Interpersonal comparison necessary for Arrovian aggregation

Abstract

While studies of social welfare functionals have shown that some interpersonal comparability such as ordinal and level comparability and cardinal and unit comparability resolves Arrow’s impossibility theorem, the kind of information necessary to solve this issue remains unclear. To address this shortcoming in the body of knowledge on this topic, the present study captures and then characterizes the features of informational structures that make available social welfare functionals satisfying Strong Pareto, Anonymity, and Independence of Irrelevant Alternatives. We know from this characterization that if utility levels are not interpersonally comparable, then transformed utility functions by a certain transformation need to be cardinal and unit comparable.

Appendix: Proofs

1.1 Welfarism Theorem

In order to prove our results, we need the following result, which is known as the welfarism theorem. We introduce some notation. A social welfare ordering (SWO) is a binary relation $\succsim$ over $V^{N}\equiv \prod \nolimits _{i\in N}V$, which is reflexive, complete, and transitive. The symmetric and asymmetric parts of $\succsim$ are denoted by $\succ$ and $\sim$, respectively.

We also define three axioms for SWOs.

Strong Pareto (SP): For each pair $u,u^{\prime }\in V^{N},$ (1) if for each $i\in N,$ $u_{i}\ge u_{i}^{\prime },$ then $u\succsim _{R}u^{\prime }$ and (2) if for each $i\in N,$ $u_{i}\ge u_{i}^{\prime }$ and for some $j\in N$ , $u_{j}>u_{j}^{\prime }$ , then $u\succ u^{\prime }$.

Anonymity (AN): For each $\left( u_{1},\ldots ,u_{n}\right) \in V^{N}$ and each permutation $\pi$ of N, $\left( u_{1},\ldots ,u_{n}\right) \sim \left( u_{\pi (1)},\ldots ,u_{\pi (n)}\right)$.

Invariance with respect to $\Phi ^{N}$ (INV- $\Phi ^{N}$ ): For each pair $u,u^{\prime }\in V^{N}$ and each $\varphi \in \Phi ^{N},$ $\left( u_{1},\ldots ,u_{n}\right) \succsim \left( u_{1}^{\prime },\ldots ,u_{n}^{\prime }\right)$ if and only if $\left( \varphi _{1}(u_{1}),\ldots ,\varphi _{n}(u_{n})\right) \succsim \left( \varphi _{1}(u_{1}^{\prime }),\ldots ,\varphi _{n}(u_{n}^{\prime })\right) .$

Lemma 1

(welfarism theorem¹²) If an SWFL R satisfies SP and IIA , then there is an SWO $\succsim _{R}$ such that for each $U\in \mathcal {U}^{N}$ and each pair $x,y\in X,$

$$\begin{aligned} x\,R_{U}\,y \textit{ if and only if } \left( U_{1}(x),\ldots ,U_{n}(x)\right) \succsim _{R}\left( U_{1}(y),\ldots ,U_{n}(y)\right) . \end{aligned}$$

Moreover, if R satisfies SP and AN and INV-$\Phi ^{N}$, then $\succsim _{R}$ also satisfies SP, AN and INV-$\Phi ^{N}$.

1.2 Algebraic structure of $\Phi$ provided by NCP

We also need to know the algebraic structure of $\Phi$ provided by NCP. This is described by Lemma 2.

Lemma 2

There is a function $G:\Phi \rightarrow \mathbb {R}$ such that for each pair $\varphi ,\psi \in \Phi$, $\left[ \varphi \le \psi ,\text { if and only if }G(\varphi )\le G(\psi ) \right]$ and $\left[ G(\varphi \circ \psi )=G(\varphi )+G(\psi ) \right]$ if and only if $\Phi$ satisfies NCP.

Lemma 2 states that $\Phi$ is order and group homomorphic to $\mathbb {R}$ if and only if $\Phi$ satisfies NCP. This algebraic structure makes it possible for the binary relation $\le$ over$\ \Phi$ to be represented by a real-valued function $G:\Phi \rightarrow \mathbb {R}$. According to this function G,  for each $\varphi \in \Phi$, a change caused by $\varphi$ is measured by $G(\varphi )\in \mathbb {R}$.

Proof of Lemma 2

We introduce some notation.

For each $n\in \mathbb {N}$ and each $\varphi \in \Phi$, define $\varphi ^{n}$ by

$$\begin{aligned} \varphi ^{n}\equiv \underset{n\, times}{\underbrace{\varphi \circ \cdots \circ \varphi }.} \end{aligned}$$

Given $\Phi$ that satisfies NCP, define $\Phi ^{+}\subseteq \Phi$ by

$$\begin{aligned} \Phi ^{+}\equiv \left\{ \varphi \in \Phi |\text { for each }u\in V, \varphi (u)\ge u\right\} . \end{aligned}$$

Step 1. For each $\varphi \in \Phi ,$ $\varphi$ is continuous.

Proof

Suppose that there is $\varphi \in \Phi \$that is not continuous. Without loss of generality, let

$$\begin{aligned} \underset{u\rightarrow u^{*}-0}{\lim }\varphi (u)<\varphi (u^{*}). \end{aligned}$$

Since $\Phi$ is a subgroup, there is $\varphi ^{-1}\in \Phi$ such that $\varphi ^{-1}\circ \varphi =id_{v}.$ Let $\widetilde{u}\in \left( \lim \nolimits _{ u\rightarrow u^{*}-0} \varphi (u),\varphi (u^{*})\right)$. Then, since for each $u^{\prime }<u^{*},$ $\widetilde{u}>\varphi (u^{\prime }),$ $\varphi ^{-1}(\widetilde{u})>\varphi ^{-1}\circ \varphi (u^{\prime })=u^{\prime }.$ Hence, $\varphi ^{-1}(\widetilde{u})\ge u^{*}.$ However, since $\widetilde{u}<\varphi (u^{*}),$ $\varphi ^{-1}( \widetilde{u})<\varphi ^{-1}(\varphi (u^{*}))=u^{*},$ which is a contradiction. $\square$

Step 2. Suppose that $\Phi$ satisfies NCP. Then, for each $\varphi \in \Phi ^{+}\backslash \left\{ id_{V}\right\}$ and each $u\in V,$

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim }\varphi ^{n}(u)=\infty . \end{aligned}$$

Proof

Suppose that there are $\varphi \in \Phi ^{+}\backslash \left\{ id_{V}\right\}$ and $u\in V,$ such that $\lim _{n\rightarrow \infty }\varphi ^{n}(u)<\infty$. Since $\left( \varphi ^{n}(u)\right) _{n\in \mathbb {N} }$ is a monotone increasing sequence, there is $u^{*}$ such that

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim }\varphi ^{n}(u)=u^{*}. \end{aligned}$$

From Step 1, $\varphi$ is continuous, so that

$$\begin{aligned} \varphi (u^{*})=\underset{n\rightarrow \infty }{\lim }\varphi (\varphi ^{n}(u))=\underset{n\rightarrow \infty }{\lim }\varphi ^{n+1}(u)=u^{*}. \end{aligned}$$

Hence, from NCP, $\varphi =$ $id_{V},$ which contradicts $\varphi \in \Phi ^{+}\backslash \left\{ id_{V}\right\} .$ $\square$

Step 3. Suppose that $\Phi$ satisfies NCP. Then, for each pair $\varphi ,\varphi ^{\prime }\in \Phi ^{+}\backslash \left\{ id_{V}\right\} ,$ there is $n\in \mathbb {N}$ such that

$$\begin{aligned} \varphi ^{n}\ge \varphi ^{\prime }. \end{aligned}$$

Proof

Let $u\in V.$ From Step 2, there is $n\in \mathbb {N}$ such that

$$\begin{aligned} \varphi ^{n}(u)\ge \varphi ^{\prime }(u). \end{aligned}$$

Then, from NCP, for each $u^{\prime }\in V,$

$$\begin{aligned} \varphi ^{n}(u^{\prime })\ge \varphi ^{\prime }(u^{\prime }). \end{aligned}$$

$\square$

Let $\varphi _{1}\in \Phi ^{+}\backslash \left\{ id_{V}\right\}$ and for each $\varphi \in \Phi ^{+},$ define $G(\varphi )\in \mathbb {R}$ by

$$\begin{aligned} G(\varphi )\equiv \sup \left\{ \frac{m}{n}|\varphi ^{n}\le \varphi _{1}^{m}, n,m\in \mathbb {N} \right\} . \end{aligned}$$

From Step 3, we can easily check that for each $\varphi \in \Phi ^{+}\backslash \left\{ id_{V}\right\} ,$ $G(\varphi )\in \mathbb {R} _{++}$.

Step 4. For each pair $\varphi ,\psi \in \Phi ^{+},$ if $\varphi \le \psi ,$ then $G(\varphi )\le G(\psi ).$

Proof

For each pair $n,m\in \mathbb {N} ,$ if $\psi ^{n}\le \varphi _{1}^{m},$ then $\varphi ^{n}\le \psi ^{n}\le \varphi _{1}^{m}.$ Hence,

$$\begin{aligned} G(\varphi )= & {} \sup \left\{ \frac{m}{n}|\varphi ^{n}\le \varphi _{1}^{m}, n,m\in \mathbb {N} \right\} \\\le & {} \sup \left\{ \frac{m}{n}|\psi ^{n}\le \varphi _{1}^{m}, n,m\in \mathbb {N} \right\} =G(\psi ). \end{aligned}$$

$\square$

Step 5. For each pair $\varphi ,\psi \in \Phi ^{+},$ $G(\varphi \circ \psi )=G(\psi \circ \varphi ).$

Proof

We distinguish two cases.

Case 1 $\min \left\{ G(\varphi )|\varphi \in \Phi _{i}^{+}\backslash \left\{ id_{V}\right\} \right\}$ exists.

Let $\varphi _{1}\in \arg \min \left\{ G(\varphi )|\varphi \in \Phi ^{+}\backslash \left\{ id_{V}\right\} \right\} .$ If there are $\varphi \in \Phi ^{+}\backslash \left\{ id_{V}\right\}$ and $n\in \mathbb {N}$ such that $\varphi _{1}^{n}<\varphi <\varphi _{1}^{n+1},$ then since

$$\begin{aligned} id_{v}=\varphi _{1}^{-n}\circ \varphi _{1}^{n}<\varphi _{1}^{-n}\circ \varphi <\varphi _{1}^{-n}\circ \varphi _{1}^{n+1}=\varphi _{1}, \end{aligned}$$

we have $\varphi _{1}^{-n}\circ \varphi \in \Phi ^{+}\backslash \left\{ id_{V}\right\}$ and $\varphi _{1}^{-n}\circ \varphi <\varphi _{1},$ which is a contradiction. Hence, for each pair $\varphi ,\psi \in \Phi ^{+}\backslash \left\{ id_{V}\right\} ,$ there are $n,n^{\prime }\in \mathbb {N}$ such that

$$\begin{aligned} \varphi =\varphi _{1}^{n},\quad { and }\quad \psi =\varphi _{1}^{n^{\prime }}. \end{aligned}$$

Therefore,

$$\begin{aligned} G(\varphi \circ \psi )=G(\varphi _{1}^{n+n^{\prime }})=G(\psi \circ \varphi ). \end{aligned}$$

$\square$

Case 2 $\min \left\{ G(\varphi )|\varphi \in \Phi ^{+}\backslash \left\{ id_{V}\right\} \right\}$ does not exist.

Suppose that $\inf \left\{ G(\varphi )|\varphi \in \Phi ^{+}\backslash \left\{ id_{V}\right\} \right\} >0.$ Then, there are $\varphi ,\psi \in \Phi ^{+}\backslash \left\{ id_{V}\right\}$ such that$\ \varphi <\psi$ and

$$\begin{aligned} G(\varphi )\le G(\psi )<2\inf \left\{ G(\varphi )|\varphi \in \Phi ^{+}\backslash \left\{ id_{V}\right\} \right\} . \end{aligned}$$

Since $\varphi ^{-1}\circ \psi >id_{V},$ there is $\xi \in \Phi ^{+}\backslash \left\{ id_{V}\right\}$ such that $\xi <min\left\{ \varphi ,\varphi ^{-1}\circ \psi \right\} .$ Then, from Step 4,

$$\begin{aligned} G(\psi )= & {} G(\varphi \circ (\varphi ^{-1}\circ \psi )) \\\ge & {} G(\xi \circ \xi ) \\= & {} 2G(\xi ) \\> & {} 2\inf \left\{ G(\varphi )|\varphi \in \Phi ^{+}\backslash \left\{ id_{V}\right\} \right\} , \end{aligned}$$

which is a contradiction.

Hence, for each $\epsilon >0$, there is $\varphi _{\epsilon }\in \Phi ^{+}\backslash \left\{ id_{V}\right\}$ such that $G(\varphi _{\epsilon })<\epsilon .$ For each pair $\varphi ,\psi \in \Phi ^{+}\backslash \left\{ id_{V}\right\} ,$ from Step 3, there are $n,n^{\prime }\in \mathbb {N}$ such that

$$\begin{aligned} \varphi _{\epsilon }^{n}\le \varphi<\varphi _{\epsilon }^{n+1}\quad \text { and }\quad \varphi _{\epsilon }^{n^{\prime }}\le \psi <\varphi _{\epsilon }^{n^{\prime }+1}. \end{aligned}$$

Then, since

$$\begin{aligned} \varphi _{\epsilon }^{n+n^{\prime }}\le \varphi \circ \psi<\varphi _{\epsilon }^{n+n^{\prime }+2}\quad \text { and }\quad \varphi _{\epsilon }^{n+n^{\prime }}\le \psi \circ \varphi <\varphi _{\epsilon }^{n+n^{\prime }+2}, \end{aligned}$$

we have

$$\begin{aligned} (n+n^{\prime })G(\varphi _{\epsilon })\le G(\varphi \circ \psi )\le (n+n^{\prime }+2)G(\varphi _{\epsilon }), \end{aligned}$$

$$\begin{aligned} (n+n^{\prime })G(\varphi _{\epsilon })\le G(\psi \circ \varphi )\le (n+n^{\prime }+2)G(\varphi _{\epsilon }), \end{aligned}$$

from which we obtain

$$\begin{aligned} \left| G(\varphi \circ \psi )-G(\psi \circ \varphi )\right| \le 2G(\varphi _{\epsilon })<2\epsilon . \end{aligned}$$

Therefore, $G(\varphi \circ \psi )=G(\psi \circ \varphi )$. $\square$

Step 6. For each pair $\varphi ,\psi \in \Phi ^{+},$ $G(\varphi \circ \psi )=G(\varphi )+G(\psi ).$

Proof

It is sufficient to show that for each pair $\varphi ,\psi \in \Phi ^{+}$ and each quadruplet$\ n,m,n^{\prime },m^{\prime }\in \mathbb {N} ,$

$$\begin{aligned} G(\varphi )\le & {} \frac{m}{n}\text { and }G(\psi )\le \frac{m^{\prime }}{ n^{\prime }}\text { together imply }G(\varphi \circ \psi )\le \frac{m}{n}+ \frac{m^{\prime }}{n^{\prime }},\text { and}\\ G(\varphi )\ge & {} \frac{m}{n}\text { and }G(\psi )\ge \frac{m^{\prime }}{ n^{\prime }}\text { together imply }G(\varphi \circ \psi )\ge \frac{m}{n}+ \frac{m^{\prime }}{n^{\prime }}. \end{aligned}$$

If $G(\varphi )\le \frac{m}{n}$ and $G(\psi )\le \frac{m^{\prime }}{ n^{\prime }}$, then from the definition of G,  we have $\varphi ^{n}\le \varphi _{1}^{m}$ and $\psi ^{n^{\prime }}\le \varphi _{1}^{m^{\prime }}.$ Then, since $\varphi ^{nn^{\prime }}\le \varphi _{1}^{n^{\prime }m}$ and $\psi ^{nn^{\prime }}\le \varphi _{1}^{nm^{\prime }},$

$$\begin{aligned} \varphi ^{nn^{\prime }}\circ \psi ^{nn^{\prime }}\le \varphi _{1}^{n^{\prime }m+nm^{\prime }}. \end{aligned}$$

Hence, from Step 5,

$$\begin{aligned} nn^{\prime }G(\varphi \circ \psi )= & {} G((\varphi \circ \psi )^{nn^{\prime }})\\= & {} G(\varphi ^{nn^{\prime }}\circ \psi ^{nn^{\prime }}) \\\le & {} G(\varphi _{1}^{n^{\prime }m+nm^{\prime }}) \\= & {} n^{\prime }m+nm^{\prime }, \end{aligned}$$

from which we obtain

$$\begin{aligned} G(\varphi \circ \psi )\le \frac{m}{n}+\frac{m^{\prime }}{n^{\prime }}. \end{aligned}$$

We can similarly show that $G(\varphi )\ge \frac{m}{n}$ and $G(\psi )\ge \frac{m^{\prime }}{n^{\prime }}$ together imply $G(\varphi \circ \psi )\ge \frac{m}{n}+\frac{m^{\prime }}{n^{\prime }}.$ $\square$

For each $\varphi \in \Phi \backslash \Phi ^{+},$ define $G(\varphi )\in \mathbb {R}$ by

$$\begin{aligned} G(\varphi )=-G(\varphi ^{-1}), \end{aligned}$$

where $\varphi ^{-1}\circ \varphi =id_{V}.$ For each $\varphi \in \Phi \backslash \Phi ^{+},$ since $\varphi ^{-1}\in \Phi ^{+},$ we have$\ G(\varphi )\le 0.$

Step 7. For each pair $\varphi ,\psi \in \Phi ,$ $G(\varphi \circ \psi )=G(\varphi )+G(\psi ).$

Proof

If $\varphi \circ \psi \in \Phi ^{+}$ and $\varphi \in \Phi \backslash \Phi ^{+}$, and then from Step 6,

$$\begin{aligned} G(\psi )= & {} G(\varphi ^{-1}\circ \varphi \circ \psi )\\= & {} G(\varphi ^{-1})+G(\varphi \circ \psi ) \\= & {} -G(\varphi )+G(\varphi \circ \psi ), \end{aligned}$$

from which we obtain $G(\varphi \circ \psi )=G(\varphi )+G(\psi ).$ We can similarly show that if $\varphi \circ \psi \in \Phi \backslash \Phi ^{+}$, then $G(\varphi \circ \psi )=G(\varphi )+G(\psi )$. $\square$

Step 8. For each pair $\varphi ,\psi \in \Phi _{i}^{{}},$ $\varphi \le \psi$ if and only if $G(\varphi )\le G(\psi ).$

Proof

Let $\varphi ,\psi \in \Phi$ be such that$\ \varphi <\psi .$ Then, since $\varphi ^{-1}\circ \psi \in \Phi ^{+}\backslash \left\{ id_{V}\right\}$, we have $G(\varphi ^{-1}\circ \psi )>0$. Then, from Step 7,

$$\begin{aligned} G(\psi )= & {} G(\varphi \circ \varphi ^{-1}\circ \psi )\\= & {} G(\varphi )+G(\varphi ^{-1}\circ \psi ) \\> & {} G(\varphi ). \end{aligned}$$

$\square$

Step 9. Suppose there is a mapping $G:\Phi \rightarrow \mathbb {R}$ such that for each pair $\varphi ,\psi \in \Phi$ , $\left[ \varphi \le \psi ,\text { if and only if }G(\varphi )\le G(\psi ) \right]$ and $\left[ G(\varphi \circ \psi )=G(\varphi )+G(\psi ) \right] ,$ then $\Phi$ satisfies NCP.

Proof

For each pair $\varphi ,\psi \in \Phi ,$ since either $G(\varphi )\le G(\psi )$ or $G(\varphi )\ge G(\psi )$, either $\varphi \le \psi$ or $\varphi \ge \psi$ also holds from the definition of G. Hence, $\Phi$ satisfies NCP. $\square$

1.3 Proof of Theorem 1

Necessity: Suppose that$\ \Phi$ does not satisfy NCP and that an SWFL R satisfies SP, AN, IIA, and INV- $\Phi .$ Then, there are $\varphi _{1},\varphi _{2}\in \Phi$ and $u_{1},u_{2}\in V$ such that

$$\begin{aligned} \left[ \varphi _{1}(u_{1})>\varphi _{2}(u_{1}),\text { and }\varphi _{1}(u_{2})\le \varphi _{2}(u_{2})\right] \text { or }\left[ \varphi _{1}(u_{1})<\varphi _{2}(u_{1}),\text { and }\varphi _{1}(u_{2})\ge \varphi _{2}(u_{2})\right] . \end{aligned}$$

Without loss of generality, suppose that $\varphi _{1}(u_{1})>\varphi _{2}(u_{1})$ and $\varphi _{1}(u_{2})\le \varphi _{2}(u_{2}).$ Let $U\in \mathcal {U}^{N}$ be such that

$$\begin{aligned} U_{1}(x)= & {} U_{2}(y)=u_{1}, U_{1}(y)=U_{2}(x)=u_{2}\text { and }\\ U_{i}(x)= & {} U_{i}(y)=u,\quad \text { for each }i\in N\backslash \left\{ 1,2\right\} . \end{aligned}$$

According to the welfarism theorem, there is $\succsim _{R}\subset V^{N}\times V^{N}$ such that for each $U\in \mathcal {U}^{N}\$and each pair $x,y\in X,$ $x\,R_{U}\,y\$if and only if $U(x)\succsim _{R}U(y).$ From the AN of $\succsim _{R},$

$$\begin{aligned} (u_{1},u_{2},u,\ldots ,u)\sim _{R}(u_{2},u_{1},u,\ldots ,u), \end{aligned}$$

so that $x\,I_{U}\,y$. From INV-$\Phi ,$ for each pair $\varphi _{1},\varphi _{2}\in \Phi ,$ we have$\ x\,I_{(\varphi _{1}\circ U_{1},\varphi _{2}\circ U_{2},U_{N\backslash \left\{ 1,2\right\} )}}\,y.$ Hence,

$$\begin{aligned} (\varphi _{1}(u_{1}),\varphi _{2}(u_{2}),u,\ldots ,u)\sim _{R}(\varphi _{1}(u_{2}),\varphi _{2}(u_{1}),u,\ldots ,u). \end{aligned}$$

From the AN of $\succsim _{R},$

$$\begin{aligned} (\varphi _{1}(u_{2}),\varphi _{2}(u_{1}),u,\ldots ,u)\sim _{R}(\varphi _{2}(u_{1}),\varphi _{1}(u_{2}),u,\ldots ,u). \end{aligned}$$

Therefore,

$$\begin{aligned} (\varphi _{1}(u_{1}),\varphi _{2}(u_{2}),u,\ldots ,u)\sim _{R}(\varphi _{2}(u_{1}),\varphi _{1}(u_{2}),u,\ldots ,u). \end{aligned}$$

However, since $\varphi _{1}(u_{1})>\varphi _{2}(u_{1})$ and $\varphi _{2}(u_{2})\ge \varphi _{2}(u_{1})$, from the SP of $\succsim _{R}$, we have

$$\begin{aligned} (\varphi _{1}(u_{1}),\varphi _{2}(u_{2}),u,\ldots ,u)\succ _{R}(\varphi _{2}(u_{1}),\varphi _{1}(u_{2}),u,\ldots ,u), \end{aligned}$$

which is a contradiction. $\square$

Sufficiency: We distinguish three cases.

Case 1 $\Phi$ satisfies RC.

Proof

We use a function $G:\Phi \rightarrow \mathbb {R}$ defined as in Lemma 2. Let $u_{0}\in V$ and $g:V\rightarrow \mathbb {R}$ be such that for each $u\in V,$

$$\begin{aligned} g(u)=G(\varphi ),\quad \text {where }\varphi (u_{0})=u\text {.} \end{aligned}$$

Since $\Phi$ satisfies NCP and RC, g is well defined. Let R be the transformed utilitarian rule associated with g. We show that R satisfies all the conditions in Theorem 1. Obviously, R satisfies AN and IIA.

First, let us show that R satisfies SP. For each pair $u,u^{\prime }\in V,$ such that$\ u<u^{\prime },$ let $\varphi ,\psi \in \Phi$ be such that $\varphi (u_{0})=u,$ $\psi (u_{0})=u^{\prime }.$ According to the definitions of g and G, 

$$\begin{aligned} g(u)=G(\varphi )<G(\psi )=g(u^{\prime }). \end{aligned}$$

Thus, g is increasing, so that R satisfies SP.

Next, we show that R satisfies INV- $\Phi$. For each pair $x,y\in X,$ each $i\in N,$ and each $\varphi \in \Phi$, $x\,R_{U}\,y$, if and only if $\sum \nolimits _{i\in N}g(U_{i}(x))\ge \sum \nolimits _{i\in N}g(U_{i}(y)).$ This is equivalent to

$$\begin{aligned} \sum \limits _{i\in N}g(U_{i}(x))+G(\varphi )\ge \sum \limits _{i\in N}g(U_{i}(y))+G(\varphi ). \end{aligned}$$

Let $\varphi ^{U_{i}(x)}\in \Phi$ be such that $\varphi ^{U_{i}(x)}(u_{0})=U_{i}(x)$. Then,

$$\begin{aligned} g(\varphi \circ U_{i}(x))= & {} g(\varphi \circ \varphi ^{U_{i}(x)}(u_{0})) \\= & {} G(\varphi \circ \varphi ^{U_{i}(x)}) \\= & {} G(\varphi )+G(\varphi ^{U_{i}(x)}) \\= & {} G(\varphi )+g( \varphi ^{U_{i}(x)}(u_{0})) \\= & {} G(\varphi )+g(U_{i}(x)). \end{aligned}$$

Similarly,

$$\begin{aligned} g(\varphi \circ U_{i}(y))=G(\varphi )+g(U_{i}(y)). \end{aligned}$$

Hence, $x\,R_{U}\,y$ if and only if

$$\begin{aligned} \sum \limits _{j\ne i}g(U_{j}(x))+g(\varphi \circ U_{i}(x))\ge \sum \limits _{j\ne i}g(U_{j}(y))+g(\varphi \circ U_{i}(y)), \end{aligned}$$

so that $x\,R_{(\varphi \circ U_{i},U_{-i})}\,y.$ Therefore, $x\,R_{U}\,y$ if and only if $x\,R_{(\varphi \circ U_{i},U_{-i})}\,y.$ $\square$

Case 2 $\Phi$ does not satisfy RC and $\min \left\{ G(\varphi )|\varphi \in \Phi ^{+}\backslash \left\{ id_{V}\right\} \right\}$ exists.

Let $\varphi _{1}=\min \left\{ G(\varphi )|\varphi \in \Phi ^{+}\backslash \left\{ id_{V}\right\} \right\}$ and $u_{0}\in V$. Define $g:V\rightarrow \mathbb {R}$ in the following way. First, for each $n\in \mathbb {Z} ,$ let

$$\begin{aligned} g(\varphi _{1}^{n}(u_{0}))\equiv n. \end{aligned}$$

Then, for $u_{0},\varphi _{1}(u_{0})\in V,$ we have $g(u_{0})=0$ and $g(\varphi _{1}(u_{0}))=1.$ Next, for each $u\in \left[ u_{0},\varphi _{1}(u_{0})\right] ,$ let

$$\begin{aligned} g(u)\equiv \frac{u-u_{0}}{\varphi _{1}(u_{0})-u_{0}}. \end{aligned}$$

Obviously, g is continuous and increasing in $\left[ u_{0},\varphi _{1}(u_{0})\right] .$ Finally, for each $u\in V,$ let $n\in \mathbb {Z}$ and $u^{\prime }\in \left[ u_{0},\varphi _{1}(u_{0})\right]$ be such that $u=\varphi _{1}^{n}(u^{\prime })$, and

$$\begin{aligned} g(u)\equiv n+g(u^{\prime }). \end{aligned}$$

We can easily check that this function $g:V\rightarrow \mathbb {R}$ is well defined.

Let R be the transformed utilitarian rule associated with g. We show that R satisfies all the conditions in Theorem 1. It is obvious that R satisfies AN and IIA.

First, let us show that R satisfies SP. To do so, it suffices to show that g is increasing. According to the definition of g,  g is increasing in $\left[ u_{0},\varphi _{1}(u_{0})\right] .$ Then, since for each pair $u,u^{\prime }\in \left[ \varphi _{1}(u_{0}),\varphi _{1}^{2}(u_{0})\right] ,$ if$\ u<u^{\prime },$ then

$$\begin{aligned} g(u)= & {} 1+g(\varphi ^{-1}(u)) \\< & {} 1+g(\varphi ^{-1}(u^{\prime })) \\= & {} g(u^{\prime }). \end{aligned}$$

Hence, g is also increasing in $\left[ \varphi _{1}(v_{0}),\varphi _{1}^{2}(v_{0})\right] .$ By taking similar steps, we can show that g is increasing everywhere.

Next, we show that R satisfies INV- $\Phi$. For each pair $x,y\in X,$ each $i\in N,$ and each $\varphi =\varphi _{1}^{n}\in \Phi ,$ $x\,R_{U}\,y$ if and only if $\sum \nolimits _{i\in N}g(U_{i}(x))\ge \sum \nolimits _{i\in N}g(U_{i}(y)).$ This is equivalent to

$$\begin{aligned} \sum \limits _{i\in N}g(U_{i}(x))+n\ge \sum \limits _{i\in N}g(U_{i}(y))+n. \end{aligned}$$

According to the definition of $\varphi ,$

$$\begin{aligned} g(\varphi \circ U_{i}(x))= & {} g(\varphi _{1}^{n}\circ U_{i}(x)) \\= & {} n+g(U_{i}(x)). \end{aligned}$$

Similarly,

$$\begin{aligned} g(\varphi \circ U_{i}(y))=n+g(U_{i}(y)). \end{aligned}$$

Hence, $x\,R_{U}\,y$ if and only if

$$\begin{aligned} \sum \limits _{j\ne i}g(U_{j}(x))+g(\varphi \circ U_{i}(x))\ge \sum \limits _{j\ne i}g(U_{j}(y))+g(\varphi \circ U_{i}(y)). \end{aligned}$$

Therefore, $x\,R_{U}\,y$ if and only if $x\,R_{(\varphi \circ U_{i},U_{-i})}\,y.$ $\square$

Case 3 $\Phi$ does not satisfy RC and $\min \left\{ G(\varphi )|\varphi \in \Phi ^{+}\backslash \left\{ id_{V}\right\} \right\}$ does not exist.

Let $G:\Phi \rightarrow \mathbb {R}$ be a mapping defined as in Lemma 2 $.\$Let $u_{0}\in V.$ For each $u\in V,$ let

$$\begin{aligned} g(u)\equiv \sup \left\{ G(\varphi )|\varphi (u_{0})\le u\right\} . \end{aligned}$$

Step 1. g is non-decreasing.

Proof

Let $u,u^{\prime }\in V$ be such that $u\le u^{\prime }.$ Since

$$\begin{aligned} \left\{ G(\varphi )|\varphi (v_{0})\le u\right\} \subseteq \left\{ G(\varphi )|\varphi (u_{0})\le u^{\prime }\right\} , \end{aligned}$$

we have

$$\begin{aligned} g(u)= & {} \sup \left\{ G(\varphi )|\varphi (u_{0})\le u\right\} \\\le & {} \sup \left\{ G(\varphi )|\varphi (u_{0})\le u^{\prime }\right\} \\\le & {} g(u^{\prime }). \end{aligned}$$

Step 2. g is continuous.

Proof

Suppose that g is not continuous at $u^{*}\in V$. Then, since

$$\begin{aligned} \underset{u\rightarrow u^{*}-0}{\lim }g(u)<\underset{u\rightarrow u^{*}+0}{\lim }g(u) \end{aligned}$$

and g is non-decreasing, there is no $u^{\prime }\in V$ such that

$$\begin{aligned} g(u^{\prime })\in \left( \underset{u\rightarrow u^{*}-0}{\lim }g(u), \underset{u\rightarrow u^{*}+0}{\lim }g(u)\right) . \end{aligned}$$

On the contrary, as shown in Step 5 of Lemma 2, $\inf \left\{ G(\varphi )|\varphi \in \Phi ^{+}\backslash \left\{ id_{V}\right\} \right\} =0$, so there is $\varphi ^{\prime }\in \Phi$ such that

$$\begin{aligned} G(\varphi ^{\prime })\in \left( \underset{u\rightarrow u^{*}-0}{\lim } g(u),\underset{u\rightarrow u^{*}+0}{\lim }g(u)\right) . \end{aligned}$$

For $\varphi ^{\prime }(u_{0})\in V,$ we have

$$\begin{aligned} g(\varphi ^{\prime }(u_{0}))= & {} \sup \left\{ G(\varphi )|\varphi (u_{0})\le \varphi ^{\prime }(u_{0})\right\} \\= & {} G(\varphi ^{\prime })\in \left( \underset{u\rightarrow u^{*}-0}{\lim } g(u),\underset{u\rightarrow u^{*}+0}{\lim }g(u)\right) , \end{aligned}$$

which is a contradiction. $\square$

Step 3. For each $u\in V\$ and each $\varphi \in \Phi ,$ $g(\varphi (u))=g(u)+G(\varphi )$.

Proof

For each $u\in V\$and each $\varphi \in \Phi ,$

$$\begin{aligned} g(\varphi (u))= & {} \sup \left\{ G(\varphi ^{\prime })|\varphi ^{\prime }(u_{0})\le \varphi (u)\right\} \\= & {} \sup \left\{ G(\varphi ^{\prime \prime }\circ \varphi )|\varphi ^{\prime \prime }\circ \varphi (u_{0})\le \varphi (u)\right\} \\= & {} \sup \left\{ G(\varphi ^{\prime \prime })+G(\varphi )|\varphi ^{\prime \prime }(u_{0})\le u\right\} \\= & {} G(\varphi )+\sup \left\{ G(\varphi ^{\prime \prime })|\varphi ^{\prime \prime }(u_{0})\le u\right\} \\= & {} G(\varphi )+g(u). \end{aligned}$$

$\square$

If g is increasing, then the transformed utilitarian rule associated with g satisfies SP and IIA. However, it is not obvious whether g is increasing, and thus we need to consider the case where g is not increasing.

We introduce the following notation. For each $x\in \mathbb {R} ,$ let $g^{-1}(x)\equiv \left\{ u^{\prime }\in V|g(u^{\prime })=x\right\} .$ Since g is non-decreasing and continuous, for each $x\in \mathbb {R} ,$ $g^{-1}(x)\$is a closed interval. Let $S\equiv \left\{ u\in V|\exists u^{\prime }\in V, u^{\prime }\ne u\text {, }g(u^{\prime })=g(u)\right\} .$ Clearly, g is increasing if and only if S is empty.

Step 4. For each $\varphi \in \Phi$ and each $x\in \mathbb {R} ,$ if $g^{-1}(x)=\left[ \widetilde{u},\overline{u}\right] ,$ then $g^{-1}(x+G(\varphi ))=\left[ \varphi (\widetilde{u}),\varphi ( \overline{u})\right] .$

Proof

For each $x\in \mathbb {R} ,$ let $\widetilde{u}=\arg \min g^{-1}(x)$ and $\overline{u}=\arg \min g^{-1}(x).$ Then, from Step 3, for each $\varphi \in \Phi ,$

$$\begin{aligned} g(\varphi (\widetilde{u}))= & {} G(\varphi )+g(\widetilde{u}) \\= & {} G(\varphi )+x \\= & {} G(\varphi )+g(\overline{u}) \\= & {} g(\varphi (\overline{u})). \end{aligned}$$

Hence, $g^{-1}(x+G(\varphi ))\supseteq \left[ \varphi (\widetilde{u} ),\varphi (\overline{u})\right] .$

For each $u^{\prime }<\varphi (\widetilde{u}),$ since $\varphi ^{-1}(u^{\prime })<\widetilde{u}$,

$$\begin{aligned} g(u^{\prime })= & {} G(\varphi )+g(\varphi ^{-1}(u^{\prime })) \\< & {} G(\varphi )+g(\widetilde{u}). \end{aligned}$$

Hence, $u^{\prime }\notin g^{-1}(x+G(\varphi ))$. We can similarly show that for each $u^{\prime }>\varphi (\overline{u}),$ $u^{\prime }\notin g^{-1}(x+G(\varphi )).$ $\square$

Define the binary relation $\approx$ over S such that for each pair $u,u^{\prime }\in S,$

$$\begin{aligned} u\approx u^{\prime }\text { if and only if there is }\varphi \in \Phi \text { such that }u^{\prime }=\varphi (u). \end{aligned}$$

From Step 4, for each $u\in S$ and each $\varphi \in \Phi ,$ $\varphi (u)\in S,$ so that $\approx$ is an equivalence relation. Let $S/\approx$ be the quotient set of S by $\approx .$ An element of $S/\approx$ is denoted by $\left[ u_{\lambda }\right] ,$ where $u_{\lambda }\in S$ denotes a representative. We denote that for each pair $u_{\lambda },u_{\lambda ^{\prime }}\in S$, if $u_{\lambda }$ is a representative and $g(u_{\lambda })=g(u_{\lambda ^{\prime }})$, then $u_{\lambda ^{\prime }}$ is a representative. Define $g_{2}:V\rightarrow \left[ 0,1\right]$ by

$$\begin{aligned} g_{2}(u)=\left\{ \begin{array}{ll} \frac{u_{\lambda }-\min g^{-1}(g(u_{\lambda }))}{\max g^{-1}(g(u_{\lambda }))-\min g^{-1}(g(u_{\lambda }))}, &{}\quad \text {if }u\in \left[ u_{\lambda } \right] \\ 0, &{} \quad \text {if }u\notin S\text {.} \end{array} \right. \end{aligned}$$

Step 5. For each $\varphi \in \Phi$ and each $u\in V,$ $g_{2}(\varphi (u))=g_{2}(u).$

Proof

From Step 4, for each $u\in V,$ if $u\notin S,$ then $\varphi (u)\notin S.$ Hence,

$$\begin{aligned} g_{2}(\varphi (u))=g_{2}(u)=0. \end{aligned}$$

For each $u\in V,$ if $u\in \left[ u_{\lambda }\right] ,$ then there is $\varphi ^{\prime }\in \Phi$ such that $\varphi ^{\prime }(u)=u_{\lambda }.$ Then, for each $\varphi \in \Phi ,$

$$\begin{aligned} \varphi ^{\prime }\circ \varphi ^{-1}(\varphi (u))=\varphi ^{\prime }(u)=u_{\lambda }, \end{aligned}$$

so that $\varphi (u)\in \left[ u_{\lambda }\right] .$ Hence,

$$\begin{aligned} g_{2}(\varphi (u))= & {} \frac{u_{\lambda }-\min g^{-1}(g(u_{\lambda }))}{\max g^{-1}(g(u_{\lambda }))-\min g^{-1}(g(u_{\lambda }))} \\= & {} g_{2}(u). \end{aligned}$$

$\square$

Step 6. Let R be an SWFL such that for each $U\in \mathcal {U}^{N}$ and each pair $x,y\in X,$ $x\,P_{U}\,y$ if and only if (1) $\sum \nolimits _{i\in N} g(U_{i}(x))>\sum \nolimits _{i\in N} g(U_{i}(y))$ or (2)$\ \sum \nolimits _{i\in N} g(U_{i}(x))=\sum \nolimits _{i\in N}g(U_{i}(y))$ and $\sum \nolimits _{i\in N}g_{2}(U_{i}(x))>\underset{i\in N}{\sum } g_{2}(U_{i}(y)).$ Then, R satisfies SP, AN, IIA, and INV-$\Phi .$

Proof

Obviously, R satisfies AN and IIA. To prove that R satisfies SP, it suffices to show that for each pair $u,u^{\prime }\in V$ such that$\ u<u^{\prime },$ either (1)$\ g(u)<g(u^{\prime })$ or (2) $g(u)=g(u^{\prime })$ and$\ g_{2}(u)<g_{2}(u^{\prime })$ holds. Since g is non-decreasing, either $g(u)<g(u^{\prime })$ or $g(u)=g(u^{\prime })$ holds. In addition, if $g(u)=g(u^{\prime }),$ then there are $\left[ u_{\lambda } \right] ,$ $\left[ u_{\lambda ^{\prime }}\right]$ and $\varphi \in \Phi$ such that $u_{\lambda }<u_{\lambda ^{\prime }},$ $\varphi (u)=u_{\lambda }$, and $\varphi (u^{\prime })=u_{\lambda ^{\prime }}.$ Hence,

$$\begin{aligned} g_{2}(u)= & {} \frac{u_{\lambda }-\min g^{-1}(g(u_{\lambda }))}{\max g^{-1}(g(u_{\lambda }))-\min g^{-1}(g(u_{\lambda }))} \\< & {} \frac{u_{\lambda ^{\prime }}-\min g^{-1}(u_{\lambda ^{\prime }})}{\max g^{-1}(g(u_{\lambda }))-\min g^{-1}(g(u_{\lambda }))} \\= & {} g_{2}(u). \end{aligned}$$

Next, we show that R satisfies INV-$\Phi .$ For each pair $x,y\in X,$ each $U\in \mathcal {U}^{N}$, each$\ i\in N,$ and each $\varphi \in \Phi$, from Step 3, $\sum \nolimits _{i\in N}g(U_{i}(x))\ge \sum \nolimits _{i\in N}g(U_{i}(y))$ if and only if $\sum \nolimits _{j\ne i}g(U_{j}(x))+g(\varphi \circ U_{i}(x))\ge \sum \nolimits _{j\ne i}g(U_{j}(y))+g(\varphi \circ U_{i}(y)).$ In addition, from Step 5, $\sum \nolimits _{i\in N}g_{2}(U_{i}(x))\ge \sum \nolimits _{i\in N}g_{2}(U_{i}(y))$ if and only if $\sum \nolimits _{j\ne i}g_{2}(U_{j}(x))+g_{2}(\varphi \circ U_{i}(x))\ge \sum \nolimits _{j\ne i}g_{2}(U_{j}(y))+g_{2}(\varphi \circ U_{i}(y)).$

Hence, $x\,R_{U}\,y$, if and only if $x\,R_{(\varphi \circ U_{i},U_{-i})}\,y.$ $\square$

1.4 Proof of Theorem 2

Suppose that distinct R and $R^{\prime }$ satisfy SP, AN, IIA, and INV- $\Phi .$ According to the welfarism theorem, the SWOs $\succsim _{R}$ and $\succsim _{R^{\prime }}$, generated by R and $R^{\prime }$, respectively, are distinct orderings. Then, there are $(u_{1},u_{2},\ldots ,u_{n})$ and $(u_{1}^{\prime },u_{2}^{\prime },\ldots ,u_{n}^{\prime })\in V^{N}$ such that $(u_{1},u_{2},\ldots ,u_{n})\succsim _{R}(u_{1}^{\prime },u_{2}^{\prime },\ldots ,u_{n}^{\prime })$ and $(u_{1}^{\prime },u_{2}^{\prime },\ldots ,u_{n}^{\prime })\succ _{R^{\prime }}(u_{1},u_{2},\ldots ,u_{n}).$ Let $u_{0}\in V$. Since $\Phi$ satisfies RC, for each $i\in N,$ there are $\varphi _{i},\psi _{i}\in \Phi$ such that $\varphi _{i}(u_{0})=u_{i}$ and $\psi _{i}(u_{0})=u_{i}^{\prime }$. Then, from INV- $\Phi$ and AN, we can show

$$\begin{aligned} (u_{1},u_{2},\ldots ,u_{n})\sim _{R}(u_{0},\varphi _{1}\circ \varphi _{2}(u_{0}),\ldots ,u_{n}), \end{aligned}$$

in a similar way to the proof of the necessity part of Theorem 1. By taking similar steps, we obtain

$$\begin{aligned} (u_{1},u_{2},\ldots ,u_{n})\sim & {} _{R}(u_{0},u_{0},\ldots ,\varphi _{1}\circ \varphi _{2}\circ \cdots \circ \varphi _{n}(u_{0})), \\ (u_{1}^{\prime },u_{2}^{\prime },\ldots ,u_{n}^{\prime })\sim & {} _{R}(u_{0},u_{0},\ldots ,\psi _{1}\circ \psi _{2}\circ \cdots \circ \psi _{n}(u_{0})), \\ (u_{1},u_{2},\ldots ,u_{n})\sim & {} _{R^{\prime }}(u_{0},u_{0},\ldots ,\varphi _{1}\circ \varphi _{2}\circ \cdots \circ \varphi _{n}(u_{0})),\text { and} \\ (u_{1}^{\prime },u_{2}^{\prime },\ldots ,u_{n}^{\prime })\sim & {} _{R^{\prime }}(u_{0},u_{0},\ldots ,\psi _{1}\circ \psi _{2}\circ \cdots \circ \psi _{n}(u_{0})). \end{aligned}$$

Hence,

$$\begin{aligned} (u_{0},u_{0},\ldots ,\varphi _{1}\circ \varphi _{2}\circ \cdots \circ \varphi _{n}(u_{0}))\succsim _{R}(u_{0},u_{0},\ldots ,\psi _{1}\circ \psi _{2}\circ \cdots \circ \psi _{n}(u_{0})), \end{aligned}$$

which implies from SP

$$\begin{aligned} \varphi _{1}\circ \varphi _{2}\circ \cdots \circ \varphi _{n}(u_{0})\ge \psi _{1}\circ \psi _{2}\circ \cdots \circ \psi _{n}(u_{0}). \end{aligned}$$

Similarly,

$$\begin{aligned} \varphi _{1}\circ \varphi _{2}\circ \cdots \circ \varphi _{n}(u_{0})<\psi _{1}\circ \psi _{2}\circ \cdots \circ \psi _{n}(u_{0}), \end{aligned}$$

which is a contradiction. Hence, there is at most one SWFL. $\square$

1.5 Proof of Theorem 3

First, let us capture the feature of equivalence relations satisfying SYM, REP, and SEP.

Lemma 3

Let $\mathcal {I}$ satisfy SYM and REP and $\Phi ^{N}$ represents $\mathcal {I}$ . $\mathcal {I}$ satisfies SEP if and only if there is $\Phi \subseteq \Psi$ , satisfying RC, such that $\prod \nolimits _{ i\in N} \Phi \subseteq \Phi ^{N}.$

Proof

We introduce the following notation. For each $i\in N$ and each $\varphi \in \Psi$, define $\varphi ^{i}=\left( \varphi _{1}^{i},\ldots ,\varphi _{n}^{i}\right) \in \Psi ^{N}$ by

$$\begin{aligned} \varphi _{j}^{i}=\left\{ \begin{array}{ll} \varphi &{} \quad \text {if }j=i \\ id_{V} &{}\quad \text {if }j\ne i \end{array} \right. \end{aligned}$$

Necessity: It suffices to show that there is a rich $\Phi \subseteq \Psi ,$ such that for each $i\in N$ and each $\varphi \in \Phi ,$ $\varphi ^{i}\in \Phi ^{N}.$ For each $i\in N$ and each pair $u,u^{\prime }\in V$, let $U,U^{\prime }\in \mathcal {U}^{N}$ be such that for each $x\in A,$ $U_{i}(x)=u$ and $U_{i}^{\prime }(x)=u^{\prime }$; and for each $j\ne i,$ $U_{j}=U_{j}^{\prime }.$ According to SEP, $U\,\mathcal {I}\,U^{\prime }$, so that there is $(\varphi _{1},\ldots ,\varphi _{n})\in \Phi ^{N}$ such that $U^{\prime }=(\varphi _{1}\circ U_{1},\ldots ,\varphi _{n}\circ U_{n}).$ Since for each $j\ne i,$ $U_{j}=U_{j}^{\prime }=\varphi _{j}\circ U_{j},$ we have for each $j\ne i,$ $\varphi _{j}=id_{V}.$ Hence, there is $\varphi \in \Psi$ such that $\varphi ^{i}\in \Phi ^{N}$ and

$$\begin{aligned} \varphi (u)=\varphi (U_{i}(x))=U_{i}^{\prime }(x)=u^{\prime }. \end{aligned}$$

$\square$

Sufficiency: Suppose that there is a rich $\Phi \subseteq \Psi$ such that $\prod \nolimits _{i\in N} \Phi \subseteq \Phi ^{N}.$ Let $U,U^{\prime }\in \mathcal {U}^{N}$ be such that there is $S\subseteq N$, such that for each $i\in S$ and each $x\in X$, $U_{i}(x)=U_{i}^{\prime }(x);$ and for each $i\in N\backslash S$ and each pair $x,y\in X$, $U_{i}(x)=U_{i}(y)$ and $U_{i}^{\prime }(x)=U_{i}^{\prime }(y)$. For each $i\in S,$ let $\varphi _{i}=id_{V}\in \Phi .$ Then, $\varphi _{i}\circ U_{i}=U_{i}=U_{i}^{\prime }.\$For each $i\in N\backslash S,$ let $\varphi _{i}\in \Phi$ be such that $\varphi _{i}\left( U_{i}(x)\right) =U_{i}^{\prime }(x).$ Then, since for each pair $x,y\in X$, $U_{i}(x)=U_{i}(y)$ and $U_{i}^{\prime }(x)=U_{i}^{\prime }(y)$, for each $y\in X$,

$$\begin{aligned} \varphi _{i}\left( U_{i}(y)\right)= & {} \varphi _{i}\left( U_{i}(x)\right) \\= & {} U_{i}^{\prime }(x)=U_{i}^{\prime }(y). \end{aligned}$$

Therefore, for $(\varphi _{1},\ldots ,\varphi _{n})\in \underset{i\in N}{ \prod }\Phi \subseteq \Phi ^{N},$ we have $U^{\prime }=(\varphi _{1}\circ U_{1},\ldots ,\varphi _{n}\circ U_{n}),$ so that $U\,\mathcal {I}\,U^{\prime }.$ $\square$

We also need the following lemma shown by Blackorby et al. (2002). This lemma specifies the feature of informational structures that permits using a transformed utilitarian rule.

Lemma 4

Let $g\in \mathcal {G}$. The transformed utilitarian rule associated with g satisfies INV-$\Phi ^{N}$ if and only if $\Phi ^{N}\subseteq \Phi ^{CUC-g}.$

Now, we are ready to prove Theorem 3.

Necessity: Let $\Phi ^{N}$ represent $\mathcal {I}$. From Lemma 3, there is a rich $\Phi \subseteq \Psi$ such that $\prod \nolimits _{i\in N} \Phi \subseteq \Phi ^{N}$. Since INV-$\Phi ^{N}$ implies INV-$\Phi$, we know from Theorems 1 and 2 and Proposition 1 that if an SWFL satisfies SP, AN, IIA, and INV-$\Phi$, then it is a transformed utilitarian rule associated with a certain transformation$\ g\in \mathcal {G}$. Hence, from Lemma 4, we must have$\ \Phi ^{N}\subseteq \Phi ^{CUC-g}.$ $\square$

Sufficiency: If $\Phi ^{N}\subseteq \Phi ^{CUC-g}$, then from Lemma 4, the transformed utilitarian rule associated with g satisfies SP, AN, IIA, and INV-$\Phi ^{N}$. $\square$