Efficient and incentive compatible exchange of real-time information

Abstract

We consider the problem of coordinating the exchange of real-time information. For example, a US Department of Transportation pilot program seeks to reduce traffic accidents by allowing each vehicle to transfer crash-relevant information (e.g. position, speed, braking status) to and from neighboring vehicles. Time is of the essence: vehicle information becomes stale quickly. Electronic files are the medium of information and corrupt/useless when partial; hence, we model them as discrete, indivisible objects. Furthermore, each agent may perfectly replicate an object in his possession and transfer it to another agent. However, replication and transfer takes time (e.g. due to bandwidth constraints), and scarcity arises due to fact that information quickly becomes valueless. How should agents transfer such objects? We study efficiency, strategy-proofness, withholding-proofness, and introduce a new axiom based on the concept of reciprocity. Our results are as follows: when each agent owns one object and consumes only one object, we identify a family of rules satisfying all four axioms. If each agent owns a bundle of objects and consumes a bundle, then the four axioms are incompatible. If agents live in a network, then for a large class of incomplete networks, efficiency is only compatible with strategy-proofness.

Appendices

Appendix 1

In Appendix 1.1, we show that $TTC-SI^{\sigma }$ satisfies non-bossiness. In Appendix 1.2, we show $TTC-SI^{\sigma }$ satisfies withholding-proofness. Let $\varphi \equiv TTC-SI^{\sigma }$.

1.1 1.1 Non-bossiness

Non-bossiness

For each $E=(R,\Omega )\in \mathcal {E}$, each $i\in N$, and each $R'_{i}\in \mathcal {R}$, if $\varphi _{i}(E)=\varphi _{i}(R'_{i},R_{-i},\Omega )$, then

$$\begin{aligned} \varphi (E)=\varphi (R'_{i},R_{-i},\Omega ). \end{aligned}$$

Let $E=(R,\Omega )\in \mathcal {E}$, and $E'=(R_{i}',R_{-i},\Omega )\in \mathcal {E}$ be s.t. $\varphi _{i}(E')=\varphi _{i}(E)$. For each $k\in \{1,\dots ,\bar{k}(E)\}$, let

• $N^{k}\equiv \{M\subseteq N:$ each $i\in M$ is a remaining agent at Step k under $E\}$,

• $W^{k}\equiv \{w\in 2^{N}:$$w=\{w_{1},\dots ,w_{z}$}, and at Step k under E, $w_{z}$ points to $w_{z-1}$ points to... $w_{1}$ points to $\Omega _{i}\}$,

• $S^{k}\equiv \{s\in 2^{N}:$s is a cycle $s_{1},\dots ,s_{z}$ at Step k under E, and there is no $t\in s$ s.t. $t\in \bigcup _{g\in \{1,\dots ,k\}}\bigcup _{w\in W^{g}}w\}$, and

• $U^{k}\equiv \{s\in 2^{N}\backslash S^{k}:$s is a cycle $s_{1},\dots ,s_{z}$ at Step k under $E\}$,

At Step k, $S^{k}$ and $U^{k}$ are the set of cycles which occur (Fig. 3). In each cycle in $S^{k}$, no agent receives $\Omega _{i}$; while in $U^{k}$, some agent receives $\Omega _{i}$. At Step k, each $s\in W^{k}$ forms a pointing “chain” of agents for which the last agent points to $\Omega _{i}$.

Let V be the Step under E at which i is assigned his object, and $W\in W^{V}$ be s.t. $i\in W$. Let Q be the first Step under E at which the cycle W forms. Let $V'$ be the Step under $E'$ at which i is assigned his object. For each $j\in N$, let $C(j,E)\subseteq N$ be the cycle j is in under E. For each $k\in \{1,\dots ,V-1\}$, since i is not assigned yet, $U^{k}=\emptyset$. For each $k\in \{V,\dots ,\bar{k}(E)\}$, either $U^{k}=\emptyset$, or there is only one cycle in $U^{k}$. Abusing notation, write $U^{k}$ as this cycle.

The proof proceeds by showing that each cycle occurring in the algorithm under E occurs under $E'$. Furthermore, for each $k\in \{1,\dots ,\bar{k}(E)\}$, each cycle occurring at Step k under E may not occur at an earlier round. This can be shown using the technique in Step 3 of Claim 4.

Case 1$V'=V$. At Step 1 under E and $E'$, each agent $j\in N\backslash \{i\}$ points to the same object, the same cycles $S^{1}$ occur, and $U^{1}=\emptyset$. Since $x^{1}$ under E and $E'$ are the same, the set of attainable objects objects at Step 2 under E and $E'$ are the same. Furthermore, $N^{2}$ agents remain at Step 2 under E and $E'$.

Iterating this reasoning for Steps $2,\dots ,\bar{k}(E)$, we have $x^{2},\dots ,x^{\bar{k}(E)}$ under E and $E'$ are the same, implying $\varphi (E')=\varphi (E)$. Note that i may point to different objects at Step $k\in \{1,\dots ,\bar{k}(E)-1\}$, but since he is not in a cycle, this does not matter.

Fig. 3

The schematic above shows the steps of the algorithm under E and $E'$ when $V'>V$. Underneath the column E, $S^{1}$ and $U^{1}$ are particular sets of cycles which occur at Step 1, and similarly for each other row. Claim 1 shows cycles $S^{V},\dots ,S^{V'}$ occur under $E'$; Claim 2 shows $U^{V'}$ occurs under $E'$

We consider the case when $V'>V$. If $V'<V$, then a symmetric reasoning holds after re-labelling.

Case 2 $V'>V$. Following the reasoning of Case 1, for each $j\in \bigcup _{g\in \{1,\dots ,V\}}\bigcup _{s\in S^{g}}s$, $\varphi _{j}(E)=\varphi _{j}(E')$. So each agent belonging to a cycle not involving i at Steps $1,\dots ,V$ receives the same assignment under E and $E'$. Since $V'>V$, the cycle C(i, E) does not occur at Step V under $E'$.

Lemma 2

If $V'>V$, then for Step V under E, there is no $h\in \bigcup _{k\in \{1,\dots ,V\}}\bigcup _{s\in W^{k}}s$ s.t. h is the earliest-priority remaining agent.

If $V'>V$, then at Step V under $E'$, i does not point to $\varphi _{i}(E)$. If there is an agent that 1) points to an object that points to an agent... that points to $\Omega _{i}$ in any Step from 1 to V under E, and 2) is the earliest-priority agent at Step V under E, then i will be assigned the object he points to at Step V under $E'$—violating the assumption that $\varphi _{i}(E)=\varphi _{i}(E')$.

Proof

Suppose by contradiction, there is $k\in \{1,\dots ,V\}$, and $h\in \bigcup _{s\in W^{k}}s$ s.t. h is the earliest-priority remaining agent at Step V under E, and hence also at Step V under $E'$. Since $V'>V$, at Step V under $E'$, i does not point at an agent in $\bigcup _{s\in W^{k}}s$ (otherwise, he is in a cycle). Then, i points to an object $a\ne \varphi _{i}(E)$ that points to $\ell \in N$, and $a\,R_{i}'\,\varphi _{i}(E)$. If $\Omega _{\ell }\ne a$, then $\ell =h$, i is in a cycle, and $a=\varphi _{i}(E')\ne \varphi _{i}(E)$—violating the assumption that $\varphi _{i}(E)=\varphi _{i}(E')$. If $\Omega _{\ell }=a$, then either 1) $\ell$ is in a cycle with i, implying $a=\varphi _{i}(E')\ne \varphi _{i}(E)$, or 2) $\ell$ is in a cycle, but not with i. First, in Step $V+1$ under $E'$ (after $\ell$ is in a cycle), for each $s\in W^{k}$, each agent in s still points to the same object as in Step V under $E'$, and hence, the same as in Step V under E. Second, since only one agent was assigned a, and no agent has been assigned $\Omega _{i}$, a is still attainable at Step $V+1$ under $E'$. Since h is the earliest-priority remaining agent at Step V under $E'$, he is so at Step $V+1$ under $E'$. Then, i points to object a points to h, and by definition of $h\in \bigcup _{s\in W^{k}}s$, $h=w_{z}$ points to $w_{z-1}$ points to... points to $w_{1}$ points to $\Omega _{i}$ points to i. Then, $a=\varphi _{i}(E')\ne \varphi _{i}(E)$. $\square$

Since $C(i,E)\in W^{V}$, Lemma 2 implies no $h\in C(i,E)$ is earliest at Step V under E. Hence, for each $h\in C(i,E)$, at Step V under $E'$, h points to an object in $\bigcup _{j\in C(i,E)}\Omega _{j}$. At Steps $V+1,\dots ,V'$, each $h\in C(i,E)\backslash \{i\}$ still points to the same agent, implying C(i, E) occurs at Step $V'$ under $E'$.

Claim 1

For each $j\in \bigcup _{k\in \{0,\dots ,V'-V\}}\bigcup _{s\in S^{V+k}}s$, $\varphi _{j}(E')=\varphi _{j}(E)$. Furthermore, for each $k\in \{0,\dots ,V'-V\}$, the set of cycles $S^{V+k}$ occurs at Step $V+k$ under E and $E'$, and for each $k\in \{1,\dots ,V'-V-1\}$, $U^{V+k}=\emptyset$.

Proof

We have shown above that for each $j\in \bigcup _{s\in S^{V}}s$ , $\varphi _{j}(E)=\varphi _{j}(E')$, and the set of cycles $S^{V}$ occurs at Step V under E and $E'$.

Step 1For each$j\in \bigcup _{k\in \{0,\dots ,V'-V\}}\bigcup _{s\in S^{V+k}}s$, $\varphi _{j}(E')=\varphi _{j}(E)$, and for each$k\in \{0,\dots ,V'-V\}$, the set of cycles$S^{V+k}$occurs at Step$V+k$underEand$E'$.

We proceed by induction. For each $k\in \{1,\dots ,V'-V-1\}$, let Case k be defined as below. For each $j\in \bigcup _{g\in \{0,\dots ,V+k-1\}}\bigcup _{s\in S^{V+g}}s$, suppose $\varphi _{j}(E)=\varphi _{j}(E')$. For each $g\in \{0,\dots ,V+k-1\}$, suppose 1) $S^{g}$ occurs at Step $V+g$ under E and $E'$, and 2) $U^{V+g}=\emptyset$.

Subcase 2.k$j\in \bigcup _{s\in S^{V+k}}s$. Then, at Step V under E and $E'$, j either points to an agent h in $S^{V}$, $S^{V+1}$,..., $S^{V+k}$, or C(i, E). Since $U^{V+1}=\cdots =U^{V+k+-1}=\emptyset$, j cannot point to an agent in $U^{V+1}$,..., or $U^{V+k-1}$.

Subcase 2.k.1h is in C(i, E). Then, h still points to i until Step $V'$ under $E'$, implying j still points to i at least until Step $V'$ under $E'$. Since $V'>V+k$, this is impossible.

Subcase 2.k.2h is in $S^{V},\dots ,S^{V+k-1}$. First, we show the set of attainable objects at $V+k$ under E and $E'$ are the same. Suppose by contradiction, there is an object b such that b is attainable at Step $V+k$ under $E'$, but not attainable at Step $V+k$ under E. At Steps $V,\dots ,V+k-1$ under $E'$, only one less cycle occurred compared to E: C(i, E). Hence, the occurrence of C(i, E) at Step V under E “causes” the unattainability of b. If for each $\ell \in C(i,E)$, $\varphi _{\ell }(E)\in \bigcup _{s\in C(i,E)}\Omega _{s}$, then each $c\in \bigcup _{s\in C(i,E)}\Omega _{s}$ is still attainable at Step $V+k$ under $E'$. Hence, there are $\ell ,\ell '\in C(i,E)$ such that $\varphi _{\ell }(E)\notin \bigcup _{s\in C(i,E)}\Omega _{s}$, and $\varphi _{\ell }(E)\ne \Omega _{\ell '}$ points to $\ell '$. Since $\ell '\in \bigcup _{s\in W^{V+k}}s$, by Lemma 2, this is impossible. So the set of attainable objects is the same at $V+k$ under E and $E'$.

Let $M\equiv N^{V+k}\cup C(i,E)$, and $\ell \equiv \arg \max _{g\in M}\sigma ^{-1}(g)$ be the earliest-priority remaining agent at $V+k$ under $E'$. Since $\ell \notin C(i,E)$, $\ell =\arg \max _{g\in M\backslash C(i,E)}\sigma ^{-1}(g)$, and $\ell$ is the earliest-priority remaining agent at $V+k$ under E. Hence, if c is in the set of attainable objects given $x^{V+k-1}$ under E, and not in $\bigcup _{h\in C(i,E)}\Omega _{h}$, then c points to the same agent $j^{c}$ at Step $V+k$ under E and $E'$.

At Step $V+k$ under E and $E'$, j cannot point to an object in $\bigcup _{g\in C(i,E)}\Omega _{g}$, because the reasoning of Case 2.k.1 would hold. So since the set of attainable objects is the same at Step $V+k$ under E and $E'$, j points to the same object c at both Steps. As shown in the previous paragraph, c points to the same agent $j^{c}$ at both Steps.

Hence, j points to the same agent at $V+k$ under E and $E'$.

Subcase 2.k.3h is in $S^{V+k}$. Then, since each object and agent still points to the same agent g until g leaves, at Step $V+k$ under E and $E'$, j points to the same agent.

Hence, for each $j\in \bigcup _{s\in S^{V+k}}s$, at Step $V+k$ under $E'$ and E, j points to the same object, which in turn points to the same agent. The same set of cycles $S^{V+k}$ occurs, and $\varphi _{j}(E')=\varphi _{j}(E)$. Subcase 2.k is complete.

Step 2$U^{V+k}=\emptyset$. Let $M\equiv N^{V+k}\cup C(i,E)$. Suppose by contradiction $U^{V+k}\ne \emptyset$, then at Step $V+k$ under E there are $u_{1},\dots ,u_{z}\in U^{V+k}$ and $j\in \{u_{1},\dots ,u_{z}\}$ s.t. $j=u_{z}$ points to $u_{z-1}$ points to... $u_{1}$ points to $\Omega _{i}$ points to j. This means, $j=\arg \min _{h\in M\backslash C(i,E)}\sigma ^{-1}(h)$. Consider Step V under $E'$. Let h be the earliest-priority remaining agent at Step V under $E'$.

If $h=i$, then since $V'\ne V$, i points to $a\ne \varphi _{i}(E)$. If the owner of a remains, then i still points to him. By finiteness of the algorithm, there is $g\in \mathbb {N}$ s.t. the owner of a is assigned at $V+g$ under $E'$, and at Step $V+g+1$ under $E'$, a points to i and $\varphi _{i}(E')=a$. If the owner of a does not remain, then i is in a cycle with a, and $\varphi _{i}(E')=a$. This contradicts the hypothesis that $\varphi _{i}(E)=\varphi _{i}(E')$.

If $h\in \bigcup _{g\in \{0,\dots ,k-1\}}\bigcup _{s\in S^{V+g}}s$, then there are two cases. At Step $V+k$ under $E'$, i points to either an object $a\ne \varphi _{i}(E)$, or $\varphi _{i}(E)$. Let $j^{a}$ be the owner of a.

Case 1 At Step $V+k$underE, ipoints to$a\ne \varphi _{i}(E)$, and$j^{a}$is a remaining agent. By finiteness of the algorithm, there is a $f\in \{k,\dots ,\bar{k}(E)-V\}$ such that at Step $V+f$ under $E'$, $j^{a}$ is in a cycle. At Step $V+k$ under $E'$, the facts that M agents remain and $j=\arg \min _{h\in M\backslash C(i,E)}\sigma ^{-1}(h)$ imply the earliest-priority remaining agent $\ell$ is such that $\ell \in C(i,E)\cup j$. Since the set of remaining agents is contained in M, $\ell$ is still the earliest-priority remaining agent at Step $V+f+1$ under $E'$. So a now points to $\ell$.

Subcase 1.1$\ell =j$. By reasoning of Subcase 2.k.2, for each $u\in U^{V+k}$, at Step $V+k$ under E and $E'$, u points to the same object, which in turn points to the same agent. Recall, by definition of j and $U^{V+k}$, at Step $V+k$ under E, $j=u_{z}$ points to $u_{z-1}$ points to... $u_{1}$ points to $\Omega _{i}$ points to j. Since i is a remaining agent, at Step $V+k$ under $E'$, we have $j=u_{z}$ points to $u_{z-1}$ points to... $u_{1}$ points to $\Omega _{i}$ points to i points to a. Each agent still points in this manner until Step $V+f+1$ under $E'$. At Step $V+f+1$ under $E'$, j is the earliest-priority remaining agent, and a points to j. Then, since i points to a, a cycle forms and $\varphi _{i}(E')=a\ne \varphi _{i}(E)$, a contradiction to the hypothesis.

Subcase 1.2$\ell \in C(i,E)$. Similarly as above, at Step $V+f+1$ under $E'$, a points to $\ell$. Hence, there is a cycle $\{u_{z},\dots ,u_{1}\}\subseteq C(i,E)$ such that $\ell =u_{z}$ points to $u_{z-1}$ points to... $u_{1}$ points to $\Omega _{i}$ points to i points to a. Then, $\varphi _{i}(E')=a\ne \varphi _{i}(E)$, a contradiction.

Case 2 At Step $V+k$under$E'$, ipoints to$\varphi _{i}(E)$. At Step V under E and $E'$, since $x^{V-1}$ under E and $E'$ are the same, each remaining agent points to the same object. Consider agents in the cycle C(i, E). Lemma 2 implies that for each $g\in C(i,E)$, g points to an object a such that $a\in \bigcup _{j\in C(i,E)}\Omega _{j}$. Hence, at Steps $V,\dots ,V+k$ under $E'$, each $g\in C(i,E)\backslash \{i\}$ still points to the same agent. Hence, if i points to $\varphi _{i}(E)$, then i is in a cycle at Step $V+k$ under $E'$. Since $V+k<V'$, this is impossible. $\square$

Claim 2

For each $j\in U^{V'}$, $\varphi _{j}(E')=\varphi _{j}(E)$.

Proof

If $U^{V'}=\emptyset$, then we are done. Let $U^{V'}\ne \emptyset$ be a cycle.

Step 1 There is an agent in uwho is the earliest-priority remaining agent at Step$V'+1$under$E'$. At Step $V'$ under E, by definition, there is an agent in $U^{V'}$ who is assigned $\Omega _{i}$. Since $V'>V$, $i\notin N^{V'}$ and $\Omega _{i}$ points to another agent $\ell \in U^{V'}$. At Step $V'+1$ under $E'$, only one extra cycle has already occurred compared to Step $V'$ under E: $S^{V'}$. Hence, the set of remaining agents at Step $V'$ under E contains the set of remaining agents at Step $V'+1$ under $E'$. Since $\ell$ is earliest at Step $V'$ under E, this implies $\ell$ is earliest at Step $V'+1$ under $E'$.

Step 2 Each $j\in U^{V'}$points to the same object at Step$V'$underEand Step$V'+1$under$E'$. Let a be the object j points to at Step $V'$ under E, and b be the object j points to at Step $V'+1$ under $E'$. Suppose by contradiction, $a\ne b$. Then at Step $V'$ under E, a is attainable given $x^{V'-1}$ under E. At Step $V'$ under E, the cycles $S^{1},\dots ,S^{V'-1},C(i,E)$ have occurred; at Step $V'+1$ under $E'$, the cycles $S^{1},\dots ,S^{V'},C(i,E)$ have occurred. Hence, at Step $V'+1$ under $E'$ the only extra set of cycles which occurs is $S^{V'}$. If each cycle $s\in S^{V'}$ is s.t. $s=u_{1},\dots ,u_{z}$ and $u_{z}$ points to $\Omega _{u_{z-1}}$ points to $u_{z-1}$ points to $\Omega _{u_{z-2}}$ points to... $\Omega _{u_{1}}$ points to $u_{1}$ points to $\Omega _{z}$, then each object which points to an agent in $S^{V'}$ is still attainable. Hence, there must be a cycle $s\in S^{V'}$ and $j\in s$ s.t. j does not point to an object in $\bigcup _{g\in s}\Omega _{g}$. This means some agent $\ell '\in s$ is the earliest-priority remaining agent at Step $V'$ under E. This is impossible, as we showed in Step 1, $\ell \in U^{V'}$ is the earliest-priority remaining agent.

Step 2implies $U^{V'}$ occurs at Step $V'+1$ under $E'$, and for each $j\in U^{V'}$, $\varphi _{j}(E')=\varphi _{j}(E)$. $\square$

For each $k\in \{V'+1,\dots ,\bar{k}(E)\}$, let

• $Q^{k}\equiv \{s\in 2^{N}:$s is a cycle at Step k under E, and for each $j\in s$, there is $h\in s$ s.t. j points to $\Omega _{h}\}$, and

• $T^{k}\equiv \{s\in 2^{N}:$s is a cycle at Step k under E, and there is $j\in T^{k}$ and $a\in \mathcal {O}$ s.t. j points to a, and $a\notin \bigcup _{h\in T^{k}}\Omega _{h}\}$.

That is, at Step k under E, $Q^{k}$ is the set of cycles which only involve the endowments of the agents in the cycle, and $T^{k}$ is the cycle for which one agent points to an object whose owner has left.

Claim 3

For each $j\in \bigcup _{s\in Q^{V'+1}\cup T^{V'+1}}s$, $\varphi _{j}(E')=\varphi _{j}(E)$. Furthermore, the cycles $Q^{V'+1}$ and $T^{V'+1}$ both occur either at $V'+1$ under $E'$ or $V'+2$ under $E'$.

Proof

If $U^{V'}=\emptyset$, then at Step $V'+1$ under E and $E'$, the same cycles $S^{1},\dots ,S^{V'},C(i,E)$ have occurred. Hence, $x^{V'}$ under E and $E'$ is the same, the set of remaing agents is the same, and each remaining agent at Step $V'+1$ points to the same object under E and $E'$. So the same cycles $Q^{V'+1}$ and $T^{V'+1}$ occur at Step $V'+1$ under $E'$.

If $U^{V'}\ne \emptyset$, then before Step $V'$ under E, the cycles $S^{1},\dots ,S^{V'-1},C(i,E)$ have occurred. At Step $V'-1$ under $E'$, only the cycles $S^{1},\dots ,S^{V'-1}$ have occurred.

Step 1 No agent assigned in Steps $V+1,\dots ,V'-1$underEis assigned an object in$\bigcup _{s\in C(i,E)}\Omega _{s}$. Suppose by contradiction, there is $j\in \bigcup _{g\in \{1,\dots ,V'-V-1\}}\bigcup _{s\in S^{V+g}}s$, $h\in C(i,E)$, and $k\in \{1,\dots ,V'-V-1\}$ s.t. j points to $\Omega _{h}$ at Step $V+k$ under E. By Claim 2, $S^{V+k}$ cycles occur at Step $V+k$ under $E'$. Since h is not assigned until $V'$, $\Omega _{h}$ still points to h until $V'$. Since $V'>V+k$, j cannot have received $\Omega _{h}$ at Step $V+k$ under $E'$.

Step 2 The set of attainable objects at Step $V'$underEand$E'$are the same. Before Step $V'$ under $E'$, the cycles $S^{1},\dots ,S^{V'-1}$ have occurred. Before Step $V'$ under E, the cycles $S^{1},\dots ,S^{V'-1},C(i,E)$ have occurred. Since each object $a\in \bigcup _{s\in C(i,E)}\Omega _{s}$ has only been assigned once (at Step V under E), each such a is attainable given $x^{V'-1}$ under E. Since C(i, E) comprises the only difference between $x^{V'-1}$ under E and $E'$, the set of attainable objects at Step $V'$ under E and $E'$ are the same.

Step 3 Each $s\in Q^{V'+1}$occurs at$V'+2$underE. Since the set of attainable objects given $x^{V'-1}$ under E and $E'$ are the same, at Step $V'$ under E and $E'$, each $j\in \bigcup _{s\in Q^{V'+1}}s$ points to the same object. As argued before, $U^{V'}\ne \emptyset$ implies there is an agent in $U^{V'}$ who is the earliest-priority remaining agent at Step $V'$ under E. This means, for each $s\in S^{V'}$, each $h\in s$ points to an object a s.t. there is $h'\in s$ with $a=\Omega _{h'}$. Hence, at Step $V'+1$ under E, each $b\in \bigcup _{s\in S^{V'}}\bigcup _{h\in s}\Omega _{h}$ is still attainable given $x^{V'}$ under E. Since each $j'\in \bigcup _{s\in Q^{V'+1}}s$ is assigned an object in the set $\bigcup _{s\in Q^{V'}}\bigcup _{h\in s}\Omega _{h}$, each $j'$ points to an agent who is not in $S^{V'}$ at Step $V'$ under E. Then, at Step $V'$ under E either $j'$ points to $\varphi _{j'}(E)$ or an agent in $U^{V'}$. There is at least one $h\in \bigcup _{s\in Q^{V'+1}}s$ s.t. h points to an agent in $U^{V'}$ at Step $V'$ under E, otherwise $Q^{V'+1}$ occurs at Step $V'$ under E.

Without loss of generality, let h be such an agent, $a\in \bigcup _{j\in U^{V'}}\Omega _{j}$, and $j^{a}$ be the owner of a. Since $j^{a}$ has not been assigned an object, a has not been assigned to any agent. There is a transfer process in which each agent in $C(j^{a},E)$ transfers to the next agent in $C(j^{a},E)$ in the first round. Then, a is still attainable given $x^{V'}$ under E: $j^{a}$ transfers to another agent in the second round. Hence, if h points to a at Step $V'$ under E, then h still points to a at Step $V'+1$ under E. By definition of $Q^{V'}+1$, there is $h'\in N^{V'+1}$ s.t. h is assigned $\Omega _{h'}$. Since $j^{a}\notin N^{V'+1}$, this is a contradiction. Hence, at Step $V'$ under E, h points to an object $a\notin \bigcup _{j\in U^{V'}}\Omega _{j}$, and a points to an agent in $U^{V'}$. Recall there is $u\in U^{V'}$ s.t. u is the earliest-priority remaining agent at Step $V'$ under E, and by definition of $U^{V'}$, $\Omega _{i}$ is assigned to some agent. In C(u, E), since $\Omega _{i}$ points to u, and $\Omega _{i}$ is the only object which does not point to its owner, at Step $V'$ under E, h must point to $\Omega _{i}$.²⁴

Since no agent in $\bigcup _{s\in S^{V'}}s$ points to $\Omega _{i}$ at Step $V'$ under E and $E'$, $\Omega _{i}$ is attainable given $x^{V'}$ under $E'$. As shown above, at Step $V'$ under E, h points to $\Omega _{i}$. Since the set of attainable objects at Step $V'+1$ under $E'$ is the same as the set of attainable objects at Step $V'$ under E, at Step $V'+1$ under $E'$, h points to $\Omega _{i}$. Since $U^{V'}\ne \emptyset$, at Step $V'+1$ under $E'$, $\Omega _{i}$ points to an agent in $U^{V'}$, and h is not assigned.

At Step $V'$ under E and $E'$, each $j\in \{\bigcup _{s\in Q^{V'+1}}s\}\backslash \{h\}$ points to $\varphi _{j}(E)$. Furthermore, at Step $V'$ under $E'$, there is $j\in \{\bigcup _{s\in Q^{V'+1}}s\}\backslash \{h\}$ which points to h. Since h is not assigned at Step $V'+1$ under $E'$, each agent in $\bigcup _{s\in Q^{V'+1}}s$ is a remaining agent at Step $V'+2$ under $E'$.

At Step $V'+2$ under $E'$, the same set of cycles $S^{1},\dots ,S^{V'},C(i,E),U^{V'}$ have occurred under E and $E'$, implying $x^{V'}$ under E and $x^{V'+1}$ under $E'$ are the same. Hence, at Step $V'+2$ under $E'$, each $j\in \bigcup _{s\in Q^{V'+1}}s$ points to the same object a which j points to at Step $V'+1$ under E. Then cycles $Q^{V'+1}$ occurs at Step $V'+2$ under $E'$, and for each $j\in \bigcup _{s\in Q^{V'+1}}s$, $\varphi _{j}(E')=\varphi _{j}(E)$.

Step 4$T^{V'+1}$occurs at Step$V'+2$under$E'$. Since the set of attainable objects given $x^{V'-1}$ under E and $E'$ are the same, at Step $V'$ under E and $E'$, each $j\in T^{V'+1}$ points to the same object.

Without loss of generality, let $h\in T^{V'+1}$ be such that at Step $V'$ under E, h points to an agent in $S^{V'}$ or $U^{V'}$. Such an h exists, otherwise $T^{V'+1}$ may occur at a step under E before $T^{V'+1}$. Since the set of attainable objects given $x^{V'-1}$ under E and $E'$ is the same, h points to the same object at Step $V'$ under E and $E'$. The only difference from Step 4 is that $h\in T^{V'+1}$ can point to an agent in $S^{V'}$.

Case 1At Step$V'$underE, hpoints to an agent in$S^{V'}$. Then $S^{V'}$ occurs at Step $V'$ under $E'$, and h is not assigned an object. Since at $V'+1$ under $E'$, $U^{V'}$ is the only cycle, h is not assigned at this step either. At Step $V'+2$ under $E'$, the same set of cycles $S^{1},\dots ,S^{V'},U^{V'},C(i,E)$ have occurred as in Step $V'+1$ under E. Hence, the set of attainable objects given $x^{V'+1}$ under $E'$ and $x^{V'}$ under E is the same. So each $j\in T^{V'+1}$ points to the same at Step $V'+2$ under $E'$ and Step $V'+1$ under E—$T^{V'+1}$ occurs at Step $V'+2$ under $E'$.

Case 2At Step$V'$underE, hpoints to an agent in$U^{V'}$. Then since $U^{V'}$ does not occur until Step $V'+1$ under $E'$, h is a remaining agent at Step $V'+2$ under $E'$ (Fig. 4).

As before $x^{V'}$ under E and $x^{V'+1}$ under $E'$ are the same, each agent $j\in T^{V'+1}$ points to the same object at Step $V'+2$ under $E'$ and $V'+1$ under E. Then, $T^{V'+1}$ occurs at Step $V'+2$ under $E'$, and for each $j\in T^{V'+1}$, $\varphi _{j}(E')=\varphi _{j}(E)$. $\square$

Fig. 4

Cycles $Q^{V'+1}$ and $T^{V'+1}$ occur at Step $V'+1$ under E, but under $E'$, when $U^{V'}\ne \emptyset$, they occur at Step $V'+2$

Claim 4

For each $k\in \{1\dots ,\bar{k}(E)-V'\}$, and each $j\in \bigcup _{s\in Q^{V'+k}\cup T^{V'+1}}s$, $\varphi _{j}(E')=\varphi _{j}(E)$.

Proof

We have proved the statement for $k=1$ above, and proceed by induction. Let the statement be true for $k=2,\dots ,k-1$. If $Q^{V'+k}=T^{V'+k}=\emptyset$, then the algorithm has terminated and we are done. If $Q^{V'+k}=\emptyset$ and $T^{V'+k}\ne \emptyset$, then proceed to Step 4 of the proof.

Let $Q^{V'+k}\ne \emptyset$.

Step 1$T^{V'+k-1}\ne \emptyset$. Suppose by contradiction, $T^{V'+k-1}=\emptyset$. By definition of $Q^{V'+k-1}$, for each $s\in Q^{V'+k-1}$, and each $h\in s$, $\varphi _{h}(E)\in \bigcup _{t\in C(h,E)}\Omega _{t}$. Let $h^{a}$ be the agent assigned a at Step $V'+k-1$ under E. For each $a\in \bigcup _{t\in C(h,E)}\Omega _{t}$, there is a transfer process and $h'\in N\backslash \{s\}$ which assigns to $h'$ object a: in R1, the owner of a replicates and transfers to $h^{a}$, and in R2, to $h'$. Since each object b attainable given $x^{V'+k-2}$ under E which is not in $\bigcup _{t\in C(h,E)}\Omega _{t}$ is not assigned to a new agent, b is still attainable given $x^{V'+k-1}$ under E. Hence, each object which is attainable given $x^{V'+k-2}$ under E is attainable given $x^{V'+k-1}$ under E. At Step $V'+k$ under E, each agent in $N^{V'+k}$ points to the same object as the previous Step, but since $s'\in Q^{V'+k}$ is a cycle, $s'$ must have happened at Step $V'+k-1$—violating the definition of the rule.

Step 2 Each $j\in \bigcup _{s\in Q^{V'+k}}s$ , jis a remaining agent at Step$V'+k$under$E'$, and each cycle$s\in Q^{V'+k}$eventually occurs under$E'$. For each $s\in Q^{V'+k}$, there is $h\in s$ s.t. at Steps $1,\dots ,V'+k-1$ under E, h points to an agent $g\notin C(h,E)$; otherwise, the cycle C(h, E) forms before Step $V'+k$ under E. At Steps $1,\dots ,V-1$ under E and $E'$, the algorithm is the same, so $h\in N^{V}$ implies h is a remaining agent at Step V under $E'$. Furthermore, since $x^{V-1}$ under E and $E'$ is the same, h points to the same object at Step V under E and $E'$. Let $G\equiv \{C(i,E),S^{V},\dots ,S^{V'},U^{V'},Q^{V'+1},\dots ,Q^{V'+k-1},T^{V'+1},\dots ,T^{V'+k-1}\}$ be the set of all cycles which occur during Steps $V,\dots ,V'+k-1$ under E.

For each $t\in \{0,\dots ,V'-V\}$, and each Step $V+t$ under E, h points to an agent in $\bigcup _{g\in G}\bigcup _{s\in g}s$. Case 1: At Step V under E, h points to an agent in C(i, E). Then, at Steps $V,\dots ,V'$ under $E'$, h points to the same agent. Case 2: There are sequences $g_{1},\dots ,g_{z}\in \{0,\dots ,V'-V\}$, and $f_{1},\dots ,f_{z}\in N$ s.t.

• $g_{1}<\cdots <g_{z}$,

• $g_{z}=V'-V$,

• for each $m\in \{1,\dots ,z\}$, $f_{m}\in \bigcup _{s\in S^{g_{m}}\cup U^{V'}}s$,

• h points to $f_{1}$ at Steps $V,\dots ,V+g_{1}$ under E, and

• for each $m\in \{2,\dots ,z\}$, h points to $f_{m}$ at Steps $V+g_{y-1}+1,\dots ,V+g_{y}$ under E.

Since h points to the same object at Step V under E and $E'$, and the set of remaining agents at Step V under E and $E'$ is the same, h points to $f_{1}$ at Step V under $E'$. We will show h is a remaining agent at Step $V'$ under $E'$.

If $f_{1}\in C(i,E)$, then at Steps $V,\dots ,V'$ under $E'$, h points to $f_{1}$.

If there is $m\in \{1,\dots ,z\}$ s.t. at Steps $V+g_{m-1}+1,\dots ,V+g_{m}$ under E, $f_{m}$ is the earliest-priority remaining agent and h points to $a\in \bigcup _{s\in C(i,E)}\Omega _{s}$ (which points to $f_{m}$), then h points to an agent in C(i, E) at Step $V'$ under $E'$.

Let $A\equiv \bigcup _{k\in \{1,\dots ,V-1\}}\bigcup _{s\in S^{k}}s$. If there is $m\in \{1,\dots ,z\}$ s.t. at Steps $V+g_{m-1},\dots ,V+g_{m}$ under E, $f_{m}$ is the earliest- priority remaining agent, h points to $a\in \bigcup _{s\in C(i,E)\cup A}\Omega _{s}$, and there is $j\in C(i,E)$ s.t. j is the earliest-priority remaining agent in the set $N^{V+g_{m-1}+1}\cup j$, then at Steps $V,\dots ,V+g_{m-1}-1$ under $E'$, h points to the same sequence of agents $f_{1},\dots ,f_{m-1}$, and at Steps $V+g_{m-1},\dots ,V'$ under $E'$, h points to j.

If the previous cases do not hold, then h points to the same sequence of agents $f_{1},\dots ,f_{z}$ at Steps $V,\dots ,V'$ under E and $E'$. In all cases, h is a remaining agent at Step $V'$ under $E'$.

Next, we show h is a remaining agent at Step $V'+k$ under $E'$.

If $U^{V'}=\emptyset$, then the same set of cycles $Q^{V'},T^{V'},Q^{V'+1},T^{V'+1},\dots ,Q^{V'+k-1},T^{V'+k-1}$ occur at Steps $V',\dots ,V'+k-1$ under E and $E'$. Hence, we can construct a sequence as above, and h points to the same sequence of agents at Steps $V',\dots ,V'+k-1$. So h is a remaining agent at Step $V'+k$ under E.

If $U^{V'}\ne \emptyset$, then as shown previously, the set of attainable objects given $x^{V'-1}$ under E and $x^{V'-1}$ under $E'$ is the same and h points to the same object at Step $V'$ under E and $E'$.

Let $u\in U^{V'}$, and $j\in u$ be s.t. j is the earliest-priority remaining agent. This means j is the earliest-priority remaining agent at Step $V'$ under E: $N^{V}\backslash \{C(i,E)\cup \{\bigcup _{s\in S^{V},\dots ,S^{V'-1}}s\}\}$. At Step $V'$ under $E'$, the set of remaining agents is $N^{V}\backslash \{\bigcup _{s\in S^{V},\dots ,S^{V'-1}}s\}$. Hence, at Step $V'$ under $E'$, either j is the earliest-priority remaining agent, or there is $j'\in C(i,E)$ s.t. $j'$ is earlier w.r.t $\sigma$.

At Step $V'$ under E, h either points to an agent in $S^{V'}$ or $U^{V'}$. If h points to an agent in $S^{V'}$, then he is a remaining agent at Step $V'$ under $E'$. If h points to an object a which points to an agent in $U^{V'}$, then at Step $V'$ under $E'$, there is either $j\in \bigcup _{s\in U^{V'}}s$ or $j'\in C(i,E)$ s.t. j or $j'$ is the earliest-priority remaining agent. If $a\notin \bigcup _{s\in C(i,E)}\Omega _{s}$ and j is the earliest, then $U^{V'}$ occurs at Step $V'+1$ under $E'$; so h still points to j at Step $V'$ and $V'+1$ under $E'$. If $a\in \bigcup _{s\in C(i,E)}\Omega _{s}$ or $j'$ is the earliest , then h points to $j'$ at Step $V'$ under $E'$, and since C(i, E) occurs at Step $V'$ under $E'$, h is not assigned an object. Furthermore, at $V'+1$ under $E'$, in either case, h points to an agent in $U^{V'}$. Since $U^{V'}$ occurs at Step $V'+1$ under $E'$, h is not assigned at $V'+1$ under $E'$.

If $U^{V'}=\emptyset$, then let $D(V'+1)=V'+1$. If $U^{V'}\ne \emptyset$, then let $D(V'+1)=V'+2$.

For each $z\in \{2,\dots ,k-1\}$, let $D(V'+z)$ be the step at which $Q^{V'+z}$ and $T^{V'+z}$ cycles occurs under $E'$ (by assumption this is true). Let $z\in \{1,\dots ,k-1\}$. Before Step $D(V'+z)$ under $E'$ and Step $V'+z$ under E, the same cycles $H\equiv \{S^{1},\dots ,S^{V'},$C(i, E), $U^{V'},$$Q^{V'+1},$$\dots ,$$Q^{V'+z-1},$$T^{V'},$$\dots$, $T^{V'+z-1}\}$ have occurred. This implies $x^{D(V'+z-1)}$ under $E'$ and $x^{V'+z-1}$ under E are the same, implying at Step $D(V'+z)$ under $E'$ and Step $V'+z$ under E, h points to the same agent. Since at Step $V'+z$ under E, h is not in a cycle, at Step $D(V'+z)$ under $E'$, h is not in a cycle either.

Hence, h is a remaining agent at Step $D(V'+k-1)+1$ under $E'$, and h points to the same agent as in Step $V'+k$ under E. Under analogous reasoning as for h, for each $j\in \bigcup _{s\in Q^{V'+k}}s$, j is a remaining agent at Step $D(V'+k-1)+1$ under $E'$, and j points to the same agent as in Step $V'+k$ under E. Since $Q^{V'+k}$ cycles occur at Step $V'+k$ under E, $Q^{V'+k}$ cycles occur at Step $D(V'+k-1)+1$ under $E'$.

Step 3$T^{V'+k}$occurs at Step$D(V'+k-1)+1$under$E'$. If $T^{V'+k}=\emptyset$, then we are done. Let $T^{V'+k}\ne \emptyset$. As before, there is $h\in \bigcup _{s\in T^{V'+k}}s$ s.t. at Steps $1,\dots ,V'+k-1$ under E, h does not point to any agent in C(h, E). The same reasoning holds as if $h\in \bigcup _{s\in Q^{V'+k}}s$. This implies (as shown above) each $j\in \bigcup _{s\in T^{V'+k}}s$ is a remaining agent at Step $D(V'+k-1)+1$ under $E'$. The same cycles H have occurred before Step $D(V'+k-1)+1$ under $E'$ and $V'+k$ under E. This means the set of remaining agents at these two steps are the same, implying the earliest-priority remaining agent is the same. Also as above, each $j\in \bigcup _{s\in T^{V'+k}}s$ points to the same object at Step $D(V'+k-1)+1$ under $E'$ and Step $V'+k$ under E. The two previous statements imply cycles $T^{V'+k}$ occur at Step $D(V'+k-1)+1$ under $E'$. $\square$

1.2 1.2 Withholding-proofness

Let Q be the first step under E that there is $j\in N$ which points to $\Omega _{i}$, $E'\equiv (R,\Omega _{i}=\emptyset ,\Omega _{-i})$, and $V'$ be the step under $E'$ at which i is assigned an object.

Case 1$Q>V$. Steps $1,\dots ,V$ under E and $E'$ are the same, so $\varphi _{i}(E)=\varphi _{i}(E')$.

Case 2$Q=V$. Steps $1,\dots ,V-1$ under E and $E'$ are the same, so at Step V under E and $E'$, the set of attainable objects are the same, and i points to the same object. If $V'=V$, then $\varphi _{i}(E)=\varphi _{i}(E')$. Let $V'>V$. For each $k\in \{V-1,\dots ,V'-1\}$, and each $j\in N$ s.t. $x_{j}^{V-1}$ under E is not $\emptyset$, $x_{j}^{k}$ under $E'$ is the same as $x_{j}^{V-1}$ under E. By Lemma 1, the set of attainable objects given $x^{V'-1}$ under E is contained in the set of attainable objects given $x^{V-1}$ under E. Hence, i points to his most preferred object in a smaller set at Step $V'$ under $E'$, implying $\varphi _{i}(E)\,R_{i}\,\varphi _{i}(E')$ (Fig. 5).

Case 3 $Q<V$.

Step 1 If iis assigned an object at Step$V\le V'$, then$\varphi _{i}(E)\,R_{i}\,\varphi _{i}(E')$. For each $k\in \{0,1,\dots ,V-1-Q\}$, let

• $J^{k}\equiv \{j\in N:$ at Step $Q+k$ under E, j points to $i\}$,

• $S^{k}\equiv \{s\subseteq 2^{N}:$s is a cycle $s_{1},\dots ,s_{z}$ at Step $Q+k$ under $E\}$,

• $T^{k}\equiv \{t\subseteq 2^{N\backslash \{J^{0}\cup \cdots \cup J^{k}\}}:$t is a cycle $t_{1},\dots ,t_{z}$ at Step $Q+k$ under $E'\}$, and

• $U^{k}\equiv \{t\subseteq 2^{N}:$t is a cycle at Step $Q+k$ under $E'$, and there is $j\in \{J^{0}\cup \cdots \cup J^{k}\}$ s.t. $j\in t\}$.

Fig. 5

At Step $Q+k$ under E the set of all cycles is $S^{k}$. At Step $Q+k$ under $E'$, the set of cycles divided into two groups: $T^{k}$ is group “unaffected” by the fact that i has withheld, and $U^{k}$ is the group “affected” by the fact that i has withheld (because there is some agent who pointed at $\Omega _{i}$)

Claim

For each $k\in \{0,\dots ,V-1-Q\}$, $S^{k}=T^{k}$.

Proof

Steps $1,\dots ,Q-1$ under E and $E'$ are the same, implying the set of attainable objects at Step Q under E and $E'$ are the same, and the earliest-priority remaining agent at Step Q under E and $E'$ is the same. Hence, each $j\in N^{Q}\backslash J^{0}$ points to the same agent $j'\in N^{Q}\backslash J^{0}$, and $S^{0}=T^{0}$.

By definition, for each $u\in U^{0}$, there is $\ell \in J^{0}$ s.t. $\ell \in u$. At Step $Q+1$ under E, $\ell$ points to $\Omega _{i}$, implying that the earliest step $\ell$ can be in a cycle is V. Hence, $\ell \notin \bigcup _{s\in S^{1}}s$. Let $j\in \bigcup _{s\in S^{1}}s$. At Step $Q+1$ under E, if j points to an agent in $\bigcup _{s\in U^{0}}s$, then he will remain until at least Step V under E—a contradiction. So $\{\bigcup _{s\in S^{1}}s\}\bigcap \{\bigcup _{s\in U^{0}}s\}=\emptyset$.

Next we show for each $s\in S^{1}$, and each $j\in s$, j points to the same object at Steps $Q+1$ under E and $E'$. The only difference between $x^{Q}$ under E and $E'$ is the assignment of agents in the set of cycles $U^{0}$. At Step Q under $E'$, if each $h\in \bigcup _{s\in U^{0}}s$ points to object in $\bigcup _{s\in U^{0}}\bigcup _{g\in s}\Omega _{g}$, then the set of attainable objects given $x^{Q}$ under E and $E'$ are the same and j points to the same object at Step $Q+1$ under E and $E'$— implying $S^{1}=T^{1}$. If instead, there is $h\in \bigcup _{s\in U^{0}}s$ s.t. h is assigned $a\notin \bigcup _{s\in U^{0}}\bigcup _{g\in s}\Omega _{g}$, then there is $h'\in \bigcup _{s\in U^{0}}s$ s.t. $h'$ is the earliest-priority remaining agent in the set $N^{Q}$, and consequently the set $N^{Q}\backslash \bigcup _{s\in S^{0}}s$. This implies $h'$ is the earliest-priority remaing agent at Step $Q+1$ under E. Since $\{\bigcup _{s\in S^{1}}s\}\bigcap \{\bigcup _{s\in U^{0}}s\}=\emptyset$, for each $s\in S^{1}$, and each $j\in s$, at Step $Q+1$ under E, the only object which points to j is $\Omega _{j}$. Hence, at Step $Q+1$ under E, each $j\in \bigcup _{s\in S^{1}}s$ points to object $a^{j}\in \bigcup _{g\in s}\Omega _{g}$. Since the set of attainable objects at Step $Q+1$ under E contains the set of attainable objects at Step $Q+1$ under $E'$, the fact that for each $j\in \bigcup _{s\in S^{1}}s$, at Step $Q+1$ under E, j points to $a^{j}$ implies at Step $Q+1$ under $E'$, j points to $a^{j}$. This implies each $s\in S^{1}$ occurs at Step $Q+1$ under $E'$, and $S^{1}=T^{1}$.

The rest of the proof is completed by iteration of the previous reasoning for $S^{2},\dots ,S^{V-1}$. The following is a demonstration for $S^{2}=T^{2}$.

Let $j\in \bigcup _{s\in S^{2}}s$, and a be the object j points to at Step $Q+2$ under E, and b be the object j points to at Step $Q+2$ under $E'$. The only difference between $x^{Q+1}$ under E and $E'$ is the assignment of agents in the set of cycles $U^{0}$and $U^{1}$.

If for each $m\in \{0,1\}$, each $s\in U^{m}$, and each $h\in s$, at Step $Q+m$ under $E'$, h points to object in $\bigcup _{g\in s}\Omega _{s}$, then the set of attainable objects given $x^{Q+1}$ under E and $E'$ are the same, and $a=b$.

If there is $h\in \bigcup _{s\in U^{0}}s$ s.t. h is assigned $d\notin \bigcup _{s\in U^{0}}\bigcup _{g\in s}\Omega _{g}$, then there is $h'\in \bigcup _{s\in U^{0}}s$ s.t. $h'$ is the earliest-priority remaining agent in the set $N^{Q}$. Since $h'$ remains at least until Step V under E, $h'$ is still the earliest-priority remaining agent at Steps $Q+1$ and $Q+2$ under E. Similarly as before, $\{\bigcup _{s\in S^{2}}s\}\bigcap \{\bigcup _{s\in U^{0}}s\}=\emptyset$. Hence, for each $s\in S^{2}$, and each $j\in s$, at Step $Q+2$ under E, the only object which points to j is $\Omega _{j}$. This implies that at Step $Q+2$ under E, for each $s\in S^{2}$ and each $j\in s$, j points to an object $a^{j}\in \bigcup _{g\in s}\Omega _{g}$. Since the set of attainable objects given $x^{Q+1}$ under E contains the set of attainable objects given $x^{Q+1}$ under $E'$, but each $a^{j}\in \bigcup _{g\in s}\Omega _{g}$ is still attainable given $x^{Q+1}$ under $E'$, for each $s\in S^{2}$, and each $j\in s$ , j points to $a^{j}$ at Step $Q+2$ under $E'$. Meaning $a=b$.

If there is $h\in \bigcup _{s\in U^{1}}s$ s.t. h is assigned $d\notin \bigcup _{s\in U^{1}}\bigcup _{j\in s}\Omega _{j}$, then there is $h'\in \bigcup _{s\in U^{0}}s$ s.t. $h'$ is the earliest-priority remaining agent in the set $N^{Q+1}$. The reasoning of above holds. $\square$

For each $i\in N$ s.t. $x_{i}^{V-1}$ under E is not $\emptyset$, either

• 1. i was assigned in $\{1,\dots ,Q-1\}$, implying $x_{i}^{V-1}$ under $E'$ and E are the same; or,

• 2. $i\in \bigcup _{k\in \{0,\dots ,V-1-Q\}}\bigcup _{s\in S^{k}}s=\bigcup _{k\in \{0,\dots ,V-1-Q\}}\bigcup _{t\in T^{k}}t$, implying $x_{i}^{V-1}$ under $E'$ and E are the same.

By Lemma 1, the set of attainable objects given $x^{V-1}$ under E contains the set of attainable objects given $x^{V-1}$ under $E'$. Since i points to his most preferred object from the former set, $\varphi _{i}(E)\,R_{i}\,\varphi _{i}(E')$. We have completed Step 1.

Now we need to show that $\varphi _{i}(E)\,R_{i}\,\varphi _{i}(E')$ when $V>V'$. For intuition, we give the proof of the case when $V'=Q+1$.

Claim If $V'=Q+1$, then $\varphi _{i}(E)\,R_{i}\,\varphi _{i}(E')$.

Proof

If $V\le V'$, then by Step 1, we are done. Let $V>V'$. Since $U^{0}$ is the only additional assignment made in $x^{Q}$ under $E'$ compared to $x^{Q}$ under E, by Lemma 1, the set of attainable objects at Step $Q+1$ under E contains the set of attainable objects at Step $Q+1$ under $E'$. Let a be the object i points to at Step $Q+1$ under E.

Case 1$a\ne \varphi _{i}(E')$. Since $U^{0}$ is the only additional set of cycles occuring before Step $Q+1$ under $E'$ compared to Step $Q+1$ under E, $U^{0}$ must have “caused” the unattainability of a after Step Q under $E'$. The two ways that a becomes unattainable are demonstrated in Fig.6.

Fig. 6

Let $j^{a}$ be the owner of a. On the left, a may be unattainable because in $x^{Q}$ under $E'$, three agents are assigned object a. On the right, since three agents are assigned $\Omega _{2}=c$, 2 must transfer in R1 and R2; similarly so for 3. This implies j must “help” 1 transfer b to 3 in R2; which further implies j may not transfer a in R2. Hence, only two agents receive a in $x^{Q}$ under $E'$, and a is unattainable

This means there is $u\in U^{0}$, and $u_{z}\in u$ such that at Step Q under $E'$ 1) $u_{z}$ is the earliest-priority remaining agent, and 2) a points to $u_{z}$.²⁵ Furthermore, by definition of $u\in U^{0}$, there is $u_{z-1},\dots ,u_{1}\in u$ such that at Step Q under $E'$, $u_{z}$ points to $u_{z-2}$ points to... $u_{1}$, and $u_{1}\in J^{0}$. Hence, at Step Q under E, a points to $u_{z}$ points to $u_{z-1}$ points to... $u_{1}$ points to $\Omega _{i}$. At Step $Q+1$ under E, each of these objects/agents still points in the same fashion. Since at Step $Q+1$ under E, i points to a, i is in a cycle. This contradicts the fact that $V>V'=Q+1$.

Case 2$a=\varphi _{i}(E')$. At Step $Q+1$ under $E'$, the fact that $\Omega _{i}'=\emptyset$ and i is in a cycle implies that i is the earliest-priority remaining agent. That is, i is earliest in the set $N^{Q}\backslash \{\bigcup _{s\in U^{0}\cup T^{0}}s\}$. This implies, at Step $Q+1$ under E, either i or an agent in $\bigcup _{s\in U^{0}}s$ is the earliest-priority remaining agent.

Subcase 2.1 At Step $Q+1$ under E, an agent in $\bigcup _{s\in U^{0}}s$ is the earliest-priority remaining agent. Let $u_{z}\in u$ be this agent.

Subcase 2.1.1 At Step $Q+1$ under E, $j^{a}$ is not a remaining agent. Then at Step $Q+1$ under E, a points to $u_{z}$. By the reasoning of Case 1, i is in a cycle at Step $Q+1$ under E.

Subcase 2.1.2 At Step $Q+1$ under E, $j^{a}$ is a remaining agent. At Step Q under $E'$, if $j^{a}$ points to an agent in $\bigcup _{s\in U^{0}}s$, then there is $u\in U^{0}$, and $u_{y},\dots ,u_{1}\in u$ such that a points to $j^{a}$ points to $u_{y}$ points to $u_{y-1}$ points to... $u_{1}$, and $u_{1}\in J^{0}$. By the reasoning of Case 1, i is in a cycle at Step $Q+1$ under E. At Step Q under $E'$, if $j^{a}$ points to either an agent in $\bigcup _{s\in T^{0}}s$ or $\varphi _{j^{a}}(E')$, then let b be the object $j^{a}$ points to at Step $Q+1$ under E.

Subcase 2.1.2.1 $b\ne \varphi _{j^{a}}(E')$. Since b is unattainable, by the reasoning in Case 1, at Step Q under E, b points to $u_{z}$. Furthermore, there is $u\in U^{0}$ and $u_{z-1},\dots ,u_{1}\in u$ such that at Step $Q+1$ under E, a points to $j^{a}$ points to b points to $u_{z}$ points to $u_{z-1}$ points to... $u_{1}$ points to $\Omega _{i}$ points to i points to a. Then, i is in a cycle at Step $Q+1$ under E.

Subcase 2.1.2.2 $b=\varphi _{j^{a}}(E')$. At Step $Q+1$ under $E'$, since i is the earliest-priority remaining agent, b either points to i or an agent $j^{b}\in c(i,E')$ where $\Omega _{j^{b}}=b$. If b points to i, then by definition of Case 2.1, at Step $Q+1$ under E, b points to $u_{z}$ and the reasoning of Case 1 holds. If b points to $j^{b}$, then we consider what $j^{b}$ points to at Step Q under $E'$ and follow the reasoning of Subcase 2.1.2, replacing $j^{a}$ with $j^{b}$. Since $c(i,E')$ is finite, there is eventually an agent j which points to i, and we are done.

Subcase 2.2 At Step $Q+1$ under E, i is the earliest-priority remaining agent. Replacing $u_{z}$ with i, the reasoning of Case 2.1 holds. $\square$

The proof for the general case involves a straightforward generalization of the proof for the case where $V'=Q+1$. Now instead of just $U^{0}$ causing unattainability of a, a sequence of sets of cycles $U^{0},U^{1},\dots ,U^{k}$ causes the unattainability of a.

Step 2$\varphi _{i}(E)\,R_{i}\,\varphi _{i}(E')$.

If $V\le V'$, then by Step 1, we are done. Let $V>V'$. Let a be the object i points to at Step $V'$ under E.

Case 1$a\ne \varphi _{i}(E')$.

Subcase 1.1At Step$V'$underE, $j^{a}$is not a remaining agent.

By Lemma 1 and Step 1, the set of attainable objects at Step $V'$ under E contains the set of attainable objects at Step $V'$ under $E'$. Since $U^{0},\dots ,U^{V'-Q-1}$ are the only extra cycles in $x^{V'-Q-1}$ under $E'$ compared to $x^{V'-Q-1}$ under E, these cycles must have “caused” the unattainability of a at some step under $E'$. That is, there is a $g\in \mathbb {N}$ such that at Step $Q+g$ under $E'$, a is attainable, but at Step $Q+g+1$ under $E'$, a is not attainable. Let $h\in U^{g}$ be the earliest-priority remaining agent at Step g under $E'$; equivalently, h is the earliest-priority remaining agent in the set $N^{Q}\backslash \{\bigcup _{s\in T^{0},\dots ,T^{g-1},U^{0},\dots ,U^{g-1}\}}s\}$. This implies that either h or an agent in $\bigcup _{s\in U^{0},\dots ,U^{g-1}}s$ is the earliest-priority remaining agent in the set $N^{Q}\backslash \{\bigcup _{s\in T^{0},\dots ,T^{g-1}}s\}$— which is the set of remaining agents at Step $Q+g$ under E. Let $h_{z}$ be this agent. By definition, only the cycles $S^{0},\dots ,S^{V'-1}$ occur at Steps $Q,\dots ,V'-1$ under E; hence, $h_{z}$ is the earliest-priority remaining agent at Steps $Q+g,\dots ,V'$ under E. Let $\bar{g}\in \{0,\dots ,g\}$ be such that Step $Q+\bar{g}$ under E is the first Step under E at which $h_{z}$ is the earliest-priority remaining agent.

For each $k\in \{0,\dots ,\bar{g}-1\}$, at Step $Q+k$ under E, if the earliest-priority remaining agent j is in the set $\bigcup _{s\in U^{0},\dots ,U^{g}}s$, then j is the earliest-priority remaining agent at Steps $Q+k,\dots ,V'$ under E. Since we have established $h_{z}\ne j$ is the earliest-priority remaining agent at Step $V'$ under E, this is not possible. Hence, for each $k\in \{0,\dots ,\bar{g}-1\}$, the earliest-priority remaining agent at Step $Q+k$ under E is in the set $\bigcup _{s\in T^{0},\dots ,T^{\bar{g}-1}}s$. This implies for each $k\in \{0,\dots ,\bar{g}-1\}$, at Steps $Q+k$ under E and $E'$, the earliest-priority remaining agent is the same.

Next, we construct a sequence of agents $h_{z-1},\dots ,h_{1}$. For Steps $Q,\dots ,Q+g$ under $E'$, $h_{z}$ points to a sequence of agents (Fig. 7).

Fig. 7

Several possible sequences of agents at which $h_{z}$ points to are shown. In the first column, at Step Q under $E'$, $h_{z}$ points to an agent in $\bigcup _{s\in T^{0}}s$. The first Step under $E'$ at which $h_{z}$ points to an agent in a “U” cycle is Step $Q+3$. Hence, if the conditions 1.a and 1.b hold, then $h_{z-1}$ is this agent, and $g_{z-1}=3$

Let $\hat{g}\in \{0,\dots ,g\}$ be such that Step $Q+\hat{g}$ under $E'$ is the first Step under $E'$ that either

• 1. for each Step $Q,\dots ,Q+\hat{g}-1$ under $E'$,

  • (a) $h_{z}$ points to an agent in $\bigcup _{s\in T^{0},\dots ,T^{\hat{g}-1}}s$,

  • (b) $h_{z}$ does not point to an object in $\bigcup _{s\in U^{0},\dots ,U^{\hat{g}-1}}\Omega _{s}$, and

  at Step $Q+\hat{g}$ under $E'$, $h_{z}$ points to an agent in $\bigcup _{s\in U^{\hat{g}}}s$, OR

• 2. for each Step $Q,\dots ,Q+\hat{g}$ under $E'$,

  • (a) $h_{z}$ points to an agent in $\bigcup _{s\in T^{0},\dots ,T^{\hat{g}}}s$, and

  • (b) at Step $Q+\hat{g}$ under $E'$, $h_{z}$ points to an object in $\bigcup _{s\in U^{0},\dots ,U^{\hat{g}-1}}\Omega _{s}$.

Such a $\hat{g}$ exists; let $g_{z-1}=\hat{g}$. If $\hat{g}$ satisfies the former condition, then let $h_{z-1}\in \bigcup _{s\in U^{\hat{g}}}s$ be the agent that $h_{z}$ points to at Step $\hat{g}$ under $E'$. If the $\hat{g}$ satisfies the latter condition, then let $h_{z-1}\in \bigcup _{s\in U^{0},\dots ,U^{\hat{g}}}s$ be the owner of the object $h_{z}$ points to at Step $Q+\hat{g}$ under $E'$.

By construction, at Step $Q+g_{z-1}$ under E, $h_{z}$ points to $h_{z-1}$. Repeat this process for $h_{z-1}$ to find $h_{z-2}$, and so forth, until we arrive at the first $h_{z-x}$ who is in $\bigcup _{s\in U^{0}}s$.

We show no agent appears twice in this sequence. Suppose by contradiction, there is a consecutive subsequence $h_{k},h_{k-1},\dots ,h_{k-x}$ such that $h_{k}=h_{k-x}$. This is only possible if each $h_{k},\dots ,h_{k-x}$ iteratively satisfies the first condition, implying $g_{k}=g_{k-1}=\cdots =g_{k-x}=g_{k}$. Then, at Step $Q+g_{k}$ under E, $h_{k},\dots ,h_{k-x}$ forms a cycle. Since each of these agents is not in $\bigcup _{s\in T^{g_{k}}}s$, this is a contradiction. By finiteness of the number of agents, we arrive at $h_{z-x}$ who is in $\bigcup _{s\in U^{0}}s$.

Consider $h_{z-x}$. By definition of $U^{0}$, $h_{z-x}$ points to an agent in $\bigcup _{s\in U^{0}}s$. Let $h_{z-x-1}$ be this agent. If $h_{z-x-1}\in J^{0}$, then we have reached the end of the sequence. Redefine z and relabel agents so that $h_{z-x-1}=h_{1}$. If $h_{z-x-1}\notin J^{0}$, then repeat this process. By the definition of $U^{0}$ and finiteness of cycles, we eventually reach an agent $h\in J^{0}$. Redefine z and relabel agents so that this agent is $h_{1}$.

By construction, for each $k\in \{0,\dots ,z\}$, at Step $Q+g_{k}$ under E, $h_{k+1}$ points to $h_{k}$. Hence, we have a sequence of agents $h_{z},\dots ,h_{1}$ such that at Step $g_{z-1}$ under E, $h_{z}$ points to $h_{z-1}$ points to... $h_{1}$ (Fig. 8). Since $h_{z}$ is the earliest-priority remaining agent at Step $V'$ under E and $j^{a}$ is not remaining, a points to $h_{z}$ points to $h_{z-1}$ points to... $h_{1}$ points to $\Omega _{i}$ points to i points to a, and i is in a cycle. This contradicts the assumption that $a\ne \varphi _{i}(E')$. We have completed Subcase 1.1.

Fig. 8

By construction, at each Step in E before Step $Q+g_{z-1}$, $h_{z}$ only points to agents in who are in cycles ( $\bigcup _{s\in T^{0},\dots ,T^{g_{z-1}-1}}s$). So Step $Q+g_{z-1}$ under E is the first Step at which he points to an agent, $h_{z-1}$, not in a cycle. We repeat this argument for $h_{z-1},\dots ,h_{1}$ until we find $h_{1}$ points to $\Omega _{i}$. Hence, at Step $Q+g_{z-1}$ under E, $h_{z}$ points to $h_{z-1}$ points to... $h_{1}$. The agents still point in this fashion until Step $V'$ under E

Subcase 1.2At Step$V'$underE, $j^{a}$is a remaining agent. Then, at Step $V'$ under $E'$, $j^{a}$ is in a cycle with i. Let $h_{z}=j^{a}$. Then, construct $h_{z-1},\dots ,h_{1}$ as in Subcase 1.1 (the sequence obtained may be different). The reasoning of Subcase 1.1 holds, implying at Step $V'$ under E, a points to $j^{a}$ points to $h_{z-1}$ points to... $h_{1}$ points to $\Omega _{i}$ points to i.

The proof for Step 2, and thereby withholding-proofness, is complete. $\Box$

Appendix 2

In sections 2.1-2.4, we show that the properties are independent. In section 2.5, we define a rule outside of the $TTC-SI^{\sigma }$ family satisfying the four properties.

1.1 2.1

For each $E\in \mathcal {E}$, and each $i\in N$, let $\varphi _{i}(E)=\emptyset$. Then, $\varphi$ satisfies group strategy-proofness, reciprocity lower bound, and withholding-proofness, but not efficiency.

1.2 2.2

Let $\bar{\mathcal {R}^{i}}\subsetneq \mathcal {R}^{i}$, and $\sigma \in \Sigma$. Let $\pi$ be a permutation of N s.t. $\pi (1)=i$, for each $k\in \{2,\dots ,\sigma ^{-1}(i)\}$, $\pi (k)=\sigma (k-1)$, and for each $k\in \{\sigma ^{-1}(i)+1,\dots ,n\}$, $\pi (k)=\sigma (k)$. Let $\pi '$ be a permutation of N s.t. $\pi (n)=i$, for each $k\in \{1,\dots ,\sigma ^{-1}(i)\}$, $\pi (k)=\sigma (k)$, and for each $k\in \{\sigma ^{-1}(i)+1,\dots ,n\}$, $\pi (k)=\sigma (k-1)$.

For each $E=(R,\Omega )\in \mathcal {E}$, and each $\sigma \in \Sigma$, let

$$\begin{aligned} \varphi ^{\sigma }(E)\equiv & {} {\left\{ \begin{array}{ll} TTC-SI^{\pi }(E) &{} \quad \text{ if } \;R_{i}\in \bar{\mathcal {R}^{i}}\\ TTC-SI^{\pi '}(E) &{} \quad \text{ if } \;R_{i}\in \mathcal {R}^{i}\backslash \bar{R^{i}} \end{array}\right. } \end{aligned}$$

Then, $\varphi ^{\sigma }$ satisfies withholding-proofness, the reciprocity lower bound, and efficiency, but not group strategy-proofness. It is straightforward to construct an economy E where i has incentive to misreport.

1.3 2.3

Let $\hat{\mathcal {E}}$ be the set of $E=(R,\hat{\Omega })\in \mathcal {E}$ s.t.

• $N=\{1,\dots ,5\}$,

• $\hat{\Omega }=(a,b,c,d,e)$,

• for each $z\in \mathcal {O}\backslash \{e\}$, $e\,R_{1}\,z$, and

• for each $i\in N\backslash \{1\}$, and each $z\in \mathcal {O}\backslash \{a\}$, $a\,R_{i}\,z$.

Let $\sigma (i)=i$, and for each $E\in \mathcal {E}$, let $R_{5}(2)$ be 5’s second-ranked object w.r.t. $R_{5}$ and

$$\begin{aligned} \varphi (E)\equiv {\left\{ \begin{array}{ll} (e,a,a,a,R_{5}(2)) &{} \quad \text{ if } \;E\in \hat{\mathcal {E}}\\ TTC-SI^{\sigma }(E) &{} \quad \text{ if } \;E\in \mathcal {E}\backslash \hat{\mathcal {E}} \end{array}\right. } \end{aligned}$$

Then, for each $E\in \hat{\mathcal {E}}$, $\varphi _{1}(E)=\Omega _{5}$, and $\Omega _{1}=a\,P_{5}\,e=\varphi _{5}(E)$, in violation of the reciprocity lower bound. It is clear that the rule satisfies efficiency and withholding-proofness.

Claim$\varphi$ satisfies group strategy-proofness.

Proof

Suppose by contradiction that there is $E=(R,\Omega )\in \mathcal {E}$, $S\subseteq N$, and $R'_{S}\in \mathcal {R}^{S}$ such that for each $i\in S$, $\varphi _{i}(R'_{S},R_{-S},\Omega )\,R_{i}\,\varphi _{i}(E)$, and for some $j\in S$, $\varphi _{j}(R'_{S},R_{-S},\Omega )\,P_{j}\,\varphi (E)$. Let $E'=(R'_{S},R_{-S},\Omega )$.

Case 1

$E=(R,\Omega )\in \hat{\mathcal {E}}$. If $E'\in \hat{\mathcal {E}}$, then $\varphi (E')=\varphi (E)$, so let $E'\in \mathcal {E}\backslash \hat{\mathcal {E}}$. For each $i\in N\backslash \{5\}$, $\varphi _{i}(E)=\arg \max R_{i}$, so 1) for each $i\in S\backslash \{5\}$, $\varphi _{i}(E)=\varphi _{i}(E')$, and 2) $5\in S$ and $\varphi _{5}(E')=a\,P_{5}\,\varphi _{5}(E)$. First, $\{2,3,4\}\subset S$ is not true, otherwise, by assumption $\varphi _{2}(E')=\varphi _{3}(E')=\varphi _{4}(E')=a$, so $\varphi _{5}(E')=a$ is not possible. Hence, there is $j\in \{2,3,4\}$ such that $j\notin S$. If each $i\in S$ top-ranks $\Omega _{j}$, then there is some agent h in S such that $\varphi _{h}(E)\,P_{h}\,\varphi _{h}(E')=\Omega _{j}$. If not every agent in S top-ranks $\Omega _{j}$, then the unique efficient allocation for $E'$ is the one where each agent is assigned his top-ranked object in $(R'_{S},R_{-S})$. Since $E'\in \mathcal {E}\backslash \hat{\mathcal {E}}$, there is $i\in S$ such that the top-ranked object in $R_{i}$ and $R_{i}'$ is not the same. If $i\ne 5$, then this violates $\varphi _{i}(E)=\varphi _{i}(E')$; if $i=5$, then i does not top-rank a, and $\varphi _{5}(E')\ne a$—violating the assumption that 5 strictly benefits.

Case 2

$E=(R,\Omega )\in \mathcal {E}\backslash \hat{\mathcal {E}}$. If $E'\in \mathcal {E}\backslash \hat{\mathcal {E}}$, then since $TTC-SI^{\sigma }$ is group-strategy-proof, existence of such a coalition S is not possible. Let $E'\in \hat{\mathcal {E}}$. If there is a unique efficient allocation, then each agent is assigned their top-ranked object and manipulation is impossible. So suppose there are multiple. This implies that there is $i,j\in N$ such that all agents $N\backslash \{i\}$ top-rank $\Omega _{i}$, $\varphi _{j}(E)\ne \Omega _{i}$, and for each $h\in N\backslash \{i,j\}$, $\varphi _{h}(E)=\Omega _{i}$. Note that j is assigned his second-ranked object, and each other agent is assigned their top-ranked object. By definition of S, $\varphi _{j}(E')=\Omega _{i}$. If $i=1$, then since $E\in \mathcal {E}\backslash \hat{\mathcal {E}}$, 1 does not top-rank e, implying $1\in S$, and $\varphi _{i}(E)\,P_{i}\,e=\varphi _{i}(E')$—a contradiction. If $i=5$, then all agents $N\backslash \{5\}$ top-rank e in R, and $\varphi _{j}(E')=e$. Since $j\ne i=5$, and $e\notin \bigcup _{h\in \{2,3,4\}}\varphi _{h}(E')$, $j=1$. Then, since 2 top-ranks e in R, but from $E'\in \mathcal {E}\backslash \hat{\mathcal {E}}$, 2 top-ranks a in $R'$, so $2\in S$ and $\varphi _{2}(E)=e\,P_{i}\,a=\varphi _{2}(E')$—a contradiction. If $i\in \{2,3,4\}$, then all agents $N\backslash \{i\}$ top-rank $\Omega _{i}$ in R. Since $\Omega _{i}\notin \bigcup _{h\in \{1,2,3,4\}}\varphi _{h}(E')$, $\varphi _{j}(E')=\Omega _{i}$ implies $j=5$. Hence, $1\in S$, and $\varphi _{1}(E)=\Omega _{i}\,P_{1}\,e=\varphi _{1}(E')$—a contradiction.$\square$

1.4 2.4

Let $\sigma \in \Sigma$. Let $\pi$ be a permutation of N s.t. $\pi (1)=i$, for each $k\in \{2,\dots ,\sigma ^{-1}(i)\}$, $\pi (k)=\sigma (k-1)$, and for each $k\in \{\sigma ^{-1}(i)+1,\dots ,n\}$, $\pi (k)=\sigma (k)$. Let $\pi '$ be a permutation of N s.t. $\pi (n)=i$, for each $k\in \{1,\dots ,\sigma ^{-1}(i)\}$, $\pi (k)=\sigma (k)$, and for each $k\in \{\sigma ^{-1}(i),\dots ,n-1\}$, $\pi (k)=\sigma (k+1)$.

For each $E=(R,\Omega )\in \mathcal {E}$, and each $\sigma \in \Sigma$, let

$$\begin{aligned} \varphi ^{\sigma }(E)\equiv & {} {\left\{ \begin{array}{ll} TTC-SI^{\pi }(E) &{} \quad \text{ if } \;\Omega _{i}\ne \emptyset \\ TTC-SI^{\pi '}(E) &{} \quad \text{ if } \;\Omega _{i}=\emptyset \end{array}\right. } \end{aligned}$$

Then, $\varphi ^{\sigma }$ satisfies group strategy-proofness, the reciprocity lower bound, and efficiency, but not withholding-proofness.

1.5 2.5

Let $\sigma ,\sigma '\in \Sigma$ be s.t. $\sigma (2)=\sigma '(3)$, $\sigma (3)=\sigma '(2)$, and for each $i\in N\backslash \{2,3\}$, $\sigma (i)=\sigma '(i)$. Let $\{A,B\}$ be a partition of $\mathcal {O}$.

For each $E\in \mathcal {E}$, let

$$\begin{aligned} \varphi (E)\equiv & {} {\left\{ \begin{array}{ll} TTC-SI^{\sigma }(E) &{} \quad \text{ if } \;TTC-SI_{1}^{\sigma }(E)\in A\\ TTC-SI^{\sigma '}(E) &{} \quad \text{ if } \;TTC-SI_{1}^{\sigma }(E)\in B \end{array}\right. }. \end{aligned}$$

Then, $\varphi$ satisfies group strategy-proofness, withholding-proofness, efficiency, and the reciprocity lower bound, but $\varphi ^{\sigma }$ is not in the $TTC-SI$ family of rules.