Social choice with approximate interpersonal comparison of welfare gains

Abstract

Suppose it is possible to make approximate interpersonal comparisons of welfare gains and losses. Thus, if $w$, $x$, $y$ and $z$ are personal states (each encoding all factors which influence the well-being of an individual), then it is sometimes possible to say “The welfare gain of the state change $w\leadsto x$ is greater than the welfare gain of the state change $y\leadsto z$”. We can represent this by the formula “$(w\leadsto x)\succeq (y\leadsto z)$”, where $(\succeq )$ is a difference preorder: an incomplete preorder on the space of all possible personal state changes. A social state change is a bundle of personal state changes. A social difference preorder (SDP) is an incomplete preorder on the space of social state changes, which satisfies Pareto and Anonymity axioms. We characterize a family of SDPs which are roughly utilitarian in nature. We also apply the model to redistributive taxation.

Appendices

Appendices

Appendix A: Background material

We begin with a lemma which will be used many times in the following proofs. It says that all ‘null changes’ are equally worthless.

Lemma 6.1

Let $(\succeq )$ be a difference preorder on ${\mathcal { X}}$. For all $x,y\in {\mathcal { X}}$, we have .

Proof

We have and because $(\succeq )$ is reflexive. Now, set $x_0=y_1=x_2=x$ and $y_0=x_1=y_2=y$ and apply (Cat*) to conclude that . $\square$

The proofs of Proposition 4.3(b,c) and Theorem 4.5 all depend on the next result.

Lemma 6.2

Let $(\succeq )$ be a preorder on ${\mathcal { X}}\times {\mathcal { X}}$ which satisfies (Inv) and (Cat). Let $x_0,x_1,\ldots ,x_N,y_0,y_1,\ldots ,y_N\in {\mathcal { X}}$.

• (a) If $(x_{n-1}\leadsto x_n) \succeq (y_{n-1}\leadsto y_n)$ for all $n\in {\left[ 1\ldots N \right]}$, then $(x_0\leadsto x_N) \succeq (y_0\leadsto y_N)$.

• (b) Suppose $(x_{n-1}\leadsto x_n) \ \succeq \ (y_{n-1}\leadsto y_n)$ for all $n\in {\left[ 1\ldots N \right]}$, and $(x_{n-1}\leadsto x_n) \ \,{\mathop {}\limits ^{\displaystyle \succ }}\,\ (y_{n-1}\leadsto y_n)$ for some $n\in {\left[ 1\ldots N \right]}$. Then $(x_0\leadsto x_N) \ \,{\mathop {}\limits ^{\displaystyle \succ }}\,\ (y_0\leadsto y_N)$.

Proof

Part (a) follows from inductive application of (Cat).

It suffices to prove part (b) in the case $N=2$; the proof for longer chains can then be derived by applying the case $N=2$ and performing induction on chain length. There are two cases.

First, suppose $(x_{0}\leadsto x_1) \ \succeq \ (y_{0}\leadsto y_1)$ and $(x_1\leadsto x_2) \ \,{\mathop {}\limits ^{\displaystyle \succ }}\,\ (y_1\leadsto y_2)$. We must show that $(x_0\leadsto x_2) \ \,{\mathop {}\limits ^{\displaystyle \succ }}\,\ (y_0\leadsto y_2)$. Axiom (Cat) implies that $(x_0\leadsto x_2) \ \succeq \ (y_0\leadsto y_2)$, so it suffices to show that $(x_0\leadsto x_2) \ {\not \!\!{\mathop {}\limits ^{\displaystyle \preceq }}} \ (y_0\leadsto y_2)$.

By contradiction, suppose $(x_0\leadsto x_2) \ {{\mathop {}\limits ^{\displaystyle \preceq }}} \ (y_0\leadsto y_2)$. Then (Inv) implies that $(x_2\leadsto x_0) \ {\,{\mathop {}\limits ^{\displaystyle \succeq }}\,} \ (y_2\leadsto y_0)$. This, together with hypothesis $(x_{0}\leadsto x_1) \ \succeq \ (y_{0}\leadsto y_1)$ and (Cat), implies that $(x_{2}\leadsto x_1) \ \succeq \ (y_{2}\leadsto y_1)$. Then (Inv) implies that $(x_{1}\leadsto x_2) \ {\mathop {}\limits ^{\displaystyle \preceq }}\ (y_{1}\leadsto y_2)$. But this contradicts the hypothesis that $(x_1\leadsto x_2) \ \,{\mathop {}\limits ^{\displaystyle \succ }}\,\ (y_1\leadsto y_2)$.

A similar proof applies if $(x_{0}\leadsto x_1) \ \,{\mathop {}\limits ^{\displaystyle \succ }}\,\ (y_{0}\leadsto y_1)$ and $(x_1\leadsto x_2) \ \succeq \ (y_1\leadsto y_2)$. $\square$

Appendix B: Proofs from Sect. 4.1

Proof of Proposition 4.1

In light of Theorem 4.9, the proof will handle both the case when ${\mathcal { I}}$ is finite and the case when it ${\mathcal { I}}$ is infinite.

(a) The reflexive property follows from axiom (Par), which, in turn, follows immediately by setting ${\mathcal { J}}_v:=\emptyset$ for all $v\in {\mathcal { V}}$, and applying (QU2) to every element of ${\mathcal { I}}$.

(Anon) Let $\pi \in \Pi$. Let ${\mathcal { J}}\subseteq {\mathcal { I}}$ be a finite subset such that $\pi (i)=i$ for all $i\in {\mathcal { I}}\setminus {\mathcal { J}}$. Let ${\mathbf { x}}\in {\mathcal { X}}^{\mathcal { I}}$, and let ${\mathbf { x}}'=\pi ({\mathbf { x}})$. Then for all $v\in {\mathcal { V}}$, we have

Here, $(*)$ is by the change of variables $j':=\pi (j)$ (because $\pi :{\mathcal { J}}{{\longrightarrow }}{\mathcal { J}}$ bijectively). If we set ${\mathcal { J}}_v:={\mathcal { J}}$, then this verifies (QU1) in both directions. Meanwhile, we obviously have $(x_i\leadsto x'_i)=(x_i\leadsto x_i)$ for all $i\in {\mathcal { I}}\setminus {\mathcal { J}}$. This verifies (QU2) in both directions. Thus , as desired.

$Transitive.$ Suppose $({\mathbf { x}}^0\leadsto {\mathbf { x}}^1)\,{\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { y}}^0\leadsto {\mathbf { y}}^1)$ and $({\mathbf { y}}^0\leadsto {\mathbf { y}}^1)\,{\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^1)$; we must show that $({\mathbf { x}}^0\leadsto {\mathbf { x}}^1)\,{\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^1)$. Let $v\in {\mathcal { V}}$, and suppose $v:{\mathcal { X}}{{\longrightarrow }}{\mathcal { R}}$, where ${\mathcal { R}}$ is some linearly ordered abelian group. By hypothesis, there exist finite subsets ${\mathcal { J}}'_v,{\mathcal { J}}''_v\subseteq {\mathcal { I}}$ such that

$$\begin{aligned} \sum _{j\in {\mathcal { J}}'_v} \left( v(x^1_j) - v(x^0_j) \right)&\ge \sum _{j\in {\mathcal { J}}'_v} \left( v(y^1_j) - v(y^0_j) \right), \end{aligned}$$

(9)

$$\begin{aligned} \sum _{j\in {\mathcal { J}}''_v} \left( v(y^1_j) - v(y^0_j) \right)&\ge \sum _{j\in {\mathcal { J}}''_v} \left( v(z^1_j) - v(z^0_j) \right), \end{aligned}$$

(10)

$$\begin{aligned} (x^0_i\leadsto x^1_i)&\succeq (y^0_i\leadsto y^1_i), \quad \text{ for } \text{ all } i \in {\mathcal { I}}\setminus {\mathcal { J}}'_v\text{, } \end{aligned}$$

(11)

$$\begin{aligned} \hbox { and }\ (y^0_i\leadsto y^1_i)&\succeq (z^0_i\leadsto z^1_i), \quad \text{ for } \text{ all } i \in {\mathcal { I}}\setminus {\mathcal { J}}''_v\text{. } \end{aligned}$$

(12)

Let ${\mathcal { J}}_v:={\mathcal { J}}'_v\cup {\mathcal { J}}''_v$. Then

$$\begin{aligned} \sum _{j\in {\mathcal { J}}_v} \left( v(x^1_j) - v(x^0_j) \right)&\ge \sum _{j\in {\mathcal { J}}_v} \left( v(y^1_j) - v(y^0_j) \right), \end{aligned}$$

(13)

$$\begin{aligned} \sum _{j\in {\mathcal { J}}_v} \left( v(y^1_j) - v(y^0_j) \right)&\ge \sum _{j\in {\mathcal { J}}_v} \left( v(z^1_j) - v(z^0_j) \right), \end{aligned}$$

(14)

$$\begin{aligned} (x^0_i\leadsto x^1_i)&\succeq (y^0_i\leadsto y^1_i), \quad \text{ for } \text{ all } i \in {\mathcal { I}}\setminus {\mathcal { J}}_v, \end{aligned}$$

(15)

$$\begin{aligned} \hbox { and }\ (y^0_i\leadsto y^1_i)&\succeq (z^0_i\leadsto z^1_i), \quad \text{ for } \text{ all } i \in {\mathcal { I}}\setminus {\mathcal { J}}_v. \end{aligned}$$

(16)

Here, (13) is obtained by combining (9), (11) and (2). Likewise, (14) is obtained by combining (10), (12) and (2). Next, (15) follows from (11), because $({\mathcal { I}}\setminus {\mathcal { J}}_v)\subseteq ({\mathcal { I}}\setminus {\mathcal { J}}'_v)$ (because ${\mathcal { J}}_v\supseteq {\mathcal { J}}'_v$). Likewise, (16) follows from (12), because $({\mathcal { I}}\setminus {\mathcal { J}}_v)\subseteq ({\mathcal { I}}\setminus {\mathcal { J}}''_v)$ (because ${\mathcal { J}}_v\supseteq {\mathcal { J}}''_v$). We conclude that

$$\begin{aligned} \sum _{j\in {\mathcal { J}}_v} \left( v(x^1_j) - v(x^0_j) \right)&\ge \sum _{j\in {\mathcal { J}}_v} \left( v(z^1_j) - v(z^0_j) \right) \end{aligned}$$

(17)

$$\begin{aligned} \hbox { and }\ (x^0_i\leadsto x^1_i)&\succeq (z^0_i\leadsto z^1_i), \quad \text{ for } \text{ all } i \in {\mathcal { I}}\setminus {\mathcal { J}}_v\text{. } \end{aligned}$$

(18)

Here, (17) is obtained be combining (13) and (14) using the transitivity of the ordering on ${\mathcal { R}}$. Meanwhile (18) is obtained from (15) and (16), by the transitivity of $(\succeq )$.

Now, (17) verifies (QU1), while (18) verifies (QU2). We can do this for any $v\in {\mathcal { V}}$; thus, $({\mathbf { x}}^0\leadsto {\mathbf { x}}^1) \,{\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { z}}^0\leadsto {\mathbf { z}}^1)$, as desired.

($\hbox {Cat}^{\trianglerighteq }$) The argument is closely analogous to the proof of Transitivity. Suppose $({\mathbf { x}}^0\leadsto {\mathbf { x}}^1)\,{\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { y}}^0\leadsto {\mathbf { y}}^1)$ and $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2)\,{\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$; we must show that $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2)\,{\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { y}}^0\leadsto {\mathbf { y}}^2)$. Let $v\in {\mathcal { V}}$, with $v:{\mathcal { X}}{{\longrightarrow }}{\mathcal { R}}$. By hypothesis, there exist finite subsets ${\mathcal { J}}'_v,{\mathcal { J}}''_v\subseteq {\mathcal { I}}$ such that

$$\begin{aligned} \sum _{j\in {\mathcal { J}}'_v} \left( v(x^1_j) - v(x^0_j) \right)&\ge \sum _{j\in {\mathcal { J}}'_v} \left( v(y^1_j) - v(y^0_j) \right), \end{aligned}$$

(19)

$$\begin{aligned} \sum _{j\in {\mathcal { J}}''_v} \left( v(x^2_j) - v(x^1_j) \right)&\ge \sum _{j\in {\mathcal { J}}''_v} \left( v(y^2_j) - v(y^1_j) \right), \end{aligned}$$

(20)

$$\begin{aligned} (x^0_i\leadsto x^1_i)&\succeq (y^0_i\leadsto y^1_i), \quad \text{ for } \text{ all } i \in {\mathcal { I}}\setminus {\mathcal { J}}'_v\text{, } \end{aligned}$$

(21)

$$\begin{aligned} \hbox { and }\ (x^1_i\leadsto x^2_i)&\succeq (y^1_i\leadsto y^2_i), \quad \text{ for } \text{ all } i \in {\mathcal { I}}\setminus {\mathcal { J}}''_v\text{. } \end{aligned}$$

(22)

Let ${\mathcal { J}}_v:={\mathcal { J}}'_v\cup {\mathcal { J}}''_v$. Then

$$\begin{aligned} \sum _{j\in {\mathcal { J}}_v} \left( v(x^1_j) - v(x^0_j) \right)&\ge \sum _{j\in {\mathcal { J}}_v} \left( v(y^1_j) - v(y^0_j) \right), \end{aligned}$$

(23)

$$\begin{aligned} \sum _{j\in {\mathcal { J}}_v} \left( v(x^2_j) - v(x^1_j) \right)&\ge \sum _{j\in {\mathcal { J}}_v} \left( v(y^2_j) - v(y^1_j) \right), \end{aligned}$$

(24)

$$\begin{aligned} (x^0_i\leadsto x^1_i)&\succeq (y^0_i\leadsto y^1_i), \quad \text{ for } \text{ all } i \in {\mathcal { I}}\setminus {\mathcal { J}}_v\text{, } \end{aligned}$$

(25)

$$\begin{aligned} \hbox { and }\ (x^1_i\leadsto x^2_i)&\succeq (y^1_i\leadsto y^2_i), \quad \text{ for } \text{ all } i \in {\mathcal { I}}\setminus {\mathcal { J}}_v\text{. } \end{aligned}$$

(26)

Here, (23) is obtained by combining (19), (21) and (2). Likewise, (24) is obtained by combining (20), (22) and (2). Next, (25) follows from (21), because $({\mathcal { I}}\setminus {\mathcal { J}}_v)\subseteq ({\mathcal { I}}\setminus {\mathcal { J}}'_v)$ (because ${\mathcal { J}}_v\supseteq {\mathcal { J}}'_v$). Likewise, (26) follows from (22), because $({\mathcal { I}}\setminus {\mathcal { J}}_v)\subseteq ({\mathcal { I}}\setminus {\mathcal { J}}''_v)$ (because ${\mathcal { J}}_v\supseteq {\mathcal { J}}''_v$). We conclude that

(27)

(28)

Here, $(*)$ comes from combining (23) and (24) and using the compatibility between the ordering and the addition operator on ${\mathcal { R}}$. Meanwhile, (28) comes from combining (25) and (26), and using axiom (Cat) for $(\succeq )$.

Now, (27) verifies (QU1), while (28) verifies (QU2). We can do this for any $v\in {\mathcal { V}}$; thus, $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2) \,{\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { y}}^0\leadsto {\mathbf { y}}^2)$, as desired.

($\hbox {Cat}_*^{\trianglerighteq }$) The proof is very similar to ($\hbox {Cat}^{\trianglerighteq }$).

($\hbox {Inv}^{\trianglerighteq }$)Let $v\in {\mathcal { V}}$. If $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2)\,{\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$, then there exists some finite ${\mathcal { J}}_v\subseteq {\mathcal { I}}$ such that $\displaystyle \sum _{j\in {\mathcal { J}}_v} \left( v(x^2_j) - v(x^1_j) \right) \ \ge \ \sum _{j\in {\mathcal { J}}_v} \left( v(y^2_j) - v(y^1_j) \right)$, while $(x^1_i\leadsto x^2_i) \succeq (y^1_i\leadsto y^2_i)$ for all $i \in {\mathcal { I}}\setminus {\mathcal { J}}_v$. But then $\displaystyle \sum _{j\in {\mathcal { J}}_v} \left( v(x^1_j) - v(x^2_j) \right) \ \le \ \sum _{j\in {\mathcal { J}}_v} \left( v(y^1_j) - v(y^2_j) \right)$. Also, since $(\succeq )$ itself satisfies (Inv), we get $(x^2_i\leadsto x^1_i) \preceq (y^2_i\leadsto y^1_i)$, for all $i\in {\mathcal { I}}\setminus {\mathcal { J}}_v$. Thus, (QU1) and (QU2) are satisfied. This holds for all $v\in {\mathcal { V}}$; we conclude that $({\mathbf { x}}^2\leadsto {\mathbf { x}}^1)\,{\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglelefteq }}\, ({\mathbf { y}}^2\leadsto {\mathbf { y}}^1)$.

(b) Suppose $(x^1_i\leadsto x^2_i)\succeq (y^1_i\leadsto y^2_i)$ for all $i\in {\mathcal { I}}$, and $(x^1_j\leadsto x^2_j)\,{\mathop {}\limits ^{\displaystyle \succ }}\,(y^1_j\leadsto y^2_j)$ for some $j\in {\mathcal { I}}$. Axiom (Par) implies that $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$. To show that $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {\scriptscriptstyle {\mathrm {u}}}\limits ^{\displaystyle \triangleright }}\,({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$, we must show that $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \ {\not \!\!\,{\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglelefteq }}\,} \ ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$.

First suppose ${\mathcal { V}}$ contains a strong utility function $v_0$. Thus $v_0(x^2_i) - v_0(x^1_i) \ge v_0(y^2_i) - v_0(y^1_i)$ for all $i\in {\mathcal { I}}$, and $v_0(x^2_j) - v_0(x^1_j) > v_0(y^2_j) - v_0(y^1_j)$. We will show that there is no finite subset ${\mathcal { J}}\subseteq {\mathcal { I}}$ which satisfies (QU1) and (QU2) for $v_0$ in the way required to show that $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglelefteq }}\, ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$.

If $j\in {\mathcal { J}}$, then $\displaystyle \sum _{j\in {\mathcal { J}}} \left( v_0(x^2_j) - v_0(x^1_j) \right) > \sum _{j\in {\mathcal { J}}} \left( v_0(y^2_j) - v_0(y^1_j) \right)$, so (QU1) is not satisfied. If $j\not \in {\mathcal { J}}$, then $j\in {\mathcal { I}}\setminus {\mathcal { J}}$, and $(x^1_j\leadsto x^2_j)\,{\mathop {}\limits ^{\displaystyle \succ }}\,(y^1_j\leadsto y^2_j)$, so (QU2) is not satisfied.

Thus, there exists at least one $v\in {\mathcal { V}}$ (namely $v_0$) such that (QU1) and (QU2) cannot both be satisfied. Thus, $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \ \not \!\!\,{\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglelefteq }}\, \ ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$, as desired.

The proof in the case when ${\mathcal { V}}$ provides a multiutility representation is similar.

(d) follows immediately from (c), which in turn follows from the definitions of $(\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq })$ and $(\,{\mathop {{\mathcal { W}}}\limits ^{\displaystyle \trianglerighteq }}\,)$. Finally, (e) follows from (b). $\square$

The proof of Theorem 4.2 uses the following result, which is of independent interest.

Proposition 6.1

Let ${\mathcal { I}}$ be finite, and let $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ be a $(\succeq )$-SDP on ${\mathcal { X}}^{\mathcal { I}}$. If $W:{\mathcal { X}}^{\mathcal { I}}{{\longrightarrow }}{\mathcal { R}}$ is a SWF for $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$, then there exists some $u\in {\mathcal { U}}(\,{\mathop {}\limits ^{\displaystyle \succeq }}\,)$ and some constant $C\in {\mathcal { R}}$ such that $W({\mathbf { x}})=\displaystyle C+ \sum _{i\in {\mathcal { I}}} u(x_i)$ for all ${\mathbf { x}}\in {\mathcal { X}}^{\mathcal { I}}$.

Proof

Let ${\mathcal { J}},{\mathcal { K}}\subset {\mathcal { I}}$ be disjoint subsets, with ${\mathcal { I}}={\mathcal { J}}\sqcup {\mathcal { K}}$. For ${\mathbf { x}},{\mathbf { y}}\in {\mathcal { X}}^{\mathcal { I}}$, we define $\left( \begin{array}{l}{{\mathbf { x}}_{\mathcal { J}}}\\ {{\mathbf { y}}_{\mathcal { K}}}\end{array}\right) \in {\mathcal { X}}^{\mathcal { I}}$ by setting $\left( \begin{array}{l}{{\mathbf { x}}_{\mathcal { J}}}\\ {{\mathbf { y}}_{\mathcal { K}}}\end{array}\right) _j:=x_j$ for all $j\in {\mathcal { J}}$, while $\left( \begin{array}{l}{{\mathbf { x}}_{\mathcal { J}}}\\ {{\mathbf { y}}_{\mathcal { K}}}\end{array}\right) _k:=y_k$ for all $k\in {\mathcal { K}}$. Fix some $o\in {\mathcal { X}}$, and define ${\mathbf { o}}\in {\mathcal { X}}^{\mathcal { I}}$ by ${\mathbf { o}}_i:=o$ for all $i\in {\mathcal { I}}$. Now fix $k\in {\mathcal { I}}$, and define $u:{\mathcal { X}}{{\longrightarrow }}{\mathcal { R}}$ by setting $u(x):=W\left( \begin{array}{l}{x_{k}}\\ {{\mathbf { o}}_{{-k}}}\end{array}\right) -W({\mathbf { o}})$ for all $x\in {\mathcal { X}}$. (It follows that $u(o)=0$.)

Claim 1

For any $x,y\in {\mathcal { X}}$, ${\mathbf { z}}\in {\mathcal { X}}^{\mathcal { I}}$, and $j\in {\mathcal { I}}$, we have $W\left( \begin{array}{l}{y_j}\\ {{\mathbf { z}}_{{-j}}}\end{array}\right) -W\left( \begin{array}{l}{x_j}\\ {{\mathbf { z}}_{{-j}}}\end{array}\right) = u(y)-u(x)$.

Proof

We have

(29)

Here, $(\dagger )$ is by (Anon), and $(*)$ is by (Par), along with the fact that Lemma 6.1 says for all $i\in {\mathcal { I}}\setminus \{j\}$. Combining (5) with (29), we obtain

$$\begin{aligned} W\left(\begin{array}{l}{y_j}\\ {{\mathbf { z}}_{{-j}}}\end{array}\right)- W\left(\begin{array}{l}{x_j}\\ {{\mathbf { z}}_{{-j}}}\end{array}\right)&= W\left(\begin{array}{l}{y_{k}}\\ {{\mathbf { o}}_{{-k}}}\end{array}\right)-W\left(\begin{array}{l}{x_{k}}\\ {{\mathbf { o}}_{{-k}}}\end{array}\right)\\&= W\left(\begin{array}{l}{y_{k}}\\ {{\mathbf { o}}_{{-k}}}\end{array}\right)-W({\mathbf { o}})+W({\mathbf { o}})-W\left(\begin{array}{l}{x_{k}}\\ {{\mathbf { o}}_{{-k}}}\end{array}\right)\\&= u(y)-u(x), \end{aligned}$$

as desired claim 1 $\diamondsuit$

Claim 2

$u\in {\mathcal { U}}(\succeq )$.

Proof

Let $x,x',y,y'\in {\mathcal { X}}$. Then

as desired. Here, $(*)$ is by (Par), $(\dagger )$ is by statement (5) and $(\diamond )$ is by Claim 1.claim 2 $\diamondsuit$

Without loss of generality, suppose ${\mathcal { I}}={\left[ 1\ldots I \right]}$. Let $C:=W({\mathbf { o}})$. Then for all ${\mathbf { x}}\in {\mathcal { X}}^{\mathcal { I}}$,

Here, $(*)$ is by Claim 1. Thus, $\displaystyle W({\mathbf { x}})=C+\sum _{i\in {\mathcal { I}}} u(x_i)$, as claimed.$\square$

Proof of Theorem 4.2

Let $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ be a $(\succeq )$-SDP on ${\mathcal { X}}^{\mathcal { I}}$.

“${\Longleftarrow }$” Clearly, if $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ is quasiutilitarian, then it admits a multiwelfare representation.

“$\Longrightarrow$” Let ${\mathcal { W}}$ be a collection of SWFs yielding a multiwelfare representation for $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$. For all $W\in {\mathcal { W}}$, Proposition 6.1 yields some $v_{_W}\in {\mathcal { U}}(\succeq )$ such that, for all ${\mathbf { x}}^1,{\mathbf { x}}^2\in {\mathcal { X}}^{\mathcal { I}}$, we have $W({\mathbf { x}}^2)-W({\mathbf { x}}^1)=\sum _{i\in {\mathcal { I}}} v_{_W}(x^2_i) - \sum _{i\in {\mathcal { I}}} v_{_W}(x^1_i)$. Let ${\mathcal { V}}:=\{v_{_W}$; $W\in {\mathcal { W}}\}$. Then, for all ${\mathbf { x}}^1,{\mathbf { x}}^2,{\mathbf { y}}^1,{\mathbf { y}}^2\in {\mathcal { X}}^{\mathcal { I}}$, formula (6) implies that $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2)\,{\mathop {}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$ if and only if (QU1) holds (with ${\mathcal { J}}_v:={\mathcal { I}}$) for all $v\in {\mathcal { V}}$. Thus, $(\mathop {}\limits ^{\displaystyle \trianglerighteq })=(\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq })$. $\square$

Appendix C: Proofs from Sect. 4.2

The proofs of Proposition 4.3(b,c) use the next result.

Lemma 6.3

Let $(\mathop {1}\limits ^{\displaystyle \trianglerighteq })$ and $(\mathop {2}\limits ^{\displaystyle \trianglerighteq })$ be two preorders on ${\mathcal { X}}^{\mathcal { I}}\times {\mathcal { X}}^{\mathcal { I}}$. If $(\mathop {2}\limits ^{\displaystyle \trianglerighteq })$ satisfies (SPar), and $(\mathop {2}\limits ^{\displaystyle \trianglerighteq })$ extends $(\mathop {1}\limits ^{\displaystyle \trianglerighteq })$, then $(\mathop {1}\limits ^{\displaystyle \trianglerighteq })$ satisfies (SPar) also.

Proof

Suppose $(x^1_i\leadsto x^2_i) {\mathop {}\limits ^{\displaystyle \preceq }}(y^1_i\leadsto y^2_i)$ for all $i \in {\mathcal { I}}$, and $(x^1_i\leadsto x^2_i) \,{\mathop {}\limits ^{\displaystyle \prec }}\,(y^1_i\leadsto y^2_i)$ for some $i \in {\mathcal { I}}$; we must show that $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \ \,{\mathop {1}\limits ^{\displaystyle \triangleleft }}\, \ ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$. Since $(\mathop {1}\limits ^{\displaystyle \trianglerighteq })$ satisfies (Par), we have $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \ \,{\mathop {1}\limits ^{\displaystyle \trianglelefteq }}\, \ ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$. We must show $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \ {\not \!\!\,{\mathop {1}\limits ^{\displaystyle \trianglerighteq }}\,} \ ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$.

By contradiction, suppose $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \ \,{\mathop {1}\limits ^{\displaystyle \trianglerighteq }}\, \ ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$. Then $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \ \,{\mathop {2}\limits ^{\displaystyle \trianglerighteq }}\, \ ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$, because $(\mathop {2}\limits ^{\displaystyle \trianglerighteq })$ extends $(\mathop {1}\limits ^{\displaystyle \trianglerighteq })$. But $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \ \,{\mathop {2}\limits ^{\displaystyle \triangleleft }}\, \ ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$ because $(\mathop {2}\limits ^{\displaystyle \trianglerighteq })$ satisfies (SPar). Contradiction. $\square$

The proof of Proposition 4.3(a) follows immediately from defining formula (7). The proofs of Proposition 4.3(b,c) require Proposition 4.8, so we prove that first.

Proof of Proposition 4.8

In light of Theorem 4.9, the proof will handle both the case when ${\mathcal { I}}$ is finite and the case when it ${\mathcal { I}}$ is infinite. Let $(\mathop {*}\limits ^{\displaystyle \trianglerighteq })$ be the SDP defined inductively by rules $(*1)$-$(*5)$. Let $(\,{\mathop {\circ }\limits ^{\displaystyle \trianglerighteq }}\,)$ be the SDP defined by formula (7). We must show that $(\mathop {*}\limits ^{\displaystyle \trianglerighteq })=(\,{\mathop {\circ }\limits ^{\displaystyle \trianglerighteq }}\,)$.

“$\supseteq$” By Proposition 4.3(a), it suffices to show that $(\mathop {*}\limits ^{\displaystyle \trianglerighteq })$ is an $(\succeq )$-SDP. It is easy to check that is reflexive and satisfies axioms (Anon) and ($\hbox {Inv}^{\trianglerighteq }$). Likewise, $(\,{\mathop {\scriptscriptstyle {\mathrm {par}}}\limits ^{\displaystyle \trianglerighteq }}\,)$ is reflexive and satisfies axioms (Par) and ($\hbox {Inv}^{\trianglerighteq }$). Thus, the binary relation is reflexive and satisfies axioms (Anon), (Par) and ($\hbox {Inv}^{\trianglerighteq }$). Thus, $(\mathop {*}\limits ^{\displaystyle \trianglerighteq })$ will also satisfy these properties, and will also be transitive and satisfy ($\hbox {Cat}^{\trianglerighteq }$) and ($\hbox {Cat}_*^{\trianglerighteq }$) by definition. Thus, $(\mathop {*}\limits ^{\displaystyle \trianglerighteq })$ is an $(\succeq )$-SDP.

“$\subseteq$” We will show that $(\mathop {*}\limits ^{\displaystyle \trianglerighteq })$ is a subrelation of every other $(\succeq )$-SDP; in particular, this implies that $(\mathop {*}\limits ^{\displaystyle \trianglerighteq })\subseteq (\,{\mathop {\circ }\limits ^{\displaystyle \trianglerighteq }}\,)$.

Let $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ be another $(\succeq )$-SDP, and let ${\mathbf { x}}^0,{\mathbf { x}}^2,{\mathbf { z}}^0,{\mathbf { y}}^2\in {\mathcal { X}}^{\mathcal { I}}$. Suppose $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2) \,{\mathop {*}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$; we must show that $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2) \,{\mathop {}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$. The proof is by induction, using the recursive definition of $(\mathop {*}\limits ^{\displaystyle \trianglerighteq })$.

$(*1)$ :

If , then because $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ satisfies (Anon).

$(*2)$ :

If $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2)\,{\mathop {\scriptscriptstyle {\mathrm {par}}}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$, then $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2)\,{\mathop {}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$, because $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ satisfies (Par).

$(*3)$ :

Suppose there exist ${\mathbf { y}}^0,{\mathbf { y}}^2$ with $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2) \,{\mathop {*}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { y}}^0\leadsto {\mathbf { y}}^2)$ and $({\mathbf { y}}^0\leadsto {\mathbf { y}}^2) \,{\mathop {*}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$. By induction, suppose we have already shown that $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2) \,{\mathop {}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { y}}^0\leadsto {\mathbf { y}}^2)$ and $({\mathbf { y}}^0\leadsto {\mathbf { y}}^2) \,{\mathop {}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$. Then $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2)\,{\mathop {}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$, because $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ is transitive.

$(*4)$ :

Suppose there exist ${\mathbf { x}}^1,{\mathbf { z}}^1$ with $({\mathbf { x}}^0\leadsto {\mathbf { x}}^1) \,{\mathop {*}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { z}}^0\leadsto {\mathbf { z}}^1)$ and $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {*}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { z}}^1\leadsto {\mathbf { z}}^2)$. By induction, suppose we have already shown $({\mathbf { x}}^0\leadsto {\mathbf { x}}^1) \,{\mathop {}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^1)$ and $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { z}}^1\leadsto {\mathbf { z}}^2)$. Then $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2)\,{\mathop {}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$, because $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ satisfies ($\hbox {Cat}^{\trianglerighteq }$).

$(*5)$ :

Suppose there exist ${\mathbf { x}}^1,{\mathbf { z}}^1$ with $({\mathbf { x}}^0\leadsto {\mathbf { x}}^1) \,{\mathop {*}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { z}}^1\leadsto {\mathbf { z}}^2)$ and $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {*}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { z}}^0\leadsto {\mathbf { z}}^1)$. By induction, suppose we have already shown $({\mathbf { x}}^0\leadsto {\mathbf { x}}^1) \,{\mathop {}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { z}}^1\leadsto {\mathbf { z}}^2)$ and $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^1)$. Then $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2)\,{\mathop {}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$, because $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ satisfies ($\hbox {Cat}_*^{\trianglerighteq }$).

By induction, we have $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2) \,{\mathop {}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$. $\square$

The next lemma is like Lemma 6.2(a), but using (Cat*) instead of (Cat).

Lemma 6.4

Let $(\succeq )$ be a difference preorder on a set ${\mathcal { X}}$. Let $x_0,x_1,x_2,y_0,y_1,y_2\in {\mathcal { X}}$, and suppose that $(x_0\leadsto x_1)\ \succeq \ (y_1\leadsto y_2)$ and $(x_1\leadsto x_2) \ \succeq \ (y_0\leadsto y_1)$, and at least one of these two preferences is strict. Then $(x_0\leadsto x_2)\ \succ \ (y_0 \leadsto y_2)$.

Proof

Suppose (i) $(x_0\leadsto x_1)\ \succeq \ (y_1\leadsto y_2)$ and (ii) $(x_1\leadsto x_2)\ \succ \ (y_0\leadsto y_1)$ (the other case is similar).

By (Cat*), we know that $(x_0\leadsto x_2)\ \succeq \ (y_0 \leadsto y_2)$. We need to only show that $(x_0\leadsto x_2)\ \not \preceq \ (y_0 \leadsto y_2)$. So, by contradiction, suppose that $(x_0\leadsto x_2)\ \preceq \ (y_0 \leadsto y_2)$. Then, by (Inv), we have $(x_2\leadsto x_0)\ \succeq \ (y_2 \leadsto y_0)$. Then, from (Cat*) and (i), we get $(x_2\leadsto x_1)\ \succeq \ (y_1 \leadsto y_0)$. Then (Inv) yields $(x_1\leadsto x_2)\ \preceq \ (y_0 \leadsto y_1)$. But this contradicts (ii). By contradiction, we cannot have $(x_0\leadsto x_2)\ \preceq \ (y_0 \leadsto y_2)$. $\square$

Proof of Proposition 4.3(b)

In light of Theorem 4.9, the proof will handle both the case when ${\mathcal { I}}$ is finite and the case when it ${\mathcal { I}}$ is infinite.

“${\Longleftarrow }$” If $(\mathop {*}\limits ^{\displaystyle \trianglerighteq })$ satisfies (SPar) and $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ properly extends $(\mathop {*}\limits ^{\displaystyle \trianglerighteq })$, then clearly $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ also satisfies (SPar).

“$\Longrightarrow$” $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ extends $(\mathop {*}\limits ^{\displaystyle \trianglerighteq })$, by Proposition 4.3(a). Thus, if $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ satisfies (SPar), then Lemma 6.3 says that $(\mathop {*}\limits ^{\displaystyle \trianglerighteq })$ satisfies (SPar).

It remains to show that $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ properly extends $(\mathop {*}\limits ^{\displaystyle \trianglerighteq })$. Suppose $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2) \,{\mathop {*}\limits ^{\displaystyle \triangleright }}\, ({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$. Then $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2) \,{\mathop {*}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$ through some sequence of applications of Steps $(*1)$-$(*5)$, but the same sequence does not yield $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2) \,{\mathop {*}\limits ^{\displaystyle \trianglelefteq }}\, ({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$. This means that in some application of Step $(*2)$, we must have a strict “$\,{\mathop {\scriptscriptstyle {\mathrm {par}}}\limits ^{\displaystyle \triangleright }}\,$” rather than “$\,{\mathop {\scriptscriptstyle {\mathrm {par}}}\limits ^{\displaystyle \trianglerighteq }}\,$”. Thus, we augment the proof of Proposition 4.8 “$\subseteq$” with the following additional observations:

$(*2)$ :

$({\mathbf { x}}^0\leadsto {\mathbf { x}}^2)\,{\mathop {\scriptscriptstyle {\mathrm {par}}}\limits ^{\displaystyle \triangleright }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$, then $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2)\,{\mathop {}\limits ^{\displaystyle \triangleright }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$, because $(\succeq )$ satisfies (SPar).

$(*3)$ :

Suppose there exist ${\mathbf { y}}^0,{\mathbf { y}}^2$ with $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2) \,{\mathop {*}\limits ^{\displaystyle \triangleright }}\, ({\mathbf { y}}^0\leadsto {\mathbf { y}}^2)$ and $({\mathbf { y}}^0\leadsto {\mathbf { y}}^2) \,{\mathop {*}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$. By induction, suppose we have already shown that $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2) \,{\mathop {}\limits ^{\displaystyle \triangleright }}\,({\mathbf { y}}^0\leadsto {\mathbf { y}}^2)$ and $({\mathbf { y}}^0\leadsto {\mathbf { y}}^2) \,{\mathop {}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$. Then $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2)\,{\mathop {}\limits ^{\displaystyle \triangleright }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$, because $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ is transitive. Likewise, if $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2) \,{\mathop {*}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { y}}^0\leadsto {\mathbf { y}}^2)$ and $({\mathbf { y}}^0\leadsto {\mathbf { y}}^2) \,{\mathop {*}\limits ^{\displaystyle \triangleright }}\, ({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$, then $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2)\,{\mathop {}\limits ^{\displaystyle \triangleright }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$.

$(*4)$ :

Suppose there exist ${\mathbf { x}}^1,{\mathbf { z}}^1$ with $({\mathbf { x}}^0\leadsto {\mathbf { x}}^1) \,{\mathop {*}\limits ^{\displaystyle \triangleright }}\, ({\mathbf { z}}^0\leadsto {\mathbf { z}}^1)$ and $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {*}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { z}}^1\leadsto {\mathbf { z}}^2)$. By induction, suppose we have already shown that $({\mathbf { x}}^0\leadsto {\mathbf { x}}^1) \,{\mathop {}\limits ^{\displaystyle \triangleright }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^1)$ and $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { z}}^1\leadsto {\mathbf { z}}^2)$. Then $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2)\,{\mathop {}\limits ^{\displaystyle \triangleright }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$ by Lemma 6.2(b). Likewise, if $({\mathbf { x}}^0\leadsto {\mathbf { x}}^1) \,{\mathop {*}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { z}}^0\leadsto {\mathbf { z}}^1)$ and $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {*}\limits ^{\displaystyle \triangleright }}\, ({\mathbf { z}}^1\leadsto {\mathbf { z}}^2)$, then $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2)\,{\mathop {}\limits ^{\displaystyle \triangleright }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$.

$(*5)$ :

Suppose there exist ${\mathbf { x}}^1,{\mathbf { z}}^1$ with $({\mathbf { x}}^0\leadsto {\mathbf { x}}^1) \,{\mathop {*}\limits ^{\displaystyle \triangleright }}\, ({\mathbf { z}}^1\leadsto {\mathbf { z}}^2)$ and $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {*}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { z}}^0\leadsto {\mathbf { z}}^1)$. By induction, suppose we have already shown that $({\mathbf { x}}^0\leadsto {\mathbf { x}}^1) \,{\mathop {}\limits ^{\displaystyle \triangleright }}\,({\mathbf { z}}^1\leadsto {\mathbf { z}}^2)$ and $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^1)$. Then $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2)\,{\mathop {}\limits ^{\displaystyle \triangleright }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$ by Lemma 6.4. Likewise, if $({\mathbf { x}}^0\leadsto {\mathbf { x}}^1) \,{\mathop {*}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { z}}^0\leadsto {\mathbf { z}}^1)$ and $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {*}\limits ^{\displaystyle \triangleright }}\, ({\mathbf { z}}^1\leadsto {\mathbf { z}}^2)$, then $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2)\,{\mathop {}\limits ^{\displaystyle \triangleright }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$.

By induction, we conclude that $({\mathbf { x}}^0\leadsto {\mathbf { x}}^2) \,{\mathop {}\limits ^{\displaystyle \triangleright }}\,({\mathbf { z}}^0\leadsto {\mathbf { z}}^2)$.

• (c) If $(\succeq )$ has a strong utility function, then Proposition 4.1(e) says that $(\mathop {\scriptscriptstyle {\mathrm {u}}}\limits ^{\displaystyle \trianglerighteq })$ satisfies (SPar). Now part (b) implies that $(\mathop {*}\limits ^{\displaystyle \trianglerighteq })$ also satisfies (SPar). $\square$

Theorem 4.4 follows from Theorem 4.5, so we will prove that first. The proofs of Theorems 4.5 and 4.10, as well as Proposition 5.1, depend upon Proposition 6.4, which will be proved in Appendix F; the reader may wish to consult that appendix before continuing.

Proof of Theorems 4.5 and 4.10

The proof will handle both the case when ${\mathcal { I}}$ is finite, and the case when ${\mathcal { I}}$ is infinite but ${\mathcal { V}}$ is finite. In either case, we will show that $(\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq })\subseteq (\mathop {\mathrm {\scriptscriptstyle {ng}}}\limits ^{\displaystyle \trianglerighteq })$. Then Proposition 6.4 implies that every SDP is an extension of $(\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq })$, which means that $(\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq })=(\mathop {*}\limits ^{\displaystyle \trianglerighteq })$. In particular, this will mean that $(\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq })\subseteq (\mathop {\scriptscriptstyle {\mathrm {u}}}\limits ^{\displaystyle \trianglerighteq })$. But Proposition 4.1(d) says that $(\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq })\supseteq (\mathop {\scriptscriptstyle {\mathrm {u}}}\limits ^{\displaystyle \trianglerighteq })$, so this will imply that $(\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq })=(\mathop {\scriptscriptstyle {\mathrm {u}}}\limits ^{\displaystyle \trianglerighteq })$.

Now, let ${\mathbf { x}}^1,{\mathbf { x}}^2,{\mathbf { y}}^1,{\mathbf { y}}^2\in {\mathcal { X}}^{\mathcal { I}}$, and suppose $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2)\,{\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$. We must show that $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2)\,{\mathop {\mathrm {\scriptscriptstyle {ng}}}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$. For any $v\in {\mathcal { V}}$, there exists some ${\mathcal { J}}_v\subseteq {\mathcal { I}}$ satisfying (QU1) and (QU2) for $v$. Let $\displaystyle {\mathcal { J}}:=\bigcup _{v\in {\mathcal { V}}} {\mathcal { J}}_v$; then ${\mathcal { J}}\subseteq {\mathcal { I}}$ satisfies (QU1) and (QU2) for all $v\in {\mathcal { V}}$. Note that ${\mathcal { J}}$ is finite, because either ${\mathcal { V}}$ is finite, or ${\mathcal { I}}$ itself is finite, by hypothesis.

Suppose ${\mathcal { J}}:=\{j_1,j_2,\ldots ,j_N\}$. Fix $w_0\in {\mathcal { X}}$. Applying solvability repeatedly, find $w_1,w_2,\ldots ,w_N\in {\mathcal { X}}$ such that

• (JK1$'$) for all $n\in {\left[ 1\ldots N \right]}$.

Fix $z_0\in {\mathcal { X}}$. Applying solvability repeatedly, find $z_1,z_2,\ldots ,z_N\in {\mathcal { X}}$ such that

• (JK2$'$) , for all $n\in {\left[ 1\ldots N \right]}$.

Claim 1

$(w_0\leadsto w_N) \ \succeq \ (z_0\leadsto z_N)$

Proof

For any $v\in {\mathcal { V}}$, we have

Here, $(\diamond )$ is by (JK1$'$) and formula (2). Next, $(*)$ is by (QU1). Finally, $(\dagger )$ is by (JK2$'$) and formula (2).

Thus, $v(w_N)-v(w_0)\ge v(z_N)-v(z_0)$ for all $v\in {\mathcal { V}}$. Thus, the multiutility representation (3) yields $(w_0\leadsto w_N) \ \succeq \ (z_0\leadsto z_N)$.$\diamondsuit$

Now, let ${\mathcal { L}}:=\{1\}$ and set ${\mathcal { I}}_0:={\mathcal { J}}_1:={\mathcal { K}}_1:={\mathcal { J}}$. Then observations (JK1$'$) and (JK2$'$) and Claim 1 together imply that $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2)\,{\mathop {{\mathcal { J}}_1,{\mathcal { K}}_1}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$; this yields (NG2). Meanwhile, (QU2) implies that $(x^1_i\leadsto x^2_i) \succeq (y^1_i\leadsto y^2_i)$ for all $i \in {\mathcal { I}}\setminus {\mathcal { I}}_0$; this yields (NG1). Thus, $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2)\,{\mathop {\mathrm {\scriptscriptstyle {ng}}}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$.   $\square$

Proof of Theorem 4.4

If $(\succeq )$ is solvable and divisible, then it has a multiutility representation (Pivato 2013a, Theorem 2.2). Thus, Theorem 4.5 implies that $(\mathop {*}\limits ^{\displaystyle \trianglerighteq })=(\mathop {\scriptscriptstyle {\mathrm {u}}}\limits ^{\displaystyle \trianglerighteq })$. $\square$

The proof of Theorem 4.7 uses the next result.

Lemma 6.5

Let ${\mathcal { V}}\subseteq {\mathcal { U}}(\succeq )$. Then, $(\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq })$ satisfies NEHIC if and only if ${\mathcal { V}}$ yields a multiutility representation for $(\succeq )$.

Proof

Let $w,w',z,z'\in {\mathcal { X}}$. Let ${\mathbf { x}}:=\left( \begin{array}{l}{w_j}\\ {{\mathbf { o}}_{-j}}\end{array}\right)$, ${\mathbf { x}}':=\left( \begin{array}{l}{w'_j}\\ {{\mathbf { o}}_{-j}}\end{array}\right)$, ${\mathbf { y}}:=\left( \begin{array}{l}{z_j}\\ {{\mathbf { o}}_{-j}}\end{array}\right)$, and ${\mathbf { y}}':=\left( \begin{array}{l}{z'_j}\\ {{\mathbf { o}}_{-j}}\end{array}\right)$. For all $v\in {\mathcal { V}}$, we have

$$\begin{aligned} \sum _{i\in {\mathcal { I}}} v(x'_i) -\sum _{i\in {\mathcal { I}}} v(x_i) = v(w')-v(w) \ \hbox { and }\ \sum _{i\in {\mathcal { I}}} v(y'_i) -\sum _{i\in {\mathcal { I}}} v(y_i) = v(z')-v(z).\nonumber \\ \end{aligned}$$

(30)

“$\Longrightarrow$” Suppose $(\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq })$ satisfies NEHIC. Then

Here, $(*)$ is by formula (30), $(\dagger )$ is by definition of $(\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq })$ and $(\diamond )$ is by NEHIC.

This holds for all $w,w',z,z'\in {\mathcal { X}}$. Thus, ${\mathcal { V}}$ yields a multiutility representation for $(\succeq )$.

“${\Longleftarrow }$” Suppose ${\mathcal { V}}$ yields a multiutility representation for $(\succeq )$. Then

Here, $(\dagger )$ is by definition of $(\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq })$ (recall that ${\mathcal { I}}$ is finite). Meanwhile, $(*)$ is by formula (30), and $(\diamond )$ is because ${\mathcal { V}}$ yields a multiutility representation for $(\succeq )$.

This holds for all $w,w',z,z'\in {\mathcal { X}}$. Thus, $(\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq })$ satisfies NEHIC. $\square$

Proof of Theorem 4.7

If $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ has a multiwelfare representation, and ${\mathcal { I}}$ is finite, then Theorem 4.2 says $(\mathop {}\limits ^{\displaystyle \trianglerighteq })=(\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq })$ for some ${\mathcal { V}}\subseteq {\mathcal { U}}(\succeq )$. If $(\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq })$ satisfies NEHIC, then Lemma 6.5 says ${\mathcal { V}}$ yields a multiutility representation for $(\succeq )$. Then, Theorem 4.5 says that $(\mathop {{\mathcal { V}}}\limits ^{\displaystyle \trianglerighteq })=(\mathop {\scriptscriptstyle {\mathrm {u}}}\limits ^{\displaystyle \trianglerighteq })=(\mathop {*}\limits ^{\displaystyle \trianglerighteq })$. $\square$

Appendix D: Proofs from Sect. 4.3

Proof of Theorem 4.9

This is clear from inspecting the proofs of Propositions 4.1, 4.3 and 4.8 (all of which explicitly handled the case when $|{\mathcal { I}}|$ infinite).$\square$

Proof of Theorem 4.10

This is clear from inspecting the proof of Theorem 4.5, which explicitly handled the case when $|{\mathcal { I}}|$ infinite.$\square$

Appendix E: Proofs from Sect. 5

Proof of Proposition 5.1

Let $r'_2 \ge r_2$, and suppose $\frac{\beta (s_1)-\beta (r_1)}{\beta (r'_2)-\beta (s_2)} > C$. (To prove part (a), it suffices to take $r'_2=r_2$.) Set ${\mathbf { x}}^1={\mathbf { y}}^1={\mathbf { y}}^2:={\left\langle {\mathbf { p}},{\mathbf { r}} \right\rangle }$ and ${\mathbf { x}}^2:={\left\langle {\mathbf { p}},{\mathbf { s}} \right\rangle }$. We must show that $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$. By Proposition 6.4 (which was proved in Appendix C), it suffices to show that $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {\mathrm {\scriptscriptstyle {ng}}}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$.

Let ${\mathcal { I}}_0:=\{1,2\}$; then condition (NG1) holds because $r_i=s_i$ for all $i\in {\mathcal { I}}\setminus \{1,2\}$. It remains to check (NG2). Let ${\mathcal { L}}:=\{1\}$ and let ${\mathcal { J}}_1:={\mathcal { K}}_1:={\mathcal { I}}_0$; we will show that $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {{\mathcal { J}}_1,{\mathcal { K}}_1}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$, by verifying conditions (JK1)-(JK3).

Define $\alpha (1):=2$ and $\alpha (2):=1$. Define $w_0:={\left\langle p_2,r_2 \right\rangle } \ ( \ =x^1_2)$, $w_1:={\left\langle p_2,s_2 \right\rangle } \ ( \ =x^2_2)$ and $w_2:={\left\langle p_2,r'_2 \right\rangle }$. Then $(w_{\alpha (2)-1}\leadsto w_{\alpha (2)})=(w_0\leadsto w_1) = (x^1_2\leadsto x^2_2)$. Meanwhile,

which verifies (JK1). Here $(*)$ is by formula (8), because $\displaystyle \frac{\beta (s_1)-\beta (r_1)}{\beta (r'_2)-\beta (s_2)}>C$.

Now define $\beta (1):=1$, $\beta (2):=2$, and let $z_0=z_1=z_2={\left\langle p_2,r_2 \right\rangle } \ ( \ =y^1_2=y^2_2)$. Then

which verifies (JK2). Here, $(*)$ is by Lemma 6.1. To check (JK3), observe that

$$\begin{aligned} (w_0\leadsto w_2) \ \ = \ \ ({\left\langle p_2,r_2 \right\rangle }\leadsto {\left\langle p_2,r'_2 \right\rangle }) \ \ \,{\mathop {(*)}\limits ^{\displaystyle \succeq }}\, \ \ ({\left\langle p_2,r_2 \right\rangle } \leadsto {\left\langle p_2,r_2 \right\rangle }) \ \ = \ \ (z_0\leadsto z_2), \nonumber \\ \end{aligned}$$

(31)

where $(*)$ is because $r'_2\ge r_2$.

Thus, $({\left\langle {\mathbf { p}},{\mathbf { r}} \right\rangle }\leadsto {\left\langle {\mathbf { p}},{\mathbf { s}} \right\rangle })\,{\mathop {\mathrm {\scriptscriptstyle {ng}}}\limits ^{\displaystyle \trianglerighteq }}\, ({\left\langle {\mathbf { p}},{\mathbf { r}} \right\rangle }\leadsto {\left\langle {\mathbf { p}},{\mathbf { r}} \right\rangle })$. Thus, if $(\succeq )$ is any SDP, then Proposition 6.4 says $({\left\langle {\mathbf { p}},{\mathbf { r}} \right\rangle }\leadsto {\left\langle {\mathbf { p}},{\mathbf { s}} \right\rangle })\,{\mathop {}\limits ^{\displaystyle \trianglerighteq }}\,({\left\langle {\mathbf { p}},{\mathbf { r}} \right\rangle }\leadsto {\left\langle {\mathbf { p}},{\mathbf { r}} \right\rangle })$. This proves part (a).

To prove (b), suppose $r'_2 > r_2$; then the relation “$\,{\mathop {(*)}\limits ^{\displaystyle \succeq }}\,$” in eqn.(31) becomes “$\,{\mathop {(*)}\limits ^{\displaystyle \succ }}\,$”. Thus, $({\left\langle {\mathbf { p}},{\mathbf { r}} \right\rangle }\leadsto {\left\langle {\mathbf { p}},{\mathbf { s}} \right\rangle })\,{\mathop {\mathrm {\scriptscriptstyle {ng}}}\limits ^{\displaystyle \triangleright }}\, ({\left\langle {\mathbf { p}},{\mathbf { r}} \right\rangle }\leadsto {\left\langle {\mathbf { p}},{\mathbf { r}} \right\rangle })$. Thus, if $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ satisfies (SPar), then Proposition 6.4 says that $({\left\langle {\mathbf { p}},{\mathbf { r}} \right\rangle }\leadsto {\left\langle {\mathbf { p}},{\mathbf { s}} \right\rangle })\,{\mathop {}\limits ^{\displaystyle \triangleright }}\,({\left\langle {\mathbf { p}},{\mathbf { r}} \right\rangle }\leadsto {\left\langle {\mathbf { p}},{\mathbf { r}} \right\rangle })$. $\square$

Appendix F: Net gain

Sections 4.1, 4.2 and 4.3 provided several compelling characterizations of quasiutilitarian SDPs. Unfortunately, these results all depend on the existence of a multiwelfare or multiutility representation, and those in Sects. 4.1 and 4.2 also required ${\mathcal { I}}$ to be finite. But not all difference preorders admit a multiutility representation. Furthermore, for applications to intertemporal social choice or social choice under uncertainty, we may want the indexing set ${\mathcal { I}}$ to be infinite. Thus, in this appendix, we will dispense with these assumptions. We will develop a method to aggregate welfare gains over arbitrary populations, without any sort of utility representation. The main result of this appendix is a necessary ingredient in the proofs of Theorems 4.5 and 4.10, and also Proposition 5.1.

For any ${\mathbf { x}}^1,{\mathbf { x}}^2,{\mathbf { y}}^1,{\mathbf { y}}^2\in {\mathcal { X}}^{\mathcal { I}}$, and any finite subsets ${\mathcal { J}},{\mathcal { K}}\subseteq {\mathcal { I}}$ with $J:=|{\mathcal { J}}|$ and $K:=|{\mathcal { K}}|$, we stipulate that $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {{\mathcal { J}},{\mathcal { K}}}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$ if there exist $w_0,w_1,\ldots ,w_J\in {\mathcal { X}}$ and $z_0,z_1,\ldots ,z_K\in {\mathcal { X}}$ and bijections $\alpha :{\mathcal { J}}{{\longrightarrow }}{\left[ 1\ldots J \right]}$ and $\beta :{\mathcal { K}}{{\longrightarrow }}{\left[ 1\ldots K \right]}$ such that:

• (JK1) $(x^1_j\leadsto x^2_j) \ \succeq \ (w_{\alpha (j)-1}\leadsto w_{\alpha (j)})$ for all $j\in {\mathcal { J}}$;

• (JK2) $(z_{\beta (k)-1}\leadsto z_{\beta (k)}) \ \succeq \ (y^1_k\leadsto y^2_k)$, for all $k\in {\mathcal { K}}$; and

• (JK3) $(w_0\leadsto w_J) \succeq (z_0\leadsto z_K)$.

Intuitively, $w_0\leadsto w_J$ aggregates the net welfare gain of the chain $w_0\leadsto w_1\leadsto w_2\leadsto \cdots \leadsto w_J$. Thus, (JK1) implies that net welfare gain for the ${\mathcal { J}}$-population induced by the change ${\mathbf { x}}^1\leadsto {\mathbf { x}}^2$ is at least as large as the net welfare gain of $w_0\leadsto w_J$. Meanwhile, (JK2) implies that the net welfare gain for the ${\mathcal { K}}$-population induced by ${\mathbf { y}}^1\leadsto {\mathbf { y}}^2$ is at most as large as $z_0\leadsto z_K$. Thus, if (JK3) holds, then the ${\mathcal { J}}$-population, in aggregate, gains more welfare from ${\mathbf { x}}^1\leadsto {\mathbf { x}}^2$ than the ${\mathcal { K}}$-population gains from ${\mathbf { y}}^1\leadsto {\mathbf { y}}^2$.

Example 6.2

Suppose ${\mathcal { I}}=\{a,b,c,\ldots ,l,m,n\}$. Let ${\mathcal { J}}:=\{a,b,c,d,e\}$ and ${\mathcal { K}}:=\{f,g,h,i\}$; and let ${\mathbf { x}}_{\mathcal { J}}^1$ and ${\mathbf { x}}_{\mathcal { J}}^2$be as shown in Figure 1(a), while ${\mathbf { y}}_{\mathcal { K}}^1$ and ${\mathbf { y}}_{\mathcal { K}}^2$ are as shown in Figure 1(c). We want to show that $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {{\mathcal { J}},{\mathcal { K}}}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$. Let $z_0,\ldots ,z_5$ and $w_0,\ldots ,w_4$ be as shown in Figure 1(b). Define $\alpha :{\mathcal { J}}{{\longrightarrow }}[1\ldots 5]$ by setting $\alpha (a):=3$, $\alpha (b):=2$, $\alpha (c):=5$, $\alpha (d)=1$ and $\alpha (e):=4$. Define $\beta :{\mathcal { K}}{{\longrightarrow }}[1\ldots 4]$ by setting $\beta (f):=3$, $\beta (g):=1$, $\beta (h):=2$ and $\beta (i):=4$. By comparing Figures 1(a) and 1(b), one can see that

$$\begin{aligned} \begin{array}{rclcl} (x^1_a\leadsto x^2_a)&{}\approx &{} (w_2\leadsto w_3)&{}=&{}(w_{\alpha (a)-1}\leadsto w_{\alpha (a)}),\\ (x^1_b\leadsto x^2_b)&{}\approx &{} (w_1\leadsto w_2)&{}=&{}(w_{\alpha (b)-1}\leadsto w_{\alpha (b)}), \\ (x^1_c\leadsto x^2_c)&{}\approx &{} (w_4\leadsto w_5)&{}=&{}(w_{\alpha (c)-1}\leadsto w_{\alpha (c)}), \\ \end{array} \end{aligned}$$

$$\begin{aligned} \begin{array}{rclcl} (x^1_d\leadsto x^2_d)&{}\approx &{} (w_0\leadsto w_1)&{}=&{}(w_{\alpha (d)-1}\leadsto w_{\alpha (d)}), \\ \hbox { and }(x^1_e\leadsto x^2_e)&{}\approx &{} (w_3\leadsto w_4)&{}=&{}(w_{\alpha (e)-1}\leadsto w_{\alpha (e)}). \end{array} \end{aligned}$$

Thus, (JK1) is satisfied. Likewise, by comparing Figures 1(b) and 1(c), one can see that

$$\begin{aligned} \begin{array}{rclcl} (y^1_f\leadsto y^2_f)&{}\approx &{} (z_2\leadsto z_3)&{}=&{}(z_{\beta (f)-1}\leadsto z_{\beta (f)}), \\ (y^1_g\leadsto y^2_g)&{}\approx &{} (z_0\leadsto z_1)&{}=&{}(z_{\beta (g)-1}\leadsto z_{\beta (g)}), \\ (y^1_h\leadsto y^2_h)&{}\approx &{} (z_1\leadsto z_2)&{}=&{}(z_{\beta (h)-1}\leadsto z_{\beta (h)}), \\ \hbox { and }(y^1_i\leadsto y^2_i)&{}\approx &{} (z_3\leadsto z_4)&{}=&{}(z_{\beta (i)-1}\leadsto z_{\beta (i)}). \end{array} \end{aligned}$$

Thus, (JK2) is satisfied. Finally, by inspecting the left and right halves of Figure 1(b), one can see that $(w_0\leadsto w_5)\succeq (z_0\leadsto z_4)$. Thus, (JK3) is satisfied. Thus, $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {{\mathcal { J}},{\mathcal { K}}}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$. A similar conclusion would hold in the more general case when $(x^1_a\leadsto x^2_a)\succeq (w_2\leadsto w_3)$, etc., while $(y^1_f\leadsto y^2_f)\preceq (z_2\leadsto z_3)$, etc. $\square$

Fig. 1

Example 6.2. In this example, ${\mathcal { J}}:=\{a,b,c,d,e\}$ and ${\mathcal { K}}:=\{f,g,h,i\}$. Here, we represent each element of ${\mathcal { X}}$ with a point in space. The height of the point visually represents its level of well-being. Thus, $x_a^1\leadsto x_a^2$ is a welfare improvement, but $x_d^1\leadsto x_d^2$ is an even larger welfare improvement

Let ${\mathcal { I}}_0\subseteq {\mathcal { I}}$ be a finite subset. A partition of ${\mathcal { I}}_0$ is a collection $\{{\mathcal { J}}_\ell \}_{\ell \in {\mathcal { L}}}$ of disjoint subsets of ${\mathcal { I}}_0$ (where ${\mathcal { L}}$ is some indexing set), such that $\displaystyle {\mathcal { I}}_0 = \bigsqcup _{\ell \in {\mathcal { L}}} {\mathcal { J}}_\ell$. We define the net gain relation $(\mathop {\mathrm {\scriptscriptstyle {ng}}}\limits ^{\displaystyle \trianglerighteq })$ as follows. For any ${\mathbf { x}}^1,{\mathbf { x}}^2,{\mathbf { y}}^1,{\mathbf { y}}^2\in {\mathcal { X}}^{\mathcal { I}}$, stipulate that $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {\mathrm {\scriptscriptstyle {ng}}}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$ if there exists some finite ${\mathcal { I}}_0\subseteq {\mathcal { I}}$ and two partitions $\{{\mathcal { J}}_\ell \}_{\ell \in {\mathcal { L}}}$ and $\{{\mathcal { K}}_\ell \}_{\ell \in {\mathcal { L}}}$ of ${\mathcal { I}}_0$ (with the same indexing set ${\mathcal { L}}$), such that

• (NG1) $(x_i^1\leadsto x_i^2) \succeq (y_i^1\leadsto y_i^2)$ for all $i\in {\mathcal { I}}\setminus {\mathcal { I}}_0$.

• (NG2) For all $\ell \in {\mathcal { L}}$, we have $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {{\mathcal { J}}_\ell ,{\mathcal { K}}_\ell }\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$.

Intuitively, condition (NG2) means we can split up ${\mathcal { I}}_0$ into disjoint subsets such that, for each $\ell \in {\mathcal { L}}$, the ‘net welfare gain’ induced by ${\mathbf { x}}^1\leadsto {\mathbf { x}}^2$ for ${\mathcal { J}}_\ell$ is demonstrably larger than the ‘net welfare gain’ induced by ${\mathbf { y}}^1\leadsto {\mathbf { y}}^2$ for ${\mathcal { K}}_\ell$ (as argued in the paragraph before Example 6.2). Thus, if we aggregate over all $\ell \in {\mathcal { L}}$, then the ‘net welfare gain’ over all of ${\mathcal { I}}_0$ must be greater for ${\mathbf { x}}^1\leadsto {\mathbf { x}}^2$ than it is for ${\mathbf { y}}^1\leadsto {\mathbf { y}}^2$. Meanwhile, condition (NG1) ensures that the people in ${\mathcal { I}}\setminus {\mathcal { I}}_0$ unanimously prefer ${\mathbf { x}}^1\leadsto {\mathbf { x}}^2$ over ${\mathbf { y}}^1\leadsto {\mathbf { y}}^2$. (If ${\mathcal { I}}$ is finite, then we can simply set ${\mathcal { I}}_0:={\mathcal { I}}$, in which case condition (NG1) becomes vacuous.)

Example 6.3

Let ${\mathcal { X}}$ be a discrete, linearly ordered set. Denote the elements of ${\mathcal { X}}$ by Roman numerals I, II, III, IV, etc., and let by the obvious ordering. (We use Roman numerals to emphasize that these states contain purely ordinal welfare information.) For any $x_1,x_2,y_1,y_2\in {\mathcal { X}}$, stipulate that $(x_1\leadsto x_2) \succeq (y_1\leadsto y_2)$ if and only if either

• and ; or

• and .

It can be shown that this relation is a difference preorder (Pivato 2013c, Proposition 9.1). But it does not admit a multiutility representation. Now let ${\mathcal { I}}:=\{1,2,\ldots ,5\}$, and define ${\mathbf { x}},{\mathbf { x}}',{\mathbf { y}},{\mathbf { y}}'\in {\mathcal { X}}^{\mathcal { I}}$ as follows:

$$\begin{aligned} \begin{array}{rclrrrr} {\mathbf { x}}&{}=&{}(III, &{} VII,&{} IV,&{} XI,&{} IX),\\ {\mathbf { x}}'&{}=&{}(IV, &{}X, &{}VII,&{} IX,&{} VII),\\ {\mathbf { y}}&{}=&{}( XII,&{} X, &{} V,&{} IV,&{} VI),\\ \hbox { and } \ {\mathbf { y}}'&{}= &{}( X, &{} VI, &{} VI,&{} V,&{} VII). \end{array} \end{aligned}$$

Let ${\mathcal { J}}_1:=\{1,2,3\}$ and ${\mathcal { K}}_1:=\{3,4,5\}$. Chaining together the transitions $x_1\leadsto x'_1$ and $x_3\leadsto x_3'$ and $x_2\leadsto x_2'$ (in that order) yields III$\leadsto$X. Chaining together $y_4\leadsto y_4'$ and $y_3\leadsto y_3'$ and $y_5\leadsto y_5'$ (in that order) yields IV$\leadsto$VII. Clearly, $($III$\leadsto$X$)\,{\mathop {*}\limits ^{\displaystyle \succ }}\, ($IV$\leadsto$VI$)$ because IIIIV and X VI. Thus, $({\mathbf { x}}\leadsto {\mathbf { x}}') \,{\mathop {{\mathcal { J}}_1,{\mathcal { K}}_1}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { y}}\leadsto {\mathbf { y}}')$.

Meanwhile, let ${\mathcal { J}}_2:=\{4,5\}$ and let ${\mathcal { K}}_2:=\{1,2\}$. Chaining together the transitions $x_4\leadsto x_4'$ and $x_5\leadsto x_5'$ yields XI$\leadsto$VII. Chaining together the transitions $y_1\leadsto y'_1$ and $y_2\leadsto y_2'$ yields XII$\leadsto$VI. Clearly, $($XI$\leadsto$VII$)\,{\mathop {*}\limits ^{\displaystyle \succ }}\, ($XII$\leadsto$VI$)$, because XIXII and VIIVI. Thus, $({\mathbf { x}}\leadsto {\mathbf { x}}') \,{\mathop {{\mathcal { J}}_2,{\mathcal { K}}_2}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { y}}\leadsto {\mathbf { y}}')$.

Clearly, $\{{\mathcal { J}}_1,{\mathcal { J}}_2\}$ is a partition of ${\mathcal { I}}$, and $\{{\mathcal { K}}_1,{\mathcal { K}}_2\}$ is another partition of ${\mathcal { I}}$. We have just established that $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {{\mathcal { J}}_\ell ,{\mathcal { K}}_\ell }\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$ for $\ell =1,2$. Thus, condition (NG2) is satisfied (and condition (NG1) is vacuous because ${\mathcal { I}}_0={\mathcal { I}}$ in this case). Thus, $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2)$ $\,{\mathop {\mathrm {\scriptscriptstyle {ng}}}\limits ^{\displaystyle \trianglerighteq }}\,$ $({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$. $\square$

If $(\succeq )$ is solvable, then it can be shown that $(\mathop {\mathrm {\scriptscriptstyle {ng}}}\limits ^{\displaystyle \trianglerighteq })=(\mathop {*}\limits ^{\displaystyle \trianglerighteq })$, and satisfies the Strong Pareto axiom (Pivato 2013c, Theorem 10.3). For our purposes, however, the following, weaker claim will be sufficient.

Proposition 6.4

If $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ is any $({\mathop {}\limits ^{\displaystyle \succeq }})$-SDP on ${\mathcal { X}}^{\mathcal { I}}$, then $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ extends $(\mathop {\mathrm {\scriptscriptstyle {ng}}}\limits ^{\displaystyle \trianglerighteq })$. Furthermore, if $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ also satisfies (SPar), then $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ properly extends $(\mathop {\mathrm {\scriptscriptstyle {ng}}}\limits ^{\displaystyle \trianglerighteq })$.

Proof

Let ${\mathbf { x}}^1,{\mathbf { x}}^2,{\mathbf { y}}^2,{\mathbf { y}}^2\in {\mathcal { X}}^{\mathcal { I}}$. Let ${\mathcal { I}}_0\subseteq {\mathcal { I}}$ be a finite subset, and let $\{{\mathcal { J}}_\ell \}_{\ell \in {\mathcal { L}}}$ and $\{{\mathcal { K}}_\ell \}_{\ell \in {\mathcal { L}}}$ be partitions of ${\mathcal { I}}_0$. Suppose $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {\mathrm {\scriptscriptstyle {ng}}}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$ via these partitions. We will show that $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$.

For all $\ell \in {\mathcal { L}}$ and $n=1,2$, let ${\mathbf { x}}^n_{{\mathcal { J}}_\ell }:=(x^n_j)_{j\in {\mathcal { J}}_\ell }\in {\mathcal { X}}^{{\mathcal { J}}_\ell }$ and ${\mathbf { y}}^n_{{\mathcal { K}}_\ell }:=(y^n_k)_{k\in {\mathcal { K}}_\ell }\in {\mathcal { X}}^{{\mathcal { K}}_\ell }$. Since ${\mathcal { I}}_0=\bigsqcup _{\ell \in {\mathcal { L}}} {\mathcal { J}}_\ell$, we can write $({\mathbf { x}}^1_{{\mathcal { I}}_0}\leadsto {\mathbf { x}}^2_{{\mathcal { I}}_0})$ as an ${\mathcal { L}}$-indexed structure $({\mathbf { x}}_{{\mathcal { J}}_\ell }^1\leadsto {\mathbf { x}}_{{\mathcal { J}}_\ell }^2)_{\ell \in {\mathcal { L}}}$. Likewise, since ${\mathcal { I}}_0=\bigsqcup _{\ell \in {\mathcal { L}}} {\mathcal { K}}_\ell$, we can write $({\mathbf { y}}^1_{{\mathcal { I}}_0}\leadsto {\mathbf { y}}^2_{{\mathcal { I}}_0})$ as an ${\mathcal { L}}$-indexed structure $({\mathbf { y}}_{{\mathcal { K}}_\ell }^1\leadsto {\mathbf { y}}_{{\mathcal { K}}_\ell }^2)_{\ell \in {\mathcal { L}}}$.

For all $\ell \!\in \!{\mathcal { L}}$, condition (NG2) says $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \ \,{\mathop {{\mathcal { J}}_\ell ,{\mathcal { K}}_\ell }\limits ^{\displaystyle \trianglerighteq }}\, \ ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$. So, let $J_\ell :=|{\mathcal { J}}_\ell |$ and $K_\ell :=|{\mathcal { K}}_\ell |$, and let $w^\ell _{0},w^\ell _{1},\ldots ,w^\ell _{J_\ell }\in {\mathcal { X}}$ and $z^\ell _{0},z^\ell _{1},\ldots ,z^\ell _{K}\in {\mathcal { X}}$, and bijections $\alpha _\ell :{\mathcal { J}}_\ell {{\longrightarrow }}{\left[ 1\ldots J_\ell \right]}$ and $\beta _\ell :{\mathcal { K}}_\ell {{\longrightarrow }}{\left[ 1\ldots K_\ell \right]}$ satisfy conditions (JK1)-(JK3) in the definition of $(\,{\mathop {{\mathcal { J}}_\ell ,{\mathcal { K}}_\ell }\limits ^{\displaystyle \trianglerighteq }}\,)$. For all $\ell \in {\mathcal { L}}$, suppose we write ${\mathcal { J}}_\ell :=\{j(\ell ,1),j(\ell ,2),\ldots ,j(\ell ,J_\ell )\}$, such that $\alpha _\ell [j(\ell ,n)]=n$ for all $n\in {\left[ 1\ldots J_\ell \right]}$. Then

(32)

where “$\,{\mathop {\scriptscriptstyle {\mathrm {par}}}\limits ^{\displaystyle \trianglerighteq }}\,$” is by (JK1), and “” is via the (finitary) permutation $\pi :{\mathcal { I}}_0{{\longrightarrow }}{\mathcal { I}}_0$ defined by $\pi [j(\ell ,1)]:=j(\ell ,J_\ell )$ and $\pi [j(\ell ,n)]:=j(\ell ,n-1)$ for all $\ell \in {\mathcal { L}}$ and $n\in {\left[ 2\ldots J_\ell \right]}$. If ${\mathbf { w}}_{{\mathcal { I}}_0}$, ${\mathbf { w}}'_{{\mathcal { I}}_0}$ and ${\mathbf { w}}''_{{\mathcal { I}}_0}$ are the elements of ${\mathcal { X}}^{{\mathcal { I}}_0}$ defined as indicated above, then formula (32) can be rewritten: . Thus, if ${\mathcal { I}}_0^\complement :={\mathcal { I}}\setminus {\mathcal { I}}_0$, then we have

Thus, since the SDP $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ satisfies axioms (Par) and (Anon), we deduce that

(33)

Next, for all $\ell \in {\mathcal { L}}$, suppose we write ${\mathcal { K}}_\ell :=\{k(\ell ,1),k(\ell ,2),\ldots ,k(\ell ,K_\ell )\}$, such that $\beta _\ell [k(\ell ,n)]=n$ for all $n\in {\left[ 1\ldots K_\ell \right]}$. Then

(34)

where “$\,{\mathop {\scriptscriptstyle {\mathrm {par}}}\limits ^{\displaystyle \trianglelefteq }}\,$” is by (JK2), and “” is via the (finitary) permutation $\pi :{\mathcal { I}}_0{{\longrightarrow }}{\mathcal { I}}_0$ defined by $\pi [k(\ell ,1)]:=k(\ell ,K_\ell )$ and $\pi [k(\ell ,n)]:=(\ell ,n-1)$ for all $\ell \in {\mathcal { L}}$ and $n\in {\left[ 2\ldots K_\ell \right]}$. If ${\mathbf { z}}_{{\mathcal { I}}_0}$, ${\mathbf { z}}'_{{\mathcal { I}}_0}$ and ${\mathbf { z}}''_{{\mathcal { I}}_0}$ are the elements of ${\mathcal { X}}^{{\mathcal { I}}_0}$ defined as indicated above, then formula (34) can be rewritten: . Combining this with the ${\mathcal { I}}_0^\complement$-coordinates, and recalling that $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ satisfies axioms (Par) and (Anon), we obtain

(35)

Now, ${\mathcal { I}}_0={\mathcal { H}}'_1\sqcup {\mathcal { H}}'_2={\mathcal { H}}''_1\sqcup {\mathcal { H}}''_2$, where

$$\begin{aligned} \begin{array}{rclcrcl} {\mathcal { H}}'_1&{} := &{} {\left\{ j(\ell ,1) \; ; \; \ell \in {\mathcal { L}} \right\} };&{}\quad &{} {\mathcal { H}}'_2&{}:=&{}{\left\{ j(\ell ,n) \; ; \; \ell \in {\mathcal { L}}\hbox { and }n\in {\left[ 2\ldots J_\ell \right]} \right\} },\\ {\mathcal { H}}''_1&{}:=&{}{\left\{ k(\ell ,1) \; ; \; \ell \in {\mathcal { L}} \right\} } &{}\hbox { and }&{} {\mathcal { H}}''_2&{}:=&{}{\left\{ k(\ell ,n) \; ; \; \ell \in {\mathcal { L}}\hbox { and }n\in {\left[ 2\ldots K_\ell \right]} \right\} }. \end{array} \end{aligned}$$

Clearly, $|{\mathcal { H}}'_1|=|{\mathcal { L}}|=|{\mathcal { H}}''_1|$, and $|{\mathcal { H}}'_2|=|{\mathcal { I}}_0|-|{\mathcal { L}}| = |{\mathcal { H}}''_2|$. Define bijection $\theta _1:{\mathcal { H}}''_1{{\longrightarrow }}{\mathcal { H}}'_1$ by $\theta _1[k(\ell ,1)]:=j(\ell ,1)$ for all $\ell \in {\mathcal { L}}$. Let $\theta _2:{\mathcal { H}}''_2{{\longrightarrow }}{\mathcal { H}}'_2$ be any bijection. Let $\theta :=\theta _1\sqcup \theta _2:{\mathcal { I}}_0{{\longrightarrow }}{\mathcal { I}}_0$. Then $\theta$ is a (finitary) permutation of ${\mathcal { I}}_0$. For all $\ell \in {\mathcal { L}}$, we have

(36)

where $(*)$ is by condition (JK3). Meanwhile, for any $\ell ,\ell '\in {\mathcal { L}}$, and any $n\in {\left[ 2\ldots K_\ell \right]}$, and $n'\in {\left[ 2\ldots J_{\ell '} \right]}$, if $\theta (k(\ell ,n))=j(\ell ',n')$, then

(37)

where $(*)$ is by Lemma 6.1. Combining (36) and (37) over all $\ell \in {\mathcal { L}}$ and $n\in {\left[ 2\ldots K_\ell \right]}$, we conclude that $\theta ({\mathbf { w}}_{{\mathcal { I}}_0}\leadsto {\mathbf { w}}''_{{\mathcal { I}}_0}) \,{\mathop {\scriptscriptstyle {\mathrm {par}}}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { z}}_{{\mathcal { I}}_0}\leadsto {\mathbf { z}}''_{{\mathcal { I}}_0})$. Meanwhile, condition (NG1) says that $({\mathbf { x}}^1_{{\mathcal { I}}_0^\complement }\leadsto {\mathbf { x}}^2_{{\mathcal { I}}_0^\complement }) \,{\mathop {\scriptscriptstyle {\mathrm {par}}}\limits ^{\displaystyle \trianglerighteq }}\,({\mathbf { y}}^1_{{\mathcal { I}}_0^\complement }\leadsto {\mathbf { y}}^2_{{\mathcal { I}}_0^\complement })$. Thus,

(38)

(39)

because the SDP $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ satisfies axioms (Par) and (Anon). Combining (33), (35) and (39) through transitivity, we conclude that $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2)\ \,{\mathop {}\limits ^{\displaystyle \trianglerighteq }}\,\ ({\mathbf { y}}^1 \leadsto {\mathbf { y}}^2)$, as desired.

• $Proper\; extension.$ Now suppose $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ satisfies (SPar); we must show $(\mathop {}\limits ^{\displaystyle \trianglerighteq })$ properlyextends $(\mathop {\mathrm {\scriptscriptstyle {ng}}}\limits ^{\displaystyle \trianglerighteq })$.

Suppose $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {\mathrm {\scriptscriptstyle {ng}}}\limits ^{\displaystyle \triangleright }}\, ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$. In other words, $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \,{\mathop {\mathrm {\scriptscriptstyle {ng}}}\limits ^{\displaystyle \trianglerighteq }}\, ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$, but $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2) \ {\not \!\!\,{\mathop {\mathrm {\scriptscriptstyle {ng}}}\limits ^{\displaystyle \trianglelefteq }}\,} \ ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$. Then there are three cases. Either

• (i) $({\mathbf { x}}^1_{{\mathcal { I}}_0^\complement }\leadsto {\mathbf { x}}^2_{{\mathcal { I}}_0^\complement }) \,{\mathop {\scriptscriptstyle {\mathrm {par}}}\limits ^{\displaystyle \triangleright }}\,({\mathbf { y}}^1_{{\mathcal { I}}_0^\complement }\leadsto {\mathbf { y}}^2_{{\mathcal { I}}_0^\complement })$; or

• (ii) one of the “$\,{\mathop {\scriptscriptstyle {\mathrm {par}}}\limits ^{\displaystyle \trianglerighteq }}\,$” in formulae (32) or (34) is actually “$\,{\mathop {\scriptscriptstyle {\mathrm {par}}}\limits ^{\displaystyle \triangleright }}\,$”; or

• (iii) one of the preferences “$\,{\mathop {(*)}\limits ^{\displaystyle \succeq }}\,$” in formula (36) is actually a “$\,{\mathop {(*)}\limits ^{\displaystyle \succ }}\,$”.

(If none of these three cases are satisfied, then we would have and for all $\ell \in {\mathcal { L}}$; which would imply that , contradicting our hypothesis.)

In case (ii), (SPar) yields a strict social preference in formulae (33) or (35). In case (i) or (iii), the “$\,{\mathop {\scriptscriptstyle {\mathrm {par}}}\limits ^{\displaystyle \trianglerighteq }}\,$” in formula (38) becomes a “$\,{\mathop {\scriptscriptstyle {\mathrm {par}}}\limits ^{\displaystyle \triangleright }}\,$”; then (SPar) yields a strict social preference in formula (39). Either way, when we combine (33), (35) and (39) through transitivity, we conclude that $({\mathbf { x}}^1\leadsto {\mathbf { x}}^2)\ \,{\mathop {}\limits ^{\displaystyle \triangleright }}\,\ ({\mathbf { y}}^1\leadsto {\mathbf { y}}^2)$, as desired. $\square$