...} is always equivalent to {..., φ, ...}. In other words, it is always safe to omit rankings which are refinements of other rankings in the set: it never changes OT-compatibility. It will be convenient for us to always bring sets of rankings we are working with to the standard form which does not include refinements. In what follows, we will assume we work with sets which are in such a standard form.

Because of how OT-compatibility works, many sets of rankings looking quite different on the surface are equivalent under our definitions. E.g.,S and T in 34 are equivalent:

(34) Let S∶= {C1≫C2≫C3, C3≫C2≫C1} and T∶=S∪ {C1≫C3≫C2}. S and T are equivalent.

The proof is given in Appendix C. Moreover, if we add toT yet another total ranking C3 ≫ C1 ≫ C2, by the same reasoning as in the proof the resulting four-ranking set will still be equivalent to S. In fact, it would be the largest set of total rankings in its equivalence class. There is a unique largest set of total rankings in every equivalence class of sets of rankings:

(35) Largest total representative lemma.

For a non-trivial equivalence class of sets of rankingsC, there always exists a unique largest total representative S^{T ot} such that 1) each φ∈S^{T ot} is total, and 2) for every T in equivalence class C consisting only of total rankings, T⊆S^{T ot}.

S^{T ot} sets play an important role: if an equivalence class of sets of rankings is a faithful grammar hypothesis for some tableau (by the end of this section we will learn that in fact any equivalence class is), then itsS^{T ot}representative contains all total rankings compatible with the data, providing the answer to the general form of the Ranking problem of OT.

Sets of total rankings within a class that arenot largest are deficient: there definitely cannot be a tableauM such that they contain all rankings compatible withM. This is an important result: not all sets of total rankings are born equal. Only some of them have direct OT significance.

Our notion of truth collapses such deficient sets into the same equivalence class with the cor- respondingS^{T ot}. Alternatively, we could have used a “weak compatibility” notion of truth which only requiresoneof the members of the set to be true in a tableau in order for the set to be declared

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true there. It is easy to see that under this notion, the setsS and U from 34 are not equivalent, and in fact no two distinct sets of total rankings are. An earlier version of this paper used this route, and it is quite possible to get all the results that way.

On the weak compatibility approach, the hard part is the unintuitive notion of truth. On the current approach, it is the coarse-grained equivalence classes of sets of rankings which are hard to understand and work with. But the main steps are the same anyway, whichever of the two routes we take. For instance, what looks like finding the right equivalence classes on the weak compatibility approach is finding the non-deficient representatives within an equivalence class on the current approach.

S^{T ot} sets only contain total rankings, and moreover they contain a maximal number of them, so they are not very manageable to work with. Luckily, we do not have to, as the following lemma will help us to find smallest, most practically convenient representatives:

(36) For an equivalence class of sets of rankingsC, for anyS∈ C and anyφ∈S, all total refinements of φare in the largest total representative S^{T ot} of C.

Thus any set of rankings in an equivalence class only contains ancestors of the total rankings in the S^{T ot} representative. This suggests where to look for the natural smallest representative of a class: it should include rankings as underspecified as possible, and at the same time cover all total rankings inS^{T ot}. We define the following notion:

(37) The minimal proper representative S^{M in} for an equivalence class of sets of rankingsC is defined as follows:

Take the largest total representative S^{T ot} of C. Add to S^{T ot} every ranking for which all total refinements are in S^{T ot}. By 15, the resulting set will also be inC: adding such rankings preserves OT-compatibility. Then subtract from that set all rankings with ancestors in the set. Again, this preserves OT-compatibility. Call the resulting setS^{M in}.

S^{M in} is clearly minimal in the sense that it only contains maximally underspecified, smallest partial rankings. But is it truly minimal, that is, can there be some set of rankings T in the same equivalence class C which is a proper subset of S^{M in}? The answer to this question is no, for there can be suchT-s, though we will only be able to show this in Section 4.3.2 after more preparatory work. Still, S^{M in} will turn out to be a useful representative as all its subsets that are also in C have smaller sets of total refinements than S^{T ot}. Thus we callS^{M in}the minimal proper representative.⁹

Our inability to say right away whetherS^{M in}is truly minimal underscores an important point: at the moment we do not have the means to work with equivalence classes of sets of rankings. For instance, given an arbitrary set of rankings, we cannot tell if it is theS^{M in}

9Later on, we will call certain sets of rankingsproper sets. The word ‘proper’ has the same sense in “the minimal proper representative: minimal proper representatives actually are proper sets.

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set for any equivalence class. Similarly, we do not know how to build from an arbitrary set of rankings the representative of its equivalence class. Such gaps cannot be filled without studying in more detail the relations between sets of rankings and tableaux.

Once we take a look at sets of rankings from the individual tableau perspective, even more questions will immediately arise, and those will guide our investigation. Earlier we have defined the notion of an M-maximal partial ranking for a tableau M: the ranking which is true in M, and no ancestor of which is. Now we define a related notion of an M-maximal set of rankings:

(38) A set of rankings S is maximal in a tableau M (is M-maximal) iff all rankings in S areM-maximal, and allM-maximal rankings are inS.

As each ranking is either maximal in a tableauM or not, for each tableau there trivially exists a unique M-maximal set of rankings. M-maximal sets are important because they record the full set of rankings compatible withM:

(39) Ranking φis true in tableauM iffφ is a refinement of someψ in the M-maximal set of rankings.

Indeed, suppose it were not so. Either the falsifying rankingφisM-maximal itself, or not. If it is, it is supposed to be in theM-maximal set. If it is not, then some ofφ’s ancestors isM-maximal, and then that ancestor is in theM-maximal set of rankings.

Thus the set of total refinements of the M-maximal set is the set of all total rankings true in M. The M-maximal set is the faithful grammar hypothesis given M. In what follows, we will prove that M-maximal sets of rankings are preciselyS^{M in} representatives of their equivalence classes, and moreover, that each S^{M in} set is the proper grammar hypothesis for some tableau.

4.2 Correspondence between tableaux and sets of rankings 4.2.1 From tableaux to their maximal sets of rankings

We will now define a procedure for building from an arbitrary tableauM theM-maximal set of rankings, the faithful grammar hypothesis for that tableau. The key to the procedure is heavy decomposition of the problem: we will define the operation of pairwise ranking- union⋓ on sets of rankings, and prove that if we apply⋓to two sets which are M1- and M2-maximal, the resulting set of rankings is the maximal for the tableau combining the two,M₁∪M₂. Then we will learn how to build the maximal sets for individual rows, and use the result about ⋓to build maximal sets for arbitrary tableaux.

S⋓T contains every ranking resulting from taking the ranking-union of a member of S and a member of T. Recall that for partial rankings φ and ψ, φ∪rψ is the smallest ranking which is true in every tableau where eitherφorψ is true. Generalization of ⋓ to an arbitrary number of arguments is straightforward.

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(40) For sets of rankingsS,T, S⋓T∶= {ξ∣ ∃φi∈S∶ ∃ψj ∈T ∶ξ=φi∪^rψj} We give an example of how⋓ works:

(41) {(C1≫C2≫C3), (C4≫C5)} ⋓ {(C1≫C5≫C4), (C3≫C4)} =

= {(C1≫C2≫C3) ∧ (C1≫C5≫C4), ∅, (C1≫C2≫C3≫C4), (C3≫C4≫C5)}

⋓ is commutative and associative since it piggy-backs on ∪r, which is both. The maximum number of rankings in S⋓T is ∣S∣ times ∣T∣, but in many cases, the actual number will be smaller: for some pairs ofφi andψj, the ranking-union is not defined, and some ranking-unions may happen to be refinements of others. E.g., in 41 the ranking-union of contradictory(C4≫C5)and(C1≫C5≫C4)is undefined, hence there are only three, not four, rankings in S⋓T.

(42) ⋓-Decomposition lemma.

For any tableaux M,M1,M2 s.t.M =M1∪M2, let the total refinements of set of rankingsS₁ be all and only total rankings true inM₁, and similarly forS₂ andM₂. Then S₁⋓S₂ is true in M, and the total refinements of S₁⋓S₂ are all and only total rankings true in M. Moreover, if S1 is maximal for M1, and S2 is maximal forM₂, then S₁⋓S₂ is maximal in M.

The proof of 42 is given in Appendix C. 42 and its proof trivially generalize to an arbitrary division of M into parts.

Note that if one of tableauxM₁,M₂ is contradictory, no ranking is true in it, and the corresponding set of rankings is ∅. A pairwise ranking-union of any set with ∅ produces

∅, in symbols ∅ ⋓S = ∅. On the other hand, if one of M1, M2 does not have a single L, its corresponding set is{Λ}, and Λ⋓S=S for any S.

Now that we have the important ⋓-decomposition lemma in 42, once we provide the way to build sets of rankings maximal for individual rows, we will have the tools for building maximal sets for an arbitrary tableau.

(43) Maximal sets of individual rows.

For an arbitrary row r, define set of rankings S_r as the biggest set of rankings (without duplicate rankings) where for any φ∈S_r two conditions hold:

∀Cj∈L(r) ∶ ∃Ci_φ,j∈W(r) ∶ (Ci_φ,j≫Cj) ∈φ ∧ ∀Ci∶ ((Ci≫Cj) ∈φ) →i=i_φ,j (1)

∀Ck/∈L(r) ∶ ¬∃Cm∶ (Cm≫Ck) ∈φ (2)

Claim: Sr is r-maximal.