On strategy-proof social choice between two alternatives

Abstract

We study strategy-proof rules for choosing between two alternatives. We consider the full preference domain which allows for indifference. In this framework, for strategy-proof rules, ontoness does not imply efficiency. We weaken the requirement of efficiency to ontoness and characterize the class of strategy-proof rules. We argue that the notion of efficiency is not desirable always. Further, we provide a simple description of the class of onto, anonymous and strategy-proof rules in this framework. The key feature of our characterization results brings out the role played by indifferent agents.

Appendices

Appendix

The Proof of Lemma 1

Proof

Note that if f is strategy-proof then it is weakly strategy-proof. So suppose that f is weakly strategy-proof, but to the contrary f is not strategy-proof. Then there exist an agent $i \in N$ and a profile $R \in {\mathcal {R}}^n$ and an $i-$deviation $R^\prime \in {\mathcal {R}}$ of R such that $R_i, R^\prime _i \in {\mathcal {P}}$ and $f(R^\prime ) P_i f(R)$. So it follows that $R_i \ne R^\prime _i$. Without loss of generality, assume that $a P_i b$ and $b P^\prime _i a$. So it follows that $f(R) = b$ and $f(R^\prime ) = a$. Now consider the profile $R^\star \in {\mathcal {R}}^n$ such that $R^\star _{N {\setminus } \{i\}} = R^\prime _{N {\setminus } \{i\}} = R_{N {\setminus } \{i\}}$, and $a I^\star _i b$. Now weak strategy-proofness for the deviation from R to $R^\star$ implies that $f(R^\star ) = b$. On the other hand weak strategy-proofness for the deviation from $R^\prime$ to $R^\star$ implies that $f(R^\star ) = a$, which contradicts the fact that $f(R^\star ) = b$ and concludes the proof. $\square$

The Proof of Theorem 1

Proof

If part. Let $f^{{\mathcal {I}}^d}_{{\mathcal {F}}_{{\mathcal {I}}^d}}$ be a GVC rule. Let ${\mathcal {I}}^d$ be the committee for indifference default $d\in \{a,b\}$ and ${\mathcal {F}}_{{\mathcal {I}}^d}$, the collection of committees for a with respect to ${\mathcal {I}}^d$. We show that $f^{{\mathcal {I}}^d}_{{\mathcal {F}}_{{\mathcal {I}}^d}}$ is onto and strategy-proof.

To prove that $f^{{\mathcal {I}}^d}_{{\mathcal {F}}_{{\mathcal {I}}^d}}$ is onto, we show that there exist $R',R''\in {\mathcal {R}}^n$ such that $f^{{\mathcal {I}}^d}_{{\mathcal {F}}_{{\mathcal {I}}^d}}(R')=a$ and $f^{{\mathcal {I}}^d}_{{\mathcal {F}}_{{\mathcal {I}}^d}}(R'')=b$. Let $R'$ and $R''$ be such that $N_a(R')=N$ and $N_b(R'')=N$ respectively. Note that $N_{A}(R')=N_b(R')=\emptyset$, $N_a(R')\in {\mathcal {F}}_{N{\setminus } N_{ab}(R'),{\mathcal {I}}^d}$ and $N_{A}(R')\notin {\mathcal {I}}^d$. Therefore, $f^{{\mathcal {I}}^d}_{{\mathcal {F}}_{{\mathcal {I}}^d}}(R')=a$. Again, since $N_{A}(R'')=N_a(R'')=\emptyset$, $N_a(R'')\notin {\mathcal {F}}_{N{\setminus } N_{ab}(R''),{\mathcal {I}}^d}$ and $N_{A}(R'')\notin {\mathcal {I}}^d$, we have $f^{{\mathcal {I}}^d}_{{\mathcal {F}}_{{\mathcal {I}}^d}}(R'')=b$.

Next we show that $f^{{\mathcal {I}}^d}_{{\mathcal {F}}_{{\mathcal {I}}^d}}$ satisfies strategy-proofness. We consider $R\in {\mathcal {R}}^n$ and $R'_i\in {\mathcal {R}}$.

First we assume that $f^{{\mathcal {I}}^d}_{{\mathcal {F}}_{{\mathcal {I}}^d}}(R)=a$. If $aR_ib$, then i can not manipulate at R via $R'_i$. If $bP_ia$ then we show that $f^{{\mathcal {I}}^d}_{{\mathcal {F}}_{{\mathcal {I}}^d}}(R'_i,R_{N {\setminus } \{i\}})=a$. The following two cases arise : (i) $aI'_ib$ and (ii) $aP'_ib$.

(i) Suppose $aI'_ib$. Let $d=a$. If $N_{A}(R)\in {\mathcal {I}}^a$, then $N_{A}(R'_i,R_{N {\setminus } \{i\}})\in {\mathcal {I}}^a$. Therefore, $f^{{\mathcal {I}}^a}_{{\mathcal {F}}_{{\mathcal {I}}^a}}(R'_i,R_{N {\setminus } \{i\}})=a$. If $N_{A}(R)\notin {\mathcal {I}}^a$, then $f^{{\mathcal {I}}^a}_{{\mathcal {F}}_{{\mathcal {I}}^a}}(R)=a$ implies that $N_a(R)\in {\mathcal {F}}_{N{\setminus } N_{A}(R),{\mathcal {I}}^a}$. Now we consider the set $N_{A}(R'_i,R_{N {\setminus } \{i\}})$. If $N_{A}(R'_i,R_{N {\setminus } \{i\}})\in {\mathcal {I}}^a$, then $f^{{\mathcal {I}}^a}_{{\mathcal {F}}_{{\mathcal {I}}^a}}(R'_i,R_{N {\setminus } \{i\}})=a$. If $N_{A}(R'_i,R_{N {\setminus } \{i\}})\notin {\mathcal {I}}^a$, then the property 2 of ${\mathcal {F}}_{{\mathcal {I}}^a}$ would imply that $N_a(R'_i,R_{N {\setminus } \{i\}})\in {\mathcal {F}}_{N{\setminus } N_{A}(R'_i,R_{N {\setminus } \{i\}}),{\mathcal {I}}^a}$. Therefore, $f^{{\mathcal {I}}^a}_{{\mathcal {F}}_{{\mathcal {I}}^a}}(R'_i,R_{N {\setminus } \{i\}})=a$.

Let $d=b$. Since $f^{{\mathcal {I}}^b}_{{\mathcal {F}}_{{\mathcal {I}}^b}}(R)=a$, $N_{A}(R)\notin {\mathcal {I}}^b$ and $N_a(R)\in {\mathcal {F}}_{N{\setminus } N_{A}(R),{\mathcal {I}}^b}$. Now we consider the set $N_{A}(R'_i,R_{N {\setminus } \{i\}})$. If $N_{A}(R'_i,R_{N {\setminus } \{i\}})\in {\mathcal {I}}^b$, then by property 1 of ${\mathcal {F}}_{{\mathcal {I}}^b}$, $i\in N_a(R)$ which is not possible. Therefore, $N_{A}(R'_i,R_{N {\setminus } \{i\}})\notin {\mathcal {I}}^b$. Since $N_a(R)\in {\mathcal {F}}_{N{\setminus } N_{A}(R),{\mathcal {I}}^b}$ and $N_a(R'_i,R_{N {\setminus } \{i\}})=N_a(R)$, by the property 2 of ${\mathcal {F}}_{{\mathcal {I}}^b}$ we have $N_a(R'_i,R_{N {\setminus } \{i\}})\in {\mathcal {F}}_{N{\setminus } N_{A}(R'_i,R_{N {\setminus } \{i\}}),{\mathcal {I}}^b}$. Therefore, $f^{{\mathcal {I}}^b}_{{\mathcal {F}}_{{\mathcal {I}}^b}}(R'_i,R_{N {\setminus } \{i\}})=a$.

(ii) Suppose $aP'_ib$. Let $d=a$. Note that $N_{A}(R'_i,R_{N {\setminus } \{i\}})=N_{A}(R)$. If $N_{A}(R)\in {\mathcal {I}}^a$, then $f^{{\mathcal {I}}^a}_{{\mathcal {F}}_{{\mathcal {I}}^a}}(R'_i,R_{N {\setminus } \{i\}})=a$. If $N_{A}(R)\notin {\mathcal {I}}^a$, then $f^{{\mathcal {I}}^a}_{{\mathcal {F}}_{{\mathcal {I}}^a}}(R)=a$ implies that $N_a(R)\in {\mathcal {F}}_{N{\setminus } N_{A}(R),{\mathcal {I}}^a}$. By monotonicity property of ${\mathcal {F}}_{N{\setminus } N_{A}(R),{\mathcal {I}}^a}$, $N_a(R'_i,R_{N {\setminus } \{i\}})\in {\mathcal {F}}_{N{\setminus } N_{A}(R),{\mathcal {I}}^a}$. Since $N_{A}(R'_i,R_{N {\setminus } \{i\}})=N_{A}(R)$, $f^{{\mathcal {I}}^a}_{{\mathcal {F}}_{{\mathcal {I}}^a}}(R'_i,R_{N {\setminus } \{i\}})=a$.

Let $d=b$. Since $f^{{\mathcal {I}}^b}_{{\mathcal {F}}_{{\mathcal {I}}^b}}(R)=a$, $N_{A}(R)\notin {\mathcal {I}}^b$ and $N_a(R)\in {\mathcal {F}}_{N{\setminus } N_{A}(R),{\mathcal {I}}^b}$. Also, since $N_{A}(R'_i,R_{N {\setminus } \{i\}})=N_{A}(R)$, $N_{A}(R'_i,R_{N {\setminus } \{i\}})\notin {\mathcal {I}}^b$. By monotonicity property of ${\mathcal {F}}_{N{\setminus } N_{A}(R),{\mathcal {I}}^b}$, $N_a(R'_i,R_{N {\setminus } \{i\}})\in {\mathcal {F}}_{N{\setminus } N_{A}(R),{\mathcal {I}}^b}$. Therefore, $f^{{\mathcal {I}}^b}_{{\mathcal {F}}_{{\mathcal {I}}^b}}(R'_i,R_{N {\setminus } \{i\}})=a$.

Now we assume that $f^{{\mathcal {I}}^d}_{{\mathcal {F}}_{{\mathcal {I}}^d}}(R)=b$. If $bR_ia$, then i can not manipulate. If $aP_ib$ then we show that $f^{{\mathcal {I}}^d}_{{\mathcal {F}}_{{\mathcal {I}}^d}}(R'_i,R_{N {\setminus } \{i\}})=b$. The following two cases arise : (i) $aI'_ib$ and (ii) $bP'_ia$.

(i) Suppose $aI'_ib$. Let $d=a$. Since $f^{{\mathcal {I}}^a}_{{\mathcal {F}}_{{\mathcal {I}}^a}}(R)=b$, $N_{A}(R)\notin {\mathcal {I}}^a$ and $N_a(R)\notin {\mathcal {F}}_{N{\setminus } N_{A}(R),{\mathcal {I}}^a}$. Now we consider the set $N_{A}(R'_i,R_{N {\setminus } \{i\}})$. If $N_{A}(R'_i,R_{N {\setminus } \{i\}})\in {\mathcal {I}}^a$, then by property 1 of ${\mathcal {F}}_{{\mathcal {I}}^a}$, $N_a(R)\in {\mathcal {F}}_{N{\setminus } N_{A}(R),{\mathcal {I}}^a}$ which is not possible. Therefore, $N_{A}(R'_i,R_{N {\setminus } \{i\}})\notin {\mathcal {I}}^a$. Since $N_a(R)\notin {\mathcal {F}}_{N{\setminus } N_{A}(R),{\mathcal {I}}^a}$ and $N_{A}(R'_i,R_{N {\setminus } \{i\}})\notin {\mathcal {I}}^a$, property 3 of ${\mathcal {F}}_{{\mathcal {I}}^a}$ would imply that $N_a(R'_i,R_{N {\setminus } \{i\}})\notin {\mathcal {F}}_{N{\setminus } N_{A}(R'_i,R_{N {\setminus } \{i\}}),{\mathcal {I}}^a}$. Therefore, $f^{{\mathcal {I}}^a}_{{\mathcal {F}}_{{\mathcal {I}}^a}}(R'_i,R_{N {\setminus } \{i\}})=b$.

Let $d=b$. If $N_{A}(R)\in {\mathcal {I}}^b$, then (by monotonicity property of ${\mathcal {I}}^b$) $N_{A}(R'_i,R_{N {\setminus } \{i\}})\in {\mathcal {I}}^b$. Therefore, $f^{{\mathcal {I}}^b}_{{\mathcal {F}}_{{\mathcal {I}}^b}}(R'_i,R_{N {\setminus } \{i\}})=b$. If $N_{A}(R)\notin {\mathcal {I}}^b$, then $f^{{\mathcal {I}}^b}_{{\mathcal {F}}_{{\mathcal {I}}^b}}(R)=b$ implies that $N_a(R)\notin {\mathcal {F}}_{N{\setminus } N_{A}(R),{\mathcal {I}}^b}$. Now we consider the set $N_{A}(R'_i,R_{N {\setminus } \{i\}})$. If $N_{A}(R'_i,R_{N {\setminus } \{i\}})\in {\mathcal {I}}^b$, then $f^{{\mathcal {I}}^b}_{{\mathcal {F}}_{{\mathcal {I}}^b}}(R'_i,R_{N {\setminus } \{i\}})=b$. If $N_{A}(R'_i,R_{N {\setminus } \{i\}})\notin {\mathcal {I}}^b$, then property 3 of ${\mathcal {F}}_{{\mathcal {I}}^b}$ would imply that $N_a(R'_i,R_{N {\setminus } \{i\}})\notin {\mathcal {F}}_{N{\setminus } N_{A}(R'_i,R_{N {\setminus } \{i\}}),{\mathcal {I}}^a}$. Therefore, $f^{{\mathcal {I}}^b}_{{\mathcal {F}}_{{\mathcal {I}}^b}}(R'_i,R_{N {\setminus } \{i\}})=b$.

(ii) Suppose $bP'_ia$. Note that $N_{A}(R'_i,R_{N {\setminus } \{i\}})=N_{A}(R)$. Let $d=a$. Since $f^{{\mathcal {I}}^a}_{{\mathcal {F}}_{{\mathcal {I}}^a}}(R)=b$, $N_{A}(R)\notin {\mathcal {I}}^a$ and $N_a(R)\notin {\mathcal {F}}_{N{\setminus } N_{A}(R),{\mathcal {I}}^a}$. Since $N_{A}(R'_i,R_{N {\setminus } \{i\}})=N_{A}(R)$, we have $N_{A}(R'_i,R_{N {\setminus } \{i\}})\notin {\mathcal {I}}^a$ and $N_a(R)\notin {\mathcal {F}}_{N{\setminus } N_{A}(R'_i,R_{N {\setminus } \{i\}}),{\mathcal {I}}^a}$. Note that $N_a(R'_i,R_{N {\setminus } \{i\}})\notin {\mathcal {F}}_{N{\setminus } N_{A}(R'_i,R_{N {\setminus } \{i\}}),{\mathcal {I}}^a}$, otherwise by monotonicity property of ${\mathcal {F}}_{N{\setminus } N_{A}(R'_i,R_{N {\setminus } \{i\}}),{\mathcal {I}}^a}$, $N_a(R)\in {\mathcal {F}}_{N{\setminus } N_{A}(R'_i,R_{N {\setminus } \{i\}}),{\mathcal {I}}^a}$ which is not possible. Therefore, $f^{{\mathcal {I}}^a}_{{\mathcal {F}}_{{\mathcal {I}}^a}}(R'_i,R_{N {\setminus } \{i\}})=b$.

Let $d=b$. If $N_{A}(R)\in {\mathcal {I}}^b$, then $N_{A}(R'_i,R_{N {\setminus } \{i\}})\in {\mathcal {I}}^b$. Therefore, $f^{{\mathcal {I}}^b}_{{\mathcal {F}}_{{\mathcal {I}}^b}}(R'_i,R_{N {\setminus } \{i\}})=b$. So, we consider that $N_{A}(R)\notin {\mathcal {I}}^b$. since $f^{{\mathcal {I}}^b}_{{\mathcal {F}}_{{\mathcal {I}}^b}}(R)=b$, $N_a(R)\notin {\mathcal {F}}_{N{\setminus } N_{A}(R),{\mathcal {I}}^b}$. Note that since $N_{A}(R'_i,R_{N {\setminus } \{i\}})=N_{A}(R)$, we have $N_a(R'_i,R_{N {\setminus } \{i\}})\notin {\mathcal {F}}_{N{\setminus } N_{A}(R'_i,R_{N {\setminus } \{i\}}),{\mathcal {I}}^a}$, otherwise by monotonicity property of ${\mathcal {F}}_{N{\setminus } N_{A}(R'_i,R_{N {\setminus } \{i\}}),{\mathcal {I}}^a}$, $N_a(R)\in {\mathcal {F}}_{N{\setminus } N_{A}(R'_i,R_{N {\setminus } \{i\}}),{\mathcal {I}}^a}$ which is not possible. Therefore, $f^{{\mathcal {I}}^b}_{{\mathcal {F}}_{{\mathcal {I}}^b}}(R'_i,R_{N {\setminus } \{i\}})=b$.

Only if part. Let f be an onto and strategy-proof SCF. Let ${\bar{R}}\in {\mathcal {R}}^n$ denotes the preference profile where all agents are indifferent between a and b. We show that if $f({\bar{R}})=a$, then there exists a committee for indifference default a, ${\mathcal {I}}^a$ and a collection of committees for a with respect to ${\mathcal {I}}^a$, ${\mathcal {F}}_{{\mathcal {I}}^a}$, such that for all $R\in {\mathcal {R}}^n$;

$$\begin{aligned} f(R) = f^{{\mathcal {I}}^a}_{{\mathcal {F}}_{{\mathcal {I}}^a}}(R). \end{aligned}$$

Similarly, if $f({\bar{R}})=b$, then there exists a committee for indifference default b, ${\mathcal {I}}^b$ and a collection of committees for a with respect to ${\mathcal {I}}^b$, ${\mathcal {F}}_{{\mathcal {I}}^b}$, such that for all $R\in {\mathcal {R}}^n$;

$$\begin{aligned} f(R) = f^{{\mathcal {I}}^b}_{{\mathcal {F}}_{{\mathcal {I}}^b}}(R). \end{aligned}$$

In the following, we consider these two cases.

Case 1: $f({\bar{R}})=a$. For each $M\subseteq N$, let $g^M_f$ be the restriction of f to $\{R\in {\mathcal {R}}^n: aI_ib \textit{ iff } i\notin M \}$. In other words, let $g^M_f : \{R\in {\mathcal {R}}^n: aI_ib \textit{ iff } i\notin M \} \longrightarrow A$ be a function defined as $g^M_f(R) = f(R)$ for all $R \in \{R\in {\mathcal {R}}^n: aI_ib \textit{ iff } i\notin M \}$. First we show the following claim.

Claim 1

For each $M\subseteq N$, either $g^{M}_f$ is a constant rule that picks a or $g^{M}_f$ is onto.

Proof

If $M=\emptyset$, then it is trivial that $g^{M}_f$ is a constant rule that picks a. Therefore, for contradiction, we assume that there exists $\emptyset \ne M'\subseteq N$ such that $g^{M'}_f$ is a constant rule that picks b. W.o.l.g. let $M'=\{1,2,\ldots ,k\}$, $k\le n$. Let $R'$ be such that $aI'_ib$ if $i\in \{k+1,\ldots ,n\}$ and $aP'_ib$ if $i\in \{1,\ldots ,k\}$. Since $g^{M'}_f$ is a constant rule that picks b, $f(R')=b$. Applying strategy-proofness, we have

$$\begin{aligned} f(R'_1,R'_2,\ldots ,R'_n)&= f(R_1,\ldots ,R_{k-1},{\bar{R}}_k,R'_{k+1},\ldots ,R'_n) \\&= f(R_1,\ldots ,R_{k-2},{\bar{R}}_{k-1},{\bar{R}}_k,R'_{k+1},\ldots ,R'_n) \\&\ \vdots \\&= f({\bar{R}}_1,\ldots ,{\bar{R}}_k,R'_{k+1},\ldots ,R'_n) \\&= f({\bar{R}})\\&= b \end{aligned}$$

This contradicts $f({\bar{R}})=a$. $\square$

Let ${\mathcal {I}}^a(f) = \{S\subseteq N: g^{N{\setminus } S}_f \textit{ is constant rule that picks a }\}$. Next we show the following fact.

Fact 1

${\mathcal {I}}^a(f)$ is a committee for indifference default a.

Proof

We show that ${\mathcal {I}}^a(f)$ satisfies following two properties.

• 1. Non-emptiness Since $N\in {\mathcal {I}}^a(f)$, ${\mathcal {I}}^a(f)\ne \emptyset$. Since f is onto and strategy-proof, $g^N_f$ is onto. Therefore, $\emptyset \notin {\mathcal {I}}^a(f)$.

• 2. Monotonicity Let $S\in {\mathcal {I}}^a(f)$ , $T\subseteq N$ and $S\subseteq T$. We show that $T\in {\mathcal {I}}^a(f)$. Since $g^{N{\setminus } S}_f$ is a constant rule that picks a, $g^{N{\setminus } T}_f$ is a constant rule that picks a. Therefore, $T\in {\mathcal {I}}^a(f)$.

$\square$

The following claim is a direct implication of Theorem 1 of Barberà et al. (1991). Hence, we omit the proof.

Claim 2

For each $M\subseteq N$, if $g^{M}_f$ is onto, then it is a voting by committee for a at M.

For each $M\subseteq N$ such that $g^{M}_f$ is onto, Claim 2 implies that $g^{M}_f$ is a voting by committee for a at M. Let ${\mathcal {F}}^{g^{M}_f}_M$ be the committee for a at M associated with $g^{M}_f$. Now, for each $M\subseteq N$, we define the set ${\mathcal {F}}_{M,{\mathcal {I}}^a(f)}$ as follows. If $g^{M}_f$ is onto, then ${\mathcal {F}}_{M,{\mathcal {I}}^a(f)}= {\mathcal {F}}^{g^{M}_f}_M$. If $g^{M}_f$ is not onto i.e. $g^{M}_f$ is a constant rule that picks a, then ${\mathcal {F}}_{M,{\mathcal {I}}^a(f)}= \emptyset$.

First we show the following fact.

Fact 2

For each $M\subseteq N$, ${\mathcal {F}}_{M,{\mathcal {I}}^a(f)}$ is a committee for a at M with respect to ${\mathcal {I}}^a(f)$.

Proof

We show that for each $M\subseteq N$, ${\mathcal {F}}_{M,{\mathcal {I}}^a(f)}$ satisfies following two properties.

• 1. Non-emptiness with respect to${\mathcal {I}}^a(f)$: If $N{\setminus } M\notin {\mathcal {I}}^a(f)$, then $g^{M}_f$ is onto. Therefore, ${\mathcal {F}}_{M,{\mathcal {I}}^a(f)}={\mathcal {F}}^{g^{M}_f}_M$. Since ${\mathcal {F}}^{g^{M}_f}_M\ne \emptyset$ and $\emptyset \notin {\mathcal {F}}^{g^{M}_f}_M$, we have ${\mathcal {F}}_{M,{\mathcal {I}}^a(f)}\ne \emptyset$ and $\emptyset \notin {\mathcal {F}}_{M,{\mathcal {I}}^a(f)}$. This follows from Claim 2 and Barberà et al. (1991). If $N{\setminus } M\in {\mathcal {I}}^a(f)$, then $g^{M}_f$ is a constant rule that picks a. Therefore, ${\mathcal {F}}_{M,{\mathcal {I}}^a(f)}=\emptyset$ by definition.

• 2. Monotonicity W.o.l.o.g. we assume that ${\mathcal {F}}_{M,{\mathcal {I}}^a(f)}\ne \emptyset$. Therefore ${\mathcal {F}}_{M,{\mathcal {I}}^a(f)}={\mathcal {F}}^{g^{M}_f}_M$. Since ${\mathcal {F}}^{g^{M}_f}_M$ satisfies monotonicity (from Claim 2 and Barberà et al. (1991)), we have that for each $S\in {\mathcal {F}}_{M,{\mathcal {I}}^a(f)}$ and $T\subseteq M$, if $S\subseteq T$, then $T\in {\mathcal {F}}_{M,{\mathcal {I}}^a(f)}$.

$\square$

Next we show that ${\mathcal {F}}_{{\mathcal {I}}^a(f)} \equiv \{{\mathcal {F}}_{M,{\mathcal {I}}^a(f)}\}_{M\subseteq N}$ satisfies the properties of a collection of committees for a with respect to ${\mathcal {I}}^a(f)$.

Fact 3

${\mathcal {F}}_{{\mathcal {I}}^a(f)} \equiv \{{\mathcal {F}}_{M,{\mathcal {I}}^a(f)}\}_{M\subseteq N}$ satisfies the properties of a collection of committees for a with respect to ${\mathcal {I}}^a(f)$.

Proof

We show that for each $M\subseteq N$ and each $i\in M$:

• 1. If $N{\setminus } M\notin {\mathcal {I}}^a(f)$ and $\{N{\setminus } M\}\cup \{i\}\in {\mathcal {I}}^a(f)$, then for all $S \subseteq M$ such that $i\in S$, $S\in {\mathcal {F}}_{M,{\mathcal {I}}^a(f)}$. Suppose not. There exist $M\subseteq N$ and $S\subseteq M$ such that $N{\setminus } M\notin {\mathcal {I}}^a(f)$, $\{N{\setminus } M\}\cup \{i\}\in {\mathcal {I}}^a(f)$ and $i\in S\notin {\mathcal {F}}_{M,{\mathcal {I}}^a(f)}$. Let $R\in {\mathcal {R}}^n$ be a preference profile such that $aI_kb$ for all $k\in \{N{\setminus } M\}$, $aP_kb$ for all $k\in S$ and $bP_ka$ for all $k\in M{\setminus } S$. Note that $g^M_f(R)=b$. Therefore $f(R)=b$. Since f is strategy-proof, $f(R'_i,R_{N {\setminus } \{i\}})=b$ where $R'_i= aI'_ib$. This contradicts with the fact that $\{N{\setminus } M\}\cup \{i\}\in {\mathcal {I}}^a(f)$.

• 2. If $S\in {\mathcal {F}}_{M,{\mathcal {I}}^a(f)}$, $i\notin S$ and $\{N{\setminus } M\}\cup \{i\}\notin {\mathcal {I}}^a(f)$, then $S\in {\mathcal {F}}_{M{\setminus }\{i\},{\mathcal {I}}^a(f)}$. Let $R\in {\mathcal {R}}^n$ be a preference profile such that $aI_kb$ for all $k\in \{N{\setminus } M\}\cup \{i\}$, $aP_kb$ for all $k\in S$ and $bP_ka$ for all $k\in M{\setminus }\{S\cup i\}$. Let $R'=(R'_i,R_{N {\setminus } \{i\}})$ where $R'_i=bP'_ia$. Since $S\in {\mathcal {F}}_{M,{\mathcal {I}}^a(f)}$, $g^M_f(R')=a$. Therefore $f(R')=a$. Then by strategy-proofness, $f(R)=a$, i.e $g^{M{\setminus } i}_f(R)=a$. Since $\{N{\setminus } M\}\cup i\notin {\mathcal {I}}^a(f)$, $S\in {\mathcal {F}}_{M{\setminus } i,{\mathcal {I}}^a(f)}$.

• 3. If $N{\setminus } M\notin {\mathcal {I}}^a(f)$, $S\cup \{i\}\notin {\mathcal {F}}_{M,{\mathcal {I}}^a(f)}$ and $\{N{\setminus } M\}\cup \{i\}\notin {\mathcal {I}}^a(f)$, then $S\notin {\mathcal {F}}_{M{\setminus }\{i\},{\mathcal {I}}^a(f)}$. Let $R\in {\mathcal {R}}^n$ be be a preference profile such that $aI_kb$ for all $k\in N{\setminus } M$, $aP_kb$ for all $k\in S\cup i$ and $bP_ka$ for all $k\in M{\setminus }\{S\cup i\}$. Let $R'=(R'_i,R_{N {\setminus } \{i\}})$ where $R'_i=bI'_ia$. Since $S \cup i\notin {\mathcal {F}}_{M,{\mathcal {I}}^a(f)}$, $g^M_f(R)=b$. Therefore $f(R)=b$. Then by strategy-proofness, $f(R')=b$, i.e $g^{M{\setminus } i}_f(R')=b$. Since $\{N{\setminus } M\}\cup i\notin {\mathcal {I}}^a(f)$, $S\notin {\mathcal {F}}_{M{\setminus } i,{\mathcal {I}}^a(f)}$.

$\square$

We complete this case by showing that for all $R\in {\mathcal {R}}^n$;

$$\begin{aligned} f(R) = h^{{\mathcal {I}}^a(f)}_{{\mathcal {F}}_{{\mathcal {I}}^a(f)}}(R). \end{aligned}$$

Consider any profile R. By Claim 1, $g^{N{\setminus } N_A(R)}_f$ is either a constant rule that picks a or it is an onto rule. Let $g^{N{\setminus } N_A(R)}_f$ be a constant rule that picks a. Therefore, $g^{N{\setminus } N_A(R)}_f(R)=a$ implies that $f(R)=a$. Which, in turn, implies that $N_{A}(R)\in {\mathcal {I}}^a(f)$; i.e.; $h^{{\mathcal {I}}^a(f)}_{{\mathcal {F}}_{{\mathcal {I}}^a(f)}}(R) = a$.

Now we assume that $g^{N{\setminus } N_A(R)}_f$ is an onto rule. Therefore, $N_{A}(R)\notin {\mathcal {I}}^a(f)$. By Claim 2, $g^{N{\setminus } N_A(R)}_f$ is a voting by committee for a at $N{\setminus } N_A(R)$. Let ${\mathcal {F}}^{g^{N{\setminus } N_A(R)}_f}_{N{\setminus } N_A(R)}$ be the committee for a at $N{\setminus } N_A(R)$ associated with $g^{N{\setminus } N_A(R)}_f$. Therefore, $g^{N{\setminus } N_A(R)}_f(R)=a$ if $N_a(R)\in {\mathcal {F}}^{g^{N{\setminus } N_A(R)}_f}_{N{\setminus } N_A(R)}$ and $g^{N{\setminus } N_A(R)}_f(R)=b$ if $N_a(R)\notin {\mathcal {F}}^{g^{N{\setminus } N_A(R)}_f}_{N{\setminus } N_A(R)}$. Since $g^{N{\setminus } N_A(R)}_f(R)=f(R)$ and

${\mathcal {F}}_{N{\setminus } N_{A}(R),{\mathcal {I}}^a(f)} = {\mathcal {F}}^{g^{N{\setminus } N_A(R)}_f}_{N{\setminus } N_A(R)}$, we are done.

Case 2: $f({\bar{R}})=b$. A similar argument (as in case 1) shows that there exists a committee for indifference default b, ${\mathcal {I}}^b$ and a collection of committees for a with respect to ${\mathcal {I}}^b$, ${\mathcal {F}}_{{\mathcal {I}}^b}$, such that for all $R\in {\mathcal {R}}^n$;

$$\begin{aligned} f(R) = f^{{\mathcal {I}}^b}_{{\mathcal {F}}_{{\mathcal {I}}^b}}(R). \end{aligned}$$

$\square$

The Proof of Theorem 2

We prove this theorem with the help of the following propositions. The first proposition is a direct implication of adding anonymity on Theorem 1. For this purpose, we introduce the following definitions. A committee for indifference default $d\in \{a,b\}$, ${\mathcal {I}}^d$, is anonymous, if $S \in {\mathcal {I}}^d$ implies $S^\prime \in {\mathcal {I}}^d$ for any $S^\prime \subseteq N$ such that $|S| = |S^\prime |$. If a committee for indifference default d is anonymous, then we refer it as anonymous committee for indifference defaultd.

Let $M\subseteq N$ and ${\mathcal {I}}^d$ be a committee for indifference default d. A committee for a at M with respect to ${\mathcal {I}}^d$, ${\mathcal {F}}_{M,{\mathcal {I}}^d}$, is anonymous, if $S \in {\mathcal {F}}_{M,{\mathcal {I}}^d}$ implies $S^\prime \in {\mathcal {F}}_{M,{\mathcal {I}}^d}$ for any $S^\prime \subseteq M$ such that $|S| = |S^\prime |$. If a committee for a at M with respect to ${\mathcal {I}}^d$ is anonymous, we refer it as anonymous committee foraatMwith respect to${\mathcal {I}}^d$.

A collection of anonymous committees forawith respect to${\mathcal {I}}^d$, is a collection of committees for a with respect to ${\mathcal {I}}^d$, satisfying following properties

• 1. For any $M\subseteq N$, ${\mathcal {F}}_{M,{\mathcal {I}}^d}$ is a anonymous committees for a at M with respect to ${\mathcal {I}}^d$.

• 2. For any $M,M'\subseteq N$, $S\subseteq M$ and $S'\subseteq M'$ where $|M|=|M'|$ and $|S|=|S'|$, if $S\in {\mathcal {F}}_{M,{\mathcal {I}}^d}$ then $S'\in {\mathcal {F}}_{M',{\mathcal {I}}^d}$.

We define generalized voting by anonymous committees (GVAC), as follows.

Definition 13

A SCF is GVAC, denoted by $f^{{\mathcal {I}}^d}_{{\mathcal {F}}_{{\mathcal {I}}^d}}$, if there exists a anonymous committee for indifference default d, ${\mathcal {I}}^d$ where $d\in A$ and a collection of anonymous committees for a with respect to ${\mathcal {I}}^d$, ${\mathcal {F}}_{{\mathcal {I}}^d}$, such that for all $R\in {\mathbb {R}}^n$;

$$\begin{aligned} f^{{\mathcal {I}}^d}_{{\mathcal {F}}_{{\mathcal {I}}^d}}(R) = \left\{ \begin{array}{l l} d &{} \quad \text { if } N_{A}(R)\in {\mathcal {I}}^d\\ a &{} \quad \text { if } N_a(R)\in {\mathcal {F}}_{N{\setminus } N_{A}(R),{\mathcal {I}}^d} \text { and }N_{A}(R)\notin {\mathcal {I}}^d\\ b &{} \quad \text { otherwise }\\ \end{array} \right. \end{aligned}$$

This brings us to the following proposition.

Proposition 4

Let $f : {\mathcal {R}}^n \longrightarrow A$ be an onto SCF. If f is anonymous and strategy-proof, then f is GVAC.

Proof

Let f be an onto, anonymous and strategy-proof SCF. Let ${\bar{R}}\in {\mathcal {R}}^n$ denotes the preference profile where all agents are indifferent between a and b. We show that if $f({\bar{R}})=a$, then there exists a anonymous committee for indifference default a, ${\mathcal {I}}^a$ and a collection of anonymous committees for a with respect to ${\mathcal {I}}^a$, ${\mathcal {F}}_{{\mathcal {I}}^a}$, such that for all $R\in {\mathcal {R}}^n$;

$$\begin{aligned} f(R) = f^{{\mathcal {I}}^a}_{{\mathcal {F}}_{{\mathcal {I}}^a}}(R). \end{aligned}$$

Similarly, if $f({\bar{R}})=b$, then there exists a anonymous committee for indifference default b, ${\mathcal {I}}^b$ and a collection of anonymous committees for a with respect to ${\mathcal {I}}^b$, ${\mathcal {F}}_{{\mathcal {I}}^b}$, such that for all $R\in {\mathcal {R}}^n$;

$$\begin{aligned} f(R) = f^{{\mathcal {I}}^b}_{{\mathcal {F}}_{{\mathcal {I}}^b}}(R). \end{aligned}$$

In the following, we consider these two cases.

Case 1: $f({\bar{R}})=a$ : As f is strategy-proof and onto, we have the following.

• For any $M \subseteq N$, let $g^M_f : \{R\in {\mathcal {R}}^n: aI_ib \textit{ iff } i\notin M \} \longrightarrow A$ be a function defined as $g^M_f(R) = f(R)$ for all $R \in \{R\in {\mathcal {R}}^n: aI_ib \textit{ iff } i\notin M \}$. Then either $g^{M}_f$ is a constant rule that picks a or $g^{M}_f$ is onto. This follows from Claim 1 in the proof of Theorem 1.

• ${\mathcal {I}}^a(f) = \{S\subseteq N: g^{N {\setminus } S}_f \text { is constant rule that picks } a\}$ is a committee for indifference default a. This follows from Fact 1 in the proof of Theorem 1.

• ${\mathcal {F}}_{{\mathcal {I}}^a(f)} \equiv \{{\mathcal {F}}_{M,{\mathcal {I}}^a(f)}\}_{M\subseteq N}$, where

  $$\begin{aligned} {\mathcal {F}}_{M,{\mathcal {I}}^a(f)} = \left\{ \begin{array}{cl} \left\{ S \subseteq M : \begin{array}{c} \exists \text { } R \in \{R\in {\mathcal {R}}^n: aI_ib \textit{ iff } i\notin M \} \\ \text { with } S = N_a(R) \subseteq M \text { and } g^{M}_f(R) = a \end{array}\right\} &{} \quad \text { if } \begin{array}{l} g^{M}_f \text { is an }\\ \text { onto function } \end{array} \\ \\ \emptyset &{} \quad \text { if } \begin{array}{l} g^{M}_f \text { is a }\\ \text { constant rule }\\ \text { that picks a} \end{array} \end{array}\right. \end{aligned}$$

  is a collection of committees for a with respect to ${\mathcal {I}}^a(f)$. This follows from Claim 2 and Facts 2 and 3 in the proof of Theorem 1.

Next we are going to show that ${\mathcal {I}}^a(f)$ is an anonymous committee for indifference default a.

Claim 3

${\mathcal {I}}^a(f)$ is an anonymous committee for indifference default a.

Proof

Consider $S, S^\prime \subseteq N$ such that $|S| = |S^\prime |$. Suppose $S \in {\mathcal {I}}^a(f)$, but to the contrary $S^\prime \notin {\mathcal {I}}^a(f)$. This implies that $g^{N {\setminus } S}_f$ is a constant rule that selects a, but $g^{N {\setminus } S^\prime }_f$ is onto. So there exists a $R \in \{R\in {\mathcal {R}}^n: aI_ib \textit{ iff } i\notin N {\setminus } S^\prime \}$ such that $g^{N {\setminus } S^\prime }_f(R) = b$. As $|S| = |S^\prime |$, we have $|N {\setminus } S| = |N {\setminus } S^\prime |$. As $N_a(R) \subseteq N {\setminus } S^\prime$, there exists $T \subseteq N {\setminus } S$ such that $|N_a(R)| = |T|$ and $|(N {\setminus } S^\prime ) {\setminus } N_a(R)| = |(N {\setminus } S) \setminus T|$. So we can define the following functions; $\sigma _1 : S^\prime \longrightarrow S$, $\sigma _2 : N_a(R) \longrightarrow T$ and $\sigma _3 : (N {\setminus } S^\prime ) {\setminus } N_a(R) \longrightarrow (N {\setminus } S) \setminus T$; which are all one-to-one and onto. Next, we define a permutation $\sigma : N \longrightarrow N$ as follows.

$$\begin{aligned} \sigma (i) = \left\{ \begin{array}{cl} \sigma _1(i) &{} \quad if i \in S^\prime \\ \sigma _2(i) &{} \quad if i \in N_a(R) \\ \sigma _3(i) &{} \quad if i \in (N {\setminus } S^\prime ) {\setminus } N_a(R) \\ \end{array}\right. \end{aligned}$$

Note that $\sigma$ is a well-defined permutation and $\sigma (R) \in \{R\in {\mathcal {R}}^n: aI_ib \textit{ iff } i\notin N {\setminus } S\}$. This implies that $g^{N {\setminus } S}_f(\sigma (R)) = a$; i.e.; $f(\sigma (R)) = a$. But this contradicts anonymity of f as $g^{N {\setminus } S^\prime }_f(R) = b$ implies $f(R) = b$. This concludes the proof of Claim 3. $\square$

Next we show that ${\mathcal {F}}_{{\mathcal {I}}^a(f)}$ is a collection of anonymous committees for a with respect to ${\mathcal {I}}^a(f)$.

Claim 4

${\mathcal {F}}_{{\mathcal {I}}^a(f)}$ is a collection of anonymous committees for a with respect to ${\mathcal {I}}^a(f)$.

Proof

First we show that for every $M \subseteq N$, ${\mathcal {F}}_{M, {\mathcal {I}}^a(f)}$ is a anonymous committees for a at M with respect to ${\mathcal {I}}^a(f)$. First note that as f is anonymous, so for any $M \subseteq N$, it follows that $g^{M}_f$ is also anonymous. Now for any $M \subseteq N$, consider $S, S^\prime \subseteq M$ such that $|S| = |S^\prime |$. Suppose for contradiction that $S \in {\mathcal {F}}_{M, {\mathcal {I}}^a(f)}$, but $S^\prime \notin {\mathcal {F}}_{M, {\mathcal {I}}^a(f)}$. This implies that there exists a profile $R \in \{R\in {\mathcal {R}}^n: aI_ib \textit{ iff } i\notin M\}$ such that $N_a(R) = S$ and $g^{M}_f(R) = a$. As $|S| = |S^\prime |$, we have $|M {\setminus } S| = |M {\setminus } S^\prime |$. So we can define the following functions; $\sigma _1 : S \longrightarrow S^\prime$ and $\sigma _2 : M {\setminus } S \longrightarrow M {\setminus } S^\prime$; which are all one-to-one and onto. Next, we define a permutation $\sigma : N \longrightarrow N$ as follows.

$$\begin{aligned} \sigma (i) = \left\{ \begin{array}{cl} i &{} \quad if i \in N {\setminus } M \\ \sigma _1(i) &{} \quad if i \in S \\ \sigma _2(i) &{} \quad if i \in M {\setminus } S \\ \end{array}\right. \end{aligned}$$

Note that $\sigma$ is a well-defined permutation and $\sigma (R) \in \{R\in {\mathcal {R}}^n: aI_ib \textit{ iff } i\notin M\}$ and $N_a(R) = S^\prime$. Now $S^\prime \notin {\mathcal {F}}_{M, {\mathcal {I}}^a(f)}$ implies that $g^{M}_f(\sigma (R)) = b$ because $g^{M}_f$ is onto. But this contradicts anonymity of $g^{M}_f$, as $g^{M}_f(R) = a$. This shows that for every $M \subseteq N$, ${\mathcal {F}}_{M, {\mathcal {I}}^a(f)}$ is a anonymous committee for a at M with respect to ${\mathcal {I}}^a(f)$. Next, consider any $M, M^\prime \subseteq N$ and $S \subseteq M$ and $S^\prime \subseteq M^\prime$ such that $|M| = |M^\prime |$ and $|S| = |S^\prime |$. We are going to show that if $S \in {\mathcal {F}}_{M, {\mathcal {I}}^a(f)}$, then $S^\prime \in {\mathcal {F}}_{M^\prime , {\mathcal {I}}^a(f)}$. So suppose for contradiction that $S \in {\mathcal {F}}_{M, {\mathcal {I}}^a(f)}$, but $S^\prime \notin {\mathcal {F}}_{M^\prime , {\mathcal {I}}^a(f)}$. As $S \in {\mathcal {F}}_{M, {\mathcal {I}}^a(f)}$, there exists a profile $R \in \{R\in {\mathcal {R}}^n: aI_ib \textit{ iff } i\notin M\}$ such that $N_a(R) = S$ and $g^{M}_f(R) = a$. As $|M| = |M^\prime |$ and $|S| = |S^\prime |$, so it follows that $|M {\setminus } S| = |M^\prime {\setminus } S^\prime |$ and $|N {\setminus } M| = |N {\setminus } M^\prime |$. So we can define the following functions; $\sigma _4 : N {\setminus } M \longrightarrow N {\setminus } M^\prime$, $\sigma _5 : S \longrightarrow S^\prime$ and $\sigma _6 : M {\setminus } S \longrightarrow M^\prime {\setminus } S^\prime$; which are all one-to-one and onto. Next, we define a permutation $\sigma ^\star : N \longrightarrow N$ as follows.

$$\begin{aligned} \sigma ^\star (i) = \left\{ \begin{array}{cl} \sigma _4(i) &{} \quad if i \in N {\setminus } M \\ \sigma _5(i) &{} \quad if i \in S \\ \sigma _6(i) &{} \quad if i \in M {\setminus } S \\ \end{array}\right. \end{aligned}$$

Note that $\sigma ^\star$ is a well-defined permutation and $\sigma ^\star (R) \in \{R\in {\mathcal {R}}^n: aI_ib \textit{ iff } i\notin M^\prime \}$ and $N_a(R) = S^\prime$. As $g^{M}_f$ is onto, so it follows that $g^{M^\prime }_f$ is also onto. Otherwise there would be a violation of Claim 3 as $|N {\setminus } M| = |N {\setminus } M^\prime |$. Then $S^\prime \notin {\mathcal {F}}_{M^\prime , {\mathcal {I}}^a(f)}$ implies that $g^{M^\prime }_f(\sigma ^\star (R)) = b$; i.e.; $f(\sigma ^\star (R)) = b$. This contradicts anonymity of f as $g^{M}_f(R) = a$ implies that $f(R) = a$. This concludes the proof of Claim 4. $\square$

We complete this case by showing that for all $R\in {\mathcal {R}}^n$;

$$\begin{aligned} f(R) = f^{{\mathcal {I}}^a(f)}_{{\mathcal {F}}_{{\mathcal {I}}^a(f)}}(R). \end{aligned}$$

This follows from the definition of ${\mathcal {I}}^a(f)$ and ${\mathcal {F}}_{{\mathcal {I}}^a(f)}$ as shown at the end of case 1 in the proof of the only if part of Theorem 1.

Case 2: $f({\bar{R}})=b$ : A similar argument (as in case 1) shows that there exists a anonymous committee for indifference default b, ${\mathcal {I}}^b$ and a collection of anonymous committees for a with respect to ${\mathcal {I}}^b$, ${\mathcal {F}}_{{\mathcal {I}}^b}$, such that for all $R\in {\mathcal {R}}^n$; $f(R) = f^{{\mathcal {I}}^b}_{{\mathcal {F}}_{{\mathcal {I}}^b}}(R).$$\square$

In the following proposition, we show that any GVAC rule can be described as either a quota rule with indifference default a or a quota rule with indifference default b.

Proposition 5

Let $f^{{\mathcal {I}}^d}_{{\mathcal {F}}_{{\mathcal {I}}^d}}$ be a GVAC rule. Then either there exists a quota rule with indifference default a ($f^{k, x}_a$) such that $f^{{\mathcal {I}}^d}_{{\mathcal {F}}_{{\mathcal {I}}^d}} \equiv f^{k, x}_a$ or a quota rule with indifference default b ($f^{k, y}_b$) such that $f^{{\mathcal {I}}^d}_{{\mathcal {F}}_{{\mathcal {I}}^d}} \equiv f^{k, y}_b$.

Proof

Let ${\mathcal {W}}$ be any collection of subsets of N. We denote $Q({\mathcal {W}})$ as the cardinality of $S \in {\mathcal {W}}$ such that S contains the least number of agents among all sets in ${\mathcal {W}}$; i.e;

$$\begin{aligned} Q({\mathcal {W}}) = \displaystyle \min _{S \in {\mathcal {W}}} |S|, \text { where } {\mathcal {W}} \subseteq 2^N. \end{aligned}$$

We prove Proposition 5 with the help of the following lemmas. $\square$

Lemma 3

For the GVAC rule $f^{{\mathcal {I}}^a}_{{\mathcal {F}}_{{\mathcal {I}}^a}}$, we have the following.

• 1. $1 \le Q({\mathcal {I}}^a)=k\le n$.

• 2. ${\mathcal {F}}_{M,{\mathcal {I}}^a}$ satisfies following conditions:

  • 2.1 For all $M\subseteq N$, if $|M|\le n-k$ then ${\mathcal {F}}_{M,{\mathcal {I}}^a}=\emptyset$.

  • 2.2 For all $M,M'\subseteq N$ such that $|M|=|M'|> n-k$, $Q({\mathcal {F}}_{M,{\mathcal {I}}^a})=Q({\mathcal {F}}_{M',{\mathcal {I}}^a})$ and ${\mathcal {F}}_{M,{\mathcal {I}}^a}\ne \emptyset$ and ${\mathcal {F}}_{M',{\mathcal {I}}^a}\ne \emptyset$.

  • 2.3 For all $M\subseteq N$ such that $|M|=n-k+l$ where $l\in \{1,\ldots ,k\}$, we have $Q({\mathcal {F}}_{M,{\mathcal {I}}^a})\in \{1,\ldots ,l\}$.

  • 2.4 For all $M,M'\subseteq N$ such that $|M'|=|M|-1> n-k$, $Q({\mathcal {F}}_{M,{\mathcal {I}}^a})\ge Q({\mathcal {F}}_{M',{\mathcal {I}}^a}) \ge Q({\mathcal {F}}_{M,{\mathcal {I}}^a}) - 1$.

Proof

As ${\mathcal {I}}^a$ is an anonymous committee for indifference default a, it follows that $N \in {\mathcal {I}}^a$ and $\emptyset \notin {\mathcal {I}}^a$ (Non-emptyness condition of ${\mathcal {I}}^a$). This implies that $1 \le Q({\mathcal {I}}^a)=k\le n$.

Next we prove statement 2.1. So consider a $M\subseteq N$ such that $|M|\le n-k$. This implies that $|N {\setminus } M| \ge k$. As $Q({\mathcal {I}}^a)=k$, monotonicity and anonymity property of ${\mathcal {I}}^a$ implies that $N {\setminus } M \in {\mathcal {I}}^a$. Then non-emptyness with respect to ${\mathcal {I}}^a$ property of ${\mathcal {F}}_{{\mathcal {I}}^a}$ implies that ${\mathcal {F}}_{M,{\mathcal {I}}^a}=\emptyset$.

Next we prove statement 2.2. So consider $M,M'\subseteq N$ such that $|M|=|M'|> n-k$. This implies that $|N {\setminus } M| = |N {\setminus } M'| < k$. As $Q({\mathcal {I}}^a)=k$, it follows that $N {\setminus } M \notin {\mathcal {I}}^a$ and $N {\setminus } M' \notin {\mathcal {I}}^a$. So from the non-emptyness with respect to ${\mathcal {I}}^a$ property of ${\mathcal {F}}_{{\mathcal {I}}^a}$, it follows that ${\mathcal {F}}_{M,{\mathcal {I}}^a} \ne \emptyset$ and ${\mathcal {F}}_{M',{\mathcal {I}}^a} \ne \emptyset$. Also as ${\mathcal {F}}_{{\mathcal {I}}^a}$ is anonymous, it follows that $Q({\mathcal {F}}_{M,{\mathcal {I}}^a})=Q({\mathcal {F}}_{M',{\mathcal {I}}^a})$ from the definition of Q.

Next we prove statement 2.3. So consider $M\subseteq N$ such that $|M|=n-k+l$ where $l\in \{1,\ldots ,k\}$. Suppose for contradiction that $Q({\mathcal {F}}_{M,{\mathcal {I}}^a}) > l$. So it follows that for all $S \in {\mathcal {F}}_{M,{\mathcal {I}}^a}$, $|S| > l$. Now we consider the situation when $l = 1$. Then $|M|=n-k+1$ implies that $|N {\setminus } M| = k - 1$. As $Q({\mathcal {I}}^a)=k$, it follows, from the definition of Q, that $N {\setminus } M \notin {\mathcal {I}}^a$. Now consider an $i \in M$ and the coalition $(N {\setminus } M) \cup \{i\}$. Note that $|(N {\setminus } M) \cup \{i\}| = k$. As ${\mathcal {I}}^a$ is an anonymous committee for indifference default a, it follows that $(N {\setminus } M) \cup \{i\} \in {\mathcal {I}}^a$. Then as ${\mathcal {F}}_{{\mathcal {I}}^a}$ is a collection of committees for a with respect to ${\mathcal {I}}^a$, it follows by using property 1 that $\{i\} \in {\mathcal {F}}_{M,{\mathcal {I}}^a}$. This contradicts our assumption that for all $S \in {\mathcal {F}}_{M,{\mathcal {I}}^a}$, $|S| > 1$. Now suppose that for all $M\subseteq N$ such that $|M|=n-k+l$ where $l\in \{1,\ldots ,k-1\}$, we have $Q({\mathcal {F}}_{M,{\mathcal {I}}^a})\in \{1,\ldots ,l\}$, but there exists a $M^\prime \subseteq N$ such that $|M^\prime |=n-k+l+1$ and $Q({\mathcal {F}}_{M^\prime ,{\mathcal {I}}^a}) > l+1$. So consider the case where $M^\prime = M \cup \{i\}$. Now consider a coalition $S \subseteq M$, such that $|S| = Q({\mathcal {F}}_{M,{\mathcal {I}}^a})$. As $Q({\mathcal {F}}_{M^\prime ,{\mathcal {I}}^a}) > l+1$, it follows that $S \cup \{i\} \notin {\mathcal {F}}_{M^\prime ,{\mathcal {I}}^a}$. Note that $|N {\setminus } M^\prime | = k - l - 1$ and $|(N {\setminus } M^\prime ) \cup \{i\}| = k - l$. As $Q({\mathcal {I}}^a)=k$, it follows that $N {\setminus } M^\prime \notin Q({\mathcal {I}}^a)$ and $(N {\setminus } M^\prime ) \cup \{i\} \notin Q({\mathcal {I}}^a)$. Then as ${\mathcal {F}}_{{\mathcal {I}}^a}$ is a collection of committees for a with respect to ${\mathcal {I}}^a$, it follows by using property 3 that $S \notin {\mathcal {F}}_{M^\prime {\setminus } \{i\},{\mathcal {I}}^a} = {\mathcal {F}}_{M,{\mathcal {I}}^a}$. This however contradicts anonymity of ${\mathcal {F}}_{{\mathcal {I}}^a}$ as $|S| = Q({\mathcal {F}}_{M,{\mathcal {I}}^a})$. Hence the proof of statement 2.3 is concluded by induction.

Next, we prove statement 2.4. So consider $M,M'\subseteq N$ such that $|M'|=|M|-1> n-k$. In view of statement 2.2, without loss of generality, it can be assumed that $M = M' \cup \{i\}$. Now suppose for contradiction that

• Case 1 : either $Q({\mathcal {F}}_{M,{\mathcal {I}}^a}) < Q({\mathcal {F}}_{M',{\mathcal {I}}^a})$,

• Case 2 : or $Q({\mathcal {F}}_{M,{\mathcal {I}}^a}) - 1 > Q({\mathcal {F}}_{M',{\mathcal {I}}^a})$.

In case 1, there exists $S \subseteq M$ such that $S \in {\mathcal {F}}_{M,{\mathcal {I}}^a}$ and $|S| = Q({\mathcal {F}}_{M,{\mathcal {I}}^a})$. Now if $S = M$, then we have a contradiction to $Q({\mathcal {F}}_{M,{\mathcal {I}}^a}) < Q({\mathcal {F}}_{M',{\mathcal {I}}^a})$ as $|M| \ge |S^\prime |$ for any $S' \subseteq M'$. So we have $S \subsetneq M$. Then it follows that there exists $S^\star \subsetneq M$ such that $|S^\star | = |S|$ and $S^\star \subseteq M^\prime$. As ${\mathcal {F}}_{{\mathcal {I}}^a}$ is a collection of anonymous committees for a with respect to ${\mathcal {I}}^a$, it follows that $S^\star \in {\mathcal {F}}_{M,{\mathcal {I}}^a}$. Note that as $S^\star \subseteq M'$, it follows that $i \notin S^\star$. Also as $|M|-1> n-k$, it follows that $|(N {\setminus } M) \cup \{i\}| < k$. As $Q({\mathcal {I}}^a) = k$, it follows that $(N {\setminus } M) \cup \{i\} \notin {\mathcal {I}}^a$. Then as ${\mathcal {F}}_{{\mathcal {I}}^a}$ is a collection of committees for a with respect to ${\mathcal {I}}^a$, it follows by using property 2 that $S^\star \in {\mathcal {F}}_{M {\setminus } \{i\},{\mathcal {I}}^a} = {\mathcal {F}}_{M',{\mathcal {I}}^a}$. This constitutes a contradiction with the definition of Q as $|S^\star | < Q({\mathcal {F}}_{M',{\mathcal {I}}^a})$.

In case 2, there exists $S \subseteq M'$ such that $|S| = Q({\mathcal {F}}_{M',{\mathcal {I}}^a})$. Note that if $S = M'$, it follows from case 2 that $Q({\mathcal {F}}_{M,{\mathcal {I}}^a}) > |M'| + 1$. This is a contradiction as for all $S \subseteq M$, we have $|S| \le |M| = |M'| + 1$. Now consider the set $S \cup \{i\} \subsetneq M$. Note that $|S \cup \{i\}| = Q({\mathcal {F}}_{M',{\mathcal {I}}^a}) + 1$. As $Q({\mathcal {F}}_{M,{\mathcal {I}}^a}) > Q({\mathcal {F}}_{M',{\mathcal {I}}^a}) + 1$, from the definition of Q, it follows that $S \cup \{i\} \notin {\mathcal {F}}_{M,{\mathcal {I}}^a}$. Also as $|M|-1> n-k$, it follows that $|(N {\setminus } M) \cup \{i\}| < k$ and $|(N {\setminus } M)|< k - 1 < k$. As $Q({\mathcal {I}}^a) = k$, it follows that $(N {\setminus } M) \cup \{i\} \notin {\mathcal {I}}^a$ and $(N {\setminus } M) \notin {\mathcal {I}}^a$. Then as ${\mathcal {F}}_{{\mathcal {I}}^a}$ is a collection of committees for a with respect to ${\mathcal {I}}^a$, it follows by using property 3 that $S \notin {\mathcal {F}}_{M {\setminus } \{i\},{\mathcal {I}}^a} = {\mathcal {F}}_{M',{\mathcal {I}}^a}$. This constitutes a contradiction with the anonymity property of ${\mathcal {F}}_{M',{\mathcal {I}}^a}$ as $|S| = Q({\mathcal {F}}_{M',{\mathcal {I}}^a})$. This completes the proof of statement 2.4 and concludes the proof of Lemma 3. $\square$

Observation 1

Given a GVAC rule $f^{{\mathcal {I}}^a}_{{\mathcal {F}}_{{\mathcal {I}}^a}}$, let $k = Q({\mathcal {I}}^a)$ and $x_l = Q({\mathcal {F}}_{M, {\mathcal {I}}^a})$, where $|M| = n - k + l$ for any $l \in \{1, 2, \ldots , k\}$. Then it follows from Lemma 3 that $f^{{\mathcal {I}}^a}_{{\mathcal {F}}_{{\mathcal {I}}^a}}(R) = f^{k, x}_a(R)$ for all $R \in {\mathcal {R}}^N$.

Lemma 4

For the GVAC rule $f^{{\mathcal {I}}^b}_{{\mathcal {F}}_{{\mathcal {I}}^b}}$, we have the following.

• 1. $1 \le Q({\mathcal {I}}^b)=k\le n$.

• 2. ${\mathcal {F}}_{M,{\mathcal {I}}^b}$ satisfies following conditions:

  • 2.1 For all $M\subseteq N$, $|M|\le n-k$ if and only if ${\mathcal {F}}_{M,{\mathcal {I}}^b}=\emptyset$

  • 2.2 For all $M,M'\subseteq N$ such that $|M|=|M'|> n-k$, $Q({\mathcal {F}}_{M,{\mathcal {I}}^b})=Q({\mathcal {F}}_{M',{\mathcal {I}}^b})$ and $Q({\mathcal {F}}_{M,{\mathcal {I}}^b}) \ne \emptyset$ and $Q({\mathcal {F}}_{M',{\mathcal {I}}^b}) \ne \emptyset$.

  • 2.3 For all $M\subseteq N$ such that $|M|=n-k+l$ where $l\in \{1,\ldots ,k\}$, we have $Q({\mathcal {F}}_{M,{\mathcal {I}}^b})\in \{n-k+1,\ldots ,n-k+l\}$

  • 2.4 For all $M,M'\subseteq N$ such that $|M'|=|M-1|> n-k$, $Q({\mathcal {F}}_{M,{\mathcal {I}}^d})\ge Q({\mathcal {F}}_{M',{\mathcal {I}}^d})\ge Q({\mathcal {F}}_{M,{\mathcal {I}}^d}) - 1$.

Proof

In view of Lemma 3, we will only show the proof of statement 2.3. So consider $M\subseteq N$ such that $|M|=n-k+l$ where $l\in \{1,\ldots ,k\}$. First consider the case, where $l = 1$. In this case, we have to show that ${\mathcal {F}}_{M,{\mathcal {I}}^b} = \{M\}$. As $|M|=n-k+1$, it follows that $|N {\setminus } M| = k - 1$. As $Q({\mathcal {I}}^b)=k$, it follows, from the definition of Q, that $N {\setminus } M \notin {\mathcal {I}}^a$. Now for every $i \in M$, consider the coalitions $(N {\setminus } M) \cup \{i\}$. Note that $|(N {\setminus } M) \cup \{i\}| = k$. As ${\mathcal {I}}^b$ is an anonymous committee for indifference default b, it follows that $(N {\setminus } M) \cup \{i\} \in {\mathcal {I}}^b$. Then as ${\mathcal {F}}_{{\mathcal {I}}^a}$ is a collection of committees for a with respect to ${\mathcal {I}}^b$, it follows by using property 1 that if $S \in {\mathcal {F}}_{M,{\mathcal {I}}^a}$ then $i \in S$. As this is true for all $i \in M$, it follows that ${\mathcal {F}}_{M,{\mathcal {I}}^b} = \{M\}$. Now suppose that for all $M\subseteq N$ such that $|M|=n-k+l$ where $l\in \{1,\ldots ,k-1\}$, we have $Q({\mathcal {F}}_{M,{\mathcal {I}}^b})\in \{n-k+1,\ldots ,n-k+l\}$, but there exists a $M^\prime \subseteq N$ such that $|M^\prime |=n-k+l+1$ and either $Q({\mathcal {F}}_{M^\prime ,{\mathcal {I}}^b}) > n-k+l+1$, or $Q({\mathcal {F}}_{M^\prime ,{\mathcal {I}}^b}) < n-k+1$. Note that $Q({\mathcal {F}}_{M^\prime ,{\mathcal {I}}^b}) > n-k+l+1$ implies for all $S \in {\mathcal {F}}_{M^\prime ,{\mathcal {I}}^a}$, we have $|S| > n-k+l+1$. This is a contradiction as $S \subseteq M^\prime$ and $|M^\prime |=n-k+l+1$. So assume that $Q({\mathcal {F}}_{M^\prime ,{\mathcal {I}}^b}) < n-k+1$. Now consider an $M \subseteq N$ such that $M^\prime = M \cup \{i\}$ for some $i \in N {\setminus } M$. Then it follows that $|M|=n-k+l$. So $Q({\mathcal {F}}_{M,{\mathcal {I}}^b})\in \{n-k+1,\ldots ,n-k+l\}$. Now consider a $S \subsetneq M$ such that $|S| = n-k$. As $Q({\mathcal {F}}_{M^\prime ,{\mathcal {I}}^b}) < n-k+1$; i.e.; $Q({\mathcal {F}}_{M^\prime ,{\mathcal {I}}^b}) \le n-k$, monotonicity and anonymity property of ${\mathcal {F}}_{M^\prime ,{\mathcal {I}}^b}$ implies that $S \in {\mathcal {F}}_{M^\prime ,{\mathcal {I}}^b}$. Note that $i \notin S$. Also $|(N {\setminus } M^\prime ) \cup \{i\}| = k-l < k$. Then from the definition of Q, it follows that $(N {\setminus } M^\prime ) \cup \{i\} \notin {\mathcal {I}}^b$. As ${\mathcal {F}}_{{\mathcal {I}}^a}$ is a collection of committees for a with respect to ${\mathcal {I}}^b$, it follows by using property 2 that $S \in {\mathcal {F}}_{M^\prime {\setminus } \{i\},{\mathcal {I}}^b} = {\mathcal {F}}_{M,{\mathcal {I}}^b}$. This contradicts the fact that $Q({\mathcal {F}}_{M,{\mathcal {I}}^b})\in \{n-k+1,\ldots ,n-k+l\}$ as $|S| = n-k$. Hence the proof of statement 2.3 is concluded by induction.

Observation 2

Given a GVAC rule $f^{{\mathcal {I}}^b}_{{\mathcal {F}}_{{\mathcal {I}}^b}}$, let $k = Q({\mathcal {I}}^b)$ and $y_l = Q({\mathcal {F}}_{M, {\mathcal {I}}^a})$, where $|M| = n - k + l$ for any $l \in \{1, 2, \ldots , k\}$. Then it follows from Lemma 4 that $f^{{\mathcal {I}}^b}_{{\mathcal {F}}_{{\mathcal {I}}^b}}(R) = f^{k, y}_b(R)$ for all $R \in {\mathcal {R}}^N$.

The proof of Proposition concludes by Observations 1 and 2. $\square$

Proof of Theorem 2

In view of Propositions 4 and 5, to prove Theorem 2, it is sufficient to show that the quota rule with indifference default a and the quota rule with indifference default b are strategy-proof, onto and anonymous. First we show that the quota rule with indifference default a ($f^{k, x}_a$) is strategy-proof, anonymous and onto. The fact that $f^{k, x}_a$ is onto and anonymous follows directly from the definition of $f^{k, x}_a$. Next, in view of Lemma 1, as $f^{k, x}_a$ is onto, it is sufficient to show that $f^{k, x}_a$ satisfies weak strategy-proofness. So consider a profile $R \in {\mathcal {R}}^N$ and an $i-$deviation $R^\prime \in {\mathcal {R}}^N$ of R. We need to show $f^{k, x}_a(R) R_i f^{k, x}_a(R^\prime )$ in the following cases.

• $a P_i b$ and $a I^\prime _i b$ : In this case, suppose $f^{k, x}_a(R) = a$. then it follows that $f^{k, x}_a(R) R_i f^{k, x}_a(R^\prime )$. So suppose that $f^{k, x}_a(R) = b$. This implies that $|N_A(R)| < k$. Also in this case we have $|N_a(R) \cup N_b(R)| = n-k+l$, for some $l \in \{2, 3, \ldots , k\}$. Otherwise, $|N_a(R) \cup N_b(R)| = n-k+1$ and $|N_a(R)| \ge 1 = x_1$ (due to the fact that $a P_i b$) would imply that $f^{k, x}_a(R) = a$, which contradict our assumption that $f^{k, x}_a(R) = b$. Also we have $|N_a(R)| < x_l$. Now $|N_a(R) \cup N_b(R)| = n-k+l$, for some $l \in \{2, 3, \ldots , k\}$ implies that $|N_A(R)| \le k-2$. So it follows that $|N_A(R^\prime )| \le k-1 < k$. Also $|N_a(R^\prime ) \cup N_b(R^\prime )| = n-k+l-1$. Note that $x_l - 1 \le x_{l-1} \le x_l$. Also $|N_a(R^\prime )| = |N_a(R)| - 1$. Now $|N_a(R)| < x_l$ implies $|N_a(R^\prime )| < x_l - 1 \le x_{l - 1}$. This shows that $f^{k, x}_a(R^\prime ) = b$ and we can conclude that $f^{k, x}_a(R) R_i f^{k, x}_a(R^\prime )$.

• $b P_i a$ and $a I^\prime _i b$ : In this case, suppose $f^{k, x}_a(R) = b$. then it follows that $f^{k, x}_a(R) R_i f^{k, x}_a(R^\prime )$. So suppose that $f^{k, x}_a(R) = a$. Now if $|N_A(R)| \ge k-1$, then it follows that $|N_A(R^\prime )| \ge k$. This implies that $f^{k, x}_a(R^\prime ) = a$. So suppose that $|N_A(R)| < k-1$ and $|N_A(R^\prime )| < k$. In this case we have $|N_a(R) \cup N_b(R)| = n-k+l$, for some $l \in \{2, 3, \ldots , k\}$ Also we have $|N_a(R)| \ge x_l$. Also $|N_a(R^\prime ) \cup N_b(R^\prime )| = n-k+l-1$. Note that $x_l - 1 \le x_{l-1} \le x_l$. Also $|N_a(R^\prime )| = |N_a(R)|$. Now $|N_a(R)| \ge x_l$ and $x_{l-1} \le x_l$ implies $|N_a(R^\prime )| \ge x_{l - 1}$. This shows that $f^{k, x}_a(R^\prime ) = a$ and we can conclude that $f^{k, x}_a(R) R_i f^{k, x}_a(R^\prime )$.

Combining these cases, it follows that $f^{k, x}_a$ satisfies weak strategy-proofness. Hence as $f^{k, x}_a$ is onto and Lemma 1 implies that $f^{k, x}_a$ is strategy-proof. In a similar way, it can be shown that $f^{k, y}_b$ is anonymous, onto and strategy-proof. This concludes the proof of Theorem 2. $\square$

The Proof of Proposition 2

Proof

Only if part. Let $f_a^{k, x}$ be a quota rule with indifference default a and it satisfies WDPR. Therefore, the length of x is either (i) $k=n$ or (ii) $k\in \{1,2,\ldots ,n-1\}$.

First we assume that $k=n$. If $x_n=1$ or n then we are done. We assume for contradiction that $x_n\in \{2,3,\ldots ,n-1\}$. Let R be a preference profile such that $N_{A}(R)=\emptyset$ and $N_a(R)=x_n$. Since $f_a^{k, x}$ be a quota rule with indifference default a, $f_a^{k, x}(R)=a$. Let $R'$ be a preference profile such that $N_{A}(R')=\emptyset$ and $N_a(R')=x_n-1$. Note that $N_a(R)$, $N_b(R)$, $N_a(R')$, $N_b(R')$$\ne$$\emptyset$. Therefore, by Lemma 2$f_a^{k, x}(R)=f_a^{k, x}(R')$. However, since $f_a^{k, x}$ be a quota rule with indifference default a, $f_a^{k, x}(R')=b$ - a contradiction. Therefore $x_n=1$ or n, which in turn imply that $x=(x_1,\ldots ,x_n)\in \{(1,1,\ldots ,1), (1,2,\ldots ,n)\}$.

Finally we assume that $k\in \{1,2,\ldots ,n-1\}$. Note that if we can show $x_i=x_{i+1}$ for all $i\in \{1,2,\ldots ,k-1\}$, then we are done. If $k=1$, we are done trivially. Therefore we assume that $k>1$. We assume for contradiction that there exists $i\in \{1,2,\ldots ,k-1\}$ such that $x_i\ne x_{i+1}$. Let $i'$ be he minimum among all $i\in \{1,2,\ldots ,k-1\}$ such that $x_i\ne x_{i+1}$. Note that $x_{i'}=1$ and $x_{i'+1}=2$. Let R and $R'$ be preference profiles such that $|N_a(R) \cup N_b(R)| = |N_a(R') \cup N_b(R')| = n - k + i'+1$. Moreover we assume that $|N_a(R)|= 2$ and $|N_a(R')|=1$. Since $N_a(R)$, $N_b(R)$, $N_a(R')$, $N_b(R')$$\ne$$\emptyset$; by Lemma 2$f_a^{k, x}(R)=f_a^{k, x}(R')$. However, since $f_a^{k, x}$ be a quota rule with indifference default a, $f_a^{k, x}(R)=a\ne b=f_a^{k, x}(R')$ - a contradiction. Therefore, $x_i=x_{i+1}$ for all $i\in \{1,2,\ldots ,k-1\}$, which in turn imply that $x=(x_1,\ldots ,x_n)=(1,1,\ldots ,1)$.

If part. We first prove the following claim.

Claim 5

Let $f:{\mathcal {R}}^n\longrightarrow A$ select the same alternative in each disagreement profile. Then f satisfies WDPR.

Proof

Let $R\in {\mathcal {R}}^n$, $i\in N$ and $R'_i\in {\mathcal {R}}$. If both R and $(R'_i,R_{-i})$ are disagreement profile then $f(R)=f(R'_i,R_{-i})$. Suppose this is not the case. Then either (i) for each $j\in N{\setminus } \{i\}$, we have $f(R)R_jf(R'_i,R_{-i})$ or (ii) for each $j\in N{\setminus } \{i\}$, we have $f(R'_i,R_{-i})R_jf(R)$. In either case WDPR is satisfied. $\square$

Let $f_a^{k, x}:{\mathcal {R}}^n\longrightarrow A$ be a quota rule with indifference default a. If x is a vector of natural numbers of length n and $x=(x_1,\ldots ,x_n)=(1,1,\ldots ,1)$, then $f_a^{k, x}$ selects a in each disagreement profile. If x is a vector of natural numbers of length n and $x=(x_1,\ldots ,x_n)=(1,2,\ldots ,n)$, then $f_a^{k, x}$ selects b in each disagreement profile. If x is a vector of natural numbers of length k, $k\in \{1,2,\ldots ,n-1\}$ and $x=(x_1,\ldots ,x_k)=(1,1,\ldots ,1)$, then $f_a^{k, x}$ selects a in each disagreement profile. Therefore, by Claim 5, all these rules satisfy WDPR. $\square$

The Proof of Proposition 3

Proof

Only if part. Let $f_b^{k, y}$ be a quota rule with indifference default b and it satisfies WDPR. Therefore, the length of y is either (i) $k=n$ or (ii) $k\in \{1,2,\ldots ,n-1\}$.

First we assume that $k=n$. If $Y_n=1$ or n then we are done. We assume for contradiction that $y_n\in \{2,3,\ldots ,n-1\}$. Let R be a preference profile such that $N_{A}(R)=\emptyset$ and $N_a(R)=y_n$. Since $f_b^{k, y}$ be a quota rule with indifference default b, $f_b^{k, y}(R)=a$. Let $R'$ be a preference profile such that $N_{A}(R')=\emptyset$ and $N_a(R')=y_n-1$. Note that $N_a(R)$, $N_b(R)$, $N_a(R')$, $N_b(R')$$\ne$$\emptyset$. Therefore, by Lemma 2$f_b^{k, y}(R)=f_b^{k, y}(R')$. However, since $f_b^{k, y}$ be a quota rule with indifference default b, $f_b^{k, y}(R')=b$ - a contradiction. Therefore $y_n=1$ or n, which in turn imply that $y=(y_1,\ldots ,y_n)\in \{(1,1,\ldots ,1), (1,2,\ldots ,n)\}$.

Finally we assume that $k\in \{1,2,\ldots ,n-1\}$. Note that if we can show $y_i\ne y_{i+1}$ for all $i\in \{1,2,\ldots ,k-1\}$, then we are done. If $k=1$, we are done trivially. Therefore we assume that $k>1$. We assume for contradiction that there exists $i\in \{1,2,\ldots ,k-1\}$ such that $y_i= y_{i+1}$. Let $i'$ be he minimum among all $i\in \{1,2,\ldots ,k-1\}$ such that $y_i= y_{i+1}$. Therefore $y_{i'}= y_{i'+1}=n-k+i'$. Let R and $R'$ be preference profiles such that $|N_a(R) \cup N_b(R)| = |N_a(R') \cup N_b(R')| = n - k + i'+1$. Moreover we assume that $|N_a(R)|= n-k+i'$ and $|N_a(R')|=n-k+i'-1$. Since $N_a(R)$, $N_b(R)$, $N_a(R')$, $N_b(R')$$\ne$$\emptyset$; by Lemma 2$f_b^{k, y}(R)=f_b^{k, y}(R')$. However, since $f_b^{k, y}$ be a quota rule with indifference default b, $f_b^{k, y}(R)=a\ne b=f_b^{k, y}(R')$ - a contradiction. Therefore, $y_i\ne y_{i+1}$ for all $i\in \{1,2,\ldots ,k-1\}$, which in turn imply that $y=(y_1,\ldots ,y_k)=(n-k+1,n-k+2,\ldots ,n)$.

If part. Let $f_b^{k, y}:{\mathcal {R}}^n\longrightarrow A$ be a quota rule with indifference default b. If y is a vector of natural numbers of length n and $y=(y_1,\ldots ,y_n)=(1,1,\ldots ,1)$, then $f_b^{k, y}$ selects a in each disagreement profile. If y is a vector of natural numbers of length n and $y=(y_1,\ldots ,y_n)=(1,2,\ldots ,n)$, then $f_b^{k, y}$ selects b in each disagreement profile. If y is a vector of natural numbers of length k, $k\in \{1,2,\ldots ,n-1\}$ and $y=(y_1,\ldots ,y_k)=(n-k+1,n-k+2,\ldots ,n)$, then $f_b^{k, y}$ selects b in each disagreement profile. Therefore, by Claim 5, all these rules satisfy WDPR. $\square$