The majority rule with arbitrators
Antonio Quesada†
Departament d’Economia, Universitat Rovira i Virgili, Avinguda de la Universitat 1, 43204 Reus, Spain
4th November 2010
169.1
Abstract
The majority rule with arbitrators is the (relative) majority rule supplemented by the indifferencebreaking rule according to which, in case of social indifference under the majority rule, the social
preference coincides with the preference of the non-indifferent individual ranked higher in an exogenous
ranking of the individuals. For the case of preferences over two alternatives, a characterization of this rule
is suggested that relies on axioms of efficiency, inter-anonymity, intra-anonymity, parity and reducibility.
It is also provided a characterization of the set of rules consisting of the majority rule and the majority
rule with arbitrators.
Keywords: Social welfare function, relative majority rule, tie-breaking rule, axiomatic characterization,
two alternatives.
JEL Classification: D71
†
E-mail address: [email protected]. Financial support from the Secretaría de Estado de Investigación of the
Spanish Ministerio de Ciencia e Innovación (research project SEJ2007-67580-C02-01) is gratefully
acknowledged. This paper is dedicated to the memory of Estela Calduch Noll.
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1. Introduction
The relative majority rule (or majority rule, for short) is the weakest majority concept
and constitutes one of the most popular procedures to aggregate preferences over two
alternatives: if the number of individuals preferring one alternative to the other is
greater than the number of individuals having the reverse preference, then the first
alternative is declared socially preferred to the second; otherwise, social indifference is
the outcome. There are several axiomatizations of the majority rule when preferences
are defined over only two alternatives. May (1952, p. 682), Fishburn (1973, p. 58; 1983,
p. 33) and Llamazares (2006, p. 319) provide characterizations for the case in which a
set of individuals is given and their preferences can vary. Aşan and Sanver (2002, p.
411), Woeginger (2003, p. 91; 2005, p. 9) and Miroiu (2004, p. 362) consider the more
general framework in which both the set of individuals and their preferences can
change. Xu and Zhong (2010, p. 120) suggest a characterization when the set of
individuals is variable but their preferences remain fixed.
Under the majority rule, the proportion of cases in which social indifference arises is, at
least for small groups, significant. For instance, with three individuals, the total number
of preference profiles over two alternatives is 27 and indifference is associated with 7
profiles. With four individuals, the proportion is 19 out of 81. This is an unappealing
feature when the aim is to make a social choice on the basis of the social preference
because indifference points to no alternative to be chosen.
An indifference-breaking rule defines a simple way of making the majority rule more
resolute. This paper considers the following such rule: there is a ranking among
individuals such that, when a social indifference results, the non-indifferent individual
appearing first in the ranking determines the social preference and, therefore, the social
choice. The individual resolving the social indifference may be viewed as an arbitrator,
that is, an individual who is given a casting vote to be exercised only when the majority
rule yields a tie. For instance, the Vice President of the United States, when acting as
President of the United States Senate, holds a casting vote.
Propositions 4.3 characterizes the majority rule with arbitrators in terms of the five
axioms E, A, R, P and S. E expresses Pareto-efficiency. A is an intra-anonymity
condition establishing conditions under which permuting the preferences of two
individuals in a given group does not alter the social preference. R is a reducibility
property according to which the social preference of a certain group I coincides with the
social preference of some subgroup J of I or with the social preference of I\J. P is a
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parity requirement: for any group consisting of two individuals, the two possible strict
preferences arise the same number of times. S represents an inter-anonymity condition:
any individual not belonging to any given group of two individuals can always replace
exactly one of the members of the group without altering the social preference.
Proposition 4.7 provides further characterizations of the majority rule with arbitrators as
a by-product of Proposition 4.6, which characterizes the set of rules given by the
majority rule and the majority rule with arbitrators in terms of the six axioms Z, C, H,
U, P and S. Z holds that the removal of an indifferent individual does not affect the
social preference. C identifies conditions under which opposite preferences can be
cancelled out without altering the social preference. H requires that, for indifferent
groups having at least three individuals, there must be a decomposition of the group into
two indifferent subgroups. U is just the unanimity principle.
Given Proposition 4.6, a parallel characterization of the majority rule and the majority
rule with arbitrators can be obtained by postulating a separating property, that is, a
condition satisfied by one of the two types of rule and not the other. Proposition 4.7
illustrates that strategy.
2. Definitions
The members of the set ℕ of positive integers are names for individuals. A society is a
finite non-empty subset of ℕ. There are two alternatives, x and y. A preference over {x,
y} is represented by a number from the set {−1, 0, 1}. If the number is 1, x is preferred
to y; if −1, y is preferred to x; if 0, x is indifferent to y. A preference profile for society I
is a function pI : I → {−1, 0, 1} assigning a preference over {x, y} to each member of I.
The set P is the set of all preference profiles pI such that I is a society. For pI ∈ P,
society J ⊆ I and i ∈ I, pJ is the restriction of pI to J and pi abbreviates pI(i). For n ∈ ℕ,
Pn is the set of all preference profiles xI such that ⏐I⏐ = n. For a ∈ {−1, 0, 1} and
society I, (aI) stands for the preference profile pI for I such that, for all i ∈ I, pi = a.
When I is a singleton {i}, ai abbreviates a{i}; for instance, (ai, bj) is the member pI of P2
such that pi = a and pj = b.
Definition 2.1. A social welfare function is a mapping f : P → {−1, 0, 1}.
A social welfare function takes as input the preferences over {x, y} of all the members
of any given society I and outputs what can be interpreted as a social preference of I
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over {x, y}. Specifically, for pI ∈ X: (i) f(pI) = 1 means that, according to f, society I
prefers x to y; (ii) f(pI) = −1, that I prefers y to x; and (iii) f(pI) = 0, that I is indifferent
between x and y. Alternatively, f(pI) may be viewed as a social decision: f(pI) = 1 means
that society I chooses x; f(pI) = −1, that I chooses y; and f(pI) = 0, that no alternative is
chosen. For pI ∈ P and a ∈ {−1, 0, 1}, na(pI) = ⏐{i ∈ I: pi = a}⏐.
Definition 2.2. The majority rule is the social welfare function μ such that, for all pI ∈
P: (i) if n1(pI) > n−1(pI), then μ(pI) = 1; (ii) if n1(pI) < n−1(pI), then μ(pI) = −1; and (iii) if
n1(pI) = n−1(pI), then μ(pI) = 0.
Definition 2.3. For linear order → on ℕ, the majority rule with arbitrators determined
by → is the social welfare function α→ such that, for all pI ∈ P: (i) if μ(pI) ≠ 0 or pI =
(0I), then α→(pI) = μ(pI); and (ii) otherwise, α→(pI) = pi, where i is the member of J = {j
∈ I: pj ≠ 0} such that either J = {i} or, for all j ∈ J\{i}, i → j.
3. Axioms
U. Unanimity. For every society I and a ∈ {−1, 0, 1}, f(aI) = a.
E. Efficiency. For all a ∈ {−1, 0, 1}, i ∈ ℕ and j ∈ ℕ\{i}, f(ai) = f(ai, aj) = f(ai, 0j) = a.
P. Parity. For every society I with ⏐I⏐ = 2, ⏐{pI ∈ P: f(pI) = 1}⏐ = ⏐{pI ∈ P: f(pI) =
−1}⏐.
Axiom P is an implication of the standard neutrality property, as defined by May’s
(1952, p. 682) Condition III. If f(pI) = a ≠ 0 means that society I chooses a given
preference profile pI, then P asserts that, in societies with two members, the two
alternatives are chosen the same number of times.
A. Anonymity. For all pI ∈ P such that f(pI) ≠ 0, if ⏐I⏐ ≥ 3 and {i ∈ I: pi = 0} = ∅, then,
for each i ∈ I satisfying f(pI) ≠ pi, there is j ∈ I such that f(pI) = pj and f(pIi↔j) = f(pI),
where pIi↔j is the preference profile qI that differs from pI only in that qi = pj and qj = pi.
S. Substitutability. For every society I with ⏐I⏐ = 2 and i ∈ ℕ\I, there is j ∈ I such that,
for all pI ∈ P, f(pI) = f(pI\{j}, (pj)i).
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Axioms A i S are anonymity-type conditions. The inter-anonymity axiom S holds that,
for societies with two members, any individual outside that society can always replace a
given member of the society without causing a change in the social preference. The
intra-anonymity axiom A asserts the following. Let I be a society with at least three
members, none of which is indifferent and such that the social preference is strict. Then,
for every member i of I with a strict preference different from the social strict
preference, there is another member j of I whose preference is the opposite of i’s and
such that the permutation of i and j’s preferences does not modify the social preference.
The standard anonymity axiom makes every permutation of every two individuals in
any given society and for every preference profile innocuous; see May’s (1952, p. 681)
Condition II.
C. Opposite preferences may cancel out. For all pI ∈ P and i ∈ I , if ⏐{k ∈ I: pk ≠ 0}⏐ ≥
3, f(pI) ≠ 0 and pi = −f(pI), then there is j ∈ I such that pj = f(pI) and f(pI\{i,j}) = f(pI).
Z. Indifference cancels out. For every society I such that⏐I⏐ ≥ 2 and i ∈ I, if pi = 0, then
f(pI) = f(pI\{i}).
Axioms C and Z are cancellability properties. They are related to Llamazares’ (2006, p.
316) cancellativeness, according to which (1i, −1j) can always be replaced by (0i, 0j)
without altering the social preference. Xu and Zhong’s (2010, p. 120) independence of
an unconcerned coalition is a more general cancellability of indifference requirement:
for disjoint societies I and J, f(pJ) = 0 implies f(pI, pJ) = f(pI). Z asserts that, in societies
with at least two members, any individual indifference can be cancelled out. C states
that, for non-indifferent societies where at least three individuals are not indifferent, if
i’s preference is the opposite of the social preference, then, for some j whose preference
agrees with the social preference, it is possible to remove i’s preference and j’s
preference without altering the social preference.
H. Indifference is hereditary. For all pI ∈ P such that ⏐I⏐ ≥ 3, if f(pI) = 0, then there is a
non-empty J ⊂ I such that f(pJ) = f(pI\J) = 0.
H requires that any indifferent society I having at least three members can be partitioned
into two indifferent subsocieties.
R. Reducibility. For every society I and every non-empty J ⊂ I, f(pI) ∈ {f(pJ), f(pI\J)}.
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R is the condition that the problem of aggregating the preferences of society I can
always be reduced to the problem of aggregating the preferences of one of two
subsocieties of I: J or I\J. The properties of weak path independence in Aşan and Sanver
(2002, p. 411), subsets decomposability in Miroiu (2004, p. 361) and reducibility to
subsocieties in Woeginger (2003, p. 90) also relate the solution of an aggregation
problem to the solution of simpler aggregation problems. Despite this, R has no direct
relationship with any of those three properties.
Remark 3.1. μ satisfies Z, C, H, U, P and S.
It should not be difficult to check that μ satisfies Z, U, P (μ treats alternatives
symmetrically) and S (μ treats individuals symmetrically). With respect to C, let μ(pI) =
a ≠ 0, ⏐I⏐ ≥ 3 and pi = −a, with i ∈ I. Since μ(pI) = a, there must be j ∈ I such that pj =
a. Therefore, μ(pI) = a implies μ(pI\{i,j}) = a. As regards H, suppose μ(pI) = 0, where ⏐I⏐
≥ 3. If, for some i ∈ I and a ∈ {−1, 1}, pi = a, then, for some j ∈ I, pj = −a. Since μ(pI) =
0, μ(pI\{i,j}) = 0. Given that μ(p{i,j}) = 0, it follows that J = {i, j} ⊂ I satisfies μ(pJ) =
μ(pI\J) = 0. On the other hand, if there are no i ∈ I and no a ∈ {−1, 1} such that pi = a,
then μ(pI) = 0 implies pI = (0I). In this case, as μ satisfies U, for any non-empty J ⊂ I,
μ(pJ) = μ(pI\J) = 0.
Remark 3.2. For any linear order → on ℕ, α→ satisfies E, Z, U, C, H, A, R, P and S.
Let → be a linear order on ℕ. It should not be difficult to verify that α→ satisfies E, Z,
U, P and H (because α→(pI) = 0 is equivalent to pI = (0I)). With respect to C, suppose
α→(pI) = a ≠ 0 and pi = −a, with the restrictions on I assumed in C. Since α→(pI) = a,
μ(pI) ∈ {0, a}. If μ(pI) = a, then there must be j ∈ I such that pj = a. As μ(pI) = a implies
μ(pI\{i,j}) = a, by definition of α→, α→(pI\{i,j}) = a. If μ(pI) = 0, then α→(pI) = a ≠ 0 means
that pk = a, where k ∈ I satisfies, for all m ∈ {r ∈ I\{k}: pr ≠ 0}, k → m. By the
assumption that at least three individuals in I have preferences different from 0, μ(pI) =
0 implies that, for some r ∈ I\{k}, pr = a. Hence, pI\{i,r} = (ar, −ai) and μ(pI) = 0 imply
μ(pI\{i,r}) = 0. Consequently, α→(pI\{i,r}) = pk = a, which proves C.
As regards A, let α→(pI) = a ≠ 0, ⏐I⏐ ≥ 3, {i ∈ I: pi = 0} = ∅ and α→(pI) ≠ pi. Therefore,
pi = −a. If μ(pI) = a, then, for all j ∈ I such that pj = a, μ(pIi↔j) = μ(pI). Since μ(pIi↔j) =
a, by definition of α→, α→(pIi↔j) = μ(pIi↔j) = a = α→(pI). If μ(pI) = 0, then let k be the
member of I ranked first in the linear order → restricted to I. Hence, α→(pI) = pk = a. As
μ(pI) = 0 and no individual in I has preference 0, ⏐I⏐ ≥ 4 and there is j ∈ I\{k} such that
pj = a. Accordingly, α→(pIi↔j) = pk = α→(pI).
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To show that R holds, observe that α→(pI) = 0 is equivalent to pI = (0I). Given this, for
any non-empty J ⊂ I, α→(pJ) = α→(pI\J) = 0. If α→(pI) = a ≠ 0, then there must be i ∈ I
such that pi = a. As a result, α→(pI) ∈ {α→(pi), α→(pI\{i})} = {a, α→(pI\{i})}. Lastly,
concerning S, let I = {i, j} and i → j. Consider any k ∉ {i, j}. If i → k then k can always
replace j in any p{i,j} ∈ P without altering the outcome and if k → i then, by the
transitivity of →, k can always replace i in any p{i,j} ∈ P without altering the outcome.
4. Results
Proposition 4.3 (based on Lemmas 4.1 and 4.2) provides a direct characterization of the
majority rule with arbitrators. Proposition 4.6 (based on Lemmas 4.4 and 4.5) offers a
characterization of the rules that are either the majority rule or the majority with
arbitrators. Proposition 4.7 presents another characterization of the majority rule with
arbitrators that goes through Proposition 4.6. Lemmas 4.4 and 4.5 have the added
interest of showing that Z, C and H are necessary and sufficient to make a social welfare
function agree with: (i) the majority rule if this rule already determines preference
aggregation in societies with one or two members; and (ii) the majority rule with
arbitrators if this rule already determines preference aggregation in societies with one or
two members. In other words, Z, C and H extend the majority rule from societies with at
most two members to all societies and also extend any majority rule with arbitrators
from societies with at most two members to all societies.
Lemma 4.1. Suppose that, for some linear order → on ℕ, f = α→ on P1 ∪ P2. Then f
satisfies A and R if and only if f = α→.
Proof. “⇐” Remark 3.2. “⇒” Taking f = α→ on P1 ∪ P2 as the base case of an induction
argument, choose n ∈ ℕ\{1, 2} and suppose that f = α→ on P1 ∪ … ∪ Pn−1. To show
that f = α→ on Pn, let pI ∈ Pn.
Case 1: μ(pI) = 0. Case 1a: for some i ∈ I, pi = 0. Case 1a1: for all j ∈ I, pj = 0. Choose
any non-empty J ⊂ I. By R, f(pI) ∈ {f(pJ), f(pI\J)} = {f(0J), f(0I\J)}. By the induction
hypothesis, f(0J) = α→(0J) = 0 and f(0I\J) = α→(0I\J) = 0. As a result, f(pI) = 0 = α→(pI).
Case 1a2: for some j ∈ I, pj = a ≠ 0. This means that α→(pI) ≠ 0. Let j ∈ I satisfy pj = a
≠ 0. By the induction hypothesis, f(pj) = α→(pj) = pj = a. Since μ(pj) = a and μ(pI) = 0,
μ(pI\{j}) = −a. Therefore, by the induction hypothesis and the definition of α→, f(pI\{j}) =
α→(pI\{j}) = μ(pI\{j}) = −a. By R, f(pI) ∈ {f(pj), f(pI\{j})} = {a, −a}. In sum, f(pI) ≠ 0. Let i
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∈ I satisfy pi = 0, so α→(pI\{i}) = α→(pI). By R, f(pI) ∈ {f(pi), f(pI\{i})}. By the induction
hypothesis, f(pi) = α→(pi) = pi = 0 and f(pI\{i}) = α→(pI\{i}) = α→(pI). Accordingly, f(pI) ∈
{0, α→(pI)}. As f(pI) ≠ 0, it must be that f(pI) = α→(pI).
Case 1b: for all i ∈ I, pi ≠ 0. It then follows from μ(pI) = 0 that ⏐I⏐ ≥ 4 and that, for
some K ⊂ I, pI = (aK, −aI\K), where ⏐K⏐ = ⏐I\K⏐. Let i ∈ I satisfy, for all j ∈ I\{i}, i →
j. Hence, α→(pI) = pi ≠ 0. It must be shown that f(pI) = pi. By R and the induction
hypothesis, f(pI) ∈ {f(aK), f(−aI\K)} = {α→(aK), α→(−aI\K)} = {a, −a}. That is, f(pI) ≠ 0.
Case 1b1: for some j ∈ I\{i} such that pj = pi and some k ∈ I such that pk = −pi, j → k.
By the induction hypothesis, f(p{j,k}) = α→(p{j,k}) = pj = pi and f(pI\{j,k}) = α→(pI\{j,k}) = pi.
By R, f(pI) ∈ {f(p{j,k}), f(pI\{j,k})} = {pi, pi}. Case 1b2: for all j ∈ I\{i} such that pj = pi
and all k ∈ I such that pk = −pi, k → j. Suppose f(pI) ≠ pi. As f(pI) ≠ 0, f(pI) = −pi.
Consider any j ∈ I\{i} such that pj = pi. By A, for some k ∈ I such that pk = −pi, f(pIi↔j)
= f(pI). This leads to case 1b1, because k → j and, in pIi↔j, k’s preference agrees with i’s,
whereas j’s preference disagrees with i’s. Consequently, by case 1b1, f(pIi↔j) = pi. This
and f(pIi↔j) = f(pI) imply f(pI) = pi: contradiction.
Case 2: μ(pI) = a ≠ 0. This means that α→(pI) = a and that na(pI) ≥ n−a(pI) + 1. It must be
shown that f(pI) = a. Case 2a: na(pI) > n−a(pI) + 1. Hence, there is a non-empty J ⊂ I
such that na(pJ) > n−a(pJ) and na(pI\J) > n−a(pI\J). By the induction hypothesis, f(pJ) =
α→(pJ) = μ(pJ) = a and f(pI\J) = α→(pI\J) = μ(pI\J) = a. By R, f(pI) ∈ {f(pJ), f(pI\J)} = {a,
a}. Case 2b: na(pI) = n−a(pI) + 1. If n−a(pI) = 0, then, by the induction hypothesis, for all i
∈ I, f(pi) = a = f(pI\{i}). By R, f(pI) ∈ {f(pJ), f(pI\J)} = {a, a}. Now, let n−a(pI) ≥ 1. Case
2b1: there are j ∈ I and k ∈ I such that pj = a, pk = −a and j → k. By the induction
hypothesis, f(p{j,k}) = α→(p{j,k}) = pj = a. Moreover, μ(pI) = a and p{j,k} = (aj, −ak) imply
μ(pI\{j,k}) = a. By the induction hypothesis, f(pI\{j,k}) = α→(pI\{j,k}) = μ(pI\{j,k}) = a. By R,
f(pI) ∈ {f(p{j,k}), f(pI\{j,k})} = {a, a}. Case 2b2: for all j ∈ I and k ∈ I such that pj = a and
pk = −a, k → j. Let J = {i ∈ I: pi = a}. By the induction hypothesis, f(pJ) = α→(pJ) =
μ(pJ) = a and f(pI\J) = α→(pI\J) = μ(pI\J) = −a, the last step because n−a(pI) ≥ 1. By R, f(pI)
∈ {f(pJ), f(pI\J)} = {a, −a}. This proves that f(pI) ≠ 0. Define K = {i ∈ I: pi = 0}. Case
2b2a: K = ∅. Suppose f(pI) ≠ a, so f(pI) = −a. Consider any j ∈ I such that pj = a. By A,
for some k ∈ I such that pk = −a, f(pIi↔j) = f(pI). Define qI = pIi↔j. Since k → j, by the
induction hypothesis, f(q{j,k}) = α→(q{j,k}) = qk = pj = a. Moreover, q{j,k} = (−aj, ak) and
μ(qI) = μ(pI) = a imply μ(qI\{j,k}) = a. By the induction hypothesis, f(qI\{j,k}) = α→(qI\{j,k})
= μ(qI\{j,k}) = a. By R, f(qI) ∈ {f(q{j,k}), f(qI\{j,k})} = {a, a}. Case 2b2b: K ≠ ∅. By R, f(pI)
∈ {f(pK), f(pI\K)}. By the induction hypothesis, f(pK) = α→(pK) = μ(pK) = 0 and f(pI\K) =
α→(pI\K) = μ(pI\K) = μ(pI) = a. Consequently, f(pI) ∈ {0, a}. Since f(pI) ≠ 0, f(pI) = a.ƒ
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Lemma 4.2. If f satisfies E, R, P and S, then, for some linear order → on ℕ, f = α→ on
P1 ∪ P2.
Proof. By E, for any linear order → on ℕ, f = α→ on P1. Choose i ∈ ℕ and j ∈ ℕ\{i}.
By E, f(0i, 0j) = 0 and, for all a ∈ {−1, 1}, f(ai, aj) = f(ai, 0j) = f(0i, aj) = a. By R and E,
for all a ∈ {−1, 1}, f(ai, −aj) ∈ {f(ai), f(−aj)} = {a, −a}. In view of all this, by P, f(ai,
−aj) = a if and only if f(−ai, aj) = −a. Therefore, either, for all p{i,j} ∈ P2, f(p{i,j}) = pi or,
for all p{i,j} ∈ P2, f(p{i,j}) = pj. Let → be the binary relation on ℕ such that, for all n ∈ ℕ
and m ∈ ℕ\{n}: (i) it is not the case that n → n; and (ii) n → m if and only if, for all
p{n,m} ∈ P2, f(p{n,m}) = pn. Hence, for the individuals i and j previously selected, either i
→ j or j → i. Since i and j were arbitrary, for all n ∈ ℕ and m ∈ ℕ\{n}, either n → m or
m → n. Consequently, for → to define a liner order on ℕ it suffices to show that → is
transitive: n → m and m → r imply n → r. Consider (an, −am). By S, f(an, −ar) = f(an,
−am) or f(ar, −am) = f(an, −am). If the latter is the case, then m → r implies f(ar, −am) =
−a, whereas n → m implies f(an, −am) = a: contradiction. Accordingly, f(an, −ar) = f(an,
−am). Given that n → m implies f(an, −am) = a, f(an, −ar) = a. Since n → r or r → n, f(an,
−ar) = a is only consistent with n → r.ƒ
Proposition 4.3. A social welfare function f satisfies E, A, R, P and S if and only if, for
some linear order → on ℕ, f = α→.
Proof. “⇐” Remark 3.2. “⇒” Lemmas 4.1 and 4.2.ƒ
Lemma 4.4. Let f = μ on P1 ∪ P2. Then f satisfies Z, C and H if and only if f = μ.
Proof. “⇐” Remark 3.1. “⇒” Taking f = μ on P1 ∪ P2 as the base case of an induction
argument, choose n ∈ ℕ\{1, 2} and suppose that, for all k ∈ {1, … , n − 1}, f = μ on Pk.
To show that f = μ on Pn, let pI ∈ Pn. Case 1: for some i ∈ I, pi = 0. By Z, f(pI) = f(pI\{i}).
By the induction hypothesis, f(pI\{i}) = μ(pI\{i}). Since pi = 0 implies μ(pI\{i}) = μ(pI), f(pI)
= μ(pI). Case 2: for all i ∈ I, pi ≠ 0. Case 2a: μ(pI) = 0. To prove that f(pI) = 0, suppose
not: f(pI) = a ≠ 0. It follows from μ(pI) = 0 that, for some J ⊂ I such that ⏐J⏐ = ⏐I\J⏐ ≥
2, pI = (aJ, −aI\J). Consider any i ∈ I\J. By C, for some j ∈ J, f(pI\{i,j}) = f(pI). By the
induction hypothesis, f(pI\{i,j}) = μ(pI\{i,j}). As μ(pI) = 0 and {pi, pj} = {−1, 1}, μ(pI\{i,j}) =
μ(pI) = 0. Therefore, f(pI) = f(pI\{i,j}) = 0: contradiction.
Case 2b: μ(pI) = a ≠ 0. To prove that f(pI) = a, suppose not: f(pI) ∈ {0, −a}. Case 2b1:
f(pI) = 0. By H, for some non-empty J ⊂ I, f(pJ) = f(pI\J) = 0. By the assumption that
μ(pI) = a ≠ 0, a ∈ {μ(pJ), μ(pI\J)}. By the induction hypothesis, a ∈ {f(pJ), f(pI\J)}:
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contradiction. Case 2b2: f(pI) = −a. Given that μ(pI) = a ≠ 0, there is i ∈ I satisfying pi =
a. By C, for some j ∈ I such that pj = −a, f(pI\{i,j}) = f(pI) = −a. This already leads to a
contradiction if pI = (aI), so let pI ≠ (aI). Since μ(pI) = a and {pi, pj} = {−1, 1}, μ(pI\{i,j})
= μ(pI) = a. By the induction hypothesis, f(pI\{i,j}) = μ(pI\{i,j}) = a: contradiction.ƒ
Lemma 4.5. Suppose that, for some linear order → on ℕ, f = α→ on P1 ∪ P2. Then f
satisfies Z, C and H if and only if f = α→.
Proof. “⇐” Remark 3.2. “⇒” Taking f = α→ on P1 ∪ P2 as the base case of an induction
argument, choose n ∈ ℕ\{1, 2} and suppose that, for all k ∈ {1, … , n − 1}, f = α→ on
Pk. To show that f = α→ on Pn, let pI ∈ Pn. Case 1: for some i ∈ I, pi = 0. By Z, f(pI) =
f(pI\{i}). By the induction hypothesis, f(pI\{i}) = α→(pI\{i}). By definition of α→, pi = 0
implies α→(pI\{i}) = α→(pI). Accordingly, f(pI) = α→(pI). Case 2: for all i ∈ I, pi ≠ 0. By
the equivalence of α→(pI) = 0 and pI = (0I), there is a ∈ {−1, 1} such that α→(pI) = a. To
prove that f(pI) = a, suppose not: f(pI) ∈ {0, −a}. Case 2a: f(pI) = 0. By H, for some nonempty J ⊂ I, f(pJ) = f(pI\J) = 0. By the induction hypothesis, f(pJ) = 0 implies α→(pJ) = 0.
But α→(pJ) = 0 is equivalent to pJ = (0J), which contradicts the assumption that, for all i
∈ I, pi ≠ 0. Case 2b: f(pI) = −a. It follows from α→(pI) = a that μ(pI) ∈ {a, 0}.
Case 2b1: μ(pI) = a. Let i ∈ I satisfy pi = a. By C, f(pI) = −a implies that, for some j ∈ I
such that pj = −a, f(pI\{i,j}) = f(pI) = −a. This already leads to a contradiction if pI = (aI),
so let pI ≠ (aI). By the induction hypothesis, f(pI\{i,j}) = α→(pI\{i,j}). Since μ(pI) = a and
{pi, pj} = {a, −a}, μ(pI\{i,j}) = a. Consequently, α→(pI\{i,j}) = a. In sum, f(pI\{i,j}) = a:
contradiction. Case 2b2: μ(pI) = 0. Then α→(pI) = a means that pi = a, where i ∈ J = {k
∈ I: pk ≠ 0} satisfies J = {i} or, for all j ∈ J\{i}, i → j. Having μ(pI) = 0 under case 2
excludes the possibility J = {i}. Case 2b2a: i is the only member of I having preference
a. As μ(pI) = 0, there is only one member j of I with preference −a. But then n ≥ 3
implies that, for some k ∈ I, pk = 0: contradiction. Case 2b2b: for some k ∈ I\{i}, pk = a.
Let pk = a. By C, for some j ∈ I such that pj = −a, f(pI\{k,j}) = f(pI) = −a. By the induction
hypothesis, −a = f(pI\{k,j}) = α→(pI\{k,j}). As {pk, pj} = {a, −a} and μ(pI) = 0, μ(pI\{k,j}) = 0.
But then α→(pI) = pi and μ(pI\{k,j}) = 0 imply α→(pI\{k,j}) = pi = a: contradiction.ƒ
Proposition 4.6. A social welfare function f satisfies Z, C, H, U, P and S if and only if f
= μ or, for some linear order → on ℕ, f = α→.
Proof. “⇐” Remarks 3.1 and 3.2. “⇒” Case 1: for some i ∈ ℕ and j ∈ ℕ\{i}, f(1i, −1j)
= 0. Suppose f(1i, −1j) = 0, for some i ∈ ℕ and j ∈ ℕ\{i}. By U, f = μ on P1. By U, for
all a ∈ {−1, 0, 1}, f(ai, aj) = a. Since f = μ on P1, by Z, for all a ∈ {−1, 1}, f(ai, 0j) =
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f(ai) = a = f(aj) = f(0i, aj). In view of this, by P, f(1i, −1j) = 0 if and only if f(−1i, 1j) = 0.
Therefore, for all p{i,j} ∈ P2, f(p{i,j}) = μ(p{i,j}). Let k ∈ ℕ\{i, j}. By S, for all p{i,k} ∈ P2,
f(p{i,k}) = μ(p{i,k}) or, for all p{k,j} ∈ P2, f(p{k,j}) = μ(p{k,j}). Since μ is symmetric on P2,
for all a ∈ {−1, 0, 1} and b ∈ {−1, 0, 1}, μ(ai, bk) = μ(ak, bj). As a result, f = μ on P1 ∪
P2. By Lemma 4.4, f = μ.
Case 2: for all i ∈ ℕ and j ∈ ℕ\{i}, f(1i, −1j) ≠ 0. Suppose f(ai, −aj) = a ≠ 0. By U, for
any linear order → on ℕ, f = α→ on P1. By U, for all a ∈ {−1, 0, 1}, f(ai, aj) = a. Since f
= μ on P1, by Z, for all a ∈ {−1, 1}, f(ai, 0j) = f(ai) = a = f(aj) = f(0i, aj). Given this, by P,
f(ai, −aj) = a if and only if f(−ai, aj) = −a. In sum, for all p{i,j} ∈ P2, f(p{i,j}) = pi. Let →
be the binary relation on ℕ such that, for all n ∈ ℕ and m ∈ ℕ\{n}: (i) it is not the case
that n → n; and (ii) n → m if and only if, for all p{n,m} ∈ P2, f(p{n,m}) = pn. Hence, the
fact that, for all p{i,j} ∈ P2, f(p{i,j}) = pi means that i → j. This and S imply that, for all n
∈ ℕ and m ∈ ℕ\{n}, either n → m or m → n. Consequently, for → to define a linear
order on ℕ it suffices to show that → is transitive: n → m and m → r imply n → r.
Consider (an, −am). By S, f(an, −ar) = f(an, −am) or f(ar, −am) = f(an, −am). If the latter is
the case, then m → r implies f(ar, −am) = −a, whereas n → m implies f(an, −am) = a:
contradiction. Accordingly, f(an, −ar) = f(an, −am). As n → m implies f(an, −am) = a, f(an,
−ar) = a. Since n → r or r → n, f(an, −ar) = a is only consistent with n → r. To recap,
there is a linear order → on ℕ such that f = α→ on P1 ∪ P2. By Lemma 4.5, f = α→.ƒ
For society I and a given social welfare function f, define n0(I) to be the number of
preference profiles pI for I such that f(pI) = 0.
Proposition 4.7. Let X ∈ {(2), ¬(3), ¬(4)}. A social welfare function f satisfies Z, C,
H, U, P, S and
(i)
(ii)
X if and only if f = μ;
¬X if and only if, for some linear order → on ℕ, f = α→.
For all societies I and J, ⏐I⏐ > ⏐J⏐ implies n0(I) > n0(J).
For every society I and every non-empty J ⊂ I, f(pI) ∈ {f(pJ), f(pI\J)}.
For every society I, f(pI) ∈ {pi}i∈I.
(1)
(2)
(3)
Proof. By Proposition 4.6, that f satisfies Z, C, H, U, P and S is equivalent to having f ∈
{μ, α→}, for some linear order → on ℕ. It should not be difficult to verify: (i) that μ
satisfies (1) and that α→ does not (since, for every society I, n0(I) = 1); (ii) that α→
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satisfies (2) and that μ does not (μ(1i, −1j) ∉ {μ(1i), μ(−1j)} = {1, −1}); and (iii) that α→
satisfies (3) and that μ does not (μ(1i, −1j) ∉ {1, −1}).ƒ
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