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Majority Voting and Higher-Order Societies

Abstract

The aim of this paper is to study the applications of social choice functions (scfs) like for example the simple majority rule μ and the consensus rule κ to higher-order societies (i.e., societies that have other societies as members). The focus is on the reducibility property of a scf: its capacity to mimic the behavior of other scfs. It is proved that although κ is not reducible to μ, conversely μ can be reduced to κ. A characterization of the class of scfs reducible to κ is given. Finally, it is proved that some scfs can be extended, in the sense that the application of the scf to a society formed of a large number of members is reducible to iterate applications of that scf to (higher-order) societies formed of only two members.

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Notes

  1. 1.

    Voting was sequential, starting with the vote of a century (perhaps selected by lot) belonging to the first class, and the results were communicated to all the members of the assembly; thus, majorities could be known before the vote of the centuries belonging to other classes (therefore, the recording of their vote was not even completed). The effect of this voting procedure was that individual votes were not equal.

  2. 2.

    To use the set-theoretical parlance, all societies are well-founded. Although non-well-founded societies might be accepted, I shall not discuss them in this paper.

  3. 3.

    A sequence \(p_{S}^{1}\), \(p_{S}^{2}\), … \(p_{S}^{m}\) of profiles of a society S is zigzag if \(p_{S}^{k}\) > \(p_{S}^{k + 1}\) or \(p_{S}^{k + 1}\) > \(p_{S}^{k}\) for each k (k = 1, … m − 1). The property NZ was introduced in Fine (1972).

  4. 4.

    Here I do not use superscripts, because f and g can have different arities.

  5. 5.

    The authors note the similarity of this hierarchical account with the work on “recursive majority” (Mossel 1998), where n-level trees are studied. See also McMorris and Powers (2008).

  6. 6.

    Note that although S is taken as a first-order society in the formulation of the lemma, in fact the result can be generalized to the cases when its members are societies.

  7. 7.

    An amusing example of such an “ideal” voter was noticed a few years ago by Der Spiegel. A man named Dan Dumitru Zamfirescu, who was a member of the European Parliament (in the period 2013 – 2014), always voted “Yes”, even when the proposals were mutually contradictory. He voted 541 times “Yes”. See http://www.spiegel.de/international/europe/suspicious-voting-record-of-romanian-mep-dumitru-zamfirescu-a-929117.html.

  8. 8.

    Fishburn (1971) appeals to a rather different property (his condition 7) to prove part (a).

  9. 9.

    The ideal voter 1+ plays the same role as a scf t (top) which at all profiles pS of S gives the value 1: t(pS) = 1. Similarly, the ideal voter 1- plays the same role as a scf b (bottom) which gives b(pS) = −1 at all profiles pS. So, theorem 1b can be rephrased as follows: the class of scfs reducible to the set formed of three scfs: the simple majority rule μ, the top function t and the bottom function b is exactly Φ{Mon}. A future interesting question would be to try to determine the classes of scfs reducible to other sets of scfs.

  10. 10.

    This is a special way to relate the simple and the absolute majority rules. In general, as remarked by Sanver (2009), relativism and absolutism are essentially incompatible conceptions of majoritarianism.

  11. 11.

    Fine (1972) argues that the reason for this is that κ also does not satisfy the non-zigzaggedness property. He gives the following example: if the group has exactly three members, then the sequence: 11–1; 1–1–1; 1–11; -1–11 is zigzag, while the value of κ for all the members of the sequence remains unchanged. The same conclusion is in Fishburn (1971). Consider Fishburn’s condition 7. Let D1 = 11–1; D2 = 1–11; D3 = -111. We have D1 > −D2, D2 > −D3, D3 > −D1. We get \(\sum\limits_{k = 1}^{3} {D_{1}^{k} }\) = 2 > 0, while κ(Di) = 0 for all i – in contradiction with his Condition 4.

  12. 12.

    Another more abstract example is this. As proved in Bartholdi et al (2020), the collection of equitable voting rules that satisfy N and R has some remarkable properties. For example, for each society S formed of n members there is some such rule with the property that it has winning coalitions of size \(2[\sqrt n ] - 1\) (their Theorem 1). Now, since R entails Mon and NZ, the rule also satisfies these two properties and therefore it is reducible by Theorem 1a to μn.

  13. 13.

    The example is not fictious. Such a rule (concerning cases in which the two Chambers voted for different forms of the bill) was in force in the Romanian Parliament when the new post-communist constitution was adopted in 1991. However, this provision was amended in 2003. See http://www.cdep.ro/pls/dic/site2015.page?den=act1_2&par1=3#t3c1s3sba74 (Article 76).

    A more complex voting rule of this type can be found in Aristotle’s Politics (VI 3, 1317a 30–1318b 1). A reconstruction of it in a similar manner is given in Miroiu and Partenie (2019).

  14. 14.

    Given neutrality, we can replace the strong Pareto property with its positive part.

  15. 15.

    However, some scfs cannot be extended: we cannot find any procedure to define them in terms of their binary parts. Two examples are the consensus rule κ2 and the absolute majority rule α2. For lack of space, in this paper I shall omit the proofs.

  16. 16.

    We can prove dual propositions in a similar way, when values 1 and -1 are mutually replaced.

  17. 17.

    For example, let S = {v1, v2, v3, v4}. Since all S−j have three members, we can apply the above procedure and get: \(\upsigma _{{S^{ - 1} }}\) = {{v4, v2}, {v2, v3}, {v4, v3}}, \(\upsigma _{{S^{ - 2} }}\) = {{v1, v4}, {v4, v3}, {v1, v3}}, \(\upsigma _{{S^{ - 3} }}\) = {{v1, v2}, {v2, v4}, {v1, v4}}and \(\upsigma _{{S^{ - 4} }}\) = {{v1, v2}, {v2, v3}, {v1, v3}}. Therefore:

    \(\sigma_{S} = \left\{ {\left\{ {\left\{ {v_{4} ,v_{2} } \right\},\left\{ {v_{2} ,v_{3} } \right\},\left\{ {v_{4} ,v_{3} } \right\}} \right\},\left\{ {\left\{ {v_{1} ,v_{4} } \right\},\left\{ {v_{4} ,v_{3} } \right\},\left\{ {v_{1} ,v_{3} } \right\}} \right\},\left\{ {\left\{ {v_{1} ,v_{2} } \right\},\left\{ {v_{2} ,v_{4} } \right\},\left\{ {v_{1} ,v_{4} } \right\}} \right\},\left\{ {\left\{ {v_{1} ,v_{2} } \right\},\left\{ {v_{2} ,v_{3} } \right\},\left\{ {v_{1} ,v_{3} } \right\}} \right\}} \right\}\)

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Appendix: proofs

Appendix: proofs

Proof of Lemma 1.

Let S = {v1, v2, … vn}. Write s for the number of members of S with the property that \(p_{{v_{j} }}\) = 1; m for the number of members of S with the property that \(p_{{v_{j} }}\) = −1, and z for the number of members of S with the property that \(p_{{v_{j} }}\) = 0. We have s + m + z = n.

For part (a), we have by definition that μn(pS) = sgn(\(\sum\limits_{k = 1}^{n} {p_{{v_{k} }} }\)). If μn(pS) = 1, it follows that \(\sum\limits_{k = 1}^{n} {p_{{v_{k} }} }\) = \(\sum\limits_{\begin{subarray}{l} k = 1 \\ k \ne j \end{subarray} }^{n} {p_{{v_{k} }} }\) + \(p_{{v_{j} }}\) ≥ 1. Note that we can have μn−1(\(p_{{S^{ - j} }}\)) < 0 only if sgn(\(\sum\limits_{\begin{subarray}{l} k = 1 \\ k \ne j \end{subarray} }^{n} {p_{{v_{k} }} }\)) = −1, i.e. \(\sum\limits_{\begin{subarray}{l} k = 1 \\ k \ne j \end{subarray} }^{n} {p_{{v_{k} }} }\) < 0; but in this case \(\sum\limits_{\begin{subarray}{l} k = 1 \\ k \ne j \end{subarray} }^{n} {p_{{v_{k} }} }\) + \(p_{{v_{j} }}\) < 1 for each value of \(p_{{v_{j} }}\) – contradiction. So we established that μn−1(\(p_{{S^{ - j} }}\)) ≥ 0 for each j = 1, …n. But we cannot have μn−1(\(p_{{S^{ - j} }}\)) = 0 for each j. To prove this, consider the following cases:

  1. (1)

    \(p_{{v_{j} }}\) = 1 for all j. Then by definition μn−1(\(p_{{S^{ - j} }}\)) = 1 for each j;

  2. (2)

    \(p_{{v_{j} }}\) = 0 for all j. Then by definition μn(pS) = 0 – contradiction;

  3. (3)

    \(p_{{v_{j} }}\) ≥ 0 for all j and \(p_{{v_{i} }}\) = 1 for some i. Then clearly μn−1(\(p_{{S^{ - i} }}\)) = 1;

  4. (4)

    there is some j such that \(p_{{v_{j} }}\) < 1. Observe that μn(pS) = 1 entails that sgn\((\sum\limits_{\begin{subarray}{l} k = 1 \\ k \ne j \end{subarray} }^{n} {p_{{v_{k} }} })=1\)and so μn−1(\(p_{{S^{ - j} }}\)) = 1.

For part (b), note first that we have s = m by definition. If \(p_{{v_{j} }}\) = 0, then clearly μn−1(\(p_{{S^{ - j} }}\)) = 0; if \(p_{{v_{j} }}\) = 1, then μ(\(p_{{S^{ - j} }}\)) = −1, because in the number of voters vi in S−j such that \(p_{{v_{i} }}\) = 1 is s − 1 < m; and if \(p_{{v_{j} }}\) = −1, then μn−1(\(p_{{S^{ - j} }}\)) = 1, because in the number of voters vi in S−j such that \(p_{{v_{i} }}\) = 1 is m – 1 < s. But the number of S−j’s with the property that μn−1(\(p_{{S^{ - j} }}\)) = 1 must be equal to the number of S−j’s with the property that μn−1(\(p_{{S^{ - j} }}\)) = 1, because s = m.

To prove part (c), suppose first that μn(pS) = 1. Then part (a) gives that that there is no member S−j of Γ such that μn−1(\(p_{{S^{ - j} }}\)) = −1 and there is a member S−j of it such μn−1(\(p_{{S^{ - j} }}\)) = 1. Then μn(pΓ) = 1. Second, suppose that μn(pS) = 0. Part (b) gives that Γ has an equal number of S−j’s such that μn−1(\(p_{{S^{ - j} }}\)) = 1 and μn−1(\(p_{{S^{ - j} }}\)) = −1. Then we also have μn(pΓ) = 0.

Finally, for part (d) suppose first again that μn(pS) = 1. Clearly, μn−1(\(p_{{\Gamma^{ - j} }}\)) ≥ 0, because μn−1(\(p_{{S^{ - i} }}\)) ≥ 0 for each member of Γ−j. Part (a) guarantees that there is some S−i with the property that μn−1(\(p_{{S^{ - i} }}\)) = 1. If it is S−j, then μ2(\(p_{{\{ \Gamma^{ - j} ,S^{ - j} \} }}\)) = 1. If it is one of the members of Γ−j, then μn−1(\(p_{{\Gamma^{ - j} }}\)) = 1and so μ2(\(p_{{\{ \Gamma^{ - j} ,S^{ - j} \} }}\)) = 1 because μn−1(\(p_{{S^{ - j} }}\)) ≥ 0. Secondly, let μn(pS) = 0. If μn−1(\(p_{{S^{ - j} }}\)) = 0, then by part (b) Γ−j has an equal number of S−i’s such that μn−1(\(p_{{S^{ - i} }}\)) = 1 and μn−1(\(p_{{S^{ - i} }}\)) = −1; therefore μn−1(\(p_{{\Gamma^{ - j} }}\)) = 0 and consequently μ2(\(p_{{\{ \Gamma^{ - j} ,S^{ - j} \} }}\)) = 0. If μn−1(\(p_{{S^{ - j} }}\)) = 1, then by part (b) the number of S−i’s in Γ−j such that μn−1(\(p_{{S^{ - i} }}\)) = −1 is larger than number of S−i’s in Γ−j such that μn−1(\(p_{{S^{ - i} }}\)) = 1; therefore μn−1(\(p_{{\Gamma^{ - j} }}\)) = −1. So μ2(\(p_{{\{ \Gamma^{ - j} ,S^{ - j} \} }}\)) = µ2(−1, 1) = 0 = μn(pS) as required.Footnote 16

Proof that α can be defined in terms of μ on domains including ideal votes.

Let S = {v1, v2, … vn}. To show this, we first construct two societies S1 and S2:

$$ \begin{aligned} S_{1} = & \left\{ {\left\{ {v_{1} ,{\mathbf{1}}^{ - } } \right\},\left\{ {v_{2} ,{\mathbf{1}}^{ - } } \right\}, \ldots \left\{ {v_{n} ,{\mathbf{1}}^{ - } } \right\},\left\{ {\left\{ {v_{1} ,{\mathbf{1}}^{ - } } \right\},{\mathbf{1}}^{ + } } \right\},\left\{ {\left\{ {v_{2} ,{\mathbf{1}}^{ - } } \right\},{\mathbf{1}}^{ + } } \right\}, \ldots \left\{ {\left\{ {v_{2} ,{\mathbf{1}}^{ - } } \right\},{\mathbf{1}}^{ + } } \right\}} \right\} \\ S_{2} = & \left\{ {\left\{ {v_{1} ,{\mathbf{1}}^{ + } } \right\},\left\{ {v_{2} ,{\mathbf{1}}^{ + } } \right\}, \ldots \left\{ {v_{n} ,{\mathbf{1}}^{ + } } \right\},\left\{ {\left\{ {v_{1} ,{\mathbf{1}}^{ + } } \right\},{\mathbf{1}}^{ - } } \right\},\left\{ {\left\{ {v_{2} ,{\mathbf{1}}^{ + } } \right\},{\mathbf{1}}^{ - } } \right\}, \ldots \left\{ {\left\{ {v_{2} ,{\mathbf{1}}^{ + } } \right\},{\mathbf{1}}^{ - } } \right\}} \right\} \\ \end{aligned} $$

Let s members of S = {v1, v2, … vn} have \(p_{{v_{j} }}\) = 1, m members have \(p_{{v_{j} }}\) = −1 and z members have \(p_{{v_{j} }}\) = 0, with s + m + z = n. Observe that when applied to S1 the function μ2n gives s times 0 and z + m times −1; and also s times 1 and z + m times 0. So μ2n(\(p_{{S_{1} }}\)) = 1 if s > z + m and μ2n(\(p_{{S_{1} }}\)) < 1 in all the other cases. But s > z + m and s + z + m = n entail that s > n/2. Analogously, we have that μ2n(\(p_{{S_{2} }}\)) = −1 when m > n/2 and μ2n(\(p_{{S_{2} }}\)) > −1 in all the other cases. Secondly, construct two other societies:

$$ \begin{aligned} S_{11} = & \left\{ {{\mathbf{1}}^{ + } ,\left\{ {{\mathbf{1}}^{ - } ,S_{1} } \right\}} \right\} \\ S_{21} = & \left\{ {{\mathbf{1}}^{ - } ,\left\{ {{\mathbf{1}}^{ + } ,S_{2} } \right\}} \right\}. \\ \end{aligned} $$

Clearly, μ2(\(p_{{S_{11} }}\)) = 1 if s > n/2 and μ2(\(p_{{S_{11} }}\)) = 0 in all the other cases; and μ2 \((p_{{S_{21} }})= -1\) if m > n/2 and μ2(\(p_{{S_{21} }}\)) = 0 in all the other cases. Finally, put

$$ \sigma_{S} = \{ S_{11} ,S_{21} \} . $$

We can easily verify that αn(pS) = μ2(\(p_{{\upsigma _{S} }}\)).

Proof of Theorem 2. The proof is by induction on the number of members of S. Suppose first that n = 2. Let S = {v1, v2}. As already noted, in this care κ2(pS) = μ2(pS) for all profiles p of S. Now let n = 3, i.e. S = {v1, v2, v3}. Clearly, there are profiles at which κ3 and μ3 do not coincide. But consider the society σS = {S−1, S−2, S−3} = {{v2, v3}, {v1, v3}, {v1, v2}}. We show that:

$$ \mu^{3} \left( {p_{S} } \right) = \kappa^{3} \left( {p_{{\sigma_{S} }} } \right)\;{\text{for}}\;{\text{all}}\;p\;{\text{of}}\;G. $$

This means that we can get μ3(pS) by the following procedure: we first calculate the value of κ3 for the three groups, each consisting in two members: κ2(\(p_{{\{ v_{1} ,v_{2} \} }}\)), κ2(\(p_{{\{ v_{2} ,v_{3} \} }}\)) and κ2(\(p_{{\{ v_{1} ,v_{3} \} }}\)). Then we calculate the value of κ3 for the group consisting in these three groups. To show that the value we get by iteratively applying κ3 is the same as μ3(pS) the only interesting cases are when: (1) for two of the members of S it is the case that \(p_{{v_{j} }}\) ≥ 0 and also \(p_{{v_{j} }}\) = 1 for at least one of these members, and for the remaining one it is the case that \(p_{{v_{j} }}\) = −1, and (2) the symmetrical ones when −1 is replaced by 1. Note that the construction of the society σS makes the order of the members of S irrelevant.

  • If \(p_{{v_{1} }}\) = \(p_{{v_{2} }}\) = 1 and \(p_{{v_{3} }}\) = −1, then we get κ2(\(p_{{\{ v_{1} ,v_{2} \} }}\)) = 1, κ2(\(p_{{\{ v_{2} ,v_{3} \} }}\)) = 0 and κ2(\(p_{{\{ v_{1} ,v_{3} \} }}\)) = 0, so κ3(\(p_{{\upsigma _{S} }}\)) = 1, in agreement with μ3(pS).

  • If \(p_{{v_{1} }}\) = 1, \(p_{{v_{2} }}\) = 0 and \(p_{{v_{3} }}\) = −1, then we get κ2(\(p_{{\{ v_{1} ,v_{2} \} }}\)) = 1, κ2(\(p_{{\{ v_{2} ,v_{3} \} }}\)) = −1 and κ2(\(p_{{\{ v_{1} ,v_{3} \} }}\)) = 0, so κ3(\(p_{{\upsigma _{S} }}\)) = 0, again in agreement with μ3(pS).

(The symmetrical cases can be dealt with in an analogous way.)

Finally let n > 3. By induction it holds for each of the n societies S−j = S − {vj} that there is some society \(\upsigma _{{S^{ - j} }}\) with the property that for all profiles \(p_{{S^{ - i} }}\) of S−j it holds that μn−1(\(p_{{S^{ - i} }}\)) = κn−1(\(p_{{\upsigma _{{S^{ - j} }} }}\)). We need to show that there is some society σS with the property that μn(pS) = κn(\(p_{{\upsigma _{S} }}\)) for all profiles pS of S. Put σS = {\(\upsigma _{{S^{ - 1} }}\), \(\upsigma _{{S^{ - 2} }}\), …\(\upsigma _{{S^{ - n} }}\)}.Footnote 17

The proof is trivial for all the cases when \(p_{{v_{i} }}\) ≥ 0 for all vi ∊ S or \(p_{{v_{i} }}\) ≤ 0 for all vi ∊ S. So suppose that \(p_{{v_{i} }}\) = 1 for some vi ∊ S and \(p_{{v_{i} }}\) = −1 for some vi ∊ S. I shall give the proof for the cases when |{ vi ∊ S: \(p_{{v_{i} }}\) = 1}| ≥|{ vi ∊ S: \(p_{{v_{i} }}\) = −1}|, i.e. when µn(pS) ≥ 0. The symmetrical cases when |{ vi ∊ S: \(p_{{v_{i} }}\) = 1}| ≤|{ vi ∊ S: \(p_{{v_{i} }}\) = −1}| can be proved in an analogous way.

There are two possibilities. First, let μn(pS) = 0. By lemma 1b It follows that |{ vi ∊ S: \(p_{{v_{i} }}\) = 1}| =|{ vi ∊ S: \(p_{{v_{i} }}\) = −1}|= m ≥ 0, where 2 m ≤ n. If m = 0, then for all vi it holds that \(p_{{v_{i} }}\) = 0 and so κn(\(p_{{\upsigma _{S} }}\)) = 0 by the definition of κ. If m > 0, then: a) since there is some vj such that \(p_{{v_{j} }}\) = 1 it follows that μn−1(\(p_{{S^{ - j} }}\)) = −1 because |{ vi ∊ S−j: \(p_{{v_{i} }}\) = −1}| >|{ vi ∊ S−j: \(p_{{v_{i} }}\) = 1}|; b) since there is some vj such that \(p_{{v_{j} }}\) = −1 it follows that μn−1(\(p_{{S^{ - j} }}\)) = 1 because |{ vi ∊ S−j: \(p_{{v_{i} }}\) = 1}| >|{ vi ∊ S−j: \(p_{{v_{i} }}\) = −1}|. By induction, for each S−j we have μn−1(\(p_{{S^{ - j} }}\)) = κn−1(\(p_{{\upsigma _{{S^{ - j} }} }}\)), so we have κn−1(\(p_{{\upsigma _{{S^{ - j} }} }}\)) = 1 in some cases and κn−1(\(p_{{\upsigma _{{S^{ - j} }} }}\)) = −1 in other cases, whence by the definition of κ we get κn−1(\(p_{{\upsigma _{S} }}\)) = 0, whence κn(\(p_{{\upsigma _{S} }}\)) = μn(pS).

Second, let μn(pS) = 1. It follows that |{ vi ∊ S: \(p_{{v_{i} }}\) = 1}|= m1 >|{ vi ∊ S: \(p_{{v_{i} }}\) = −1}|= m2. Let vj be some member of S. As shown in the proof of lemma 1a we have: (i) if \(p_{{v_{j} }}\) = −1, then μn−1(\(p_{{S^{ - j} }}\)) = 1, since clearly |{ vi ∊ S−j: \(p_{{v_{i} }}\) = 1}| >|{ vi ∊ S−j: \(p_{{v_{i} }}\) = −1}|; consequently, by induction κn−1(\(p_{{\upsigma _{{S^{ - j} }} }}\)) = 1. Similarly, observe that: (ii) if \(p_{{v_{j} }}\) = 0, then μn−1(\(p_{{S^{ - j} }}\)) = 1 and so by induction κn−1(\(p_{{\upsigma _{{S^{ - j} }} }}\)) = 1. Finally, (iii) if \(p_{{v_{j} }}\) = −1, then we must have μn−1(\(p_{{S^{ - j} }}\)) ≥ 0 and thus κn−1(\(p_{{\upsigma _{{S^{ - j} }} }}\)) ≥ 0. There are two subcases:

  • If m2 > 0 (i.e. there is at least a vj with the property that \(p_{{\upsigma _{{S^{ - j} }} }}\) = −1), then there is some S−j such that μ(\(p_{{S^{ - j} }}\)) = 1, which entails κn−1(\(p_{{\upsigma _{{S^{ - j} }} }}\)) = 1, and so κn−1(\(p_{{\upsigma _{S} }}\)) = 1 = μn(pS).

  • If m2 = 0 (i.e. there is no vj with the property that \(p_{{v_{j} }}\) = −1), it follows that there is some vj ∊ S such that \(p_{{v_{j} }}\) = 1 and for all the other vj ∊ S it holds that \(p_{{v_{j} }}\) ≥ 0. If \(p_{{v_{j} }}\) = 1 for all vj ∊ S, then given that n > 3, we must have m1 – 1 > 0 and thus μn−1(\(p_{{S^{ - j} }}\)) = 1 for all S−j. Consequently, κn−1(\(p_{{\upsigma _{{S^{ - j} }} }}\)) = 1, and so κn(\(p_{{\upsigma _{S} }}\)) = 1 = μn(pS). But if \(p_{{v_{j} }}\) = 0 for some vj ∊ S, then by induction (see the argument in the second case above) μn−1(\(p_{{S^{ - j} }}\)) = 1. We can again conclude that κn−1(\(p_{{\upsigma _{{S^{ - j} }} }}\)) = 1, whence κn(\(p_{{\upsigma _{S} }}\)) = 1 = μn(pS).

Remark on theorem 3: if Neu and Mon hold, then NZ entails SP. Suppose that fn satisfies NZ. Then if at a profile pS we have that fn(pS) = 0, there must be a zigzag sequence \(p_{S}^{1}\), \(p_{S}^{2}\), … \(p_{S}^{m}\) of profiles of S such that fn(\(p_{S}^{k}\)) = 1 for some profile \(p_{S}^{k}\) in this sequence (this is so because fn satisfies Neu). Then by Mon we get that fn(\(p_{S}^{^{\prime}}\)) = 1 at a profile \(p_{S}^{^{\prime}}\) where \(p_{{v_{j} }}\) ≥ 0 for all vj ∈ S and \(p_{{v_{j} }}\) = 1 for some vj ∈ S.

Proof of Theorem 4. We need to show that for each S there is some (higher-order) binary society σS such that fn(pS) = f2(\(p_{{\sigma_{S} }}\)). The society σS is recursively constructed as follows:

  1. (i)

    if S = {i1, i2, i3}, then σS = {{i1, i2}, {i3}};

  2. (ii)

    if S = {i1, … in}, then σS = {\(\upsigma _{{S - \{ i_{n} \} }}\), {in}}.

We can check that since the scfs Min, Max and υ are associative, they can all be extended to the n-ary case. Consider the unanimity rule υ. We can easily see that it satisfies associativity: υn(\(p_{{\{ i_{1} ,i_{2} ,...i_{n - 1} ,i_{n} \} }}\)) = υ2n−1(\(p_{{\{ i_{1} ,i_{2} ,...i_{n - 1} \} }}\)),\(p_{{i_{n} }}\)). Suppose that υn(\(p_{{\{ i_{1} ,i_{2} ,...i_{n - 1} ,i_{n} \} }}\)) = 1. By definition, this means that \(p_{{i_{k} }}\) = 1 for all k. But then υn−1(\(p_{{\{ i_{1} ,i_{2} ,...i_{n - 1} \} }}\)) = 1 and also \(p_{{i_{n} }}\) = 1. Therefore υ2n−1(\(p_{{\{ i_{1} ,i_{2} ,...i_{n - 1} \} }}\)),\(p_{{i_{n} }}\)) = υ2(1, 1) = 1. The case when υn(\(p_{{\{ i_{1} ,i_{2} ,...i_{n - 1} ,i_{n} \} }}\)) = −1 is similar. If υn(\(p_{{\{ i_{1} ,i_{2} ,...i_{n - 1} ,i_{n} \} }}\)) = 0, we must have at least two individuals ik and ik' such that \(p_{{i_{k} }}\) ≠ \(p_{{i_{k^{\prime}} }}\). If both are in {i1, … in−1}, then υn−1(\(p_{{\{ i_{1} ,i_{2} ,...i_{n - 1} \} }}\)) = 0 and so υn(\(p_{{\{ i_{1} ,i_{2} ,...i_{n - 1} ,i_{n} \} }}\)) = υ2n−1(\(p_{{\{ i_{1} ,i_{2} ,...i_{n - 1} \} }}\)),\(p_{{i_{n} }}\)) = υ2n−1(0,\(p_{{i_{n} }}\)) = 0. If one of them, e.g. ik', is \(p_{{i_{n} }}\), then we have three cases:

  1. (1)

    \(p_{{i_{n} }}\) = 1. Then if \(p_{{i_{k} }}\) = −1 we may have either υn−1(\(p_{{\{ i_{1} ,i_{2} ,...i_{n - 1} \} }}\)) = −1 or υn−1(\(p_{{\{ i_{1} ,i_{2} ,...i_{n - 1} \} }}\)) = 0. But in both subcases υ2n−1(\(p_{{\{ i_{1} ,i_{2} ,...i_{n - 1} \} }}\)),\(p_{{i_{n} }}\)) = 0. If \(p_{{i_{k} }}\) = 0, then υn−1(\(p_{{\{ i_{1} ,i_{2} ,...i_{n - 1} \} }}\)) = 0 and again υ2n−1(\(p_{{\{ i_{1} ,i_{2} ,...i_{n - 1} \} }}\)),\(p_{{i_{n} }}\)) = υ2(0, 1) = 0.

  2. (2)

    \(p_{{i_{n} }}\) = −1. This case is analogous to the first case.

  3. (3)

    \(p_{{i_{n} }}\) = 0. Then we always get υ2n−1(\(p_{{\{ i_{1} ,i_{2} ,...i_{n - 1} \} }}\)), 0) = 0.

Proof of Theorem 5. We show that for each society S ⊆ G formed of n members there is some (higher-order) society σS such that (1) σS is binary; and (2) for all profiles pS of S we have: μn(pS) = μ2(\(p_{{\sigma_{S} }}\)). The proof of the theorem is by induction on the number of members of S. For n = 2, the proof is trivial. For n = 3, we have S−1 = {v2, v3}, S−2 = {v1, v3}, S−3 = {v1, v2}. Further, put Γ−3 = {S−1, S−2} = {{v2, v3}, {v1, v3}}. Finally, let σS = {Γ−3, S−3} = {{{v2, v3}, {v1, v3}}, {v1, v2}}. Since σS is clearly binary, the function μ2 can be iteratively applied to it. Lemma 1d gives μ2(\(p_{{\sigma_{S} }}\)) = μ2(\(p_{{\{ \Gamma^{ - j} ,S^{ - j} \} }}\)) = μ3(pS).

Now consider the case when S contains n members. Put Γn = {S−1, S−2, … S(n−1)}. By definition, μn−1(\(p_{{\Gamma^{ - n} }}\)) = μn−1n−1(\(p_{{S^{ - 1} }}\)), μn−1(\(p_{{S^{ - 2} }}\)), … μn−1(\(p_{{S^{ - (n - 1)} }}\))). By induction, the property we want to prove holds when μ applies to societies formed of at most n−1 members. So, for each S−j there is some binary society \(\upsigma _{{S^{ - 1} }}\) with the property that μn−1(\(p_{{S^{ - j} }}\)) = μ2(\(p_{{\sigma_{{S^{ - j} }} }}\)). So we get: μn−1(\(p_{{\Gamma^{ - n} }}\)) = μn−12(\(p_{{\sigma_{{S^{ - 1} }} }}\)), μ2(\(p_{{\sigma_{{S^{ - 2} }} }}\)), … μ2(\(p_{{\sigma_{{S^{ - (n - 1)} }} }}\))). Moreover, the society \(\Gamma_{\upsigma }^{ - n}\) = {\(\upsigma _{{S^{ - 1} }}\), \(\upsigma _{{S^{ - 2} }}\), …\(\upsigma _{{S^{ - (n - 1)} }}\)} has n – 1 members and so by induction there is some binary society \(\upsigma _{{\Gamma_{\upsigma }^{ - n} }}\) with the property that μn−1(\(p_{{\Gamma_{\upsigma }^{ - n} }}\)) = μ2(\(p_{{\upsigma _{{\Gamma_{\upsigma }^{ - n} }} }}\)). Finally, let σS = {\(\upsigma _{{S^{ - n} }}\), \(\upsigma _{{\Gamma_{\upsigma }^{ - n} }}\)}. By lemma 1d we have that μn(pS) = μ2(\(p_{{\{ \Gamma^{ - n} ,S^{ - n} \} }}\)) = μ2n−1(\(p_{{\Gamma^{ - n} }}\)), μn−1(\(p_{{S^{ - n} }}\))) and so μn(pS) = μ22(\(p_{{\upsigma _{{\Gamma_{\upsigma }^{ - n} }} }}\)), μ2(\(p_{{\sigma_{{S^{ - n} }} }}\))), which proves that σS is the society we wanted to construct.

Proof of Theorem 6. Notice first that µ2 satisfies the three axioms. Conversely, we need to show that if a scf f satisfies these axioms, then it must be exactly µ2. Let S = {v1, v2}. We show that f2(\(p_{{\{ v_{1} ,v_{2} \} }}\)) = sgn(\(p_{{v_{1} }}\) + \(p_{{v_{2} }}\)) in all possible cases. We have: (i) for profiles (1, 1) and (−1, −1) axiom BU gives f2(\(p_{{\{ v_{1} ,v_{2} \} }}\)) = 1, respectively f2(\(p_{{\{ v_{1} ,v_{2} \} }}\)) = −1; (ii) for profiles (1, −1) and (−1, 1) axiom SET gives f2(\(p_{{\{ v_{1} ,v_{2} \} }}\)) = 0; (iii) for profiles (−1, 0) and (0, −1) axiom IIS gives f2(\(p_{{\{ v_{1} ,v_{2} \} }}\)) = −1; similarly, for profiles (1, 0) and (0, 1) IIS gives f2(\(p_{{\{ v_{1} ,v_{2} \} }}\)) = 1; iv) for the profile (0, 0) axiom IIS gives f2(\(p_{{\{ v_{1} ,v_{2} \} }}\)) = 0. □

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Miroiu, A. Majority Voting and Higher-Order Societies. Group Decis Negot 30, 983–999 (2021). https://doi.org/10.1007/s10726-021-09744-z

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Keywords

  • Higher-order societies
  • Majority rule
  • Consensus
  • Reducibility
  • Binary societies