Multilevel multidimensional consistent aggregators

Abstract

This paper examines the structure of consistent, multidimensional, multilevel aggregators in two distinct models- one, where the set of evaluations is the unit interval and the other, where it is finite. In the first model, we characterize a class of separable rules called component-wise $\alpha$-median rules. In the finite model, separability is no longer guaranteed. In addition to consistency, stronger notions of unanimity and anonymity are required to characterize a class of separable rules called bipartite rules.

Appendix

Proof of Theorem 1

It is easy to verify that component-wise $\alpha$-median aggregators satisfy anonymity, unanimity, continuity and monotonicity. We show that is satisfies consistency.

Consistency: Let $f^{\alpha }$ be a component-wise $\alpha$-median aggregator. In view of the separability of component-wise aggregators it clearly suffices to show that it satisfies consistency for any arbitrary issue.

Pick a profile $v\in {A}$ and issue j. Then $f^{\alpha }_{j}(v) \in \{\min _{i\in N}(v_{ij}), \alpha _{j}, \max _{i\in N}(v_{ij})\}$. If $f^{\alpha }_{j}(v)=\min _{i\in N}(v_{ij})$ i.e $\alpha _{j} <\min _{i\in N}$, consistency follows from the same argument used to show that the min aggregator is consistent. Likewise, the arguments used to show that the max aggregator is consistent can be used to show that $f^{\alpha }$ is consistent when $f^{\alpha }(v)=\max _{i\in N}(v_{ij})$ i.e $\alpha _{j}>\max _{i\in N}(v_{ij})$.

If $\alpha _{j} \in [\min _{i\in N}(v_{i}), \max _{i\in N}(v_{i})]$, then $f_{j}^{\alpha }(v)=\alpha _{j}$. Let $I=\{N_{i},\ldots ,N_{K}\}$ be any partition of N. There exists a set $N_{k} \in I$ such that for some $i\in N_{k}$, $v_{ij}\le \alpha _{j}$. By definition, $f_{j}^{\alpha }(v_{N_{k}})\le \alpha _{j}$. Therefore, $\alpha _{j} \ge \min _{N_{l}\in I}\left( f_{j}^{\alpha }(v_{N_{k}})\right) .$ Similarly, there exists $N_{k'} \in I$ such that for some $i' \in N_{k'}$, $v_{i'j}\ge \alpha _{j}$. By definition, $f_{j}^{\alpha }(v_{N_{k'}})\ge \alpha _{j}$. Therefore, $\alpha _{j} \le \max _{N_{l}\in I}\left( f_{j}^{\alpha }(v_{N_{l}})\right) .$ By the above arguments we have, $\alpha _{j} \in [\min _{N_{l}\in I}\left( f_{j}^{\alpha }(v_{N_{l}})\right) , \max _{N_{l}\in I}\left( f_{j}^{\alpha }(v_{N_{l}})\right) ].$ By definition, $f_{j}^{\alpha }(f^{\alpha }(v_{N_{1}}),\ldots ,f^{\alpha }(v_{N_{K}}))=\alpha _{j}$. Therefore, component-wise $\alpha$-median aggregators are consistent.

Let f be an aggregator which satisfies consistency, unanimity, anonymity, monotonicity and continuity. Observe that f is actually a collection of rules $\{f^k\}$, $k=1, \ldots , |\mathbf {N}|$ where $f^{k}$ is an aggregator for any k-size collection of voter opinions. The next lemma shows that f can be constructed by a repeated application of the function $f^{2}$.$\square$

Lemma 1

Let $N =\{i_{1},\ldots ,i_{n}\}$ $\subseteq \mathbf N$ and let $v_{i_{k}} \in {A}$ for $k=1,\ldots n$. Then $f^{n}(v_{i_{1}},\ldots ,v_{i_{n}}) =f^{2}(\ldots f^{2}(f^{2}(v_{i_{1}},v_{i_{2}}),v_{i_{3}})\ldots v_{i_{n}}).$

This lemma follows directly by the application of consistency. For instance, if $N=\{1,2, 3,4\}$, then

$$\begin{aligned} f^{4}(v_{1},v_{2}, v_{3},v_{4})=f^{2}(f^{3}(v_{1},v_{2},v_{3}),v_{4}) =f^{2}(f^{2}(f^{2}(v_{1},v_{2}),v_{3}),v_{4}). \end{aligned}$$

By Lemma 1 we can restrict attention to $f^{2}$.

Applying Lemma 1 we can restrict attention to the two voter aggregator $f^{2}$. From now onwards, we simply write f in place of $f^{2}$ for simplicity of notation. In some cases, we will revert back to $f^{2}$ where necessary. We introduce the notion of two evaluations being ordered. Let $v,v'\in {A}^{n}$, $N\in \mathbf{N}$. If either $v_{j}\ge v_{j}'$ or $v_{j}\le v_{j}'$ for all $j\in X$ then v is ordered with $v'$. We define a Box as follows. Let $v_{i},v_{k}$ be a pair of voter opinions. Then

$$\begin{aligned} Box(v_{i},v_{k}) = \bigg \{v_{t} \in {A}^{2}\quad \bigg |\quad v_{ij} \in \big [\min _{i\in N}(v_{ij}),\max _{i\in N}(v_{kj})\big ] \quad \forall j \in X\bigg \}. \end{aligned}$$

Lemma 2

Let $v_{i},v_{k} \in {A}$. Then $f(v_{i},v_{k}) \in Box(v_{i},v_{k})$.

Proof

We consider two cases.

• Case 1: $v_{i}$ and $v_{k}$ are ordered. Assume w.l.o.g. $v_{i} \ge v_{k}$. The case where $v_{k} \ge v_{i}$ can be dealt with by using a symmetric argument. Applying monotonicity,

  $$\begin{aligned} f(v_{i},v_{k}) \ge f(v_{k},v_{k}) =v_{k}. \end{aligned}$$

  The last inequality holds due to unanimity. Similarly,

  $$\begin{aligned} f(v_{i},v_{k}) \le f(v_{i},v_{i}) =v_{i}. \end{aligned}$$

  Therefore $f(v_{i},v_{k}) \in Box(v_{i},v_{k})$.

• Case 2: Case 1 does not hold. Let $\underline{v}$ be such that,

  $$\begin{aligned} \underline{v}_{j} =\min (v_{ij}, v_{kj})\quad \forall j\in X. \end{aligned}$$

  Similarly, let $\overline{v}$ is such that,

  $$\begin{aligned} \overline{v}_{j} =\max (v_{ij}, v_{kj})\quad \forall j\in X. \end{aligned}$$

  Note that $Box(v_{i},v_{k})=Box(\underline{v},\overline{v})$. By monotonicity,

  $$\begin{aligned} f(v_{i}, v_{k}) \ge f(\underline{v},\underline{v})=\underline{v}. \end{aligned}$$

  Similarly,

  $$\begin{aligned} f(v_{i}, v_{k}) \le f(\overline{v},\overline{v}) = \overline{v}. \end{aligned}$$

  Therefore, $f(v_{i}, v_{k}) \in Box(v_{i}, v_{k})$. $\square$

The next lemma is illustrated in Fig. 1.

Fig. 1

Illustration for Lemma 3

Lemma 3

Let $v_{i}, v_{k} \in {A}$ be ordered (assume w.l.o.g. $v_{i}\le v_{k}$ and $f(v_{i},v_{k})= v_{t}$. Then for all $v_{r},v_{u} \in {A}$ such that $v_{r}\in Box(v_{i},v_{t})$ and $v_{u}\in Box(v_{t}, v_{k})$,

$$\begin{aligned} f(v_{r}, v_{u}) =v_{t},\quad f(v_{r},v_{t})=v_{t},\quad f(v_{t},v_{u})=v_{t}. \end{aligned}$$

Proof

By Lemma 1 and unanimity,

$$\begin{aligned} f(v_{i},v_{t})= & {} f^{2}(v_{i}, f^{2}(v_{i},v_{k})) = f^{3}(v_{i}, v_{i}, v_{k})\\= & {} f^{2}(f^{2}(v_{i},v_{i}),v_{k})=f(v_{i},v_{k})=v_{t}. \end{aligned}$$

By an analogous argument $f(v_{t},v_{k})=v_{t}$. By monotonicity,

$$\begin{aligned} f(v_{r}, v_{u}) \le f(v_{t},v_{u})\le f(v_{t},v_{k}) =v_{t}. \end{aligned}$$

Similarly,

$$\begin{aligned} f(v_{r}, v_{u}) \ge f(v_{i},v_{u})\ge f(v_{i},v_{t}) =v_{t}. \end{aligned}$$

Therefore $f(v_{r},v_{u}) =v_{t}$. Again by monotonicity,

$$\begin{aligned} f(v_{r}, v_{t}) \le f(v_{t},v_{k}) =v_{t}. \end{aligned}$$

Also,

$$\begin{aligned} f(v_{r}, v_{t}) \ge f(v_{i},v_{t}) =v_{t}. \end{aligned}$$

Therefore $f(v_{r},v_{t}) =v_{t}$. By a similar argument it follows that $f(v_{t},v_{u})=v_{t}.$ $\square$

Lemma 4

Let $v_{i},v_{k},v_{i}',v_{k}'$ be such that (i) $v_{i}$ is ordered with $v_{k}$, $v_{i}'$ is ordered with $v_{k}'$ (ii) $f^{2}(v_{i},v_{k})=v_{t} \in \text {int}Box(v_{i},v_{k})$ ⁹ and (iii) $f(v_{i}',v_{k}')=v_{t}' \in \text {int}Box(v_{i}',v_{k}')$. Then $v_{i}'<v_{t}$ and $v_{k}>v_{t}'$ both cannot hold.

Proof

We prove this by contradiction. So suppose $v_{i}'<v_{t}$ and $v_{k}>v_{t}'$ hold. Then by applying Lemma 3 on $Box(v_{i},v_{k})$ we have $f(v_{t},v_{t}')=v_{t}$ and by applying Lemma 3 on $Box(v_{i}',v_{k}')$ we have $f(v_{t},v_{t}')=v_{t}'$. This is a contradiction. Therefore both $v_{i}'<v_{t}$ and $v_{k}>v_{t}'$ cannot be true. $\square$

Let $v_{t} \in {A}$. The box $MBox(v_{t})=Box(\bar{v}_{i},\bar{v}_{k})$ is a maximal box for $v_{t}$ if there does not exist $v_{i}'<v_{i}$ and $v_{k}'>v_{k}$ such that $f(v_{i}',v_{k}')=v_{t}$. Suppose $v_t$ is in the range of f. Then $MBox(v_t)$ exists by the virtue of continuity of f. Note that a maximal set may not be unique. By similar arguments as in Lemma 3 we can prove the following Lemma.

Lemma 5

Let $MBox(v_{t})$ be a maximal box for $v_{t}$. Let $v_{r},v_{u} \in {A}$ such that $v_{r}\in Box(\overline{v}_{i},v_{t})$ and $v_{u}\in Box(v_{t}, \overline{v}_{k})$. Then,

• (i) $f(v_{r}, v_{u}) =v_{t},\;\; f(v_{r},v_{t})=v_{t},\;\;f(v_{t},v_{u})=v_{t}.$

• (ii) Let $\{v_{t}^{q}\}_{q=1}^{\infty }$ be a sequence such that $\lim _{n \rightarrow \infty }v_{t}^{q} =v_{t}$. Then,

  $$\begin{aligned} \lim _{n \rightarrow \infty }MBox(v_{t}^{q}) = MBox(v_{t}). \end{aligned}$$

Proof

The first part of the proof is proved analogously as in the previous Lemma 3. The second part is an implication of the continuity of $f^{2}$. $\square$

Lemma 6

Let $v_{i}, v_{k}$ and $v_{i}', v_{k}', v_{t}'$ be such that (i) $v_{i}$ is ordered with $v_{k}$, $v_{i}'$ is ordered with $v_{k}'$ and $v_{t}$ is ordered with $v_{t}'$ (ii) $f(v_{i},v_{k})=v_{t} \in \text {int} Box(v_{i},v_{k})$ and (iii) $f(v_{i}',v_{k}')=v_{t}' \in \text {int} Box(v_{i}',v_{k}')$. Then $\exists$ $v_{i}'',v_{k}''$ and $v_{t}''$ such that (a) $v_{i}'',v_{k}'',v_{t}'' \in Box(v_{t},v_{t}')$ (b) $f(v_{i}'',v_{k}'')=v_{t}''$ and $v_{t}'' \in \text {int} Box(v_{i}'',v_{k}'')$ (c) $v_{t}'' \notin \{v_{t},v_{t}'\}$.

Proof

W.l.o.g. let $v_{i}\le v_{k}$, $v_{i}'\le v_{k}'$ and $v_{t} \le v_{t}'$. By Lemma 4 we have¹⁰

$$\begin{aligned} Box(v_{i},v_{k}) \cap Box(v_{t},v_{t}') \ne \emptyset \quad \text {and}\quad Box(v_{i},v_{k})^{C} \cap Box(v_{t},v_{t}') \ne \emptyset . \\ \text {or}\quad Box(v_{i}',v_{k}') \cap Box(v_{t},v_{t}') \ne \emptyset \quad \text {and}\quad Box(v_{i}',v_{k}')^{C} \cap Box(v_{t},v_{t}') \ne \emptyset . \end{aligned}$$

Therefore assume w.l.o.g.

$$\begin{aligned} Box(v_{i}',v_{k}') \cap Box(v_{t},v_{t}') \ne \emptyset \quad \text {and}\quad Box(v_{i}',v_{k}')^{C} \cap Box(v_{t},v_{t}') \ne \emptyset .\;\;\;\;(\#) \end{aligned}$$

Pick $v_{r}\in Box(v_{i},v_{t})$ and $v_{u}\in Box(v_{t}', v_{k}')$. By applying Lemma 3 to $Box(v_{i},v_{k})$ and $Box(v_{i}',v_{k}')$ we have $f(v_{r},v_{u}) \ge f(v_{i},v_{t})=v_{t}$ and $f(v_{r},v_{u}) \le f(v_{t}',v_{k}') =v_{t}'$ respectively. If $f(v_{r},v_{u}) \notin \{v_{t},v_{t}'\}$ then the Lemma holds with $v_{i}''=v_{r}, v_{k}''=v_{u}$ and $v_{t}''=f(v_{r},v_{u})$. So suppose $f(v_{r},v_{u}) \in \{v_{t},v_{t}'\}$. We consider two cases.

Case 1: $f(v_{r},v_{u}') =v_{t}$. Consider an increasing sequence $\{v_{r}^{q}\}$ such that $\lim _{n\rightarrow \infty }v_{r}^{q} =v_{t}'$. In view of (#) there exists a q such that $v_{r}^{q}$ is on the boundary of $Box(v_{i}',v_{k}')$ and is in $Box(v_{t},v_{t}')$. By continuity $\lim _{n\rightarrow \infty }f(v_{t}^{r},v_{t}')=v_{t}'$. By choosing a point $q'$ sufficiently close to $v_{t}^{r}$ we can satisfy the conditions of the Lemma.

Case 2: $f(v_{r},v_{u}) =v_{t}'$. Suppose $v_{k} \ge v_{t}'$. By applying Lemma 3 to $Box(v_{i},v_{k})$ and $Box(v_{i}',v_{k}')$ we have $f(v_{r},v_{u})=v_{t}$ and $f(v_{r},v_{r})=v_{t}'$. This is a contradiction. Hence $v_{k}\le v_{t}'$. Now by repeating the arguments in Case 1 the Lemma holds with $v_{i}''=v_{r}, v_{k}''=v_{u}$ and $v_{t}''=f(v_{r},v_{u})$. $\square$

The next Lemma states that there exists at most one element in the range of f which is in the interior of its relevant box.

Lemma 7

There do not exist $v_{i}, v_{k}, v_{i}', v_{k}'$ such that (i) $v_{i}$ is ordered with $v_{k}$ and $v_{i}'$ is ordered with $v_{k}'$ (ii) $f(v_{i},v_{k}) \in \text {int}Box(v_{i},v_{k})$ (iii) $f(v_{i}',v_{k}')\in \text {int}Box(v_{i}',v_{k}')$ and (iv) $f(v_{i},v_{k}) \ne f(v_{i}',v_{k}')$.

Proof

We prove the Lemma by contradiction i.e there exist $v_{i},v_{k},v_{i}',v_{k}'$ as specified in the statement of Lemma 7. Let $f(v_{i},v_{k})=v_{t}$ and $f(v_{i}',v_{k}')=v_{t}'$.

Case 1: Suppose $v_{t}, v_{t}'$ are ordered. Assume w.l.o.g. $v_{t}\le v_{t}'$. By Lemma 5 there exists $v_{i}'',v_{k}''$ such that $f(v_{i}'',v_{k}'')=v_{t}''$ and $v_{t}'' \in \text {int} Box(v_{i}'',v_{k}'')$. In fact, by applying the Lemma repeatedly we can construct a sequence $\{v_{t}^{q}\}_{q=1}^{\infty }$ such that $f(v_{i}^{q},v_{k}^{q})=v_{t}^{q} \in \text {int} Box(v_{i}^{q},v_{k}^{q})$ for all q and $\lim v_{t}^{q}= v_{t}'$.

Let $\{\tilde{v_{t}}^{q}\}_{q=1}^{\infty }$, be a subsequence of $\{v_{t}^{q}\}$ such that $\tilde{v_{t}}^{q} \in \text {M}Box(v_{t})$ for all q. Note that $Box(v_{i'},v_{t}')\cap Box(\tilde{v_{i}}^{q},\tilde{v_{t}}^{q}) \ne \emptyset$. We claim that $\tilde{v_{k}}^{q}\ge v_{t}'$ cannot hold. Suppose contrariwise that $\tilde{v_{k}}^{q} \ge v_{t}'$. Pick $v_{r} \in Box(v_{i}',v_{t}')\cap Box(\tilde{v_{i}}^{q},\tilde{v_{t}}^{q})$ and $v_{u} \in Box(v_{t}',v_{k}')\cap Box(\tilde{v_{t}}^{q},\tilde{v_{k}}^{q}).$ Applying Lemma 3 to the boxes $Box(\tilde{v_{i}}^{q},\tilde{v_{t}}^{q})$ and $Box(v_{i}',v_{k}')$ we have $f(v_{r},v_{u})=\tilde{v_{t}}^{q}$ and $f(v_{r},v_{u})=v_{t}'$ respectively. This is a contradiction. Therefore $\tilde{v_{k}}^{q}\ge v_{t}'$ cannot hold and we have

$$\begin{aligned} \lim _{n \rightarrow \infty }\tilde{v_{t}}^{q}= v_{t}' \Rightarrow \lim _{n \rightarrow \infty }\tilde{v_{k}}^{q} = v_{t}'. \end{aligned}$$

Since $\tilde{v_{t}}^{q}\rightarrow v_{t}'$ we know by Lemma 4 that $\lim _{n \rightarrow \infty }MBox(\tilde{v_{t}}^{q})= MBox(v_{t}')$. Hence $\lim _{n \rightarrow \infty }\text {M}Box(\tilde{v_{t}}^{q}) = \text {M}Box(v_{t}')=Box(\bar{v_{i}}^{q},v_{t}')$ where $\bar{v_{i}}^{q}=\lim _{n \rightarrow \infty }\tilde{v_{i}}^{q}$, i.e $v_{t}' \notin \text {int}MBox(v_{t}')=Box(\bar{v_{i}}^{q},v_{t}')$. However $v_{t}' \in \text {int}Box(v_{i}', v_{k}')$ implies $v_{t}' \in \text {int}MBox(v_{t}')$ by assumption. Thus we have a contradiction.

Case 2: $v_{t}$ and $v_{t}'$ are not ordered. Pick $v_{r} \in Box(\mathbf{0},v_{t}) \cap Box(\mathbf{0},v_{t})$.¹¹ By Case 1 $f(v_{r},v_{t}) \notin \text {int} Box(v_{r},v_{t})$ and $f(v_{r},v_{t}') \notin \text {int} Box(v_{r},v_{t}')$.

Fig. 2

Illustration for Lemma 7

We claim that $f(v_{r},v_{t}) = v_{t}$. Suppose this is false. By virtue of the fact that $f(v_{r},v_{t}) \notin \text {int}Box(v_{r},v_{t})$, $f^{2}(v_{r},v_{t})$ must lie on the boundary of $Box(v_{r},v_{t})$ but not equal to $v_{t}$. By constructing a sequence $\{v_{r}^{q}\}_{q=1}^{\infty } \rightarrow v_{t}$ and using arguments from Lemma 5 we obtain a contradiction. Therefore $f(v_{r},v_{t})=v_{t}$. By an identical argument $f(v_{r}, v_{t}')=v_{t}'$ (Fig. 2).

Pick $v_{u} \in Box(v_{t},\mathbf{1}) \cap Box(v_{t}',\mathbf{1})$. Using the same arguments as in the previous paragraph, we have $f(v_{u}, v_{t})=v_{t}$ and $f(v_{u}, v_{t}')=v_{t}'$. Applying Lemma 3 and monotonicity,

$$\begin{aligned} f(v_{r},v_{u})\ge f(v_{r},v_{t})=v_{t}.\\ f(v_{r},v_{u}) \le f(v_{t},v_{u})=v_{t}. \end{aligned}$$

Therefore $f(v_{r},v_{u})=v_t$. However, the same argument with $v_{t}'$ substituted for $v_t$ yields $f(v_{r},v_{u})=v_{t}'$. We have a contradiction. $\square$

Lemma 8

Let $v_{i},v_{k}$ be ordered and $f(v_{i},v_{k})=v_{t}$. Then

$$\begin{aligned} \big [ v_{r},v_{u} \in Box(v_{i},v_{t}), v_{r} \le v_{u} \big ]\Rightarrow & {} \big [f(v_{r},v_{u})=v_{u}\big ].\\ \big [ v_{r},v_{u} \in Box(v_{t},v_{k}), v_{r}\le v_{u} \big ]\Rightarrow & {} \big [f(v_{r},v_{u})=v_{r}\big ]. \end{aligned}$$

Proof

Suppose $v_{r},v_{u}\in Box(v_{i},v_{t})$, $v_{r}\le v_{u}$ and $f(v_{r},v_{u})\ne v_{u}$. Suppose $f(v_{r},v_{u})=v_{t}'$. By Lemma 7, $v_{t}' \notin \text {int}Box(v_{r},v_{u})$. By applying Lemma 3 on $Box(v_{r},v_{u})$ we have $f(v_{t}',v_{s})=v_{t}'$ for all $v_{s}\in Box(v_{r},v_{t}')$. Similarly, by applying Lemma 3 on $Box(v_{i},v_{t})$ we have $f(v_{r},v_{t})=v_{t}$. This implies that there exists $v_{k}'\ge v_{u}$ such that $f(v_{r},v_{k}')>v_{t}'$ and $f(v_{r},v_{k}')\in Box(v_{r},v_{t}')$. By applying Lemma 3 on $Box(v_{r},v_{t}')$ we have $f(f(v_{r},v_{k}'),v_{t}')=v_{t}'$. However, by Lemma 3 on $Box(f(v_{t},v_{k}'),v_{k}')$ we have $f(v_{t}',f(v_{r},v_{k}'))=f(v_{r},v_{k}')$. This is a contradiction. Therefore, $f(v_{r},v_{u})=v_{u}$.

The case where $v_{r}\le v_{u}$ with $v_{r},v_{u}\in Box(v_{t},v_{k})$ can be proved by an argument similar to the one above. $\square$

Lemma 9

Pick any $v_{i},v_{k} \in {A}$. Then $f(v_{i},v_{k})=f(\underline{v},\overline{v})$ where $\underline{v}$ and $\overline{v}$ are as defined before.

Proof

There is nothing to prove in the case where $v_{i}$ and $v_{k}$ are ordered. Therefore assume that $v_{i},v_{k}$ are not ordered. Let $f(\underline{v},\overline{v}) =v_{t}$. For each $j \in X$ we have $v_{t}^{j} ,v_{t}'^{j}$ such that

• (i) $v_{tj}^{j}=v_{tj},\; v_{tj'}^{j}= \min (v_{ij'},v_{kj'})\quad \forall j' \in X.$

• (ii) $v_{tj}'^{j}=v_{tj},\; v_{tj'}'^{j}= \max (v_{ij'},v_{kj'})\quad \forall j' \in X.$

Note that $v_{t}^{j} \in Box(\underline{v}, v_{t})$ and $v_{t}'^{j}\in Box(v_{t},\overline{v})$ for all j. By applying Lemma 8 to $Box(\underline{v}, v_{t})$ and $Box(v_{t},\overline{v})$ and using monotonicity we have

$$\begin{aligned} f(v_{i},v_{k})\ge & {} f(v_{t}^{j},v_{t}') =v_{t}^{j}. \\ f(v_{i},v_{k})\le & {} f(v_{t}'^{j},v_{t}')=v_{t}'^{j} . \end{aligned}$$

This implies $f(v_{i},v_{k})=f(\underline{v},\overline{v})=v_{t}$. $\square$

As an implication of Lemma 9 we can restrict attention to any ordered pair $v_{i},v_{k}$. Our final Lemma proves the theorem.

Lemma 10

There exists $\alpha \in {A}$ such that for all $v_{i},v_{k} \in {A}$

$$\begin{aligned} f_{j}(v_{i},v_{k}) =\text {med}(\min _{i\in N}\{v_{ij}\},\max _{i\in N}\{v_{ij}\}, \alpha _{j}) \quad \forall j\;\in X. \end{aligned}$$

Proof

Let $f(\mathbf{0},\mathbf{1})=v_{t}^{*}$. We show that f is an $\alpha$-median rule with $\alpha =v_{t}^{*}$. Let $v_{i},v_{k} \in {A}^{n}$. By Lemma 9 we only need to consider the case where they are ordered. W.l.o.g. assume $v_{i} \le v_{k}$.

• 1. Case 1: Suppose $v_{i},v_{k}$ are both ordered with respect to $v_{t}^{*}$. We show that f is an $\alpha$-median rule with $\alpha =v_{t}^{*}$. By Lemma 3, $f(v_{i},v_{k}) =v_{t}^{*}$ for all $v_{i} \in Box(\mathbf{0},v_{t}^{*})$ and $v_{k} \in Box(v_{t}^{*},\mathbf{1})$. By Lemma 8, $f(v_{k},\mathbf{1})=v_{k}$ for all $v_{k} \in Box(v_{t}^{*},\mathbf{1})$ and $f(\mathbf{0},v_{i})=v_{i}$ for all $v_{i} \in Box(\mathbf{0},v_{t}^{*}).$ By Lemmas 8 and 9,

  $$\begin{aligned} f(v_{i},v_{k})= & {} f(\underline{v},\overline{v})=\overline{v} \quad \forall \; v_{i},v_{k} \in Box(\mathbf{0},v_{t}^{*}).\\ f^{2}(v_{i},v_{k})= & {} f^{2}(\underline{v},\overline{v}) =\underline{v}\quad \forall \; v_{i},v_{k} \in Box(v_{t}^{*},\mathbf{1}). \end{aligned}$$

  Therefore $v_{t}^{*}$ is the $\alpha$-median for all $v_{i}$ and $v_{k}$ ordered such that either $v_{i},v_{k}$ $\in Box(\mathbf{0},v_{t}^{*})$ or $v_{i},v_{k}$ $\in Box(v_{t}^{*},\mathbf{1})$. If $v_{i},v_{k} \in Box(\mathbf{0},v_{t}^{*})$ are not ordered then by Lemmas 8 and 9, $f^{2}(v_{i},v_{k})=\overline{v}$. Similarly if $v_{i},v_{k} \in Box(v_{t}^{*},\mathbf{1})$ are not ordered then by applying Lemmas 8 and 9, $f^{2}(v_{i},v_{k})=\underline{v}$. Therefore, in both the cases f picks the component-wise $\alpha$-median for $j\in X$ with $\alpha =v_{t}^{*}.$

• 2. Case 2: Suppose $v_{i}$ is ordered with $v_{t}^{*}$ but $v_{k}$ is not ordered with $v_{t}^{*}$. Pick $v_{t}^{\gamma } \in Box(v_{i},v_{k})\cap Box(\mathbf{0},v_{t}^{*})$ such that $v_{tj}^{\gamma }=\text {med}(v_{ij},v_{kj},\alpha _{j})$ for all $j\in X$. By Lemmas 3 and 8 and monotonicity, $f(\mathbf{0},v_{t}^{*})=v_{t}^{\gamma } \le f(v_{i},v_{k})$ and $f(v_{t}^{*},\mathbf{1})=v_{t}^{*} \le f(v_{i},v_{k})$. This implies $f(v_{i},v_{k})=v_{t}^{\gamma }.$ The same arguments hold for the case when $v_{i}$ is not ordered with $v_{t}^{*}$ but $v_{k}$ is ordered with $v_{t}^{*}$.

• 3. Case 3: Neither $v_{i}$ nor $v_{k}$ is ordered with respect to $v_{t}^{*}$. Pick $\underline{v}_{i},\underline{v}_{k},\overline{v}_{i},\overline{v}_{k}$ such that $\underline{v}_{ij}=\min (v_{ij},v_{tj}^{*})$, $\underline{v}_{kj}=\min (v_{kj},v_{tj}^{*})$, $\overline{v}_{ij}=\max (v_{ij},v_{tj}^{*})$ and $\overline{v}_{kj}=\max (v_{kj},v_{tj}^{*})$. By applying Lemma 8 to $Box(\mathbf{0},v_{t}^{*})$ and $Box(v_{t}^{*},\mathbf{1})$ and using monotonicity ,

  $$\begin{aligned} f(\underline{v}_{i},\underline{v}_{k})= & {} \underline{v}_{k} \le f(v_{i},v_{k}).\\ f(\overline{v}_{i},\overline{v}_{k})= & {} \overline{v}_{i}\ge f(v_{i},v_{k}). \end{aligned}$$

  This implies $f(v_{i},v_{k})=\text {med}(v_{ij},v_{kj},\alpha _{j}).$ $\square$

Let $f(\mathbf{0},\mathbf{1})=v_{t}^{*}$ such that $v_{t}^{*} \in {A}$. We have proved that $f^{2}$ is a component-wise $\alpha$-median aggregator with $\alpha =v_{t}^{*}$. Note that $f^{k}$ is also a component-wise aggregator i.e the aggregation over an issue is independent of the opinions over other issues. We show that $f^{k}$ is a component-wise $\alpha$-median rule for $k=1,2,\ldots ,n.$

Let $v \in {A}^{k}$, $k\in N$ be a profile. We show that

$$\begin{aligned} f_{j}(v)= \text {med}(\min _{i=1,\ldots ,k}v_{ij},\max _{i=1,\ldots ,k}v_{ij},\alpha _{j}) \end{aligned}$$

for all $j\in X$. There are several cases to consider. Pick $j\in X$. Suppose $v_{ij}\le \alpha _{j}$ for all $i \in N$. Since $f^{2}$ is a component-wise $\alpha$-median aggregator $f^{2}(v_{ij},v_{i'j})=\max (v_{ij},v_{i'j})$ for all $i,i'$. Therefore,

$$\begin{aligned} f^{k}(v_{1j},\ldots ,v_{kj})= & {} f^{2}(\ldots f^{2}(f^{2}(v_{1j},v_{2j}),\ldots ,v_{kj})\\= & {} \max (\ldots \max (\max _{}(v_{1j},v_{2j})\ldots ,v_{kj}) \\= & {} \max (v_{1j},\ldots ,v_{kj})\\= & {} f^{k}(v_{1j},\ldots ,v_{kj})\\= & {} \text {med}(\min _{i}(v_{ij}),\max _{i}(v_{ij}), \alpha _{j}). \end{aligned}$$

Suppose $v_{ij}\ge \alpha _{j}$ for all $i\in N$. An argument analogous to the previous one gives $f^{k}(v_{1j},\ldots ,v_{kj})=\min (v_{1j},\ldots ,v_{kj})=\text {med}(\min _{i}(v_{ij}),\max _{i}(v_{ij}), \alpha _{j})$.

Finally consider the case where $\alpha _{j}\in (\min _{i}(v_{ij}),\max _{i}(v_{ij}))$. Let

$$\begin{aligned} f^{2}(v_{1j},v_{2j})= & {} z^{1}.\\ f^{2}(f^{2}(v_{1j},v_{2j}),v_{3j})= & {} z^{2}.\\ \quad \vdots \quad \quad&\\ f^{2}(\ldots (f^{2}(v_{1j},v_{2j}),\ldots v_{kj}))= & {} z^{k-1}. \end{aligned}$$

In view of the nature of $f^{2}$ there must exist q such that $z^{q}=\alpha _{j}$ and $z^{q'}=\alpha _{j}$ for all $q'\ge q$. Therefore $f^{k}(v_{1j},\ldots ,v_{kj})=\text {med}(\min _{i}(v_{ij}),\max _{i}(v_{ij}), \alpha _{j})$. This completes the proof.$\square$

We show that the axioms used in Theorem 1 are independent. We consider each axiom in turn and show that there exists an aggregator that satisfies the other axioms.

Consistency The median aggregator satisfies all the axioms except consistency.

Unanimity Constant aggregators satisfy all the axioms except unanimity.

Anonymity We define an aggregator that specifies a dictator for every subset of the voters and outputs the vector of evaluations of the dictator for all profile. We proceed as follows. Let $\underline{i}(N)=\min _{i\in N}\#i$. Then $f^{D}$ is a sequential dictator aggregator if $f^{D}=v_{\underline{i}(N)}$ for all $N\in \mathbf{N}$ for all $v\in {A}^{n}$.

The aggregator is consistent as we show below. Consider a profile $v\in {A}^{n}$. Then by definition of the aggregator, $f^{D}(v_{1},\ldots ,v_{n})=v_{1}$. Consider any partition $I=\{N_{1},\ldots ,N_{K}\}$. By applying the rule to the sub-groups we have,

$$\begin{aligned} f^{D}(f^{D}(v_{N_{1}}),\ldots ,f^{D}(v_{N_{K}}))=f(v_{\underline{i} (N_{1})},\ldots ,v_{\underline{i}(N_{K})})=v_{\underline{i}(N)} =f^{D}(v)=v_{1}. \end{aligned}$$

The sequential dictatorship clearly violates anonymity.

Continuity We have shown earlier that the L-min aggregator satisfies all the axioms other than continuity.

Monotonicity We define an aggregator for the case when the number of issues is two. The construction can be easily generalized to an arbitrary number of issues.

Define $f^{2}$ as follows. Pick $\bar{v}\in {A}$ with $\bar{v}_{2}>0$. The aggregator will be separable. For the first component, $f^2$ picks the smaller of the first component of the two voter evaluations, i.e $f_{1}(v_{i},v_{k})=\min (v_{i1},v_{k1})$ for all $v_{i},v_{k}\in {A}$. For the second component, there are three cases:

• (i) $\max (v_{i2},v_{k2}) \le \bar{v}_{2}$. Then $f_{2}(v_{i},v_{k})= \text {max}(v_{i2},v_{k2})$ .

• (ii) $\min (v_{i2},v_{k2}) \ge \bar{v}_{2}$. Then $f_{2}(v_{i},v_{k})= \text {min}(v_{i2},v_{k2})$.

• (iii) $\min (v_{i2},v_{k2})<\bar{v}_{2}$ and $\max (v_{i2}, v_{k2}) > \bar{v}_{2}$. Then

  $$\begin{aligned} f_{1}(v_{i},v_{k})=\max \big (\min (v_{i2},v_{k2}),\bar{v}_{2}-|\bar{v}_{2}-\max (v_{i2},v_{k2})|\big ). \end{aligned}$$

The aggregator $f^{k}$, $k\in \{1,\ldots ,n\}$ can be obtained from $f^{2}$ in the following way. For any $v\in {A}^{k}$,

$$\begin{aligned} f(v_{1},\ldots ,v_{k})=\max \big (\min _{i}(v_{i2}, \bar{v}_{2})-|\bar{v_{2}}-\max _{i}v_{i2}|\big ). \end{aligned}$$

Fig. 3

Violation of monotonicity

We show that the rule is not monotonic. In Fig. 3 $v_{r}\in {A}$ satisfies $v_{rj}<\bar{v}_{j}$, $j\in \{1,2\}$. For the profile $v=(v_{r},\bar{v})$ we have $f(v_{r},\bar{v})=\bar{v}$. Pick $v_{u}$ such that $v_{uj}>\bar{v}_{j}$, $j\in \{1,2\}$ and $f(v_{r},v_{u})=v_{t}$ where $v_{t2}<\bar{v}_{2}$. Therefore, the rule violates monotonicity.

The aggregator is consistent for any profile $v\in {A}^{n}$. Suppose $\bar{i}$ such that $v_{\bar{i}2}=\max _{i\in N}v_{i2}.$ Consider a partition $I=\{N_{1},\ldots ,N_{K}\}$. Suppose $\bar{i}\in N_{k}$ for some $k\in \{1,\ldots ,K\}$. Note that $\min _{i\in N_{k}}v_{i2} \ge \min _{i\in N_{k}}v_{i2}$. Therefore,

$$\begin{aligned} \max \big (\min _{i\in N_{k}}(v_{i2}, \bar{v}_{2})-|\bar{v_{2}}-\max _{i\in N_{k}}v_{i2}|\big )\ge \max \big (\min _{i\in N_{k'}}(v_{i2}, \bar{v}_{2})-|\bar{v_{2}}-\max _{i\in N_{k'}}v_{i2}|\big ) \end{aligned}$$

for all $N_{k'} \in I.$ Therefore,

$$\begin{aligned} f(f(v_{N_{1}},\ldots ,f(v_{N_{K}}))=\max \big (\min _{i\in N}(v_{i2}, \bar{v}_{2})-|\bar{v_{2}}-v_{\bar{i}2}|\big )=f(v). \end{aligned}$$

Figure 4 shows the continuity of the aggregator. Continuity is an issue only for sequences of the following kind: $\{(v^{q}_{1},v^{q}_{2})\}$, $q=1,2 \ldots$ such that (i) $v_{1}^{q}=v_{rj}\le \hat{v}_{j}$ for all q and for $j\in \{1,2\}$ and (ii) $\{v^{q}_{2}\} \rightarrow \hat{v}$. In this case, $f(v_{1}^{q},v^{q}_{2})\rightarrow v_{t}$ and $f(v_{r},\hat{v})=v_{t}$ so that f is continuous.

Fig. 4

Continuity of the aggregator

Proof of Theorem 2

Suppose an aggregator satisfies consistency, component unanimity and component anonymity. Define the order ($\preceq$) on ${A}^{n}$ as follows.

$$\begin{aligned} v_{i} \preceq v_{k}\quad \text {if}\quad f(v_{i},v_{k})=v_{i}\quad \text {for all}\quad v_{i},v_{k} \in {A}. \end{aligned}$$

We show that the order ($\preceq$) is a partial order i.e it satisfies the following three properties.

• (i) Reflexivity: Pick any $v_{i}\in {A}$. By component unanimity, $f(v_{i},v_{i})=v_{i}$. Therefore, $v_{i}\preceq v_{i}$ for all $v_{i} \in {A}$.

• (ii) Anti-symmetry: Suppose $v_{i},v_{k}\in {A}$ such that $v_{i} \preceq v_{k}$ and $v_{k}\preceq v_{i}$. Then by definition, $f(v_{i},v_{k})=v_{i}$ and $f(v_{k},v_{i})=v_{k}$. By component anonymity, $f(v_{i},v_{k})=f(v_{k},v_{i})=v_{i}=v_{k}$.

• (iii) Transitivity: Suppose $v_{i},v_{k},v_{t}\in {A}$ such that $v_{i}\preceq v_{k}$ and $v_{k}\preceq v_{t}$. By definition, $f(v_{i},v_{k})=v_{i}$ and $f(v_{k},v_{t})=v_{k}$. Therefore, by consistency and component unanimity,

  $$\begin{aligned}&f(v_{i},v_{t})= f^{2}(f(v_{i},v_{k}),f(v_{k},v_{t}))=f^{4}(v_{i},v_{k},v_{k},v_{t}).\\&=f^{3}(v_{i},f(v_{k},v_{k}),v_{t})=f^{3}(v_{i},v_{k},v_{t})=f^{2}(v_{i},f(v_{k},v_{t}))=f(v_{i},v_{k})=v_{t}. \end{aligned}$$

Therefore, the ordering ($\preceq$) is a partial order. We claim the following. Suppose $v_{i}\preceq v_{k}$ for some $v_{i},v_{k} \in {A}^{n}$. Then $f(v_{i},v_{t})\preceq f(v_{k},v_{t})$ for all $v_{t} \in {A}^{2}$.

By consistency and component unanimity we have,

$$\begin{aligned} f^{2}(f(v_{i},v_{k}),f(v_{k},v_{t})= & {} f^{4}(v_{i},v_{k},v_{k},v_{t}) =f^{3}(v_{i},v_{k},v_{t}).\\= & {} f^{2}(f(v_{i},v_{k}),v_{t})=f(v_{i},v_{t}). \end{aligned}$$

Therefore, the aggregator is increasing in the order ($\preceq$). We claim that $f(v_{i},v_{k}) \preceq v_{i}$ and $f(v_{i},v_{k}) \preceq v_{k}$. By consistency, component anonymity and component unanimity,

$$\begin{aligned} f^{2}(f(v_{i},v_{k}),v_{i})=f^{3}(v_{i},v_{k},v_{i})= f^{2}(f(v_{i},v_{i}),v_{k})=f(v_{i},v_{k}). \end{aligned}$$

Therefore, by the definition of $(\preceq )$ we have $f(v_{i},v_{k})\preceq v_{i}$. Similarly, we can show that $f(v_{i},v_{k}) \preceq v_{k}$. Therefore, the aggregator outputs a vector of evaluations which is a lower bound according to $(\preceq )$. We finally show that the aggregator must select the unique greatest lower bound vector of opinions for any pair of voter opinions.

Suppose $f(v_{i},v_{k})=v_{t}$. We have shown that $v_{t}$ must be a upper bound of $v_{i}$ and $v_{k}$. We claim that $v_{t}$ is the unique greatest lower bound. We prove this by contradiction. Suppose $v_{t}'$ is another lower bound. By definition,

$$\begin{aligned} f(v_{i},v_{t}')=v_{t}'\quad \text {and}\quad f(v_{k},v_{t}' )=v_{t}' . \end{aligned}$$

Therefore, by consistency,

$$\begin{aligned} f(v_{i},v_{t}')= & {} f^{2}(v_{i},f(v_{k},v_{t}'))=f^{3}(v_{i},v_{k},v_{t}').\\= & {} f^{2}(f(v_{i},v_{k}), v_{t}')=f(v_{t},v_{t}'). \end{aligned}$$

Since $f(v_{i},v_{t}')=v_{t}'$ we have $f(v_{t},v_{t}')=v_{t}'$. Therefore, $v_{t}' \preceq v_{t}=f(v_{i},v_{k})$. Therefore, $f(v_{i},v_{k})$ is the unique greatest lower bound of $v_{i}$ and $v_{k}$.

We show that the aggregator is invariant to permutations of opinions over an issue.¹² We claim the following. Let $\pi :N\times X\rightarrow N$ be a bijection. Suppose v, $v'\in {A}^{2}$ such that $v_{ij}'=v_{\pi (ij)j}$ for some $j\in X$ and $v_{ij'}'=v_{ij'}$ for all $j'\in X, j'\ne j$. Then $f(v)=f(v').$

We prove the above claim by contradiction. Consider a profile $v\in {A}^{2}$ and an issue $j\in X$. The claim is trivially true if $j\in L(v)\cup W(v)$. Suppose $j\notin L(v)\cup W(v)$. Let $v=(v_{i},v_{k})$ and $v'=(v_{i}',v_{k}')$ such that $v_{ij}=v_{kj}'$, $v_{kj}=v_{ij}$, $v_{ij'}'=v_{ij'}$ and $v_{kj'}'=v_{kj'}$ for all $j'\ne j$, $f_{j}(v)=f_{j}(v')$ and $f_{j'}(v)\ne f_{j'}(v')$. Therefore, the bijection $\pi$ is such that $\pi (i,j)=k$ and $\pi (k,j)=i$ and $\pi (i',j')=i'$ for all $i'\in N$ and $j'\in X$, $j'\ne j$.

We claim that f(v) must be ordered with $f(v')$. Suppose contrariwise, that f(v) is not ordered with $f(v')$. Then $f(f(v),f(v'))=v''$ where $v''\notin \{f(v),f(v')\}$. W.l.o.g assume that $v_{j'}''=f_{j'}(v)$. By definition of $(\preceq )$, we have $f_{j'}(v_{i},v_{i}')=f_{j'}(v')$. This is a violation of component unanimity. Therefore, f(v) is ordered with $f(v')$.

W.l.o.g suppose $f(v)\preceq f(v')$. By the definition of $(\preceq )$, we have $f_{j'}(v_{i},f(v'))=f_{j'}(v).$ This is a contradiction to component anonymity since by our construction $f_{j'}(v)=f_{j'}(v')=1-f_{j'}(v)$. Similar arguments can be made when $f(v')\preceq f(v)$. The final claim proves separability.

We claim the following. For all $v,v'\in {A}^{2}$, $\big [v_{j}=v_{j}'\big ]\Rightarrow \big [f_{j}(v)=f_{j}(v')\big ]$ for all $j\in X$.

Let $\bar{v}=(v_{1},v_{2})\in {A}^{2}$ be a profile such that $\bar{v}_{1j}+\bar{v}_{2j}=1$. To prove the claim it is sufficient to show that for all $v\in {A}^{2}$, $\big [v_{1j}+v_{2j}=1\big ]\Rightarrow \big [f_{j}(v)=f_{j}(\bar{v})\big ]$ for all $j\in X$. So pick any $j\in X$ and $v\in {A}^{2}$ such that $v_{1j}+v_{2j}=1$. By definition $f^{4}(\bar{v},\bar{v})=f^{2}(f^{2}(\bar{v}),f^{2}(\bar{v}))$. By the property of $\bar{v}$ there exists a profile $\hat{v}=(\hat{v}_{1},\hat{v}_{2})\in {A}^{2}$ such that $f^{4}(\bar{v},\bar{v})=f^{4}(v,\hat{v})$. We construct $\hat{v}$ as follows: (1) $\hat{v}_{ij}=v_{ij}$ (ii) $\hat{v}_{ij'}=v_{ij'}$ for all $j'\notin L(v)\cup W(v)$, $j'\ne j$ (iii) $\hat{v}_{ij'}=1-v_{ij'}$ for all $j'\in L(v)\cup W(v)$, $j'\ne j$ for $i\in \{1,2\}$. Therefore, $(v,\hat{v})$ is constructed by permutations of component values in the profile $(\bar{v},\bar{v})$.

By our previous claim and consistency, we have $f^{4}(\bar{v},\bar{v})=f^{4}(v,\hat{v})=f^{4}(f(v),f(v),\hat{v})$. Now, we construct a profile $\tilde{v}=(\tilde{v}_{1},\tilde{v}_{2})\in {A}^{2}$ such that $f^{4}(v,\hat{v})=f^{4}(\tilde{v},v)$. We construct $\tilde{v}$ as follows: (1) $\tilde{v}_{ij}=f_{j}(v)$ (ii) $\tilde{v}_{ij'}=v_{ij'}$ for all $j'\notin L(v)\cup W(v)$, $j'\ne j$ (iii) $\tilde{v}_{ij'}=1-v_{ij'}$ for all $j'\in L(v)\cup W(v)$, $j'\ne j$ for $i\in \{1,2\}$. Therefore, $(\tilde{v},v)$ is constructed by permutations of component values in the profile $(f(v), f(v),\hat{v})$.

By our previous claim, we have $f(v,\hat{v})=f(\tilde{v},v)$. Also, note that $f_{j}(\tilde{v})=f_{j}(v)$. By consistency, component anonymity and component unanimity, we have $f_{j}^{4}(\bar{v},\bar{v})=f_{j}^{4}(v,\hat{v})=f_{j}^{3}(f(v),\hat{v})=f^{4}(f(v),f(v),\hat{v})=f_{j}^{4}(\tilde{v},v)=f_{j}^{3}(f_{j}(v),v)=f_{j}^{2}(v)$.

Therefore, our claim is true and $f^{2}$ is a bipartite rule where an issue $j\in X$ is in the favoured set $F\subset X$ if $f_{j}(0,1)=1$ and it is in the non-favoured set $F^{C}$ if $f_{j}(0,1)=0$.

We show that if $f^{2}$ is a bipartite rule then $f^{k}$ is a bipartite rule, $k\in \{1,\ldots ,N\}$. To see this, take any profile $v\in {A}^{K}$. Then we have $f(v)=f^{2}(\ldots f^{2}(f^{2}(v_{1},v_{2}),v_{3}),\ldots ,v_{K})$. Since $f^{2}$ is separable, we can focus our attention to any arbitrary issue j. Suppose $j\in L(v)$. Then by component unanimity $f_{j}(v)=0$. Similarly $f_{j}(v)=1$ if W(v). Suppose $j\notin L(v)\cup W(v)$. Let

$$\begin{aligned} f^{2}(v_{1j},v_{2j})= & {} z^{1}.\\ f^{2}(f^{2}(v_{1j},v_{2j}),v_{3j})= & {} z^{2}.\\ \quad \vdots \quad \quad&\\ f^{2}(\ldots (f^{2}(v_{1j},v_{2j}),\ldots v_{kj}))= & {} z^{k-1}. \end{aligned}$$

In view of the nature of $f^{2}$ there must exist q such that $z^{q}\in \{0,1\}$ such that $z^{q'}=z^{q}$ for all $q'\ge q$. Therefore, $f^{K}$ is a bipartite rule with $j\in X$ in the favoured set F if $f_{j}(0,1)=1$ or j in the non-favoured set $F^{C}$ if $f_{j}(0,1)=0$. This completes the proof. $\square$

We show independence of the axioms below.

Component unanimity Constant Rules satisfy all axioms except component unanimity.

Component anonymity L-min aggregators satisfy all axioms except component anonymity.

Consistency The following aggregator satisfies all the axioms except consistency. An aggregator $f^{P}$ is a parity aggregator if for any profile $v\in {A}^{n}$, $N\in \mathbf{N}$: (i) $f_{j}^{P}(v)=\min _{i\in N}(v_{ij})$ if N is odd and (ii) $f^{P}_{j}(v)=\max _{i\in N}(v_{ij})$ if N is even. This aggregator satisfies the component unanimity and component anonymity but is not consistent. We show that it violates consistency. Note that the aggregator is separable so it is sufficient to show its violation of consistency for some arbitrary issue j. Suppose $v_{j}=(0,1,1)$ is the vector of opinions of voters 1, 2 and 3 for an issue j. By definition we have $f_{j}^{P}(v_{j})=\min (0,1,1)=0$. Consider the partition $I=\{\{1,2\},\{3\}\}$. By applying the aggregator to the subgroups we have $f_{j}^{P}(f_{j}^{P}(0,1),1)=f^{P}(1, 1)=1$. Therefore, $f_{j}^{P}(v_{j})\ne f^{P}_{j}(f^{P}_{j}(v_{\{1,2\}}),f^{P}_{j}(v_{3}))$.