Consistent strategy-proof assignment by hierarchical exchange

Abstract

We characterize the family of efficient, consistent, and strategy-proof rules in house allocation problems. These rules follow an endowment inheritance and trade procedure as in Pápai’s hierarchical exchange rules (Pápai in Econometrica 68, 1403–1433, 2000) and closely resemble Ergin’s priority rules (Ergin in Econometrica 70, 2489–2497, 2002). We prove that if there are at least four objects, these are the only rules that are efficient in two-agent problems, $2$-consistent, and strategy-proof. A corollary is that these three basic properties together imply the full requirements of efficiency, consistency, group strategy-proofness, and reallocation-proofness.

Appendices

Appendix 1: Implications operator

In this appendix, we define the “implications operator” described in Sect. 5. Recall that $F^{\varphi }\equiv \left\{ \succsim _{xy}^{\varphi }:x,y\in \mathcal X ,x\not =y\right\} \subseteq \mathcal X ^{2}\times \mathcal A ^{2}$ is the family of binary relations on $\mathcal{A }$ defined as follows: for each $\{i,j\}\subseteq \mathcal{A }$ and each $\{x,y\}\subseteq \mathcal X$ such that $x\not =y$, $i\succ _{xy}^{\varphi }j$ if and only if $\varphi _i(xP_{i}y,xP_{j}y)=x$.

Let $X=\left\{ x,y,z\right\} \subseteq \mathcal X$ and $R^{xyz}$ denote the preference $x\,P\,y\,P\,z$. Using this notation, $\mathcal R \left( X\right) =\left\{ R^{xyz},R^{xzy},R^{yxz},R^{yzx},R^{zxy},R^{zyx}\right\}$.

Let $N\in \mathcal N$, $X\subseteq \mathcal{X }$ such that $|X|=|N|$, and $R\in \mathcal R ^{N}(X)$. For each $h\subseteq \mathcal X ^{2}\times \mathcal A ^{2}$, let $A\left( h,R\right) \subseteq Z\left( R\right)$ be the set:

$$\begin{aligned} A\left( h,R\right) \equiv \left\{ \mu \in Z\left( R\right) : \begin{array}{l} \mathrm{there \, is \, no} \,\left\{ i,j\right\} \subseteq N\mathrm{such \, that \, either\, (i) }\, \mu _j\,P_i\,\mu _i\\ \mathrm{and }\,\mu _i\,P_j\,\mu _j,\mathrm{or \, (ii) \, } \, \mu _i\,P_{i}\,\mu _j,\,\mu _i\,P_{j}\,\mu _j,\mathrm{and}\\ \{j\succ _{\mu _i\mu _j}i\}\subseteq h \end{array} \right\} \!. \end{aligned}$$

The set $A\left( h,R\right)$ contains the allocations in $Z\left( R\right)$ such that for each rule $\varphi$ with $h\subseteq F^{\varphi }$, if $\varphi \left( R\right) \not \in A\left( h,R\right)$, then $\varphi$ violates either 2-efficiency or $2$-consistency.²⁰ For example, the unique allocation $\mu$ in $A\left( h,R\right)$ where $h=\left\{ i\succ _{xy}j,\,j\succ _{xy}k\right\}$ and $R=\left( R_{i}^{zxy},R_{j}^{xyz},R_{k}^{xzy}\right)$, is $\mu _i=z$, $\mu _j=x$, and $\mu _k=y$. That is, if $\varphi$ is such that $h\subseteq F^{\varphi }$, then if $\varphi \left( R\right) \not \in \left\{ \mu \right\}$, $\varphi$ violates either $2$-consistency, or 2-efficiency, and $2$-consistency.

For each $N\in \mathcal{N }$, $X\subseteq \mathcal{X }$ such that $|X|=|N|$, $R\in \mathcal R ^{N}(X)$, and $\mu \in Z\left( R\right)$, let $\mathcal H \left( R,\mu \right) \subseteq \mathcal X ^{2}\times \mathcal A ^{2}$ be the set $\left\{ i\succ _{\mu _i\mu _j}j\in \mathcal X ^{2}\times \mathcal A ^{2}:\mu _i\,P_i\mu _j\,\mathrm{and }\,\mu _i\,P_j\,\mu _j\right\}$. In words, if $\varphi$ is a $2$-consistent rule and $\varphi (R)=\mu$, then the set $\mathcal H \left( R,\mu \right)$ is contained in $F^\varphi$. For example, if $R=\left( R_{i}^{zxy},R_{j}^{xyz},R_{k}^{xzy}\right)$, $\mu _i=z$, $\mu _j=x$, and $\mu _k=y$, then $\mathcal H \left( R,\mu \right) =\left\{ i\succ _{zy}k,j\succ _{xy}k\right\}$.

For each $h\subseteq \mathcal X ^{2}\times \mathcal A ^{2}$ and each $R\in \mathcal R ^{N}$, let $\mathcal I (h,R)$ be the set:²¹

$$\begin{aligned} \mathcal I (h,R)\equiv \left\{ \begin{array}{l@{\quad }l}h &{} \mathrm{if\ }|A(h,R)|>1,\\ h\cup \mathcal H (R,A(h,R))&{} \mathrm{if\ }|A(h,R)|=1,\\ \mathcal X ^{2}\times \mathcal A ^{2}&{}\mathrm{if\ }|A(h,R)|=0. \end{array}\right. \end{aligned}$$

We call $\mathcal I (h,R)$ the set of implications of $h$ by analyzing problem $R$. The following lemma justifies our choice for this name.

Lemma 2

Let $h\subseteq \mathcal X ^{2}\times \mathcal A ^{2}$ and $R\in \mathcal{R }$. Let $\varphi$ be a 2-efficient and $2$-consistent rule. If $h\subseteq F^{\varphi }$, then $\mathcal I \left( h,R\right) \subseteq F^\varphi$. If $h'\subseteq \mathcal X ^{2}\times \mathcal A ^{2}$ is such that $h\subseteq h'$, then $\mathcal I \left( h,R\right) \subseteq \mathcal I \left( h',R\right)$.

Proof

Let $h\subseteq \mathcal X ^{2}\times \mathcal A ^{2}$ and $R\in \mathcal{R }$. Observe that $\varphi (R)\in A(R,e)$ and thus $A(h,R)\ne \emptyset$. If $|A(h,R)|=1$ and since $\varphi$ is 2-efficient and $2$-consistent, then $\mathcal H (R,A(h,R))\subseteq F^\varphi$. Thus, $\mathcal I \left( h,R\right) \subseteq F^\varphi$. Otherwise, $\mathcal I \left( h,R\right) =h\subseteq F^\varphi$. Now, let $h'\subseteq \mathcal X ^{2}\times \mathcal A ^{2}$ be such that $h\subseteq h'$. Observe that $A(h',R)\subseteq A(h,R)$. Hence, $\mathcal I \left( h,R\right) \subseteq \mathcal I \left( h',R\right)$. $\square$

Observe that a consequence of Lemma 2 is that, if $h\subseteq \mathcal X ^{2}\times \mathcal A ^{2}$ and $R\in \mathcal{R }$ are such that $\mathcal I (h,R)=\mathcal X ^{2}\times \mathcal A ^{2}$, then for each 2-efficient and $2$-consistent rule $\varphi$, $h\nsubseteq F^\varphi$.²²

The following example illustrates the computation of $\mathcal{I }$.

Example 5

Let $h\equiv \{i\succ _{xy}j,j\succ _{xy}k\}$ and $R\equiv \left( R_i^{zxy},R_j^{xyz},R_k^{xzy}\right)$. Let $\mu :\{i,j,k\}\rightarrow \{x,y,z\}$ be defined by $\mu _i=z$, $\mu _j=x$, and $\mu _k=y$. Then, $A(h,R)=\{\mu \}$ and $\mathcal H (R,\mu )=\{i\succ _{zy}k,j\succ _{xy}k\}$. Thus, $\mathcal{I }(h,R)=h\cup \{i\succ _{zy}k,j\succ _{xy}k\}=h\cup \{i\succ _{zy}k\}$.

Let $h\subseteq \mathcal{X }^2\times \mathcal{A }^2$ and $\{R^s\}_{s=1}^S$ be a finite ordered list of problems. We define now the set of implications of $h$ by analyzing $\{R^s\}_{s=1}^S$. We denote it by $\mathcal I (h,\{R^s\}_{s=1}^S)$. We do so by means of the following algorithm.

Algorithm 1

Calculating the implications of a set $h\subseteq \mathcal{X }^2\times \mathcal{A }^2$ by analyzing a finite list of problems $\{R^s\}_{s=1}^S$.²³

Input: $(h,\{R^s\}_{s=1}^S)$.

Output: $\mathcal I (h,\{R^s\}_{s=1}^S)$.

Step 1: Let $h_{S+1}\leftarrow h$.

Step 2: Let $t\leftarrow 1$ and $h_1\leftarrow h_{S+1}$.

Step 3: While $t\le S$, let $h_{t+1}\leftarrow \mathcal I (h_t,R^t)$ and $t\leftarrow t+1$.

Step 4: If $h_{K+1}\setminus h_1\ne \emptyset$, then go to Step 2. Otherwise go to Step 5.

Step 5: Let $\mathcal I (h,\{R^s\}_{s=1}^S)\leftarrow h_{K+1}$.

Let $h\subseteq \mathcal{X }^2\times \mathcal{A }^2$ and $\{R^s\}_{s=1}^S$ be a finite list of problems. Algorithm 1 with input $(h,\{R^s\}_{s=1}^S)$ finds the implications of $h$ by analyzing the list of problems $\{R^s\}_{s=1}^S$. It proceeds as follows. First, it computes the implications of $h_1=h$ by analyzing $R^1$. This set is $h_2$. Then, it computes the implications of $h_2$ by analyzing $R^2$ and so on. After finding the implications of $h_{S}$ by analyzing $R^S$, $h_{S+1}$, the algorithm verifies if $h_{S+1}\setminus h_1\ne ~\emptyset$. If this is the case, it returns to the previous step. That is, it redefines $h_1$ as $h_{S+1}$ and analyzes the implications of $h_1$ by analyzing $R^1$ and so on. Otherwise, the algorithm terminates and the set of implications of $h$ by analyzing the list of problems $\{R^s\}_{s=1}^S$ is the last set calculated in the algorithm, i.e., $h_{S+1}$.²⁴

The following remark guarantees that for each $h\subseteq \mathcal{X }^2\times \mathcal{A }^2$ and each finite list of problems $\{R^s\}_{s=1}^S$, $\mathcal I (h,\{R^s\}_{s=1}^S)$ is well defined.

Remark 1

Let $h\subseteq \mathcal{X }^2\times \mathcal{A }^2$ and $\{R^s\}_{s=1}^S$ be a finite list of problems. Then Algorithm 1 with $h$ and $\{R^s\}_{s=1}^S$ as input stops.

Proof

Observe that the algorithm terminates if and only if Step 4 is reached a finite number of times. Now, if the algorithm reaches Step 4 and $h_{S+1}=\mathcal{X }^2\times \mathcal{A }^2$, then from Lemma 2, it can return to Step 4 at most once. Otherwise, Step 4 is reached at most $\sum _{s=1}^S|R^s|(|R^s|-1)$ times, where $|R^s|$ is the number of agents in problem $R^s$.²⁵ $\square$

Lemma 3

Let $h\subseteq \mathcal X ^{2}\times \mathcal A ^{2}$ and $\{R^s\}_{s=1}^S$ be a finite list of problems. Let $\varphi$ be a 2-efficient and $2$-consistent rule. If $h\subseteq F^{\varphi }$, then $\mathcal I \left( h,\{R^s\}_{s=1}^S\right) \subseteq F^\varphi$.

Proof

Let $h\subseteq \mathcal X ^{2}\times \mathcal A ^{2}$ and $\{R^s\}_{s=1}^S$ be a finite list of problems. Let $\varphi$ be a 2-efficient and $2$-consistent rule. We claim that if $h\subseteq F^{\varphi }$, then $\mathcal I \left( h,\{R^s\}_{s=1}^S\right) \subseteq F^\varphi$. We use an inductive argument. Observe that at the first time that the algorithm (with $h$ and $\{R^s\}_{s=1}^S$ as input) reaches Step 3, $t=1$ and $h_t=h\subseteq F^\varphi$. Suppose now that at some step in the algorithm, $t\le S$ and $h_t\subseteq F^\varphi$. By Lemma 2, $h_{t+1}=\mathcal I \left( h_t,R^t\right) \subseteq F^\varphi$. Hence, at the last time that the algorithm reaches Step 4, $h_{S+1}\subseteq F^\varphi$. Thus, $\mathcal I \left( h,\{R^s\}_{s=1}^S\right) \subseteq F^\varphi$.

Observe that a consequence of Lemma 3 is that, if $h\subseteq \mathcal X ^{2}\times \mathcal A ^{2}$ and the finite list of problems $\{R^s\}_{s=1}^S$ are such that $\mathcal I (h,\{R^s\}_{s=1}^S)=\mathcal X ^{2}\times \mathcal A ^{2}$, then for each 2-efficient and $2$-consistent rule $\varphi$, $h\nsubseteq F^\varphi$.

The following lemma states that $\mathcal I$ is invariant under permutations of lists (second argument).

Lemma 4

Let $h\subseteq \mathcal X ^{2}\times \mathcal A ^{2}$ and $\{R^s\}_{s=1}^S$ a finite list of problems. If $\{\bar{R}^s\}_{s=1}^S$ is a permutation of $\{R^s\}_{s=1}^S$, then $\mathcal I (h,\{R^s\}_{s=1}^S)=\mathcal I (h,\{\bar{R}_s\}_{s=1}^S)$.

Proof

Let $h\subseteq \mathcal X ^{2}\times \mathcal A ^{2}$. Let $\{R^s\}_{s=1}^S$ and $\{\bar{R}^s\}_{s=1}^S$ be two finite lists of problems satisfying the hypothesis of the lemma. We claim that $\mathcal I (h,\{R^s\}_{s=1}^S)=\mathcal I (h,\{\bar{R}_s\}_{s=1}^S)$. We prove first that $\mathcal I (h,\{R^s\}_{s=1}^S)\subseteq \mathcal I (h,\{\bar{R}^s\}_{s=1}^S)$. We use an inductive argument. Observe that at the first time that the algorithm (with $h$ and $\{R^s\}_{s=1}^S$ as input) reaches Step 2, $h_1=h\subseteq \mathcal I (h,\{\bar{R}^s\}_{s=1}^S)$. Now, suppose that at some step in the algorithm (with $h$ and $\{R^s\}_{s=1}^S$ as input), $t\le S$ and $h_t\subseteq \mathcal I (h,\{\bar{R}^s\}_{s=1}^S)$. As the algorithm (with $h$ and $\{\bar{R}^s\}_{s=1}^S$ as input) reaches Step 4, then it has to be the case that $\mathcal I (\mathcal I (h,\{\bar{R}^s\}_{s=1}^S),R^t)=\mathcal I (h,\{\bar{R}^s\}_{s=1}^S)$. From Lemma 2, $h_{t+1}\subseteq \mathcal I (h,\{\bar{R}^s\}_{s=1}^S)$.²⁶ Thus, $\mathcal I (h,\{R^s\}_{s=1}^S)\subseteq \mathcal I (h,\{\bar{R}^s\}_{s=1}^S)$. A symmetric argument shows that $\mathcal I (h,\{\bar{R}^s\}_{s=1}^S)\subseteq \mathcal I (h,\{R^s\}_{s=1}^S)$.   $\square$

For each $h\subseteq \mathcal X ^{2}\times \mathcal A ^{2}$ and each $C\subseteq \mathcal{R }^N$ such that $|C|<\infty$, the set of implications of $h$ by analyzing $C$ is $\mathcal I \left( h,C\right) \equiv \mathcal I \left( h,\{R^s\}_{s=1}^S\right)$ for some list $\{R^s\}_{s=1}^S$ such that $C=\bigcup _{s=1}^SR^s$. From Lemma 4, $\mathcal I \left( h,C\right)$ is well defined. From Lemma 3, for each $h\subseteq \mathcal X ^{2}\times \mathcal A ^{2}$ and each $C\subseteq \mathcal{R }$ such that $|C|<\infty$, if $h\subseteq F^\varphi$, then $\mathcal I \left( h,C\right) \subseteq F^\varphi$. Besides, if $\mathcal I \left( h,C\right) =\mathcal X ^{2}\times \mathcal A ^{2}$, then $h\nsubseteq F^\varphi$.

For each $N\in \mathcal{N }$ and $C\subseteq \mathcal{R }^N$ such that $|C|<\infty$, let $C^X_N$ be the set of problems $C^X_N\equiv \{R\in \mathcal{R }^{N'}:N'\subseteq N,\,X'\subseteq X,\,\mathrm{and\ }|N'|=|X'|=3\}$.

Example 6

If $h_0\equiv \{i\succ _{xy}j,j\succ _{xy}k,j\succ _{xz}i\}$, then $\mathcal I (h_0, C^{\{x,y,z,w\}}_{\{i,j,k\}})=\mathcal X ^2\times \mathcal A ^2$. Table 1 contains a “step by step” computation of $\mathcal I (h_0,C^{\{x,y,z,w\}}_{\{i,j,k\}})$.²⁷

Table 1 Computation of $\mathcal I (h_0,C^{\{x,y,z,w\}}_{\{i,j,k\}})$ in Example 6

Appendix 2: Proofs

Proof of Lemma 1

The family of CHE-1 rules belongs to the family of EP rules (Ergin 2002) in our domain. It is well known that EP rules are consistent (Ergin 2002).

Let $f^\varGamma$ be a CHE-2 rule associated with $(\varPi _2,T_2,Q)$ where $\varPi _2\equiv (\pi _l)_{l=1}^m$. Let $N\in \mathcal{N }$, $X\subseteq \mathcal{X }$ such that $|N|=|X|$, $R\in \mathcal{R }^N(X)$, and $\mu \equiv f^{\varGamma (N,X)}(R)$. Let $M\subsetneq N$, $Z\equiv \bigcup _{i\in M}\mu _i$, and $\tilde{\mu }\equiv f^{\varGamma (M,Z)}(R_M|Z)$. We claim that $\tilde{\mu }=\mu |_M$. Let $g^\varGamma$ be a CHE-1 rule associated with a partition $\varPi '_1\equiv (\pi '_l)_{l=1}^{m+1}$ such that for each $l<m$, $\pi '_l=\pi _l$; and tiebreaker function $T_1$ such that for each $l=1,...,m-1$, $T_1|\mathcal{X }\times \varPi _2\setminus \{\pi _m\}=T_2$. That is, $g^\varGamma$ is a CHE-1 rule for which the inheritance among the agents in $\pi _1,...,\pi _{m-1}$ coincides with that of $f^\varGamma$. Then, $\mu |M\setminus \pi _m=g^\varGamma (R)|M\setminus \pi _m$ and $\tilde{\mu }|M\setminus \pi _m=g^\varGamma (R_M|Z)|M\setminus \pi _m$. Since $g^\varGamma$ is consistent, $g^\varGamma (R)|M\setminus \pi _m=g^\varGamma (R_M|Z)|M\setminus \pi _m$. Thus, $\mu |M\setminus \pi _m=\tilde{\mu }|M\setminus \pi _m$. Since for each $l<m$ and each $i\in \pi _{l}$, $\tilde{\mu }_i=\mu _i$, then $\{\tilde{\mu }_{k}:k\in \pi _m\cap M\}=\{\mu _k:k\in \pi _m\cap M\}$. Thus, if $\pi _m\cap M$ is the singleton $\{i\}$, then $\tilde{\mu }_i=\mu _i$. Suppose then that $\pi _m\cap M$ is the pair $\{i,j\}$. If $\mathrm{Top}(R_i,\{\mu _i,\mu _j\})=\mu _i$ and $\mathrm{Top}(R_j,\{\mu _i,\mu _j\})=\mu _j$, then since $f^\varGamma$ is efficient, we have that $\tilde{\mu }_i=\mu _i$ and $\tilde{\mu }_j=\mu _j$. If $\mathrm{Top}(R_i,\{\mu _i,\mu _j\})=\mathrm{Top}(R_j,\{\mu _i,\mu _j\})=\mu _j$, then by C2-3, $Q(\mu _j,\mu _i)=j$. Thus, when calculating $\tilde{\mu }$, agent $j$ inherits $\mu _j$ before agent $i$. Thus, $\tilde{\mu }_j=\mu _j$. Thus, $\tilde{\mu }_i=\mu _i$. If $\mathrm{Top}(R_i,\{\mu _i,\mu _j\})=\mathrm{Top}(R_j,\{\mu _i,\mu _j\})=\mu _i$, the symmetric argument shows that $\tilde{\mu }_i=\mu _i$ and $\tilde{\mu }_j=\mu _j$.

Let $f^\varGamma$ be a CHE-3 rule associated with $(\varPi _3,T_3,w,Y)$ where $\varPi _3\equiv (\pi _l)_{l=1}^m$. Let $N\in \mathcal{N }$, $X\subseteq \mathcal{X }$ such that $|N|=|X|$, $R\in \mathcal{R }^N(X)$, and $\mu \equiv f^{\varGamma (N,X)}(R)$. Let $M\subsetneq N$, $Z\equiv \bigcup _{i\in M}\mu _i$, and $\tilde{\mu }\equiv f^{\varGamma (M,Z)}(R_M|Z)$. Let $(\varepsilon _t,M_t,X_t)_{t=1,\dots ,T}$ and $(\tilde{\varepsilon }_t,\tilde{M}_t,\tilde{X}_t)_{t=1\dots ,\tilde{T}}$ be the lists of endowment functions, agent sets, and endowment sets in the calculation of $\mu$ and $\tilde{\mu }$, respectively. We claim that $\tilde{\mu }=\mu |_M$. For simplicity, suppose that $\pi _{m-2}=\{1\}$, $\pi _{m-1}=\{2\}$, $\pi _{m}=\{3\}$. The same argument in the proof that CHE-2 rules are consistent shows that $\mu |M\setminus \{1,2,3\}=\tilde{\mu }|M\setminus \{1,2,3\}$ ($g^\varGamma$ here is for instance the CHE-1 rule associated with $(\varPi _3,T_3)$). Let $t$ be the minimal index for which $\bigcup _{i\in \{1,2,3\}\cap M}\tilde{\varepsilon }_t(i)\ne \emptyset$. Since for each $l<m-2$ and each $j\in \pi _{l}\cap M$, $\tilde{\mu }_j=\mu _j$, then by C1-1, $\bigcup _{i\in \{1,2,3\}\cap M}\tilde{\varepsilon }_t(i)=\tilde{X}_t=\bigcup _{i\in \{1,2,3\}\cap M}\{\mu _i\}$. There are four cases. Case 1 $\{1,2,3\}\cap M=\{1,2\}$. By C3-3 and C3-4, $\tilde{\varepsilon }_t(1)=\{\mu _1,\mu _2\}\setminus \{w\}$ and $\tilde{\varepsilon }_t(2)=\{\mu _1,\mu _2\}\cap \{w\}$. If $\mu _1\ne w$, then $\mathrm{Top}(R_1,\tilde{X}_t)\in \tilde{\varepsilon }_t(1)$. Thus $\tilde{\mu }_1=\mu _1$. Thus, $\tilde{\mu }_2=\mu _2$. If $\mu _1=w$, then by C3-4, when calculating $\mu$, agent $2$ inherits $w$ before agent $1$. Thus, $\mu _2\,P_2\,\mu _1$. Since $f^\varGamma$ is efficient, then $\tilde{\mu }_1=\mu _1$ and $\tilde{\mu }_2=\mu _2$. Case 2 $\{1,2,3\}\cap M=\{1,3\}$. By C3-5, $\tilde{\varepsilon }_t(1)=\{\mu _1,\mu _3\}$. Thus, $\tilde{\mu }_1=\mu _1$. Thus, $\tilde{\mu }_3=\mu _3$. Case 3 $\{1,2,3\}\cap M=\{2,3\}$. If $\mathrm{Top}(R_2,\{\mu _2,\mu _3\})=\mu _2$ and $\mathrm{Top}(R_3,\{\mu _2,\mu _3\})=\mu _3$, then since $f^\varGamma$ is efficient, $\tilde{\mu }_3=\mu _2$ and $\tilde{\mu }_3=\mu _3$. If $\mathrm{Top}(R_2,\{\mu _2,\mu _3\})=\mathrm{Top}(R_3,\{\mu _2,\mu _3\})=\mu _2$, then $\mu _2\,P_3\,\mu _3$. Thus, when calculating $\mu$, agent $2$ inherits $\mu _2$ before agent $3$. Thus, $w\not \in \{\mu _2,\mu _3\}$, or $\mu _3=w$ and $\mu _2\not \in Y$. Thus, $\mu _2\in \tilde{\varepsilon }_t(2)$. Thus, $\tilde{\mu }_2=\mu _2$. Thus, $\tilde{\mu }_3=\mu _3$. If $\mathrm{Top}(R_2,\{\mu _2,\mu _3\})=\mathrm{Top}(R_3,\{\mu _2,\mu _3\})=\mu _3$, then $\mu _3\,P_2\,\mu _2$. Thus, when calculating $\mu$, agent $3$ inherits $\mu _3$ before agent $2$. Thus, $\mu _2=w$ and $\mu _3\in Y$. Thus, $\mu _3\in \tilde{\varepsilon }_t(3)$. Thus, $\tilde{\mu }_3=\mu _3$. Thus, $\tilde{\mu }_2=\mu _2$. Case 4 $\{1,2,3\}\cap M=\{1,2,3\}$. Let $t'$ be the minimal index for which $\bigcup _{i\in \{1,2,3\}\cap M}\varepsilon _{t'}(i)\ne \emptyset$. By C3-1, $M_{t'}=\tilde{M}_t$. Since for each $l<m-2$ and each $i\in \pi _{l}\cap M$, $\tilde{\mu }_i=\mu _i$, then $X_t=\tilde{X}_t$. By C3-3, C3-4, and C3-5, $\varepsilon _{t'}(1)=\tilde{\varepsilon }_{t}(1)=X_t\setminus \{w\}$, $\varepsilon _{t'}(2)=\tilde{\varepsilon }_{t}(2)=X_t\cap \{w\}$, and $\varepsilon _{t'}(3)=\tilde{\varepsilon }_{t}(3)=\emptyset$. Thus, $\tilde{\mu }_1=\mu _1$, $\tilde{\mu }_2=\mu _2$, and $\tilde{\mu }_3=\mu _3$. $\square$

We now present three preliminary results to the proof of Theorem 1. They state the restrictions on rules implied by 2-efficiency and $2$-consistency. Let us first introduce some definitions.

Let $x\in \mathcal{X }$. The binary relation $\succsim _x^\varphi$ on $\mathcal{A }$ is defined as follows: for each $\{i,j\}\subseteq \mathcal A$ and each $x\in \mathcal X$, $i\succ _{x}^{\varphi }j$ if and only if for each $y\in \mathcal X \backslash \left\{ x\right\}$, $i\succ _{xy}^{\varphi }j$. In words, $i\succ _{x}^{\varphi }j$ if $\varphi$ gives priority to agent $i$ over agent $j$ in problem $\left( x\,P_{i}\,y,x\,P_{j}\,y\right)$, independently of $y$. Observe that for each rule $\varphi$, $\succsim _x^\varphi$ is antisymmetric, but not necessarily complete.

Let $\succsim ^{\varphi }$ be the binary relation on $\mathcal A$ defined as follows: for each $\{i,j\}\subseteq \mathcal A$, $i\succ ^{\varphi }j$ if and only if for each $x\in \mathcal X$, $i\succ _{x}^{\varphi }j$. In words, $i\succ ^{\varphi }j$ if $\varphi$ gives priority to agent $i$ over agent $j$ in every two-agent problem. Observe that for each $\varphi$, the relation $\succsim ^{\varphi }$ is antisymmetric, but not necessarily complete.

Lemma 5

Let $\varphi$ be a 2-efficient and $2$-consistent rule and $\{i,j,k\}\subseteq \mathcal A$. If for some $\{x,y\}\subseteq \mathcal X$ such that $x\not =y$, $i\succ _{xy}^{\varphi }j$ and $j\succ _{xy}^{\varphi }k$, then for each $z\in \mathcal X \backslash \left\{ x,y\right\}$, $i\succ _{zy}^{\varphi }k$.

Proof

Let $z\in \mathcal X \backslash \left\{ x,y\right\}$. Suppose that $i\succ _{xy}^{\varphi }j$ and $j\succ _{xy}^{\varphi }k$. We prove that $i\succ _{zy}^{\varphi }k$. Let $R\equiv (R^{zxy}_i,R^{xyz}_j,R^{xzy}_k)$ and $h\equiv \{i\succ _{xy}j,j\succ _{xy}k\}$. By Example 5, $\mathcal{I }(h,R)=h\cup \{ i\succ _{zy}k\}$. Thus, $i\succ _{zy}^{\varphi }k$. $\square$

Lemma 6

Assume $\left| \mathcal X \right| \ge 4$. Let $\varphi$ be a 2-efficient and $2$-consistent rule and $\{i,j,k\}\subseteq \mathcal A$. If for some $\{x,y\}\subseteq \mathcal X$ such that $x\not =y$, $i\succ _{xy}^{\varphi }j$ and $j\succ _{xy}^{\varphi }k$, then $i\succ _{x}^{\varphi }\left\{ j,k\right\}$.

Proof

Assume that $|\mathcal{X }|\ge 4$, $i\succ _{xy}^{\varphi }j$, and $j\succ _{xy}^{\varphi }~k$. We claim that, $i\succ _{xy}^{\varphi }k$. Suppose by contradiction that $k\succ _{xy}^{\varphi }i$. Let $\{z,w\}\subseteq \mathcal{X }\setminus \{x,y\}$ be such that $z\ne w$. From Lemma 5, $h\equiv \{j\succ _{zy}^{\varphi }i,\ i\succ _{zy}^{\varphi }k,\ k\succ _{wy}^{\varphi }j\}\subseteq F^\varphi$. Observe that $\mathcal I (h,(R^{zyw}_i,R^{wzy}_j,R^{zwy}_k))=\mathcal{X }^2\times \mathcal{A }^2\subseteq F^\varphi$. This is a contradiction. Now, we claim that $i\succ _{xz}^{\varphi }k$. To prove this, let $h'\equiv \{i\succ _{xy}j,\,j\succ _{xy}k,\,i\succ _{xy}k\}$ and observe that $\{i\succ _{xz}k\}\subseteq \mathcal I (h',(R^{xyz}_i,R^{xyz}_j,R^{xzy}_k))$. It is left to prove that $i\succ _{xz}^{\varphi }j$. To do this, suppose by contradiction that $j\succ _{xz}^{\varphi }i$. Let $h_0\equiv \{i\succ _{xy}j,\,j\succ _{xy}k,\,j\succ _{xz}i\}$. From Example 6, $\mathcal I (h_0,C^{\{x,y,z,w\}}_{\{i,j,k\}})=\mathcal{X }^2\times \mathcal{A }^2\subseteq F^\varphi$. This is a contradiction. $\square$

Lemma 7

Assume $\left| \mathcal A \right| \ge 4$. Let $\varphi$ be a 2-efficient and $2$-consistent rule and $\{i,j,k,l\}\subseteq \mathcal A$. If for some $\{x,y\}\subseteq \mathcal X$ such that $x\not =y$, $i\succ _{xy}^{\varphi }j\succ _{xy}^{\varphi }k\succ _{xy}^{\varphi }l$, then $i\succ ^{\varphi }\left\{ k,l\right\}$.

Proof

Assume $|\mathcal A |\ge 4$. Let $\varphi$ be a 2-efficient and $2$-consistent rule and $\{i,j,k,l\}\subseteq \mathcal A$. Suppose that for some $\{x,y\}\subseteq \mathcal X$ such that $x\not =y$, $i\succ _{xy}^{\varphi }j\succ _{xy}^{\varphi }k\succ _{xy}^{\varphi }l$. We want to prove $i\succ ^{\varphi }\left\{ k,l\right\}$. From Lemma 6, $i\succ _{x}^{\varphi }j\succ _{x}^{\varphi }\left\{ k,l\right\}$. Let $\{z,v\}\subseteq \mathcal X \backslash \left\{ x\right\}$ be such that $z\not =v$. Hence, $i\succ _{xz}^{\varphi }j\succ _{xz}^{\varphi }\{k,l\}$, and from Lemma 1, $i\succ _{vz}^{\varphi }\{k,l\}$. It remains to prove that for each $z\in X\setminus \left\{ x\right\}$, $i\succ _{zx}^{\varphi }\left\{ k,l\right\}$.

Let $z\in X\backslash \left\{ x,y\right\}$ and $h_{1}\equiv \left\{ i\succ _{zy}^{\varphi }k,k\succ _{xy}^{\varphi }l\right\}$. Observe that $\left\{ i\succ _{zx}^{\varphi }k\right\} \subseteq$ $\mathcal I \left( h_{1},\left( R_{i}^{zyx},R_{k}^{zxy},R_{l}^{xyz}\right) \right)$. Now, let $h_{2}\equiv \left\{ i\succ _{zx}^{\varphi }k,i\succ _{yz}^{\varphi }l\right\}$. Observe that $\left\{ i\succ _{yx}^{\varphi }l\right\}$ $=\mathcal I \left( h_{2},\left( R_{i}^{yzx},R_{k}^{zxy},R_{l}^{yxz}\right) \right)$. Now, we claim that for each $z\in X\backslash \left\{ x,y\right\}$, $i\succ _{zx}^{\varphi }l$. To prove this, suppose by contradiction that $l\succ _{zx}^{\varphi }i$. By Lemma 5 and since $i\succ _{zx}^{\varphi }k$, then $l\succ _{yx}^{\varphi }k$. Now, let $h_{3}\equiv \left\{ i\succ _{zy}^{\varphi }l,l\succ _{zx}^{\varphi }i\right\}$. Observe that $\left\{ k\succ _{yx}^{\varphi }i\right\} \subseteq \mathcal I \left( h_{3},\left( R_{i}^{zyx},R_{k}^{yxz},R_{l}^{zxy}\right) \right)$. Thus, $k\succ _{yx}^{\varphi }i\succ _{yx}^{\varphi }l\succ _{yx}^{\varphi }k$. This contradicts Lemma 2. Finally, we claim that $i\succ _{yx}^{\varphi }k$. To prove this, suppose by contradiction that $k\succ _{yx}^{\varphi }~i$.

Let $h_{5}\equiv \left\{ k\succ _{yx}^{\varphi }i,i\succ _{zx}^{\varphi }l,i\succ _{yz}^{\varphi }l,i\succ _{yz}^{\varphi }k\right\}$. This leads to a contradiction because $\mathcal I \left( h_{5},\left( R_{i}^{yzx},R_{k}^{yxz},R_{l}^{yzx}\right) \right) =\mathcal X ^2\times \mathcal A ^2$. $\square$

Proof of Theorem 1

HE rules are efficient and strategy-proof (Pápai 2000). By Lemma 1, each CHE rule is consistent. Thus, it is left to prove that if $\varphi$ is a $2$-efficient and $2$-consistent rule, then it is a CHE rule. Let $\varphi$ be such a rule. First, we construct a partition $\widetilde{\varPi }$ and a function $T$ associated with $\varphi$. Then we show that there is a CHE rule $f^\varGamma$ such that $\varphi =f^\varGamma$.

$\bullet$ Constructing $\varvec{\widetilde{\varPi }}$ and $\varvec{T}$: Let $\{x,y\}\subseteq \mathcal{X }$ be such that $x\ne y$. We inductively construct a partition of $\mathcal{A }$. Suppose that we have defined the list of disjoint sets of at most two agents $(\tilde{\pi }_l)_{l=1}^{t-1}$ satisfying that (i) for each $1\le l<l'\le t-1$, each $i\in \tilde{\pi }_l$, and each $j\in \tilde{\pi }_{l'}$, $i\succ ^\varphi j$, and (ii) $\mathcal{A }\setminus \bigcup _{l=1}^{t-1}\tilde{\pi }_l\ne \emptyset$. Suppose that for each $l\in \{1,\dots ,t-1\}$ such that $|\tilde{\pi }_l|=2$ and each $z\in \mathcal{X }$, there is an agent $T(z,\tilde{\pi }_l)\in \tilde{\pi }_l$ such that $T(z,\tilde{\pi }_l)\succ ^\varphi _z\tilde{\pi }_l\setminus \{T(z,\tilde{\pi }_l)\}$. Let $\mathcal{A }'\equiv \mathcal{A }\setminus \bigcup _{l=1}^{t-1}\tilde{\pi }_l$. We define $\tilde{\pi }_t$. There are four cases.

Case 1: $\mathcal{A }'$ is the singleton $\{i\}$. Then, let $\tilde{\pi }_t\equiv \{i\}$.

Case 2: $\mathcal{A }'$ is the pair $\{i,j\}$. If there is an agent, say $i$, such that $i\succ ^\varphi j$, then let $\tilde{\pi }_t\equiv \{i\}$. Otherwise, let $\tilde{\pi }_t\equiv \{i,j\}$.

Case 3: $\mathcal{A }'$ is the triple $\{i,j,k\}$. If there is an agent, say $i$, such that $i\succ ^\varphi \{j,k\}$, then let $\tilde{\pi }_t\equiv \{i\}$. If there is an agent, say $i$, such that $\{j,k\}\succ ^\varphi i$, then let $\tilde{\pi }_t\equiv \{j,k\}$. By Lemma 6, for each $z\in \mathcal{X }$, there is an agent $T(z,\tilde{\pi }_t)\in \tilde{\pi }_t$ such that $T(z,\tilde{\pi }_t)\succ ^\varphi _z\tilde{\pi }_t\setminus \{T(z,\tilde{\pi }_t)\}$. If there is no agent, say $i\in \mathcal{A }'$, such that either $i\succ ^\varphi \{j,k\}$ or $\{j,k\}\succ ^\varphi i$, then let $\tilde{\pi }_t\equiv \mathcal{A }'$.

Case 4: $|\mathcal{A }'|\ge 4$. Since $|\mathcal{X }|\ge 4$, then by Lemma 6, $\succsim ^\varphi _{xy}$ is transitive. Let $\{i,j\}\subseteq \mathcal{A }'$ be such that $i\succ _{xy}^\varphi j\succ _{xy}^\varphi \mathcal{A }'\setminus \{i,j\}$. By Lemma 7, $i\succ ^\varphi \mathcal{A }'\setminus \{i,j\}$. If $i\succ ^\varphi j$, then let $\tilde{\pi }_t=\{i\}$. Otherwise, let $\tilde{\pi }_t\equiv \{i,j\}$. Since it is not the case that $i\succ ^\varphi j$, then there are objects $\{z,w\}\subseteq \mathcal{X }$ such that $z\ne w$ and $j\succ _{zw}^\varphi i$. Thus, $j\succ _{zw}^\varphi i\succ ^\varphi \mathcal{A }'\setminus \{i,j\}$. By Lemma 7, $j\succ ^\varphi \mathcal{A }'\setminus \{i,j\}$. Let $\{z,w\}\in \mathcal{X }$. Suppose that $i\succ ^\varphi _{zw}j$. Since, $j\succ ^\varphi \mathcal{A }'\setminus \{i,j\}$, then $i\succ ^\varphi _{zw}j\succ ^\varphi \mathcal{A }'\setminus \{i,j\}$. Since $\mathcal{A }'\setminus \{i,j\}\ne \emptyset$, then by Lemma 6, $i\succ ^\varphi _{z}j$. Thus, for each $z\in \mathcal{X }$, there is an agent $T(z,\tilde{\pi }_t)\in \tilde{\pi }_t$ such that $T(z,\tilde{\pi }_t)\succ ^\varphi _z\tilde{\pi }_t\setminus \{T(z,\tilde{\pi }_t)\}$.

Since $|\mathcal{X }|\ge 4$, then by Lemma 6, $\succsim ^\varphi$ is transitive. Thus, for each $1\le l<l'\le t$, each $i\in \tilde{\pi }_l$, and each $j\in \tilde{\pi }_{l'}$, $i\succ ^\varphi j$.

$\bullet$ Claim 1: Let $N\in \mathcal{N }$, $X\subseteq \mathcal{X }$ such that $|N|=|X|$, and $R\in \mathcal{R }^N(X)$. We claim that for each pair $\{l,l'\}$ such that $l\ne l'$, each $i\in \tilde{\pi }_l$, and each $j\in \tilde{\pi }_{l'}$ such that $\{i,j\}\subseteq N$, if $\varphi _j(R)\,P_i\,\varphi _i(R)$, then $j\succ ^\varphi i$ and $l>l'$.

Let $R$, $i$, $j$, be as in the statement of the claim. Since $\varphi$ is $2$-consistent, then $\varphi (R_{\{i,j\}}|\{\varphi _i(R),\varphi _j(R)\})=\varphi (R)|_{\{i,j\}}$. Since $\varphi _j(R)\,P_i\,\varphi _i(R)$ and $\varphi$ is $2$-efficient, then $\varphi _j(R)\,P_j\,\varphi _i(R)$. Thus, $j\succ ^\varphi _{\varphi _j(R)\varphi _i(R)}i$. Since $l\ne l'$, then $j\succ ^\varphi i$. Thus $l>l'$.

We complete the proof of the theorem by analyzing three cases.

$\bullet$ Case 1: The last component of $\varvec{\widetilde{\varPi }}$ is a singleton. Let $\varPi _1\equiv (\pi _l)_{l=1}^m\equiv \widetilde{\varPi }$ and $T_1\equiv T$. Let $f^\varGamma$ be the CHE-1 rule associated with $(\varPi _1,T_1)$. We prove that $\varphi =f^\varGamma$. Let $N\in \mathcal{N }$, $X\subseteq \mathcal{X }$ such that $|N|=|X|$, and $R\in \mathcal{R }^N(X)$. Let $\mu \equiv f^{\varGamma (N,X)}(R)$. We claim that $\varphi (R)=\mu$. Let $(\varepsilon _t,M_t,X_t)_{t=1,\dots ,T}$ be the lists of endowment functions, agent sets, and endowment sets in the calculation of $\mu$. Let $l\in \{1,\dots ,m\}$ and suppose that for each $l'<l$ and each $i\in \pi _{l'}\cap N$, $\varphi _i(R)=\mu _i$. We claim that $\varphi (R)|_{\pi _l\cap N}=\mu |_{\pi _l\cap N}$. Let $t$ be the first stage at which $\bigcup _{i\in \pi _{l}\cap N}\varepsilon _t(i)\ne \emptyset$. By C1-1, $\bigcup _{i\in \pi _{l}\cap N}\varepsilon _t(i)=X_t$. Since for each $l'<l$ and each $i\in \pi _{l'}\cap N$, $\varphi _i(R)=\mu _i$, then $X_t=\{\mu _i:l'\ge l,\,i\in \pi _{l'}\}$. There are two cases.

Case 1-1: $\pi _l\cap N$ is the singleton $\{i\}$. Since $\mathrm{Top}(R_i,X_t)\in X_t$, then $\mu _i=\mathrm{Top}(R_i,X_t)$. We claim that $\varphi _i(R)=\mu _i$. Suppose by means of contradiction that there is $l'>l$ and $j\in \pi _{l'}$ such that $\varphi _j(R)=\mu _i$. Thus, $\varphi _j(R)\,P_i\,\varphi _i(R)$. By Claim 1, $l>l'$. This is a contradiction.

Case 1-2: $\pi _l\cap N$ is the pair $\{i,j\}$. Since $\varepsilon _t(i)\cup \varepsilon _t(j)=X_t$, then by C1-1, when calculating $\mu$, both agents in $\{i,j\}$ are in a TTC at $t$, or one agent is in a TTC at $t$ and the other agent is in a TTC with herself at $t+1$. Thus, $\mu _i\,P_i\,X_t\setminus \{\mu _i,\mu _j\}$ and $\mu _j\,P_j\,X_t\setminus \{\mu _i,\mu _j\}$. We claim that $\{\varphi _i(R),\varphi _j(R)\}=\{\mu _i,\mu _j\}$. Suppose by contradiction that there is $l'>l$ and $k\in \pi _{l'}$ such that $\varphi _k(R)\in \{\mu _i,\mu _j\}$. Suppose without loss of generality that $\varphi _k(R)=\mu _i$. Thus, $\varphi _k(R)\,P_i\,\varphi _i(R)$. By Claim 1, $l>l'$. This is a contradiction. If $\mathrm{Top}(R_i,X_t)\ne \mathrm{Top}(R_j,X_t)$, then when calculating $\mu$, agents $i$ and $j$ are in a TTC at $t$. Thus, $\mu _i=\mathrm{Top}(R_i,X_t)$ and $\mu _j=\mathrm{Top}(R_j,X_t)$. Since $\varphi$ is $2$-consistent, then $\varphi (R_{\{i,j\}}|\{\mu _i,\mu _i\})=\varphi (R)|_{\{i,j\}}$. Since $\varphi$ is $2$-efficient, then $\varphi (R_{\{i,j\}}|\{\mu _i,\mu _i\})=\mu |_{\{i,j\}}$. Thus, $\varphi _i(R)=\mu _i$ and $\varphi _j(R)=\mu _j$. Suppose now that $\mathrm{Top}(R_i,X_t)=\mathrm{Top}(R_j,X_t)=\mu _i$. By C1-2, $i=T_1(\mu _i,\pi _l)$. By definition of $T_1$, $i\succ ^\varphi _{\mu _i}j$. Thus, $\varphi (R_{\{i,j\}}|\{\mu _i,\mu _i\})=\mu |_{\{i,j\}}$. Since $\varphi (R_{\{i,j\}}|\{\mu _i,\mu _i\})=\varphi (R)|_{\{i,j\}}$, then $\varphi _i(R)=\mu _i$ and $\varphi _j(R)=\mu _j$.

$\bullet$ Case 2: The last component of $\varvec{\widetilde{\varPi }}$ is a pair. Let $\varPi _2\equiv (\pi _l)_{l=1}^m\equiv \widetilde{\varPi }$ and $T_2\equiv T$. Suppose without loss of generality that $\pi _m=\{1,2\}$. Let $Q:\mathcal{X }\times \mathcal{X }\setminus \{(x,x):x\in X\}\rightarrow \pi _m$ be the function defined by: for each $\{x,y\}$ such that $x\ne y$, $Q(x,y)$ is the agent who receives her preferred object at $(R_1^{xy},R_2^{xy})$. Let $f^\varGamma$ be the CHE-2 rule associated with $(\varPi _2,T_2,Q)$. We prove that $\varphi =f^\varGamma$. Let $N\in \mathcal{N }$, $X\subseteq \mathcal{X }$ such that $|N|=|X|$, and $R\in \mathcal{R }^N(X)$. Let $\mu \equiv f^{\varGamma (N,X)}(R)$. We claim that $\varphi (R)=\mu$. Let $l\in \{1,\dots ,m\}$ be such that $\pi _l\cap N\ne \emptyset$ and suppose that for each $l'<l$ and each $i\in \pi _{l'}\cap N$, $\varphi _i(R)=\mu _i$. If $l<m$, then by the same argument as in Case 1, $\varphi (R)|_{\pi _l\cap N}=\mu |_{\pi _l\cap N}$. We prove that $\varphi (R)|_{\pi _m\cap N}=\mu |_{\pi _m\cap N}$. Since for each $l'<l$ and each $i\in \pi _{l'}\cap N$, $\varphi _i(R)=\mu _i$, then $\bigcup _{i\in \pi _m\cap N}\{\varphi _i(R)\}=\bigcup _{i\in \pi _m\cap N}\{\mu _i\}$. Thus, if $\pi _m\cap N$ is a singleton, then $\varphi (R)|_{\pi _m\cap N}=\mu |_{\pi _m\cap N}$. Suppose that $\pi _m\cap N=\{1,2\}$. Since $\varphi$ is $2$-consistent, then $\varphi (R_{\{i,j\}}|\{\mu _i,\mu _i\})=\varphi (R)|_{\{i,j\}}$. Suppose that $\mathrm{Top}(R_1,\{\mu _1,\mu _2\})\ne \mathrm{Top}(R_2,\{\mu _1,\mu _2\})$. Since both $\varphi$ and $f^\varGamma$ are $2$-efficient, then $\varphi (R_{\{i,j\}}|\{\mu _i,\mu _i\})=f^{\varGamma (\{1,2\},\{\mu _1,\mu _2\})}(R_{\{1,2\}}|\{\mu _1,\mu _2\})$. Since $f^\varGamma$ is consistent, then $f^{\varGamma (\{1,2\},\{\mu _1,\mu _2\})}(R_{\{1,2\}}|\{\mu _1,\mu _2\})=\mu |_{\pi _l\cap N}$. Thus, $\varphi (R)|_{\pi _m\cap N}=\mu |_{\pi _l\cap N}$. Suppose now that $\mathrm{Top}(R_1,\{\mu _1,\mu _2\})=\mathrm{Top}(R_2,\{\mu _1,\mu _2\})=\varphi _1(R)$. Thus, by definition of $Q$, $Q(\mu _1,\mu _2)=1$. By C2-3, $\mu _1=\varphi _1(R)$. Thus, $\mu _2=\varphi _2(R)$. The case when $\mathrm{Top}(R_1,\{\mu _1,\mu _2\})=\mathrm{Top}(R_2,\{\mu _1,\mu _2\})=\varphi _2(R)$ is symmetric.

$\bullet$ Case 3: The last component of $\varvec{\widetilde{\varPi }}$ is a triple. Suppose without loss of generality that $\widetilde{\varPi }$ has $m-2$ components, i.e., $\widetilde{\varPi }=(\tilde{\pi }_l)_{l=1}^{m-2}$. Suppose also without loss of generality that $\tilde{\pi }_{m-2}=\{1,2,3\}$.

For each $i\in \{1,2,3\}$, let $O\left( i\right) \equiv \left\{ x\in \mathcal X :i\succ _{x}^{\varphi }\{1,2,3\}\setminus \left\{ i\right\} \right\}$. Since for each $x\in \mathcal{X }$, $\succ ^\varphi _x$ is antisymmetric, then for each $\{i,j\}\subseteq \{1,2,3\}$ such that $i\ne j$, $O\left( i\right) \bigcap O\left( j\right) =\emptyset$. We claim that $\bigcup _{i\in \{1,2,3\} }O\left( i\right) =\mathcal{X }$. To prove this, let $x\in \mathcal{X }$ and $z\in \mathcal{X }\backslash \left\{ x\right\}$. Since $|\mathcal{X }|\ge 4$, then by Lemma 6, $\succ _{xz}^{\varphi }$ is transitive. Thus, there is $i_{x}\in \{1,2,3\}$ such that $i_{x}\succ _{xz}^{\varphi }\{1,2,3\}\setminus \left\{ i_{x}\right\}$. From Lemma 6, $i_{x}\succ _{x}^{\varphi }\{1,2,3\}\setminus \left\{ i_{x}\right\}$ and $x\in \bigcup _{i\in \{1,2,3\}}O(i)$. We claim that $|\left\{ i\in \{1,2,3\}:O( i)\ne \emptyset \right\} |=2$. Since $|\tilde{\pi }_{m-2}|=3$, then there is no $i\in \tilde{\pi }_{m-2}$ such that either $i\succ ^\varphi \pi _{m-2}\setminus \{i\}$ or $\pi _{m-2}\setminus \{i\}\succ ^\varphi i$. This implies that $|\left\{ i\in \{1,2,3\}:O( i)\ne \emptyset \right\} |\ge 2$. We claim that $|\left\{ i\in \{1,2,3\}:O( i)\ne \emptyset \right\} |\le 2$. Suppose means of contradiction that there are $\{x,y,z\}\subseteq \mathcal{X }$ such that $x\in O(1)$, $y\in O(2)$, and $z\in O(3)$. Let $w\in \mathcal{X }\setminus \{ x,y,z\}$ and $h\equiv \left\{ 1\succ _{xw}2,1\succ _{xw}3,\right.$ $\left. 2\succ _{yw}1\right\}$. Direct computation yields $\left\{ 2\succ _{xw}3\right\} \subseteq \mathcal I (h,( R_{1}^{yxw},R_{2}^{xyw},R_{3}^{xyw}))$. Thus, $1\succ _{x}^\varphi 2\succ _{xw}^{\varphi }3$. By Lemma 5, $1\succ _{zw}^{\varphi }3$. This contradicts $z\in O(3)$. Suppose without loss of generality that $\left\{ i\in \{1,2,3\}:O(i)\ne \emptyset \right\} =\left\{ 1,2\right\}$. Since it is not true that $\left\{ 1,2\right\} \succ ^{\varphi }\left\{ 3\right\}$, then there are $i\in \left\{ 1,2\right\}$ and $\{x,w\}\subseteq \mathcal{X }$ with $x\ne w$, such that $3\succ _{xw}^{\varphi }i$. Suppose without loss of generality that $i=2$. Let $Y\equiv \left\{ z\in \mathcal{X }:3\succ _{zw}^{\varphi }2\right\}$. Clearly, $Y\not =\emptyset$. We claim that $O(1)=\mathcal{X }\setminus \{w\}$ and $O(2)=\{w\}$. To prove this, let $x\in Y$ and $z\in \mathcal{X }\setminus \{x,w\}$. Since $x\in Y$, then $3\succ ^\varphi _{xw}2$. Thus, $x\not \in O(2)$. Thus, $x\in O(1)$. Since $1\succ ^\varphi _{xw}3\succ ^\varphi _{xw}2$, then from Lemma 5, $1\succ ^\varphi _{zw}2$. Thus, $z\in O(1)$. Thus, $O(1)\supseteq \mathcal{X }\setminus \{w\}$. Since $O(2)\ne \emptyset$, then $O(2)=\{w\}$.

We claim that for each $z\in \mathcal{X }\setminus \left\{ w\right\}$, $1\succ _{w z}^{\varphi }3$ and for each $\{v,z\}\subseteq \mathcal{X }\setminus \left\{ w\right\}$ such that $v\ne z$, $2\succ _{vz}^{\varphi }3$. Let $h\equiv \left\{ 1\succ _{vz}2,1\succ _{vz}3,2\succ _{w z}\{1,2\}\right\}$. The direct calculation yields that $\mathcal I (h,(R_{1}^{w vz},R_{2}^{vw z},R_{3}^{vw z}) )=h\cup \left\{ 1\succ _{w z}3,2\succ _{vz}3\right\}$. Thus, $1\succ ^\varphi 3$.

Let $\varPi _3\equiv (\pi _l)_{l=1}^m$ be the partition of $\mathcal{A }$ defined as follows. For each $l<m-2$, let $\pi _l\equiv \tilde{\pi }_l$; let $\pi _{m-2}\equiv \{1\}$, $\pi _{m-1}\equiv \{2\}$, and $\pi _{m}\equiv \{3\}$. Let $T_3\equiv T$. Let $f^\varGamma$ be the CHE-3 rule associated with $(\varPi _3,T_3,w,Y)$. We prove that $\varphi =f^\varGamma$. Let $N\in \mathcal{N }$, $X\subseteq \mathcal{X }$ such that $|N|=|X|$, and $R\in \mathcal{R }^N(X)$. Let $\mu \equiv f^{\varGamma (N,X)}(R)$. We claim that $\varphi (R)=\mu$. Let $l\in \{1,\dots ,m\}$ be such that $\pi _l\cap N\ne \emptyset$ and suppose that for each $l'<l$ and each $i\in \pi _{l'}\cap N$, $\varphi _i(R)=\mu _i$. If $l<m-2$, then by the same argument as in Case 1, $\varphi (R)|_{\pi _l\cap N}=\mu |_{\pi _l\cap N}$. We prove that $\varphi (R)|_{\{1,2,3\}\cap N}=\mu |_{\{1,2,3\}\cap N}$. Since for each $l'<l$ and each $i\in \pi _{l'}\cap N$, $\varphi _i(R)=\mu _i$, then $\bigcup _{i\in \{1,2,3\}\cap N}\{\varphi _i(R)\}=\bigcup _{i\in \{1,2,3\}\cap N}\{\mu _i\}$. There are three cases.

Case 3-1: $\{1,2,3\}\cap N$ is the singleton $\{i\}$. Then $\varphi _i(R)=\mu _i$.

Case 3-2: $\{1,2,3\}\cap N$ is the pair $\{i,j\}$. Since $\varphi$ is $2$-consistent, then $\varphi (R_{\{i,j\}}|\{\mu _i,\mu _j\})=\varphi (R)|_{\{i,j\}}$. If $\mathrm{Top}(R_i,\{\mu _i,\mu _j\})\ne \mathrm{Top}(R_j,\{\mu _i,\mu _j\})$, then since $\varphi$ is $2$-efficient, we have that $\varphi (R_{\{i,j\}}|\{\mu _i,\mu _j\})=\mu |_{\{i,j\}}$. Suppose that $\mathrm{Top}(R_i,\{\mu _i,\mu _j\})= \mathrm{Top}(R_j,\{\mu _i,\mu _j\})$. Suppose that $\{i,j\}=\{1,2\}$. Since $2\succ ^\varphi _w 1$ and for each $x\in \mathcal{X }\setminus \{w\}$, $1\succ ^\varphi _x 2$, then by C3-3, $\varphi (R_{\{1,2\}}|\{\mu _1,\mu _2\})=\mu |_{\{1,2\}}$. Suppose that $\{i,j\}=\{1,3\}$. Since $1\succ ^\varphi 3$ then by C3-4 we have that $\varphi (R_{\{1,3\}}|\{\mu _1,\mu _3\})=\mu |_{\{1,3\}}$. Suppose that $\{i,j\}=\{2,3\}$. Since $2\succ ^\varphi _{zv} 3$ if and only if $w\in \{z,v\}$ and $\{z,v\}\setminus \{w\}\subseteq Y$, then by C3-5, $\varphi (R_{\{2,3\}}|\{\mu _2,\mu _3\})=\mu |_{\{2,3\}}$. In all cases $\varphi (R)|_{\{i,j\}}=\mu |_{\{i,j\}}$.

Case 3-3: $\{1,2,3\}\cap N$ is the triple $\{1,2,3\}$. If there is $i\in \{1,2,3\}$ such that $\varphi _i(R)=\mu _i$, then the same argument as in Cases 3-2 shows that $\varphi (R)|_{\{1,2,3\}\setminus \{i\}}=\mu |_{\{1,2,3\}\setminus \{i\}}$. Thus, it is enough to prove that $\varphi (R)$ and $f^\varGamma$ coincide for a member of $\{1,2,3\}$. If $\mathrm{Top}(R_1,\{\mu _1,\mu _2,\mu _3\})\ne w$, then by C3-3 and C3-4, $\mu _1=\mathrm{Top}(R_1,\{\mu _1,\mu _2,\mu _3\})$. Recall that $1\succ ^\varphi _{\mu _1}\{2,3\}$. Thus, since $\varphi$ is $2$-consistent and $2$-efficient, $\varphi _1(R)=\mu _1$. If $\mathrm{Top}(R_2,\{\mu _1,\mu _2,\mu _3\})=w$, then by C3-3 and C3-5, $\mu _2=w$. Recall that $2\succ ^\varphi _{w}\{1,3\}$. Thus, since $\varphi$ is $2$-consistent and $2$-efficient, $\varphi _2(R)=\mu _2$. Suppose then that $\mathrm{Top}(R_1,\{\mu _1,\mu _2,\mu _3\})=w$ and $\mathrm{Top}(R_2,\{\mu _1,\mu _2,\mu _3\})\ne w$. By C3-3, C3-4, and C3-5, $\mu _1=w$ and $\mu _2=\mathrm{Top}(R_2,\{\mu _1,\mu _2,\mu _3\})$. Denote $\{x,z\}\equiv \{\mu _1,\mu _2,\mu _3\}\setminus \{w\}$. There are two cases.

Case 3-3a. $w\,P_1\,\mu _2\,P_1\,\mu _3$. We claim that $\mu _1=w$. Suppose by contradiction that $\mu _1\ne w$. Recall that $1\succ ^\varphi _w3$. Since $\varphi$ is $2$-consistent and $2$-efficient, then $\varphi _3(R)\ne w$ and $\varphi _1(R)=\mu _2$. Thus, $\varphi _2(R)=w$. Since $\varphi$ is $2$-consistent then $\varphi _1(R_{\{1,2\}}|\{\mu _2,w\})=\mu _2$ and $\varphi _2(R_{\{1,2\}}|\{\mu _2,w\})=w$. This contradicts $2$-efficiency of $\varphi$.

Case 3-3b. $w\,P_1\,\mu _3\,P_1\,\mu _2$. We claim that $\mu _1=w$. By strategy-proofness of $\varphi$, $\varphi _1(R)\,R_1\,\varphi (R_{-1},R_1^{w\mu _2\mu _3})$. By Case 3-3a, $\varphi (R_{-1},R_1^{w\mu _2\mu _3})=w$. Thus, $\varphi _1(R)=w$. $\square$

Proof of Theorem 2

Let $\varphi$ be a $2$-efficiency, $2$-consistent, and conversely consistent rule. Our proof of Theorem 1 goes through for each rule satisfying $2$-efficiency and $2$-consistency, except for Case 3 (after Claim 1), which requires strategy-proofness. We claim that if in addition $\varphi$ is conversely consistent, then this case never holds. Suppose by contradiction that $O(1)=\mathcal{X }\setminus \{\omega \}$, $O(2)=\{\omega \}$, and $3\succ ^\varphi _{z\omega }2$ for some $z\in X\setminus \{\omega \}$. Consider the problem $R\equiv (\omega \,P_1\,x\,P_1\,z,$ $z\,P_2\,\omega \,P_2\,x, z\,P_3\,x \,P_3\,\omega )$. Let $\mu$ be the allocation defined by $\mu _1(R)\equiv x$, $\mu _2(R)\equiv \omega$, and $\mu _3(R)\equiv z$. Let $\mu '\equiv D^{1\succ 2\succ 3}(R)$. Both allocations $\mu$ and $\mu '$ are such that their restrictions for each two-agent reduced problem that results when one agent leaves are selected by the rule for the reduced problems (this is clear from the characterization of allocations in two-agent problems in the proof of Case 3 (after Claim 1), Theorem 1). Since $\varphi$ is conversely consistent, then $\mu =\varphi (R)=\mu '$. This is a contradiction. Thus, $\varphi$ is a rule that is described in Cases 1 and 2 (after Claim 1) in the proof of Theorem 1, i.e., $\varphi$ is a CHE-1 or a CHE-2 rule. One can easily prove the converse statement, i.e., that CHE-1 and CHE-2 rules are conversely consistent. $\square$

Example 7

(Efficient and conversely consistent rule that is not 2-consistent) Let $\mathcal{A }\equiv \{1,...,n\}$ with $n\ge 3$ and $\mathcal{X }$ such that $|\mathcal{X }|\ge 4$. We define rule $\varphi$. We do it first in two-agent problems. Since we define an efficient rule, it is enough to specify $F^\varphi$. Let $z^*\in \mathcal{X }$. $F^\varphi$ coincides with $n\succ n-1\succ \dots \succ 1$ for problems in which $z^*$ is not in the endowment. For each $\{i,j\}\subseteq \mathcal{A }$ such that $i> j$ and $i>3$, $i\succ ^\varphi j$; $3\succ ^\varphi 2$; and for each $x\ne z^*$ and each $y\in \mathcal{X }\setminus \{x\}$, $3\succ ^\varphi _{xy}1$, $1\succ ^\varphi _{z^*x}\{2,3\}$, $1\succ ^\varphi _{xz^*}2$, and if $y\ne z^*$, $2\succ ^\varphi _{xy}1$. Let $R\in \mathcal{D }$, $R\equiv (R_i)_{i\in N}\in \mathcal{R }^N(X)$. One can easily see that if $|N\cap \{1,2,3\}|< 3$ or $z^*$ is not in $X$, then there is a unique $\mu \in Z(R)$ from which no two agents can gain from trading and such that for each $\{i,j\}\subseteq N$, if $\mu _i\,P_i,\mu _j$ and $\mu _i\,P_j\,\mu _j$, then $i\succ ^\varphi _{\mu _i\mu _j}j$. Moreover, such a $\mu$ is efficient for $R$. Thus, in order to prove that the two-agent rule so defined can be extended to a conversely consistent rule, we have to prove that for each problem $R\in \mathcal{R }^{\{1,2,3\}}(x,y,z^*)$ there is at most one $\mu \in Z(R)$ such that no two agents can gain from trading their consumptions at $\mu$ and for each $\{i,j\}\subseteq N$, if $\mu _i\,P_i,\mu _j$ and $\mu _i\,P_j\,\mu _j$, then $i\succ ^\varphi _{\mu _i\mu _j}j$. Moreover, we have to prove that if such a $\mu$ exists, it is efficient for $R$. Since for each $x\ne z^*$ and each $y\in \mathcal{X }\setminus \{x\}$, $3\succ ^\varphi _{xy}1$ and $3\succ ^\varphi 2$, then it is enough to restrict to the case where $z^*\,R_3\,x\,R_3\,y$ and assume that if there is such a $\mu$, then $\mu _3\in \{z^*,x\}$. There is at most one $\mu$ satisfying the required conditions and such that $\mu _3=z^*$. Moreover, if such a $\mu$ exists, it is efficient. Then, suppose that such a $\mu$ and exists. We prove that no $\mu '$ such that $\mu '_3=x$ satisfies the required conditions. Since $\mu _3=z^*$ and for each $x\ne z^*$, $1\succ ^\varphi _{z^*x}\{2,3\}$, then $\mu _1\,P_1\,z^*$. Now, since $3\succ ^\varphi 2$, then $\mu '_2\ne z^*$. Since $\mu '_3=x$, then $\mu '_1=z^*$ and $\mu '_2=y$. Then, $\mu _1=y$, for otherwise agents $1$ and $3$ would benefit by trading their consumption at $\mu '$. Then, either agents $2$ and $1$ can benefit by trading their consumption at $\mu '$ or $\mu '_2\,P_1\,\mu '_1$, $\mu '_2\,P_2\,\mu '_1$, and $1\succ ^\varphi _{\mu '_2\mu '_1}2$. Thus, $\mu '$ does not satisfy the required conditions. Now, there is at most one $\mu '$ satisfying the required conditions and such that $\mu '_3=x$. Thus, it is only left to prove that if $\mu '\in Z(R)$ is such that $\mu '_3=x$ satisfies the required conditions, then $\mu '$ is efficient. Since $3\succ ^\varphi 2$, then $\mu '_2\ne z^*$. Since $\mu '_3=x$, then $\mu '_1=z^*$, and $\mu '_2=y$. Then, $z^*\,P_1\,x$, for otherwise agents $1$ and $3$ can benefit by trading their consumptions at $\mu '$. If $y\,P_1\,z^*$, then either agents $2$ and $1$ can benefit by trading their consumption at $\mu '$ or $\mu '_2\,P_1\,\mu '_1$, $\mu '_2\,P_2\,\mu '_1$, and $1\succ ^\varphi _{\mu '_2\mu '_1}2$. Thus, $z^*\,P_1\,y$. Thus, agent $1$ is receiving her top choice in $X$ at $\mu '$. Thus, since agents $1$ and $2$ cannot benefit by trading from $\mu '$, $\mu '$ is efficient for $R$. $\square$

Appendix 3: Efficiency and consistency

We define a family of rules for a domain with three potential agents, say $\mathcal{A }\equiv \{1,2,3\}$. In order to simplify our presentation, we introduce the serial priority (SP) rule associated with an order $i\succ j\succ k$, denoted $D^\succ$. In each problem, $D^\succ$ recommends the allocation at which the highest ranked agent with respect to $\succ$ receives her top choice, the second highest ranked agent receives her top choice among the remaining objects, and so on.

With some exceptions a tail rule coincides with $D^\succ$ for a priority, say $1\succ 2\succ 3$. The exceptions are given by the presence of an object in the endowment that may reverse the priority between agents $3$ and $2$ and between agents $2$ and $1$.

A tail rule is parameterized by a list of four elements, $(\gamma ,\omega _2,Y,E)$, which we call a “tail structure.” The first component, $\gamma$, is a renaming of the agents, i.e., a bijection from $\{1,2,3\}$ to $\mathcal{A }$. The second component is an object $\omega _2\in \mathcal{X }$. The third component is a non-empty set of objects $Y\subseteq \mathcal{X }\setminus \{\omega _2\}$. Let $\mathcal{R }(\gamma ,\,\omega _2,Y)\subseteq \mathcal{R }^{\mathcal{A }}$ be the set of problems

$$\begin{aligned} \mathcal{R }(\gamma ,\,\omega _2,Y)\equiv \left\{ R\in \mathcal{R }^\mathcal{A }(X):\begin{array}{l} X=\{\omega _2,z,x\}\ ,z\in Y,\ \omega _2\,P_{\gamma (1)}\,x\,P_{\gamma (1)}z,\\ z\,P_{\gamma (2)}\,\omega _2\,P_{\gamma (2)}\,x,\ z\,P_{\gamma (3)}\,\omega _2,\ z\,P_{\gamma (3)}\,x.\end{array} \right\} . \end{aligned}$$

The last element of the tail structure, i.e., $E$, is an arbitrary subset of $\mathcal{R }(\gamma ,\,\omega _2,Y)$. The tail rule associated with ${(\gamma ,\omega _2,Y,E)}$, denoted $\mathcal T \langle \gamma ,\omega _2,Y,E\rangle$, coincides with the serial priority $D^{\gamma (1)\succ \gamma (2)\succ \gamma (3)}$, except in the five cases listed in Table 2.

Table 2 Exceptions in the definition of the tail rule $\mathcal{T }\langle \gamma ,\omega _2,Y,E\rangle$

Lemma 8

Tail rules are efficient and consistent.

Proof

Suppose that $\mathcal A =\{1,2,3\}$. Let $\left( \gamma ,\omega _2,Y,E\right)$ be a tail structure such that $\gamma (1)=1$, $\gamma (2)=2$, and $\gamma (3)=3$. We want to prove that $\mathcal T \langle \gamma ,\omega _2,Y,E\rangle$ is efficient and consistent. We use the shorthand $\mathcal T$ for $\mathcal T \langle \gamma ,\omega _2,Y,E\rangle$ hereafter. Let $R\in \mathcal{R }$. Since $\mathcal T (R)=D^{\succ _R}$ for some linear order $\succ _R$, then $\mathcal T (R)\in P(R)$. We prove $\mathcal T$ is consistent. We have to prove that for each $X\subseteq \mathcal{X }$ such that $|X|=3$, $R\in \mathcal{R }^\mathcal{A }$, and each $N\in \mathcal{N }$, $|N|=2$, $\mathcal T (R)|_{N}=\mathcal T (R_{N}|\bigcup _{i\in N}\mathcal T _i(R))$. Observe first that this is true in case $R$ falls in the exceptions listed in Table 2 (this can be seen by inspection of the allocations in each case). Otherwise, $\mathcal T (R)=D^{\succ }(R)$, where the order $\succ$ is given by $3\succ 2\succ 1$. By definition of $\mathcal T$, $\mathcal T (R_{\{3,1\}}|X')= D^{\succ }(R_{\{3,1\}}|X')$ where $X'=\bigcup _{i\in \{3,1\}}\mathcal T _i(R)$. Now, we claim that $\mathcal T (R_{\{3,2\}}|X'')=D^{\succ }(R_{\{3,2\}}|X'')$ where$X''=\bigcup _{i\in \{3,2\}}\mathcal T _i(R)$. Suppose by contradiction that this is not true. Thus, $\omega _2\in X$, i.e., $X=\{x,\omega _2,z\}$; besides, $R_{\{3,2\}}|X''=(\omega _2\,P_{\{3,2\}}\,z)$, and $\mathcal T _{3}(R_{\{3,2\}}|X'')=z$. By definition of $\mathcal T$, agent $2$ never gets an object that is worse than $\omega _2$. Thus, $\mathcal T _{2}(R)\,R_{2}\,\omega _2$; consequently $\mathcal T _{2}(R)=\omega _2$ and $\mathcal T _{3}(R)=z$.²⁸ This is a contradiction, because $D^{\succ }_{3}(R)\,R_{3}\,D^{\succ }_{2}(R)$. Let $X'''\equiv \bigcup _{l\in \{2,1\}}\mathcal T _l(R)$. To prove that $\mathcal T (R_{\{2,1\}}|X''')= D^{\succ }(R_{\{2,1\}}|X''')$, observe that if $\mathcal T (R_{\{2,1\}}|X''')=D^{1\succ 2}(R_{\{2,1\}}|X''')$, then: $\omega _2\in X$, $X=\{x,\omega _2,z\}$, $x\,P_{3}\,X\setminus \{x\}$, and $R_{\{2,1\}}|\{\omega _2,z\}=(x\,P_{\{2,1\}}\,z)$. Thus, $R$ falls in case IV in Table 2, which is a contradiction. $\square$

Example 8

(A tail rule)  Let $\mathcal A \equiv \left\{ 1,2,3\right\}$ and $\mathcal X \equiv \left\{ a,b,c,d\right\}$. Let $\gamma$ be the bijection defined as follows: $\gamma \left( 1\right) \equiv 1$, $\gamma \left( 2\right) \equiv 2$, and $\gamma \left( 3\right) \equiv 3$. Let $\omega _2\equiv a$, $Y\equiv \left\{ b,c\right\}$, and $E$ the set that contains the two problems $(a\, P_{1}\,b\, P_{1}\,c,c\, P_{2}\,a\, P_{2}\,b,$ $c\, P_{3}\,a\, P_{3}\,b)$ and $(a\, P_{1}\,d\, P_{1}\,b,b\, P_{2}\,a\, P_{2}\,d,$ $b\, P_{3}\,d\, P_{3}\,a)$. We illustrate $\mathcal T \langle \gamma ,a,Y,E\rangle$ by applying it to five sample three-agent problems (Table 3). Each problem is depicted with its respective recommendation in brackets. The first problem, from left to right, does not fall into the exceptional cases in Table 2. The second problem falls into Case III, the third problem falls into Case IV, and the fourth problem falls into Case V.

Table 3 Example 8

Remark 2

Tail rules are not strategy-proof: agent $\gamma (2)$ can gain by misstating her preferences at each $R\in E$. In Example 8, if $R=(a\, P_{1}\,d\, P_{1}\,b,b\, P_{2}\,a\, P_{2}\,d,$ $b\, P_{3}\,d\, P_{3}\,a)$, then agent $2$ can gain by reporting $b\, P_{2}'\,d\, P_{2}'\,a$.

Example 9

( A consistent HE rule that is not a CHE rule when $|\mathcal{X }|=3$ ) Let $\mathcal{A }\equiv \{1,2,3\}$ and $\mathcal{X }\equiv \{x_1,x_2,x_3\}$. Let $\varphi =f^\varGamma$ be the HE rule defined as follows: $\varGamma (\{1,2,3\})$ is the inheritance tree in which each $i\in \mathcal{A }$ labels the root of the three for $x_i$. Since each object is initially assigned to a different agent, there is no need to specify further inheritance (Pápai 2000). For the two-agent problems, for each $i\in \mathcal{A }$, each $a\in \mathcal{X }\setminus \{x_i\}$, and each $j\ne i$, $i\succ ^\varphi _{x_ia}j$ and $j\succ _{ax_i}^\varphi i$. We prove that $\varphi$ is consistent. Since $\varphi$ is efficient, then in each three-agent problem at least one agent gets her first choice (Abdulkadiroğlu and Sönmez 1998). If exactly one or three agents get their first choice, there is clearly no violations of consistency. If only two agents get their first choice, then either each is getting her own object, or they are trading in the first stage of the algorithm. Since for each $i\in \mathcal{A }$, each $a\in \mathcal{X }\setminus \{x_i\}$, and each $j\ne i$, $j\succ _{ax_i}^\varphi i$, then there is no violation of consistency.