Bargaining on monotonic social choice environments

Abstract

Applying the solutions defined in the axiomatic bargaining theory to actual bargaining problems is a challenge when the problem is not described by its Utility Possibility Set (UPS) but as a social choice environment specifying the set of alternatives and utility profile underlying the UPS. It requires computing the UPS, which is an operational challenge, and then identifying at least one alternative that actually achieves the bargained solution’s outcome. We introduce the axioms of Independence of Non-Strongly-Efficient Alternatives (resp. Weakly) and Independence of Redundant Alternatives. A solution satisfying these axioms can be applied to a simplified problem based on any reduced set of alternatives generating the strong (resp. weak) Pareto frontier of the initial problem, without changing the outcome, making the application of bargaining solutions to actual problems easier. We compare our axioms to usual independence axioms, and discuss their consistency with usual bargaining solutions. Then, we introduce monotonicity conditions corresponding to the existence of an interest group, i.e., agents ranking the alternatives in the same order. For such monotonic social choice environments, we provide a parameterized family of alternatives that generates the Pareto frontier of the bargaining problem, in line with our previous results. Our analysis illustrates that an axiomatic approach can be useful to foster the application of bargaining solutions, in complement to usual computational methods.

Appendix A: Proofs of propositions

1.1 A.1 Proof of Proposition 1

Before proving the proposition, we provide technical results on the characterization of strongly-efficient alternatives.

Lemma 1

Given a social choice environment $\xi =\langle {\mathbb {A}} \rangle$, the set ${\mathbb {A}}^s$ of strongly-efficient alternatives, as in Definition 6, is characterized as follows

$$\begin{aligned} \begin{aligned} A \subset {\mathbb {A}}\,\hbox { and }\, U(A) \subset {\mathcal {E}}^s\big (U({\mathbb {A}})\big ) \,\hbox { and }\, U({\mathbb {A}}\backslash A) \cap {\mathcal {E}}^s\big (U({\mathbb {A}})\big ) = \emptyset \\ \iff A = {\mathbb {A}}^s \; .\end{aligned} \end{aligned}$$

(5)

Furthermore, one has that

$$\begin{aligned} A \subset {\mathbb {A}}\,\hbox { and }\, {\mathcal {E}}^s\big (U({\mathbb {A}})\big ) \subset U(A) \Rightarrow {\mathcal {E}}^s\big (U({\mathbb {A}})\big ) \subset U({A} \cap {{\mathbb {A}}^s} ) \; .\end{aligned}$$

(6)

Finally, the following equalities hold true

$$\begin{aligned} U({\mathbb {A}}^s) = {\mathcal {E}}^s\big (U({\mathbb {A}})\big ) \,\hbox { and }\, U( {\mathbb {A}}\backslash {\mathbb {A}}^s ) = U( {\mathbb {A}}) \backslash {\mathcal {E}}^s\big (U({\mathbb {A}})\big ) \; .\end{aligned}$$

(7)

Proof of Lemma 1

Observe that if  $A \subset {\mathbb {A}}$ is such that $U(A) \subset {\mathcal {E}}^s\big (U({\mathbb {A}})\big )$, then $A \subset {\mathbb {A}}^s$ from Definition 6 of ${\mathbb {A}}^s$. On the other hand, if  $A \subset {\mathbb {A}}$ is such that $U({\mathbb {A}}\backslash A) \cap {\mathcal {E}}^s\big (U({\mathbb {A}})\big ) = \emptyset$, one can directly check that ${\mathbb {A}}^s \subset A$. Thus, implication $\Rightarrow$ in (5) is proven. The other implication $\Leftarrow$ is straightforward.

For proving (6), let $A \subset {\mathbb {A}}$ be such that ${\mathcal {E}}^s\big (U({\mathbb {A}})\big ) \subset U(A)$. Therefore, any element of ${\mathcal {E}}^s\big (U({\mathbb {A}})\big )$ can be written as $U(a)$, where $a\in A$. Now, by the definition of ${\mathbb {A}}^s$ (Definition 6), we deduce that $a\in {\mathbb {A}}^s$. We conclude that $a\in A \cap {\mathbb {A}}^s$ and that (6) holds true.

Equation (7) is made of two equalities. First, we prove that ${\mathcal {E}}^s\big (U({\mathbb {A}})\big ) = U({\mathbb {A}}^s)$ by two inclusions. On the one hand, by the definition of ${\mathbb {A}}^s$ (Definition 6), we have that $U({\mathbb {A}}^s) \subset {\mathcal {E}}^s\big (U({\mathbb {A}})\big )$. On the other hand, the reverse inclusion ${\mathcal {E}}^s\big (U({\mathbb {A}})\big ) \subset U({\mathbb {A}}^s)$ is a consequence of (6) with $A= {\mathbb {A}}$. Now we prove that $U( {\mathbb {A}}\backslash {\mathbb {A}}^s ) = U( {\mathbb {A}}) \backslash {\mathcal {E}}^s\big (U({\mathbb {A}})\big )$ by two inclusions. On the one hand, any element in $U( {\mathbb {A}}\backslash {\mathbb {A}}^s )$ of course belongs to $U( {\mathbb {A}})$. If this element is also in ${\mathcal {E}}^s\big (U({\mathbb {A}})\big )$ we obtain a contradiction with (5), because in this case $U( {\mathbb {A}}\backslash {\mathbb {A}}^s ) \cap {\mathcal {E}}^s\big (U({\mathbb {A}})\big ) \ne \emptyset$. Therefore, we obtain $U( {\mathbb {A}}\backslash {\mathbb {A}}^s ) \subset U( {\mathbb {A}}) \backslash {\mathcal {E}}^s\big (U({\mathbb {A}})\big )$. On the other hand, consider $a\in {\mathbb {A}}$ such that $U(a) \not \in {\mathcal {E}}^s\big (U({\mathbb {A}})\big )$. By the definition of ${\mathbb {A}}^s$ (Definition 6), we deduce that $a\not \in {\mathbb {A}}^s$. Therefore, $a\in {\mathbb {A}}\backslash {\mathbb {A}}^s$ and $U(a) \in U( {\mathbb {A}}\backslash {\mathbb {A}}^s )$. We conclude that $U( {\mathbb {A}}) \backslash {\mathcal {E}}^s\big (U({\mathbb {A}})\big ) \subset U( {\mathbb {A}}\backslash {\mathbb {A}}^s )$.

Hence, we have obtained (7).

We now provide the proof of Proposition 1.

In the following, for two vectors $u= (u_1,\ldots ,u_H)$ and $u' = (u_1',\ldots ,u_H')$ in ${{\mathbb {R}}}^H$, we will use the distance $\delta (u,u')= \sum _{h=1}^H|u_h- u_h'|$.

The proof is in two steps. In the first part, we prove that, for every $u\in U({\mathbb {A}})$, there exists $a^* \in {\mathbb {A}}$ such that $u^*=U(a^*) \in U({\mathbb {A}}) \cap (u+ {{\mathbb {R}}}^H_+)$ and

$$\begin{aligned} \delta (u,u^*)= \alpha _u= \sup _{u' \in U({\mathbb {A}}) \cap (u+ {{\mathbb {R}}}^H_+)} \delta (u,u') \; .\end{aligned}$$

(8)

For this purpose, we consider a sequence $(u^k)_{k \in {{\mathbb {N}}^*}}$ in $U({\mathbb {A}}) \cap (u+ {{\mathbb {R}}}^H_+)$ such that

$$\begin{aligned} \delta (u,u^k) \rightarrow \alpha _u\,\hbox { as }\, k \rightarrow \infty \; .\end{aligned}$$

(9)

Since $u^k \in U({\mathbb {A}}) \cap (u+ {{\mathbb {R}}}^H_+)$, then, for each $k \in {{\mathbb {N}}^*}$, there exist $a^k \in {\mathbb {A}}$ and $v^k = (v^k_1, \ldots ,v^k_H) \in {{\mathbb {R}}}^H_+$ such that

$$\begin{aligned} u^k = u+ v^k = U(a^k)= \big (U_1(a^k),\ldots ,U_H(a^k)\big ) \; .\end{aligned}$$

(10)

As the set ${\mathbb {A}}$ is metric compact, there exist $a^* \in {\mathbb {A}}$ and a subsequence $(a^{k_j})_{j \in {{\mathbb {N}}^*}}$ such that

$$\begin{aligned} a^{k_j} \rightarrow a^* \,\hbox { as }\, j \rightarrow \infty \; .\end{aligned}$$

(11)

We put

$$\begin{aligned} u^*= (u^*_1,\ldots ,u^*_H) \,\hbox { where }\, u^*_h= U_h(a^*) \; , \forall h= 1, \ldots , H\; ,\end{aligned}$$

(12)

and we now prove that $u^* \in U({\mathbb {A}}) \cap (u+ {{\mathbb {R}}}^H_+)$ and solves (8). Observe that, by (12), we have that $u^* \in U({\mathbb {A}})$. We also have that $u^* \in (u+ {{\mathbb {R}}}^H_+)$. Indeed, from (11) and the upper semicontinuity of the functions $U_h: {\mathbb {A}}\rightarrow {{\mathbb {R}}}$, for $h=1,\ldots ,H$, we obtain that

$$\begin{aligned} \limsup _{j \rightarrow \infty } U_h(a^{k_j} ) \le U_h(a^*) = u^*_h\; , \forall h= 1, \ldots , H\; . \end{aligned}$$

(13)

Therefore, by (10) where $v^k = (v^k_1, \ldots ,v^k_H) \in {{\mathbb {R}}}^H_+$, we deduce that

$$\begin{aligned} 0 \le \limsup _{j \rightarrow \infty } v^{k_j}_h= \limsup _{j \rightarrow \infty } U_h(a^{k_j} ) -u_h\le u^*_h-u_h\; ,\end{aligned}$$

(14)

for all $h= 1, \ldots , H$, that is, $u^* \in (u+ {{\mathbb {R}}}^H_+)$.

$$\begin{aligned} \alpha _u&= \lim _{j \rightarrow \infty } \delta (u,u^{k_j}) \,\hbox { by (9)}\, \\&= \limsup _{j \rightarrow \infty } \delta (u,u^{k_j}) = \limsup _{j \rightarrow \infty } \sum _{h=1}^H|u_h- u_h^{k_j}| = \limsup _{j \rightarrow \infty } \sum _{h=1}^H|v^{k_j}_h| \,\hbox { by (10)}\, \\&= \limsup _{j \rightarrow \infty } \sum _{h=1}^Hv^{k_j}_h\,\hbox { because }\, v^k = (v^k_1, \ldots ,v^k_H) \in {{\mathbb {R}}}^H_+ \\&\le \sum _{h=1}^H\limsup _{j \rightarrow \infty }v^{k_j}_h\le \sum _{h=1}^H(u^*_h-u_h) \,\hbox { by (14)}\, \\&= \delta (u, u^*) \; .\end{aligned}$$

Thus, we obtain that $\delta (u,u^*) \ge \alpha _u= \sup _{u' \in U({\mathbb {A}}) \cap (u+ {{\mathbb {R}}}^H_+)} \delta (u,u')$. Now, as $u^* = U(a^*) \in U({\mathbb {A}}) \cap (u+ {{\mathbb {R}}}^H_+)$ by (13), we conclude that (8) holds true.

In the second part, we prove that, for any $u\in U({\mathbb {A}})$ and $u^* \in U({\mathbb {A}}) \cap (u+ {{\mathbb {R}}}^H_+)$ such that $\alpha _u= \delta (u,u^*)$ ($\alpha _u$ defined in (8)), then $u^* \in {\mathcal {E}}^s\big (U({\mathbb {A}})\big )$.

The proof is obtained by contradiction. Suppose $u^* \notin {\mathcal {E}}^s\big (U({\mathbb {A}})\big )$. Then, there exists $\tilde{u}\in U({\mathbb {A}}) \cap (u^* + {{\mathbb {R}}}^H_+{\setminus } \{0\})$. Expressing that $u^* \in (u+ {{\mathbb {R}}}^H_+)$ and that ${{\tilde{u}}} \in U({\mathbb {A}}) \cap (u^* + {{\mathbb {R}}}^H_+{\setminus } \{0\})$, we get that $u^* = u+ v$ and $\tilde{u}= u^* + w$ for some $v=(v_1,\ldots ,v_H) \in {{\mathbb {R}}}^H_+$ and $w = (w_1,\ldots ,w_H) \in {{\mathbb {R}}}^H_+{\setminus } \{0\}$ (i.e., $w_h> 0$ for some $h\in \{1,\ldots ,H\}$). We easily deduce that ${{\tilde{u}}} \in U({\mathbb {A}}) \cap (u+ {{\mathbb {R}}}^H_+)$. Now, we have that $\alpha _u= \delta (u,u^*) = \sum _{h=1}^Hv_h < \sum _{h=1}^H(v_h + w_h) = \delta (u,\tilde{u})$, which contradicts the definition of $\alpha _u$ (Eq. 8).

We have shown that, for every $u\in U({\mathbb {A}})$, there exists $a^* \in {\mathbb {A}}$ such that $U(a^*) = u^* \in {\mathcal {E}}^s\big (U({\mathbb {A}})\big ) \cap (u+ {{\mathbb {R}}}^H_+)$. As a consequence, the set ${\mathbb {A}}^s=\left\{ a\in {\mathbb {A}}\mid U(a)\in {\mathcal {E}}^s\big (U({\mathbb {A}})\big ) \right\}$ is not empty.

1.2 A.2 Proof of Proposition 2

The proof of Proposition 2 will be based on the following results that characterize the efficient alternatives of reduced social choice environments.

Lemma 2

Let $(\xi ', \xi )$ be a couple of social choice environments such that

$$\begin{aligned} {\mathbb {A}}'&\subset {\mathbb {A}} \end{aligned}$$

(15)

$$\begin{aligned} {\mathcal {E}}^s\big (U({\mathbb {A}})\big )&\subset U({\mathbb {A}}') \; . \end{aligned}$$

(16)

Suppose that the set ${\mathbb {A}}$ of alternatives is a compact metric space and the utility profile $U: {\mathbb {A}}\rightarrow {{\mathbb {R}}}^H$ is upper semicontinuous. Then, we have that ${\mathbb {A}}'^s \subset {\mathbb {A}}^s$ and $U({\mathbb {A}}'^s)=U({\mathbb {A}}^s)$.

Proof of Lemma 2

Consider a couple $(\xi ', \xi )$ of social choice environments such that ${\mathbb {A}}' \subset {\mathbb {A}}$ and ${\mathcal {E}}^s\big (U({\mathbb {A}})\big ) \subset U({\mathbb {A}}')$. First, we prove that

$$\begin{aligned} U({\mathbb {A}}'\cap {\mathbb {A}}^s)= U({\mathbb {A}}^s) \; . \end{aligned}$$

(17)

From (7), we have that ${\mathcal {E}}^s(U({\mathbb {A}}))=U({\mathbb {A}}^s)$. As, by assumption (16), ${\mathcal {E}}^s(U({\mathbb {A}}))\subset U({\mathbb {A}}')$, we deduce from (6) that ${\mathcal {E}}^s(U({\mathbb {A}}))\subset U({\mathbb {A}}'\cap {\mathbb {A}}^s)$. Linking these two results, we obtain $U({\mathbb {A}}^s) = {\mathcal {E}}^s(U({\mathbb {A}})) \subset U({\mathbb {A}}'\cap {\mathbb {A}}^s) \subset U({\mathbb {A}}^s)$, and we conclude that all terms are equal, so that $U({\mathbb {A}}'\cap {\mathbb {A}}^s)= U({\mathbb {A}}^s)$.

Second, we prove that

$$\begin{aligned} U({\mathbb {A}}^s) \subset U({\mathbb {A}}'^s) \; . \end{aligned}$$

(18)

For this purpose, we will use the following property, which follows from the definition of Pareto sets:

$$\begin{aligned} {{\bar{{\mathcal {U}}}}} \subset {\mathcal {U}}\subset {{\mathbb {R}}}^H\Rightarrow {\mathcal {E}}^s({\mathcal {U}}) \cap {{\bar{{\mathcal {U}}}}} \subset {\mathcal {E}}^s({{\bar{{\mathcal {U}}}}}) \; . \end{aligned}$$

(19)

We have

$$\begin{aligned} U({\mathbb {A}}^s)&= U({\mathbb {A}}'\cap {\mathbb {A}}^s) \,\hbox { by (18)}\, \\&\subset U({\mathbb {A}}') \cap U({\mathbb {A}}^s) = U({\mathbb {A}}') \cap {\mathcal {E}}^s(U({\mathbb {A}})) \,\hbox { by (7)}\, \\&\subset {\mathcal {E}}^s\big (U({\mathbb {A}}')\big ) \,\hbox { by (20)}\, \end{aligned}$$

since ${\mathbb {A}}' \subset {\mathbb {A}}$, by assumption (15), hence $U({\mathbb {A}}') \subset U({\mathbb {A}})$. Therefore, we conclude that $U({\mathbb {A}}^s) \subset {\mathcal {E}}^s\big (U({\mathbb {A}}')\big ) =U({\mathbb {A}}'^s)$ by (7).

Third, for concluding the proof, we need to show the inclusions ${\mathbb {A}}'^s \subset {\mathbb {A}}^s$ and $U({\mathbb {A}}'^s)\subset U({\mathbb {A}}^s)$. For this purpose, it is sufficient to prove that ${\mathbb {A}}'^s={\mathbb {A}}'\cap {\mathbb {A}}^s$, an equality that we will deduce using the equivalence (5), that requires three conditions. We establish that these three conditions hold true.

• 1. We establish that ${\mathbb {A}}'\cap {\mathbb {A}}^s \subset {\mathbb {A}}'$. This is obvious.

• 2. We establish that $U({\mathbb {A}}'\cap {\mathbb {A}}^s) \subset {\mathcal {E}}^s\big (U({\mathbb {A}}')\big )$. Indeed,

  $$\begin{aligned} U({\mathbb {A}}'\cap {\mathbb {A}}^s)&= U({\mathbb {A}}^s) \,\hbox { by (18)}\, \\&\subset U({\mathbb {A}}'^s) \,\hbox { by (19)}\, \\&= {\mathcal {E}}^s\big (U({\mathbb {A}}')\big ) \,\hbox { by (7).}\, \end{aligned}$$

• 3. We establish that $U({\mathbb {A}}'\backslash ({\mathbb {A}}'\cap {\mathbb {A}}^s)) \cap {\mathcal {E}}^s\big (U({\mathbb {A}}')\big ) = \emptyset$. This is where the assumptions that the set ${\mathbb {A}}$ of alternatives is compact and that the utility profile $U: {\mathbb {A}}\rightarrow {{\mathbb {R}}}^H$ is upper semicontinuous play a role, as they make it possible to use Proposition 1. Consider an alternative $a\in {{\mathbb {A}}'\backslash ({\mathbb {A}}'\cap {\mathbb {A}}^s)}$. By Proposition 1, there exists an alternative ${{\tilde{a}}} \in {\mathbb {A}}$ such that $U({{\tilde{a}}})\gneq U(a)$, with $U({{\tilde{a}}}) \in {\mathcal {E}}^s(U({\mathbb {A}}))$. As ${\mathcal {E}}^s(U({\mathbb {A}}))\subset U({\mathbb {A}}')$, we deduce that there exists an alternative ${{\bar{a}}} \in {\mathbb {A}}'$ such that $U({{\bar{a}}})=U({{\tilde{a}}}) \gneq U(a)$. As a consequence, $a\not \in {\mathcal {E}}^s\big (U({\mathbb {A}}')\big )$.

  We have thus proven that $U({\mathbb {A}}'\backslash ({\mathbb {A}}'\cap {\mathbb {A}}^s)) \cap {\mathcal {E}}^s\big (U({\mathbb {A}}')\big ) = \emptyset$.

We now provide the proof of Proposition 2.

Consider a couple $(\xi , \xi ')$ of social choice environments such that ${\mathbb {A}}' \subset {\mathbb {A}}$ and ${\mathcal {E}}^s\big (U({\mathbb {A}})\big ) \subset U({\mathbb {A}}')$. We use the corresponding strongly-efficient reduced environments $\xi ^s=\langle {\mathbb {A}}^s \rangle$ and ${\xi '}^s=\langle {\mathbb {A}}'^s \rangle$ (Definition 6). The assumptions of Lemma 2 are satisfied, as ${\mathbb {A}}' \subset {\mathbb {A}}$ and ${\mathcal {E}}^s\big (U({\mathbb {A}})\big ) \subset U({\mathbb {A}}')$ correspond to equations (15) and (16). From Lemma 2, we obtain that ${\mathbb {A}}'^s\subset {\mathbb {A}}^s$ and $U({\mathbb {A}}'^{s})=U({\mathbb {A}}^s)$. As the solution $\mu$ is IRA, we deduce from Axiom 2.3 that

$$\begin{aligned} U\big (\mu (\xi ^s)\big )= U\big (\mu (\xi '^s)\big ) \; .\end{aligned}$$

(20)

Since the solution $\mu$ is INSEA and ${\mathbb {A}}^s \subset {\mathbb {A}}$, we deduce from Axiom 2.3 that

$$\begin{aligned} \mu (\xi ^s)=\mu (\xi ) \; .\end{aligned}$$

(21)

Repeating the argument with the environment ${\xi '}^s$, we obtain that

$$\begin{aligned} \mu (\xi '^s)=\mu (\xi ') \; .\end{aligned}$$

(22)

Therefore, from (20), (21), and (22), we conclude that $U\big (\mu (\xi )\big ) = U\big (\mu (\xi ^s)\big ) =U\big (\mu (\xi '^s)\big ) = U\big (\mu (\xi ')\big )$.

1.3 A.3 Proof of the weak version of Proposition 2

Consider a couple $(\xi , \xi ')$ of social choice environments such that ${\mathbb {A}}' \subset {\mathbb {A}}$ and ${\mathcal {E}}^w\big (U({\mathbb {A}})\big ) \subset U({\mathbb {A}}')$. We aim at proving that, if a bargaining solution $\mu$ satisfies the INWEA and IRA axioms, then $U\big (\mu (\xi )\big ) = U\big (\mu (\xi ')\big )$.

For this purpose, we consider the corresponding weakly-efficient reduced social choice environments, $\xi ^w=\langle {\mathbb {A}}^w \rangle$ and ${\xi '}^w=\langle {\mathbb {A}}'^w \rangle$ (Definition 5), and shall establish the following equalities

• $\mu (\xi ^w)=\mu (\xi )$ (from the INWEA axiom), and thus $U(\mu (\xi ^w))=U(\mu (\xi ))$.

• $\mu (\xi '^w)=\mu (\xi ')$ (from the INWEA axiom), and thus $U(\mu (\xi '^w))=U(\mu (\xi '))$.

• $U({\mathbb {A}}^w)=U({\mathbb {A}}'^w)$ which implies, from the IRA axiom, that $U(\mu (\xi ^w))=U(\mu (\xi '^w))$.

The two first equalities are direct consequences of the definition of the INWEA axiom. To prove the last equality, we need to show that the conditions of the IRA axiom are satisfied.

We first prove that ${\mathbb {A}}'^w\subset {\mathbb {A}}^w$. Consider a weakly-efficient alternative of $\xi '$, i.e., $a\in {\mathbb {A}}'^w \subset {\mathbb {A}}' \subset {\mathbb {A}}$. Assume that $a\notin {\mathbb {A}}^w$. Then, by the definition of the set of weakly-efficient alternatives ${\mathbb {A}}^w$, there would be an alternative $b\in {\mathbb {A}}$ such that $U(b) > U(a)$. Given the existence of strongly-efficient alternatives in ${\mathbb {A}}$ (Proposition 1), there would then exist an alternative $b^s\in {\mathbb {A}}^s \subset {\mathbb {A}}^w$ such that $U(b^s)\ge U(b) > U(a)$. As ${\mathcal {E}}^w\big (U({\mathbb {A}})\big ) \subset U({\mathbb {A}}')$, one would have $U(b^s)\in U({\mathbb {A}}')$, meaning that $a$ would be strongly dominated in ${\mathbb {A}}'$, and thus not part of ${\mathbb {A}}'^w$, a contradiction. We thus conclude that any alternative in ${\mathbb {A}}'^w$ is also in ${\mathbb {A}}^w$, meaning that ${\mathbb {A}}'^w\subset {\mathbb {A}}^w$.

We now prove that $U({\mathbb {A}}'^w)=U({\mathbb {A}}^w)$, by a double inclusion. From the previous step, we deduce $U({\mathbb {A}}'^w)\subset U({\mathbb {A}}^w)$. To prove the opposite inclusion, consider an outcome $u\in U({\mathbb {A}}^w)$, i.e., $u\in {\mathcal {E}}^w\big (U({\mathbb {A}})\big )$. From the assumption ${\mathcal {E}}^w\big (U({\mathbb {A}})\big ) \subset U({\mathbb {A}}')$, one has $u\in U({\mathbb {A}}')$. There is thus an alternative $a\in {\mathbb {A}}'$ such that $U(a)=u$. Assume that this alternative is not weakly-Pareto-efficient in ${\mathbb {A}}'$, i.e., $a\notin {\mathbb {A}}'^w$. Then, there would be an alternative $b\in {\mathbb {A}}'$ such that $U(b)>U(a)$. As ${\mathbb {A}}'\subset {\mathbb {A}}$, the alternative b would be in ${\mathbb {A}}$, meaning that $u< U(b)$, a contradiction with the assumption that $u$ is weakly-efficient in ${\mathbb {A}}$. We thus have $u\in U({\mathbb {A}}^w) \Rightarrow u\in U({\mathbb {A}}'^w)$, i.e., $U({\mathbb {A}}^w)\subset U({\mathbb {A}}'^w)$. We conclude that $U({\mathbb {A}}^w)=U({\mathbb {A}}'^w)$. The two conditions of the IRA axiom are thus satisfied, proving the third equality.

Combining the three established equalities, we obtain $U(\mu (\xi ))=U(\mu (\xi ^w))=U(\mu (\xi '^w))=U(\mu (\xi '))$, proving the Proposition.

1.4 A.4 Proof of Proposition 5

We need a preliminary Lemma.

Lemma 3

Consider a given $(i,o)$-monotonic social choice environment $\xi =\langle {\mathbb {A}} \rangle$, as in Definition 7. Then, for all $( u^{o}, {\bar{u}}^{o} ) \in {\mathcal {U}}^{o} \times {\mathcal {U}}^{o}$ and all ${{\bar{a}}} \in {\mathbb {A}}$, we have

$$\begin{aligned} u^{o} \le U^{o}( {{\bar{a}}} )&\Rightarrow U^{i}( {{\bar{a}}} ) \le U^{i}\big ( a^* \big )\quad \forall ~a^* \in a^{o}(u^{o})\; , \end{aligned}$$

(23a)

$$\begin{aligned} {{\bar{a}}}^* \in a^{o}({{\bar{u}}}^{o}) ,~u^{o} \le U^{o}\big ( \bar{a}^*\big )&\Rightarrow U^{i}\big ( \bar{a}^*\big ) \le U^{i}\big ( a^* \big )\quad \forall ~a^* \in a^{o}(u^{o}) \; , \end{aligned}$$

(23b)

$$\begin{aligned} u^{o} \le {\bar{u}}^{o}&\Rightarrow U^{i}\big ({{\bar{a}}}^* \big ) \le U^{i}\big ( a^* \big )\quad \forall ~\bar{a}^* \in a^{o}({{\bar{u}}}^{o}),~\forall ~a^* \in a^{o}(u^{o}) \; . \end{aligned}$$

(23c)

Proof of Lemma 3

Let $( u^{o}, {\bar{u}}^{o} )\in {\mathcal {U}}^{o} \times {\mathcal {U}}^{o}$ and ${{\bar{a}}} \in {\mathbb {A}}$.

• 1. Suppose that $u^{o} \le U^{o}({{\bar{a}}})$. By Eq. (3) in Definition 8, defining the set $a^{o}(u^{o})$ composed by minima of the set $\{ a\in {\mathbb {A}}\mid U^{o}(a) \ge u^{o} \}$ (see (1) in Definition 7), we deduce that $a^* \preceq {{{\bar{a}}}}$ for all $a^* \in a^{o}(u^{o})$. Now, the mapping $U^{i}: {\mathbb {A}}\rightarrow {{{\mathbb {R}}}^{|i|}}$ is nonincreasing by item I in the Definition 7 of MonE. We deduce that $U^{i}({{\bar{a}}}) \le U^{i}\big (a^*\big )$, for all $a^* \in a^{o}(u^{o})$. Hence, (23a) is proven.

• 2. Use (23a) with ${{\bar{a}}} = {{\bar{a}}}^* \in a^{o}({{\bar{u}}}^{o})$ to derive (23b).

• 3. Let $u^{o} \le {\bar{u}}^{o}$. As ${{\bar{u}}}^{o} \in {\mathcal {U}}^{o}$, by definition of the set ${\mathcal {U}}^{o}$ of feasible payoffs for outsiders (Eq. 2), and by Eq. (3), for ${{\bar{a}}}^* \in a^{o}({{\bar{u}}}^{o})$ we have that ${{\bar{u}}}^{o} \le U^{o}\big ( {{\bar{a}}}^* \big )$. Therefore, we get that $u^{o} \le U^{o}\big ({{\bar{a}}}^* \big )$ and we use (23b) to obtain (23c).

 

Proof of Proposition 5

First, we prove that

$$\begin{aligned} \big ( u^{i}, u^{o} \big ) \in U({\mathbb {A}}) \Rightarrow u^{o} \in {\mathcal {U}}^{o} \; , u^{i} \le U^{i}\big ( a^*\big ) \; , u^{o} \le U^{o}\big ( a^* \big ) \quad \forall ~a^* \in a^{o}(u^{o}) \; . \end{aligned}$$

(24)

By definition of $U({\mathbb {A}})$, there exists an alternative ${{\bar{a}}} \in {\mathbb {A}}$, such that $\big ( u^{i}, u^{o} \big ) = u= U({{\bar{a}}}) = \big ( U^{i}({{\bar{a}}}), U^{o}({{\bar{a}}}) \big )$. On the one hand, as $u^{o} = U^{o}({{\bar{a}}})$, we deduce from item II in the Definition 7 of MonE (Eq. 2) that $u^{o} \in {\mathcal {U}}^{o}$, and that $u^{i} = U^{i}( {{\bar{a}}} ) \le U^{i}\big (a^*\big )$ for all $a^* \in a^{o}(u^{o})$ by (23a). On the other hand, by definition of the set ${\mathcal {U}}^{o}$ of feasible payoffs for outsiders (Eq. 2), and by Eq. (3), we obtain that $u^{o} \in {\mathcal {U}}^{o} \Rightarrow u^{o} \le U^{o}\big (a^*\big )$ for all $a^* \in a^{o}(u^{o})$.

Second, we prove that ${\mathcal {E}}^s\big (U({\mathbb {A}})\big ) \subset U({\mathbb {A}}^{o})$. Let $u\in {\mathcal {E}}^s\big (U({\mathbb {A}})\big )$. By the definition of a Pareto set, we have that $u\in {\mathcal {E}}^s\big (U({\mathbb {A}})\big ) \subset U({\mathbb {A}})$. By (24), we obtain that $u^{i} \le U^{i}\big (a^*\big )$ and that $u^{o} \le U^{o}\big ( a^*\big )$ for all $a^* \in a^{o}(u^{o})$. Putting ${\bar{u}} = \big ( U^{i}\big (a^*\big ), U^{o}\big ( a^* \big ) \big )$ for some (fixed) $a^* \in a^{o}(u^{o})$, we deduce that $u\le {\bar{u}}$. As ${\bar{u}} \in U({\mathbb {A}})$ and $u\in {\mathcal {E}}^s\big (U({\mathbb {A}})\big )$, we obtain that ${\bar{u}} = u$, by definition of a (strong) Pareto set. By definition of ${\mathbb {A}}^{o}$ (Eq. 4), we conclude that $u= {\bar{u}}= \big ( U^{i}\big (a^*\big ), U^{o}\big ( a^* \big ) \big ) \in U({\mathbb {A}}^{o})$.

Third, we prove that $U({\mathbb {A}}^{o}) \subset {\mathcal {E}}^w\big (U({\mathbb {A}})\big )$. The proof is obtained by contradiction. Let $u\in U({\mathbb {A}}^{o})$ and suppose that there exists ${\bar{u}} \in U({\mathbb {A}})$ such that $u< {\bar{u}}$ (i.e, a strict inequality component by component). On the one hand, by definition of ${\mathbb {A}}^{o}$ (Eq. 4), there exists ${\hat{u}}^{o} \in {\mathcal {U}}^{o}$ such that $u= U\big ({{\hat{a}}} \big )$ for some $\hat{a}\in a^{o}({\hat{u}}^{o})$. On the other hand, we have that ${{\bar{u}}} \le U\big (\hat{a}\big )$, by (24). We deduce that

$$\begin{aligned} U\big ({{\hat{a}}}\big ) = u< {{\bar{u}}} \le U\big ( {{\bar{a}}}^*\big ) \quad \forall ~{{\bar{a}}}^* \in a^{o}({{\bar{u}}}^{o}) \; . \end{aligned}$$

(25)

As already seen, we have that ${\hat{u}}^{o} \le U^{o}\big ({{\hat{a}}} \big )$. Together with inequality (25), this yields ${\hat{u}}^{o} < U^{o}\big ( \bar{a}^*\big )$ for all ${{\bar{a}}}^* \in a^{o}({{\bar{u}}}^{o})$. Now fix ${{\bar{a}}}^* \in a^{o}({{\bar{u}}}^{o})$. By (23b), we deduce that $U^{i}\big ( {{\bar{a}}}^*\big ) \le U^{i}\big ({{\hat{a}}}\big )$. Now, using inequality (25), we get that ${{\bar{u}}}^{i} \le U^{i}\big (\bar{a}^*\big ) \le U^{i}\big ({{\hat{a}}}\big ) = u^{i}$. However, this contradicts $u< {\bar{u}}$. Therefore, no such ${\bar{u}}$ exists and we conclude that $u\in {\mathcal {E}}^w\big (U({\mathbb {A}})\big )$.