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


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.

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Fig. 1


  1. The classical assumption allows an individual who has made a policy proposal to join an opposing coalition in order to defeat that proposal, and this makes little sense in real-life politics.

  2. Note that the initial decision-making framework proposed by Shenoy (1980) is sequential, whereas our framework is reminiscent of a one-shot game. The one-core as a solution concept is better justified by our framework because in a sequential framework, players must be farsighted to be rational (Chwe 1994; Fodouop Fotso et al. 2014).

  3. Irreflexive: for any \(x\in A\), \(not(x~\succ _i~x)\). Antisymmetric: for any \(x,y\in A\), if \(x~\succ _i~y\) and \(y~\succ _i~x\), then \(x~=~y\). Transitive: for any \(x,y,z\in A\), if \(x~\succ _i~y\) and \(y~\succ _i~z\), then \(x~\succ _i~z\).

  4. In this expression, \(\prod ^{N_i}\) stands for the set of all partitions of \(N_i\) given a player i.


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Correspondence to Joseph Armel Momo Kenfack.

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I am grateful to Marcus Pivato, Mathieu Martin, Bertrand Tchantcho, Roland Pongou, Sébastien Courtin, Ashley Piggins and the two anonymous referees for helpful suggestions. This research was partially supported by IDEX Paris-Seine AAP-2018, France.



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.


\(\Rightarrow /\) 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, \( \mu (n,q)\ge 3\).


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 \(0 \le \frac{\eta }{\mu } < 1\).


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

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. (1)

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

  2. (2)

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


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. (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. (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:

figure c

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.

figure d

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, \(i\in T_1\).


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


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

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,

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


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

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:

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


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


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

    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. \(\square \)

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Momo Kenfack, J.A. Committees under qualified majority rules: the one-core stability index. Soc Choice Welf 59, 401–422 (2022).

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