Committees under qualified majority rules: the one-core stability index

Abstract

A policy proposal introduced by a committee member is either adopted or abandoned in favor of a new proposal after lengthy deliberations. If a proposal is abandoned, the committee member who introduced it does not cooperate in any effort to replace it. For a player, not cooperate means to vote against a proposal when the rule identifies him or her as one of those who are entitled to make a decision. The one-core is a solution concept that captures that idea. It is never empty if the committee has less than five individuals, but might be empty if there are five or more individuals. I identify a necessary and sufficient condition for the non-emptiness of the one-core no matter the number of alternatives, the preference profile or the number of players in a committee game, under any qualified majority rule.

Appendix

Recall that a qualified majority rule with quota q ($\frac{n}{2}< q\le n$) is a mapping from $2^{N}$ to $\{0,1\}$ such that any coalition S is winning if and only if $|S|\ge q$. Recall also that:

• $\mathcal {W} =\{S\in 2^{N}:|S|\ge q \}$ is the set of winning coalitions

• for each $i\in N$, $\mathcal {W}(i)=\{S\in \mathcal {W}:i\notin S\}$, is the set of winning coalitions to which i does not belong, and the associated game is

• $\Gamma (i)=\left( N_i,A,q,(\succeq _{j})_{j\in N_i}\right)$

• for any nonempty $B\subseteq A$, the restricted game on B is the tuple $(N,B,q,(\succ ^{B} _{i})_{i\in N})$ where, $\succ ^{B}_{i}$ stands for the restriction of $\succ _{i}$ over B.

The result below characterizes the stability of a game relative to the size of the set of alternatives and links the initial game to all its restricted games.

Proposition 4.1

Let N be an n-individual set, A a set of alternatives and q a voting rule. $\mathcal {C}_1(N,A,q,R)$ is nonempty for each profile $R\in \mathcal {L}^N(A)$ if and only if $\mathcal {C}_1(N,B,q,Q)$ is nonempty for each $B\in 2^A$ and each profile $Q\in \mathcal {L}^N(B)$.

According to this result, as soon as no stable proposal can be obtained in a smaller set of alternatives for some profile of preferences, enlarging the set will not allow for stability for every profile of preferences. Moreover, if there is no stable proposal in a large set of alternatives, it could also be the case in a properly selected subset of alternatives. However, once a stable proposal does exist, it will remain even if the size of alternatives reduces.

The existence of a stable proposition could therefore be conditioned by determining the appropriate size of alternatives guaranteeing it.

Proof

$\Rightarrow /$ Assume that there exists $R\in \mathcal {L}^N(A)$ such that $\mathcal {C}_{1}(N,A,q, R)=\emptyset$, then it suffices to consider $B=A$ and the result follows.

$\Leftarrow /$ Conversely, assume that there exists $B\in 2^A$ and a profile $Q\in \mathcal {L}^N(B)$ such that $\mathcal {C}_{1}(N,B,q, Q) = \emptyset$. If $B=A$ the result holds. If $B \subsetneq A$ denoted $A\smallsetminus B=\{a_1, \ldots , a_p\}$ and $B_k = B\cup \{a_1, \ldots , a_k\}$, $1\le k\le p$.

• If $p=1$, then $A=B\cup \{a_1\}$. Define the profile $R_1$ on A by : $\left\{ \begin{array}{l} {R_{1}}_{/_B}= Q; \\ \forall i \in N, \forall x\in B, xR_{1} ^i a_1~ and ~not(a_1 R_{1} ^i x) \end{array} \right.$ where ${R_{1}}_{/_B}$ is the restriction of profile $R_1$’s preference relations on the set B. Thus, each $x\in B$ is dominated with respect to $R_{1}$, by some $y \in B$. Moreover, each $x\in B$ dominates $a_{1}$ with respect to $R_{1}$ via N. Now, let $i\in N$ and $x\in B$. Since $\mathcal {C}_{1}(N,B,q,Q) = \emptyset$, x is dominated by an alternative say $y\in B$ with respect to Q, via a winning coalition S such that $i\notin S$. Then, y dominates $a_{1}$ with respect to $R_{1}$ via S. That is $\hat{C}^{i}(N,A,q, R_{1}) = \emptyset$ for each $i\in N$.

• If $p\ge 2$, we have $B_{k+1} = B_{k}\cup \{a_{k+1}\}$, $0\le k\le p-1$ where $B_0=B$. Then, for all $k\in \{0,\ldots ,p-1\}$, there exists $R_{k+1}\in \mathcal {L}^N(B_{k+1})$ such that $\mathcal {C}_{1}(N,B_{k+1},q, R_{k+1} ) = \emptyset$. Now $B_{p} = B_{p-1}\cup \{a_{p}\}=A$, thus $\mathcal {C}_{1}(N,A,q, R_{p}) = \emptyset$.

$\square$

Proof of Proposition 3.1

Let N be an n-individual set, A a set of policy alternatives, q the quota such that $q> n-2$ and $i\in N$ a society member. $N_i$ is the unique winning coalition in the game $\Gamma (i)$ and there cannot exists a domination cycle in $\Gamma (i)$. Therefore there is necessarily an alternative $x\in A$ which is undominated in $\Gamma (i)$ and the set $\hat{C}^{i}(\Gamma )$ is nonempty. That is, $\mathcal {C}_1(\Gamma )\ne \emptyset$. $\square$

Denote by $\mu (n,q)$ or simply $\mu$ (for short) the number defined by:

$$\begin{aligned} \mu (n,q)= \left\lceil \frac{n-1}{n-1-q} \right\rceil ; \end{aligned}$$

that is, the smallest integer greater than or equal to $\frac{n-1}{n-1-q}$. The committee being with at least five individuals, $\mu (n,q)\ge 3$ as showed by the following lemma.

Lemma 4.1

Let $\Gamma$ be a qualified majority game with $|N|\ge 5$ such that $\frac{n+1}{2} \le q \le n-2$. Then, $\mu (n,q)\ge 3$.

Proof

Since $\frac{n+1}{2} \le q \le n-2$, the following holds:

$$\begin{aligned} \begin{array}{ccccccc} \frac{n+1}{2} \le q \le n-2&{} \iff &{} 1 &{} \le &{} n-1-q &{}\le &{} \frac{n-3}{2} \\ &{} \iff &{} \frac{2}{n-3} &{}\le &{} \frac{1}{n-1-q} &{} \le &{} 1 \\ &{} \iff &{} \frac{2(n-1)}{n-3} &{} \le &{} \frac{n-1}{n-1-q} &{} \le &{} n-1 \end{array} \end{aligned}$$

In addition, $~ \frac{2(n-1)}{n-3} =\frac{2(n-3)+4}{n-3} = 2 + \frac{4}{n-3}$.

Now, $n\ge 5$ implies that $0<\frac{4}{n-3}\le 2$. Therefore, $~\lceil \frac{n-1}{n-1-q}\rceil \ge 3$. $\square$

The case $\frac{n-1}{n-1-q}\notin \mathbb {N}$

Proof of Proposition 3.2

$\Leftarrow /$ Let N be an n-player committee, A a set of alternatives and q the quota such that $\frac{n-1}{n-1-q}\notin \mathbb {N}$. The case $|A|<\mu (n,q)$ is straightforward.

Indeed, assume that the one-core is empty. This means that for each individual $i\in N$, the set $\hat{C}^i(\Gamma )$ of undominated proposals is empty. That is, for each $i\in N$, there are |A| winning coalitions in $(N_i,A,q,\succ )$ that have an empty intersection. It cannot be the case since, the smallest number of winning coalitions in $\mathcal {W}(i)$ having an empty intersection is $\mu (n,q)$ (see Theorem 2.1).

$\Rightarrow /$ Now assume $|A|\ge \mu (n,q)$. To prove the result, it suffices to construct a profile for $|A|=\mu (n,q)$, since the result would be deduced by applying Proposition 4.1.

Let $A=\{a_1,a_2,a_3,\ldots ,a_{\mu }\}$

$$\begin{aligned} \begin{array}{lcclcll} \frac{n-1}{n-1-q}\notin \mathbb {N} \text { and } \mu =\lceil \frac{n-1}{n-1-q}\rceil &{} \iff &{}\mu -1 &{}< &{} \frac{n-1}{n-1-q} &{}< &{}\mu \\ &{} \iff &{} \frac{1}{\mu } &{}< &{} \frac{n-1-q}{n-1} &{}< &{} \frac{1}{ \mu -1}\\ &{} \iff &{} \frac{n-1}{\mu }&{}< &{} n-1-q &{}< &{} \frac{n-1}{ \mu -1} \\ &{} \iff &{} - \frac{\mu -1}{\mu } (n-1)&{}< &{} -q &{}< &{} - \frac{\mu -2}{ \mu -1} (n-1) \\ &{} \iff &{} \frac{\mu -2}{\mu -1} (n-1)&{}< &{} q &{} < &{} \frac{\mu -1}{ \mu }(n-1) \end{array} \end{aligned}$$

There exists $(\alpha ,\eta )\in \mathbb {N}\times \mathbb {N}$ such that $n-1= \alpha \mu +\eta$, with $0\le \eta <\mu$. Thus $(\mu -1) (n-1) = \alpha \mu (\mu -1) +\eta \mu - \eta$ and $\frac{\mu -1}{\mu } (n-1)= \alpha (\mu -1) +\eta - \frac{\eta }{\mu }$ with $0 \le \frac{\eta }{\mu } < 1$.

Hence,

$$\begin{aligned} \left\{ \begin{array}{lcc} q< \frac{\mu -1}{ \mu }(n-1) &{} ~ &{} ~ \\ q \in \mathbb {N} &{} \Rightarrow &{} q < \alpha (\mu -1) +\eta - \frac{\eta }{\mu } \\ \frac{\mu -1}{\mu } (n-1)= \alpha (\mu -1) +\eta - \frac{\eta }{\mu } &{} ~ &{} ~ \end{array} \right. \end{aligned}$$

Let $j \in N$. Recall that $N_j= N \smallsetminus \{j\}$.

Case 1: $\frac{\mu -1}{\mu } (n-1)\in \mathbb {N}.$

In this case, $\eta =0$ and $q \le \alpha (\mu -1)-1$. Consider $\{B_1,B_2,\ldots ,B_{\mu }\}$ a partition of $N_j$ such that $|B_k|=\alpha$ for each $k\in \{1,\ldots ,\mu \}$. Define a profile by:

$$\begin{aligned} \forall i\in N,~ \begin{array}{lll} \succ _{i} ~: &{}a_{\mu }\cdots a_m a_{m-1}\cdots a_{1} &{} i\in B_1, ~m=2,\ldots ,\mu \\ \succ _{i} ~: &{}a_{1}a_{\mu }\cdots a_m a_{m-1}\cdots a_{2} &{} i\in B_2\cup j, ~m=3,\ldots ,\mu \\ \succ _{i} ~: &{}a_{k-1}\cdots a_{1}a_{\mu }\cdots a_{k} &{} i\in B_k, ~k=3,\ldots ,\mu \end{array} \end{aligned}$$

Then, for each $k\in \{1,\ldots ,\mu \}$, $a_k~dom_{N_j\smallsetminus B_k}~a_{k-1}$ (with $a_0=a_{\mu }$) and $|N_j\smallsetminus B_k | = n-1-\alpha = \alpha \mu - \alpha$ $= \alpha (\mu -1) > q$; therefore we get the result.

Case 2: $\frac{\mu -1}{\mu } (n-1)\notin \mathbb {N}.$

Since $\frac{\mu -1}{\mu } (n-1)\notin \mathbb {N}$, then $0<\eta$ and $q \le \alpha (\mu -1)+\eta -1$. Consider $\{B_1,B_2,\ldots ,B_{\mu }, B'\}$ a partition of $N_j$ such that $|B'|=\eta$ and $|B_k|=\alpha , \text { for each } k\in \{1,\ldots ,\mu \}$. Denote $B'=\{i_1,\ldots ,i_{\eta }\}$.

• If $\eta = 1$ consider the profile below defined by :

  $$\begin{aligned} \forall i\in N,~ \begin{array}{lll} \succ _{i} ~: &{}a_{\mu }\cdots a_m a_{m-1}\cdots a_{1} &{} i\in B_1\cup i_1, ~m=2,\ldots ,\mu \\ \succ _{i} ~: &{}a_{1}a_{\mu }\cdots a_m a_{m-1}\cdots a_{2} &{} i\in B_2\cup j, ~m=3,\ldots ,\mu \\ \succ _{i} ~: &{}a_{k-1}\cdots a_{1}a_{\mu }\cdots a_{k} &{} i\in B_k, ~k=3,\ldots ,\mu \end{array} \end{aligned}$$

  The following holds: $a_1 ~ dom_{N\smallsetminus \{B_1\cup i_1\}}~ a_{\mu }$, $a_2 ~ dom_{N\smallsetminus \{B_2 \cup j\}}~ a_{1}$ and $a_k~dom_{N\smallsetminus B_k }~ a_{k-1}$, for $k=3,\ldots ,\mu$; $|N\smallsetminus \{B_1\cup i_1\}|=|N\smallsetminus \{B_2 \cup j\}|=n-1-\alpha =\alpha \mu + \eta - \alpha$ $= \alpha (\mu -1) + \eta >q$; and $|N\smallsetminus B_k |> |N\smallsetminus \{B_1\cup i_1\}| >q$. Therefore we get the result.

• If $\eta \ge 2$ consider the profile defined as follows:

  $$\begin{aligned} \forall i\in N,~ \begin{array}{llll} \succ _{i} &{} : &{}a_{k}\cdots a_{1}a_{\mu }\cdots a_{k+1} &{} i\in B_k\cup i_k, k=1,\ldots ,\eta \\ \succ _{i} &{} : &{}a_{k}\cdots a_{1}a_{\mu }\cdots a_{k+1} &{} i\in B_k, k=\eta +1,\ldots ,\mu -1 \\ \succ _{i} &{} : &{}a_{\mu } \cdots a_{m}a_{m-1}\cdots a_{1}&{} i\in B_{\mu }\cup j, ~2\le m \le \mu \end{array} \end{aligned}$$

  It follows that: $a_{k+1} ~ dom_{N\smallsetminus (B_k\cup \{i_k\})}~ a_{k}$, for each $k \in \{1,\ldots ,\eta \}$; $a_{k+1} ~ dom_{N\smallsetminus B_k }~ a_{k}$, for each $k\in \{\eta +1,\ldots ,\mu -1\}$; and $a_{1}~ dom_{N\smallsetminus (B_{\mu }\cup \{j\}) }~ a_{\mu }$. It is then easy to check that for each $k \in \{1,\ldots ,\eta \}$ and each $l\in \{1,\ldots ,\mu \}$,$|N_j\smallsetminus (B_{k}\cup \{i_{k}\})| = n-1-(\alpha +1)=\alpha \mu + \eta - \alpha -1= \alpha (\mu -1) + \eta -1 \ge q$$|N\smallsetminus (B_{\mu }\cup \{j\})|=|N_j\smallsetminus B_{l}| > |N_j\smallsetminus (B_{k}\cup \{i_{k}\})| \ge q$$|N\smallsetminus B_l |>|N\smallsetminus (B_k\cup \{i_k\})| = n-1-(\alpha )= \alpha (\mu -1) + \eta > q;$therefore the result holds.

$\square$

If $\frac{n+1}{2} \le q\le n-2$ and $\frac{n-1}{n-1-q}\in \mathbb {N}$, $\mu +1$ is a stability index.

In this case, the number $\mu$ fails to ensure a characterization result. Meanwhile, it enables us to set another number which will help to obtain a complete characterization of the non-emptiness of the one-core for symmetric quota committee games. First of all, let us observe what follows.

Remark 4.1

If $\frac{n-1}{n-1-q}\in \mathbb {N}$, then $\mu =\frac{n-1}{n-1-q}$ and $q = \frac{\mu -1}{\mu }(n-1)$.

In order to construct the second main result, some preliminary results will be useful.

Lemma 4.2

Let N be an n-player committee, A a set of alternatives and q the quota such that $\frac{n-1}{n-1-q}\in \mathbb {N}$. For all $i\in N$ and all $S_1,\ldots ,S_{\mu }\subset N_{i}$ satisfying $|S_k|=q$ for each $k\in \{1,\ldots ,\mu \}$ and $\bigcap _{k=1}^{\mu }S_k=\emptyset$, the following holds:

• (1) $\forall k=1,\ldots ,\mu , \left| {\bigcap\limits_{t=1/ t\ne k}^{{\mu }}}S_t\right| =n-1-q$;

• (2) $\left\{ {\bigcap\limits_{t=1/ t\ne k}^{\mu }}S_t: k=1,\ldots ,\mu \right\} \in \prod ^{N_i}$.⁴

The interpretation of this result is that, for each individual i, (1) any intersection of $\mu -1$ distinct minimal winning coalitions belonging to $\mathcal {W}(i)$ has exactly $n-q-1$ individuals and (2) the collection of all such intersections for any given set of $\mu$ minimal winning coalitions of $\mathcal {W}(i)$ is a partition of $N_i$.

Proof

Let N be an n-individual set, A a set of alternatives and q the quota such that $\frac{n-1}{n-1-q}\in \mathbb {N}$.

Let $i\in N$ and $S_1,\ldots ,S_{\mu }\subset N_i$ satisfying $|S_k|=q$ and ${\bigcap _{k=1}^{{\mu }}}S_k=\emptyset$.

• (1) Let $k\in \{1,\ldots ,\mu \}$. To prove that $\left| {\bigcap\limits_{ \begin{subarray}{c} t=1 \\ t\ne k \end{subarray}}^{\mu }}S_t\right| =n-1-q$, it is important to notice that $\left| {\bigcap\limits_{ \begin{subarray}{c} t=1 \\ t\ne k \end{subarray}}^{\mu }}S_t\right| \ge (\mu -1)q-(\mu -2)(n-1)$. Indeed, it suffices to remark that for any two coalitions $S_1$ and $S_2$ satisfying the hypothesis, $|S_1\cap S_2|=|S_1|+|S_2|-|S_1\cup S_2| \ge 2q-(n-1)$ since $S_1\cup S_2\subseteq N_i$. Applying the above observation, it follows that

  $$\begin{aligned} \begin{array}{rcl} \left| \underset{ \begin{array}{c} t=1 \\ t\ne k \end{array}}{\overset{\mu }{\bigcap }}S_t\right| &{} = &{}|S_1|+ \left| \underset{\begin{subarray}{c} t=2 \\ t\notin \{1,k\} \end{subarray}}{\overset{\mu }{\bigcap }}S_t\right| - \left| S_1\bigcup \left( \underset{\begin{subarray}{c} t=2 \\ t\notin \{1,k\} \end{subarray}}{\overset{\mu }{\bigcap }}S_t\right) \right| \\ ~&{} = &{}|S_1|+|S_2|+ \left| \underset{ \begin{subarray}{c} t=3 \\ t\notin \{1,2,k\} \end{subarray}}{\overset{\mu }{\bigcap }}S_t\right| - \left| S_2\bigcup \left( \underset{ \begin{subarray}{c} t=3 \\ t\notin \{1,2,k\} \end{subarray}}{\overset{\mu }{\bigcap }}S_t\right) \right| - \left| S_1\bigcup \left( \underset{\begin{subarray}{c} t=2 \\ t\notin \{1,k\} \end{subarray}}{\overset{\mu }{\bigcap }}S_t\right) \right| \\ ~&{} \vdots &{} ~ \\ ~&{} = &{} \underset{\begin{subarray}{c} t=1 \\ t\ne k \end{subarray}}{\overset{\mu }{\sum }}|S_t| - \underset{\begin{subarray}{c} t=1 \\ t\ne k \end{subarray}}{\overset{\mu -1}{\sum }}\left| S_t \bigcup \left( \underset{\begin{subarray}{c} t'=t+1 \\ t'\ne k \end{subarray}}{\overset{\mu }{\bigcap }}S_{t'}\right) \right| \\ \end{array} \end{aligned}$$

  Since for each t, $|S_t|=q$ and $\left| S_t \bigcup \left({\bigcap\limits_{ \begin{subarray}{c} t'=t+1 \\ t'\ne k \end{subarray}}^{{\mu }}}S_{t'}\right) \right| \le n-1$ as $S_t \bigcup \left( \underset{ \begin{subarray}{c} t'=t+1 \\ t'\ne k \end{subarray}}{\overset{\mu }{\bigcap }}S_{t'}\right) \subseteq N\smallsetminus \{i\}$ it easily happens that,

  $$\begin{aligned} \begin{array}{rcl} \left| \underset{ \begin{subarray}{c} t=1 \\ t\ne k \end{subarray}}{\overset{\mu }{\bigcap }}S_t\right| &{} \ge &{} (\mu -1)q-(\mu -2)(n-1) \\ ~&{} = &{} (\mu -1)q-(\mu -1)(n-1)+ n-1 \\ ~&{} = &{} (\mu -1)(n-1)(\frac{q}{n-1}-1)+ n-1\\ ~&{} = &{} (\mu -1)(n-1)(\frac{\mu -1}{\mu }-1)+ n-1,{ since q =\frac{\mu -1}{\mu }(n-1);\, (see \,remark \,4.1).}\\ ~&{} = &{} n-1-\frac{\mu -1}{\mu }(n-1)\\ ~&{} = &{} n-1-q. \end{array} \end{aligned}$$

  But $|S_k|=q$ and $S_{k}\bigcap \left( \underset{ \begin{subarray}{c} t=1 \\ t\ne k \end{subarray}}{\overset{\mu }{\bigcap }}S_t\right) = \emptyset$ imply $\underset{ \begin{subarray}{c} t=1 \\ t\ne k \end{subarray}}{\overset{\mu }{\bigcap }}S_t \subset N\smallsetminus (\{i\}\cup S_k)$. This yields $\left| \underset{ \begin{subarray}{c} t=1 \\ t\ne k \end{subarray}}{\overset{\mu }{\bigcap }}S_t\right| \le n-1-q$ and we get the result.

• (2) For each $k\in \{1,\ldots ,\mu \}$ denote $M_k=\underset{\begin{subarray}{c} t=1 \\ t\ne k \end{subarray}}{\overset{\mu }{\bigcap }}S_t$. For $k,k'\in \{1,\ldots ,\mu \}$ such that $k\ne k'$, it follows that: If $M_k\cap M_{k'}\ne \emptyset$, then there exists $j\in M_k$ such that $j\in M_{k'}$. Therefore, $j\in S_{k'}$ and $j\in M_{k'}$, that is $S_{k'}\cap M_{k'}\ne \emptyset$ which is a contradiction. Hence for each $k,k'\in \{1,\ldots ,\mu \}$ with $k\ne k'$, $M_k\cap M_{k'}= \emptyset$. Moreover thanks to (1) above,

  $$\begin{aligned} \begin{array}{rcl} \left| \underset{k=1}{\overset{\mu }{\bigcup }}M_k\right| &{} = &{} \underset{k=1}{\overset{\mu }{\sum }}\left| M_k\right| \\ ~&{} = &{} \mu (n-1-\frac{\mu -1}{ \mu }(n-1)) \\ ~&{} = &{} \mu (n-1)-(\mu -1)(n-1) \\ ~&{} = &{} n-1 \end{array} \end{aligned}$$

  therefore, $\left\{ M_k: k=1,\ldots ,\mu \right\} \in \prod ^{N_i}$.

$\square$

The next proof shows that the one-core is always nonempty whenever $|A|=\mu (n,q)$.

Proof of Proposition 3.3

Let $A=\{a_1,a_2,a_3,\ldots ,a_{\mu }\}$ and assume that there exists a profile $\succ$ for which the one-core is empty. Then for each $i\in N$, $C^i(N,A,q,\succ )=\emptyset$. Let $i\in N$, there exists a sequence $(S_{k})_{1\le k\le \mu }$ of coalitions such that:

Now for each $k=1,\ldots ,\mu$, let $M_k=\underset{\begin{subarray}{c} t=1 \\ t\ne k \end{subarray}}{\overset{\mu }{\bigcap }}S_t$. Then, $M_k\ne \emptyset$ for each k and thanks to Lemma 4.2, $\underset{ k=1}{ \overset{k=\mu }{\bigcup }} M_k = N_i$. Therefore, any $l\in N_i$ has a preference relation defined as follows:

$$\begin{aligned} \begin{array}{lll} \succ _{l} ~: &{}a_{\mu }\cdots a_m a_{m-1}\cdots a_{1} &{} l\in M_1, ~m=2,\ldots ,\mu \\ \succ _{l} ~: &{}a_{k-1}\cdots a_{1}a_{\mu }\cdots a_{k} &{} l\in M_k,~k=2,\ldots ,\mu \end{array} \end{aligned}$$

This means that the preferences of players in $N_i$ are completely known and, it suffices to determine the preference relation of player i to obtain the whole profile which will enable us to set the contradiction.

Let $j\in N_i$. Since for each $k=1,\ldots ,\mu , ~\{M_k,S_k\}\in \prod ^{N_i}$, there exists $k_0\in \{1,\ldots ,\mu \}$ such that $j\in M_{k_0}$ and $j\notin S_{k_0}$. Without loss of generality, let $k_0 = \mu$, then $j\notin S_{\mu }$. That is $S_{\mu }\subset N_j$ and $a_{\mu }~dom_{S_{\mu }}~a_{\mu -1}$. Moreover, $C^j(N,A,q,\succ )=\emptyset$ then there exists a cycle on $N_j$ whose length equals $\mu$. In that cycle, $a_{\mu } \text { dominates } a_{\mu -1} \text { via coalition } S_{\mu }$ and $a_{\mu }$ is dominated. We are going to show that $a_{\mu }$ is dominated by $a_1$ via coalition $\{i\} \cup S_1\smallsetminus \{j\}$.

Assume that there is an alternative say $a_x\ne a_1$ and a winning coalition $T_1 \subset N_j$ such that $a_x$ dominates $a_{\mu }$ via $T_1$. Then, $a_x \in A\smallsetminus \{a_1, a_{\mu -1},a_{\mu }\}$. If $i\notin T_1$, then $T_1 \subset N_i$ and since $T_1$ is a winning coalition, a new domination arrow appears in the domination graph obtained over $N_i$ as shown in the figure below.

Hence there are at least 2 cycles: the initial one ($a_{k}~dom_{S_{k}}~a_{k-1}$, $~k=1,\ldots ,\mu$, with $a_{0}=a_{\mu }$) of length $\mu$ and the new one ($a_{x}~dom_{T_{1}}~a_{\mu }$ and $a_{k}~dom_{S_{k}}~a_{k-1}$, $~k=x+1,\ldots ,\mu$) length of which is $\mu -x +1$, which is smaller than $\mu$ since $x\ge 2$. Therefore, there is in $N_i$ a cycle length of which is smaller than $\mu$. This contradicts the minimality of $\mu$ (see Theorem 2.1). Thus, $i\in T_1$.

Now,

$$\begin{aligned} \left\{ \begin{array}{l} a_x~dom_{T_1}~a_{\mu } \\ a_x~dom_{S_x}~a_{x-1}~dom \cdots dom~a_1~dom_{S_1}~a_{\mu } \\ a_{\mu }~dom_{S_{\mu }}~a_{\mu -1}~dom_{S_{\mu -1}}~ \cdots dom~a_{x+1}~dom_{S_{x+1}}~a_x \\ j\in S_k,~~ \forall k=1,\ldots ,\mu -1 \end{array}\right. \Longrightarrow \left\{ \begin{array}{l} \underset{k=1}{\overset{k=x}{\bigcap }}(S_k\smallsetminus \{j\}) \subset T_1 \\ \left( \underset{k=x+1}{\overset{k=\mu }{\bigcap }}S_k\right) \bigcap T_1 =\emptyset \end{array}\right. \end{aligned}$$

Therefore,

$$\begin{aligned} \left| \underset{k=1}{\overset{k=x}{\bigcap }}S_k\right| -1\le \left| T_1\right| \le n-1-\left| \underset{k=x+1}{\overset{k=\mu }{\bigcap }}S_k\right| \end{aligned}$$

Since for each k, $S_k=\underset{\begin{subarray}{c} t=1 \\ t\ne k \end{subarray}}{\overset{\mu }{\bigcup }}M_t$, the following holds:

$$\begin{aligned} \begin{array}{rclcl} \left| \underset{k=1}{\overset{k=x}{\bigcap }}S_k\right| &{} = &{} \left| \underset{k=1}{\overset{k=x}{\bigcap }}\left( \underset{\begin{subarray}{c} t=1 \\ t\ne k \end{subarray}}{\overset{\mu }{\bigcup }}M_t \right) \right| &{} = &{} \left| \underset{t=x+1}{\overset{\mu }{\bigcup }}M_t \right| \\ &{} ~ &{} ~ &{} = &{} (\mu -x)(n-1-q) \end{array} \end{aligned}$$

As well,

$$\begin{aligned} \begin{array}{rclcl} \left| \underset{k=x+1}{\overset{k=\mu }{\bigcap }}S_k\right| &{} = &{} \left| \underset{k=x+1}{\overset{k=\mu }{\bigcap }}\left( \underset{\begin{subarray}{c} t=1 \\ t\ne k \end{subarray}}{\overset{\mu }{\bigcup }}M_t \right) \right| &{} = &{} \left| \underset{t=1}{\overset{x}{\bigcup }}M_t \right| \\ &{} ~ &{} ~ &{} = &{} x(n-1-q) \end{array} \end{aligned}$$

Thus,

$$\begin{aligned} \begin{array}{rclcl} (\mu -x)(n-1-q)-1\le & {} \left| T_1 \right|\le & {} n-1-x(n-1-q). \end{array} \end{aligned}$$

That is,

$$\begin{aligned} \begin{array}{rclcl} (\mu -x)(n-1-q)-1 &{} \le &{} \left| T_1 \right| &{} \le &{} \mu (n-1-q)-x(n-1-q) \\ ~ &{} ~ &{} ~ &{} ~ &{} \text { since }\mu =\frac{n-1}{n-1-q} \\ ~ &{} ~ &{} ~ &{}= &{} (\mu -x)(n-1-q). \end{array} \end{aligned}$$

On the other side, since $2 \le x \le \mu -2$ then,

$$\begin{aligned} \begin{array}{rcl} 2(n-q-1) \le (\mu -x)(n-1-q)&{} \le &{}(\mu -2)(n-q-1) \\ &{} < &{}(\mu -1)(n-q-1) \\ &{} = &{} q. \end{array} \end{aligned}$$

Hence, if $a_x\ne a_1$, then $|T_1|<q$; that is $T_1$ is not a winning coalition. Therefore, $a_x= a_1$ and $S_1\smallsetminus \{j\} \subset T_1$. Moreover,

$$\begin{aligned} \begin{array}{lcrcl} \left( \underset{k=2}{\overset{k=\mu }{\bigcap }}S_k\right) \bigcap T_1 =\emptyset &{} \Rightarrow &{} |T_1| &{} \le &{} n-1-(n-q-1) \\ ~&{} ~ &{} ~ &{} = &{} q \end{array} \end{aligned}$$

Then we get, $q-1 \le |T_1| \le q$.

If $i\notin T_1$, then $a_{\mu }$ is undominated and the one-core is non-empty; this is a contradiction. Thus $i\in T_1$, $T_1=\{i\} \cup S_1\smallsetminus \{j\}$ and $a_1~dom_{T_1}~a_{\mu }$.

Now let $k\in \{2, \ldots , \mu -1\}$; we are going to show that $a_k~dom_{T_k}~a_{k-1}$ where $T_k=\{i\} \cup S_k\smallsetminus \{j\}$. Since $a_{k-1}$ is dominated, let $a_x\in A$ such that $a_x~dom_{T}~a_{k-1}$ where $T\subset N_j$. Then $a_x\in A\smallsetminus \{a_{k-1},a_{k-2}\}$.

Assume that $a_x\ne a_k$ then, $a_x \in A\smallsetminus \{a_k, a_{k-1},a_{k-2}\}$. If $i\notin T$, then $T \subset N_i$ and there exists in $N_i$ a cycle length of which is smaller than $\mu$, which is a contradiction; therefore $i\in T$ and either $x<k-2$ or $x>k$.

• If $x<k-2$, then

  $$\begin{aligned} \begin{array}{lcl} a_{k-1}~dom_{S_{k-1}}~a_{k-2}~dom~ \cdots dom~a_{x+1}~dom_{S_{x+1}}~a_x &{} \Rightarrow &{} T\cap \left( \underset{t=x+1}{\overset{t=k-1}{\bigcap }}S_t\right) = \emptyset \\ ~ &{} \Rightarrow &{} \left| T\right| \le \left| N_j \smallsetminus \left( \underset{t=x+1}{\overset{t=k-1}{\bigcap }}S_t\right) \right| \\ ~ &{} \Rightarrow &{} \left| T\right| \le n-1-\left| \underset{t=x+1}{\overset{t=k-1}{\bigcap }}S_t\right| \end{array} \end{aligned}$$

  As indicated above, since $S_t=\underset{\begin{subarray}{c} p=1 \\ p\ne t \end{subarray}}{\overset{\mu }{\bigcup }}M_p$ for each t, then:

  $$\begin{aligned} \begin{array}{rclcl} \left| \underset{t=x+1}{\overset{t=k-1}{\bigcap }}S_t\right| &{} = &{} \left| \underset{t=x+1}{\overset{t=k-1}{\bigcap }}\left( \underset{\begin{subarray}{c} p=1 \\ p\ne t \end{subarray}}{\overset{\mu }{\bigcup }}M_p \right) \right| &{} = &{} \left| \left( \underset{p=1}{\overset{p=x}{\bigcup }}M_p\right) \bigcup \left( \underset{p=k}{\overset{p=\mu }{\bigcup }}M_p\right) \right| \\ &{} ~ &{} ~ &{} = &{} x(n-1-q)+(\mu -k+1)(n-1-q)\\ &{} ~ &{} ~ &{} = &{} (\mu +x-k+1)(n-1-q) \end{array} \end{aligned}$$

  Thus,

  $$\begin{aligned} \begin{array}{rcl} n-1-\left| \underset{t=x+1}{\overset{t=k-1}{\bigcap }}S_t\right| &{} = &{} n-1 - (\mu +x-k+1)(n-1-q)\\ ~ &{} = &{}\mu (n-1-q)-(\mu +x-k+1)(n-1-q),\\ ~ &{} ~ &{}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \text { since } n-1 = \mu (n-1-q)\\ ~ &{} = &{}(k-x-1)(n-1-q) \end{array} \end{aligned}$$

  Yet,

  $$\begin{aligned} \begin{array}{rcrccll} x<k\le \mu -1&{} \Leftrightarrow &{} 0 &{} < &{} k-x &{} \le &{} \mu -1-x \\ ~ &{} \Leftrightarrow &{} 0 &{} \le &{} k-x-1&{} \le &{} \mu -2-x \\ ~ &{} \Rightarrow &{} 0 &{} \le &{} (k-x-1)(n-1-q) &{} \le &{} (\mu -2-x)(n-1-q) \end{array} \end{aligned}$$

  and $\left| T \right| \le (\mu -2-x)(n-1-q) < (\mu -1)(n-1-q)=q$ since $x\ge 1$. This means that T is not a winning coalition and the domination does not hold.

• If $x>k$, since $a_{k-1}~dom_{S_{k-1}}~a_{k-2}~dom~ \cdots dom~a_{1}~dom_{S_{1}}~a_{\mu } \cdots ~dom_{S_{x+1}}~a_x$ we have $T\cap \left( \left( \underset{t=1}{\overset{t=k-1}{\bigcap }}S_t\right) \cap \left( \underset{t=x+1}{\overset{t=\mu }{\bigcap }}S_t\right) \right) = \emptyset$ and as above,

  $$\begin{aligned} \begin{array}{lcl} T\subset N_j \smallsetminus \left( \left( \underset{t=1}{\overset{t=k-1}{\bigcap }}S_t\right) \cap \left( \underset{t=x+1}{\overset{t=\mu }{\bigcap }}S_t\right) \right) &{} \Rightarrow &{} T \subset N_j \smallsetminus \left( \underset{p=k}{\overset{p=x}{\bigcup }}M_p\right) \\ ~ &{} \Rightarrow &{} \left| T\right| \le \left| N_j \smallsetminus \left( \underset{p=k}{\overset{p=x}{\bigcup }}M_p\right) \right| \\ ~ &{} \Rightarrow &{} \left| T\right| \le n-1-(x-k+1)(n-1-q)\\ ~ &{} \Rightarrow &{} \left| T\right| \le (\mu -(x-k)-1)(n-1-q) \\ ~ &{} \Rightarrow &{} \left| T\right| < (\mu -1)(n-1-q)=q, \\ ~ &{} ~ &{} ~~~~~~~~~~~~~~~~~~~~~~\text { indeed } x-k\ge 1. \end{array} \end{aligned}$$

  This means that T is not a winning coalition and the domination does not hold.

Hence, if $a_x\ne a_k$, then $|T|<q$ and T is not a winning coalition. Consequently, $a_x= a_k$ and $S_k\smallsetminus \{j\} \subset T$ since $a_k~dom_{S_k}~a_{k-1}$ and $j\in S_k$. Moreover,

$$\begin{aligned} \left( \underset{\begin{subarray}{c} t=1 \\ t\ne k \end{subarray}}{\overset{t=\mu }{\bigcap }}S_t\right) \bigcap T =\emptyset . \end{aligned}$$

This implies, $|T| \le n-1-(n-q-1) = (\mu -1)(n-q-1) = q$

and we get, $q-1 \le |T| \le q$.

If $i\notin T$, then $a_{k-1}$ is undominated and the one-core is non-empty; this is a contradiction. Thus $i\in T$, $T=\{i\} \cup S_k\smallsetminus \{j\} = T_k$ and $a_k~dom_{T_k}~a_{k-1}$.

To summarize, we proved that $a_{\mu }~dom_{S_{\mu }}~a_{\mu -1}$ and $a_k~dom_{T_k}~a_{k-1}$, $\forall k\in \{1,\ldots ,\mu -1 \}$ where $T_k=\{i\} \cup S_k\smallsetminus \{j\}$ with $a_0=a_{\mu }$. This enables us to state that player i’s preference relation is exactly the same as player’s j one; that is

$$\begin{aligned} \begin{array}{lcc} \succ _i&:&a_{\mu -1}a_{\mu -2}\cdots a_{1}a_{\mu } \end{array} \end{aligned}$$

Hence, the whole profile is defined by $\forall l\in N,$

$$\begin{aligned} \begin{array}{llll} \succ _{l} &{} : &{}a_{\mu }\cdots a_m a_{m-1}\cdots a_{1} &{} l\in M_1, ~2\le m\le \mu \\ \succ _{l} &{} : &{}a_{k-1}\cdots a_{1}a_{\mu }\cdots a_{k} &{} l\in M_k, ~2 \le k\le \mu , \\ \succ _{l} &{} : &{} a_{\mu -1}\cdots a_{1}a_{\mu } &{} l = i \end{array} \end{aligned}$$

Now, pick any player k in $S_{\mu }$ it is easy to verify that $a_{\mu -1}$ is undominated in $\mathcal {W}(k)$, implying that $C^k(N,A,q,\succ )\ne \emptyset$ and this later is a contradiction. $\square$

We can now prove the result that establishes the nonemptiness of the one-core whenever $\frac{n-1}{n-1-q}\in \mathbb {N}$.

Proof of Proposition 3.4

$\Leftarrow /$Let N be an n-individual set, A a set of policy alternatives and q the quota such that $\frac{n-1}{n-1-q}\in \mathbb {N}$. The case $|A|< \mu (n,q)$ is obvious since it uses the same arguments developed in the necessity part of the proof of Proposition 3.2 for this very case. The case $|A|= \mu (n,q)$ is also true thanks to Lemma 3.3.

$\Rightarrow /$ Assume $|A|\ge \mu +1$. To prove the result, it suffices to construct a profile for $|A|=\mu +1$, since the result would be deduced by applying Proposition 4.1. Let $A=\{a_0,a_1,a_2,a_3,\ldots ,a_{\mu }\}$. Since $\frac{\mu -1}{ \mu }(n-1)=q\in \mathbb {N}$, let $p=\frac{n-1}{ \mu }, p\in \mathbb {N}$ that is $(n-1=p\mu )$ and $j\in N$. Consider $\{B_1,B_2,\ldots ,B_{\mu }\}$ a partition of $N_j$ such that $|B_k|=p, ~\forall k=1,\ldots ,\mu$. Define the profile $\succ$ as follows:

$$\begin{aligned} \forall i\in N, \begin{array}{lll} \succ _{i} ~: &{}a_{\mu }\cdots a_m a_{m-1}\cdots a_{0} &{} i\in B_1, ~m=1,\ldots ,\mu \\ \succ _{i} ~: &{}a_{k-1}\cdots a_{0}a_{\mu }\cdots a_{k} &{} i\in B_k, ~k=2,\ldots ,\mu ,\\ \succ _{i} ~: &{}a_{0}a_{\mu }\cdots a_m a_{m-1}\cdots a_{1} &{} i=j, ~m=2,\ldots ,\mu \end{array} \end{aligned}$$

Now, $N_j$ and all the $N\smallsetminus B_k$ are winning coalitions since $|N_j| \ge |N\smallsetminus B_k | = p(\mu -1)+1 = q+1>q$. Moreover, the following dominations hold: $a_{1 ~} dom_{N_j~} a_{0}$, $a_{0 ~} dom_{N\smallsetminus B_1~} a_{\mu }$ and for each $k\in \{2,\ldots ,\mu \}, a_{k~}dom_{N\smallsetminus B_{k}~} a_{k-1}$. This means that the one-core is empty. $\square$