Fair social orderings with other-regarding preferences

Abstract

We study the construction of social ordering functions in a multidimensional allocation problem where agents have heterogeneous other-regarding preferences (ORP). We show that there exists leximin social ordering functions satisfying equality and efficiency principles. When equality is defined as equality of resources, and ORP are only taken into account by efficiency principles, some of these social ordering functions are independent of the other-regarding part of preferences. When ORP are also taken into account by equality principles, results depend on the degree of resourcism of the social planner. If the planner still worries about equality of resources, some of the social ordering functions satisfying equality and efficiency remain independent of ORP. If the planner departs from a resourcist notion of equality, then social-ordering functions satisfying equality and efficiency must use information on the other-regarding part of preferences.

Appendix

1.1 Proof of Proposition 1

1.1.1 No resource destruction unanimity implies social monotonicity on ${\mathcal {R}}$

Take any $z_N \in Z$ and any $\bar{w} \in {\mathbb {R}}^l_{++}$. We will only need to consider the distributions of $\bar{w}$ in which each agent receives a share $\sigma _i \ge 0$ of $\bar{w}$, with $\sum _{i\in N} \sigma _i = 1$. That is the vector of additional resources for any of these distributions $\sigma$ is the Kronecker product $\sigma \otimes \bar{w} = (\sigma _1 \bar{w}, \dots , \sigma _n \bar{w})$, and the resulting allocation is $(z_N + \sigma \otimes \bar{w})$.

Let $\varSigma$ be the $n-1$-dimensional simplex, i.e. the set of all distributions $\sigma$. Let us define the set of $\sigma \in \varSigma$ which lead to allocations that i strictly prefers to $z_N$,

$$\begin{aligned} B_i{:=}\{\sigma \in \varSigma | (z_N + \sigma \otimes \bar{w})~P_i~z_N\}. \end{aligned}$$

Similarly, let us define the indifference counter-part of $B_i$,

$$\begin{aligned} E_i{:=}\{\sigma \in \varSigma | (z_N + \sigma \otimes \bar{w})~I_i~z_N\}. \end{aligned}$$

Given the above notation, a sufficient condition for social monotonicity to hold is $\cap _{i \in N} B_i\ne \varnothing$ and no resource destruction unanimity implies that $\sigma \in \cup _{i \in N} B_i$ for any $\sigma \in \varSigma$.

The proof is by induction. It relies on two classes of properties called subgroup social monotonicity-$k (SSM^k)$ and Subgroup Indifference-$(k-1) (SI^{k-1})$. In the induction basis, we prove $SSM^2$ and $SI^1$. The induction step then consists in proving that $SSM^k$ and $SI^{k-1}$ hold if $SSM^{k-1}$ and $SI^{k-2}$ hold. Finally, noticing that $SSM^{|N|}$ implies social monotonicity will complete the proof.

We first introduce $SSM^k$, which is a version of social monotonicity in which the new resources are only distributed to a subset of agents $N' \subseteq N$, with $|N| =k$. For any subset $N' \subseteq N$ with $|N| =k$, let $\varSigma ^{N'}$ be the $k-1$-simplex, that is the distributions $\sigma \in \varSigma$ such that $\sigma _j = 0$ for all $j\in N\backslash N'$.

Definition 4

(Subgroup social monotonicity-k) For all $N'\subseteq N$ with $\vert N'\vert =k$, for all $z_N \in Z$, and for all $\bar{w} \in {\mathbb {R}}^l_{++}$, there exists a distribution $\sigma ^* \in \varSigma ^{N'}$ such that,

$$\begin{aligned} (z_N + \sigma ^* \otimes \bar{w})~P_i~z_N,\quad \text {for all } i\in N'. \end{aligned}$$

Given the above notations, subgroup social monotonicity-k holds if $\cap _{i \in N'} B_i\ne \varnothing$ for all $N' \subseteq N$ with $\vert N'\vert =k$.

Second, we introduce $SI^{k-1}$, which states that there is a way to distribute the complete extra resource among a subset of k agents leaving $k-1$ of them indifferent to the initial allocation, while the last agent receives a strictly positive amount of resources.

Definition 5

[Subgroup indifference-(k-1)] For all $N'\subseteq N$ with $\vert N'\vert =k$, for all $z_N \in Z$, for all $\bar{w} \in {\mathbb {R}}^l_{++}$, and for any $j\in N'$, there exists a distribution $\sigma ^*\in \varSigma ^{N'}$ with $\sigma _j>0$ such that

$$\begin{aligned} (z_N + \sigma ^* \otimes \bar{w})~I_i~z_N,\quad \text {for all } i\in N'\backslash \{j\}. \end{aligned}$$

Given the above notations, $SI^{k-1}$ holds if for all $N' \subseteq N$ with $\vert N'\vert =k$ and any $j \in N'$, there exists a $\sigma$ with $\sigma _j >0$ such that $\sigma \in \cap _{i \in N'\backslash \{j\}} E_i$.

• 1. Induction basis We show that both $SSM^2$ and $SI^1$ hold under the assumptions of our domain. First we prove $SI^1$. Take any two agents $j,k \in N$. For $SI^1$ to hold, it is enough to construct a $\sigma ' \in \varSigma ^{\{j,k\}}$ with $\sigma '_k>0$ and $\sigma ' \in E_j$. Notice that by strict monotonicity in own consumption, $\varSigma ^{\{j\}} \in B_j$. Also, by no altruism, $\varSigma ^{\{k\}} \notin B_j$. Thus consider the continuous path that goes from $\varSigma ^{\{j\}}$ to $\varSigma ^{\{k\}}$ along the edge $\varSigma ^{\{j,k\}}$. By continuity, there must exists some $\sigma '' \in \varSigma ^{\{j,k\}}$ with $\sigma '' \in E_j$. In order to derive a contradiction, assume $\sigma ''_k = 0$. Then by no altruism and strict monotonicity in own consumption, $\sigma '' \notin B_k$. Because $\sigma \in \varSigma ^{\{j,k\}}, \sigma _h'' =0$ for all $h\ne j,k \in N$. Thus by no altruism and strict monotonicity in own consumption, $\sigma '' \notin B_h$ too. But this means $\sigma '' \notin B_i$ for all $i\in N$, contradicting no resource destruction unanimity. Hence we must have $\sigma ''_k >0$, and we found the desired distribution. Second we prove $SSM^2$. Take any two agents $j,k \in N$. By $SI^1$, there exists $\sigma ' \in E_j \cap \varSigma ^{\{j,k\}}$ with $\sigma _k'>0$. As argued above, $\sigma ' \notin B_h$ for all $h\ne j,k \in N$ because $\sigma ' \in \varSigma ^{\{j,k\}}$. But then by no resource destruction unanimity, $\sigma ' \in B_k$, as otherwise no-one is strictly better at $\sigma '$. By continuity, $B_k \cap \varSigma ^{\{j,k\}}$ is open in $\varSigma ^{\{j,k\}}$. Thus there exists a 1 dimensional ball $b \in \varSigma ^{\{j,k\}}$ centered in $\sigma '$ such that for all $\sigma \in b$ we have $\sigma \in B_k$. In particular, there exists $\sigma '' \in B_k \cap \varSigma ^{\{j,k\}}$ with $\sigma _k''<\sigma _k'$, and hence $\sigma _j'<\sigma _j''$. By strict monotonicity in own consumption, no altruism, and because the initial $\sigma ' \in E_j$, we have $\sigma ''~P_j~\sigma '$ which implies $\sigma '' \in B_j\cap B_k$, the desired result.

• 2. Induction step If $SSM^h$ and $SI^{h-1}$ hold for all $h < k$, then $SSM^k$ and $SI^{k-1}$ hold.

First we prove $SI^{k-1}$. Take any $N' \subseteq N$ with $\vert N'\vert =k$ and any $j \in N'$.

Take $N'=\{1,\dots ,k\}$. We chose $\{1,\dots ,k\}$ for notational convenience and without loss of generality. Consider any distribution

$$\begin{aligned} \underline{\sigma } \in \varSigma ^{N'\backslash \{1\}}\quad \hbox {with } \underline{\sigma }_k>0\quad \hbox {and}\quad \underline{\sigma } \in \cap _{i \in \{2,\dots ,k-1\}}E_i. \end{aligned}$$

By $SI^{k-2}, \underline{\sigma }$ exists. We prove $SI^{k-1}$ by showing the existence of a continuous path that lies in the intersection of $\varSigma ^{N'}$ and $\cap _{h \in \{2,\dots ,k-1\}}E_h$ and connects $\underline{\sigma }$ to some distribution in the non-empty set

$$\begin{aligned} \bar{\varSigma } {:=}\{ \bar{\sigma } \in \varSigma ^{N'\backslash \{k\}}|\bar{\sigma }_1>0\hbox { and }\bar{\sigma } \in \cap _{i \in \{2,\dots ,k-1\}}E_i\}. \end{aligned}$$

If such path exists, following the same argument as in the inductions basis, it must cross $E_1$ at some distribution $\sigma '\in \varSigma ^{N'}$, as $\underline{\sigma } \in \varSigma \backslash B_1$ whereas $\bar{\sigma } \in B_1$ and preferences are continuous. Again, if $\sigma '_k=0$, then no resource destruction unanimity is violated as we have $\sigma ' \in I_i$ for all $i \in N'\backslash \{k\}, \sigma _k'=0$ and $\sigma _j'=0$ for all $j \in N\backslash N'$. Thus $\sigma '$ has the desired properties and $SI^{k-1}$ holds.

There remains to prove the existence of such path. We construct it as the limit of a sequence of sequences $\{\gamma ^n\}_{n=1}^C$, where each sequence corresponds to a different value of C. For any $C, \gamma ^n \in \varSigma ^{N'}$, $\gamma ^{1}{:=}\underline{\sigma }$ and $\gamma ^C \in \bar{\varSigma }$. These sequences are constructed in such a way that all $\gamma ^n \in \cap _{i \in \{2,\dots ,k-1\}}E_i$. In a nutshell, moving from $\gamma ^{n}$ to $\gamma ^{n+1}$ is done as follows. Let $\gamma ^{n+1}_k=\gamma ^n_k-\alpha \gamma ^1_k$ for some $\alpha \in (0,1)$, and distribute the fraction of resource $\alpha \gamma ^1_k$ among agents in $\{1, \dots , k-1\}$ in a way that leaves all agents in $\{2, \dots , k-1\}$ indifferent with $\gamma ^n$. Observe that the existence of such distribution is not implied by $SI^{k-2}$ as the resources are not added to the economy, but rather taken from k. If we can prove the existence of such $\gamma ^{n+1}$, the properties of our domain imply that $\gamma ^{n+1}_1>\gamma ^{n}_1$. By choosing $\alpha = \frac{1}{C}$, we have $\gamma ^{C}_k=0$, and hence $\gamma ^{C}\in \bar{\varSigma }$. The number C of sequences can be made arbitrarily large, which will make $\alpha$ arbitrarily small. Thus the sequence of sequences tends to the desired continuous path.

There remains to prove for all n that it is possible to distribute the resource $\alpha \gamma ^1_k$ among agents in $\{1, \dots , k-1\}$ in such a way that $\gamma ^{n+1}_1>\gamma ^{n}_1$ and $\gamma ^{n+1} \in \cap _{i \in \{2,\dots ,k-1\}}E_i$. The desired distribution $\gamma ^{n+1}$ is constructed using the following procedure.

• Set $\gamma ^{n+1}_k{:=}\gamma ^{n}_k-\alpha \gamma ^{1}_k$.

• Let $\hat{\rho }^n$ be such that $\hat{\rho }^n_i{:=}\gamma ^n_i$ for all $i \in N\backslash \{k\}$ and $\hat{\rho }^n_k{:=}\gamma ^{n+1}_k$. By no altruism, we have $\hat{\rho }^n~\in B_i \cup E_i$ for all $i \in N\backslash \{k\}$.

• By $SSM^{k-2}$, there exists a way to distribute the share of resources taken from k among agents in $\{2,\dots ,k-1\}$ so as to leave them all strictly better off than in $\hat{\rho }^n$. Formally, there exists a $\rho \in \varSigma ^{N'}$ with

  • $\rho _i > \gamma ^n_i$ for all $i\in \{2,\dots ,k-1\}$ and $\sum _{i \in \{2,\dots ,k-1\}}\rho _i - \gamma ^n_i=\alpha \gamma ^{1}_k$,

  • $\rho _k=\gamma ^{n+1}_k$,

  • $\rho _j = {\gamma }^n_j$ for all $j\in N\backslash \{2,\dots ,k\}$,

  such that $\rho \in B_j$ for all $j\in \{2,\dots ,k-1\}$.

• Now consider the set of distributions $\rho ^{\tau }$ obtained by setting $\rho _j^{\tau } = \rho _j - \tau _j$ with $\tau _j \in (-\infty ,\rho _j]$ for all $j \in \{2,\dots ,k-1\}$, and transferring all the resources to agent 1, that is $\rho ^{\tau }_1 = \rho _1 + \sum _{j\in \{2,\dots ,k-1\}} \tau _j$. For any $\tau {:=}(\tau _2,\dots ,\tau _{k-1})$, let $\rho ^{\tau }$ denote the resulting allocation.

• We consider a particular subset of such distributions $\rho ^{\tau }$. Let

  $$\begin{aligned} T&{:=}\left\{ \tau ~\Big | \sum _{j\in \{2,\dots ,k-1\}} \tau _j > 0,\right. \\&\qquad \left. \text { and } \rho ^{\tau } \in B_j \cup E_j \text { for all } j\in \{2,\dots ,k-1\}\right\} \end{aligned}$$

  By continuity, because $\rho \in B_j$ for all $j\in \{2,\dots ,k-1\}$, there exists $\underline{\tau }\gg 0$ small enough in every dimension to have $\rho ^{\underline{\tau }}\in B_j$ for all $j\in \{2,\dots ,k-1\}$. Thus T is non-empty.

• Now let

  $$\begin{aligned} \tau ^* {:=}\arg \sup _{\tau \in T}~ \sum _{j\in \{2,\dots ,k-1\}} \tau _j. \end{aligned}$$

  By continuity and the finiteness of $\{2,\dots ,k-1\}, \cap _{j\in \{2,\dots ,k-1\}} (B_j \cup E_j)$ is closed, hence $\rho ^{\tau ^*} \in \cap _{j\in \{2,\dots ,k-1\}} (B_j \cup E_j)$. In fact, as we show below, we must have $\rho ^{\tau ^*} \in E_j$ for all $j \in \{2,\dots ,k-1\}$. Then setting $\gamma ^{n+1} = \rho ^{\tau ^*}$ completes the argument.

• We show that $\rho ^{\tau ^*} \in E_j$ for all $j \in \{2,\dots ,k-1\}$ by contradiction. Assume $\rho ^{\tau ^*} \in B_j$ for some $j \in \{2,\dots ,k-1\}$. Then by continuity, there exists $\tilde{\tau }_j$ (small enough) such that, starting from $\rho ^{\tau ^*}$, if we take $\tilde{\tau }_j$ away from j’s resource and distribute it in any way among $\{1,\dots ,j-1,j+1,\dots ,k-1\}$, the resulting allocation remains in $B_j$. By $SSM^{k-2}$ and no altruism, we can choose a redistribution $(\tilde{\tau }_1,\dots ,\tilde{\tau }_{j-1},\tilde{\tau }_{j+1},\dots ,\tilde{\tau }_{k-1})$ of $\tilde{\tau }_j$ that makes every agent in $\{1,\dots ,j-1,j+1,\dots ,k-1\}$ strictly better off at $\rho ^{\tilde{\tau }}$ than at $\rho ^{\tau ^*}$. Because $\rho ^{\tau ^*} \cap _{j\in \{2,\dots ,k-1\}} (B_j \cup E_j)$, we get $\rho ^{\tilde{\tau }} \in \cap _{j\in \{2,\dots ,k-1\}} B_j$. Notice also that

  $$\begin{aligned} \sum _{j\in \{2,\dots ,k-1\}} \tilde{\tau }_j \ge \sum _{j\in \{2,\dots ,k-1\}} \tau _j^*. \end{aligned}$$

  (4)

  Then by the above argument, we can take away some more resources ${\underline{\tilde{\tau }}}\gg 0$ from all agent in $\{2,\dots ,k-1\}$ and redistribute them to agent 1 while remaining in $\cap _{j\in \{2,\dots ,k-1\}} B_j$ (for ${\underline{\tau }^*}\gg 0$ small enough). But this means we just found some $\hat{\tau } \in T$ such that

  $$\begin{aligned} \sum _{j\in \{2,\dots ,k-1\}} \hat{\tau }_j= & {} \sum _{j\in \{2,\dots ,k-1\}} \tilde{\tau }_j + \sum _{j\in \{2,\dots ,k-1\}} \tilde{\underline{\tau }}_j> & {} \sum _{j\in \{2,\dots ,k-1\}} \tilde{\tau }_j, \end{aligned}$$

  (5)

  which combined with (4) contradicts the fact that $\tau ^*$ was the $\arg \sup _{\tau \in T} \sum _{j\in \{2,\dots ,k-1\}} \tau _j$.

This completes the proof of $SI^{k-1}$.

Second we show $SSM^{k}$ holds. Take any $N' \subseteq N$ with $\vert N'\vert =k$ and any $j \in N'$. By $SI^{k-1}$, there exists $\sigma ' \in \cap _{i\in N'\backslash \{j\}} I_i \cap \varSigma ^{N'}$ with $\sigma _j'>0$. Since $B_j \cap \varSigma ^{N'}$ is open in $\varSigma ^{N'}$, there exists a ball $b \in \varSigma ^{N'}$ centered in $\sigma '$ such that for all $\sigma \in b$ we have $\sigma \in B_j$. Take a fraction of resource from $\sigma _j'$ sufficiently small such that all its distributions among agents in $N'\backslash \{j\}$ leave j in $B_j$. Again, by no altruism and $SSM^{k-1}$, there exists $\sigma ''\in b$ such that $\sigma '' \in \cap _{i \in N'}B_i$. This proves the theorem.

1.1.2 Social monotonicity implies no resource destruction unanimity on ${\mathcal {R}}^{WBE}$

The proof is by contradiction. Take any $z_N^{00} \in Z$. Assume that there exists $z_N^{0} \in Z$ such that

$$\begin{aligned}&z_N^{0}~R_i~z_N^{00}, \quad \text {for all } i \in N ,\end{aligned}$$

(6)

$$\begin{aligned}&z_j^{0} < z_j^{00} , \quad \text {for all } j \in S \subseteq N ,\end{aligned}$$

(7)

$$\begin{aligned}&z_k^{00}=z_k^{0} , \quad \text {for all } k \in N{\backslash S} , \end{aligned}$$

(8)

so that no resource destruction unanimity is violated. We show that if social monotonicity holds, the existence of such $z_N^{0}$ implies a contradiction .

The intuition of the proof relies on the idea that by social monotonicity, starting from $z_N^0$, we can start redistributing additional resources in arbitrarily small increments while making everyone better off. If we do so in the appropriate way and given the contradiction assumption (6), we can reach an allocation $z_N^{\pi }$ in which for some $S_{n^*} \subset S$,

$$\begin{aligned}&z_N^{\pi }~P_j~z_N^{00}, \quad \text {for all } j\in S_{n^*}, \end{aligned}$$

(9)

$$\begin{aligned}&z_j^{00}~I_j^{int}~z_j^\pi , \quad \text {for all } j \in S_{n^*},\nonumber \\ \text {and}\quad&z_k^\pi ~P_k^{int}~z_k^{00}, \quad \text {for all } k \in N\backslash S_{n^*}. \end{aligned}$$

(10)

Then by no altruism and well-being externality we find a series of allocations which eventually bring agents in $N\backslash S_{n^*}$ back to the bundle they had in $z_N^{00}$, while preserving (9). By well-being externality again, we can do the same with agents in $S_{n^*}$, which means everyone is back at $z_N^{00}$. But this induces a contradiction because we then have $z_N^{00}~P_j~z_N^{00}$ for all $j\in S_{n^*}$.

By social monotonicity, for any $r >0$, there exists some $(w_1,\dots , w_n) \in Z_+$ such that $\sum _{i \in N} w_i = (r,\dots ,r)$ and $(z_N^{0}+w)~P_i~z_N^{0}$ for all $i \in N$. Now consider Fig. 6. By a standard argument (see for instance Mas-Collel et al. 1995, Proposition 3.C.1), because preferences satisfy strict monotonicity in own consumption and continuity, there exists $z_N^{\nearrow }$ such that for all $i \in N, z_i^{\nearrow }~I_i^{int}~(z^{0}_i + w_i)$ and $z_i^{\nearrow }= z^{0}_i + (\gamma _i,\dots ,\gamma _i)$ for some $\gamma _i \in {\mathbb {R}}$. Because $w_i > 0$ by assumption and preferences satisfy strict monotonicity in own consumption, we have $\gamma _i > 0$, for all $i\in N$.

It will be convenient to use the $\gamma _i$ constructed above to define an internal utility function for agent i. For all $z_N \in Z$ and for all $i\in N$, let

$$\begin{aligned} m_i ({z}_i) {:=}{\gamma }_i. \end{aligned}$$

This internal utility function is defined for the particular reference allocation $z^0$. Notice that because preferences satisfy separability, $m_i$ is indeed a utility representation of i’s internal preferences. Also $m_i$ is continuous (again, see Mas-Collel et al. 1995, Proposition 3.C.1).

Fig. 6

An internal utility function starting from $z^0$

The next lemma shows that starting from any allocation, if we distribute $(r,\dots ,r)$ additional resources following social monotonicity, there is an upper-bound to the amount of extra internal utility that can be obtained, and that this upper-bound is strictly decreasing in r.

To state the lemma, we need some additional notation. For any $z_N \in Z$, let $m_N(z_N) {:=}(m_1(z_1), \dots ,m_n(z_n) )$. Also, for any $r>0$ , let $M(r,z_N)$ be the set of internal utility vectors $m_N {:=}(m_1, \dots , m_n)$ which can be obtained from $z_N$ by distributing $(r,\dots ,r)$ and making everyone strictly better off than at $z_N$. Formally

$$\begin{aligned} \begin{aligned} M(r,z_N)&{:=}\left\{ m_N \in {\mathbb {R}}^n| m_N = m_N(\hat{z}_N) \text { for some } \hat{z}_N \in Z \text { such that}\right. \\&\qquad \; \hat{z}_N = (z_N+w), \text { for some } w \in Z_+ \text { with } \sum _{i \in N} w_i = (r,\dots ,r),\\&\qquad \;\left. \text {and for all } i\in N,~ \hat{z}_N ~P_i~z_N \right\} . \end{aligned} \end{aligned}$$

(11)

By social monotonicity, for any $r>0$ and any $z_N \in Z$, the set M is non-empty.

Lemma 1

(Upper-bound to extra utility strictly decreasing in r) For any $z_N \in Z$, any $r>0$, and any $m_N \in {M}(r,z_N), m_N \le \left( m_N(z_N) + (r,\dots ,r) \right)$.

Proof

By definition of $M(r,z_N)$ for all $m_N \in M(r,z_N)$, there exists some $\hat{z}_N \in Z$ such that $m_N = m_N(\hat{z}_N)$ and

$$\begin{aligned} \hat{z}_i \le z_i + (r,\dots ,r), \quad \text {for all } i\in N \end{aligned}$$

(12)

Now assume there exists $\tilde{m}_N \in M(r,z_N)$ such that for some $j\in N, \tilde{m}_j > \left( m_j(z_j) + r \right)$. This implies that $\hat{z}_j~P^{int}_j~\big (z_j + (r,\dots ,r)\big )$. But by strict monotonicity in own consumption, this means that there is some good h such that $\hat{z}_{jh} > z_{jh} + r$, contradicting (12). $\square$

The next lemma shows that for every allocation $z_N$ in which every agent $i\in S$ has strictly lower internal utility than at $z_N^{00}$, there exists another allocation $z_N'$ which everyone prefers to $z_N$, in which all agents have strictly higher internal utility than at $z_N$, but in which agents $i\in S$ still have strictly lower internal utility than at $z_N^{00}$.

From now on, we will denote $z_S {:=}(z_i)_{i\in S}$ the allocation $z_N$ restricted to the subset of agents in S. Similarly $m_S(z_S) {:=}(m_i(z_i))_{i\in S}$ and $m_S {:=}(m_i)_{i\in S}$. Also let $m_N^{00} {:=}m_N(z_N^{00})$.

Lemma 2

(Internal utility increasing but still lower than $m_N^{00}$) Take any $z_N\in Z$ such that $m_S(z_S) \ll m_S^{00}$. There exists $z_N' \in Z$ such that

$$\begin{aligned}&z_N'~P_i~z_N, \quad \text {for all } i\in N, \end{aligned}$$

(13)

$$\begin{aligned}&m_i(z_i) < m_i(z_i'), \quad \text {for all } i\in N, \end{aligned}$$

(14)

$$\begin{aligned} \text {and}\quad&m_j(z_j') < m_j^{00}, \quad \text {for all } j\in S, \end{aligned}$$

(15)

Proof

Let

$$\begin{aligned} r^* {:=}\frac{\min _{s\in S} (m_s^{00} - m_s(z_s))}{2}, \end{aligned}$$

and let $m_n' {:=}m_N(z_N')$ be any element of $M(r^*,z_N)$. This construction is illustrated in Fig. 7 for the case $z_N = z_N^0$ with two agents. By assumption $m_S(z_S) \ll m_S^{00}$, which implies $\min _{s\in S} \big (m_s^{00} - m_s(z_s)\big ) > 0$ and $r^* > 0$, so that $M(r^*,z_N)$ is non-empty.

By definition of $M(r^*,z_N)$ and because preferences satisfy strict monotonicity in own consumption, there exists some $z_N'$ constructed as in (11) such that conditions (13) and (14) are satisfied. There only remains to show that at $m_n'$ condition (15) is satisfied.

Because $m_N' \in M(r^*,z_N)$ we can apply Lemma 1 to get

$$\begin{aligned} m_N '&\le m_N(z_N) + (r^*,\dots ,r^*), \\ m_S '&\le m_S(z_S) + (r^*,\dots ,r^*), \\ m_S'&\le m_S(z_S) + \left( \frac{\min _{s\in S} \big (m_s^{00} - m_s(z_s)\big )}{2},\dots ,\frac{\min _{s\in S} \big (m_s^{00} - m_s(z_s)\big )}{2}\right) \\&\qquad [\text {by construction of }r^*],\\ m_S' - m_S(z_S)&\le \frac{m_S^{00} - m_S(z_S)}{2}\\ m_S' - m_S(z_S)&\ll m_S^{00} - m_S(z_S) \quad [\text {by }m_S^{00} \gg m_S(z_S)],\\ m_S'&\ll m_S^{00}, \end{aligned}$$

the desired result. $\square$

Fig. 7

Choosing an allocation to satisfy Lemma 2

We are now ready to prove the general result.

By construction of $z_N^0$ and given that preferences satisfy strict monotonicity in own consumption, $m_S(z_S^0) \ll m_S(z_S^{00})$. So starting from $z_N^0$, Lemma 2 applies and we can obtain $z_N^*$ satisfying the conditions (13)–(15) with respect to $z_N^0$. In particular

$$\begin{aligned} z_N^*~P_i~z_N^0, \quad \text {for all } i\in N, \end{aligned}$$

(16)

Notice that $z_N^*$ also satisfies the conditions of Lemma 2. Thus we can apply the lemma again to $z_N^*$ and get a nonempty set of allocations $z_N$ such that

$$\begin{aligned}&z_N~R_i~z_N^*, \quad \text {for all } i\in N, \end{aligned}$$

(17)

$$\begin{aligned}&m_i(z_i^*) \le m_i(z_i), \quad \text {for all } i\in N, \nonumber \\ \text {and}\quad&m_j(z_j) \le m_j^{00}, \quad \text {for all } j\in S, \end{aligned}$$

(18)

were we voluntarily turned strict inequalities and preference relations into weak ones. In particular, the last two inequalities imply

$$\begin{aligned} 0 \le m_j^{00} - m_j(z_j) \le m_j^{00} - m_j(z_j^*), \quad \text {for all } j\in S, \end{aligned}$$

(19)

which in turn implies

$$\begin{aligned} 0 \le \max _{j\in S} (m_j^{00} - m_j(z_j)) \le \max _{j\in S} (m_j^{00} - m_j(z_j^*)) . \end{aligned}$$

(20)

Next we define C, the set of $\max _{j\in S} (m_j^{00} - m_j(z_j)) \in [0, \max _{j\in S} (m_j^{00} - m_j(z_j^*)) ]$ which can be obtained via some $z_N$ satisfying conditions (17)–(19). Possibles $z_N$ are illustrated in Fig. 8 for the case of three agents. Formally

$$\begin{aligned} C {:=}&\left\{ t \in \left[ 0, \max _{j\in S} (m_j^{00} - m_j(z_j^*)) \right] |\right. \\&\quad t = \max _{j\in S} (m_j^{00} - m_j(z_j)) \text { for some } z_N \in Z \text { satisfying }\\&\quad z_N~R_j~z_N^*, \quad \text {for all } i \in N , \\&\quad m_i(z_i^*) \le m_i(z_i), \quad \text {for all } i\in N, \\ \text { and}&\quad \left. 0 \le m_j^{00} - m_j(z_j) \le m_j^{00} - m_j(z_j^*),\quad \text {for all } j \in S \right\} . \end{aligned}$$

The rest of the argument follows the intuition we gave at the beginning of the proof, using the set C to construct the appropriate allocations. We will be interested in $c {:=}\inf {C}$. We will consider two cases. In case 1, $c=0$. We will show that given the assumptions on our domain, this means that all agents in S are back to their initial internal utility level $m_S^{00}$, which will lead to a contradiction.

Fig. 8

Possible allocations associated with $t \in C$, where $S=\{1,2\}$

In case two, $c>0$, and there will be two subcases. In the subcase 1, we will assume $(m_j^{00} - m_j(z_j^{**})) > 0$ for all $j\in S$ in which case we will be able to apply Lemma 2 again and show that c could not have been the infimum of C. In subcase 2, we will assume that $(m_j^{00} - m_j(z_j^{**})) > 0$ for some subset of $S_1\subset S$ only. Then we will be able to repeat the whole argument several times up to the point where $S_{n^*}$ brings us back to case 1.

In order to solve those two cases, we will need to associate c with an allocation that everyone weakly prefers to $z_N^*$. Because c is the infimum of C, and C might not be closed, we have no guaranteed that $c\in C$, and such allocation might not exist. However, we will be able to construct it under the assumptions on our domain. The next part of the proof describes this construction.

By (17)–(19), C is nonempty. Because C is also bounded, c is well-defined. Because c is the infimum of C, there is a sequence $\{c^g\}_{g=1}^{\infty }$ such that for all $g \in {\mathbb {N}}, c^g \in C$, and $c^g \rightarrow c$. By definition of C this means that there is a corresponding sequence $\{\tilde{z}_N^g\}_{g=1}^{\infty }$ such that for all $g \in {\mathbb {N}}$,

$$\begin{aligned}&c^g = \max _{j\in S} (m_j^{00} - m_j(\tilde{z}_j^g)), \end{aligned}$$

(21)

$$\begin{aligned}&\tilde{z}_N^g~R_j~z_N^*, \quad \text {for all } i \in N , \end{aligned}$$

(22)

$$\begin{aligned}&m_i(z_i^*) \le m_i(\tilde{z}_i^g), \quad \text {for all } i\in N, \end{aligned}$$

(23)

$$\begin{aligned} \text { and}\quad&0 \le m_j^{00} - m_j(\tilde{z}_j^g) \le m_j^{00} - m_j(z_j^*),\quad \text {for all } j \in S. \end{aligned}$$

(24)

The next lemma shows that $\{\tilde{z}_N^g\}_{g=1}^{\infty }$ can be turned into a bounded sequence having similar properties for $j\in N$. By the Bolzano–Weierstrass theorem [Rudin 1976, Theorem 3.6(b)], we will then be able to find a converging subsequence which will allow us to apply continuity.

Lemma 3

($\{\tilde{z}_N^g\}_{g=1}^{\infty }$can be turned into an equivalent bounded sequence) Based on $\{\tilde{z}_N^g\}_{g=1}^{\infty }$, we can construct yet another sequence $\{\hat{z}_N^g\}_{g=1}^{\infty }$, such that for all $g\in {\mathbb {N}}$,

$$\begin{aligned}&\hat{z}_N^g~R_j~\tilde{z}_N^g, \quad \text {for all } j\in S,\end{aligned}$$

(25)

$$\begin{aligned}&m_j(\hat{z}_j^g)~=~m_j(\tilde{z}_j^g), \quad \text {for all } j\in S,\end{aligned}$$

(26)

$$\begin{aligned} \text {and}\quad&\{\hat{z}_N^g\}_{g=1}^{\infty } \hbox {is bounded} . \end{aligned}$$

(27)

Proof

The construction goes as follows.

$$\begin{aligned} \hat{z}^g_j&{:=}z_j^0 + (m_j(\tilde{z}^g_j),\dots , m_j(\tilde{z}^g_j)), \quad \text {for all } j\in S,\end{aligned}$$

(28)

$$\begin{aligned} \hat{z}^g_k&{:=}z^{00}_k, \quad \text {for all } k\in N\backslash S. \end{aligned}$$

(29)

By construction of $\hat{z}_N$ and by definition of $m_j$, (26) is immediate. Then by (24) it follows directly that $\text {for all } j \in S$ and for all $g\in {\mathbb {N}}$,

$$\begin{aligned} 0&\le m_j^{00} - m_j(\tilde{z}_j^g) \le m_j^{00} - m_j(z_j^*),\\ 0&\le m_j(z_j^*) \le m_j(\tilde{z}_j^g). \end{aligned}$$

But then by construction $\{\hat{z}_S^g\}_{g=1}^{\infty }$ is bounded. Then, because $\{\hat{z}_{N\backslash S}^g\}_{g=1}^{\infty }$ is constant, $\{\hat{z}_N^g\}_{g=1}^{\infty }$ is bounded.

We now prove (25). By construction we have

$$\begin{aligned} \hat{z}_j^g~I_j^{int}~\tilde{z}_j^g, \quad \text {for all } j\in S. \end{aligned}$$

(30)

By separability, this implies

$$\begin{aligned} (\hat{z}^g_S,\tilde{z}^g_{N\backslash S})~I_j~ \tilde{z}^g_{N}, \quad \text {for all } j\in S, \end{aligned}$$

(31)

Now by (23), given that $z_N^*$ satisfies (15) with respect to $z_N^0$ and that for all $k \in N\backslash S, z_k^{0} = z_k^{00}$ by assumption, we have

$$\begin{aligned} m_k (\tilde{z}_k^g) > m_k^{00}, \quad \text {for all } k \in N\backslash S. \end{aligned}$$

(32)

Therefore by no altruism and well-being externality

$$\begin{aligned} (\hat{z}^g_S,\hat{z}^g_{N\backslash S})~R_j~ (\hat{z}^g_S,\tilde{z}^g_{N\backslash S}), \quad \text {for all } j\in S, \end{aligned}$$

(33)

which by transitivity implies

$$\begin{aligned} (\hat{z}^g_S,\hat{z}^g_{N\backslash S})~R_j~ \tilde{z}^g_{N}, \quad \text {for all } j\in S, \end{aligned}$$

(34)

the desired result. $\square$

Now by transitivity and (22), Lemma 3 implies that for all $g\in {\mathbb {N}}$,

$$\begin{aligned} \hat{z}^g_{N}~R_j~ z_N^*, \quad \text {for all } j\in S, \end{aligned}$$

(35)

where $\{\hat{z}_N^g\}_{g=1}^{\infty }$ is as constructed in the lemma. Because $\{\hat{z}_{N\backslash S}^g\}_{g=1}^{\infty }$ is bounded, by the Bolzano-Weierstrass theorem, the sequence has a converging subsequence, say $\{\hat{z}_{N\backslash S}^{t(g)}\}_{g=1}^{\infty }$. Let $z_N^{**}$ be the limit of $\{\tilde{z}_{N\backslash S}^{t(g)}\}_{g=1}^{\infty }$. By continuity,

$$\begin{aligned} {z}_N^{**}~R_j~ z_N^*, \quad \text {for all } j\in S. \end{aligned}$$

(36)

By transitivity, (16), and given the contradiction assumption (6), we then have

$$\begin{aligned} {z}_N^{**}~P_j~ z_N^{00}, \quad \text {for all } j\in S. \end{aligned}$$

(37)

By (24) and (26) we have that for all $g\in {\mathbb {N}}$

$$\begin{aligned} 0 \le m_j^{00} - m_j(\hat{z}_j^{t(g)}) , \quad \text {for all } j \in S. \end{aligned}$$

(38)

Thus by continuity of $m_j$ we get

$$\begin{aligned} 0 \le m_j^{00} - m_j({z}_j^{**}) , \quad \text {for all } j \in S. \end{aligned}$$

(39)

Also, by (21) and (26), for all $g\in {\mathbb {N}}$,

$$\begin{aligned} c^{t(g)} = \max _{j\in S} (m_j^{00} - m_j(\tilde{z}_j^{t(g)})) = \max _{j\in S} (m_j^{00} - m_j(\hat{z}_j^{t(g)})). \end{aligned}$$

So by continuity of m

$$\begin{aligned} c = \max _{j\in S} (m_j^{00} - m_j({z}_j^{**})), \end{aligned}$$

Remember that by definition of C, we have $c \in [0, \max _{j\in S} (m_j^{00} - m_j(z_j^*)) ]$. Finally notice that because $\hat{z}_k^{t(g)} = z_k^{00}$ for all $k \in N\backslash S$ and every $g\in {\mathbb {N}}$, we also have

$$\begin{aligned} z_k^{**} = z_k^{00}, \quad \text {for all } k \in N\backslash S. \end{aligned}$$

(40)

We are now ready to study the two cases mentioned above.

• Case 1 $\max _{j\in S} (m_j^{00} - m_j(z_j^{**})) = 0$. This is equivalent to

  $$\begin{aligned} 0 \ge m_j^{00} - m_j({z}_j^{**}) , \quad \text {for all } j \in S. \end{aligned}$$

  (41)

  By (39) and (41) we have

  $$\begin{aligned} 0 = m_j^{00} - m_j({z}_j^{**}) , \quad \text {for all } j \in S. \end{aligned}$$

  (42)

  By well-being externality, this means

  $$\begin{aligned} ({z}_S^{00}, z_{N\backslash S}^{**})~I_j~ z_N^{**}, \quad \text {for all } j\in S. \end{aligned}$$

  (43)

  But by (40),

  $$\begin{aligned} ({z}_S^{00}, z_{N\backslash S}^{**}) = ({z}_S^{00}, z_{N\backslash S}^{00}) = z_N^{00}. \end{aligned}$$

  (44)

  Thus (43) can be rewritten as

  $$\begin{aligned} z_N^{00}~I_j~ z_N^{**} , \quad \text {for all } j\in S \end{aligned}$$

  (45)

  contradicting (37).

• Case 2 $\max _{j\in S} (m_j^{00} - m_j(z_j^{**})) > 0$. There are two subcases.

  • Subcase 1: $(m_j^{00} - m_j(z_j^{**})) > 0$ for all $j\in S$. Then notice that the assumptions of Lemma 2 hold at $z_N^{**}$. So we can apply the lemma once again and obtain an allocation $z_N^{***}$ which is associated with some $r^{***} \in C$ such that $r^{***} < c$, contradicting the fact that c is the infimum of C.

  • Subcase 2: there exists a nonempty $\tilde{S}_1 \subset S$ with $(m_j^{00} - m_j(z_j^{**})) = 0$ for all $j \in \tilde{S}_1$. Slightly abusing the notation, let $\tilde{S}_1$ be the largest such set. Then for any $j\in S^1 {:=}S\backslash \tilde{S}_1, (m_j^{00} - m_j(z_j^{**})) > 0$. Thus we can repeat the former steps.

    By well-being externality, we have

    $$\begin{aligned} (z_{S_1}^{**}, {z}_{\tilde{S}_1}^{00}, z_{N\backslash S}^{**})~I_h~ z_N^{**} , \quad \text {for all } h\in S. \end{aligned}$$

    (46)

    Thus by (32), well-being externality, and no altruism we have

    $$\begin{aligned} (z_{S_1}^{**}, {z}_{\tilde{S}_1}^{00}, z_{N\backslash S}^{00})~R_h~ (z_{S_1}^{**}, {z}_{\tilde{S}_1}^{00}, z_{N\backslash S}^{**}) , \quad \text {for all } h\in S_1 \end{aligned}$$

    (47)

    which by transitivity yields

    $$\begin{aligned} (z_{S_1}^{**},z^{00}_{N\backslash S_1})~R_h~ z_N^{**} , \quad \text {for all } h\in S_1, \end{aligned}$$

    (48)

    and

    $$\begin{aligned} (z_{S_1}^{**},z^{00}_{N\backslash S_1})~P_h~ z_N^{00} , \quad \text {for all } h\in S_1. \end{aligned}$$

    (49)

    Notice that this brings us back to an allocation $(z_{S_1}^{**},z^{00}_{N\backslash S_1})$ very similar to $z_N^0$, except that the relevant set of agents is now $S_1 \subset S$ instead of S. Starting from $(z_{S_1}^{**},z^{00}_{N\backslash S_1})$ we can repeat the whole argument as many times as we want. Every time we do so, we get smaller and smaller sets $S_n \subset \dots \subset S_1 \subset S$.

    Because N is finite, either we reach subcase 1 directly for some $S_{n^*}$ and get a contradiction, or there is some $S_{n^{**}}$ with a single agent. But if $S_{n^{**}}$ contains a single agent, we again reach Case 1 and get a contradiction. Hence we are done.

1.2 Proof of Proposition 2

On ${\mathcal {R}}^{WBE-NRDU}$, the social ordering function $\varvec{R}^{\varOmega lex}$ satisfies Equal-split transfer and strong Pareto.

1.2.1 Equal-split transfer

Take any $z_N',z_N \in Z$ s.t. $z_N'~\varvec{R}(R_N)~z_N$ by virtue of Equal-split transfer. By definition of Equal-split transfer we have $z_j\gg z_j'\gg \frac{\varOmega }{|N|} \gg z_k' \gg z_k$. By strict monotonicity in own consumption, this implies

$$\begin{aligned}&z_j~P^{int}_j~z_j'~P^{int}_j~\frac{\varOmega }{|N|}~P^{int}_j~z_k'~P^{int}_j~z_k,\quad \text {and}\\&z_j~P^{int}_k~z_j'~P^{int}_k~\frac{\varOmega }{|N|}~P^{int}_k~z_k'~P^{int}_k~z_k, \end{aligned}$$

which in turn means that

$$\begin{aligned} u_j^{\varOmega int}(z_j,R_j)>u_j^{\varOmega int}(z_j',R_j)>\frac{1}{|N|}>u_k^{\varOmega int}(z_k',R_k)>u_k^{\varOmega int}(z_k,R_k). \end{aligned}$$

As the $u_i^{\varOmega int}$ depend only on the internal preferences, the value of $u_i^{\varOmega int}$ is equal in $z_N$ and $z_N'$ for all $i\ne j,k$. Hence $z_N'~\varvec{R}^{\varOmega lex}(R_N)~z_N$.

1.2.2 Strong Pareto

$\varvec{R}^{\varOmega lex}$ satisfies strong Pareto INT. By Proposition 3, on ${\mathcal {R}}^{WBE-NRDU}$ strong Pareto INT implies strong Pareto. So $\varvec{R}^{\varOmega lex}$ satisfies strong Pareto.

1.3 Proof of Proposition 6

Let the economy be composed of two goods $z_1,z_2$ and four agents, $g,h,j,k \in N$. Agents j and k share the same internal preferences represented by the internal utility function $m=m_j=m_k$ and so do agents g and h: $m'=m_g=m_h$. Their preferences are represented by the following global utility functions:

$$\begin{aligned}&U_{j}(z_N) = m(z_j) - \frac{8}{11} m(z_h),\\&U_{k}(z_N) = m(z_k) - \frac{8}{11} m(z_g),\\&U_{g}(z_N) = m'(z_{g}),\\&U_{h}(z_N) = m'(z_{h}). \end{aligned}$$

This preference profile respects both separability and no resource destruction unanimity. Suppose the equal split bundle is (5, 5) and consider the following serie of allocations represented in Fig. 9 (the level of internal utility for these allocations are as represented in the figure):

$$\begin{aligned}&z_N^1 = (\underbrace{(9,9)}_{j},\underbrace{(1,1)}_{k},\underbrace{(1,9)}_{h},\underbrace{(9,1)}_{g}), \\&z_N^2 = (\underbrace{(8,8)}_{j},\underbrace{(2,2)}_{k},\underbrace{(1,9)}_{h},\underbrace{(9,1)}_{g}), \\&z_N^3 = (\underbrace{(3,3)}_{j},\underbrace{(7,7)}_{k},\underbrace{(9,1)}_{h},\underbrace{(1,9)}_{g}), \\&z_N^4 = (\underbrace{(4,4)}_{j},\underbrace{(6,6)}_{k},\underbrace{(9,1)}_{h},\underbrace{(1,9)}_{g}). \end{aligned}$$

This profile of preferences violates well-being externality, and induces the following cycle showing that strong Pareto and equal-split transfer are not compatible.

• $z_N^2~R(R_N)~z_N^1$ by equal-split transfer;

• $z_N^3~I(R_N)~z_N^2$ by strong Pareto;

• $z_N^4~R(R_N)~z_N^3$ by equal-split transfer;

• $z_N^1~P(R_N)~z_N^4$ by strong Pareto, since agent k strictly prefers $z_N^1$.

Fig. 9

On ${\mathcal {R}}^{NRDU-OPE}$, no SOF satisfy strong Pareto and equal-split transfer

1.4 Proof of Proposition 7

Let the economy be composed of two goods $z_1,z_2$ and four agents, g, h, j and k. Consider the following internal utility function,

$$\begin{aligned} m(z_i) = z_{i1}+ z_{i2}, \end{aligned}$$

(50)

where $z_{im}$ is the quantity of the m-th good in i’s bundle. Suppose the four agents have the following preferences.

$$\begin{aligned}&U_{j}(z_N) = m(z_{j}) - \beta _{j} \sum _{s \ne j \in N} \frac{m(z_s)}{m(z_s) + 1} ,\\&U_{k}(z_N) = m(z_{k}) - \beta _{k} \sum _{s \ne k \in N} \frac{m(z_s)}{m(z_s) + 1} ,\\&U_{g}(z_N) = m(z_{g}),\\&U_{h}(z_N) = m(z_{h}), \end{aligned}$$

where $\beta _{j}, \beta _{k} \ge 0$. This profile satisfies well-being externality but not necessarily no resource destruction unanimity. As we show hereafter, for $\beta _{j}$ and $\beta _{k}$ sufficiently large, there exist allocations in which j and k would agree together to destroy part of their resources. Assume the equal split bundle is (3, 3) and consider the two following allocations

$$\begin{aligned}&z_N = (\underbrace{(1,0)}_{j},\underbrace{(0,1)}_{k},\underbrace{(6,5)}_{h},\underbrace{(5,6)}_{g}), \\&z_N' = (\underbrace{(2,\epsilon )}_{j},\underbrace{(\epsilon ,2)}_{k},\underbrace{(5,5-\epsilon )}_{h},\underbrace{(5-\epsilon ,5)}_{g}), \end{aligned}$$

for some $\epsilon >0$ arbitrarily small.¹⁸ Applying equal-split transfer twice, we have $z_N'~\varvec{R}(R_N)~z_N$. But we also have

$$\begin{aligned}&U_{j} (z_N') - U_{j}(z_N) \approx 1 - \beta _{j} \underbrace{\left[ \left( \frac{2}{3} + \frac{10}{11} + \frac{10}{11}\right) - \left( \frac{1}{2} + \frac{11}{12} + \frac{11}{12} \right) \right] }_{= t > 0}, \\&U_{k} (z_N') - U_{k}(z_N) \approx 1 - \beta _{k} \underbrace{\left[ \left( \frac{2}{3} + \frac{10}{11} + \frac{10}{11}\right) - \left( \frac{1}{2} + \frac{11}{12} + \frac{11}{12} \right) \right] }_{= t > 0}, \end{aligned}$$

where the approximation is arbitrarily accurate as $\epsilon$ tends to zero. So for $\beta _{j},\beta _{k} > \frac{1}{t}, z_N~P_{j}~z_N'$ and $z_N~P_{k}~z_N'$. As $z_N~P_{g}~z_N'$ and $z_N~P_{h}~z_N'$, we have $z_N~\varvec{P}(R_N)~z_N'$ by strong Pareto, a contradiction.

1.5 Proof of Proposition 8

To prove Proposition 8, we first prove that Reference distribution $\varOmega$-equivalent utility is a utility representation of the preferences.

Lemma 4

For any $i\in N, u_i^{RD}$ is a utility representation of $R_i$.

Proof

Take any $z_N,z_N' \in Z$ such that $z_N'~R_i~z_N$.

By definition of $u_i^{RD}, \left( \lambda _i'\frac{\varOmega }{|N|},\frac{1}{\lambda _i'} \frac{\varOmega }{|N|},\dots , \frac{1}{\lambda _i'}\frac{\varOmega }{|N|}\right) ~R_i~\left( \lambda _i\frac{\varOmega }{|N|},\frac{1}{\lambda _i}\frac{\varOmega }{|N|},\dots ,\frac{1}{\lambda _i}\frac{\varOmega }{|N|}\right)$. By strict monotonicity in own consumption and no altruism this implies that $\lambda _i' \ge \lambda _i$, and hence $u_i^{RD}(z_N',R_N) \ge u_i^{RD}(z_N,R_N)$. $\square$

1.5.1 Strong Pareto

This is a direct consequence of Lemma 4 and the definition of the leximin operator.

1.5.2 Neutral equal-split redistribution

Let $z_N'~\varvec{R}(R_N)~z_N$ by virtue of neutral equal-split redistribution. By definition we have $z_N~P_j~z_N'$ and $z_N'~P_k~z_N$ so that $u_j^{RD}(z_N,R_N)>u_j^{RD}(z_N',R_N)$ and $u_k^{RD}(z_N')>u_k^{RD}(z_N)$ by lemma 4.

Because of the neutral character of the axiom we have that $u_i^{RD}(z_N,R_N) = u_i^{RD}(z_N',R_N) \hbox {for all } i \ne j,k \in N$. Also $z_N'~P_j~(\frac{\varOmega }{|N|}, \dots ,{\tiny } \frac{\varOmega }{|N|}) \Rightarrow u_j^{RD}(z_N',R_N) > {1}$ by strict monotonicity in own consumption and no altruism. On the other hand, $(\frac{\varOmega }{|N|}, \dots \frac{\varOmega }{|N|})~P_k~z_N' \Rightarrow u_k^{RD}(z_N') < {1}$ for the same reason. So $u_j^{RD}(z_N',R_N)>u_j^{RD}(z_N,R_N)>{1}>u_k^{RD}(z_N)>u_k^{RD}(z_N')$, the desired result.

1.6 Proof of Proposition 9

The proof is by contradiction. Assume there exists $\varvec{R}$ satisfying strong Pareto, independence of other-regarding features.

1.6.1 Neutral equal-split redistribution

Consider a profile with two agents $j,k\in N$ having the same preferences represented by the following utility functions:

$$\begin{aligned}&U_{j} (z_N) = {\left\{ \begin{array}{ll} m(z_j) - m(z_k),\qquad &{} \text { if } m(z_j) < m(z_k)\\ m(z_j), &{} \text { if } m(z_j) \ge m(z_k) \end{array}\right. }\\&U_{k} (z_N) = {\left\{ \begin{array}{ll} m(z_k) - m(z_j),\qquad &{} \text { if } m(z_k) < m(z_j)\\ m(z_k), &{} \text { if } m(z_k) \ge m(z_j). \end{array}\right. } \end{aligned}$$

Notice that the induced profile satisfies no resource destruction unanimity. Let the values of the internal utility function $m_i$ at $z_N$ and $z_N'$ be as represented in Fig. 10, where $z_k' - z_k = z_j - z_j'$. We have $z_N'~\varvec{R}(R_N)~z_N$ by neutral equal-split redistribution. Observe that even if k is internally worse-off after the redistribution, j’s internal utility loss is sufficient for k’s global utility to increase. Also $z_N''~\varvec{P}(R_N)~z_N'$ by strong Pareto, so $z_N''~\varvec{P}(R_N)~z_N$ by transitivity. Now consider $R_N'$ where $j,k\in N$ have self-centered preferences

$$\begin{aligned}&u_{j}(z_N) = m (z_{j}), \\&u_{k}(z_N) = m (z_{k}). \end{aligned}$$

By strong Pareto, $z_N~\varvec{P}(R_N')~z_N''$, which contradicts independence of other-regarding features.

Fig. 10

On ${\mathcal {R}}^{WBE-NRDU}$, no SOF satisfies strong Pareto, independence of other-regarding features and any of neutral equal-split redistribution

1.7 Domain of profiles given in Example 2

The domain of profiles given in Example 2 belongs to ${\mathcal {R}}^{WBE-NRDU}$. It satisfies no altruism since $\alpha _i \ge 0$, separability because of the additively separable form of agents’ ORP and well-being externality as the utility functions is of the form $U_i(z_N)=U_i(m_i(z_i),m_j(z_j),m_k(z_k),\dots )$. There remains to prove that no resource destruction unanimity is satisfied.

Remember that the condition means that for any $z_N \in Z$, and any $w=(w_1,w_2,\dots ,w_N)$ with $w_i \in {\mathbb {R}}_+^l$ and $\sum _i w_i = \bar{w} >0$, we have $(z_N + w) ~P_j~z_N$, for some $j \in N$.

Let $\varGamma _i{:=}m_i(z_i+w_i)-m_i(z_i)$ be the internal well-being gain obtained by agent i from the distribution of w. We have that for all $i \in N, U_i(z_N+w)-U_i(z_N)=\varGamma _i-\frac{\alpha _i}{n-1} \sum _{j \ne i}\varGamma _j$. No resource destruction unanimity is violated if and only if we have $\varGamma _i \le \frac{\alpha _i}{n-1} \sum _{j \ne i}\varGamma _j$ for all $i \in N$. We show that the last inequality cannot hold by contradiction. Summing the n previous inequality yields

$$\begin{aligned} \sum _{i \in N} \varGamma _i \le \sum _{i \in N} \left[ \frac{\alpha _i}{n-1} \sum _{j \ne i}\varGamma _j \right] . \end{aligned}$$

By expending the sum on the right-hand side of this inequality, on can see that

$$\begin{aligned} \sum _{i \in N} \left[ \frac{\alpha _i}{n-1} \sum _{j \ne i}\varGamma _j \right] = \sum _{i \in N} \left[ \varGamma _i \sum _{j \ne i} \frac{\alpha _j}{n-1} \right] , \end{aligned}$$

Let us denote the term in the parenthesis of the right-hand side $\widetilde{\varGamma _i}{:=}\varGamma _i\sum _{j \ne i} \frac{\alpha _j}{n-1}$. Since by assumption $\alpha _i < 1$ for all $i \in N$, we have $\widetilde{\varGamma _i} < \varGamma _i$. But then the inequality cannot hold, therefore our profile must respect no resource destruction unanimity.