(34) Let S∶= {C1≫C2≫C3, C3≫C2≫C1} and T ∶=S∪ {C1≫C3≫C2}. Then S and T are equivalent.
Proof. Suppose towards a contradiction there is a rowr whereS is true, butT is not.
This means that inr, the two rankings from S are true, but the ranking C1≫C3≫C2 from T is not. This can only be if C1 ≫C3≫C2 lacks some atomic ranking crucial in accounting forr. What could this atomic ranking be? It cannot beC3≫C1 orC2≫C1, for the first ranking in S lacks them as well. It also cannot be C2≫C3, for the second ranking inS lacks it. But all other 3 atomic rankings possible for a set of 3 constraints are already in C1≫C3≫C2. So there can be no such r in which S is true, but T is not.⊣
(35) Largest total representative lemma.
For a non-trivial equivalence class of sets of rankingsC, there always exists a unique largest total representative ST ot such that 1) each φ∈ST ot is total, and 2) for every T in equivalence class C consisting only of total rankings, T⊆ST ot.
Proof. Existence of at least one member of the class which only contains total rankings follows from 15. Suppose there is a non-trivial classCof sets of rankings, pick some member U of it which contains a non-total ranking. By 15, that partial ranking is true iff all of its total refinements are true, so if we replace it with the set of such refinements, the resulting U′ will be in the same equivalence class. By induction on the non-total rankings inU, we can build a member of C containing only total rankings.
For uniqueness, suppose someS andT are two largest distinct members of classCthat contain only total rankings. As they are distinct, their unionS∪T has larger cardinality.
As every ranking inS∪T was either inSor inT,S∪T is inCas well. SoS andT were not largest sets of total rankings inC, contrary to assumption. In general,ST ot is the union of all sets of total rankings inC.⊣
(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 ST ot of C.
Proof. Suppose there are S, φ for which 36 does not hold. Fix some total ranking ψ which is a refinement ofφand is not inST ot. By definition, for any tableau whereSis true, φand therefore by 15 ψ have to be true. But then ST ot∪ {ψ} is in the same equivalence classC, andST ot was not the largest representative, contrary to assumption. ⊣
(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. Claim 1: S1⋓S2 is true in M, and the total refinements of S1⋓S2 are all and only total rankings true in M.
Claim 2: if S1 is maximal for M1, and S2 is maximal for M2, then S1⋓S2 is maximal in M.
Proof. For any i, j, φi⋓ψj, if defined, is compatible with both M1 and M2, and therefore withM as a whole. Thus all rankings in S1⋓S2 are true inM, and thus the set as a whole is.
Furthermore, suppose towards a contradiction that some total rankingξ is true inM, but is not a refinement for any ranking inS1⋓S2. Asξ is true inM, it is also true in its partsM1 andM2. By assumption, all total rankings true in either are refinements of some φfromS1 and someψfromS2. But thenξ is also a refinement ofφ∪rψwhich is inS1⋓S2
if defined. It cannot be that the ranking-union is defined for any pair of rankings from S1 and S2 which are ancestors toξ: that would mean that ξ itself contains contradictory atomic rankings. Thus there can be no suchξ.
For the last claim, observe that any M-maximal ranking has to be a ranking-union of anM1-maximal φand M2-maximalψ. ⊣
(43) Maximal sets of individual rows.
For an arbitrary row r, define set of rankings Sr as the biggest set of rankings (without duplicate rankings) where for any φ∈Sr two conditions hold:
∀Cj∈L(r) ∶ ∃Ciφ,j∈W(r) ∶ (Ciφ,j≫Cj) ∈φ ∧ ∀Ci∶ ((Ci≫Cj) ∈φ) →i=iφ,j
(3)
∀Ck/∈L(r) ∶ ¬∃Cm∶ (Cm≫Ck) ∈φ (4)
Claim: Sr is r-maximal.
Proof. The proof essentially spells out of what the two conditions on φ-s inSr say.
Take an arbitrary row r. If there are no W-s in it, no φsatisfies the conditions, thus Sr= ∅. In a W-less row, no ranking is true, so indeed∅ is the set of rankings maximal for it.
Suppose there is at least one W inr. Then it is easy to find a rankingφsatisfying the two conditions: fix someCk∈W(r), and buildφas⋀{Ck≫Cl∣Cl∈L(r)}. Clearly such aφ cannot be contradictory, and thusSr will always be non-trivial if there is at least one W inr.
We first show that every ranking φ satisfying the two conditions is r-maximal, and secondly, that every r-maximal ranking satisfies the two conditions and thus is in Sr.
The first condition guarantees that anyφ∈Srhas for each L an atomic ranking covering it with a W, soφ must be true atr. Suppose someφ satisfying the two conditions is not r-maximal. Then there should be an atomic ranking which we can subtract fromφwithout making it false atr. By the second condition, only constraints fromL(r)can be dominated inφ. Take an arbitraryCj∈L(r). By the first condition, there is only one atomic ranking where Cj is dominated. If we subtract that ranking, there will be no way to account for the L in Cj. As Cj was arbitrary, no atomic ranking can be subtracted from φ without making it false atr, soφhas to ber-maximal.
Take an arbitrary r-maximal ψ. We show that then it satisfies both conditions, and thus is in Sr. Clearly an r-maximal ψ has to satisfy the first part of the first condition:
∀Cj ∈ L(r) ∶ ∃Ciψ,j ∈ W(r) ∶ (Ciψ,j ≫ Cj) ∈ ψ, for otherwise ψ would be false in r.
Suppose ψ does not satisfy the second part of the first condition. Then ∃Cj ∈ L(r) ∶
∃Ck, Cl∈W(r) ∶k≠l ∧ (Ck≫Cj) ∈ψ ∧ (Cl≫Cj) ∈ψ. But then we can subtract either (Ck≫Cj) or (Cl≫Cj) without makingψ false at r, so ψ is not maximal. Thus ψ has to validate the first condition. Suppose ψ does not validate the second condition. Then, similarly, we find (Cm≫Ck) ∈ψ where Ck/∈L(r), and note that subtracting it will not affectψ’s truth at r, so again ψ is not maximal, contrary to the assumption. Thus every r-maximal ranking satisfies the two conditions, and has to be inSr. ⊣
(51) Set of alternatives theorem.
For a Ci-alternative set SCi, there existsOCi= {r1, ..., rn}s.t. Sr1⋓...⋓Srn =SCi. (That is, SCi is OCi-maximal.) Moreover, (for finite Con) there is an effective function computingOCi from an arbitrarySCi.
Proof. We prove 51 for the finite case by actually providing the effective function which computes OCi from an arbitrary SCi. The same construction works for the infinite case, but of course the function is not effective in that case.
First, we make sure that every combination of constraints dominating Ciin different φk in SCi has some useful job in the OCi we are constructing. For that, we form all combinatorially possible sets of constraintsT such that for eachφk, there is one dominator constraint from φk in T. Then we build for each such set T a rowr with W(r) =T. If
we combine those rows together, eachφk will have to be maximal in the resulting tableau:
for each atomic ranking withinφk, there will be at least one row where that ranking is the only one inφk which can account for the row.
It is instrumental to give an example. Let Con ∶= {C1, ..., C6}. Consider the set of three C6-alternatives S ∶= {(C1 ≫C6) ∧ (C2 ≫ C6), (C3 ≫C6) ∧ (C4 ≫ C6), (C1 ≫ C6) ∧ (C4 ≫C6) ∧ (C5≫C6)}. Below we build the full set of T combinations for that S. (Recall that eachT-set contains one dominator constraint from each of the 3 rankings in S; as some constraints are dominating in more than one ranking, some T sets will be equal, but we spell out all 2×2×3=12 combinations.)
T1= {C1, C3, C1} = {C1, C3} T2= {C1, C3, C4} T3= {C1, C3, C5} T4= {C1, C4, C1} = {C1, C4} T5= {C1, C4, C4} = {C1, C4}
T6= {C1, C4, C5}
T7= {C2, C3, C1} = {C1, C2, C3} T8= {C2, C3, C4}
T9= {C2, C3, C5}
T10= {C2, C4, C1} = {C1, C2, C4} T11= {C2, C4, C4} = {C2, C4}
T12= {C2, C4, C5}
For each Tl, we build a row r s.t. L(r) ∶=Ci, W(r) ∶=Tl. Obviously we can omit the rows built from thoseTi-s which are supersets of someTj, as well as full duplicates — they will be superfluous when we combine the built rows into a single tableau. In our example, the row forT1 will entail the rows forT2,T3,T7; the rows forT4 andT5 are duplicate, and also entail the rows for T6 and T10. T11 entails T8,T10 (also entailed by T4/T5), and T12. The row forT9 is independent and does not entail any other row we are supposed to build from T-s. This leaves us with T1, T4, T9 and T11. So here is what those four T sets are, and the tableau built from them:
(85) T1= {C1, C3} (T2,T3 and T7 are supersets of this one) T4= {C1, C4} (T5 is a duplicate,T6 and T10 are supersets) T9= {C2, C3, C5} (no duplicates or supersets)
T11= {C2, C4} (T12is a superset)
(86)
C1 C2 C3 C4 C5 C6
W e W e e L
W e e W e L
e W e W e L
e W W e W L
It is easy to check that each ranking in the original set S ∶= {(C1 ≫ C6) ∧ (C2 ≫ C6), (C3≫C6) ∧ (C4≫C6), (C1≫C6) ∧ (C4≫C6) ∧ (C5≫C6)} is maximal in the tableau in 86. In fact, ithad to be maximal, the way we constructed 86. We give the proof for it now.
Take an arbitraryTl set. For each φk∈SCi, there is some Cj∈Tl s.t. (Cj≫Ci) ∈φk, by construction ofTl. So if we build a row withW(r) ∶=Tl,L(r) ∶=Ci, everyφk∈SCi will be true in this row.
Thus when we combine all rows built fromTl sets into a tableauO, clearly all rankings inSCi will be true in O. What we need to show is that they not only are true, but also are maximal there, and that allO-maximal rankings are in SCi.
Suppose towards a contradiction that there exists a ranking ψ true in O such that ψ is not a refinement of any φk ∈ SCi. (If there is an O-maximal ranking not in S, then there must exist such a ψ.) Then for each φk ∈ SCi there is at least one atomic ranking Cj≫Ci which is in φk, but not in ψ. We fix one such Cj≫Ci for each φk, and gather their dominator constraints into a set. This set is in fact one of the Tl sets we built when constructing O. So there must be a row rl in O built from this set of constraints. But because of how we picked Cj constraints, they do not dominate Ci in ψ, so ψ cannot account for that rowrl, and hence for the tableauO as a whole, contrary to assumption.22 Thus by construction ofO, allψ-s compatible withOhave to be entailed by one ofφk∈SCi. From this fact, all ancestors of any φk ∈ SCi which is not a refinement of any other ranking inSCi have to be false inO. Thus eachφk isO-maximal. Any ranking not inSCi cannot beO-maximal because if it is not a refinement of someφk∈SCi, then it is false at O. ThusSCi is indeed maximal for the constructed O. ⊣
(52) Corollary to 51.
There is a one-one correspondence between normal form tableauxOCi with L-s in the same constraint in all rows, and sets of rankings which are sets ofCi-alternatives (as defined in 50), such that the corresponding SCi is OCi-maximal.
Proof. The set of alternatives theorem 51 provides us with a function from sets SCi to the domain of tableaux with L-s in a single constraint Ci, such that the value of the
22We omitted from the resulting tableauOtherl rows entailed by some other row inO, but that does not makeψ’s life any better. The entailer row will have a subset ofrl’s W-s, andψwill not be able to use any of those W-s to explain the L asψdoes not feature the relevantCj constraints as dominators.
function is the tableau for which SCi is maximal. We need to show that the function is both into and onto that domain.
First, we prove injectivity. Suppose two sets SCi1 and SCi2 are mapped to the same tableau OCi. By 51, each ranking ψ true in OCi is either in SCi1 or a refinement of a ranking in it; and same forSCi2 . It follows that SCi1 =SCi2 .
Surjectivity is not much harder to prove. Take an arbitrary OCi and compute the pairwise ranking-union of the maximal sets for its rows SOCi, then trim the refinements.
By 42, the resulting set S is OCi-maximal. As any row in OCi only has an L in Ci, rankings in S only contain atomic rankings of the form X ≫Ci. Thus S is a proper set ofCi-alternatives, and is taken to anO tableau for which it is maximal by the function in 52. SinceS is maximal for bothOCi and O, the two tableaux are equivalent,OCi=O. As OCi was arbitrary, the range of the function in 52 is the whole domain ofOCi tableaux. ⊣
(59) Proper set theorem
Set of rankings S is proper iffS=SP+. Proof.
(⇒) Suppose S is proper. There must be a tableau MS witnessing that. MS can be divided into the non-disjunctive partMSCo, the representative tableau for the core ranking SCo of S, and the disjunctive part MS ∖MSCo. The disjunctive part may be further subdivided intoOCi tableaux. By 52, there are SCi sets maximal for thoseOCi tableaux.
By 42,S= {SCo} ⋓(⋓Ci∈Con{SCi}).
Now we need to show that TCi sets extracted from S as defined in the procedure in 58 are equal to SCi sets maximal for the OCi tableaux. Fix a Ci. From the assumption that S is maximal for MS, we have that for any φj in the residue part of S, no atomic ranking can be subtracted fromφj without making it false in theOCi subtableau of MS. Asφj∖φCij is irrelevant for OCi, not having any atomic rankings whereCi is dominated, it follows thatφCij is OCi-maximal.23
Suppose there exists a ranking ξCi maximal in OCi which was not in the TCi set, and thus in anyφj. Pick someφk, and derive the ranking(φk∖φCik ) ∪ξCi. It must be maximal inMS by 42, as(φk∖φCik ) is maximal in(MS∖MSCo) ∖OCi, andξCi is maximal inOCi. But then(φk∖φCik ) ∪ξCi must have been inS, and ξCi, inTCi, contrary to assumption.
Thus the recovered TCi sets are maximal for OCi parts of the corresponding tableau, and SP+ is maximal forMS, but that can only be when S=SP+.
(⇐) In the other direction, supposeS=SP+. Given thatSP+= {SCo} ⋓(⋓Ci∈Con{TCi}), we can build MSCo and a number of little OCi tableaux for which the sets whose pairwise ranking-union we took are maximal. By 42, the set of rankingsSP+=S is maximal for the union of those tableaux.⊣
23By our definition,φCij sets only contain meaningful rankings fromφj, omitting those derived by tran-sitivity. A row for which such an omitted atomic ranking would be crucial for would have to be entailed by MS, and thus superfluous there. IfMS is in normal form, then itsOCipart would never have such a row.
(64) For a tableauM and its corresponding set of tableaux σM, a ranking is true inM iff it is true in all N ∈σM.
Proof. Suppose some φ is true in M, but is false in some tableau N ∈ σM. But the normal form ofN is a subset of MEn, and thus if φ is false inN, it is also false in MEn, and thus in M, contrary to assumption. In the other direction, M itself is in σM, so if a ranking is false inM, it cannot be true in all tableaux in σM. ⊣
(65) σ set completeness theorem
For a tableau M, there is no tableau where all rankings true in M are true which is not in σM.
Proof. Suppose towards a contradiction there is a tableauN such that all rankings true inM are also true in N, and yetN is not inσM. Then there must be a row r∈N which is not inMEn, and theM-maximal set of rankingsSM must be true in that r as it is true inN as a whole by assumption.
Consider that fixedr. Without loss of generality, we can assume r has a single L. Any ranking inSM must put one of the constraints inW(r)on top of the constraintCiinL(r). If there is a single W in r, then the corresponding atomic ranking putting that W on top of the L must be in the core ranking of SM. But thenM contains precisely the same rowr, contrary to assumption.
If there are multiple W-s in r, any ranking in SM must put one of those W-s on top of that L in Ci. If it is the same W in any ranking in SM, that means r is entailed by a row in the representative tableau of the core ranking ofSM. If it is different W-s, then the SCi set recovered fromSM minus its core ranking must be such that each alternative in it contains an atomic ranking putting one of the W-s inW(r) on top of Ci. If so, then the correspondingOCi contains a row entailingr.
Thus there can be no row not in MEn where all the rankings true in M were true.
Therefore anyN where all rankings true in M are true is bound to be inσM. ⊣
(66) Proper set S is the minimal proper representative SM inof its equivalence classC. Proof. For an arbitrary proper set S, there exists a set of tableauxσMS which contains all and only tableaux S is true in. Any set of rankings in the equivalence class C of S is true in exactly the same set of tableauxσMS.
We compare the set S′ of all total refinements ofS and the ST ot representative of C (see 35). It cannot be that (S′∖ST ot) ≠ ∅, for then ST ot were not the largest set in C containing only total rankings. It cannot also be that(ST ot∖S′) ≠ ∅: if that were the case, there would be a rankingφ∈ST ot true in all tableaux inσMS yet not inS′. In particular, suchφwould have to be true inMS. But then it would be a total refinement of one of the rankings inS, and thus included inS′, contrary to assumption. Thus S′=ST ot.
After having established that S’s set of total refinements is ST ot, we consider SM in (see 37). SM incontains all the rankings for which all total refinements are in ST ot. Those rankings are precisely the rankings maximal for the tableauMS, by definition of tableau-maximal rankings. Furthermore, SM in does not contain any other rankings than MS -maximal ones. HenceS=SM in.⊣
(67) A minimal proper representative SM in is always a proper set of rankings.
Proof. Suppose towards a contradiction there is some SM in set which is not proper.
Fix the set of tableaux in whichSM in is true asτ. It is clear thatτ cannot be theσM set of tableaux for anyM, for otherwise SM inwould have been proper.
We gather all the rows present in at least one normal form tableau in τ into a single tableau Mτ (which, in the general case, would not be a normal form tableau). Clearly SM inis true in Mτ, and Mτ itself is inτ.
The set ST ot of all total refinements of SM in is the set of all total rankings true in Mτ. Therefore SM in is MτN o-maximal. But thenτ is after all the set σMτN o, contrary to assumption.⊣
(69) In each equivalence class C, there exists exactly one set of rankings being the min-imal proper extension of any other set.
Proof. Each equivalence class of sets of rankings has its minimal proper representative SM in. As SM in is proper by 67, it is equal to its own minimal proper extension by 59.
Hence there always exists oneSP+per class.
For uniqueness, suppose there are two distinct minimal proper extensions in some C. By 59, both must be proper. Then by 66, both are equal to the SM in representative of classC. But then they are equal, contrary to assumption. ⊣
(70) For any set of rankings S and its minimal proper extension SP+, S and SP+ are equivalent sets of rankings.
Proof. Suppose towards contradiction that were not so. Fix some SandSP+which are not equivalent. By construction ofSP+,S ⊆SP+, and thus SP+⊧S. So it must be that S /⊧SP+.
If S /⊧SP+, it must be witnessed by some tableauM whereS is true, butSP+ is not.
We fix a particular row r of M whereS, but not SP+, is true. W.l.o.g., we assume r has a single L in Ci. First, r cannot be true in S by virtue of being accounted for by SCo, for SP+ has the same core. Second, suppose r is accounted for by the non-core part of S. Then in eachφ∈S, there is an atomic ranking accounting for the L in Ciin r. Each rankingψ∈SP+ contains the partφCij from some φj∈S. Therefore each ψhas to account
We fix a particular row r of M whereS, but not SP+, is true. W.l.o.g., we assume r has a single L in Ci. First, r cannot be true in S by virtue of being accounted for by SCo, for SP+ has the same core. Second, suppose r is accounted for by the non-core part of S. Then in eachφ∈S, there is an atomic ranking accounting for the L in Ciin r. Each rankingψ∈SP+ contains the partφCij from some φj∈S. Therefore each ψhas to account