Proportionality, equality, and duality in bankruptcy problems with nontransferable utility

Abstract

This paper studies bankruptcy problems with nontransferable utility as a generalization of bankruptcy problems with monetary estate and claims. Following the theory on TU-bankruptcy, we introduce a duality notion for NTU-bankruptcy rules and derive several axiomatic characterizations of the proportional rule and the constrained relative equal awards rule.

Appendix

Proof of Theorem 2

• (i) Let $(E,c)\in \mathrm{BR}^N$ and let $t\in (0,1)$. Then

  $$\begin{aligned} \mathrm{Prop}(tE,c)=\frac{1}{\lambda ^{tE,c}}c=t\frac{1}{\lambda ^{E,c}}c=t\mathrm{Prop}(E,c). \end{aligned}$$

  Hence, the proportional rule satisfies path linearity. Let f be a bankruptcy rule satisfying path linearity. Let $(E,c)\in \mathrm{BR}^N$. Then

  $$\begin{aligned} f(E,c)=f\left( \frac{1}{\lambda ^{E,c}}\lambda ^{E,c}E,c\right) =\frac{1}{\lambda ^{E,c}}f(\lambda ^{E,c}E,c)=\frac{1}{\lambda ^{E,c}}c=\mathrm{Prop}(E,c). \end{aligned}$$

  Hence, $f=\mathrm{Prop}$.

• (ii) By Theorem 1(i), the proportional rule satisfies self-duality. By Theorem 2(i), the proportional rule satisfies path composition down. Let f be a bankruptcy rule satisfying path composition down and self-duality. Let $(E,c)\in \mathrm{BR}^N$. By path continuity, there is a $t\in (0,\lambda ^{E,c})$ such that $\sum _{i\in N}f_i(tE,c)=\frac{1}{2}\sum _{i\in N}c_i$. By self-duality, $c-f(tE,c)=f(\lambda ^{E,c-f(tE,c)}E,c)$. This means that

  $$\begin{aligned} \sum _{i\in N}f_i\Big (\lambda ^{E,c-f(tE,c)}E,c\Big )=\sum _{i\in N}c_i-\sum _{i\in N}f_i(tE,c)=\frac{1}{2}\sum _{i\in N}c_i. \end{aligned}$$

  By path monotonicity, $t=\lambda ^{E,c-f(tE,c)}$ and $f(tE,c)=f(\lambda ^{E,c-f(tE,c)}E,c)$. This means that $t=\frac{1}{2}\lambda ^{E,c}$ and $f(tE,c)=\frac{1}{2}c$. Similarly, $f(\frac{1}{2}\lambda ^{E,\frac{1}{2}c}E,\frac{1}{2}c)=\frac{1}{4}c$. By path composition down,

  $$\begin{aligned} f\Big (\tfrac{1}{4}\lambda ^{E,c}E,c\Big )= & {} f\Big (\tfrac{1}{4}\lambda ^{E,c}E,f\Big (\tfrac{1}{2}\lambda ^{E,c}E,c\Big )\Big )=f\Big (\tfrac{1}{4}\lambda ^{E,c}E,\tfrac{1}{2}c\Big )\\= & {} f\Big (\tfrac{1}{2}\lambda ^{E,\tfrac{1}{2}c}E,\tfrac{1}{2}c\Big )=\tfrac{1}{4}c. \end{aligned}$$

  By self-duality,

  $$\begin{aligned} f\Big (\tfrac{3}{4}\lambda ^{E,c}E,c\Big )= & {} f\Big (\lambda ^{E,\tfrac{3}{4}c}E,c\Big )=f\left( \lambda ^{E,c-f\Big (\tfrac{1}{4}\lambda ^{E,c}E,c\Big )}E,c\right) \\= & {} c-f\Big (\tfrac{1}{4}\lambda ^{E,c}E,c\Big )=\tfrac{3}{4}c. \end{aligned}$$

  Continuing this reasoning, $f(\frac{m}{2^n}\lambda ^{E,c}E,c)=\frac{m}{2^n}c$ for all $m,n\in \mathbb {N}$ with $m\le 2^n$. By path continuity, $f(t\lambda ^{E,c}E,c)=tc$ for all $t\in (0,1)$. Hence, $f(E,c)=\mathrm{Prop}(E,c)$.

• (iii) By Theorem 1(i), the proportional rule satisfies self-duality. By Theorem 2(i), the proportional rule satisfies path composition up. Let f be a bankruptcy rule satisfying path composition up and self-duality. By Lemma 2(ii), f satisfies path composition down and self-duality. By Theorem 2(ii), $f=\mathrm{Prop}$.

Proof of Theorem 3

• (i) Let $(E,c)\in \mathrm{BR}^N$ and let $i,j\in N$ such that $\frac{c_i}{u_i^E}=\frac{c_j}{u_j^E}$. Then

  $$\begin{aligned} \frac{\mathrm{CREA}_i(E,c)}{u_i^E}= & {} \frac{\min \{c_i,\alpha ^{E,c}u_i^E\}}{u_i^E} = \min \left\{ \frac{c_i}{u_i^E},\alpha ^{E,c}\right\} \\= & {} \min \left\{ \frac{c_j}{u_j^E},\alpha ^{E,c}\right\} = \frac{\min \{c_j,\alpha ^{E,c}u_j^E\}}{u_j^E}=\frac{\mathrm{CREA}_j(E,c)}{u_j^E}. \end{aligned}$$

  Hence, the constrained relative equal awards rule satisfies relative symmetry. Let $(E,c)\in \mathrm{BR}^N$. For all $i\in N$,

  $$\begin{aligned} \mathrm{CREA}_i(E,\hat{c}^E)= & {} \min \{\hat{c}_i^E,\alpha ^{E,\hat{c}^E}u_i^E\} = \min \{\min \{c_i,u_i^E\},\alpha ^{E,\hat{c}^E}u_i^E\} \\= & {} \min \{c_i,u_i^E,\alpha ^{E,\hat{c}^E}u_i^E\} = \min \{c_i,\alpha ^{E,\hat{c}^E}u_i^E\}. \end{aligned}$$

  Since E is nonleveled, this means that $\mathrm{CREA}(E,c)=\mathrm{CREA}(E,\hat{c}^E)$. Hence, the constrained relative equal awards rule satisfies truncation invariance. Let $(E,c)\in \mathrm{BR}^N$ and let $t\in (0,1)$. For all $i\in N$, $\mathrm{CREA}_i(tE,c)=\min \{c_i,t\alpha ^{tE,c}u_i^E\}$. This means that $\mathrm{CREA}(tE,c)\le \mathrm{CREA}(E,c)$. Denote $d=\lambda ^{E,\mathrm{CREA}(E,c)-\mathrm{CREA}(tE,c)}$. Suppose that $d\alpha ^{dE,c-\mathrm{CREA}(tE,c)}\le \alpha ^{E,c}-t\alpha ^{tE,c}$. For all $i\in N$ with $\mathrm{CREA}_i(tE,c)=c_i$,

  $$\begin{aligned} \mathrm{CREA}_i(dE,c-\mathrm{CREA}(tE,c))=0=c_i-c_i=\mathrm{CREA}_i(E,c)-\mathrm{CREA}_i(tE,c). \end{aligned}$$

  For all $i\in N$ with $\mathrm{CREA}_i(tE,c)=t\alpha ^{tE,c}u_i^E$,

  $$\begin{aligned} \mathrm{CREA}_i(dE,c-\mathrm{CREA}(tE,c))= & {} \min \left\{ c_i-\mathrm{CREA}_i(tE,c),\alpha ^{dE,c-\mathrm{CREA}(tE,c)}u_i^{dE}\right\} \\= & {} \min \left\{ c_i-t\alpha ^{tE,c}u_i^E,d\alpha ^{dE,c-\mathrm{CREA}(tE,c)}u_i^E\right\} \\\le & {} \min \left\{ c_i-t\alpha ^{tE,c}u_i^E,\alpha ^{E,c}u_i^E-t\alpha ^{tE,c}u_i^E\right\} \\= & {} \min \left. \{c_i,\alpha ^{E,c}u_i^E\}-t\alpha ^{tE,c}u_i^E \right. \\= & {} \mathrm{CREA}_i(E,c)-\mathrm{CREA}_i(tE,c). \end{aligned}$$

  Since E is nonleveled, $\mathrm{CREA}(dE,c-\mathrm{CREA}(tE,c))\in \mathrm{P}(dE)$, and $\mathrm{CREA}(E,c)-\mathrm{CREA}(tE,c)\in \mathrm{P}(dE)$, this means that

  $$\begin{aligned} \mathrm{CREA}(dE,c-\mathrm{CREA}(tE,c))=\mathrm{CREA}(E,c)-\mathrm{CREA}(tE,c). \end{aligned}$$

  Clearly, similar arguments apply to the case $d\alpha ^{dE,c-\mathrm{CREA}(tE,c)}>\alpha ^{E,c}-t\alpha ^{tE,c}$. Hence, the constrained relative equal awards rule satisfies path composition up. Let f be a bankruptcy rule satisfying relative symmetry, truncation invariance, and path composition up. Let $(E,c)\in \mathrm{BR}^N$. Suppose that $f(tE,c)\ne \mathrm{CREA}(tE,c)$ for some $t\in [0,\lambda ^{E,c}]$. Define $\hat{t}=\inf \{t\in [0,\lambda ^{E,c}]\mid f(tE,c)\ne \mathrm{CREA}(tE,c)\}$. By path continuity, $\hat{t}\in [0,\lambda ^{E,c})$ and $f(\hat{t}E,c)=\mathrm{CREA}(\hat{t}E,c)$. Denote $N=\{1,\ldots ,|N|\}$ such that $\frac{c_1}{u_1^E}\le \cdots \le \frac{c_{|N|}}{u_{|N|}^E}$. Let $k\in N$ be such that $f_i(\hat{t}E,c)=c_i$ for all $i<k$, and $f_i(\hat{t}E,c)=\hat{t}\alpha ^{\hat{t}E,c}u_i^E<c_i$ for all $i\ge k$. Define $m=\min \{\Vert x\Vert \mid x\in \mathrm{P}(E)\}$. Let $\varepsilon \in (0,m(\frac{c_k}{u_k^E}-\frac{f_k(\hat{t}E,c)}{u_k^E}))$. By path continuity, there is a $\delta >0$ such that $\Vert f(tE,c)-f(\hat{t}E,c)\Vert <\varepsilon$ for all $t\in (\hat{t},\min \{\hat{t}+\delta ,\lambda ^{E,c}\})$. Let $t\in (\hat{t},\min \{\hat{t}+\delta ,\lambda ^{E,c}\})$. By path monotonicity, $\lambda ^{E,f(tE,c)-f(\hat{t}E,c)}\in (0,\lambda ^{E,c})$. Denote $d=\lambda ^{E,f(tE,c)-f(\hat{t}E,c)}$. By path composition up,

  $$\begin{aligned} m\left( \frac{c_k}{u_k^E}-\frac{f_k(\hat{t}E,c)}{u_k^E}\right)>\varepsilon >\Vert f(tE,c)-f(\hat{t}E,c)\Vert =\Vert f(dE,c-f(\hat{t}E,c))\Vert \ge dm. \end{aligned}$$

  This means that $d<(\frac{c_k}{u_k^E}-\frac{f_k(\hat{t}E,c)}{u_k^E})$. Define $\tilde{u}^{dE}\in \mathbb {R}_+^N$ by

  $$\begin{aligned} \tilde{u}_i^{dE}= {\left\{ \begin{array}{ll} 0&{}{{\text {for}}\;{\text {all}}\;i < k;} \\ {u_i^{dE}}&{}{{\text {for}}\;{\text {all}}\;i \ge k.} \end{array}\right. } \end{aligned}$$

  For all $i<k$,

  $$\begin{aligned} \tilde{u}_i^{dE}=0=c_i-c_i=c_i-f_i(\hat{t}E,c)=c_i-\mathrm{CREA}_i(\hat{t}E,c). \end{aligned}$$

  For all $i\ge k$,

  $$\begin{aligned} \tilde{u}_i^{dE}= & {} u_i^{dE}=du_i^E < \left( \frac{c_k}{u_k^E}-\frac{f_k(\hat{t}E,c)}{u_k^E}\right) u_i^E \le \left( \frac{c_i}{u_i^E}-\frac{\hat{t}\alpha ^{\hat{t}E,c}u_k^E}{u_k^E}\right) u_i^E \\= & {} c_i-\hat{t}\alpha ^{\hat{t}E,c}u_i^E = c_i-f_i(\hat{t}E,c) = c_i-\mathrm{CREA}_i(\hat{t}E,c). \end{aligned}$$

  By truncation invariance and relative symmetry,

  $$\begin{aligned} f\big (dE,c-f(\hat{t}E,c)\big )= & {} f\big (dE,\tilde{u}^{dE}\big ) = \frac{1}{\lambda ^{dE,\tilde{u}^{dE}}}\tilde{u}^{dE} \\= & {} \mathrm{CREA}\big (dE,\tilde{u}^{dE}\big ) = \mathrm{CREA}\big (dE,c-\mathrm{CREA}(\hat{t}E,c)\big ). \end{aligned}$$

  By composition up,

  $$\begin{aligned} f(tE,c)= & {} f(\hat{t}E,c)+f(dE,c-f(\hat{t}E,c)) \\= & {} \mathrm{CREA}(\hat{t}E,c)+\mathrm{CREA}(dE,c-\mathrm{CREA}(\hat{t}E,c)) \\= & {} \mathrm{CREA}(tE,c). \end{aligned}$$

  This contradicts the definition of $\hat{t}$. Hence, $f(tE,c)=\mathrm{CREA}(tE,c)$ for all $t\in [0,\lambda ^{E,c}]$.

• (ii) Let $(E,c)\in \mathrm{BR}^N$ and let $i\in N$ with $c_i'>c_i$. If $\alpha ^{E,(c_i',c_{N\setminus \{i\}})}\ge \alpha ^{E,c}$, then

  $$\begin{aligned} \mathrm{CREA}_i(E,(c_i',c_{N\setminus \{i\}}))=\min \{c_i',\alpha ^{E,(c_i',c_{N\setminus \{i\}})}u_i^E\}\ge \min \{c_i,\alpha ^{E,c}u_i^E\}=\mathrm{CREA}_i(E,c). \end{aligned}$$

  Suppose that $\alpha ^{E,(c_i',c_{N\setminus \{i\}})}<\alpha ^{E,c}$. For all $j\in N\setminus \{i\}$,

  $$\begin{aligned} \mathrm{CREA}_j(E,(c_i',c_{N\setminus \{i\}}))= & {} \min \{c_j,\alpha ^{E,(c_i',c_{N\setminus \{i\}})}u_j^E\}\le \min \{c_j,\alpha ^{E,c}u_j^E\}\\= & {} \mathrm{CREA}_j(E,c). \end{aligned}$$

  Since E is nonleveled, this means that $\mathrm{CREA}_i(E,c')\ge \mathrm{CREA}_i(E,c)$. Hence, the constrained relative equal awards rule satisfies claim monotonicity. Let $(E,c)\in \mathrm{BR}^N$ and let $i\in N$ with

  $$\begin{aligned} \left( \min \left\{ \frac{c_i}{u_i^E},\frac{c_j}{u_j^E}\right\} u_j^E\right) _{j\in N}\in E. \end{aligned}$$

  Suppose that $\mathrm{CREA}_i(E,c)=\alpha ^{E,c}u_i^E$. For all $j\in N$,

  $$\begin{aligned} \mathrm{CREA}_j(E,c)=\min \{c_j,\alpha ^{E,c}u_j^E\}=\min \left\{ \alpha ^{E,c},\frac{c_j}{u_j^E}\right\} u_j^E\le \min \left\{ \frac{c_i}{u_i^E},\frac{c_j}{u_j^E}\right\} u_j^E. \end{aligned}$$

  Since E is nonleveled, this means that $\mathrm{CREA}_i(E,c)=c_i$. Hence, the constrained relative equal awards rule satisfies conditional full compensation. Let f be a bankruptcy rule satisfying claim monotonicity and conditional full compensation. Let $(E,c)\in \mathrm{BR}^N$. Let $i\in N$ with $\mathrm{CREA}_i(E,c)=c_i$. For all $j\in N$,

  $$\begin{aligned} \min \left\{ \frac{c_i}{u_i^E},\frac{c_j}{u_j^E}\right\} u_j^E\le \min \left\{ \alpha ^{E,c},\frac{c_j}{u_j^E}\right\} u_j^E=\min \{c_j,\alpha ^{E,c}u_j^E\}=\mathrm{CREA}_j(E,c). \end{aligned}$$

  Since E is comprehensive, this means that

  $$\begin{aligned} \left( \min \left\{ \frac{c_i}{u_i^E},\frac{c_j}{u_j^E}\right\} u_j^E\right) _{j\in N}\in E. \end{aligned}$$

  By conditional full compensation, $f_i(E,c)=c_i$. Suppose that $f(E,c)\ne \mathrm{CREA}(E,c)$. Since E is nonleveled, there is a $k\in N$ with $f_k(E,c)<\mathrm{CREA}_k(E,c)=\alpha ^{E,c}u_k^E<c_k$. Define $c_k'=\alpha ^{E,c}u_k^E$. For all $j\in N$,

  $$\begin{aligned} \min \left\{ \frac{c_k'}{u_k^E},\frac{c_j}{u_j^E}\right\} u_j^E=\min \left\{ \alpha ^{E,c},\frac{c_j}{u_j^E}\right\} u_j^E=\min \{c_j,\alpha ^{E,c}u_j^E\}=\mathrm{CREA}_j(E,c). \end{aligned}$$

  This means that

  $$\begin{aligned} \left( \min \left\{ \frac{c_k'}{u_k^E},\frac{c_j}{u_j^E}\right\} u_j^E\right) _{j\in N}\in E. \end{aligned}$$

  By conditional full compensation, $f_k(E,(c_k',c_{N\setminus \{k\}}))=c_k'$. Then $f_k(E,(c_k',c_{N\setminus \{k\}}))>f_k(E,c)$, contradicting claim monotonicity. Hence, $f(E,c)=\mathrm{CREA}(E,c)$.

• (iii) By Theorem 3(ii), the constrained relative equal awards rule satisfies conditional full compensation. Let $(E,c)\in \mathrm{BR}^N$ and let $t\in (0,1)$. For all $i\in N$, $\mathrm{CREA}_i(tE,c)=\min \{c_i,t\alpha ^{tE,c}u_i^E\}$. This means that $\mathrm{CREA}(tE,c)\le \mathrm{CREA}(E,c)$. Suppose that $\alpha ^{tE,\mathrm{CREA}(E,c)}\le \alpha ^{tE,c}$. For all $i\in N$,

  $$\begin{aligned} \mathrm{CREA}_i(tE,\mathrm{CREA}(E,c))= & {} \min \{\mathrm{CREA}_i(E,c),\alpha ^{tE,\mathrm{CREA}(E,c)}u_i^{tE}\} \\\le & {} \min \{\min \{c_i,\alpha ^{E,c}u_i^E\},\alpha ^{tE,c}u_i^{tE}\} \\= & {} \min \{\min \{c_i,\alpha ^{E,c}u_i^E\},\min \{c_i,\alpha ^{tE,c}u_i^{tE}\}\} \\= & {} \min \{\mathrm{CREA}_i(E,c),\mathrm{CREA}_i(tE,c)\} \\= & {} \mathrm{CREA}_i(tE,c). \end{aligned}$$

  Since E is nonleveled, $\mathrm{CREA}(tE,\mathrm{CREA}(E,c))\in \mathrm{P}(tE)$, and $\mathrm{CREA}(tE,c)\in \mathrm{P}(tE)$, this means that $\mathrm{CREA}(tE,\mathrm{CREA}(E,c))=\mathrm{CREA}(tE,c)$. Clearly, similar arguments apply to the case $\alpha ^{tE,\mathrm{CREA}(E,c)}>\alpha ^{tE,c}$. Hence, the constrained relative equal awards rule satisfies path composition down. Let f be a bankruptcy rule satisfying conditional full compensation and path composition down. Let $(E,c)\in \mathrm{BR}^N$. Let $i\in N$ with $\mathrm{CREA}_i(E,c)=c_i$. For all $j\in N$,

  $$\begin{aligned} \min \left\{ \frac{c_i}{u_i^E},\frac{c_j}{u_j^E}\right\} u_j^E\le \min \left\{ \alpha ^{E,c},\frac{c_j}{u_j^E}\right\} u_j^E=\min \{c_j,\alpha ^{E,c}u_j^E\}=\mathrm{CREA}_j(E,c). \end{aligned}$$

  Since E is comprehensive, this means that

  $$\begin{aligned} \left( \min \left\{ \frac{c_i}{u_i^E},\frac{c_j}{u_j^E}\right\} u_j^E\right) _{j\in N}\in E. \end{aligned}$$

  By conditional full compensation, $f_i(E,c)=c_i$. Suppose that $f(E,c)\ne \mathrm{CREA}(E,c)$. Since E is nonleveled, there is a $k\in N$ with $f_k(E,c)<\mathrm{CREA}_k(E,c)=\alpha ^{E,c}u_k^E<c_k$. By path monotonicity and path continuity, there is a $t\in (1,\lambda ^{E,c})$ such that $f_k(tE,c)=\alpha ^{E,c}u_k^E$. For all $j\in N$,

  $$\begin{aligned} \min \left\{ \frac{f_k(tE,c)}{u_k^E},\frac{f_j(tE,c)}{u_j^E}\right\} u_j^E\le \min \left\{ \alpha ^{E,c},\frac{c_j}{u_j^E}\right\} u_j^E=\mathrm{CREA}_j(E,c). \end{aligned}$$

  Since E is comprehensive, this means that

  $$\begin{aligned} \left( \min \left\{ \frac{f_k(tE,c)}{u_k^E},\frac{f_j(tE,c)}{u_j^E}\right\} u_j^E\right) _{j\in N}\in E. \end{aligned}$$

  By conditional full compensation, $f_k(E,f(tE,c))=f_k(tE,c)$. By path composition down, $f_k(E,c)=f_k(tE,c)$. This is a contradiction. Hence, $f(E,c)=\mathrm{CREA}(E,c)$.

• (iv) By Theorem 3(i), the constrained relative equal awards rule satisfies relative symmetry. Let $(E,c)\in \mathrm{BR}^N$ and let $i,j\in N$ with $i\ne j$ and $\frac{c_j'}{u_j^E}>\frac{c_j}{u_j^E}\ge \frac{c_i}{u_i^E}$. Suppose that $\alpha ^{E,(c_j',c_{N\setminus \{j\}})}\ge \alpha ^{E,c}$. For all $k\in N$,

  $$\begin{aligned} \mathrm{CREA}_k(E,(c_j',c_{N\setminus \{j\}}))\ge \min \{c_k,\alpha ^{E,c}u_k^E\}=\mathrm{CREA}_k(E,c). \end{aligned}$$

  Since E is nonleveled, this means that $\mathrm{CREA}(E,(c_j',c_{N\setminus \{j\}}))=\mathrm{CREA}(E,c)$. Now suppose that $\alpha ^{E,(c_j',c_{N\setminus \{j\}})}<\alpha ^{E,c}$. For all $k\in N\setminus \{j\}$,

  $$\begin{aligned} \mathrm{CREA}_k(E,(c_j',c_{N\setminus \{j\}}))\le \min \{c_k,\alpha ^{E,c}u_k^E\}=\mathrm{CREA}_k(E,c). \end{aligned}$$

  Suppose that $\mathrm{CREA}_i(E,(c_j',c_{N\setminus \{j\}}))<\mathrm{CREA}_i(E,c)$. Then $\alpha ^{E,(c_j',c_{N\setminus \{j\}})}<\frac{c_i}{u_i^E}$. Since E is nonleveled, $\mathrm{CREA}_j(E,(c_j',c_{N\setminus \{j\}}))>\mathrm{CREA}_j(E,c)$. Then $\alpha ^{E,(c_j',c_{N\setminus \{j\}})}>\frac{c_j}{u_j^E}$. This is a contradiction. Hence, $\mathrm{CREA}_i(E,(c_j',c_{N\setminus \{j\}}))=\mathrm{CREA}_i(E,c)$ and the constrained relative equal awards rule satisfies independence of larger relative claims. Let f be a bankruptcy rule satisfying relative symmetry and independence of larger relative claims. Let $(E,c)\in \mathrm{BR}^N$. Denote $N=\{1,\ldots ,|N|\}$ such that $\frac{c_1}{u_1^E}\le \cdots \le \frac{c_{|N|}}{u_{|N|}^E}$. Let $k\in N$ be such that $\mathrm{CREA}_i(E,c)=c_i$ for all $i<k$, and $\mathrm{CREA}_i(E,c)=\alpha ^{E,c}u_i^E<c_i$ for all $i\ge k$. For all $i<k$, by independence of larger relative claims,

  $$\begin{aligned} f_i(E,c)=f_i(E,\mathrm{CREA}(E,c))=\mathrm{CREA}_i(E,c). \end{aligned}$$

  For all $i\ge k$, define $c^i\in \mathbb {R}_+^N$ by

  $$\begin{aligned} c_j^i= {\left\{ \begin{array}{ll} {{c_j}}&{}{{\text {for}}\;{\text {all}}\;j \le i;} \\ {\frac{{{c_i}}}{{u_i^E}}u_j^E}&{}{{\text {for}}\;{\text {all}}\;j > i.} \end{array}\right. } \end{aligned}$$

  By independence of larger relative claims and relative symmetry,

  $$\begin{aligned} f_k(E,c)=f_k(E,c^k)=\alpha ^{E,c^k}u_k^E=\mathrm{CREA}_k(E,c^k)=\mathrm{CREA}_k(E,c). \end{aligned}$$

  Next, these arguments apply to claimant $k+1$. Continuing this reasoning, $f_i(E,c)=\mathrm{CREA}_i(E,c)$ for all $i\ge k$. Hence, $f(E,c)=\mathrm{CREA}(E,c)$.