3.2 Preliminary results for the proof of Theorem 1
In this Subsection, we introduce additional notation and obtain some preliminary results that will be useful to prove Theorem
1.
Given \(x\in {\mathcal {M}},\) we denote by \(u^{x}:{\mathcal {M}}\cup \{\emptyset \}\rightarrow {\mathbb {R}}\) any function such that \(u^{x}(x)>\left| u^{x}(y)\right| >u^{x}(\emptyset )=0\) for all \(y\in {\mathcal {M}}\setminus \{x\}\), and if \(x\in T\) and \(x\notin T^{\prime }\) then \(\sum _{y\in T}u^{x}(y)>\sum \nolimits _{y\in T^{\prime }}u^{x}(y)\); accordingly, if \(u^{x}\) represents \(P_{i}\), then \(x\in T\) and \(x\notin T^{\prime }\) imply \(T\, P_{i}\,T^{\prime }\). Similarly, given \(y\in {\mathcal {M}}\), we denote by \(u_{y}:{\mathcal {M}}\cup \{\emptyset \}\rightarrow {\mathbb {R}}\) any function such that \(\left| u_{y}(y)\right|>\left| u_{y}(x)\right|>u_{y}(\emptyset )=0>u_{y}(y)\) for all \(x\in {\mathcal {M}}{\setminus } \{y\}\), and if \(y\in T\) and \(y\notin T^{\prime }\) then \(\sum _{x\in T}u_{y}(x)<\sum \nolimits _{x\in T^{\prime }}u_{y}(x)\); accordingly, if \(u_{y}\) represents \(P_{i}\), then \(y\in T\) and \(y\notin T^{\prime }\) imply \(T^{\prime }\,P_{i}\,T\).
We first show that in our context, false-name-proofness, strategy-proofness, and ontoness imply anonymity.
Proof of Lemma 1
\((\Rightarrow )\) To obtain a contradiction, let \(S\in {\mathcal {W}}_{N\cup N'\,,\,x}\) and assume that \(S\notin {\mathcal {W}}_{N\,,\,x}\) holds. Then, there exists a profile \(P\in {\mathcal {D}}^N\) such that \(x\in b(P_i)\) for all \(i\in S\) and \(x\notin f_N(P)\). Consider voter \(i\in S\) and any additive preference \(P^{\prime }_i\in {\mathcal {A}}\) such that (i) \(b(P^{\prime }_i)=b(P_i)\) and (ii) \(P^{\prime }_i\) is represented by \(u^x\). Since \(f_N\) is voting by committees, it depends only on the profile of tops. Accordingly, \(x\notin f_N(P^{\prime }_i,P_{N{\setminus }\{i\}})=f_N(P)\). Consider the profile \({\widehat{P}}\in {\mathcal {D}}^{N^{\prime }}\) where, for each \(j\in N^{\prime }\), \({\widehat{P}}_j=P_i^{\prime }\). Since f is false-name-proof, \(f_N(P^{\prime }_i,P_{N{\setminus }\{i\}})\,R_i^{\prime }\,f_{N\cup N'}(P^{\prime }_i,P_{N{\setminus }\{i\}},{\widehat{P}})\) which, by (ii) in the definition of \(P^{\prime }_i\), implies that \(x\notin f_{N\cup N'}(P^{\prime }_i,P_{N{\setminus }\{i\}},{\widehat{P}})\). Define \(P''=(P^{\prime }_i,P_{N{\setminus }\{i\}},{\widehat{P}})\in {\mathcal {D}}^{N\cup N^{\prime }}\). Hence, at profile \(P''\), \(S\subseteq \{j\in N\cup N'\mid x\in b(P''_j)\}\notin {\mathcal {W}}_{N\cup N^{\prime }\,,\,x}\). Coalition monotonicity of the committee implies that \(S\notin {\mathcal {W}}_{N\cup N',\,x}\), which is a contradiction.
\((\Leftarrow )\) To obtain a contradiction, let \(S\in {\mathcal {W}}_{N\,,\,x}\) and assume that \(S\notin {\mathcal {W}}_{N\cup N'\,,\,x}\) holds. Then, there exists a profile \(P'\in {\mathcal {D}}^{N\cup N'}\) such that \(x\in b(P'_i)\) for all \(i\in S\) and \(x\notin f_{N\cup N'}(P')\). Consider the profile \(P\in {\mathcal {D}}^N\) such that \(P_i=P'_i\) for all \(i\in S\) and \(P_j\) is represented by \(u_x\) for all \(j\in N\backslash {S}\). Observe that the assumption \(S\subsetneq N\) implies \(N\backslash {S}\ne \emptyset\). Since \(S\in {\mathcal {W}}_{N\,,\,x}\), \(x\in f_N(P)\) holds. Consider now the profile \(P^*\in {\mathcal {D}}^{N\cup N'}\) such that \(P^*_i=P'_i=P_i\) for all \(i\in S\) and each \(j\in (N\cup N'){\setminus } S\) has the same preference \(P^*_j\in {\mathcal {A}}\) represented by \(u_x\). Since f is false-name-proof, we have that \(x\in f_{N\cup N'}(P^*)\), otherwise an agent \(j\in N\backslash {S}\ne \emptyset\) has the incentive to replicate its vote under the identities of voters in \(N'\). Now, for a given \(j\in (N\cup N')\backslash {S}\), we have by strategy-proofness that \(x\in f_{N\cup N'}(P^*_{-j},P'_j)\). Otherwise, agent j has the incentive to misreport her vote. By repeating this procedure for each \(j\in (N\cup N')\backslash {S}\) and by strategy-proofness, we obtain that \(x\in f_{N\cup N'}(P')\), a contradiction. \(\square\)
We now proceed with the proof of Proposition
3.
Suppose f is not anonymous. Then, there exists a bijection \(\sigma :Z_+\rightarrow Z_+\), a society \(N\in {\mathcal {N}}\) and a profile \(P\in {\mathcal {D}}^N\) such that \(f_N(P)\ne f_{\sigma (N)}(\sigma (P))\). Since \(\sigma\) is a bijection we can assume, without loss of generality, that there exist \(x\in {\mathcal {M}}\) and \(S\in N\) such that (i) \(x\in f_N(P)\) and \(x\notin f_{\sigma (N)}(\sigma (P))\). Hence, there is \(S\in {\mathcal {W}}_{N,x}\) such that \(T:=\sigma (S)\notin {\mathcal {W}}_{\sigma (N),x}\). As the whole society always is a winning coalition, we have that \(T\subsetneq \sigma (N)\), and thus \(S\subsetneq N\). Define \(N^*=\sigma (N)\cup N\) and \(N'=N^*{\setminus }{S}\). We proceed by distinguishing between two cases.
Case 1: \(T\ne S\).
Since \(\sigma\) is a bijection, \(T{\setminus } S\ne \emptyset\). To obtain a contradiction, suppose \(T{\setminus } S\notin {\mathcal {W}}_{N^*,x}\). As \(T{\setminus } S\subsetneq (\sigma (N)\cup N){\setminus } S\) we have, by Lemma 1, that \(T{\setminus } S\notin {\mathcal {W}}_{N'\,,\,x}\). This means that there is a profile \(P'\in {\mathcal {D}}^{N'}\) such that \(x\in b(P'_t)\) for all \(t\in T{\setminus } S\) and \(x\notin f_{N'}(P')\). By Lemma 1, as \(S\in {\mathcal {W}}_{N\,,\,x}\) and \(S\subsetneq N\), we have that \(S\in {\mathcal {W}}_{N^*\,,\,x}\). As \(S\subsetneq (T{\setminus } S)\cup S\), by coalitional monotonicity, we have that \((T{\setminus } S)\cup S\in {\mathcal {W}}_{N^*\,,\,x}\). Now, consider a profile \(P''\in {\mathcal {D}}^{N'}\) such that \(b(P''_t)=b(P'_t)\) and \(P''_t\) is represented by \(u^x\) for all \(t\in T{\setminus } S\), and \(P''_k=P'_k\) for all \(k\notin T{\setminus } S\). As \(f_{N'}\) is voting by committees, it depends only on the profile of tops, so \(x\notin f_{N'}(P'')=f_{N'}(P')\). But then any agent \(t\in T{\setminus } S\) has the incentive to replicate her vote under the identities of voters in S, to obtain \(x\in f_{N^*}(P'',P_{S})\), where \(P_i=P''_t\) for all \(i\in S\), violating false-name-proofness. So we have that \(T{\setminus } S\in {\mathcal {W}}_{N^*\,,\,x}\). As \(T{\setminus } S\ne \emptyset\), by Lemma 1, we obtain that \(T{\setminus } S\in {\mathcal {W}}_{\sigma (N)\,,\,x}\) and thus, by coalition monotonicity, \(T=\sigma (S)\in {\mathcal {W}}_{\sigma (N),\,x}\), a contradiction with our initial hypothesis.
Case 2: \(T=S\)
Since we have that
\(S\subsetneq N\), by Lemma
1,
\(S\in {\mathcal {W}}_{N^*,\,x}\), Then, as
\(T\subsetneq \sigma (N)\), and again by Lemma
1, we obtain that
\(T\in {\mathcal {W}}_{\sigma (N),\,x}\), a contradiction with our initial hypothesis that
f is not anonymous.
This finishes with the proof of Proposition
3\(\square\)
The second preliminary result shows the effect that false-name-proofness has on the quotas of a voting rule.
Proof of the Lemma 2
To obtain a contradiction, suppose that \(q_{N\,,\,x}>q_{N\cup N'\,,\,x}\) for an object \(x\in {\mathcal {M}}\). Hence, \(q_{N,\,x}>1\). Consider \(i\in N\) and \(P\in {\mathcal {D}}^{N}\) such that \(P_i\in {\mathcal {A}}\) is additively represented by \(u^{x}\) and \(x_{N}(P)=q_{N,\,x}-1\). Observe that \(x\in b(P_i)\) by definition of \(u^{x}\), and \(x\notin f_N(P)\) by definition of voting by quota \(q_{N}\). Consider now a profile \(P'\in {\mathcal {D}}^{N'}\) such that \(P'_j=P_i\) for all \(j\in N'\). Since \(N\cap N'=\emptyset\), \(x_{N\cup N'}(P, P')=x_{N}(P)+x_{N'}(P')=q_{N\,,\,x}-1+n'>q_{N\cup N'\,,\,x}\), by our contradiction hypothesis and \(n'\ge 1\). Since \(f_{N\cup N'}\) is voting by quota \(q_{N\cup N'}\), \(x\in f_{N\cup N'}(P, P')\) and, by definition of \(u^{x}\), \(f_{N\cup N'}(P, P')\,P_{i}\,f_{N}(P)\), a contradiction with false-name-proofness.\(\Box\)
To proceed with the proof of Proposition
4, first observe that by the definition of voting by quota, anonymity, and Lemma
2, for a given
x,
\(q_{1\,,\,x}=1\le q_{2\,,\,x}\le 2\) holds, where
\(q_{k\,,\,x}=q_{S\,,\,x}\) for any
\(S\in {\mathcal {N}}\) with
\(|S |=k\). Fix
\(x\in {\mathcal {M}}\). We distinguish between the two possible cases and, for each of them, we show that the statement of Proposition
4 holds by induction on
n (the number of voters).
Case 1: \(q_{1,\,x}= q_{2,\,x}=1\).
Induction hypothesis: Assume \(q_{1\,,\,x}=\dots =q_{n\,,\,x}=1\).
Consider the case \(n+1\), where \(n\ge 2\). We want to show that \(q_{n+1,\,x}=1\). To obtain a contradiction, suppose \(q_{n+1,\,x}>q_{n,\,x}=1\). Consider \(i\in N\) and a profile \(P\in {\mathcal {D}}^N\) such that \(P_i\in {\mathcal {A}}\) is additively represented by \(u_x\) and \(x_N(P)=1\). Observe that \(x\notin b(P_i)\) by definition of \(u_x\), and \(x\in f_N(P)\) by definition of voting by quota \(q_{n,\,x}\). Let \(j\notin N\) and \(P_j=P_i\). Then, \(x_{N\cup \{j\}}(P, P_j)=1<q_{n+1,\,x}\) and, accordingly, \(x\notin f_{N\cup \{j\}}(P, P_{j})\), which is a contradiction with false-name-proofness because, by definition of \(u_x\), \(f_{N\cup \{j\}}(P, P_{j})\,P_{i}\,f_N(P)\).
Case 2: \(q_{1,\,x}= 1<2=q_{2,\,x}\).
Induction hypothesis: Assume \(q_{x,\,t}=t=|T|\) for every \(T\in {\mathcal {N}}\) such that \(1\le t\le n\).
Consider the case
\(n+1\), where
\(n\ge 2\). We want to show that
\(q_{n+1,\,x}=n+1\). By the definition of quota, Lemma
2, and the induction hypothesis we have that
\(n\le q_{n+1\,,\,x}\le n+1\). Suppose that
\(q_{n+1\,,\,x}=n\). Consider
\(i\in N\),
\(x\in {\mathcal {M}}\) and a profile
\(P\in {\mathcal {D}}^N\) such that
\(P_i\in {\mathcal {A}}\) is additively represented by
\(u^x\) and
\(x_N(P)=q_{n,\,x}-1=n-1\). Observe that
\(x\in b(P_i)\) by definition of
\(u^{x}\), and
\(x\notin f_N(P)\) by definition of voting by quota
\(q_{n}\). Let
\(j\notin N\) and
\(P_j=P_i\). Then,
\(x_{N\cup \{j\}}(P, P_j)=n-1+1=q_{n+1,\,x}\) and, accordingly,
\(x\in f_{N\cup \{j\}}(P, P_{j})\), which is a contradiction with false-name-proofness because, by definition of
\(u^x\),
\(f_{N\cup \{j\}}(P,P_{j})\,P_{i}\,f_N(P)\).
Thus, f is voting by quota \(q=(q_{N\,,\,x})_{N\in {\mathcal {N}}\,,\,x\in {\mathcal {M}}}\) where, for every \(x\in {\mathcal {M}}\), either \(q_{N\,,\,x}=1\) for every \(N\in {\mathcal {N}}\) or \(q_{N,\,x}=n\) for every \(N\in {\mathcal {N}}\). \(\square\)
The third preliminary result states that voting by quota in this subclass satisfies false-name-proofness.
Proof
Let \(f=\{f_{N}:{\mathcal {D}}^{N}\rightarrow 2^{{\mathcal {M}}}\}_{N\in {\mathcal {N}}}\) be a voting by quota \(q=(q_{N,\,x})_{N\in {\mathcal {N}},\,x\in {\mathcal {M}}}\). Assume first that, for a given \(x\in {\mathcal {M}}\), \(q_{N\,,\,x}=n\) for all \(N\in {\mathcal {N}}\). We show that f is strongly false-name-proof, which would imply that f is false-name-proof. Let \((N,P)\in {\mathcal {N}}\times {\mathcal {D}}^{N}\), \(i\in N\), and \((N^{\prime },P^{\prime })\in {\mathcal {N}}\times {\mathcal {D}}^{N^{\prime }}\) be arbitrary, and assume \(N\cap N'=\emptyset\). As \(f_{N}\) and \(f_{N\cup N^{\prime }}\) are quota n and \(n+n^{\prime }\), respectively, \(f_{N}(P)=\cap _{j\in N}b(P_{j})\supseteq \cap _{j\in N\cup N^{\prime }}b(P_{j})=f_{N\cup N^{\prime }}(P, P^{\prime })\). Therefore, \(f_{N\cup N^{\prime }}(P, P^{\prime })=f_N(P){\setminus } B\) for some \(B\subseteq b(P_{i})\). By iteratively applying separability to each object in B and transitivity of \(P_{i}\), we obtain that \(f_N(P)\;R_{i}\;f_{N\cup N^{\prime }}(P, P^{\prime })\) and so f is strongly false-name-proof. Therefore f is also false-name-proof.
Assume now that, for a given \(x\in {\mathcal {M}}\), \(q_{N\,,\,x}=1\) for all \(N\in {\mathcal {N}}\). We show that f is strongly false-name-proof, which would imply that f is false-name-proof. Let \((N,P)\in {\mathcal {N}}\times {\mathcal {D}}^{N}\), \(i\in N\), and \((N^{\prime },P^{\prime })\in {\mathcal {N}}\times {\mathcal {D}}^{N^{\prime }}\) be arbitrary, and assume \(N\cap N'=\emptyset\). As \(f_{N}\) and \(f_{N\cup N^{\prime }}\) are both quota one, \(b(P_{i})\subseteq f_{N}(P)=\cup _{j\in N}b(P_{j})\subseteq \cup _{j\in N\cup N^{\prime }}b(P_{j})=f_{N\cup N^{\prime }}(P, P^{\prime })\). Therefore, \(f_{N\cup N^{\prime }}(P, P^{\prime })=f_N(P)\sqcup W\) for some \(W\subseteq w(P_{i})\), where \(\sqcup\) stands for the disjoint union. By iteratively applying separability to each object in W and transitivity of \(P_{i}\), we obtain that \(f_N(P)\,R_{i}\,f_{N\cup N^{\prime }}(P, P^{\prime })\) and so f is strongly false-name-proof. Therefore, f is also false-name-proof. \(\square\)
3.4 Additional results
In this subsection, we present additional results. We first show that if objects only need one vote or a unanimous vote to be chosen, then the rule must verify participation. Hence, participation follows from false-name-proofness, strategy-proofness, and ontoness.
For the case of a unique object, which corresponds to the general setting where there are only two alternatives (identified as
\(\{x\}\) and
\(\emptyset\)), participation implies strategy-proofness.
8
It is easy to see that if there are two or more objects, participation does not imply strategy-proofness: the Borda count, combined with a tie-breaking that selects one among the potentially many subsets of winners, is an example of a voting rule that satisfies participation and it is not strategy-proof.
We now show that in
any setting (and, in particular, in
any domain of preferences) strong false-name-proofness, anonymity, and participation imply strategy-proofness.
9 Let
A be any set of social alternatives, let
\({\mathcal {U}}\) be any set of complete and transitive preferences over
A, and adapt the properties of a rule
\(f_N:{\mathcal {U}}^N\rightarrow A\) for
N as previously defined for our setting. Denote by
\(R_i\) a generic (and possibly weak) preference of voter
i over
A in
\({\mathcal {U}}\) and let
\(R=(R_i)_{i\in N}\in {\mathcal {U}}^N\) be a profile.
Therefore, Corollary
3 below follows from Corollary
1 and Propositions
3,
6 and
8.
Example
1 below shows that the statement of Proposition
8 does not hold if strong false-name-proofness is replaced by false-name-proofness, even if the domain of the voting rule is restricted to be the set of separable preferences.