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 higherorder 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 (higherorder) societies formed of only two members.
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Notes
 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.
To use the settheoretical parlance, all societies are wellfounded. Although nonwellfounded societies might be accepted, I shall not discuss them in this paper.
 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.
Here I do not use superscripts, because f and g can have different arities.
 5.
 6.
Note that although S is taken as a firstorder society in the formulation of the lemma, in fact the result can be generalized to the cases when its members are societies.
 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/suspiciousvotingrecordofromanianmepdumitruzamfirescua929117.html.
 8.
Fishburn (1971) appeals to a rather different property (his condition 7) to prove part (a).
 9.
The ideal voter 1^{+} plays the same role as a scf t (top) which at all profiles p_{S} of S gives the value 1: t(p_{S}) = 1. Similarly, the ideal voter 1^{} plays the same role as a scf b (bottom) which gives b(p_{S}) = −1 at all profiles p_{S}. 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.
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.
Fine (1972) argues that the reason for this is that κ also does not satisfy the nonzigzaggedness 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 D^{1} = 11–1; D^{2} = 1–11; D^{3} = 111. We have D^{1} > −D^{2}, D^{2} > −D^{3}, D^{3} > −D^{1}. We get \(\sum\limits_{k = 1}^{3} {D_{1}^{k} }\) = 2 > 0, while κ(D^{i}) = 0 for all i – in contradiction with his Condition 4.
 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 μ^{n}.
 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 postcommunist 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.
Given neutrality, we can replace the strong Pareto property with its positive part.
 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 κ^{2} and the absolute majority rule α^{2}. For lack of space, in this paper I shall omit the proofs.
 16.
We can prove dual propositions in a similar way, when values 1 and 1 are mutually replaced.
 17.
For example, let S = {v_{1}, v_{2}, v_{3}, v_{4}}. Since all S^{−j} have three members, we can apply the above procedure and get: \(\upsigma _{{S^{  1} }}\) = {{v_{4}, v_{2}}, {v_{2}, v_{3}}, {v_{4}, v_{3}}}, \(\upsigma _{{S^{  2} }}\) = {{v_{1}, v_{4}}, {v_{4}, v_{3}}, {v_{1}, v_{3}}}, \(\upsigma _{{S^{  3} }}\) = {{v_{1}, v_{2}}, {v_{2}, v_{4}}, {v_{1}, v_{4}}}and \(\upsigma _{{S^{  4} }}\) = {{v_{1}, v_{2}}, {v_{2}, v_{3}}, {v_{1}, v_{3}}}. 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 = {v_{1}, v_{2}, … v_{n}}. 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 μ^{n}(p_{S}) = sgn(\(\sum\limits_{k = 1}^{n} {p_{{v_{k} }} }\)). If μ^{n}(p_{S}) = 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 μ^{n−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 μ^{n−1}(\(p_{{S^{  j} }}\)) ≥ 0 for each j = 1, …n. But we cannot have μ^{n−1}(\(p_{{S^{  j} }}\)) = 0 for each j. To prove this, consider the following cases:

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

(2)
\(p_{{v_{j} }}\) = 0 for all j. Then by definition μ^{n}(p_{S}) = 0 – contradiction;

(3)
\(p_{{v_{j} }}\) ≥ 0 for all j and \(p_{{v_{i} }}\) = 1 for some i. Then clearly μ^{n−1}(\(p_{{S^{  i} }}\)) = 1;

(4)
there is some j such that \(p_{{v_{j} }}\) < 1. Observe that μ^{n}(p_{S}) = 1 entails that sgn\((\sum\limits_{\begin{subarray}{l} k = 1 \\ k \ne j \end{subarray} }^{n} {p_{{v_{k} }} })=1\)and so μ^{n−1}(\(p_{{S^{  j} }}\)) = 1.
For part (b), note first that we have s = m by definition. If \(p_{{v_{j} }}\) = 0, then clearly μ^{n−1}(\(p_{{S^{  j} }}\)) = 0; if \(p_{{v_{j} }}\) = 1, then μ(\(p_{{S^{  j} }}\)) = −1, because in the number of voters v_{i} in S^{−j} such that \(p_{{v_{i} }}\) = 1 is s − 1 < m; and if \(p_{{v_{j} }}\) = −1, then μ^{n−1}(\(p_{{S^{  j} }}\)) = 1, because in the number of voters v_{i} in S^{−j} such that \(p_{{v_{i} }}\) = 1 is m – 1 < s. But the number of S^{−j}’s with the property that μ^{n−1}(\(p_{{S^{  j} }}\)) = 1 must be equal to the number of S^{−j}’s with the property that μ^{n−1}(\(p_{{S^{  j} }}\)) = 1, because s = m.
To prove part (c), suppose first that μ^{n}(p_{S}) = 1. Then part (a) gives that that there is no member S^{−j} of Γ such that μ^{n−1}(\(p_{{S^{  j} }}\)) = −1 and there is a member S^{−j} of it such μ^{n−1}(\(p_{{S^{  j} }}\)) = 1. Then μ^{n}(p_{Γ}) = 1. Second, suppose that μ^{n}(p_{S}) = 0. Part (b) gives that Γ has an equal number of S^{−j}’s such that μ^{n−1}(\(p_{{S^{  j} }}\)) = 1 and μ^{n−1}(\(p_{{S^{  j} }}\)) = −1. Then we also have μ^{n}(p_{Γ}) = 0.
Finally, for part (d) suppose first again that μ^{n}(p_{S}) = 1. Clearly, μ^{n−1}(\(p_{{\Gamma^{  j} }}\)) ≥ 0, because μ^{n−1}(\(p_{{S^{  i} }}\)) ≥ 0 for each member of Γ^{−j}. Part (a) guarantees that there is some S^{−i} with the property that μ^{n−1}(\(p_{{S^{  i} }}\)) = 1. If it is S^{−j}, then μ^{2}(\(p_{{\{ \Gamma^{  j} ,S^{  j} \} }}\)) = 1. If it is one of the members of Γ^{−j}, then μ^{n−1}(\(p_{{\Gamma^{  j} }}\)) = 1and so μ^{2}(\(p_{{\{ \Gamma^{  j} ,S^{  j} \} }}\)) = 1 because μ^{n−1}(\(p_{{S^{  j} }}\)) ≥ 0. Secondly, let μ^{n}(p_{S}) = 0. If μ^{n−1}(\(p_{{S^{  j} }}\)) = 0, then by part (b) Γ^{−j} has an equal number of S^{−i}’s such that μ^{n−1}(\(p_{{S^{  i} }}\)) = 1 and μ^{n−1}(\(p_{{S^{  i} }}\)) = −1; therefore μ^{n−1}(\(p_{{\Gamma^{  j} }}\)) = 0 and consequently μ^{2}(\(p_{{\{ \Gamma^{  j} ,S^{  j} \} }}\)) = 0. If μ^{n−1}(\(p_{{S^{  j} }}\)) = 1, then by part (b) the number of S^{−i}’s in Γ^{−j} such that μ^{n−1}(\(p_{{S^{  i} }}\)) = −1 is larger than number of S^{−i}’s in Γ^{−j} such that μ^{n−1}(\(p_{{S^{  i} }}\)) = 1; therefore μ^{n−1}(\(p_{{\Gamma^{  j} }}\)) = −1. So μ^{2}(\(p_{{\{ \Gamma^{  j} ,S^{  j} \} }}\)) = µ^{2}(−1, 1) = 0 = μ^{n}(p_{S}) as required.^{Footnote 16}
Proof that α can be defined in terms of μ on domains including ideal votes.
Let S = {v_{1}, v_{2}, … v_{n}}. To show this, we first construct two societies S_{1} and S_{2}:
Let s members of S = {v_{1}, v_{2}, … v_{n}} 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 S_{1} the function μ^{2n} gives s times 0 and z + m times −1; and also s times 1 and z + m times 0. So μ^{2n}(\(p_{{S_{1} }}\)) = 1 if s > z + m and μ^{2n}(\(p_{{S_{1} }}\)) < 1 in all the other cases. But s > z + m and s + z + m = n entail that s > n/2. Analogously, we have that μ^{2n}(\(p_{{S_{2} }}\)) = −1 when m > n/2 and μ^{2n}(\(p_{{S_{2} }}\)) > −1 in all the other cases. Secondly, construct two other societies:
Clearly, μ^{2}(\(p_{{S_{11} }}\)) = 1 if s > n/2 and μ^{2}(\(p_{{S_{11} }}\)) = 0 in all the other cases; and μ^{2} \((p_{{S_{21} }})= 1\) if m > n/2 and μ^{2}(\(p_{{S_{21} }}\)) = 0 in all the other cases. Finally, put
We can easily verify that α^{n}(p_{S}) = μ^{2}(\(p_{{\upsigma _{S} }}\)).
Proof of Theorem 2. The proof is by induction on the number of members of S. Suppose first that n = 2. Let S = {v_{1}, v_{2}}. As already noted, in this care κ^{2}(p_{S}) = μ^{2}(p_{S}) for all profiles p of S. Now let n = 3, i.e. S = {v_{1}, v_{2}, v_{3}}. Clearly, there are profiles at which κ^{3} and μ^{3} do not coincide. But consider the society σ_{S} = {S^{−1}, S^{−2}, S^{−3}} = {{v_{2}, v_{3}}, {v_{1}, v_{3}}, {v_{1}, v_{2}}}. We show that:
This means that we can get μ^{3}(p_{S}) by the following procedure: we first calculate the value of κ^{3} for the three groups, each consisting in two members: κ^{2}(\(p_{{\{ v_{1} ,v_{2} \} }}\)), κ^{2}(\(p_{{\{ v_{2} ,v_{3} \} }}\)) and κ^{2}(\(p_{{\{ v_{1} ,v_{3} \} }}\)). Then we calculate the value of κ^{3} for the group consisting in these three groups. To show that the value we get by iteratively applying κ^{3} is the same as μ^{3}(p_{S}) 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 σ_{S} 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 κ^{2}(\(p_{{\{ v_{1} ,v_{2} \} }}\)) = 1, κ^{2}(\(p_{{\{ v_{2} ,v_{3} \} }}\)) = 0 and κ^{2}(\(p_{{\{ v_{1} ,v_{3} \} }}\)) = 0, so κ^{3}(\(p_{{\upsigma _{S} }}\)) = 1, in agreement with μ^{3}(p_{S}).

If \(p_{{v_{1} }}\) = 1, \(p_{{v_{2} }}\) = 0 and \(p_{{v_{3} }}\) = −1, then we get κ^{2}(\(p_{{\{ v_{1} ,v_{2} \} }}\)) = 1, κ^{2}(\(p_{{\{ v_{2} ,v_{3} \} }}\)) = −1 and κ^{2}(\(p_{{\{ v_{1} ,v_{3} \} }}\)) = 0, so κ^{3}(\(p_{{\upsigma _{S} }}\)) = 0, again in agreement with μ^{3}(p_{S}).
(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 − {v_{j}} that there is some society \(\upsigma _{{S^{  j} }}\) with the property that for all profiles \(p_{{S^{  i} }}\) of S^{−j} it holds that μ^{n−1}(\(p_{{S^{  i} }}\)) = κ^{n−1}(\(p_{{\upsigma _{{S^{  j} }} }}\)). We need to show that there is some society σ_{S} with the property that μ^{n}(p_{S}) = κ^{n}(\(p_{{\upsigma _{S} }}\)) for all profiles p_{S} of S. Put σ_{S} = {\(\upsigma _{{S^{  1} }}\), \(\upsigma _{{S^{  2} }}\), …\(\upsigma _{{S^{  n} }}\)}.^{Footnote 17}
The proof is trivial for all the cases when \(p_{{v_{i} }}\) ≥ 0 for all v_{i} ∊ S or \(p_{{v_{i} }}\) ≤ 0 for all v_{i} ∊ S. So suppose that \(p_{{v_{i} }}\) = 1 for some v_{i} ∊ S and \(p_{{v_{i} }}\) = −1 for some v_{i} ∊ S. I shall give the proof for the cases when { v_{i} ∊ S: \(p_{{v_{i} }}\) = 1} ≥{ v_{i} ∊ S: \(p_{{v_{i} }}\) = −1}, i.e. when µ^{n}(p_{S}) ≥ 0. The symmetrical cases when { v_{i} ∊ S: \(p_{{v_{i} }}\) = 1} ≤{ v_{i} ∊ S: \(p_{{v_{i} }}\) = −1} can be proved in an analogous way.
There are two possibilities. First, let μ^{n}(p_{S}) = 0. By lemma 1b It follows that { v_{i} ∊ S: \(p_{{v_{i} }}\) = 1} ={ v_{i} ∊ S: \(p_{{v_{i} }}\) = −1}= m ≥ 0, where 2 m ≤ n. If m = 0, then for all v_{i} it holds that \(p_{{v_{i} }}\) = 0 and so κ^{n}(\(p_{{\upsigma _{S} }}\)) = 0 by the definition of κ. If m > 0, then: a) since there is some v_{j} such that \(p_{{v_{j} }}\) = 1 it follows that μ^{n−1}(\(p_{{S^{  j} }}\)) = −1 because { v_{i} ∊ S^{−j}: \(p_{{v_{i} }}\) = −1} >{ v_{i} ∊ S^{−j}: \(p_{{v_{i} }}\) = 1}; b) since there is some v_{j} such that \(p_{{v_{j} }}\) = −1 it follows that μ^{n−1}(\(p_{{S^{  j} }}\)) = 1 because { v_{i} ∊ S^{−j}: \(p_{{v_{i} }}\) = 1} >{ v_{i} ∊ S^{−j}: \(p_{{v_{i} }}\) = −1}. By induction, for each S^{−j} we have μ^{n−1}(\(p_{{S^{  j} }}\)) = κ^{n−1}(\(p_{{\upsigma _{{S^{  j} }} }}\)), so we have κ^{n−1}(\(p_{{\upsigma _{{S^{  j} }} }}\)) = 1 in some cases and κ^{n−1}(\(p_{{\upsigma _{{S^{  j} }} }}\)) = −1 in other cases, whence by the definition of κ we get κ^{n−1}(\(p_{{\upsigma _{S} }}\)) = 0, whence κ^{n}(\(p_{{\upsigma _{S} }}\)) = μ^{n}(p_{S}).
Second, let μ^{n}(p_{S}) = 1. It follows that { v_{i} ∊ S: \(p_{{v_{i} }}\) = 1}= m_{1} >{ v_{i} ∊ S: \(p_{{v_{i} }}\) = −1}= m_{2}. Let v_{j} be some member of S. As shown in the proof of lemma 1a we have: (i) if \(p_{{v_{j} }}\) = −1, then μ^{n−1}(\(p_{{S^{  j} }}\)) = 1, since clearly { v_{i} ∊ S^{−j}: \(p_{{v_{i} }}\) = 1} >{ v_{i} ∊ S^{−j}: \(p_{{v_{i} }}\) = −1}; consequently, by induction κ^{n−1}(\(p_{{\upsigma _{{S^{  j} }} }}\)) = 1. Similarly, observe that: (ii) if \(p_{{v_{j} }}\) = 0, then μ^{n−1}(\(p_{{S^{  j} }}\)) = 1 and so by induction κ^{n−1}(\(p_{{\upsigma _{{S^{  j} }} }}\)) = 1. Finally, (iii) if \(p_{{v_{j} }}\) = −1, then we must have μ^{n−1}(\(p_{{S^{  j} }}\)) ≥ 0 and thus κ^{n−1}(\(p_{{\upsigma _{{S^{  j} }} }}\)) ≥ 0. There are two subcases:

If m_{2} > 0 (i.e. there is at least a v_{j} with the property that \(p_{{\upsigma _{{S^{  j} }} }}\) = −1), then there is some S^{−j} such that μ(\(p_{{S^{  j} }}\)) = 1, which entails κ^{n−1}(\(p_{{\upsigma _{{S^{  j} }} }}\)) = 1, and so κ^{n−1}(\(p_{{\upsigma _{S} }}\)) = 1 = μ^{n}(p_{S}).

If m_{2} = 0 (i.e. there is no v_{j} with the property that \(p_{{v_{j} }}\) = −1), it follows that there is some v_{j} ∊ S such that \(p_{{v_{j} }}\) = 1 and for all the other v_{j} ∊ S it holds that \(p_{{v_{j} }}\) ≥ 0. If \(p_{{v_{j} }}\) = 1 for all v_{j} ∊ S, then given that n > 3, we must have m_{1} – 1 > 0 and thus μ^{n−1}(\(p_{{S^{  j} }}\)) = 1 for all S^{−j}. Consequently, κ^{n−1}(\(p_{{\upsigma _{{S^{  j} }} }}\)) = 1, and so κ^{n}(\(p_{{\upsigma _{S} }}\)) = 1 = μ^{n}(p_{S}). But if \(p_{{v_{j} }}\) = 0 for some v_{j} ∊ S, then by induction (see the argument in the second case above) μ^{n−1}(\(p_{{S^{  j} }}\)) = 1. We can again conclude that κ^{n−1}(\(p_{{\upsigma _{{S^{  j} }} }}\)) = 1, whence κ^{n}(\(p_{{\upsigma _{S} }}\)) = 1 = μ^{n}(p_{S}).
Remark on theorem 3: if Neu and Mon hold, then NZ entails SP. Suppose that f^{n} satisfies NZ. Then if at a profile p_{S} we have that f^{n}(p_{S}) = 0, there must be a zigzag sequence \(p_{S}^{1}\), \(p_{S}^{2}\), … \(p_{S}^{m}\) of profiles of S such that f^{n}(\(p_{S}^{k}\)) = 1 for some profile \(p_{S}^{k}\) in this sequence (this is so because f^{n} satisfies Neu). Then by Mon we get that f^{n}(\(p_{S}^{^{\prime}}\)) = 1 at a profile \(p_{S}^{^{\prime}}\) where \(p_{{v_{j} }}\) ≥ 0 for all v_{j} ∈ S and \(p_{{v_{j} }}\) = 1 for some v_{j} ∈ S.
Proof of Theorem 4. We need to show that for each S there is some (higherorder) binary society σ_{S} such that f^{n}(p_{S}) = f^{2}(\(p_{{\sigma_{S} }}\)). The society σ_{S} is recursively constructed as follows:

(i)
if S = {i_{1}, i_{2}, i_{3}}, then σ_{S} = {{i_{1}, i_{2}}, {i_{3}}};

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

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

(2)
\(p_{{i_{n} }}\) = −1. This case is analogous to the first case.

(3)
\(p_{{i_{n} }}\) = 0. Then we always get υ^{2}(υ^{n−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 (higherorder) society σ_{S} such that (1) σ_{S} is binary; and (2) for all profiles p_{S} of S we have: μ^{n}(p_{S}) = μ^{2}(\(p_{{\sigma_{S} }}\)). 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} = {v_{2}, v_{3}}, S^{−2} = {v_{1}, v_{3}}, S^{−3} = {v_{1}, v_{2}}. Further, put Γ^{−3} = {S^{−1}, S^{−2}} = {{v_{2}, v_{3}}, {v_{1}, v_{3}}}. Finally, let σ_{S} = {Γ^{−3}, S^{−3}} = {{{v_{2}, v_{3}}, {v_{1}, v_{3}}}, {v_{1}, v_{2}}}. Since σ_{S} is clearly binary, the function μ^{2} can be iteratively applied to it. Lemma 1d gives μ^{2}(\(p_{{\sigma_{S} }}\)) = μ^{2}(\(p_{{\{ \Gamma^{  j} ,S^{  j} \} }}\)) = μ^{3}(p_{S}).
Now consider the case when S contains n members. Put Γ^{−n} = {S^{−1}, S^{−2}, … S^{−(n−1)}}. By definition, μ^{n−1}(\(p_{{\Gamma^{  n} }}\)) = μ^{n−1}(μ^{n−1}(\(p_{{S^{  1} }}\)), μ^{n−1}(\(p_{{S^{  2} }}\)), … μ^{n−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 μ^{n−1}(\(p_{{S^{  j} }}\)) = μ^{2}(\(p_{{\sigma_{{S^{  j} }} }}\)). So we get: μ^{n−1}(\(p_{{\Gamma^{  n} }}\)) = μ^{n−1}(μ^{2}(\(p_{{\sigma_{{S^{  1} }} }}\)), μ^{2}(\(p_{{\sigma_{{S^{  2} }} }}\)), … μ^{2}(\(p_{{\sigma_{{S^{  (n  1)} }} }}\))). 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 μ^{n−1}(\(p_{{\Gamma_{\upsigma }^{  n} }}\)) = μ^{2}(\(p_{{\upsigma _{{\Gamma_{\upsigma }^{  n} }} }}\)). Finally, let σ_{S} = {\(\upsigma _{{S^{  n} }}\), \(\upsigma _{{\Gamma_{\upsigma }^{  n} }}\)}. By lemma 1d we have that μ^{n}(p_{S}) = μ^{2}(\(p_{{\{ \Gamma^{  n} ,S^{  n} \} }}\)) = μ^{2}(μ^{n−1}(\(p_{{\Gamma^{  n} }}\)), μ^{n−1}(\(p_{{S^{  n} }}\))) and so μ^{n}(p_{S}) = μ^{2}(μ^{2}(\(p_{{\upsigma _{{\Gamma_{\upsigma }^{  n} }} }}\)), μ^{2}(\(p_{{\sigma_{{S^{  n} }} }}\))), which proves that σ_{S} is the society we wanted to construct.
Proof of Theorem 6. Notice first that µ^{2} satisfies the three axioms. Conversely, we need to show that if a scf f satisfies these axioms, then it must be exactly µ^{2}. Let S = {v_{1}, v_{2}}. We show that f^{2}(\(p_{{\{ v_{1} ,v_{2} \} }}\)) = sgn(\(p_{{v_{1} }}\) + \(p_{{v_{2} }}\)) in all possible cases. We have: (i) for profiles (1, 1) and (−1, −1) axiom BU gives f^{2}(\(p_{{\{ v_{1} ,v_{2} \} }}\)) = 1, respectively f^{2}(\(p_{{\{ v_{1} ,v_{2} \} }}\)) = −1; (ii) for profiles (1, −1) and (−1, 1) axiom SET gives f^{2}(\(p_{{\{ v_{1} ,v_{2} \} }}\)) = 0; (iii) for profiles (−1, 0) and (0, −1) axiom IIS gives f^{2}(\(p_{{\{ v_{1} ,v_{2} \} }}\)) = −1; similarly, for profiles (1, 0) and (0, 1) IIS gives f^{2}(\(p_{{\{ v_{1} ,v_{2} \} }}\)) = 1; iv) for the profile (0, 0) axiom IIS gives f^{2}(\(p_{{\{ v_{1} ,v_{2} \} }}\)) = 0. □
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Miroiu, A. Majority Voting and HigherOrder Societies. Group Decis Negot 30, 983–999 (2021). https://doi.org/10.1007/s1072602109744z
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Keywords
 Higherorder societies
 Majority rule
 Consensus
 Reducibility
 Binary societies