Two Factor Additive Conjoint Measurement With One Solvable Component

Transcription

1 Two Factor Additive Conjoint Measurement With One Solvable Component Christophe Gonzales LIP6 Université Paris 6 tour 46-0, ème étage 4, place Jussieu 755 Paris Cedex 05, France ChristopheGonzales@lip6fr fax: (33) Proofs should be sent to: Christophe GONZALES See the address above I am greatly indebted to Jean-Yves Jaffray, Peter Fishburn and Peter Wakker for their very useful comments and suggestions 1

2 Abstract This paper addresses conditions for the existence of additive separable utilities It considers mainly two-dimensional Cartesian products in which restricted solvability holds wrt one component, but some results are extended to n-dimensional spaces The main result shows that, in general, cancellation axioms of any order are required to ensure additive representability More precisely, a generic family of counterexamples is provided, proving that the (m + 1)st-order cancellation axiom cannot be derived from the mth order cancellation axiom when m is even However, a special case is considered, in which the existence of additive representations can be derived from the independence axiom alone Unlike the classical representation theorems, these representations are not unique up to strictly positive affine transformations, but follow Fishburn s (1981) uniqueness property

3 1 Introduction Additive conjoint measurement has been extensively studied since 1960 and Debreu s celebrated paper For instance, Jaffray (1974a, 1974b) give necessary and sufficient conditions ensuring the existence of additive representations, which are closely related to cancellation axioms (see Krantz et al (1971, p )) Other necessary and sufficient conditions can be found in Krantz et al (1971, p430) and in Fishburn (199) However, the lack of testability of high order cancellation axioms led researchers to look for other axioms that were nonnecessary but easier to test Thus, Luce and Tukey (1964), Krantz (1964), Luce (1966), Fishburn (1970, p58) and Krantz et al (1971, p57) show sufficient conditions of an algebraic nature, as well as uniqueness properties of the additive representations Debreu (1960) and Wakker (1989, p49) give their counterpart in a topological framework Fuhrken and Richter (1991) gave necessary and sufficient conditions for the existence of continuous additive representations on connected topological spaces All these papers assume that some structural nonnecessary conditions called unrestricted or restricted solvability in the former approach and connectedness of the topological spaces in the latter hold wrt every component of the Cartesian product But these assumptions are particularly restrictive because they simply do not hold in many practical situations For instance, in economics, the first component may describe a waiting time and the second one the object waited for (in this case, using an exponential transform, one gets a multiplicative model) In medical decision making, the first component may refer to the duration of a treatment, and the second one to the level of quality of life induced by it Another example can be found in Gonzales and Jaffray (1997), in which certainty effects make a restricted solvability assumption unrealistic Therefore, it is of interest to study Cartesian products in which solvability or connectedness does not hold wrt every component Gonzales (1996a, 1996b) describes testable conditions ensuring additive representability when unrestricted or restricted solvability holds wrt at least two components But, to my knowledge, the only paper dealing with spaces in which restricted solvability holds wrt exactly one component is Fishburn (1981), which studies the uniqueness property of additive utilities In this paper, no conditions are mentioned ensuring the existence of such representations Those are addressed in the present paper The choice of using solvability rather than connectedness was recommended by Wakker (1988) In section, axioms needed in the paper are given Section 3 presents a generic family of counterexamples showing that, in general, cancellation axioms of any order are needed to entail additive representability on two-dimensional Cartesian products in which unrestricted solvability holds wrt only one component This can be seen as a complement to Fishburn (1997a, 1997b, 1997c), which gives the same result for finite sets, and as an extension of Scott and Suppes (1958) The result is trivially extended to n-dimensional Cartesian products in which only one component satisfies unrestricted solvability Section 3 also studies the relationship between cancellation axioms of different orders In particular, it is proved that the second order cancellation axiom is equivalent to the conjunction of independence and the Thomsen condition; this result was already well known when restricted solvability holds wrt all the components (see eg, Krantz et al (1971, p57)) or when unrestricted solvability holds wrt at 3

4 least two components (see Gonzales (1996a)) or even when a component is finite and the other one is binary (see Krantz et al (1971, p45)) Section 4 describes how a weak structural assumption wrt the second component, namely equal spacedness, combined with restricted solvability wrt the first component, independence and an Archimedean axiom, implies additive representability This is an extension of Krantz et al (1971, p45) and Wakker (1991) All proofs are given in the appendix The axiomatic system Consider a Cartesian product X = X 1 X and a weak order over X, ie, is complete (for all x, y X, x y or y x) and transitive Let x y [x y and y x], x y [x y and Not(y x)], and x y y x is representable by an additive utility function u : X R if there exist mappings u 1 : X 1 R and u : X R such that: u(x 1, x ) = u 1 (x 1 ) + u (x ) for all (x 1, x ) X and (x 1, x ) (y 1, y ) u(x 1, x ) u(y 1, y ) for all (x 1, x ), (y 1, y ) X The following four axioms are well known to be necessary for additive representability on any set X: Axiom 1 (independence) for all x, y X, (x 1, x ) (x 1, y ) (y 1, x ) (y 1, y ) and (x 1, x ) (y 1, x ) (x 1, y ) (y 1, y ) By axiom 1, weak orders 1 and, induced by respectively on X 1 and X, can be defined for all x 1, y 1 X 1, x, y X as: x 1 1 y 1 [there exists c X such that (x 1, c ) (y 1, c )], and x y [there exists c 1 X 1 such that (c 1, x ) (c 1, y )] Axiom (Thomsen condition) For every x 1, y 1, z 1 X 1, x, y, z X, if (x 1, z ) (z 1, y ) and (z 1, x ) (y 1, z ), then (y 1, y ) (x 1, x ) The Thomsen condition can be generalized by substituting indifference relations by weak preference relations, thus resulting in the following axiom: Axiom 3 (Generalized Thomsen condition) For every x 1, y 1, z 1 X 1, x, y, z X, if (x 1, z ) (z 1, y ) and (z 1, x ) (y 1, z ), then (y 1, y ) (x 1, x ) Axiom 4 ((C ) : second order cancellation axiom)consider (x i 1, xi ) and (yi 1, yi ), i {1,, 3}, such that (x 1 j, x j, x3 j ) is a permutation of (y1 j, y j, y3 j ) for j = 1, If (x 1 1, x1 ) (y1 1, y1 ) and (x 1, x ) (y 1, y ), then (x3 1, x3 ) (y3 1, y3 ) The following two structural conditions are usually assumed wrt every component of X, but, in what follows, they will only be wrt one component Axiom 5 (unrestricted solvability wrt the first component) For every y X and every x X, there exists x 1 X 1 such that y (x 1, x ) 4

5 Axiom 6 (restricted solvability wrt the first component) For every y X, if (a 1, x ) y (b 1, x ), then there exists x 1 X 1 such that y (x 1, x ) Clearly, when unrestricted solvability holds, then restricted solvability holds as well, so that representation theorems involving restricted solvability are more general than the ones involving unrestricted solvability These two axioms are structural conditions because: i) they are nonnecessary for additive representability; and ii) they strengthen axioms 1 through 4 to the point that they induce additive representability Throughout, when f and g are functions, f g( ) = f(g( )); if, moreover, p is a strictly positive integer, f p = f f f (p times), and f 1 denotes the inverse function of f, ie, f f 1 is the identity function 3 Counterexamples 31 Independence and the Thomsen conditions Lemma 1 Let be a weak order on a two-dimensional Cartesian product X Then ( ) the independence axiom (axiom 1) the second order + the generalized Thomsen condition (axiom 3) cancellation axiom (C ) This lemma is rather general since, unlike Krantz et al (1971), it requires no structural condition such as finiteness of X i or solvability wrt X 1 However, when the latter holds, the generalized Thomsen condition can be substituted by the classical Thomsen condition, hence the following lemma: Lemma Let be a weak order on a two-dimensional Cartesian product X satisfying unrestricted solvability wrt the first component (axiom 5) Then ( + the independence axiom (axiom 1) the Thomsen condition (axiom ) ) the second order cancellation axiom (C ) This lemma is a slight generalization of property P1* P1 of theorem 53 in Fishburn (1970) (because, in lemma, solvability is not required wrt the second component) However, the proof I present here is completely different from the one written by Fishburn It is a classical result that in two-dimensional Cartesian products, the second order cancellation axiom is not implied by independence alone: one must add the Thomsen condition to get this implication A well known counterexample (see eg, Wakker (1989, p71) 1 ) is the following one: X = R R and is defined by the utility function U(x 1, x ) = x 1 + x + min{x 1, x } 1 in this example, the Thomsen condition is substituted by the hexagon condition, which is somewhat less restrictive But according to Karni and Safra (1997), under restricted solvability wrt both components and independence, the Thomsen condition is equivalent to the hexagon condition 5

6 3 Relations between different cancellation axioms In this subsection, a new condition, called a symmetric cancellation axiom and denoted (S m+1 ), is introduced This is a particular symmetric case of the classical cancellation axiom (C m+1 ) (defined below) As we shall see, it fills the gap between (C m ) and (C m+1 ) when unrestricted solvability holds wrt one component of X Axiom 7 ((C m ) : mth order cancellation axiom) Let m Consider m+1 elements (x i 1, xi ) of X, i = 1,, m+1, and m+1 elements (yi 1, yi ) such that (y1 1,, ym+1 1 ) and (y 1,, ym+1 ) are permutations of (x 1 1,, xm+1 1 ) and (x 1,, xm+1 ) respectively If (x i 1, xi ) (yi 1, yi ) for all i = 1,,, m, then (xm+1 1, x m+1 ) (y1 m+1, y m+1 ) Note that the order of the cancellation axiom follows the notation of Krantz et al (1971, p343) (Fishburn (1970) and Jaffray (1974a) would have called it the (m + 1)st order cancellation axiom) Axiom 8 ((S m+1 ) : (m + 1)st order symmetric cancellation axiom) Suppose that m is an even number Let k = (m + )/ Consider elements x 1 1,, xm+ 1 X 1, x 1,, xk X and y 1,, yk X such that, for all i k and j > k, x i 1 xj 1 and for all i, j < k, x i yj Then: (x 1 1, x1 ) (xk 1, y1 ) (x 1, x ) (x1 1, y ) (x 3 1, x3 ) (x 1, y3 ) (x k 1, xk ) (xk 1 1, y k) (x m+ (x k+1 1, y 1 ) (xm+ 1, x 1 ) 1, y) k (x m+1 1, x k ) (x k+ 1, y ) (xk+1 1, x ) (x m+1 1, y k 1 ) (x m 1, xk 1 ) Note the symmetry between the left hand side and the right hand side of preference relations, and between the elements above the horizontal line and those under the line Note also that when the s are substituted by s, the third order symmetric cancellation axiom corresponds to the hexagon condition (see Wakker (1989, p47 48)) Theorem 1 Let be a weak order on X = X 1 X Assume that unrestricted solvability wrt the first component (axiom 5) holds Then, for any m, { if m is odd, then (Cm ) (C m+1 ) if m is even, then (C m ) + (S m+1 ) (C m+1 ) 33 Second and third order cancellation axioms In this subsection, an example is given which shows that, when unrestricted solvability holds wrt only one component, (C ) may hold without (C 3 ) holding Suppose that is represented on X = R {0,, 4, 6} by the following utility function: { x1 + x U(x 1, x ) = if x 4, 05(x 1 mod ) (1) + x 1 / + 65 if x = 6 6

7 Remark that for a fixed value x of X, U(x, x ) henceforth denoted U x (x) is a continuous function on R Hence, working in X 1 R, where R represents the set of values that can be taken by U(x 1, x ) over X, is very attractive because in this space properties of can be easily visualized (see figure 1) Now, the problem is to translate the definition of independence and the Thomsen condition into this new space (x 1, x ) (y 1, x ) (x 1, y ) (y 1, y ) is equivalent to U x (x 1 ) U x (y 1 ) U y (x 1 ) U y (y 1 ) This condition clearly holds since functions U x ( ) are strictly increasing for every x X (x 1, x ) (x 1, y ) (y 1, x ) (y 1, y ) is equivalent to U x (x 1 ) U y (x 1 ) U x (y 1 ) U y (y 1 ) So, since we consider only continuous functions, it comes to the non intersection of the graph of U x by that of U y This condition also holds because 05(x mod ) + x/ + 65 > x + 4 for every x R Therefore, as represented by (1), satisfies the independence axiom Let us now translate the Thomsen condition Figure 1 helps in understanding the process: [(x 1 1, x ) (x 1, x1 ) and (x1 1, x3 ) (x3 1, x1 )] (x 1, x3 ) (x3 1, x ) So, in terms X x 3 x x 1 C A E B F D R C A E B U x 3 (x 1 ) U x (x 1 ) F U x 1 (x 1 ) D x 1 1 x 1 x 3 1 X 1 x 1 1 x 1 x 3 1 X 1 Figure 1: Translation of the Thomsen condition of utility functions, U x (x 1 1 ) = U x 1(x 1 ), U x 3(x1 1 ) = U x 1(x3 1 ) and U x 3(x 1 ) = U x (x3 1 ) By the first equality, x 1 = U 1 U x 1 x (x 1 1 ), and by the second equality, x3 1 = U 1 U x 1 x 3 (x 1 1 ) Hence, the third equality can be written as U x 3 U 1 x 1 But this must be true for all x 1 1 for all x 1, x, x 3 X and all x R, U x 3 U 1 x 1 U x 3 (x 1 1 ) U x (x 1 1 ) = U x U 1 x 1 Hence the Thomsen condition can be written as: U x (x) = U x U 1 x 1 U x 3 (x) Now, to show that, as represented by (1), satisfies the Thomsen condition, let me introduce the following lemma: Lemma 3 Let be a weak order on a X = X 1 X Assume unrestricted solvability wrt the first component (axiom 5) and independence (axiom 1) Suppose that for every x 1 x x 3 and every x1 1, x 1, x3 1 X 1, [ (x 1 1, x ) (x 1, x1 ) and (x1 1, x3 ) (x3 1, x1 )] (x 1, x3 ) (x3 1, x ) () Then satisfies the Thomsen condition (axiom ) 7

8 Thus, to prove that satisfies the Thomsen condition, it is sufficient to prove that: for all x 1 x x 3 and all x X 1, U x 3 U 1 x 1 U x (x) = U x U 1 x 1 U x 3 (x) (3) Suppose that x 3 6 Then (3) is obviously equivalent to ((x + x ) x1 ) + x3 = ((x + x 3 ) x1 ) + x ; this relation is clearly satisfied by U, as defined by (1) Now suppose that x 3 = 6; then if x = 6, (3) trivially holds, otherwise (3) is equivalent to: U 6 (x + x x1 ) = U 6(x) + x x1 Since x x1 are multiples of, this can be summarized as: U 6 (x + ) = U 6 (x) + By (1), U obviously satisfies this condition Hence the Thomsen condition holds for represented by U By subsection 31, we can conclude that the second order cancellation axiom holds However, cannot be represented by an additive utility function because the third order cancellation axiom does not hold Indeed, U(1, 6) = U(3, 4) = 7, U(3, 0) = U(1, ) = 3, U(15, ) = U(35, 0) = 35 and U(35, 4) = 75 < U(15, 6) = Cancellation axioms of order m + 1 and m + In this subsection we shall see an example proving that the (m + 1)st order cancellation axiom does not imply the (m + )nd order cancellation axiom for m odd when unrestricted solvability holds wrt only one component Consequently, no cancellation axiom is sufficient to ensure additive representability Let be a weak order on X = R {0,, 4,, m}, with m 3 odd, represented by the following utility function: { x1 + x U(x 1, x ) = m m + 5 if x < m, 05(x 1 mod ) (4) + x 1 / if x = m Then the following lemma holds: Lemma 4 (C i ) holds for all i {0, 1,, m + 1} However, (C m+ ) cannot hold because: ((m 1) + 1, 0) (1, (m 1)) (4(m 1) + 1, 0) ((m 1) + 1, (m 1)) ((m + 1)(m 1) + 1, 0) ((m 1)(m 1) + 1, (m 1)) (1, m) ((m + 1)(m 1) + 1, (m 1)) (0, (m 1)) ((m 1), 0) ((m 1), (m 1)) (4(m 1), 0) ((m 1)(m 1), (m 1)) ((m + 1)(m 1), 0) ((m + 1)(m 1), (m 1)) (0, m) 35 An extension to n-dimensional spaces The preceding results can be extended rather straightforwardly to n-dimensional Cartesian products Thus, the mth-order cancellation axiom, the (m + 1)st-order symmetric cancellation axiom and unrestricted solvability wrt the first component can be extended respectively as follows: 8

9 Axiom 9 ((C m ) for n-dimensional spaces) Let m Let (x i 1,, xi n) and (y1 i,, yi n), i = 1,, m + 1, be some elements of X such that (yj 1,, ym+1 j ) are permutations of (x 1 j,, xm+1 j ) for all j {1,, n} If (x i 1,, xi n) (y1 i,, yi n) for all i = 1,,, m, then (x m+1 1,, x m+1 n ) (y1 m+1,, yn m+1 ) Axiom 10 ((S m+1 ) for n-dimensional spaces) Let m be an even number and let k = (m + )/ Let x 1 1,, xm+ 1 X 1, and x 1 i,, xm+ i, yi 1,, ym+ i X i, i =,, n, be such that: for all j k and all p > k, x j 1 xp 1 ; for all i {,, n}, there exists an index h i {0,, k} such that x j i = yp i { p = j if j hi or if k < j k + h i ; p = j + k if h i < j k; at least one index h i is equal to 0 Then: (x 1 1, x1,, x1 n) (x k 1, y1,, y1 n) (x 1, x,, x n) (x 1 1, y,, y n) (x 3 1, x3,, x3 n) (x 1, y3,, y3 n) (x k 1, xk,, xk n) (x k 1 1, y k,, yk n) (x k+1 1, x k+1,, x k+1 n ) (x m+ 1, y k+1,, yn k+1 ) (x k+ 1, x k+,, x k+ n ) (x k+1 1, y k+,, yn k+ ) (x m+1 1, x m+1,, x m+1 n ) (x m 1, ym+1,, yn m+1 ) (x m+ 1, x m+,, x m+ n ) (x m+1 1, y m+,, y m+ n ) Axiom 11 (unrestricted solvability wrt the first component) For all y X and all x j X j, j =,, n, there exists x 1 X 1 such that y (x 1, x,, x n ) Theorem 1 is extended by the following theorem: Theorem Let be a weak order on X = n i=1 X i and assume unrestricted solvability wrt the first component (axiom 11) Then, for any m, { if m is odd, then (Cm ) (C m+1 ) if m is even, then (C m ) + (S m+1 ) (C m+1 ) The generic family of counterexamples of subsection 34 can also be extended to n-dimensional Cartesian products, thus showing that, in general, no cancellation axiom is sufficient to ensure additive representability: 9

10 Let m be any odd number greater than or equal to 3 Let be a weak order on X = R {0,, 4,, m} {0, } n represented by the following utility function: H + x 1 + x + x i if x < m, U(x 1,, x n ) = (5) V (x 1 ) + x i if x = m where Then, the following lemma holds: H = m mn 3n + 85, V (x 1 ) = 5(x 1 mod ) + x 1 / Lemma 5 (C i ) holds for all i {0, 1,, m + 1} However, (C m+ ) cannot hold because: (1 + (m + n 3), 0, 0,, 0) (1, (m 1),,, ) (1 + 4(m + n 3), 0, 0,, 0) (1 + (m + n 3), (m 1),,, ) (1 + (m + 1)(m + n 3), 0, 0,, 0) (1 + (m 1)(m + n 3), (m 1),,, ) (1, m, 0,, 0) (1 + (m + 1)(m + n 3), (m 1),,, ) (0, (m 1),,, ) ((m + n 3), 0, 0,, 0) ((m + n 3), (m 1),,, ) (4(m + n 3), 0, 0,, 0) ((m 1)(m + n 3), (m 1),,, ) ((m + 1)(m + n 3), 0, 0,, 0) ((m + 1)(m + n 3), (m 1),,, ) (0, m, 0,, 0) 4 Representation theorem The preceding section provided arguments showing that additive representability required cancellation axioms of every order when the only structural assumption made is solvability wrt one component This section describes how this can be circumvented by adding a weak assumption wrt the second component, namely equal spacedness To provide results as general as possible, restricted solvability is assumed wrt the first component of X (instead of unrestricted solvability, as in the preceding section) Moreover, X is assumed to be a sequence, (x i ) i N, where N is a set of consecutive integers (positive or negative, finite or infinite) such that Card(N) 3 For convenience, assume that x i+1 x i, for all i, i + 1 N Cancellation axioms, though necessary, are not generally sufficient to ensure additive representability: the latter still requires an Archimedean property Classically, in additive conjoint measurement, this is stated in terms of standard sequences: When X is a finite set, the set of all cancellation axioms is sufficient to ensure additive representability; see Scott (1964) or Adams (1965) 10

11 Definition 1 (standard sequence wrt the first component) For any set M of consecutive integers, a set {x k 1 X 1, k M} is a standard sequence wrt the first component if there exist x 0, x1 X such that Not((x 0 1, x0 ) (x0 1, x1 )) and (xk+1 1, x 0 ) (x k 1, x1 ) for all k, k + 1 M {x0, x1 } is called the mesh of the standard sequence Axiom 1 (Archimedean axiom wrt the first component) Any bounded standard sequence wrt the first component is finite, ie, if {x k 1 X 1, k M} is a standard sequence with mesh {x 0, x1 } and if there exist y, z X such that y (xk 1, x0 ) z for all k M, then M is finite Obviously independence and the Archimedean axiom are required However they are still not sufficient Indeed, assume that X = C {0, 1}, where C is the set of complex numbers Let on X be such that for all x 1, y 1 C, (x 1, 0) (y 1, 1), for all x 1 + iy 1, z 1 + it 1 C and all x {0, 1}, (x 1 + iy 1, x ) (z 1 + it 1, x ) x 1 z 1 or (x 1 = z 1 and y 1 t 1 ) Then is not even representable by a utility function, it is not a fortiori representable by an additive utility function Hence, we will also require that be representable by a utility function on X This set of axioms is not yet sufficient to ensure additive representability because the structure of the second component is too poor Hence we also require that X be equally spaced and the lemma below follows Definition X is equally spaced if, for all x 1, y 1 X 1 and all i, i + 1, i + N: (x 1, x i ) (y 1, x i+1 ) (x 1, x i+1 ) (y 1, x i+ ) (6) Lemma 6 Let be a weak order on X = X 1 X, where X = {x i, i N} Assume restricted solvability wrt X 1 (axiom 6), independence (axiom 1) and the Archimedean axiom (axiom 1) Suppose that is representable by a utility function, that x i+1 x i for all i, i+1 N, and that X is equally spaced (definition ) Then is representable by an additive utility function 3 Lemma 6 can be viewed as a slight generalization of Wakker (1991): in addition to equal spacedness 4 wrt every component of X, the latter indeed requires restricted solvability wrt every component Of course, the existence of a utility function representing is not a very appealing assumption However, according to a well known theorem (see Debreu (1954, 1964) or Krantz et al (1971, p40)), if there exists a set A X, at most denumerable, such that x y there exist a, b A such that x a b y, then is representable by a 3 This lemma is important because it can serve as a basis to prove the classical representation theorem for two-dimensional Cartesian products see Fishburn (1970, theorem 5, p58) using neither topology nor algebra But this is beyond the scope of this paper and is explained in full detail in Gonzales (1996b) 4 In fact, the definition of equal spacedness stated in Wakker (1991) is different from definition, but together with restricted solvability and weak separability (which are assumed by Wakker (1991)) both definitions are equivalent 11

12 utility function In the present framework, the assumption that there exists a utility function representing can be substituted by the following condition: there exists a denumerable set A 1 X 1 such that x 1 1 y 1 there exist a 1, b 1 A 1 such that x 1 1 a 1 1 b 1 1 y 1 As was mentioned in Wakker (1991), although equal spacedness may have been considered complicated because it is of a combinatorial nature, representation theorems are in fact simpler than in the classical frameworks (in Krantz et al (1971) and Fishburn (1970), the Thomsen condition must be assumed to derive the existence of additive utilities in two-dimensional Cartesian products) 5 Appendix: Proofs Proof of lemma 1: It is straightforward to see that (C ) implies independence and the generalized Thomsen condition Let us prove the converse (C ) can be written as: (x 1, x ) (x 1, x ) } (y 1, y ) (y 1, y ) (z 1, z ) (z 1, z ), (7) where (x i, y i, z i ) are permutations of (x i, y i, z i ) for i {1, } By permutation of the above left hand side (LHS) relations, it is not restrictive to assume that either x or y is equal to x Assume that y = x If y 1 = x 1, then the right hand side (RHS) relation is deduced from the generalized Thomsen condition If y 1 = y 1, then substituting y 1 and y 1 by x 1 and applying the generalized Thomsen condition and then independence gives the RHS relation If y 1 = z 1, then either x 1 = x 1 and, substituting x 1 and x 1 by y 1 and applying the generalized Thomsen condition and independence, one gets the RHS relation, or x 1 = y 1 and, by transitivity, one gets the RHS relation A similar reasoning holds when x = x Hence, the conjunction of independence and the generalized Thomsen condition implies the second order cancellation axiom Proof of lemma : By lemma 1, to prove lemma it is sufficient to show that independence (axiom 1), unrestricted solvability wrt the first component (axiom 5) and the Thomsen condition (axiom ) imply the generalized Thomsen condition (axiom 3) Suppose that the generalized Thomsen condition does not hold Then, there exist x 1, y 1, z 1 X 1 and x, y, z X such that (x 1, z ) (z 1, y ), (z 1, x ) (y 1, z ) and (y 1, y ) (x 1, x ), with at least one strict preference relation By unrestricted solvability wrt the first component, there exists t 1 X 1 such that (x 1, z ) (t 1, y ), and by transitivity, (t 1, y ) (z 1, y ) Therefore, by independence, (t 1, x ) (z 1, x ) (y 1, z ) Hence (x 1, z ) (t 1, y ), (t 1, x ) (y 1, z ) and (y 1, y ) (x 1, x ), with at least one strict preference relation Again, by unrestricted solvability wrt the first component, there exists s 1 X 1 such that (t 1, x ) (s 1, z ) and (x 1, z ) (t 1, y ), (t 1, x ) (s 1, z ) and (s 1, y ) (x 1, x ) But the above implication is impossible because it contradicts the Thomsen condition (axiom ) Hence, by contradiction, it entails lemma 1

13 Proof of lemma 3: Assume that () holds Let x 1 1, x 1, x3 1 X 1 and x 1, x, x3 X be arbitrary elements such that (x 1 1, x ) (x 1, x1 ) and (x1 1, x3 ) (x3 1, x1 ) By symmetry, it is not restrictive to assume that x x 3 Let us show that (x 1, x3 ) (x3 1, x ) If x1 x x 3, then, by (), (x 1, x3 ) (x 3 1, x ) Now assume that either x x 3 x 1 or x x 1 x 3, and that Not((x 1, x 3 ) (x 3 1, x )) (8) By unrestricted solvability wrt the first component, there exists g 1 X 1 such that (g 1, x ) (x 1, x3 ) and, by (8), Not((g 1, x ) (x 3 1, x )) (9) But, then, (x 1, x3 ) (g1, x ), (x 1, x1 ) (x1 1, x ) and either x x 3 x 1 or x x 1 x 3 Hence, by (), (g1, x 1 ) (x1 1, x3 ), and so (g1, x 1 ) (x3 1, x1 ) But this is impossible according to (9) and independence Hence (8) cannot hold Proof of theorems 1 and : Theorem 1 is proved when fixing n = (C m+1 ) (C m ) is a classical result and since (S m+1 ) is a subset of (C m+1 ), the implication from right to left is obvious The proof of the converse is adapted from a proof given in Jaffray (1974a) and Jaffray (199) We first note that (C m ) can be stated as follows: Condition (C m ): Suppose that, for all j {1,, n}, (x 1 j,, xm+1 j ) is a permutation of (yj 1,, ym+1 j ) Then [(x i 1,, xi n) (y1 i,, yi n) for all i {1,, m + 1}] [(x i 1,, xi n) (y1 i,, yi n) for all i {1,, m + 1}] Hereafter we will use this definition instead of axiom 9 in subsection 35 Suppose that (C m ) holds as well as (S m+1 ) if m is even and consider the following sequence of preferences: [(x i 1,, x i n) (y i 1,, y i n) for all i {1,, m + }] (10) First case: there exists i 0 and j 0 such that x i 0 = y j 0 : Substitute x i 0 y i 0 and x j 0 y j 0 by x i 0 y j 0 and x j 0 y i 0 Since x i 0 = y j 0, by removing the line x i 0 y j 0, the conditions of application of (C m ) obtain, and x j 0 y i 0 and x i y i for all i i 0, j 0 But x j 0 y j 0 = x i 0 y i 0, hence x i 0 y i 0 and x j 0 y j 0 Therefore (C m+1 ) holds If the previous case does not obtain, change the order of the preference relations so as to get cycles wrt the first component, ie, a sequence of the form: (x 1 1, x1,, x1 n) (x k 1, y1,, y1 n), (x i 1, xi,, xi n) (x i 1 1, y i,, yi n) for all i {,, k} (11) Second case: k = m + : We know that there exists i such that y i = x 1 If i 1 then, by unrestricted solvability wrt X 1, there exists g i 1 () such that (g i 1 (), y i 1, y i 3,, y i n ) (x i 1 1, y i,, y i n ) But then, by (C ), the following holds: 13

14 (g i 1 (), y i 1, y i 3,, y i n ) (x i 1 1, y i,, y i n ) (x i 1 1, x i 1, x i 1 3,, x i 1 n ) (x i 1, y i 1,, y i 1 n ) }{{} (x i 1, y i, y i 1 3,, y i 1 n ) (g i 1 (), x i 1,, x i 1 n ) So, the lines corresponding to indices i 1 and i in (11) can be substituted by (g i 1 (), x i 1, x i 1 3,, x i 1 n ) (x i 1, y i, y i 1 3,, y i 1 n ) (x i 1, x i, x i 3,, x i n ) (g i 1 (), y i 1, y i 3,, y i n ) (1) (13) Repeat the process with indices i 1, i,, 1, so that y i moves upward till it reaches the first preference relation Thus we get a cycle of the following form: (g 1 (), x 1,, x1 n) (x m+ 1, x 1, y1 3,, y1 n), (g i (), x i, xi n) (g i 1 (), y σ (i), y i 3,, yi n) for all i {,, m + }, where σ is the permutation such that σ (i) = i 1 for all i i and σ (i) = i for all i > i Similarly, there exists i 3 such that y i 3 3 = x 1 3 Repeat with the above cycle the same process, ie, find elements g i (3) to move y i 3 3 upward to the first preference relation By induction, repeat the process for all j 4, hence resulting in the cycle: (g 1 (n), x 1,, x1 n) (x m+ 1, x 1, x1 3,, x1 n), (g i (n), x i, xi n) (g i 1 (n), y σ (i), y σ 3(i) 3,, yn σn(i) ) for all i {,, m + }, (14) where σ j (i) = i 1 for all i i j and σ j (i) = i for all i > i j The formulas for modifying the (r 1)th and the rth lines of the pth cycle constructed are as follows: (i) before the modification, the cycle is 14

15 (g 1 (p 1), x 1,, x1 n) (x m+ 1, x 1,, x1 p 1, y1 p,, yn), 1 (g i (p 1), x i,, xi n) (g i 1 (p 1), y σ (i),, y σ p 1(i) p 1, yp, i, yn) i for all i r 1, (g r (p), x r,, xr n) (g r 1 (p 1), y σ (r),, y σ p 1(r) p 1, yp ip, yp+1 r,, yr n), (g i (p), x i,, xi n) (g i 1 (p), y σ (i),, yp σp(i), yp+1 i,, yi n) for all i > r (ii) g r 1 (p) is defined by (g r 1 (p), y σ (r),, y σ p 1(r) p 1, yp r 1, yp+1 r,, yr n) (g r 1 (p 1), y σ (r),, y σ p 1(r) p 1, yp ip, yp+1 r,, yr n); (iii) formula (13) is generalized by: (g r 1 (p), y σ (r),, y σ p 1(r) p 1, yp r 1, yp+1 r,, yr n) (g r 1 (p 1), y σ (r),, y σ p 1(r) p 1, yp ip, yp+1 r,, yr n) } (g r 1 (p 1), x r 1,, x r 1 n ) (g r (p 1), y σ (r 1),, y σ p 1(r 1) p 1, yp r 1,, yn r 1 ) {{ } (g r (p 1), y σ (r 1),, y σ p 1(r 1) p 1, yp ip, yp+1 r 1 n ) (g r 1 (p), x r 1,, x r 1 n ) and (iv) the (r 1)th and rth lines of the cycle are substituted by: (g r 1 (p), x r 1,, x r 1 n ) (g r (p 1), y σ (r 1),, y σ p 1(r 1) p 1, yp ip, yp+1 r 1,, yr 1 n ), (g r (p), x r,, xr n) (g r 1 (p), y σ (r),, y σ p 1(r) p 1, yp r 1, yp+1 r,, yr n) Now, by independence, the first line of (14) can be substituted by (g 1 (n), x m+,, x m+ n ) (x m+ 1, x m+,, x m+ n ) Then the first case of this proof can be applied and s can be replaced by s in (14) Now to propagate indifference relations back to the initial cycle (11), consider the converse of (1), namely, for the (r 1)th and the rth lines of the cycle generated by the g ( ) (p), the following implications: (g r 1 (p), x r 1,, x r 1 n ) (g r (p 1), y σ (r 1),, y σ p 1(r 1) p 1, yp ip, yp+1 r 1,, yr 1 n ) (g r 1 (p 1), y σ (r),, y σ p 1(r) p 1, yp ip, yp+1 r,, yr n) (g r 1 (p), y σ (r),, y σ p 1(r) p 1, yp r 1, yp+1 r, yr n) }{{} (g r (p 1), y σ (r 1),, y σ p 1(r 1) p 1, yp r 1,, yn r 1 ) (g r 1 (p 1), x r 1,, x r 1 n ) and (g r 1 (p 1), y σ (r),, y σ p 1(r) p 1, yp ip, yp+1 r,, yr n) (g r 1 (p), y σ (r),, y σ p 1(r) p 1, yp r 1, yp+1 r,, yr n) } (g r 1 (p), y σ (r),, y σ p 1(r) p 1, yp r 1, yp+1 r,, yr n) (g r (p), x r,, xr n) {{ } (g r (p), x r,, xr n) (g r 1 (p 1), y σ (r),, y σ p 1(r), yp ip, yp+1 r,, yr n) But, then the (r 1)th and the rth lines can be substituted back to: (g r 1 (p 1), x r 1,, x r 1 n ) (g r (p 1), y σ (r 1),, y σ p 1(r 1) p 1, yp r 1,, yn r 1 ), (g r (p), x r,, xr n) (g r 1 (p 1), y σ (r),, y σ p 1(r) p 1, yp ip, yp+1 r,, yr n) By induction this leads to replacing all the s in (11) by s Hence (C m+1 ) holds Third case: there exist exactly cycles of length k = (m + )/: p 1 15

16 If there exist i, j such that i m+ < j and x i 1 = xj 1, then the second case can be applied since both cycles can be aggregated to form one cycle of length m + Of course, x j 1 appears in several lines, but the process described in the second case can still be applied, hence ensuring that (C m+1 ) holds Assume now that there exist no i, j such that i m+ < j and x i 1 = xj 1 Using transformation (1), each element yj i, i k, such that there exists p k such that x p j = yi j, can be moved in the list (11) such that xp j = yp j A transformation similar to (1) would enable to move x p j along the list instead of yi j If for all j {,, n}, there exist i, p k such that x p j = yi j, then transformations described above enable to modify the initial list (11) so that x 1 j = y j for all j Thus, the first case of this proof can be applied and s can be substituted by s Inverse transformations entail that this result also holds for the initial list, so that (C m+1 ) holds If there exists an index j {,, n} such that, for all i k, there exists no p k such that x p j = yi j, then we are in the conditions of the (m + 1)st order symmetric cancellation axiom, and so (C m+1 ) holds Fourth case: there exists a cycle of length k < m+ : Suppose that y k+1 = x 1 and that the corresponding relation is (xk+1 1, x k+1, x k+1 3,, x k+1 n ) (x p 1, x1, yk+1 3,, yn k+1 ) Then, by unrestricted solvability wrt the first component, there exists g 1 X 1 such that (g 1, x 1, x1 3,, x1 n) (x p 1, x1, yk+1 3,, yn k+1 ) Again, by unrestricted solvability wrt the first component, there exist g,, g k such that, for all i {1,, k 1}, (g i, y i, yi 3,, yi n) (g i+1, x i+1, x i+1 3,, x i+1 n ) and (C k 1 ) implies that (g k, y k, yk 3,, yk n) (g 1, x 1, x1 3,, x1 n) (x p 1, x1, yk+1 3,, yn k+1 ) (x k+1 1, x k+1, x k+1 3,, x k+1 n ) So one can substitute in the initial list the lines (g 1, x 1, x1 3,, x1 n) (g k, y k, yk 3,, yk n) and (x k+1 1, x k+1, x k+1 3,, x k+1 n ) (x p 1, x1, yk+1 3,, yn k+1 ) by (x k+1 1, x k+1, x k+1 3,, x k+1 n ) (g k, y k, yk 3,, yk n) and (g 1, x 1, x1 3,, x1 n) (x p 1, x1, yk+1 3,, yn k+1 ) Hence cycle (1 k) where (1 k) denotes the list of relations located on rows 1 through k and cycle (k + 1 p) now form only one cycle 1 p One can go on with the same process until either the initial list (11) is transformed into a list with a cycle of length m + (with some x j 1 possibly in several lines), or is transformed into a list containing two cycles of length m+ (with some x j 1 possibly in several relations) The former corresponds to case and so all the s can be substituted by s and one can trace the s back to the initial list The latter case can be split into two cases: either no cycle contains several lines with the same x j 1, and we are in the conditions of the third case, or at least one cycle contains several identical x j 1 s So now we have two cycles of length m+, ie, cycles (1 k) and (k + 1 m + ), of which at least one contains several identical x j 1 s Consider that the latter is cycle (1 k) If the identical x j 1 s are not on a same line, ie, the cycle (1 k) can be split into several cycles of length strictly shorter than m+, then the process described above can be used to aggregate one of these small cycles with the second cycle of length m+, hence creating a new cycle of overall length m + ; then using case, we can deduce that (C m+1 ) holds If all the identical x j 1 s are on the same lines, ie, on some lines like (g i, x p, xp 3,, xp n) (g i, y h, y h 3,, y h n), (15) 16

17 then, removing these lines from the cycle 1 k, we get a cycle of length < m+, that we can aggregate with cycle k + 1 m + Then, using independence, lines (15) can be substituted by (x m+ 1, x p, xp 3,, xp n) (x m+ 1, y h, yh 3,, yh n) But then we have created a cycle of length m + and the second case ensures that (C m+1 ) holds Proof of lemmas 4 and 5: Lemma 4 is proved by assigning n = in the rest of the proof The proof is organized in three steps: first, it is shown that independence holds; then the second order cancellation axiom is shown to hold, and, finally, using theorem, the other cancellation axioms are studied First step: independence holds Clearly, by definition of U, independence holds wrt components 3 to n Independence also holds wrt the second component: suppose that (x 1, m, x 3,, x n ) (y 1, m, y 3,, y n ) Then, by unrestricted solvability wrt the first component, there exists g 1 such that (g 1, m, x 3,, x n ) (y 1, m, y 3,, y n ) or, equivalently: 05 (g 1 mod ) + g 1 / + x 3 = 05 (y 1 mod ) + y 1 / + and so 05 (g 1 mod ) 05 (y 1 mod ) = g 1 / + y 1 / + y 3 (y 3 x 3 ) The right hand side being a multiple of, it is clear that there exists an integer k (positive or negative) such that y 1 = g 1 + k But then g 1 / + y 1 / + (y 3 x 3 ) = 0 = g 1 + y 1 + (y 3 x 3 ), or, equivalently, for all x {0,, (m 1)}, g 1 + x + n (x 3) + H = y 1 + x + n (y 3)+H Therefore, since U represents, (g 1, x, x 3,, x n ) (y 1, x, y 3,, y n ), and since 05 (x 1 mod ) + x 1 / 05 (g 1 mod ) + g 1 /, (x 1, x, x 3,, x n ) (y 1, x, y 3,, y n ) Conversely, if (g 1, x, x 3,, x n ) (y 1, x, y 3,, y n ), x m, then, since all the x i s and all the y i s, i, are multiples of, there exists an integer k (positive or negative) such that g 1 = y 1 + k Then 05 (g 1 mod ) + g 1 / + n x 3 = 05 (y 1 mod ) + y 1 / + n y 3 and so (g 1, m, x 3,, x n ) (y 1, m, y 3,, y n ) Independence axiom also holds wrt X 1 Indeed, assume that (x 1, x, x 3,, x n ) (x 1, y, y 3,, y n ) If x = y or if x m and y m, then, clearly, by definition of U, (y 1, x, x 3,, x n ) (y 1, y, y 3,, y n ) for any y 1 R Now, y = m and x m is impossible because 05 (x 1 mod ) + x 1 / + n y 3 x 1 05 and x i + H x 1 + (m 1) + (n ) + H x 1 (m 1) n(m + 1) + 35 i=1 and so, since m 3 and n, n i=1 x i + H x 1 85 Therefore, assume that (x 1, m, x 3,, x n ) (x 1, y, y 3,, y n ) Then, clearly, for any y 1 R, 05 (y 1 mod ) + y 1 / + n y 3 > n i=1 y i + H, and so (y 1, m, x 3,, x n ) (y 1, y, y 3,, y n ) Hence independence holds 17

18 Second step: (C ) holds Assume that (C ) does not hold, ie, that there exist x j i, yj i, i {1,, n}, j {1,, 3}, such that (yi 1, y i, y3 i ) are permutations of (x1 i, x i, x3 i ) and such that (x 1 1, x1, x1 3,, x1 n) (y1 1, y1, y1 3,, y1 n) (x 1, x, x 3,, x n) (y1, y, y 3,, y n) (16) (x 3 1, x3, x3 3,, x3 n) (y1 3, y3, y3 3,, y3 n) with at least one strict preference Let us prove that (16) cannot hold If x j = m for all j {1,, 3}, then, by independence, one can substitute all the x j, yj by 0; translating these preference relations in terms of utility U, and summing the three inequalities thus obtained, one gets 3 j=1 ( n i=1 xj i + H) > 3 j=1 ( n i=1 yj i + H) But the y j i s are permutations of the xj i s, so that the last inequality cannot hold Similarly, if x j m for all j {1,, 3}, then (16) cannot hold If there exist i, j, i j, such that x i = xj = m, then at least one preference relation has a m term on its right and left hand sides By independence, one can substitute the m s by 0 s So, one need only study the following case: there exists a unique index i such that x i = m, and, moreover, yi m Without loss of generality, by reordering the preference relations, x 1 = m and y = m Hence the following relations: (x 1 1, m, x1 3,, x1 n) (y1 1, y1, y1 3,, y1 n) (x 1, x, x 3,, x n) (y1, m, y 3,, y n) (x 3 1, x3, x3 3,, x3 n) (y1 3, y3, y3 3,, y3 n) with at least one strict preference If y1 = x1 1, then, summing the utilities over the three preference relations, one gets: 05 ( x 1 1 mod ) + x 1 1 / + x 1 + x3 1 x1 + ( 3 n ) j=1 i= xj i + H > 05 ( x 1 1 mod ) + x 1 1 / + x 1 + x3 1 x1 + 3 j=1 ( n i= ) xj i + H Hence y1 = x1 1 is impossible If y 1 = x 1, then, translating the second preference relation into its utility counterpart, one gets: x 1 +x + n x i +H V (x 1 )+ n y i But this is impossible because V (x 1 ) x 1 05 and, since m 3 and n, x 1 +x + n x i +H x 1 85 Hence y 1 = x3 1 Consequently, if (16) holds, then it can be written as: (x 1 1, m, x1 3,, x1 n) (y1 1, y1, y1 3,, y1 n) (x 1, x, x 3,, x n) (x 3 1, m, y 3,, y n) (x 3 1, x3, x3 3,, x3 n) (y1 3, y3, y3 3,, y3 n) with at least one strict preference If y1 3 = x 1, then, in terms of utility U, the last two preference relations are equivalent to x 1 V (x3 1 ) + n (y i x i ) x H and x3 1 x 1 + n i= (y3 i x3 i ) Therefore x 3 1 V (x 3 1) + (yi x i ) + (yi 3 x 3 i ) + y 3 x 3 x H 18

19 But since the y j i s are permutations of the xj i s, n (y i x i ) + n (y3 i x3 i ) = n (x1 i y1 i ) And since m 3 and n, H = [(m ) + n(m + 1) 45] + (n ) + 4(m 1) (yi 1 x 1 i ) + x 3 + x + [(m ) + n(m + 1) 45] (yi 1 x 1 i ) + x 3 + x + 45, one gets: x 3 1 V (x3 1 ) + 45, which is impossible since V (x3 1 ) x Therefore, y1 3 = x1 1 and (16) can be written as: (x 1 1, m, x1 3,, x1 n) (x 1, y1, y1 3,, y1 n) (x 1, x, x 3,, x n) (x 3 1, m, y 3,, y n) (x 3 1, x3, x3 3,, x3 n) (x 1 1, y3, y3 3,, y3 n) with at least one strict preference Now, by unrestricted solvability wrt the first component, there exists g R such that (g, x, x 3,, x n) (x 3 1, m, y 3,, y n) and there exists g 1 R such that (g 1, m, x 1 3,, x1 n) (g, y 1, y1 3,, y1 n) But then, (g 1, m, x 1 3,, x1 n) (g, y 1, y1 3,, y1 n) (g, x, x 3,, x n) (x 3 1, m, y 3,, y n) (x 3 1, x3, x3 3,, x3 n) (g 1, y 3, y3 3,, y3 n) Therefore, in terms of utility function U, V (g 1 ) + x 1 i = g + y 1 + yi 1 + H and g + x + x i + H = V (x 3 1) + yi Replacing in the first equality g by its value given by the second equality, V (g 1 ) + n x1 i = V (x3 1 ) + n (y i x i ) x + y1 + n y1 i or, equivalently, V (g 1 ) = V (x 3 1) + (yi x i ) x + y 1 + (yi 1 x 1 i ) (17) Now, since the y j i s are permutations of xj i s, m+x +x3 = y1 +m+y3 and n (y1 i x 1 i ) + n (y i x i ) + n (y3 i x3 i ) = 0 So, (17) is equivalent to V (g 1 ) + y 3 + yi 3 = V (x 3 1) + x 3 + x 3 i Since x j i and yj i, i, are multiples of, there exists an integer k (positive or negative) such that g 1 = x k, and so g1 + y 3 + n y3 i = x3 1 + x3 + n x3 i But this equality corresponds to the preference relation (g 1, y 3,, y3 n) (x 3 1, x3,, x3 n) So (16) cannot occur and, consequently, (C ) holds 19

20 Third step: (C i ), i 3, holds According to theorem, it is sufficient to show that (S i ) holds for any i 3 Let k (m + 1)/ and assume that (S k 1 ) does not hold, ie, that there exist elements x i j, yi j such that xi 1 xp 1 for all i k < p and either yi j = xi j or yi j = xi j + k, and such that (x k+1 1, x k+1 (x k+j 1, x k+j (x 1 1, x1,, x1 n) (x k 1, y1,, y1 n), (x i 1, xi,, xi n) (x i 1,, x k+1 n,, x k+j n 1, y i,, yi n), i {,, k}, n ),,, yn k+j ), j {,, k}, ) (x k 1, yk+1,, y k+1 ) (x k+j 1 1, y k+j (18) with at least one strict preference relation If there exists no j such that x j = m or y j = m, then (18) cannot occur because it would violate a cancellation axiom and because is represented on this subspace by an additive utility function So, suppose that there exists at least one index j {1,, k} such that x j = m or yj = m When x j = yj, by independence substitute xj and yj by 0 s Now, by symmetry, one can suppose that there exist p indices {i 1, i,, i p }, i j k, such that y i j = m Using transformation (1) as in the proof of theorem, we can furthermore suppose that i 1,, i p are the first p indices This transformation has no consequence on the rest of the proof except to simplify notation Now translating the k first preference relations into their U counterpart, we get: Thus, k i=1 j=1 i=1 p 1 x i j + kh V (x i 1) + V (x k 1) + i=1 i=1 k i=p+1 k p 1 x i [ [ ] V (x i 1 ) x i 1] + V (x k 1) x k 1 + (x i y i + H) + k i=p+1 y i ph + Now by (5), V (x) x 05, so the inequality above implies that k x i 05p + i=1 k i=p+1 But k m+1, so, expanding H, we get i=1 k i=1 j=3 k i=1 j=3 yj i y i ph k(n ) 05p ph k(n ) k x i 05p + p[m + mn + 3n 85] m + 1 (n ) ( pm + p 1 ) ( mn + m + 3p 1 ) n 9p + 1 Since m 3, n and p 1, k i=1 (yj i x i j) ( x i pm + 6 p 1 ) ( p 1 ) 9p + 1 pm + 3p m + 3 0

21 But this is impossible because k m+1 and x i (m 1) since xi yi j = m, so that k i=1 x i m + 1 (m 1) = m 1 Hence (18) cannot hold Thus, (S i ) holds for any i {3,, m}, and consequently (C m ) also holds Since m is odd, by theorem, (C m+1 ) also holds In order to prove lemma 6, I introduce the following lemma: Lemma 7 Let be a weak order on X = X 1 X Assume restricted solvability wrt X 1 (axiom 6), independence (axiom 1) and the Archimedean axiom (axiom 1) Suppose that is representable by a utility function, and that Card(X ) = Then is representable by an additive utility function Proof of lemma 7: Let X = {a, b} If a b, the proof is obvious Assume henceforth that b a Let X1 be the set of equivalent classes of 1 The existence of an additive utility function on X 1 {a, b} is equivalent to the existence of an additive utility function on X 1 {a, b} The advantage of working in the latter set rather than in X 1 {a, b} is that the utility function is one-to-one in this space So, let us work in X 1 {a, b} By hypothesis, is representable by a utility function, say u(, ) Let u a ( ) = u(, a) and u b ( ) = u(, b) To prove the existence of an additive utility, it is sufficient to prove that there exists a strictly increasing function, say ϕ( ), transforming u( ) into an additive function In other words, there exists α R + such that ϕ u b (x) = ϕ u a (x) + α, for all x X 1 (19) Suppose that (x 1, a) (y 1, b) for all x 1, y 1 X 1 Then, in terms of utility functions, u a ( X 1 ) u b ( X 1 ) = Let ϕ( ) be defined by: { arctan(x) if x ua ( ϕ(x) = X 1 ), arctan(u a (u 1 b (x))) + π if x u b ( X 1 ) Then it is obvious that ϕ u(, ) is an additive utility representing on X 1 {a, b} Now, suppose that u a ( X 1 ) u b ( X 1 ) (19) is equivalent to the following equation: ϕ (u b u 1 a )(x) = ϕ(x) + α, for all x u a ( X 1 ) (0) Note that the value of α is not important, as long as it remains strictly positive Indeed, if it were replaced by β > 0, ϕ( ) would be replaced by β α ϕ( ) Let x0 u a ( X 1 ) The value of ϕ(x 0 ) can be taken arbitrarily; indeed, adding a constant to ϕ( ) has no consequence on the validity of (0) Now, let us construct a function ϕ( ) satisfying (0) We first define ϕ( ) on [x 0, u b u 1 a (x 0 )] as: ϕ(x) = α(x x 0 ) u b u 1 a (x 0 ) x + 0 ϕ(x0 ), for all x [x 0, u b u 1 a (x 0 )] (1) 1

22 Note that since (x, a) (x, b) for every x X 1, u b u 1 a (x 0 ) > x 0, so that ϕ( ), as defined by equation (1), is well defined and is strictly increasing Extend ϕ( ) as follows: for all x [x 0, u b u 1 a (x 0 )] u a ( X 1 ), equation (0) can be applied Let x 1 = sup{x : x [x 0, u b u 1 a (x 0 )] u a ( X 1 )} Then, applying (0) for all x [x 0, x 1 [ u a ( X 1 ) if x 1 is not reached (resp [x 0, x 1 ] u a ( X 1 ) otherwise), we define ϕ( ) on [u b u 1 a (x 0 ), u b u 1 a (x 1 )[ u b ( X 1 ) (resp [u b u 1 a (x 0 ), u b u 1 a (x 1 )] u b ( X 1 )) In fact, restricted solvability wrt X 1 ensures that we define ϕ( ) on [u b u 1 a (x 0 ), u b u 1 a (x 1 )[ (u a ( X 1 ) u b ( X 1 )) (resp [u b u 1 a (x 0 ), u b u 1 a (x 1 )] (u a ( X 1 ) u b ( X 1 ))) Moreover, ϕ( ) strictly increases because, by independence, u b u 1 a (x) > x for all x u a ( X 1 ) and α > 0 Now, two cases can occur In the first one, x 1 = u b u 1 a (x 0 ) but x 1 u a ( X 1 ), or x 1 < u b u 1 a (x 0 ) Then, by restricted solvability, there exists no element, say y X 1, such that u a (y) > x 1 otherwise, by definition of x 1, u a (y) > u b u 1 a (x 0 ), which would imply that (y, a) (u 1 a (x 0 ), b) (u 1 a (x 0 ), a) and would ensure the existence of t X 1 such that (t, a) (u 1 a (x 0 ), b) Therefore, ϕ( ) has been defined on {x u a ( X 1 ) u b ( X 1 ) such that x x 0 } In the second case, x 1 = u b u 1 a (x 0 ) and x 1 u a ( X 1 ) Then ϕ( ) has been defined on [x 0, (u b u 1 a ) (x 0 )] (u a ( X 1 ) u b ( X 1 )) Let x = sup{x : x [x 0, (u b u 1 a ) (x 0 )] u a ( X 1 )} A process similar to that of the preceding paragraphs can be applied with the set [x 1, x ] instead of [x 0, x 1 ] By continuing the same process, we construct a sequence of points (x i ) n i=0, where n is either an integer or the infinity ϕ( ) is then defined either on [x 0, x n ] (u a ( X 1 ) u b ( X 1 )) or on [x 0, x n [ (u a ( X 1 ) u b ( X 1 )), depending on whether x n is reached or not If n is finite, by restricted solvability wrt the first component, ϕ( ) has been defined on the set {x u a ( X 1 ) u b ( X 1 ) : x x 0 } Problems arise when n is infinite In such a case, ϕ( ) has been defined on [x 0, (u b ) n (x 0 )[ (u a ( X 1 ) u b ( X 1 )) Now, suppose that there exists x u b ( X 1 ) such that, for every integer i, (u b u 1 a ) i (x 0 ) < x By construction, ϕ((u b u 1 a ) i (x 0 )) = ϕ(x 0 )+iα; u 1 a hence, this should tend toward + when i tends toward + But, then, the problem is that ϕ(x) cannot be finite, and so ϕ( ) is not well defined on u a ( X 1 ) u b ( X 1 ) Fortunately, this case cannot arise For all integers i, i + 1 0, x i+1 = u b u 1 a (x i ) Now, by the process of construction, x i, x i+1 u a ( X 1 ); so, there exists a sequence of elements of X1, say (y i ) i 0, such that x i = u a (y i ) for all i But, then, u a (y i+1 ) = u b (y i ) In terms of preference relations, (y i+1, a) (y i, b) Therefore, (y i ) is a standard sequence So, by the Archimedean axiom, if x existed such that x is greater than (u b u 1 a ) i (x 0 ) for all i, then {i} would be finite Hence, the case described in the previous paragraph would not arise Consequently, ϕ( ) has been well defined on {x u a ( X 1 ) u b ( X 1 ) : x x 0 } Now, using (0) conversely, ie, ϕ (u a u 1 b )(x) = ϕ(x) α, for all x u b ( X 1 ), and applying a process similar to that of the preceding paragraphs, staring with the set [x 0, (u b u 1 a )(x 0 )], one can construct ϕ( ) on {x u a ( X 1 ) u b ( X 1 ) : x 0 x} And, of course, as shown above, ϕ( ) is strictly increasing Hence, ϕ( ) exists on u a ( X 1 ) u b ( X 1 ) Note that ϕ( ) is not unique up to positive affine transformations because ϕ( ) can be defined with a certain degree of freedom in equation (1)