Conjoint axiomatization of the Choquet integral for heterogeneous product sets
aa r X i v : . [ q -f i n . E C ] M a r Conjoint axiomatization of the Choquet integral forheterogeneous product sets
Mikhail TimoninMarch 28, 2018
Abstract
We propose an axiomatization of the Choquet integral model for the generalcase of a heterogeneous product set X = X × . . . × X n . In MCDA elementsof X are interpreted as alternatives, characterized by criteria taking values fromthe sets X i . Previous axiomatizations of the Choquet integral have been givenfor particular cases X = Y n and X = R n . However, within multicriteria contextsuch identicalness, hence commensurateness, of criteria cannot be assumed a priori.This constitutes the major difference of this paper from the earlier axiomatizations.In particular, the notion of “comonotonicity” cannot be used in a heterogeneousstructure, as there does not exist a “built-in” order between elements of sets X i and X j . However, such an order is implied by the representation model. Our approachdoes not assume commensurateness of criteria. We construct the representationand study its uniqueness properties. The Choquet integral is widely used in decision analysis and, in particular, MCDAGrabisch and Labreuche [2008], although its use is still somewhat restricted due to bothmethodological problems and difficulties in practical implementation. Rank-dependentmodels first appeared in the axiomatic decision theory in reply to the criticism of Savage’spostulates of rationality Savage [1954]. The renowned Ellsberg paradox Ellsberg [1961]has shown that people can violate Savage’s axioms and still consider their behaviourrational. First models accounting for the so-called uncertainty aversion observed in thisparadox appeared in the 1980s, in the work Quiggin [1982] and others (see Wakker [1991b]for a review). One particular generalization of the expected utility model (EU) charac-terized by Schmeidler Schmeidler [1989] is the Choquet expected utility (CEU), whereprobability is replaced by a non-additive set function (called capacity) and integration isperformed using the Choquet integral.Since Schmeidler’s paper, various versions of the same model have been characterizedin the literature (e.g. Gilboa [1987], Wakker [1991a]). CEU has gained some momentumin both theoretical and applied economic literature, being used mainly for analysis ofproblems involving Knightian uncertainty. At the same time, rank-dependent models, inparticular the Choquet integral, were adopted in multiattribute utility theory (MAUT)1eeney and Raiffa [1976]. Here the integral gained popularity due to the tractability ofnon-additive measures in this context (see Grabisch and Labreuche [2008] for a review).The model permitted various preferential phenomena, such as criteria interaction, whichwere impossible to reflect in the traditional additive models.The connection between MAUT and decision making under uncertainty has beenknown for a long time. In the case when the number of states is finite, which is as-sumed hereafter, states can be associated with criteria. Accordingly, acts correspond tomulticriteria alternatives. Finally, the sets of outcomes at each state can be associatedwith the sets of criteria values. However, this last transition is not quite trivial. It iscommonly assumed that the set of outcomes is the same in each state of the world Savage[1954], Schmeidler [1989]. In multicriteria decision making the opposite is true. Indeed,consider preferences of consumers choosing cars. Each car is characterized by a numberof features (criteria), such as colour, maximal speed, fuel consumption, comfort, etc. Ap-parently, sets of values taken by each criterion can be completely different from those ofthe others. In such context the ranking stage of rank-dependent models, which in decisionunder uncertainty involves comparing outcomes attained at various states, would amountto comparing colours to the level of fuel consumption, and maximal speed to comfort.Indeed, the traditional additive model Debreu [1959], Krantz et al. [1971] only impliesmeaningful comparability of units between goods in the bundle, but not of their absolutelevels. However, in rank-dependent models such comparability seems to be a necessarycondition.We propose a representation theorem for the Choquet integral model in the MCDAcontext. Binary relation < is defined on a heterogeneous product set X = X × . . . × X n .In multicriteria decision analysis (MCDA), elements of the set X are interpreted as al-ternatives, characterized by criteria taking values from sets X i . Previous axiomatiza-tions of the Choquet integral model have been given for the special cases of X = Y n (see K¨obberling and Wakker [2003] for a review of approaches) and X = R n (seeGrabisch and Labreuche [2008] for a review). One related result is the recent axiomatiza-tion of the Sugeno integral model (Greco et al. [2004], Bouyssou et al. [2009]). Anotherapproach using conditions on the utility functions was proposed in Labreuche [2012]. The“conjoint” axiomatization of the Choquet integral for the case of a general X was an openproblem in the literature. The crucial difference with the previous axiomatizations is thatthe notion of “comonotonicity” cannot be used in the heterogeneous case, due to the factthat there does not exist a meaningful “built-in” order between elements of sets X i . Newaxioms and modifications of proof techniques had to be introduced to account for that.Our first axiom shows, roughly, how the set X can be partitioned into subsets basedon properties necessary for existence of an additive representation. The axiom ( A3 )we introduce is similar to the “2-graded” condition previously used for characterizingof MIN/MAX and the Sugeno integral (Greco et al. [2004], Bouyssou et al. [2009]). Atevery point z ∈ X for every pair of coordinates i, j ∈ N it is possible to build two“rectangular cones” - one made up of points from X i which are “greater” than z i andpoints from X j which are “less” than z j , and the second for the opposite case. The axiomstates that triple cancellation for < restricted to i, j must then hold on at least one ofthese cones. This allows to partition X into subsets by using intersection of such conesfor various pairs i, j .The second property is that the additive representations on different subsets are inter-2elated, in particular “trade-offs” between criteria values are consistent across partitionelements both within the same dimension and across different ones. This is reflected bytwo axioms ( A4, A5 ), similar to the ones used in Wakker [1991a] and Krantz et al. [1971](section 8.2). One, roughly speaking, states that triple cancellation holds across subsets,while the other says that ordering of intervals on any dimension must be preserved whenthey are projected onto another dimension by means of equivalence relations. Theseaxioms are complemented by a new condition called bi-independence ( A6 ) and weak sep-arability ( A2 ) Bouyssou et al. [2009] - which together reflect the monotonicity propertyof the integral, and also the standard essentiality, “comonotonic” Archimedean axiomand restricted solvability ( A7,A8,A9 ). Finally, < is supposed to be a weak order ( A1 ),and X is order dense. Definition 1.
Let N = { , . . . , n } be a finite set and N its power set. Capacity (non-additive measure, fuzzy measure) is a set function ν : 2 N → R + such that:1. ν ( ∅ ) = 0 ;2. A ⊆ B ⇒ ν ( A ) ≤ ν ( B ) , ∀ A, B ∈ N .In this paper, it is also assumed that capacities are normalized, i.e. ν ( N ) = 1 . Definition 2.
The Choquet integral of a function f : N → R with respect to a capacity ν is defined as C ( ν, f ) = ∞ Z ν ( { i ∈ N : f ( i ) ≥ r } ) dr + Z −∞ [ ν ( { i ∈ N : f ( i ) ≥ r } ) − dr Denoting the range of f : N → R as { f , . . . , f n } , the definition can be written down as: C ( ν, ( f , . . . , f n )) = n X i =1 ( f ( i ) − f ( i − ) ν ( { j ∈ N : f j ≥ f ( i ) } ) where f (1) , . . . , f ( n ) is a permutation of f , . . . , f n such that f (1) ≤ f (2) ≤ · · · ≤ f ( n ) , and f (0) = 0 . On of the most useful tools for analysis of the capacity is the so-called M¨obius trans-form. It’s a linear transformation of the capacity which is given by: m ( A ) = X B ⊂ A ( − | A \ B | ν ( B ) . The Choquet integral can be written in a very convenient form using the M¨obiustransform coefficients: C ( ν, f ) = X A ∈ N m ( A ) min i ∈ A ( f i ) . .1 The model Let < be a binary relation on the set X = X × . . . × X n . ≻ , ≺ , , ∼ , are defined inthe usual way. In MCDA, elements of set X are interpreted as alternatives characterizedby criteria from the set N = { , . . . , n } . Set X i contains criteria values for criterion i .We say that < can be represented by a Choquet integral, if there exists a capacity ν andfunctions f i : X i → R , called value functions, such that: x < y ⇐⇒ C ( ν, ( f ( x ) , . . . , f n ( x n )) ≥ C ( ν, ( f ( y ) , . . . , f n ( y n )) . As seen in the definition of the Choquet integral, its calculation involves comparisonof f i ’s to each other. It is not immediately obvious how this operation can have anymeaning in the MCDA decision framework. It is well-known that direct comparison ofvalue functions for various attributes is meaningless in the additive model Krantz et al.[1971] (recall that the origin of each value function can be changed independently). Inthe homogeneous case X = Y n this problem is readily solved, as we have a single set of“consequences” Y (in the context of decision making under uncertainty). The requiredorder is either assumed as given Wakker [1991b] or is readily derived from the orderingof “constant” acts ( y, . . . , y ) Wakker [1991a]. Since there is a single “consequence” set,we also only have one value function U : Y → R , and thus comparing U ( y i ) to U ( y j ) isperfectly sensible, since U represents the order on the set Y . None of these methods canbe readily applied in the heterogeneous case. Below are given some important properties of the Choquet integral:1. Functions f : N → R and g : N → R are comonotonic if for no i, j ∈ N holds f ( i ) > f ( j ) and g ( i ) < g ( j ). For all comonotonic f the Choquet integral reducesto a usual Lebesgue integral. In the finite case, the integral is accordingly reducedto a weighted sum.2. Particular cases of the Choquet integral (e.g. Grabisch and Labreuche [2008]). • If m ( { } ) = . . . = m ( { n } ) = 1, then C ( ν, ( f , . . . , f n )) = max( f , . . . , f n ). • If m ( N ) = 1 , m ( A ) = 0 , A = N , then C ( ν, ( f , . . . , f n )) = min( f , . . . , f n ). • If m ( A ) = 0, for all A ⊂ N : | A | ≥
2, then C ( ν, ( f , . . . , f n )) = P i ∈ N ν ( { i } ) f i Property 1 states that the set X can be partitioned into subsets corresponding toparticular ordering of the value functions. There are n ! such sets. Since the integral oneach of the sets is reduced to a weighted sum, i.e. an additive representation, we shouldexpect many of the axioms of the additive conjoint model to be valid on this subsets.This is the intuition behind several of the axioms given in the following section.4 Axioms and definitions
Definition 3.
Given i, j ∈ N , a relation < on X × . . . × X n satisfies ij -triple cancellation( ij-3C ) , if for all a i , b i , c i , d i ∈ X i , p j , q j , r j , s j ∈ X j , and all z − ij ∈ X − ij holds: a i p j z − ij b i q j z − ij a i r j z − ij < b i s j z − ij c i p j z − ij < d i q j z − ij ⇒ c i r j z − ij < d i s j z − ij . A1 - Weak order. < is a weak order. A2 - Weak separability.
For all i , if a i x − i ≻ b i x − i for some a i , b i ∈ X i , x − i ∈ X − i , then a i y − i < b i y − i for all y − i ∈ X − i .Note, that from this follows, that for any a i , b i ∈ X i either a i x − i < b i x − i or b i x − i < a i x − i for all x − i ∈ X − i . This allows to introduce the following definition: Definition 4.
For all a i , b i ∈ X i define < i as a i < i b i ⇐⇒ a i x − i < b i x − i for all x − i ∈ X − i . Definition 5.
For any z ∈ X define SE zij = { x i x j z − ij ∈ X : x i < i z i , z j < j x j } , and NW zij = { x i x j z − ij ∈ X : z i < i x i , x j < j z j } . A3 - Coordinate Ordering Completeness.
For any z ∈ X , and all i, j ∈ N , ij -triplecancellation holds either on SE zij or on NW zij .This new property would allow us to divide X into subsets without the need to use thenotion of comonotonicity. We can introduce the following binary relations: Definition 6.
We write:1. i R z j if ij -triple cancellation holds on the set SE zij .2. i S z j if [NOT j R z i ].3. i E z j if [ i R z j AND j R z i ]. Note that R z is complete (which is why we have called axiom A3 “Coordinate Or-dering Completeness”) and S z is partial. Since N is finite, there is only a finite numberof various partial orders S z , so we can index them ( S a , S b , . . . ) and drop the superscriptswhen not needed. Also, each of the partial orders S k uniquely defines the corresponding R k - i R k j if [NOT j S k i ].In contrast to the case with two variables, this property alone is not sufficient toconstruct a representation. Comparing value functions for different attributes suggestssome sort of transitivity. For example, f i ( x i ) > f j ( x j ) and f j ( x j ) > f k ( x k ) imply f i ( x i ) >f k ( x k ). The property we introduce is weaker - it is acyclicity. if it is empty for all z , other axioms entail the existence of an additive representation on X For all z ∈ X , S z is acyclic. Inother words, i S z j S z . . . S z k ⇒ i R z k. This axiom effectively defines how the set X is partitioned. It is required for the Choquetintegral representation to exist.We also introduce the following notions: Definition 7.
Define SE ij as a union of the following three sets: • All z ∈ X such that i R z j , if z i is not maximal and z j is not minimal; • All z ∈ X such that z i is maximal and for no x j , y j ∈ X j : z j < j x j < j y j we have j R x j z − j i and NOT j R y j z − j i ; • All z ∈ X such that z j is minimal and for no x i , y i ∈ X i : y i < i x i < i z i we have j R x i z − i i and NOT j R y i z − i i .Define NW ij as a union of the following three sets: • All z ∈ X such that j R z i , if z j is not maximal and z i is not minimal; • All z ∈ X such that z i is minimal and for no x j , y j ∈ X j : y j < j x j < j z j we have i R x j z − j j and NOT i R y j z − j j ; • All z ∈ X such that z j is maximal and for no x i , y i ∈ X i : z i < i x i < i y i we have i R x i z − i j and NOT i R y i z − i j . Presence of maximal and minimal points significantly complicates the definitions of SE ij and NW ij , since at such points some of the sets SE zij and NW zij become degenerateand condition trivially holds. If sets X i and X j do not contain minimal or maximalpoints, we can drop the corresponding conditions in each definition and simply state that SE ij = { z : i R z j } and NW ij = { z : j R z i } .Partial orders S i define subsets of the set X as follows. Definition 8.
We write X S i = T ( k,j ): k R i j SE kj It is well known that the sufficient property for an additive representation to exist ona Cartesian product is strong independence Krantz et al. [1971]. In the X = Y n case, theChoquet integral was previously axiomatized using comonotonic strong independence (orcomonotonic trade-off consistency Wakker [1991a]). In this paper we will be using sets X S i to formulate a similar condition. Definition 9.
We say that i ∈ N is essential on A ⊂ X if there exist x i x − i , y i x − i ∈ A ,such that x i x − i ≻ y i x − i . A4 - Intra-coordinate trade-off consistency a i x − i b i y − i a i w − i < b i z − i c i x − i < d i y − i ⇒ c i w − i < d i z − i , provided that either: 6) Exists X S j such that a i x − i , b i y − i , a i w − i , b i z − i , c i x − i , d i y − i , c i w − i , d i z − i ∈ X S j b) Exist X S j , X S k such that a i x − i , b i y − i , a i w − i , b i z − i ∈ X S j , i is essential on X S j ,and c i x − i , d i y − i , c i w − i , d i z − i ∈ X S k , or;c) Exist X S j , X S k such that a i x − i , b i y − i , c i x − i , d i y − i ∈ X S j , i is essential on X S j ,and a i w − i , b i z − i , c i w − i , d i z − i ∈ X S k .Informally, the meaning of the axiom is that ordering between preference differences(“intervals”) is preserved irrespective of the “measuring rods” used to measure them.However, contrary to the additive case this does not hold on all X , but only when eitherpoints involved in all four relations lie in the same “3C-set” X S j , or points involved intwo relations lie in one such set and those involved in the other two in another. A5 - Inter-coordinate trade-off consistency a i x − i b i y − i c i x − i < d i y − i a i y − i ∼ p j x − j b i y − i ∼ q j x − j c i y − i ∼ r j x − j d i y − i ∼ s j x − j p j e − j < q j f − j ⇒ r j e − j < s j f − j for all a i x − i , b i y − i , c i x − i , d i y − i ∈ X S j provided i is essential on X S j , a i y − i , b i y − i , c i y − i , d i y − i ∈ X S k , p j x − j , q j x − j , r j x − j , s j x − j ∈ X S l provided j is es-sential on X S l , p j e − j , q j f − j , r j e − j , s j f − j ∈ X S m .The formal statement of the A5 is rather complicated, but it simply means that theordering of the “intervals” is preserved across dimensions. Together with A4 the con-ditions are similar to Wakker’s trade-off consistency condition Wakker [1991b] . Theaxiom bears even stronger similarity to Axiom 5 (compatibility) from section 8.2.6 ofKrantz et al. [1971]. Roughly speaking, it says that if the “interval” between c i and d i is “larger” than that between a i and b i , then “projecting” these intervals onto anotherdimension by means of the equivalence relations must leave this order unchanged. Weadditionally require the comparison of intervals and “projection” to be consistent - mean-ing that each quadruple of points in each part of the statement belongs to the same X S i .Another version of this axiom, which is used frequently in proofs, can be formulated interms of standard sequences (Lemma 15). A6 - Bi-independence
Let a i x − i , b i x − i , c i x − i , d i x − i ∈ X S i and a i x − i ≻ b i x − i . If forsome y − i ∈ X − i we have c i y − i ≻ d i y − i , then c i x − i ≻ d i x − i for all i ∈ N .This axiom is similar to “strong monotonicity” in Wakker [1991b]. We analyze its neces-sity and the intuition behind it in section A.2. A7 - Essentiality
All coordinates are essential on X .7 If a i x − i < y < b i x − i , then there exists c : c i x − i ∼ y for i ∈ N . A9 - Archimedean axiom
Every bounded standard sequence contained in some X S i is finite, and in the case of only one essential coordinate, there exists a countableorder-dense subset of X S i .Finally, we can introduce a notion of interacting coordinates. Definition 10.
Coordinates i and j are interacting if exists z ∈ X , such that i S z j or j S z i . We call a set A ⊂ N an interaction clique if for each i, j ∈ A we can build achain of coordinates i, k, . . . , j , such that every two subsequent coordinates in the chainare interacting. Interaction cliques play an important role in the uniqueness properties of the repre-sentation. In what follows we will be considering only cliques of maximal possible size ifnot specified otherwise.
The following additional assumptions are made. The reasoning behind each one is ex-plained below. They are not required for the construction of the representation in general. “Collapsed” equivalent points along dimensions.
For no i ∈ N and no a i , b i ∈ X i holds a i x − i ∼ b i x − i for all x − i ∈ X − i .If this wasn’t true, we could have value functions assigning the same value to severalpoints in the same set X i . To simplify things we exclude such case, however, it can beeasily reconstructed once the representation is built. Density.
We assume that for all i ∈ N , whenever a i x − i ≻ b i x − i , there exists c i ∈ X i suchthat a i x − i ≻ c i x − i ≻ b i x − i ( X is order dense). “Closedness”. For every i and j , if there exist x i x j z − ij such that i S x i x j z − ij j and y i x j z − ij such that j S y i x j z − ij i , then exists z i ∈ X i such that i E z i x j z − ij j .This assumption says that sets SE ij and NW ij are “closed”. In the representationthis translates into existence of the inverse for all points where value functions f i and f j are equal, provided i and j are interacting. This is a technical simplifying assumptionand the proof can be done without it. Geometry of X . For every clique of interacting variables A ⊂ N , there exist at leasttwo points r A , r A ∈ X such, that for every pair i, j ∈ A , we have i E r A j and i E r A j .Again, this is a simplifying assumption, making the proof somewhat less general andcloser to the homogeneous case. Without it we can have a situation, where the smallestvalue of f i : X i is larger then the greatest value of f j : X j for some i, j ∈ N . This in turndoes not allow to construct the capacity in a unique way. Another way to stating thisassumption, is to say that X must contain points corresponding to all possible acyclicpartial orders on N , generated by interacting pairs i S j . Work to remove this assumptionis still in progress. 8 Representation theorem
As follows from the definition of the Choquet integral (Section 2), every point x ∈ X uniquely corresponds to a set of weights p xi : p xi ≥ , P i ∈ N p xi = 1. This notation is usedto simplify the statement of the following theorems. Theorem 1.
Let < be an order on X and the structural assumption hold. Then, ifaxioms A1 - A9 are satisfied, there exists a capacity ν and value functions f : X → R , . . . , f n : X n → R , such that < can be represented by the Choquet integral: x < y ⇐⇒ C ( ν, ( f ( x ) , . . . , f n ( x n ))) ≥ C ( ν, ( f ( y ) , . . . , f n ( y n ))) , (1) for all x, y ∈ X . Capacity and value functions have the following uniqueness properties. Let I = { A , . . . , A k } be a partition of N , such that m ( B ) = 0 for all B ⊂ N such that B ∩ A i = ∅ , B ∩ A j = ∅ . If no such partition exists, let I = { N } . Theorem 2.
Let g : X → R , . . . , g n : X n → R be such that (1) holds with f i substitutedby g i . Then, at all x i ∈ X i , such that for some z − i we have p x i z − i i > , and also p x i z − i j > , j = i , value functions f i and g i are related in the following way: f i ( x i ) = α A j g i ( x i ) + β A j , Capacity changes as follows m ′ ( B ) = α A j m ( B ) P C ⊂ A i ,A i ∈I α A i m ( C ) . At the remaining points of X , i.e. for x i such that for any z − i ∈ X − i we have p x i z − i i = 1 , and p x i z − i j = 0 for all j = i , value functions f i have the following uniquenessproperties : f i ( x i ) = ψ i ( g i ( x i )) , where ψ i is an increasing function, and for all j ∈ N, j = i , such that exists A ∈ N : i, j ∈ A, m ( A ) > , we additionally have f i ( x i ) = f j ( x j ) ⇐⇒ g i ( x i ) = g j ( x j ) . . X S a We start by removing maximal and minimal elements from the sets X i . The representa-tion will be extended to these points in Section 8.Similar to [Wakker, 1991b] we will be covering the sets X S a with “rectangular” subsets.Given a point z ∈ X S a we construct a “rectangular” set X z ( S a ) in the following way: Due to our assumption that for no a i , b i we have a i z − i ∼ b i z − i for all z − i , we can’t have p x i z − i = 0for all z − i . If j is minimal in S a , then X z ( S a ) j = x j ∈ X j : z j < j x j . • If j is maximal in S a , then X z ( S a ) j = x j ∈ X j : x j < j z j . • If j is neither maximal not minimal, then X z ( S a ) j = [ x j ∈ X j : x j < j z j , x j z − j ∈ X S a ]. • If for no k we have j S a k or k S a j , then X z ( S a ) j = X j . X z ( S a ) We assume that X S a has at least two essential coordinates. By Lemma 18, all sets X z ( S a ) therefore have at least two essential coordinates. Moreover, the essential coordinates arethe same across all sets. Theorem 3.
For any z ∈ X S a there exists an additive representation of < on X z ( S a ) : x < y ⇔ n X i =1 V zi ( x i ) ≥ n X i =1 V zi ( y i ) , for all x, y ∈ X z ( S a ) .Proof. X z ( S a ) is a Cartesian product, < is a weak order on X z ( S a ) , < satisfies generalizedtriple cancellation on X z ( S a ) , < satisfies Archimedean axiom on X z ( S a ) , at least twocoordinates are essential. It remains to show that < satisfies restricted solvability on X z ( S a ) .Assume that for some x i z − i , w, y i z − i ∈ X z ( S a ) , we have x i z − i < w < y i z − i , hence exists z i ∈ X i : z i z − i ∼ w . We need to show that z i z − i ∈ X S a . If w ∼ x i z − i or w ∼ y i z − i ,then the conclusion is immediate (since either point belongs to X z ( S a ) ). Hence, assume x i z − i ≻ z i z − i ≻ y i z − i . This means that x i < i z i < i y i . Since z i is “sandwiched” between x i and y i we conclude that for any j ∈ N \ i , i S x i z − i j and i S y i z − i j imply also i S z i z − i j , andsymmetrically j S x i z − i i and j S y i z − i i imply j S z i z − i i . Hence, it is also in X S a .Therefore all conditions for the existence of an additive representation are met[Wakker, 1991a]. V S a on X S a This section is based on [Wakker, 1991b] with some modifications.
Theorem 4.
There exists an additive interval scale V S a ( z ) = P ni =1 V S a i ( z i ) on X S a ,which represents < on every X z ( S a ) with z ∈ X S a .Proof. Choose the reference set - pick any r ∈ X S a such that X r ( S a ) i contains more thantwo points for any X S a -essential i . Choose a “zero” point - any r ∈ X r ( S a ) , and a “unitmark” - a point r k r − k ∈ X r ( S a ) , such that: • k is essential on X S a , • r k < k r k . 10et V ri ( r i ) = 0 for all i ∈ N and V rk ( r k ) = 1. This uniquely defines unit and locations ofall V ri , i ∈ N .In the following we assume that sets X z ( S a ) i , X z ( S a ) k each contain at least two points,otherwise, alignment is trivial.Assume X r ( S a ) i ∩ X z ( S a ) i = ∅ and X r ( S a ) k ∩ X z ( S a ) k = ∅ (variations are all coveredby the below procedure). We will construct two auxiliary points z ′ and r ′ such that X z ′ ( S a ) i ⊂ X z ( S a ) i , X z ′ ( S a ) i ⊂ X r ( S a ) i , X r ′ ( S a ) k ⊂ X z ( S a ) k , X r ′ ( S a ) k ⊂ X r ( S a ) k . It would allow usto align first V r and V r ′ , then V r ′ and V z ′ , and finally V z ′ and V z .Construct the point z ′ by taking coordinate-wise maxima of r and z for coor-dinates j such that j R a i , not including i itself, and coordinate-wise minima of r and z for coordinates j , such that i R a j and i itself. In the short notation thefirst point is z ′ := max( r j , z j ) j : j R a i min( r j , z j ) j = i,j : i R a j . The second point r ′ is con-structed by taking coordinate-wise maxima of r and z for coordinates j such that j R a k , not including k itself, and coordinate-wise minima of r and z for coordinates j , such that k R a j and k itself. In the short notation the second point looks like r ′ := max( r j , z j ) j : j R a k min( r j , z j ) j = k,j : k R a j .Note that both points are in X S a since relations j R a l remain intact for all pairs j, l .Note also, that X z ′ ( S a ) i contains both X z ( S a ) i and X r ( S a ) i , and X r ′ ( S a ) k contains both X z ( S a ) k and X r ( S a ) k .Now we have that sets ( X z ( S a ) i × X z ( S a ) k ) ∩ ( X z ′ ( S a ) i × X z ′ ( S a ) k ), ( X z ′ ( S a ) i × X z ′ ( S a ) k ) ∩ ( X r ′ ( S a ) i × X r ′ ( S a ) k ), ( X r ′ ( S a ) i × X r ′ ( S a ) k ) ∩ ( X r ( S a ) i × X r ( S a ) k ) are all non-empty, and each di-mension contains more than two points. Relation < i,k on these sets satisfies Archimedeanaxiom, restricted solvability, and A4 . Hence we can apply standard uniqueness propertiesof additive representations. We first align V r ′ k with V rk and V r ′ i with V ri , then V z ′ k with V r ′ k and V z ′ i with V r ′ i , and finally V zk with V z ′ k and V zi with V z ′ i by changing the commonunit and locations of corresponding value functions.Having aligned like this V zi and V zk with V ri and V rk for all z ∈ X S a we can performthe same alignment operation for all remaining essential coordinates j , using pairs V zj and V zk . At this stage, functions V zk are already aligned, hence have a correct unit andlocation. As above, uniqueness properties of additive representations of relation < j.k imply that the unit of functions V zj is already aligned with that of V rj and only locationchange has to be performed. This can also be done as above.Once such alignment has been performed for all essential coordinates, we can verifythat this is done consistently throughout X S a . In particular, for any s and t from X S a wemust be able to show that for any essential j ∈ N , we have V sj = V tj on X s ( S a ) j ∩ X t ( S a ) j .To show this a following argument can be used. During the initial alignment of V sj and V tj , auxiliary points t ′ and s ′ were used, such that X s ′ ( S a ) j includes X s ( S a ) j and X r ( S a ) j ,and X t ′ ( S a ) j includes X t ( S a ) j and X r ( S a ) j . Hence, functions V s ′ j and V t ′ j coincide with V rj on X r ( S a ) j . To show that they coincide on all common domain, including X s ( S a ) j ∩ X t ( S a ) j ,we just need to follow the same procedure as before and construct a point that contains X s ′ ( X a ) k and X t ′ ( X a ) k for some essential k . Then a uniqueness argument can be evoked onceagain, and since V s ′ j and V t ′ j coincide on X r ( S a ) j , they would necessarily coincide also onthe remaining common domain, which includes X s ( S a ) j ∩ X t ( S a ) j . Finally, since V sj = V s ′ j on X s ( S a ) j , and V tj = V t ′ j on on X t ( S a ) j , we get that V sj = V tj on X s ( S a ) j ∩ X t ( S a ) j .11t this point we can drop the superscripts and define functions V S a i which coincidewith V z ( S a ) i for all z ∈ X S a on the corresponding domains. By the above argument, thesefunctions are well-defined. V S a is globally representing on X S a Lemma 1.
For all X S a -essential i ∈ N , V S a i represents < i on X S a i .Proof. Let α i , β i ∈ X S a i be such that α i < i β i . Similarly to the construction of r ′ and z ′ inthe proof of theorem 4, we can show that always exists x i such that α i x − i , β i x − i ∈ X S a .The conclusion follows. Theorem 5.
Representation V S a obtained in Theorem 4 is globally representing on X S a .Proof. We need to show that x < y ⇐⇒ V S a ( x ) ≥ V S a ( y ). • If exists z such that x, y ∈ X z ( S a ) then the result is immediate. • If the above is not true, we will show that exists x ′ ∼ x such that V S a ( x ) = V S a ( x ′ )and x ′ i < i y i for all i .The procedure is identical to Wakker [1991a] with some minor modifications.1. Find i such that y i ≻ i x i and x k < k y k for all k such that k S a i . We have y i x − i ∈ X S a (since for all k ∈ N such that k S a i we have x k < y k , hence k R y i y − i i implies k R y i x − i i , whereas for all t ∈ N such that i S a t we have i R x i x − i t , hence i R y i x − i ).2. Similarly, find j such that x j ≻ j y j and y k < k x k for all k such that j S a k . By similarreasoning, y j x − j ∈ X S a .3. We are increasing x i and decreasing x j and thus move in the direction of y .4. Note, that x − ij y i y j ∈ X S a .5. If x − ij y i y j < x , then by restricted solvability ( x − ij y i y j < x < x − ij x i y j ) exists x ′ := x − ij x ′ i y j ∼ x , where y i < i x ′ i < i x i . If x ≻ x − ij y i y j , then by restricted solvability( x − ij y i x j < x ≻ x − ij y i y j ) exists x ′ := x − ij y i x ′ j ∼ x , and x j < j x ′ j < J y j .6. In both cases, the resulting point x ′ is in X S a , moreover x ′ , x ∈ X z ( S a ) where z := x ij x i y j , hence x ′ has the same V S a -value as x , but one more coordinate becomesidentical to that of y .7. After repeating the procedure unless x ′ i < i y i (at most n times), we get the resultby Lemma 1.8. Moreover, if x ∼ y , we at the end of the procedure we would necessarily arrive to y itself (by strong monotonicity as in Lemma 18, and structural assumption SA1 ).Hence we get x ≻ y ⇒ V S a ( x ) > V S a ( y ) and x ∼ y ⇒ V S a ( x ) = V S a ( y ), whichimplies that x < y ⇐⇒ V S a ( x ) ≥ V S a ( y )12 Aligning cardinal representations for different X S a There can be several cases depending on what variables are essential on various sets X S i .We start with the case where exist X S a and X S b having at least two essential variableseach. X S i with at least two essential coor-dinates Theorem 6.
Assume that at least two coordinates are essential on X S a and X S b . Forany i ∈ N that is essential on both areas, it holds V S a i ( z i ) = λ abi V S b i ( z i ) for all z i fromthe common domain of V S a i ( z i ) and V S b i ( z i ) , if a common location is chosen for bothfunctions.Proof. If the common domain of V S a i ( z i ) and V S b i ( z i ) is empty or contains just one point,the result is trivial. Assume that i, j are essential on X S a and i, l are essential on X S b . First, we will establish that a standard sequence on coordinate i in X S a is alsoa standard sequence in X S b (provided all points of the sequence lie within a commondomain of V S a i ( z i ) and V S b i ( z i )). This follows from A4 . Build any standard sequence X S a i , say { α ki : α ki v j x − ij ∼ α k +1 i w j x − ij } . Then, { α ki : α ki t l x − il ∼ α k +1 i u l x − il } is a standardsequence in S b , i.e. if exist t l , u l ∈ X l such that α ki t l x − il ∼ α k +1 i u l x − il for some k , then by A4 : α ki v j x − ij ∼ α k +1 i w j x − ij α ki t l x − il ∼ α k +1 i u l x − il α k +1 i v j x − ij ∼ α k +2 i w j x − ij ⇒ α k +1 i t l x − il ∼ α k +2 i u l x − il Pick two points r i and r i in the common domain and set V S a i ( r i ) = V S b i ( r i ) = 0.Assume we now have V S a i ( r i ) = v a and V S b i ( r i ) = v b . We need to show that for anypoint z i from the common domain of V S a i and V S b i we have V S a i ( z i ) = λ abi V S a i ( z i ), where λ abi = v b v a .Build standard sequences from r i to r i , and from r i to z i . We have V S a i ( r i ) − V S a i ( r i ) ≈ n [ V S a j ( v j ) − V S a j ( w j )] V S a i ( z i ) − V S a i ( r i ) ≈ m [ V S a j ( v j ) − V S a j ( w j )] .V S a i ( r i ) = 0, hence V S a i ( z i ) ≈ mV S a i ( r i ) n . Such n and m exist by the Archimedean axiom. By the argument above we get V S b i ( r i ) − V S b i ( r i ) ≈ n [ V S b j ( t l ) − V S b j ( u l )] V S b i ( z i ) − V S b i ( r i ) ≈ m [ V S b j ( t l ) − V S b j ( u l )] . Obviously, the common domain is not empty if we assume
SA4 , in which case r i and r i are in thecommon domain by assumption V S b i ( z i ) ≈ mV S b i ( r i ) n . By density, we can pick an arbitrary small step of the standard sequences, so the ratio mn converges to a limit. Thus, finally V S a i ( z i ) = V S a i ( r i ) V S b i ( r i ) V S b i ( z i ) = v a v b V S b i ( z i ) = α abi V S b i ( z i ) . We proceed by picking common locations for all value functions. Since r belongs toall X S a , we can set V ai ( r i ) = 0. At this point we can drop superscripts and say that wehave representations λ ai V i + . . . + λ an V n on each X S a , defining also λ ai := 0 for variables i that are inessential on the set X S a . By assumption, we have two points two points r and r such that i E r j and i E r j forevery interacting i, j . From this follows, that both points belong to every X S i . We canassume that r i < i r i for all i ∈ N (for variables not interacting with others we can takeit to be so, for others see results in Section A.5). Set V i ( r i ) = 0 for all i ∈ N . Choosesome j ∈ N , such that j is essential on at least one X S a , which has two or more essentialvariables (including j ). Set V j ( r j ) = 1. This sets unit and location for all functions V i such, that i is essential on some X S a where at least one more coordinate is essential.For each such V i , we now have V i ( r i ) = k i (thus k j = 1). Define φ i := V i k i , for all i ∈ N .Additive representations on various X S a now have the form λ a k φ ( x )+ . . . + λ an k n φ n ( x n ).Finally, re-scale one more time by dividing everything by the sum of coefficients: λ a k P ni =1 λ ai k i φ ( x ) + . . . + λ an k n P ni =1 λ ai k i φ n ( x n )Denoting α aj = λ aj k j P ni =1 λ ai k i , we arrive to: φ a ( x ) := α a φ ( x ) + . . . + α an φ n ( x n ) , note that P ni =1 α i = 1.Note that here we set φ i ( r i ) = 0 and φ i ( r i ) = 1 for all i . As will be shown in theSection 9, this can be relaxed - origin and scaling factors can be chosen individually foreach clique. X At this stage we can show that representations φ a ( x ) = α a φ ( x ) + . . . + α an φ n ( x n ) assignthe same value to equivalence classes of < in all X S i . To simplify the construction in themain theorem of this section, we introduce the following lemma.14 emma 2. For every X S a and every z ∈ X S a , such that r ≻ z , we can find z ′ , such that z ′ ∼ z , z ′ ∈ X S a , and r i < i z i . Likewise, for every y ∈ X S a , such that y ≻ r , we can find y ′ , such that y ′ ∼ y , y ′ ∈ X S a , and y i < i r i .Proof. We can use the same procedure as was used in the proof of Theorem 5. For thecase y ≻ r the procedure is exactly the same, while for r ≻ z it is symmetric, as we aremoving z this time, and not r .Notice that as a result of the rescaling made in Section 6.2, points r and r have thesame values (0 and 1) in all X S a (since values of all φ i are equal, weights α ai sum up to1, and weights of all inessential variables are zero). Theorem 7.
Let each of X S a and X S b have at least two essential variables. Then forany x ∈ X S a , y ∈ X S b we have x < y iff φ a ( x ) < φ b ( y ) .Proof. First take x ∼ y , such that x ∈ X S a , and y ∈ X S b . If x ∼ y ∼ r or x ∼ y ∼ r , theconclusion is immediate, so assume otherwise. Let x ≻ r . Using Lemma 2 we construct x ′ ∈ X S a and y ′ ∈ X S b , such that x ∼ y ∼ x ′ ∼ y ′ and x ′ i < i r i , while y ′ i < i r i .Next, build equispaced sequences from r to r in X S a and X S b , such that first stepsof each sequence are equivalent (see details in Section A.3). By A5 the number of stepsin both sequences is equal.Finally, build sequences from r to x ′ and y ′ (coordinate-wise dominance simplifiesconstruction of the sequences). The number of steps again must be equal, hence theratios between the number of steps it takes to reach r and x ′ , and between the numberof steps it takes to reach r and y ′ are equal, and hence taking the limit, we get φ a ( x ) = φ a ( x ′ ) = φ b ( y ′ ) = φ b ( y ).The same approach applies for x ≻ y . By A5 the number of steps in the equispacedsequence from r to x must be greater, than in the sequence from r to y . Hence also φ a ( x ) > φ b ( x ).We now have x ∼ y ⇒ φ a ( x ) = φ b ( y ) and x ≻ y ⇒ φ a ( x ) > φ b ( y ). This implies that x < y ⇐⇒ φ a ( x ) ≥ φ b ( y ).At this point we can define value functions on the areas with a single essential variable.Theorem 7 establishes that all areas with two or more essential coordinates assign thesame value to points from the same equivalence class. Define the value assigned to anequivalence class belonging to an area with a single essential coordinate to be the same,as it has in some area with two or more essential coordinates. Such equivalence classesmust exist (e.g. those containing points r and r ). Finally, define α ai = 1 for the essentialcoordinate i and α aj = 0 for all other j . If after this procedure there remain points insome X i , for which φ i is not yet defined, then we have some equivalence classes to whichnone of the representations φ k assign any value. Since all equivalence classes found in thesets X S k , which have two or more essential variables, by now have a defined value, suchclasses are entirely within sets, that have only a single essential variable. Hence, we cantrivially extend the representations, and get also φ i ( x i ) > φ i ( y i ) iff x i < i y i (see Lemma 7which shows that functions are well-defined).15 emma 3. Given S a and S b , such that exists A ⊂ N , for which we have i R a j iff i R b j for all i ∈ A, j ∈ N \ A , the following is true X i ∈ A α ai = X i ∈ A α bi . Proof.
Consider r A r − A , which belongs both to X S a and X S b . By Theorem 7 and theabove construction of value functions for the areas with a single essential coordinate,we have φ a ( r A r − A ) = φ b ( r A r − A ), hence P i ∈ A α ai φ ( r i ) = P i ∈ A α bi φ ( r i ), from which theconclusion follows as φ i ( r i ) = 1 for all i ∈ N .Now we can proceed with construction of a unique capacity ν : 2 N → R from coeffi-cients α ai which exist on various X S a . As shown in Wakker [1989], the condition of Lemma3 is a necessary requirement for this. Capacity ν also has a unique M¨obius transform m : 2 N → R (see definition in Section 2.1).We can now construct a representation very similar to the Choquet integral. In orderto so, let us define the following function first: Φ ∧ ( x, A ) := φ i ( x i ) for i such that j R x i for all j ∈ A \ i , in case when this is true for several i , any can be chosen. We can nowconstruct a global value function (cf. Section 2.1): φ ( x ) := X A ∈ N m ( A )Φ ∧ ( x, A ) . (2)It is easy to see that for each x ∈ X and every X S i , such that x ∈ X S i , we have φ a ( x ) = φ ( x ). Now we can show that φ i ( x i ) = φ j ( x j ) whenever i E x j , providing i, j interact. Lemma 4.
For any non-extreme x ∈ X it holds: i E x j ⇒ φ i ( x i ) = φ j ( x j ) , unless i and j do not interact.Proof. Assume x ∈ X S a , x ∈ X S b such that kS a l whenever kS b l for all k, l ∈ N apartfrom i, j , for which we have iS a j and jS b i . By Theorem 7, φ a ( x ) = φ b ( x ) and by Lemma3, it is trivial to show that α ak = α bk for all k = i, j , and α ai + α aj = α bi + α bj .We have α ai φ i ( x i ) + α aj φ j ( x j ) = α bi φ i ( x i ) + α bj φ j ( x j ) (other sum components cancelout). Dividing by α ai + α aj = α bi + α bj , we get a convex combination of φ i ( x i ) and φ j ( x j )on both sides. From this follows that either φ i ( x i ) = φ j ( x j ) or α ai = α bi and α aj = α bj .Assume the latter. Repeating this operation for all possible combinations of X S k and X S l would lead us to the conclusion that m ( B ) = 0 for all B ⊃ { i, j } , as weights α ki , α li , α kj , α lj do not change when we move from X S k to X S l , and, accordingly, from φ k to φ l . The conclusion results from equation (2).Finally, we can show that this implies that i and j do not interact. This means that ij -triple cancellation - a i p j z − ij b i q j z − ij a i r j z − ij < b i s j z − ij c i p j z − ij < d i q j z − ij ⇒ c i r j z − ij < d i s j z − ij , a i , b i , c i , d i ∈ X i , p j , q j , r j , s j ∈ X j , and all z − ij ∈ X − ij . To show this, useequation (2) to write the values for all involved points, grouping the sum components asfollows. For example, for a i p j z − ij : φ ( a i p j z − ij ) = X A ⊃ iA j m ( A )Φ ∧ ( a i p j z − ij , A ) + X A ⊃ jA i m ( A )Φ ∧ ( a i p j z − ij , A )++ X A i,j m ( A )Φ ∧ ( a i p j z − ij , A ) + X A ⊃ i,j m ( A )Φ ∧ ( a i p j z − ij , A ) . Notice, that due to the above argument, we have P A ⊃ i,j m ( A )Φ ∧ ( a i p j z − ij , A ) =0. Also notice that P A ⊃ jA i m ( A )Φ ∧ ( a i p j z − ij , A ) does not depend on a i , and P A i,j m ( A )Φ ∧ ( a i p j z − ij , A ) does not depend on a i or p j .Assume, towards a contradiction, that d i s j z − ij ≻ c i r j z − ij . Writing values for all points,and summing the first and the third, and the second and the fourth inequalities gives: X A ⊃ iA j m ( A )[Φ ∧ ( a i p j z − ij , A ) + Φ ∧ ( d i p j z − ij , A )] ≤ X A ⊃ iA j m ( A )[Φ ∧ ( b i p j z − ij , A ) + Φ ∧ ( c i p j z − ij , A )] X A ⊃ iA j m ( A )[Φ ∧ ( a i p j z − ij , A ) + Φ ∧ ( d i p j z − ij , A )] > X A ⊃ iA j m ( A )[Φ ∧ ( b i p j z − ij , A ) + Φ ∧ ( c i p j z − ij , A )] , which is a contradiction. Hence, ij -triple cancellation holds for all a i , b i , c i , d i ∈ X i , p j , q j , r j , s j ∈ X j , and all z − ij ∈ X − ij , and thus i and j do not interact. Lemma 5.
If for some z ∈ X we have i S z j , then φ i ( z i ) > φ j ( z j ) .Proof. It is easy to verify that (due to the “Closedness” assumption), there exists x ij z − ij ,such that i E x ij z − ij j and z i < i x i , whereas x j < j z j . Since φ i represents < i , by Lemma 4,and the fact that < i is asymmetric (due to the structural assumption), we have φ i ( z i ) >φ i ( x i ) = φ j ( x j ) > φ j ( z j ).Now we have that [ i E x j ] ⇒ [ φ i ( x i ) = φ j ( x j )] for interacting i, j , and [ i S x j ] ⇒ [ φ i ( x i ) > φ j ( x j )], hence we conclude that [ i R x j ] ⇐⇒ [ φ i ( x i ) < φ j ( x j )] for all interacting i, j , which allows us to finally rewrite (2) as the Choquet integral: φ ( x ) = X A ⊂ N m ( A ) min i ∈ A f i ( x i ) . To summarize the results of this section we state the following lemma:
Lemma 6.
Let X ′ i := X \ { maximal and minimal elements of X } . Let X ′ := X ′ × . . . × X ′ n . Assume that at least one of the sets X ′ S a , defined as previously, has more than twoessential variables. Then for every x, y ∈ X ′ we have x < y ⇐⇒ φ ( x ) ≥ φ ( y ) . .1 Case with a single essential variable on every X S a For this case we only need A3 to construct the representation. Since < is a weak orderand each X S i has a countable order-dense subset, there exists a function F : X ⇒ R ,such that x < y ⇐⇒ F ( x ) ≥ F ( y ). To perform the construction of the value functionswe need the following lemma. Lemma 7.
Let x i z − i ∈ X S a and y i z − i ∈ X S b . Let also i be the only essential coordinateon X S a and X S b . Then, x i z − i ∼ y i z − i .Proof. The idea of the proof is to “trace a path” from X S a to X S b by constructing asequence of points and subsets X S j . We will keep x i unchanged, but will move theremaining coordinates, in order to show that each point in the sequence belongs to thecurrent and the subsequent subsets, moreover i will be the only essential coordinate onall X S j .There are several steps. We start with pairs of coordinates j, k such that i S a j, k , and j S a k but k S b j (or vice versa). Note that both j, k are inessential in X S a . Let j S a k ,and j, k be subsequent in S a , i.e. there is no m such that j S a m S a k . Such a pair mustobviously exist. Using a simple argument, we can show that we can construct a point x i z − i by changing z j and z k , such that k E x i z ′− i j . Call this new ordering S . By a densityargument, we can show that if both j and k were not essential on X S a , they will notbe essential on X S either. Essentiality of all other coordinates is also not affected, so i remains the only essential coordinate on X S . Note, that because k E x i z ′− i j , we have x i z − i ∈ X S a , x i z − i ∈ X S . Proceeding like this, we build a sequence x i z j − i until eventuallyall pairs j, k such that i S a j , i S a k are ranked in the same order as in S b . Using the sameapproach, we repeat it for j, k such that j S a i , k S a i . Eventually, the sequence ends withan order S n .It remains to switch the order of coordinates which change the relative position with i in S a and S b . By using the closedness structural assumption, we can construct a pointwhich will belongs both to X S n and X S b , and since i is the only essential coordinate onboth sets, we obtain the result.Using this lemma, we can now define value functions φ i ( x i ) = F ( x ) by picking x ∈ X S a where i is the essential coordinate. Lemma shows that the functions are well-defined. Itremains to construct a capacity. We can do so, by letting α ai = 1 for essential i and α aj = 0for the remaining coordinates. Points r and r can be used to show a result similar toLemma 3. This means, that there exist a capacity ν , and the preference relation canbe represented by a Choquet integral with respect to this capacity and value functions,defined as above.An alternative construction is given in Section A.6.18 Extending the representation to the extremepoints of X X i Due to a larger number of possible cases we can’t easily show that all value functions arebounded on X i as in Wakker [1991b]. Instead, we can show which functions are boundedon X S a i . There can be two cases in which we do not know if φ i is bounded on X S a i . Formaximal points these cases are: • max X S a i = max X i , or • M ai := max X S a i and M ai z − i ∈ X S a i only if for some j = i , we have z j = max X j . Lemma 8.
Let i be essential on X S a which has two or more essential variables. Let M ai be the maximal element of X S a i . Then, φ i is bounded on X S a i if either:1. Exists y and z − i , such that y, M ai z − i ∈ X S a and y < M i z − i , or2. i is in the same interaction clique as variable j , for which the first option applies.Proof. We start from the first case. We would need to construct y ′ and z ′− i , such thatthey do not contain any minimal or maximal points, and still y ′ < M ai z ′− i . Assume z − i contain some maximal points. All coordinates can’t be maximal by monotonicity. Find S a -minimal j , such that z j is maximal in X S a j (in can be also maximal in X j or not), butfor all coordinates k , such that j S M ai z − i k (note the superscript, point can belong to morethan one X S a in case if some coordinates are in E ), we have that z k is not maximal in X S a k .We can slightly decrease z j , finding z ′ j j z j , such that M i z − ij z ′ j ∈ X S a . By monotonicitystill y < M i z − ij z ′ j . Proceeding in a similar way we can construct z ′− i not containing anymaximal points. If it contains minimal points as well, we can increase them, starting withthe S a -maximal one, staying in X S a and keeping the relation y < M i z − ij z ′ j (see Lemma16). Similarly, we can replace all maximal and minimal coordinates of y .Now we have y ′ < M i z ′− i and neither y ′ , nor z ′− i contain any extreme coordinates.We can therefore conclude, that by monotonicity for every w i ∈ X S a i , we have φ i ( w i ) <φ ( y ′ ) − P j = i α j φ j ( z ′ j ), which shows that φ i is bounded from above on X S a i .The second case relies on Lemma 4. It is easy to see that if y , such that y < M i z − i ,does not exist, we must not be able to increase any of coordinates z − i , otherwise doingso would give us such y by monotonicity. It also implies, that all other essential variablesare in the same interaction clique as i , otherwise we would be able to change them andagain obtain a y . Finally, it means that S a -maximal coordinate has a maximal value,and relations between all interacting coordinates are E (since we are not able neither todecrease, nor to increase any coordinate). Thus, we can use Lemma 4 and conclude that φ i has an upper bound on X S a i .Having proved Lemma 8, we can now define φ ( M i ) := lim z i → M i φ i ( z i ), and φ ( m i ) :=lim z i → m i φ i ( z i ) for all i ∈ N . Assigning values to minimal elements of X i can be done ina similar manner. Finally we can prove the final theorem.19 .2 Global representation on the whole X Theorem 8.
For any x, y ∈ X , we have x < y ⇐⇒ φ ( x ) < φ ( y ) .Proof. We proceed as in Wakker [1991b] (Lemma 21) with some modifications. For pointsthat do not contain any maximal or minimal coordinates, this has been already proved(Section 7). Thus, assume that x or y contain maximal or minimal coordinates. Let x < y , and let x ∈ X S a , y ∈ X S b . Find S a -minimal j such that x j is maximal. We canalso assume that for all k , such that j S a k , we have j S x k . If this is not the case, then x belongs to several X S i (by definition of these sets), and there must be one where thiscondition holds. By Lemma 16 we can find x ′ j : x j < j x ′ j such that x ′ j x − j ∈ X S a and still x ′ j x − j ≻ y . Proceeding like this we get x ′ which does not contain any maximal coordinates,and x ′ ≻ y . We now need to show that φ ( x ′ ) > φ ( y ). Similarly, we can replace minimalcoordinates of y , and so it is now required to show that φ ( x ′ ) > φ ( y ′ ). So we can assumethat x has no maximal and y has no minimal coordinates. x must have a non-minimal X S a -essential coordinate, find a S a -maximal one i . Again, we can assume that for all k , such that j S a k , we have j S x k . ;By Lemma 16 we can decrease it slightly and find x ′ i : x i ≻ x ′ i and x ′ i x − i ∈ X S a and still x ′ := x ′ i x − i ≻ y . So we need to show now onlythat φ ( x ′ ) ≥ φ ( y ). If we replace all minimal coordinates of x ′ by non-minimal ones andall maximal coordinates of y by non-maximal ones, then by monotonicity the preferencebetween them is not affected, and by Lemma 6, we have φ ( x ′ ) > φ ( y ). Thus any smallincrease of minimal and small decrease of maximal values leads to a strict inequality.By definition of φ i at extreme elements of X i , we have that φ ( x ′ ) is the infimum of allsuch φ -values, and φ ( y ) is the supremum. Hence, φ ( x ′ ) < φ ( y ). x < y ⇒ φ ( x ) < φ ( y ) alsoimplies φ ( x ) ≥ φ ( y ) ⇒ x < y .Now let φ ( x ) > φ ( y ). x cannot have all it’s essential coordinates minimal, so find S a -maximal j , such that x j is not minimal. By denserangedness of φ j , we can find anon-minimal x ′ j : x j ≻ j x ′ j and still φ ( x ′ j x − j ) > φ ( y ). By the above argument, we have x ′ j x − j < y , and by strict monotonicity we have x ≻ y . When defining functions φ i we set φ i ( r i ) = 0 and φ i ( r i ) = 1 for all i . This can be relaxed,in fact for every interaction clique A ∈ N , we can choose the origin and scaling factorindependently.Changing the origin alone would not alter the capacity, but changing the scalingfactor would. For example, let us define, as previously, φ ′ i ( r i ) = 0 for all i ∈ N , but set φ ′ i ( r i ) = 1 for i A and φ ′ j ( r j ) = t A for j ∈ A , so now α a ′ j = t A α aj for some j ∈ A .Accordingly, when normalizing coefficients α a ′ i in the additive representations, we will bedividing by P i A α ai + t A P i ∈ A α ai .It is not hard to see, that the A − N A lemma (Lemma 3) is still intact - indeed,sums of α a ′ i within each clique remain the same, just scaled by some common factor -( P i A α ai + t A P i ∈ A α ai ) for α a ′ i , i A , and t A ( P i A α ai + t A P i ∈ A α ai ) for α a ′ j , j ∈ A .It is also not hard to show that equivalence classes would still have identical valuesin different X S a after such operation. The following Lemmas prepare this.20 emma 9. Let A , A , ... be interaction cliques of N . Then, for any B : B ∩ A i = ∅ , B ∩ A j = ∅ , we have m ( B ) = 0 . Also, if for two sets A and A we have m ( B ) = 0 forall B : B ∩ A = ∅ , B ∩ A = ∅ , then coordinates from A do not interact with coordinatesfrom A .Proof. It’s enough to show this for singletons. Let i ∈ A and j ∈ A . We need to showthat for any B : i, j ∈ B , we have m ( B ) = 0, and vice versa, if m ( B ) = 0 for everysuch B , then i and j do not interact. Assume that for some such B we have m ( B ) = 0.Then, we can find x ij z − ij ∈ X S a , y ij z − ij ∈ X S b , such that α ak = α bk for all k = i, j , and α ai = α bi , α aj = α bj . This implies that ij -trade-off consistency does not hold on all X ij ,hence the variables interact. To show the reverse, note that we have α ai = α bi , α aj = α bj forall possible points in X , which implies ij -trade-off consistency on all X ij .The following two Lemmas follow trivially. Lemma 10. P B ⊂ A i m ( B ) ≥ for all interaction cliques A i . Lemma 11.
Let A , A , ... be interaction cliques of N . Then the Choquet integral wrta corresponding ν ,, can be written as a sum of integrals wrt “sub-capacities”, defined onsets of all subsets of A i . Consequently, we can substitute φ i for k A ψ i for all i ∈ A , and re-normalize thecapacity by multiplying every m ( A ), hence also ν ( A ) by P B A m ( B )+ k A P B ∈ Am ( B ).Apparently, this implies that points from different X S a belonging to the same equivalenceclass would still have identical values.Uniqueness properties of the value functions are similar to those obtained in thehomogeneous case X = Y n , but are modified to accommodate for the heterogeneousstructure of the set X in this paper. Because of our general setup, value functionsmight admit “ordinal” transformations at certain points, and “cardinal” at the others,even within the same coordinate. In particular, if a point x i belongs to some X S a i , and X S a has two or more essential coordinates, one of which is i , then φ i ( x i ) admits only acardinal transformation, unless an extreme case applies, when x i z − i ∈ X S a for a single z − i . Two dimensional case is much more transparent to understanding, and the generalidea remains intact.Let I = { A , . . . , A k } be a set of interaction cliques of N . Obviously, I is a partitionof N . Lemma 12.
Let g : X → R , . . . , g n : X n → R be such that (1) holds with f i substitutedby g i . Then, at all x i ∈ X i , such that for more than one z − i we have x i z − i ∈ X S a , where X S a has two essential coordinates, one of which is i , value functions f i and g i are relatedin the following way: f i ( x i ) = α A j g i ( x i ) + β A j , Capacity changes as follows m ′ ( B ) = α A j m ( B ) P C ⊂ A i ,A i ∈I α A i m ( C ) . At the remaining points of X , value functions f i have the following uniqueness prop-erties: f i ( x i ) = ψ i ( g i ( x i )) , here ψ i is an increasing function, and for all j ∈ N, j = i , such that exists A ∈ N : i, j ∈ A, m ( A ) > , we additionally have f i ( x i ) = f j ( x j ) ⇐⇒ g i ( x i ) = g j ( x j ) . Proof.
Mostly follows from uniqueness properties of additive and ordinal representationsand Lemma 4. The only complication is the special case, when x i z − i ∈ X S a for a unique z − i . This effectively means that x i z − i is the only representative of its equivalence classin X S a , hence it must be maximal or minimal. It also implies, that the transformation of φ i ( x i ) does not have to be the same as for all other points in X S a i , as the only condition ithas to satisfy is that φ ( x i z − i ) has to be greater (or less) than values of all other equivalenceclasses in X S a . AppendixA.1 Technical lemmas
Lemma 13. If < satisfies triple cancellation then it is independent.Proof. a i p − i a i p − i , a i q − i < a i q − i , a i p − i < b i p − i ⇒ a i q − i < b i q − i . Lemma 14. X = S X S i .Proof. Immediate by A3 . Definition 11.
For any set I of consecutive integers, a set { g ki : g ki ∈ X i , k ∈ I } isa standard sequence on coordinate i if exist z − ij , y j , y j such that y j j y j and for all i, i + 1 ∈ I we have g ki y j z − ij ∼ g k +1 i y j z − ij . Further, we say that { g ki : k ∈ I } is containedin X S a if z − ij , y j , y j can be chosen in such a way, that all resulting points are in X S a . Lemma 15.
Axiom A5 implies the following condition. Let { g ki : k ∈ I } and { h kj : k ∈ I } be two standard sequences, each contained in some X S a . Assume also, that for some m ∈ I , g mi y l z − il ∼ h mj w n x − jn and g m +1 i y l z − il ∼ h m +1 j w n x − jn . Then, for all k ∈ I , g ki y l z − il ∼ h kj w n x − jn .Proof. The proof is very similar to the one from [Krantz et al., 1971] (Lemma 5 in sec-tion 8.3.1). Assume wlog that { g ki : k ∈ I, g ki y l z − il ∼ g k +1 i y l z − il } is an increasing stan-dard sequence on i , which is contained in X S a , whereas { h kj : h kj w n x − jn ∼ h k +1 w n x − jn } is an increasing standard sequence on j , and lies entirely in X S b . We will show that g m +2 i y l z − il ∼ h m +2 j w n x − jn from which everything follows by induction.Assume h m +2 j w n x − jn ≻ g m +2 i y l z − il . Then, by restricted solvability exists ˆ h j ∈ X j such22hat ˆ h j w n x − jn ∼ g m +2 i y l z − il . By A5 , g mi y l z − il ∼ g m +1 i y l z − il g m +1 i y l z − il ∼ g m +2 i y l z − il g mi y l z − il ∼ h mj w n x − jn g m +1 i y l z − il ∼ h m +1 j w n x − jn g m +1 i y l z − il ∼ h m +1 j w n x − jn g m +2 i y l z − il ∼ ˆ h j w n x − jn h mj w n x − jn ∼ h m +1 j w n x − jn ⇒ h m +1 j w n x − jn ∼ ˆ h j w n x − jn . By definition of { h kj } , we have h m +1 j w n x − jn ∼ h m +2 j w n x − jn . Thus, h m +2 j w n x − jn ∼ ˆ h j w n x − jn , hence also g m +2 i y l z − il ∼ h m +2 j w n x − jn , a contradiction. The other cases aresymmetrical. Lemma 16.
Let there be x, y, i such that x ≻ y , x ∈ X S a , i S x k for all k : i S a k , and x i is non-minimal if i is minimal in S a . Then exists z i , such that z i < i x i , i S z i x − i k forall k : i S a k , and z i x − i ≻ y . Similarly, let there be x, y, i such that y ≻ x , x ∈ X S a , k S x i for all k : k S a i , and x i is non-maximal if i is maximal in S a . Then exists z i , such that x i < i z i , k S z i x − i i for all k : k S a i , and y ≻ z i x − i .Proof. By restricted solvability and monotonicity. See Wakker [1991b] Lemma 11.
A.2 Essentiality and monotonicity
In what follows the essentiality of coordinates within various X S i is critical. The centralmechanism to guarantee consistency in the number of essential coordinates within various X S i is bi-independence, which is closely related to comonotonic strong monotonicity ofWakker [1989].In the Choquet integral representation problem for a heterogeneous product set X = X × . . . × X n , strong monotonicity is actually a necessary condition. It is directly impliedby A6 - bi-independence, together with the structural assumption. Lemma 17.
Pointwise monotonicity. If for all i ∈ N it holds x i < i y i , then x < y .Proof. x < y x − < y x − < . . . < y . Lemma 18. If i is essential on X S a , then a i < i b i iff a i x − i ≻ b i x − i for all a i x − i , b i x − i ∈ X S a .Proof. If a i < i b i then by the structural assumption exists a i z − i ≻ b i z − i . The result followsby bi-independence ( A6 ).Conceptually, Lemma 18 implies that if a coordinate i is essential on some subset of X S a , then it is also essential on the whole X S a . This allows us to make statements like“coordinate i is essential on X S a ”. 23 .3 Equispaced sequences A usual standard sequence goes along a single dimension as defined above. In this paperwe often require to move along several dimensions, one at a time, maintaining the incre-ment between steps constant in some sense. In order to achieve this we will introducethe concept of equispaced sequences . Figure 1 illustrates the process. X j X i r α i r γ kj α ki α k +1 i γ k +1 j γ k +2 j Figure 1: Equispaced sequences in two dimensionsAssume that r , r are such that i R r j and i R r j . We would like to build a sequencefrom r to r staying in the area where i R j . We can choose the size of the sequencestep arbitrarily. However, the problem is that r does not have an equivalent point withthe second coordinate equal to r j , so we cannot build a “normal” standard sequence toachieve that. Our aim is to maintain the sequence within set where i R j . We also assumethat X i and X j do not have maximal or minimal elements (or they have been removed).By density and the absence of maximal and minimal elements, we can find α ki , α k +1 i such that α k +1 i < i r i < i α ki . We need to change the direction of the sequence from thedimension i to the dimension j at r i . We construct a point equivalent to a ki and a pointequivalent to a k +1 i such that their i ’s coordinate is r i ( points γ kj and γ k +1 j ). Since we canchoose the step of the sequence arbitrarily, by density and absence of maximal elements,we can move on and construct a standard sequence on the coordinate j using these twopoints.Remarkably the spacing between subsequent members of the equispaced sequence α , . . . , α k − , γ k , γ k +1 , . . . stays in a certain sense the same, no matter along which dimen-sion we are moving. Once an additive interval scale is constructed, the vague notion ofthe equal spacing will convert into a clear constant difference of values for subsequentmembers of the sequence. See also [Bouyssou and Marchant, 2010] for a similar idea xtension of A5 to equispaced sequences Construction of equispaced sequencesallows us to extend the statement of A5 , and more precisely that of Lemma 15 to equi-spaced sequences. Lemma 19. If g k and h k are two equispaced sequences entirely lying in X S a and X S b correspondingly, and for some i we have g i ∼ h i and g i +1 ∼ h i +1 , then for all j such thatexist g j and h j we have g j ∼ h j .Proof. Without loss of generality assume that g i := g ik g − k and g i +1 := g i +1 k g − k , while h i := h il h − l and h i +1 := h i +1 l h − l , i.e. in both cases the points are from subsequences on thesame dimensions. Assume further, that h i +2 := h i +2 l h − l and g i +2 := g i +2 m g ′− m , i.e. there isa change of dimension in the sequence g k . We will show that this entails that the followingsteps of the equispaced sequence g k will be equivalent to the corresponding steps of thesequence h k as long as both keep going along dimensions m and l correspondingly. If eitherof them changes its direction the technique can be applied again. We have by construction(see Figure 1) g i +2 k g − k ∼ g i +2 m g ′− m and g i +3 k g − k ∼ g i +3 m g ′− m . Therefore, g i +2 m g ′− m ∼ h i +2 l h − l and g i +3 m g ′− m ∼ h i +3 l h − l . The statement follows by Lemma 15 ( A5 ). A.4 Necessity of axioms
Lemma 20. A3 is necessary.Proof. At any point z ∈ X and for every i, j , we must have either f i ( z i ) ≥ g i ( z i ) or g i ( z i ) ≥ f i ( z i ). From this everything follows trivially (write the condition using Mobiusrepresentation of the integral). Lemma 21. A6 is necessary.Proof. Assume a i x − i , b i x − i , c i x − i , d i x − i ∈ X S a and a i x − i ≻ b i x − i , c i x − i ∼ d i x − i , c i y − i ≻ d i y − i .There can be three cases:1. c i y − i , d i y − i ∈ X S b . We have additive representations on X S a and X S b , so α i f i ( a i ) + X j ∈ N \ i α j f j ( x j ) > α i f i ( b i ) + X j ∈ N \ i α j f j ( x j ) α i f i ( c i ) + X j ∈ N \ i α j f j ( x j ) = α i f i ( d i ) + X j ∈ N \ i α j f j ( x j ) β i f i ( c i ) + X j ∈ N \ i β j f j ( y j ) > β i f i ( d i ) + X j ∈ N \ i β j f j ( y j ) . The first inequality entails α i = 0. From this and the following equality follows f i ( c i ) = f i ( d i ), which contradicts with the last inequality. Thus c i y − i ≻ d i y − i implies c i x − i ≻ d i x − i but only in the presence of a i x − i ≻ b i x − i in the same X S a (the case when c i y − i and d i y − i are not both in the same X S b can be reduced to this one. This is also the reason behindthe name we gave to this condition - “weak bi-independence”.2. c i y − i , d i y − i ∈ X S a . In this case we get α i f i ( c i ) > α i f i ( d i ) and α i f i ( c i ) = α i f i ( d i ), acontradiction.3. c i y − i ∈ X S a , d i y − i ∈ X S b . As above α i = 0, so it follows that f i ( a i ) = f i ( b i ). But thenwe must have d i y − i ∈ X S a (value functions are all equal to those for c i y − i ), and hencethe conclusion follows as in the previous case.25 .5 Shape of { z ij : i E z j } Shape of the boundary between subsets of X ij where i R j and j R i is an interestingand important question. Axiom A3 only guarantees that this boundary is in a certainsense “quasiconvex”, i.e. an increase along i cannot be matched by a decrease along j .Strengthening this statement requires invoking other axioms. A.5.1 Every X S i has one essential variable Assume that every X S i has only one essential variable. We will show that an increasealong i must be matched by an increase along j . This is actually required to constructa representation (see Section 7.1). The main axiom, required to show this in addition to A3 is strong monotonicity ( A6 ). Lemma 22.
Let a i p j be such that i E a i p j z − ij j . Then, unless i and j do not interact, forno b i we can have i E b i p j z − ij j .Proof. Assume such b i exists. Moreover, assume, wlog, that a i < i b i , and we took maximal a i and minimal b i for which this holds.1. If b i is minimal in X i and a i is maximal in X i , then i, j do not interact by A3 , soassume that b i is not minimal (the other case is symmetric).2. Since we took the smallest b i for which i E b i p j z − ij j , we can assume wlog (other casesare similar) that exists c i i b i , such that b i q j z − ij ≻ c i q j z − ij , and b i p j z − ij ∼ c i p j z − ij (a violation of ij -independence, required by the presence of interaction).3. By density assumption there must exist s j : i E c i s j z − ij .4. We have b i p j z − ij ≻ c i s j z − ij , hence c i p j z − ij ≻ c i s j z − ij .5. Now we need to extend X S a ⊃ c i q j z − ij “to the right”, so that X S a i ⊃ d i : d i ≻ i b i .We also need to extend X S b ⊃ b i p j z − ij , so that X S b j ⊃ t j : t j ≻ j p j . This can be doneby adjusting particular coordinates of z ij . Assume, that z ij is already such, thatthe above conditions hold. In the case when such adjustment cannot be performed,we might need to perform a similar extension “to the left” from d i p j z − ij .6. Note that on SE c i q j z − ij ij we can either have i essential, or neither i nor j , whereason NW a i p j z − ij ij it can either be j , or none as well.7. The point of the extensions is to show that by strong monotonicity A6 , we musthave d i p j z − ij ≻ b i p j z − ij , as i is essential on SE c i q j z − ij ij , but also d i p j z − ij ∼ b i p j z − ij ,as i is inessential on NW a i p j z − ij ij . If an extension “to the left” was performed, wemust get c i s j z − ij ∼ b i p j z − ij , but also b i p j z − ij ≻ c i s j z − ij , as j stays essential.26 .5.2 X S i have two or more essential variables In section 7 we have shown, that in the representation value functions for sets X i and X j are equal for points z where i E z j . Theorem 9 provides a qualitative version of thisstatement. The assumption we must make is that i and j are essential on X S a and X S b ,such that S a and S b differ only with respect to order of i and j . In the below proof, weassume that i, j are S a and S b -maximal, but this can be easily changed, by starting fromsome r A r − A instead of r . Theorem 9.
Let r : i E r j, r : i E r j and a ki and b kj are two standard sequences suchthat a i r − i ∼ r i r − i and b j r − j ∼ r j r − j and r i r − i ∼ a mi r − i whereas r j r − j ∼ b mj r − j . Assume r is such that i E r j and r i r − i ∼ a ni r − i . Then r j r − j ∼ b nj r − j .Proof. Build two equispaced sequences from r to r ij r − ij : • e k starting from r i via r i r − i , and • w k starting from r j via r j r − j ,such that e i r − i ∼ w j r − j . By Lemma 15 ( A5 ) it follows then that all corresponding stepsof two sequences are equivalent, in other words, e k ∼ w k for all k . Consequently, there isthe same number of steps both sequences make between r and r ij r − ij , say K .For some s < K we have r i r − i lying between e s and e s +1 , i.e. e s +1 ≻ r i r − i < e s .Similarly, for some t < K we have w t +1 ≻ r j r − j < w t . We can write:[ r , r i r − i ] ≈ na ki ≈ se k , which means: r i r i lies between a ni r − i and a n +1 i r − i and also between e s and e s +1 . Similarly,[ r , r j r − j ] ≈ nb kj ≈ tw k [ r i r − i , r ij r − ij ] ≈ nb kj ≈ ( K − s ) e k [ r j r − j , r ij r − ij ] ≈ na ki ≈ ( K − t ) w k . Two last statements are possible because by density we can can get arbitrarily close topoints r j r − j and r i r − i by choosing finer sequences e k and w k .For point r ij r − ij we have: [ r , r i r − i ] ≈ ma ki ≈ mn se k [ r j r − j , r ij r − ij ] ≈ ma ki ≈ mn ( K − t ) w k . Assume that the number of steps on two other segments is different:[ r , r j r − j ] ≈ lb kj ≈ ln tw k [ r i r − i , r ij r − ij ] ≈ lb kj ≈ ln ( K − s ) e k . r ij r − ij we get ms + l ( K − s ) n e k for SE ij and m ( K − t )+ ltn w k for NW ij . By Lemma 15 ( A5 ) the number of steps must be identical,so: ms + l ( K − s ) n = m ( K − t ) + ltn , or m ( s + t − K ) = l ( s + t − K ) . There are two possible solutions: • m = l , and • t = K − s , which means that trade-offs are consistent throughout X , hence i and j do not interact, a contradiction.The result follows. Corollary 1. If a i p j : i E a i p j z − ij j and b i p j : i E b i p j z − ij j , then x i y j : i E x i y j z − ij j for all x i ∈ X i , y j ∈ Y j . Corollary 2. If a i p j : i E a i p j z − ij j , then • for any b i , such that b i < i a i we have i S b i p j z − ij j , • for any b i , such that a i < i b i we have j S b i p j z − ij i , • for any q j , such that q j < j p j we have j S a i q j z − ij i , • for any q j , such that p j < j q j we have i S a i q j z − ij j .or i, j do not interact. A.6 Alternative treatment of the case with single es-sential variables
For this case we can also construct the representation as follows. We will define valuefunctions for all sets X i in accordance with < i . We would additionally require that φ i ( x i ) ≥ φ j ( x j ) iff i R x i x j z − ij j . Finally, we will prove a lemma, similar to Lemma 3,which would allow us to construct a capacity and the Choquet integral.We start by considering if we can define value functions on X i according to the rulesdefined in the previous paragraph. Since < i is a weak order, we can obviously definefunctions such that φ i ( x i ) ≥ φ j ( x j ) iff x i < i y i . However, the second condition is morecomplicated. One particular case, when this would be impossible, is if exist x i x j z − ij and y i x j z − ij such that i E x i x j z − ij j and i E y i x j z − ij j . However, this would eventually imply thatonce the representation is constructed we have f i ( x i ) = f i ( y i ), and hence C ( ν, f ( x i z − i )) = C ( ν, f ( y i z − i )) for any z − i , which implies x i z − i ∼ y i z − i for all z − i . This in turn contradictsa structural assumption that we have made. We state the following lemma: Lemma 23.
Assume that i and j interact. Then, if i E x i x j z − ij j , then for no y i x j z − ij holds i E y i x j z − ij j . roof. By A3 and strong monotonicity A6 . See Lemma 22 in the appendix for details.Next, we will use this to show that for interacting variables R is transitive. Lemma 24.
Let i, j and j, k interact. Then, i R z j and j R z k imply i R z k .Proof. If i and k do not interact, the the result is immediate, hence, assume that theyinteract. Also, if we have i S z j and j S z , then by acylicity i R z k , so one of the relationsmust be a E z .Assume first that i E z j, j S z k . Assume also k S z i . We want to increase z i slightly,so that for some x i < i z i we have i S x i z − i j , but still k S x i z − i i , which leads to a violationof acyclicity. This is possible by density, unless z i is maximal. In this case, decrease z j slightly, so that for x j : z j < x j we have i S x j z − j j but still i S x j z − j k . This is againpossible unless z j is minimal, in which case we conclude ( z i is maximal, z j is minimal),that ij − C holds for all x i , x j , hence i and j do not interact, a contradiction.Similarly, assume i S z j, j E z k, k S z i . Increase z i or decrease z k to violate acyclicity,or otherwise conclude that j and k do not interact, as above.Finally, assume i E z j, j E z k, k S z i . Increase z i and decrease z k to get i S x ik z − ik j, j S x ik z − ik k, k S x ik z − ik i , which violates acyclicity. If z i is maximal, decrease z k to get i E x k z − k j, j S x k z − k k, k S x k z − k i , which is the first case considered above. If z i canbe increased, but z k is minimal, we obtain he second case above. Finally, if z i is maximaland z k is minimal, then by definition i R z k .Since R is transitive it follows that for interacting variables E and S would also betransitive.Finally, we can construct value functions φ i : X i → R . To do so, we start by con-structing the function on any variable X j . Since < i is a weak order, this can obviouslybe done. Next, we construct the value function on X k , such that j and k interact. Thistime, the function would also have to satisfy the second constraint - φ k ( x k ) ≥ φ j ( x j ) iff k R x k x j z − kj j . Then the construction proceeds to some X l , such that l and j or l and k interact. We continue like this, until all variables in the interaction clique, containing j do not have value functions defined. After this, variables from all other interactioncliques can be defined in the same way.Assume, that we have already defined some value functions, in a manner, consistentwith the constraints above, and want to define φ i . A contradiction can occur, if at somestage we get i R z j but φ j ( z j ) > φ i ( z i ). This can happen, if i is interacting with morethan one variable, and we used some other value function φ k to define φ i . However, wecan update value functions throughout the clique to resolve this violation.For any φ i ( x i ) we have either:1. φ i ( x i ) = φ k ( x k ) for some x k in X k , or2. φ i ( x i ) > φ k ( x k ) for all x k in X k , or3. φ i ( x i ) < φ k ( x k ) for all x k in X k .Same holds for every other interacting pair of variables, apart from i, j , for which thevalue functions have already been constructed.29ince we have φ j ( z j ) > φ i ( z i ), but i R z j , in order to resolve the conflict, we need toincrease φ i ( z i ) and decrease φ j ( z j ). Since value functions are ordinal, we can do this byapplying some increasing transformation to them. If it is the case that φ i ( z i ) > φ k ( x k )for all x k in X k , this can be done straightforwardly. However, if it is the case that φ i ( z i ) = φ k ( z k ) or φ i ( z i ) < φ k ( x k ) for all x k in X k , we will have to change φ k as well.Eventually, either the conflict is resolved, i.e. value functions are sufficiently adjusted,or we find some common “predecessor” - a coordinate m and build a chain like the onebelow: φ i ( z i ) ≤ φ k ( x k ) ≤ φ l ( x l ) ≤ . . . ≤ φ o ( x o ) ≤ φ m ( x m ) , where each point is taken so that it is minimal with respect to the predecessor, for examplein pair φ i ( z i ) ≤ φ k ( x k ) we pick x k such that for no y k we have φ i ( z i ) ≤ φ k ( y k ) ≤ φ k ( x k ).Similarly, for φ j ( z j ) we get the following chain: φ j ( z j ) ≥ φ s ( w s ) ≥ φ t ( w t ) ≥ . . . ≥ φ m ( w m ) . If now φ m ( w m ) > φ m ( x m ), we get φ j ( z j ) ≥ φ s ( w s ) ≥ φ t ( w t ) ≥ . . . ≥ φ m ( w m ) > φ m ( x m ) ≥ φ o ( x o ) ≥ . . . ≥ φ l ( x l ) ≥ φ k ( x k ) ≥ φ i ( z i ) , and hence at q := z j w s w t . . . w m x o . . . x l x k z i z − jst...mo...lki we have j R q s R q t R q . . . R q m S q o R q . . . R q l R q R q k R q i, where every variable interacts with a subsequent one. Hence, as shown previously, wemust have j S q i , hence j S z i , which is a contradiction. References
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