An axiomatic characterization of a class of rank mobility measures

Roberto Ghiselli Ricci

doi:10.1007/s00355-018-01171-5

Social Choice and Welfare

pp 1–33 | Cite as

An axiomatic characterization of a class of rank mobility measures

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Roberto Ghiselli Ricci

Original Paper

First Online: 05 January 2019

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Abstract

In this paper we provide an axiomatic characterization of a total preorder for assessing the mobility associated with any pair of rankings relative to a same group by means of a new class of rank-mobility measures. These measures constitute a strong generalization of Spearman’s $\rho $ index. The problem of dealing with mobility of subgroups, which is a pressing issue in several and various applicative contexts ranging from the use of socio-economic indicators to information retrieval, is addressed in terms of partial permutations which include standard permutations as a special case. Literature is lacking in a theoretical discussion for comparing mobility of variable-size subgroups: we show that our axiomatic approach fills this gap.

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Notes

Acknowledgements

We would like to thank the anonymous referees for their careful reading and valuable comments that have been helpful for providing a much improved version of the present work.

Appendix A: Proof of Proposition 1

A.1 Preliminary lemmata

Lemma 2

Let $f:{\mathbb {N}}_{0}\rightarrow {\mathbb {R}}$. Then f is strictly convex if, and only if,

$$\begin{aligned} f(u+1)-f(u)>f(v+1)-f(v) \end{aligned}$$

(12)

for all $\,u,v\in {\mathbb {N}}_{0}$ such that $u>v$.

Proof

Let f be strictly convex. Given any $v\in {\mathbb {N}}_{0}$, let $E_v=\{u\in {\mathbb {N}}: u>v$ and $\,f(x+1)-f(x)>f(v+1)-f(v)\,$ for all $x\in \,\,]v,u]\cap {\mathbb {N}}\}$. Let us show that $E_v$ is non-empty: indeed, $v+1\in E_v$ due to Eq. (2) with $n=v+1$. If $u\in E_v$ also $u+1\in E_v$, because $f(u+2)-f(u+1)>f(u+1)-f(u)$ is assured by Eq. (2) with $n=u+1$, while $f(u+1)-f(u)>f(v+1)-f(v)$ follows from the fact that $u\in E_v$. Thus $E_v=\,]v,+\infty [\,\cap \, {\mathbb {N}}$ and the necessity is proved. The converse is immediate. $\square $

Lemma 3

Let $f:{\mathbb {N}}_{0}\rightarrow {\mathbb {R}}$. Then f is strictly convex if, and only if,

$$\begin{aligned} f(s+p)-f(s)>f(t+p)-f(t) \end{aligned}$$

(13)

for all $\,t,s\in {\mathbb {N}}_{0}$ such that $s>t$ and for every $p\in {\mathbb {N}}$.

Proof

Let f be strictly convex. Given any $p\in {\mathbb {N}}$ and any $t\in {\mathbb {N}}_{0}$, let $E_{t,p}=\{s\in {\mathbb {N}}: s>t$ and $\,f(x+p)-f(x)>f(t+p)-f(t)\,$ for all $x\in \,\,]t,s]\cap {\mathbb {N}}\}$. Let us show that $E_{t,p}$ is non-empty: indeed, $t+1\in E_{t,p}$ due to Eq. (12) with $v=t$ and $u=t+p$. If $s\in E_{t,p}$ also $s+1\in E_{t,p}$, because $f(s+1+p)-f(s+1)>f(s+p)-f(s)$ is assured by Eq. (12) with $v=s$ and $u=s+p$, while $f(s+p)-f(s)>f(t+p)-f(t)$ follows from the fact that $s\in E_{t,p}$. Thus $E_{t,p}=\,]s,+\infty [\,\cap \, {\mathbb {N}}$ and the necessity is proved. The converse is immediate. $\square $

A.2 Proof of Proposition 1

Proof

Let f be strictly convex and strictly increasing. We divide the proof into two main cases with different sub-cases each.

Case $r\ge 0$: trivially, we have $j-i+r>0$ and $r>i-j+k$. Particularly, the last inequality clearly shows that the set of all admissible r is $[r_{m},+\infty [\,\cap \,{\mathbb {N}}$, where

$$\begin{aligned} r_{m}={\left\{ \begin{array}{ll} i-j+k+1,&{} \quad {\text {if }} \ i-j+k\ge 0;\\ 0,&{}\quad {\text {if }} \ i-j+k<0. \end{array}\right. } \end{aligned}$$

In both the sub-cases $i-j+k\ge 0$ and $i-j+k<0$, Eq. (3) goes to

$$\begin{aligned} f(|k|)-f(|i-j+k|)<f(r+p)-f(r), \end{aligned}$$

(14)

where $p=j-i$. According to Lemma 3, it suffices to show Eq. (14) for $r=r_{m}$. In the first sub-case $i-j+k\ge 0$, Eq. (14) with $r=r_m=i-j+k+1$ becomes

$$\begin{aligned} f(i-j+k+1)-f(i-j+k)<f(k+1)-f(k), \end{aligned}$$

which is true due to Eq. (12) with $v=i-j+k$ and $u=k$. In the second sub-case $i-j+k<0$, consider first the situation $0\le k<j-i$. Then, Eq. (14) with $r=r_m=0$ boils down to

$$\begin{aligned} f(k)+f(0)<f(j-i)+f(j-i-k), \end{aligned}$$

which is assured by the strict monotonicity of f, since by assumption we have $k<j-i$. Now, consider the second sub-case in the remaining situation $k<0$: then, Eq. (14) with $r=r_m=0$ goes to

$$\begin{aligned} f(|k|)+f(0)<f(j-i)+f(j-i+|k|), \end{aligned}$$

which is certainly valid owing to the strict monotonicity of f and $j>i$, so closing the case $r\ge 0$.

Case $r<0$: the first sub-case under consideration is $j-i+r<0$, which implies $i-j+k<r<0$ and $k<j-i+r<0$. Particularly, the last inequality clearly shows that the set of the absolute values of all admissible k is $[|k|_{m},+\infty [\,\cap \,{\mathbb {N}}$, where $|k|_{m}=i-j+|r|+1$. In this sub-case, Eq. (3) becomes

$$\begin{aligned} f(|r|)-f(i-j+|r|)<f(|k|+p)-f(|k|), \end{aligned}$$

(15)

where $p=j-i$. According to Lemma 3, it suffices to show Eq. (15) for $|k|=|k|_{m}$, so obtaining

$$\begin{aligned} f(i-j+|r|+1)-f(i-j+|r|)<f(|r|+1)-f(|r|), \end{aligned}$$

which is true appealing to Eq. (12) with $v=i-j+|r|$ and $u=|r|$. The second sub-case is $j-i+r\ge 0$, which still implies $i-j+k<r<0$: now, consider first the situation $k\ge 0$. Then, Eq. (3) goes to

$$\begin{aligned} f(k)+f(|r|)<f(j-i+r)+f(j-i-k), \end{aligned}$$

which is assured by the strict monotonicity of f, since by assumption we have $k<j-i+r$. In the remaining situation $k<0$ of the second sub-case, Eq. (3) boils down to

$$\begin{aligned} f(|r|)-f(j-i+r)<f(|k|+p)-f(|k|), \end{aligned}$$

(16)

where $p=j-i$. Repeating the above argument, it suffices to show Eq. (16) for $|k|=|k|_{m}$. It is easy to see that the set of all admissible r is $\{-1,-2,\ldots ,i-j\}$, hence, combining $k<r+j-i$ with our assumption $k<0$, we get $k\le -1$ and consequently $|k|_{m}=1$. Rewriting Eq. (16) for $|k|=|k|_{m}=1$ yields

$$\begin{aligned} f(1)+f(|r|)<f(j-i+1)+f(j-i+r). \end{aligned}$$

(17)

Note that the above equation is fulfilled for $r=i-j$, because it reduces to Eq. (12) with $v=0$ and $u=j-i$. Instead, when $r\in \{-1,-2,\ldots , i-j+1\}$, Eq. (17) is assured by the strict monotonicity of f, since $1\le j-i+r$, so closing the necessity of this proof. Conversely, assume that f satisfies Eq. (3). Particularly, given any $n\in {\mathbb {N}}$, if we apply Eq. (3) with $i=1$, $j=2$ and $k=r=n$, we exactly get Eq. (2), so proving the strict convexity of f. With regard to the strict increasing monotonicity of f, given any $n\in {\mathbb {N}}$, it suffices to apply Eq. (3) with $i=1$, $j=2$, $k=-n$ and $r=n$. $\square $

Appendix B: Proof of Proposition 2

B.1 Preliminary lemma

Lemma 4

Given any $(m,n)\in {\mathcal {A}}$ with $m\ge 2$ and any $\sigma \in {\mathcal {F}}_{m,n}(D,R)$, there exist $m-1$ inversions $\nu _1,\ldots ,\nu _{m-1}$ on R such that $\,\nu _{m-1}\circ \cdots \circ \nu _1\circ \sigma =\sigma _{\pi (D,R)}$.

Proof

We present a constructive proof. First of all, we obviously have that $\sigma (x_1)\le \sigma _{\pi (D,R)}(x_1)$, hence the inversion $\nu _1:=\nu (\sigma (x_1),\sigma _{\pi (D,R)}(x_1))$ is well-defined. Since $(\nu _1\circ \sigma )(x_2)\ne (\nu _1\circ \sigma )(x_1)=\sigma _{\pi (D,R)}(x_1)$, we deduce that $(\nu _1\circ \sigma )(x_2)<\sigma _{\pi (D,R)}(x_1)$ which in turn entails $(\nu _1\circ \sigma )(x_2)\le \sigma _{\pi (D,R)}(x_2)$. Thus, the inversion $\nu _2:=\nu ((\nu _1\circ \sigma )(x_2),\sigma _{\pi (D,R)}(x_2))$ is well-defined. Note that $(\nu _1\circ \sigma )(x_2)<\sigma _{\pi (D,R)}(x_1)$ also implies $(\nu _2\circ \nu _1\circ \sigma )(x_1)=\sigma _{\pi (D,R)}(x_1)$ and that, by definition, $(\nu _2\circ \nu _1\circ \sigma )(x_2)=\sigma _{\pi (D,R)}(x_2)$. Repeating the same argument for a finite number of steps, for any $s\in \{3,\ldots ,m-1\}$ the inversion $\nu _s:=\nu ((\nu _{s-1}\circ \cdots \circ \nu _1)(\sigma (x_s)),\sigma _{\pi (D,R)}(x_s))$ is well-defined and satisfies the property $(\nu _{s}\circ \cdots \circ \nu _1\circ \sigma )(x_i)=\sigma _{\pi (D,R)}(x_i)$ for all $i=1,\ldots ,s$. Particularly, for $s=m-1$, we easily derive the claim. $\square $

B.2 Proof of Proposition 2

Proof

By definition of $\nu =\nu (\sigma (x_{l}),\sigma (x_{l'}))$, it is not difficult to show that

$$\begin{aligned} S_f(\nu \circ \sigma )=\sum _{s\ne l,l'}f(\delta _{\sigma }(x_s))+f(|x_{l}-\sigma (x_{l'})|)+f(|x_{l'}-\sigma (x_{l})|). \end{aligned}$$

As straightforward consequence, we obtain that $S_f(\nu \circ \sigma )>S_f(\sigma )$ if, and only if,

$$\begin{aligned} f(\delta _{\sigma }(x_{l}))+f(\delta _{\sigma }(x_{l'}))<f(|x_{l}-\sigma (x_{l'})|)+f(|x_{l'}-\sigma (x_{l})|). \end{aligned}$$

After the assignments $i:=x_{l}$, $j:=x_{l'}$, $k:=\sigma (x_{l})-x_{l}$ and $r:=\sigma (x_{l'})-x_{l'}$, it is immediate to see that $i,j,k,r\in {\mathbb {Z}}$, with $i<j$ and $i+k<j+r$. Now, we can apply Proposition 1 and the claim follows by observing that the above equation simply reduces to Eq. (3). $\square $

Appendix C: Proofs of Propositions 3 and 4

C.1 Proof of Proposition 3

Proof

Without loss of generality, suppose that $m\ge 2$. Fix any $\sigma \in {\mathcal {F}}_{m,n}(D,R)$, with $\sigma \ne \sigma _{\pi (D,R)}$. Appealing to Lemma 4, there exist $m-1$ inversions $\nu _1,\ldots ,\nu _{m-1}$ on R such that $\,\nu _{m-1}\circ \cdots \circ \nu _1\circ \sigma =\sigma _{\pi (D,R)}$. Let us focus on $\nu _1$, defined as $\nu _1=\nu (\sigma (x_1),\sigma _{\pi (D,R)}(x_1))$. Obviously, there exists a unique index in $\{1,\ldots ,m\}$, say $j_1$, such that $\sigma _{\pi (D,R)}(x_1)=\sigma (x_{j_1})$. Note that if $j_1=1$, then $\nu _1$ reduces to the identity function, otherwise we can apply Proposition 2 to $\sigma $, with $\nu =\nu _1$, $l=1$ and $l'=j_1$, so obtaining $S_f(\nu _1\circ \sigma )>S_f(\sigma )$. Therefore, we can state that $S_f(\nu _1\circ \sigma )\ge S_f(\sigma )$, remarking that the inequality is strict if, and only if, $\nu _1\ne id_R$. Resorting again to Lemma 4, let us focus now on the inversion $\nu _2=\nu ((\nu _1\circ \sigma )(x_2),\sigma _{\pi (D,R)}(x_2))$. Since we know that $(\nu _1\circ \sigma )(x_1)=\sigma _{\pi (D,R)}(x_1)$, it is immediate to derive that there exists a unique index in $\{2,\ldots ,m\}$, say $j_2$, such that $\sigma _{\pi (D,R)}(x_2)=(\nu _1\circ \sigma )(x_{j_2})$. Note that if $j_2=2$, then $\nu _2$ reduces to the identity function, otherwise we can apply Proposition 2 to $\nu _1\circ \sigma $, with $\nu =\nu _2$, $l=2$ and $l'=j_2$, so obtaining $S_f(\nu _2\circ \nu _1\circ \sigma )>S_f(\nu _1\circ \sigma )$. Accordingly, we can assert that $S_f(\nu _2\circ \nu _1\circ \sigma )\ge S_f(\nu _1\circ \sigma )$, remarking that the inequality is strict if, and only if, $\nu _2\ne id_R$. Combining this assertion with the previous statement, we get $S_f(\nu _2\circ \nu _1\circ \sigma )\ge S_f(\sigma )$, observing that the inequality is strict if at least one between $\nu _1$ and $\nu _2$ does not reduce to the identity function. Repeating the same argument for a finite number of steps, one finds $S_f(\nu _{m-1}\circ \cdots \circ \nu _1\circ \sigma )\ge S_f(\sigma )$. Thus, the claim follows by recalling that $\nu _{m-1}\circ \cdots \circ \nu _1\circ \sigma =\sigma _{\pi (D,R)}$ and by observing that the assumption $\sigma \ne \sigma _{\pi (D,R)}$ necessarily implies that at least one among $\nu _1,\ldots ,\nu _{m-1}$ is not the identity function. $\square $

C.2 Preliminary lemmata

Lemma 5

Given any $(m,n)\in {\mathcal {A}}$, with m even, suppose that $\{z_1,\ldots ,z_{m/2}\}$ is a subset of ${\mathbb {N}}_n$ such that $z_i<z_{i+1}$ for every $i\in {\mathbb {N}}_{m/2-1}$. Then we have that $x^{*}_i\le z_i$ for all $i\in {\mathbb {N}}_{m/2}$.

Proof

Trivially, we have that $z_1\ge x^{*}_1=1$: fixing any $i\in [2,m/2]\cap {\mathbb {N}}$, assume that $z_j\ge x^{*}_j\,$ for every $j\in {\mathbb {N}}_{i-1}$, and let us prove that $z_i\ge x^{*}_i$. Suppose ab absurdo that $z_i<x^{*}_i$: consequently, we easily derive that $x^{*}_{i-1}\le z_{i-1}<z_i<x^{*}_i$, with the evident contradiction that the natural number $z_i$ lies between the two consecutive natural numbers $x^{*}_{i-1}$ and $x^{*}_i$, so concluding the proof. $\square $

Lemma 6

Given any $(m,n)\in {\mathcal {A}}$, with m even, suppose that $\{z_1,\ldots ,z_{m}\}$ is a subset of ${\mathbb {N}}_n$ such that $z_i<z_{i+1}$ for every $i\in {\mathbb {N}}_{m-1}$. Then we have that $z_{i}< x^{*}_{m+1-i}$ for all $i\in {\mathbb {N}}_{m/2}$.

Proof

By assumption, we know that $z_i<z_{m/2}$ and $x^{*}_{1+m/2}< x^{*}_{1+m-i}$ for all $i\in {\mathbb {N}}_{m/2-1}$, so the claim is equivalent to showing that $z_{m/2}< x^{*}_{1+m/2}$. Suppose ab absurdo that $z_{m/2}\ge x^{*}_{1+m/2}$. This immediately implies that $E_{m/2}\subseteq {\mathbb {N}}_n\setminus {\mathbb {N}}_{1+n-m/2}$, where $E_{m/2}:=\{z_{1+m/2},\ldots ,z_m\}$, with the evident contradiction that $|E_{m/2}|=m/2>m/2-1=|{\mathbb {N}}_n\setminus {\mathbb {N}}_{1+n-m/2}|$, so concluding the proof.

The next results are analogous to the previous ones, so we will omit the proofs.

Lemma 7

Given any $(m,n)\in {\mathcal {A}}$, with m even, suppose that $\{z_1,\ldots ,z_{m/2}\}$ is a subset of ${\mathbb {N}}_n$ such that $z_i>z_{i+1}$ for every $i\in {\mathbb {N}}_{m-1}$. Then we have that $z_{i}\le x^{*}_{m+1-i}$ for all $i\in {\mathbb {N}}_{m/2}$.

Lemma 8

Given any $(m,n)\in {\mathcal {A}}$, with m even, suppose that $\{z_1,\ldots ,z_{m}\}$ is a subset of ${\mathbb {N}}_n$ such that $z_i>z_{i+1}$ for every $i\in {\mathbb {N}}_{m-1}$. Then we have that $z_{i}> x^{*}_{i}$ for all $i\in {\mathbb {N}}_{m/2}$.

Finally, observe that we will make use of the immediate property

$$\begin{aligned} \sigma _{\pi (n)}(x^{*}_i)=x^{*}_{m+1-i},\quad {\text {for all }} \ i\in {\mathbb {N}}_m. \end{aligned}$$

(18)

C.3 Proof of Proposition 4

Proof

Based on Proposition 3, we deduce that the claim is equivalent to showing that, fixing any $(m,n)\in {\mathcal {A}}$,

$$\begin{aligned} S_f(\sigma _{\pi (D^{n}_m,D^{n}_m)})>S_f(\sigma _{\pi (D,R)}) \end{aligned}$$

(19)

for every pair (D, R) of subsets of ${\mathbb {N}}_n$ of cardinality m. Let (D, R) be such a pair, with $D=\{x_1,\ldots ,x_m\}$: we assert that $\delta _{\sigma _{\pi (D,R)}}(x_i)\le \delta _{\sigma _{\pi (n)}}(x^{*}_i)$ for all $i\in {\mathbb {N}}_{m}$ and equality holds if, and only if, $\sigma _{\pi (D,R)}$ coincides with the restriction of $\sigma _{\pi (n)}$ to $D^{n}_m$. We only prove the assertion for the case $i\in {\mathbb {N}}_{m/2}$, since the other case is analogous. First of all, setting $z_i:=x_i$ for all $i\in {\mathbb {N}}_{m}$, it is easy to see that Lemma 5 can be applied to $z_1,\ldots ,z_{m/2}$, so getting $x_i\ge x^{*}_i$ for all $i\in {\mathbb {N}}_{m/2}$. Moreover, owing to Lemma 6 and Eq. (18), we deduce that $x_i<\sigma _{\pi (n)}(x^{*}_{i})$ for all $i\in {\mathbb {N}}_{m/2}$. Therefore, we have shown that $x^{*}_i\le x_i<\sigma _{\pi (n)}(x^{*}_{i})$ for every $i\in {\mathbb {N}}_{m/2}$. By the same token, after the assignment $z_i:=\sigma _{\pi (D,R)}(x_{i})$ for every $i\in {\mathbb {N}}_m$, by Lemma 7, applied to $z_1,\ldots ,z_{m/2}$, and Lemma 8, one may easily prove that $x^{*}_i<\sigma _{\pi (D,R)}(x_{i})\le \sigma _{\pi (n)}(x^{*}_{i})$ for all $i\in {\mathbb {N}}_{m/2}$. As a straightforward consequence, the assertion is shown. Finally, based upon the assertion, Eq. (19) directly follows from the evident fact that the restriction of $\sigma _{\pi (n)}$ to $D^{n}_m$ is equal to $\sigma _{\pi (D^{n}_m,D^{n}_m)}$. $\square $

Appendix D: Proof of Corollary 1

We need some preliminary notions. Let $\sigma $ be a partial permutation. We say that a pair $(\sigma _{1},\sigma _{2})$ of partial permutations, usually denoted by $p(\sigma )$, is a partition of $\sigma $ if $\sigma _{1}$ and $\sigma _{2}$ are disjoint and $\sigma =\sigma _1\cup \sigma _2$. In the sequel, we will call the natural partition of a partial permutation $\sigma $ associated with a non-void proper subdomain D of $D_{\sigma }$ the pair $p(\sigma )=(\sigma _{1},\sigma _{2})$ given by the restrictions of $\sigma $ to D and $D_{\sigma }\setminus D$, respectively.

Let $\sigma ,\tau \in {\mathcal {F}}_{m,n}$ and let $(p(\sigma ),p(\tau ))$ be a pair of partitions of $\sigma $ and $\tau $, respectively. We say that $(p(\sigma ),p(\tau ))$ is compatible if, and only if, $|D_{\sigma _{1}}|=|D_{\tau _{1}}|=m_1$ for some $m_1\in [1,m[\,\,\cap \,\,{\mathbb {N}}$. Evidently, if $(p(\sigma ),p(\tau ))$ is compatible, we have that $|D_{\sigma _{2}}|=|D_{\tau _{2}}|=m_2$, with $m_2=m-m_1$.

Finally, fixing any $k\in {\mathbb {N}}_0$, we will denote by ${\mathcal {L}}^k$ the family of partial permutations given by ${\mathcal {L}}^k=\{\sigma \in {\mathcal {F}}\,:\delta _{\sigma }(x)=k\,$ for all $x\in D_{\sigma }\}$.

D.1 Preliminary results

Lemma 9

Let $\sigma ,\tau \in {\mathcal {F}}_{m,n}$. Assume that there exists an injection $s:D_{\sigma }\rightarrow D_{\tau }$ such that $\delta _{\sigma }(x)\le \delta _{\tau }(s(x))$ for every $x\in D_{\sigma }$. Then we have that $\sigma \preceq ^{m}_{n}\tau $.

Proof

Given any $n\in {\mathbb {N}}^1$, the claim directly follows from the first axiom if $m=1$, so suppose that $m=2$. Let $D_{\sigma }=\{x_1,x_2\}$ and $D_{\tau }=\{s(x_1),s(x_2)\}$ for some $1\le x_1<x_2\le n$: note in passing that $D_{\tau }$ might not be arranged in increasing order, but this does not affect our proof. By assumption, we have that $\delta _{\sigma }(x_i)\le \delta _{\tau }(s(x_i))$ for $i=1,2$. Consider now the natural partitions of $\sigma $ and $\tau $ associated with $\{x_1\}$ and $\{s(x_1)\}$, respectively. By Axiom 1 we know that $\sigma _{i}\preceq ^{1}_{n}\tau _{i}$ for $i=1,2$ and immediately, by means of Axiom 2, we get the claim for $m=2$. If $n=2$, the proof is concluded, otherwise let p be the maximum integer in ${\mathbb {N}}_n$ such that the claim is true when $\sigma ,\tau \in {\mathcal {F}}_{m,n}$ for every $m\in {\mathbb {N}}_p$: we have just shown that $p\ge 2$. The proof is concluded if we show that $p=n$. Suppose ab absurdo that $p<n$ and consider $\sigma ,\tau \in {\mathcal {F}}_{p+1,n}$, with $D_{\sigma }=\{x_1,\ldots ,x_{p+1}\}$ and $D_{\tau }=\{s(x_1),\ldots ,s(x_{p+1})\}$. From now on, if we simply mimic the same line of reasoning as in the case $m=2$, we get $\sigma \preceq ^{p+1}_{n}\tau $, contradicting our assumption about p.

As immediate consequences of the preceding lemma, we have the following results.

Corollary 2

Let $\sigma ,\tau \in {\mathcal {F}}_{m,n}$ and assume that $\sigma \in {\mathcal {L}}^0$. Then $\sigma \preceq ^{m}_{n}\tau $.

Corollary 3

Let $\sigma ,\tau \in {\mathcal {F}}_{m,n}\cap {\mathcal {L}}^k$. Then $\sigma \sim ^{m}_{n}\tau $.

D.2 Proof of Corollary 1

Proof

Denote by $\lambda _{1}(\sigma )$ and $\lambda _{2}(\sigma )$ the restrictions of $\lambda (\sigma )$ to $D_{\sigma }$ and $D_{\lambda (\sigma )}\setminus D_{\sigma }$, respectively, and similarly for $\mu _{1}(\sigma )$ and $\mu _{2}(\sigma )$. By the properties of any outer extension, we derive that $\lambda _{1}(\sigma )=\mu _{1}(\sigma )=\sigma \in {\mathcal {F}}_{m,n^{\prime }}$ and that both $\lambda _{2}(\sigma )$ and $\mu _{2}(\sigma )$ reduce to the identity function. Accordingly, we get $\lambda _{1}(\sigma )\sim ^{m}_{n^{\prime }}\mu _{1}(\sigma )$, while $\lambda _{2}(\sigma )\sim ^{m^{\prime }-m}_{n^{\prime }}\mu _{2}(\sigma )$ directly follows from Corollary 3 with $k=0$. Thus, by Axiom 2, the claim is assured. $\square $

Appendix E: Proofs of Theorems 1 and 2

We warn the reader that the proof of Theorem 1 is preceded by a long series of preparatory lemmata and crucial propositions.

E.1 Preliminary results

Lemma 10

Let $\sigma ,\tau \in {\mathcal {F}}_{m,n}$ be such that $\sigma \preceq ^{m}_{n}\tau $. Then $\sigma \preceq ^{m}_{n^{\prime }}\tau $ for every $n^{\prime }>n$.

Proof

Suppose ab absurdo that $\tau \prec ^{m}_{n^{\prime }}\sigma $ for some $n^{\prime }>n$. Then, by Axiom 3, there exists a $(m^{\prime \prime },n^{\prime \prime })>(m,n^{\prime })$ and an outer extension $\lambda (\tau )$ such that

$$\begin{aligned} \lambda (\tau )\sim ^{m^{\prime \prime }}_{n^{\prime \prime }}I(\sigma ). \end{aligned}$$

(20)

Since $(m^{\prime \prime },n^{\prime \prime })>(m,n)$, by Axiom 3 we also deduce that Eq. (20) implies $\tau \prec ^{m}_{n}\sigma $ against the hypothesis. $\square $

In the sequel, we will denote the partition of a simple reproduction of $\sigma $ as $p(I(\sigma ))=(I_1(\sigma ),I_2(\sigma ))$.

Corollary 4

Let $\sigma ,\tau \in {\mathcal {F}}_{m,n}$ be such that $\sigma \preceq ^{m}_{n}\tau $. Then $I(\sigma )\preceq ^{m^{\prime }}_{n^{\prime }}I(\tau )$ for every $(m^{\prime },n^{\prime })>(m,n)$.

Proof

Consider the compatible pair $(p(I(\sigma )),p(I(\tau )))$ given by the natural partitions of $I(\sigma )$ and $I(\tau )$ associated with $D_{\sigma }$ and $D_{\tau }$, respectively. It is immediate to see that $I_1(\sigma )=\sigma \in {\mathcal {F}}_{m,n^{\prime }}$ and $I_1(\tau )=\tau \in {\mathcal {F}}_{m,n^{\prime }}$. Thus, the assumption $\sigma \preceq ^{m}_{n}\tau $, combined with Lemma 10, leads to $I_1(\sigma )\preceq ^{m}_{n^{\prime }}I_1(\tau )$, while $I_2(\sigma )\sim ^{m^{\prime }-m}_{n^{\prime }}I_2(\tau )$ directly follows from Corollary 3 with $k=0$. Therefore, by virtue of Axiom 2, we derive the validity of the claim. $\square $

Lemma 11

Let $\sigma ,\tau \in {\mathcal {F}}_{m,n}$ be such that $\sigma \sim ^{m}_{n}\tau $. Assume that there exists a compatible pair $(p(\sigma ),p(\tau ))$ such that $\sigma _{1}\sim ^{m_1}_{n}\tau _{1}$. Then $\sigma _{2}\sim ^{m_2}_{n}\tau _{2}$.

Proof

Suppose ab absurdo that $\sigma _{2}\not \sim ^{m_2}_{n}\tau _{2}$. We may limit ourselves to treat the case $\sigma _{2}\prec ^{m_2}_{n}\tau _{2}$, since the other one is similar and is left to the reader. Then, by Axiom 3, there exists a $(m^{\prime },n^{\prime })>(m_2,n)$ and an outer extension $\lambda (\sigma _{2})$ such that $\lambda (\sigma _{2})\sim ^{m^{\prime }}_{n^{\prime }}I(\tau _{2})$. By Lemma 10, we know that $\sigma _{1}\sim ^{m_1}_{n^{\prime }}\tau _{1}$. Moreover, we know that $D_{\sigma _1}\subsetneq {\mathbb {N}}_n$, while $D_{\lambda (\sigma _2)}\cap {\mathbb {N}}_n=D_{\sigma _2}$ follows from property (i) of the outer extensions, hence we derive that $\lambda (\sigma _{2})$ is disjoint with $\sigma _{1}$ and that their union is an extension of $\sigma $, hereafter denoted by $\mu (\sigma )$. Similarly, it can be seen that $I(\tau _{2})$ is disjoint with $\tau _{1}$ and their union is $I(\tau )$: consequently, by Axiom 2, we get $\mu (\sigma )\sim ^{m^{\prime }+m_1}_{n^{\prime }}I(\tau )$. Thus, since $(m^{\prime }+m_1,n^{\prime })>(m,n)$, invoking Axiom 3 we conclude that $\sigma \prec ^{m}_{n}\tau $ against the hypothesis. $\square $

Lemma 12

Let $\sigma ,\tau \in {\mathcal {F}}_{m,n}$ be such that $\sigma \sim ^{m}_{n}\tau $. Assume that there exists a compatible pair $(p(\sigma ),p(\tau ))$ such that $\sigma _{1}\prec ^{m_1}_{n}\tau _{1}$. Then $\sigma _{2}\succ ^{m_2}_{n}\tau _{2}$.

Proof

According to Lemma 11, it suffices to prove that $\sigma _{2}\prec ^{m_2}_{n}\tau _{2}$ leads to a contradiction. Indeed, in this case, by Axiom 3, there would exist a $(m^{\prime },n^{\prime })>(m_2,n)$ and an outer extension $\lambda (\sigma _{2})$ such that $\lambda (\sigma _{2})\sim ^{m^{\prime }}_{n^{\prime }}I(\tau _{2})$. According to Lemma 10, we know that $\sigma _1\preceq ^{m_1}_{n^{\prime }}\tau _1$: in the trivial case $\sigma _1\sim ^{m_1}_{n^{\prime }}\tau _1$, we can simply mimic the proof as in Lemma 11 obtaining the contradiction $\sigma \prec ^{m}_{n}\tau $, so we can suppose that $\sigma _1\prec ^{m_1}_{n^{\prime }}\tau _1$. Then, by Axiom 3, there exists a $(m^{\prime \prime },n^{\prime \prime })>(m_1,n^{\prime })$ and an outer extension $\mu (\sigma _1)$ such that $\mu (\sigma _1)\sim ^{m^{\prime \prime }}_{n^{\prime \prime }}I(\tau _1)$. Owing to Lemma 10, we have also that $\lambda (\sigma _{2})\sim ^{m^{\prime }}_{n^{\prime \prime }}I(\tau _{2})$. By the property (i) of the outer extensions, with regard to $\lambda (\sigma _{2})$, we have that $D_{\lambda (\sigma _2)}\cap {\mathbb {N}}_n=D_{\sigma _2}$ and $D_{\lambda (\sigma _2)}\setminus D_{\sigma _2}\subseteq \{n+1,\ldots ,n^{\prime }\}$. By the same token, we also find that $D_{\mu (\sigma _1)}\cap {\mathbb {N}}_n=D_{\sigma _1}$ and $D_{\mu (\sigma _1)}\setminus D_{\sigma _1}\subseteq \{n^{\prime }+1,\ldots ,n^{\prime \prime }\}$. Accordingly, $\mu (\sigma _1)$ is disjoint with $\lambda (\sigma _{2})$ and their union is an extension of $\sigma $, hereafter denoted by $\alpha (\sigma )$. Similarly, it can be seen that $I(\tau _{1})$ is disjoint with $I(\tau _{2})$ and their union is $I(\tau )$. Therefore, by Axiom 2, we derive that $\alpha (\sigma )\sim ^{m^{\prime }+m^{\prime \prime }}_{n^{\prime \prime }}I(\tau )$. Thus, since $(m^{\prime }+m^{\prime \prime },n^{\prime \prime })>(m,n)$, invoking Axiom 3 we conclude that $\sigma \prec ^{m}_{n}\tau $ contradicting the hypothesis. $\square $

Corollary 5

Let $\sigma ,\tau \in {\mathcal {F}}_{m,n}$ be such that $\sigma \prec ^{m}_{n}\tau $. Assume that $\sigma \prec ^{m}_{n^{\prime }}\tau $ for every $n^{\prime }>n$. Then $I(\sigma )\prec ^{m^{\prime }}_{n^{\prime }}I(\tau )$ for every $(m^{\prime },n^{\prime })>(m,n)$.

Proof

According to Corollary 4, it suffices to show that $I(\sigma )\sim ^{m^{\prime }}_{n^{\prime }}I(\tau )$ leads to a contradiction. Consider the compatible pair $(p(I(\sigma )),p(I(\tau )))$ given by the natural partitions of $I(\sigma )$ and $I(\tau )$ associated with $D_{\sigma }$ and $D_{\tau }$, respectively. It is immediate to see that $I_1(\sigma )=\sigma \in {\mathcal {F}}_{m,n^{\prime }}$ and $I_1(\tau )=\tau \in {\mathcal {F}}_{m,n^{\prime }}$. Thus, by assumption, we have that $I_1(\sigma )\prec ^{m}_{n^{\prime }}I_1(\tau )$. Due to Lemma 12, this fact, combined with $I(\sigma )\sim ^{m^{\prime }}_{n^{\prime }}I(\tau )$, implies that $I_2(\sigma )\succ ^{m^{\prime }-m}_{n^{\prime }}I_2(\tau )$, which clearly contradicts Corollary 3 with $k=0$, so showing the validity of the claim. $\square $

Lemma 13

Let $\sigma ,\tau \in {\mathcal {F}}_{m,n}$ be such that $\sigma \in {\mathcal {L}}^0$ and $\tau \notin {\mathcal {L}}^0$. Then $\sigma \prec ^{m}_{n}\tau $.

Proof

Owing to Axiom 1, we can suppose that $m>1$. According to Corollary 2, it suffices to show that $\sigma \sim ^{m}_{n}\tau $ leads to a contradiction. Indeed, let $y_1\in D_{\tau }$ be such that $\delta _{\tau } (y_1)>0$ and let $(p(\sigma ),p(\tau ))$ be the compatible pair given by the natural partitions of $\sigma $ and $\tau $ associated with $\{y_1\}$ and $\{x_1\}$, respectively, where $x_1$ is an arbitrary element of $D_{\sigma }$. By Axiom 1, we get $\sigma _{1}\prec ^{1}_{n}\tau _{1}$, hence by applying Lemma 12 yields $\sigma _{2}\succ ^{m-1}_{n}\tau _{2}$, which clearly contradicts Corollary 2, seeing that $\sigma _{2}\in {\mathcal {L}}^0$. $\square $

Now, we are ready to reverse Lemma 10 and Corollary 4 with a weakened assumption.

Lemma 14

Let $\sigma ,\tau \in {\mathcal {F}}_{m,n}$ be such that $\sigma \preceq ^{m}_{n_1}\tau $ for at least a $n_1>n$. Then $\sigma \preceq ^{m}_{n}\tau $.

Proof

Suppose ab absurdo that $\tau \prec ^{m}_{n}\sigma $. By Lemma 10, this implies that $\tau \preceq ^{m}_{n_1}\sigma $, hence, due to the assumption $\sigma \preceq ^{m}_{n_1}\tau $, we obtain $\sigma \sim ^{m}_{n_1}\tau $, which in its turn, by Lemma 10, leads to

$$\begin{aligned} \sigma \sim ^{m}_{n^{\prime }}\tau \quad {\text {for all }} \ n^{\prime }>n_1. \end{aligned}$$

(21)

Relying upon Axiom 3, $\tau \prec ^{m}_{n}\sigma $ guarantees the existence of a $(m^{\prime },n^{\prime })>(m,n)$ and of an outer extension $\lambda (\tau )$ such that

$$\begin{aligned} \lambda (\tau )\sim ^{m^{\prime }}_{n^{\prime }}I(\sigma ). \end{aligned}$$

(22)

Without any loss of generality, we can suppose $n^{\prime }>n_1$. Consider now the compatible pair $(p(\lambda (\tau )), p(I(\sigma )))$ given by the natural partitions of $\lambda (\tau )$ and $I(\sigma )$ associated with $D_{\tau }$ and $D_{\sigma }$, respectively. It is immediate to see that $\lambda _{1}(\tau )=\tau \in {\mathcal {F}}_{m,n^{\prime }}$ and $I_1(\sigma )=\sigma \in {\mathcal {F}}_{m,n^{\prime }}$. Thus, Eq. (21) holds true: moreover, due to Eq. (22), Lemma 11 applies, with the consequence that $\lambda _{2}(\tau )\sim ^{m^{\prime }-m}_{n^{\prime }}I_2(\sigma )$. However, this contradicts Lemma 13, because $\lambda _{2}(\tau )\notin {\mathcal {L}}^0$, while $I_2(\sigma )\in {\mathcal {L}}^0$, so closing the proof. $\square $

Corollary 6

Let $\sigma ,\tau \in {\mathcal {F}}_{m,n}$. Then $\sigma \preceq ^{m}_{n}\tau $ if, and only if, there exists at least a $(m^{\prime },n^{\prime })>(m,n)$ such that $I(\sigma )\preceq ^{m^{\prime }}_{n^{\prime }}I(\tau )$.

Proof

In view of Corollary 4, it suffices to show sufficiency, so assume that there exists at least a $(m^{\prime },n^{\prime })>(m,n)$ such that $I(\sigma )\preceq ^{m^{\prime }}_{n^{\prime }}I(\tau )$. Consider the compatible pair $(p(I(\sigma )),p(I(\tau )))$ given by the natural partitions of $I(\sigma )$ and $I(\tau )$ associated with $D_{\sigma }$ and $D_{\tau }$, respectively. Observe that $I_2(\sigma )\sim ^{m^{\prime }-m}_{n^{\prime }}I_2(\tau )$ directly follows from Corollary 3 with $k=0$. Consequently, if $I(\sigma )\sim ^{m^{\prime }}_{n^{\prime }}I(\tau )$, then $I_1(\sigma )\sim ^{m}_{n^{\prime }}I_1(\tau )$ by Lemma 11, otherwise $I_1(\sigma )\succ ^{m}_{n^{\prime }}I_1(\tau )$ is ruled out by Axiom 2. Thus, there is at least a $n^{\prime }>n$ such that $I_1(\sigma )\preceq ^{m}_{n^{\prime }}I_1(\tau )$. It is immediate to see that $I_{1}(\sigma )=\sigma \in {\mathcal {F}}_{m,n^{\prime }}$ and $I_1(\tau )=\tau \in {\mathcal {F}}_{m,n^{\prime }}$, hence Lemma 14 applies and the proof is concluded. $\square $

Corollary 7

Let $\lambda ,\mu \in {\mathcal {L}}^1$, with $\lambda \in {\mathcal {F}}_{m_1,n_1}$ and $\mu \in {\mathcal {F}}_{m_2,n_2}$. Suppose that $m_1>m_2$. Then $I(\mu )\prec ^{m_1}_{n}\lambda $ for any sufficiently large n.

Proof

Let D be any subset of $D_{\lambda }$ of cardinality $m_2$. Fixing any sufficiently large n, consider the natural partitions of $I(\mu )\in {\mathcal {F}}_{m_1,n}$ and $\lambda \in {\mathcal {F}}_{m_1,n}$ associated with $D_{\mu }$ and D, respectively. Remark that $I_1(\mu ),\lambda _{1},\lambda _{2}\in {\mathcal {L}}^1$, while $I_2(\mu )\in {\mathcal {L}}^0$. In view of Corollary 3, we know that $I_1(\mu )\sim ^{m_2}_{n}\lambda _{1}$, while $I_2(\mu )\prec ^{m_1-m_2}_{n}\lambda _{2}$ due to Lemma 13. Thus, applying Axiom 2 and Lemma 11 immediately leads to the claim. $\square $

Remark 6

Let $\sigma _j\in {\mathcal {F}}_{1,q}$ for $j\in {\mathbb {N}}_{p^{\prime }}$, with $p^{\prime }\ge 2$. Let $m_1,\ldots ,m_{p^{\prime }}\in {\mathbb {N}}^1$. Fix any $\,n^{\prime }\ge q-p^{\prime }+\sum _{j=1}^{p^{\prime }}m_j$. We claim that it is always possible to construct mutually disjoint simple reproductions $I(\sigma _{j})\in {\mathcal {F}}_{m_j,n^{\prime }}$ for $j=1,\ldots ,p^{\prime }$. To this purpose, given any $j\in {\mathbb {N}}_{p^{\prime }}$, let $D_{\sigma _j}=\{x_j\}$, with $x_j<x_{j+1}$, and let $\varphi _{p^{\prime }}:{\mathbb {N}}_{p^{\prime }}\rightarrow {\mathbb {N}}_{n^{\prime }}$ be the function described as

$$\begin{aligned} \varphi _{p^{\prime }}(j)={\left\{ \begin{array}{ll} q+1,&{} \quad {\text {if }} \ j=1;\\ q+2-j+\sum _{r=1}^{j-1}m_r,&{} \quad {\text {if }} \ j>1. \end{array}\right. } \end{aligned}$$

It is not difficult to check that $\varphi _{p^{\prime }}$ is a strictly increasing mapping. Hence, if we choose as domain of $I(\sigma _{j})$ the set

$$\begin{aligned} D_{I(\sigma _{j})}=\{x_j,\varphi _{p^{\prime }}(j),\varphi _{p^{\prime }}(j)+1,\ldots , \varphi _{p^{\prime }}(j)+m_j-2\}, \end{aligned}$$

as straightforward consequence of the strict increasing monotonicity of $\varphi _{p^{\prime }}$ we derive that $\{D_{I(\sigma _{j})}:j=1,\ldots , p^{\prime }\}$ is made of mutually disjoint sets, which is enough to prove the claim.

Lemma 15

Let $\alpha _{j}\in {\mathcal {F}}_{m_j,n_j}\cap {\mathcal {L}}^k$ for $j=1,\ldots ,p$. Let $n^{\prime }\ge \sum _{j=1}^{p} n_j$. Then there exists a set $\{\lambda _1,\ldots ,\lambda _p\}$ of mutually disjoint partial permutations belonging to ${\mathcal {L}}^k$ such that $\lambda _{j}\sim ^{m_j}_{n^{\prime }}\alpha _j$ for every $j\in {\mathbb {N}}_p$.

Proof

Let $\lambda _1=\alpha _1$: trivially, $\lambda _1\in F_{m_1,n_1}\cap {\mathcal {L}}^k$ and $\lambda _{1}\sim ^{m_1}_{n^{\prime }}\alpha _1$. Let $\lambda _2:D_{\lambda _2}\rightarrow {\mathbb {N}}_{n_1+n_2}$ be defined as $\lambda _2(i)=\alpha _2(i-n_1)+n_1$, where $D_{\lambda _2}=\{x+n_1:x\in D_{\alpha _2}\}$. First of all, $\lambda _2$ is clearly injective and its domain and range are subsets of $\{n_1+1,\ldots ,n_1+n_2\}$. Accordingly, we derive at the same time that $\lambda _2\in F_{m_2,n_1+n_2}$ and that $\lambda _1$ and $\lambda _2$ are disjoint. Moreover, it is not difficult to check that $\delta _{\lambda _2}(x+n_1)=\delta _{\alpha _2}(x)$ for every $x\in D_{\alpha _2}$, hence $\lambda _2\in {\mathcal {L}}^k$, which in its turn implies that $\lambda _{2}\sim ^{m_2}_{n^{\prime }}\alpha _2$ by Corollary 3. Now, if we define $\lambda _3:D_{\lambda _3}\rightarrow {\mathbb {N}}_{n_1+n_2+n_3}$ as $\lambda _3(i)=\alpha _3(i-n_1-n_2)+n_1+n_2$ on the domain $D_{\lambda _3}=\{x+n_1+n_2:x\in D_{\alpha _3}\}$, following a similar line of reasoning as before, we conclude that $\lambda _3$ belongs to $F_{m_3,n_1+n_2+n_3}\cap {\mathcal {L}}^k$, it is disjoint with $\lambda _1$ and $\lambda _2$: further, we have that $\lambda _{3}\sim ^{m_3}_{n^{\prime }}\alpha _3$ again by Corollary 3. Repeating the same argument for a finite number of steps, the claim is shown. $\square $

In the following, we will implicitly make use of the evident property $I(I(\sigma ))=I(\sigma )$, valid for any partial permutation $\sigma $.

Proposition 5

Let $\preceq $ be a total preorder over ${\mathcal {F}}$ satisfying Axioms 1–3. Given any $k\ge 2$ and any $n>1$, let $\,\tau \in {\mathcal {F}}_{1,n}\cap {\mathcal {L}}^k$. Then there exists a $(m^{\prime },n^{\prime })>(1,n)$ and a partial permutation belonging to ${\mathcal {F}}_{m^{\prime },n^{\prime }}\cap {\mathcal {L}}^1$ which is equivalent to $I(\tau )$ with respect to $\sim ^{m^{\prime }}_{n^{\prime }}$.

Proof

We prove the proposition by induction on the index k: for $k=2$, let $D_{\tau }=\{x_1\}$ and consider an arbitrary $\sigma \in {\mathcal {F}}_{1,n}\cap {\mathcal {L}}^1$, with $D_{\sigma }=\{y_1\}$ for some $x_1,y_1\in {\mathbb {N}}_n$. Owing to Axiom 1, we know that $\sigma \prec ^{1}_{n} \tau $, hence, by Axiom 3, there exists an outer extension $\mu (\sigma )$ such that

$$\begin{aligned} \mu (\sigma )\sim ^{p}_{q}I(\tau )\quad {\text {for some }} \ (p,q)>(1,n). \end{aligned}$$

(23)

We assert that $\delta _{\mu (\sigma )} (y)\le 1$ for every $y\in D_{\mu (\sigma )}$. Suppose ab absurdo there exists a $y^{*}\in D_{\mu (\sigma )}$ such that $\delta _{\mu (\sigma )} (y^{*})\ge 2$. Now, let $(p(\mu (\sigma )),p(I(\tau )))$ be the compatible pair given by the natural partitions of $\mu (\sigma )$ and $I(\tau )$ associated with $\{y^{*}\}$ and $\{x_1\}$, respectively. Then, by Axiom 1, we obtain $\mu _{1}(\sigma )\sim ^{1}_{q}I_1(\tau )$ or $\mu _{1}(\sigma )\succ ^{1}_{q}I_1(\tau )$ according as $\delta _{\mu (\sigma )} (y^{*})$ is equal to 2 or greater than 2. In the first case, by applying Lemma 11 yields $\mu _{2}(\sigma )\sim ^{p-1}_{q}I_2(\tau )$, while in the second case $\mu _{2}(\sigma )\prec ^{p-1}_{q}I_2(\tau )$ directly follows from Lemma 12. However, both these conclusions contradict Lemma 13, seeing that $I_2(\tau )\in {\mathcal {L}}^0$, while $\mu _2(\sigma )\notin {\mathcal {L}}^0$ because of $y_1$. By means of this assertion, we have shown that $\mu (\sigma )\in {\mathcal {L}}^0\cup {\mathcal {L}}^1$. If $\mu (\sigma )\in {\mathcal {L}}^1$, the proof of the case $k=2$ is finished. Otherwise, let $p^{\prime }=|N_{\mu (\sigma )}|$: trivially, we have that $1\le p^{\prime }\le p-2$, hence $p-p^{\prime }>1$. In this case, we state that the searched partial permutation is the $(p-p^{\prime })\times q$ partial permutation given by the restriction of $\mu (\sigma )$ to $D_{\mu (\sigma )}\setminus N_{\mu (\sigma )}$. Indeed, by Lemma 11, it is not difficult to check that such partial permutation, clearly belonging to ${\mathcal {L}}^1$, is equivalent to $I(\tau )$ with respect to $\sim ^{p-p^{\prime }}_{q}$, so definitely closing the case $k=2$. Suppose now that the proposition holds when $\tau \in {\mathcal {F}}_{1,n}\cap {\mathcal {L}}^j$ for every $j=2,\ldots ,k$ and let us show that it holds also for $j=k+1$. If we simply mimic the same argument as in the case $k=2$, replacing $\sigma \in {\mathcal {F}}_{1,n}\cap {\mathcal {L}}^1$ with $\sigma \in {\mathcal {F}}_{1,n}\cap {\mathcal {L}}^k$, we can assume without loss of generality that there exists a $p\times q$ partial permutation $\mu (\sigma )\in \bigcup _{r=1}^{k}{\mathcal {L}}^r$ which satisfies Eq. (23). According to Remark 1, we can represent $\mu (\sigma )$ as $\mu (\sigma )=\bigcup _{j=1}^{p} \mu _{j}(\sigma )$ for $j=1,\ldots ,p$. Since obviously $\mu (\sigma )$ does not belong to ${\mathcal {L}}^1$, there exists a (possibly empty) proper subset $J\subsetneq {\mathbb {N}}_p$ such that $\mu _{j}(\sigma )\in {\mathcal {F}}_{1,q}\cap {\mathcal {L}}^1$ if, and only if, $j\in J$ or, equivalently, $\mu _{j}(\sigma )\in {\mathcal {F}}_{1,q}\cap \bigcup _{r=2}^{k}{\mathcal {L}}^r$, if, and only if, $j\in {\mathbb {N}}_p\setminus J$. By means of a suitable rearrangement, we can always suppose that ${\mathbb {N}}_p\setminus J=\{1,\ldots ,p^{\prime }\}$ for some $p^{\prime }$ such that $1\le p^{\prime }\le p$. Given any $j\in {\mathbb {N}}_{p^{\prime }}$, by the induction assumption there exists a pair $(m_j,n_j)>(1,q)$ and a suitable $m_j\times n_j$ partial permutation $\alpha _{j}\in {\mathcal {L}}^1$ such that

$$\begin{aligned} \alpha _{j}\sim ^{m_j}_{n_j}I(\mu _{j}(\sigma )). \end{aligned}$$

(24)

Consider now the set of partial permutations belonging to ${\mathcal {L}}^1$ given by $\{\alpha _j\in {\mathcal {F}}_{m_j,n_j}:j\in {\mathbb {N}}_{p^{\prime }}\} \cup \{\mu _j(\sigma )\in {\mathcal {F}}_{1,q}:j\in J\}$ and fix any sufficiently large $n^{\prime }$: based upon Lemma 15, there exists a set $\{\lambda _1,\ldots ,\lambda _p\}$ of mutually disjoint partial permutations belonging to ${\mathcal {L}}^1$ such that

$$\begin{aligned} \lambda _{j}\sim ^{m_j}_{n^{\prime }}\alpha _j\quad {\text {for }} \ j=1,\ldots ,p^{\prime } \end{aligned}$$

(25)

and

$$\begin{aligned} \lambda _{j}\sim ^{1}_{n^{\prime }}\mu _j(\sigma )\quad {\text {for }} \ j=p^{\prime }+1,\ldots ,p. \end{aligned}$$

(26)

If $p^{\prime }>1$, due to Remark 6, we may assume that the set $\{I(\mu _{j}(\sigma ))\in {\mathcal {F}}_{m_j,n^{\prime }}:j\in {\mathbb {N}}_{p^{\prime }}\}$ is made of mutually disjoint partial permutations. Therefore, whatever is $p^{\prime }$, it is trivial to check that also the set $T(\mu (\sigma ))=\{ I(\mu _{j}(\sigma ))\in {\mathcal {F}}_{m_j,n^{\prime }}:j\in {\mathbb {N}}_{p^{\prime }} \}\cup \{ \mu _j(\sigma )\in {\mathcal {F}}_{1,n^{\prime }}:j\in J\}$ is made of mutually disjoint partial permutations. Well, it is not difficult to see that the union of all mutually disjoint partial permutations of the set $T(\mu (\sigma ))$ coincides with $I(\mu (\sigma ))$. Analogously, the union of all mutually disjoint partial permutations of the set $\{ \lambda _1,\ldots ,\lambda _p\}$, hereafter denoted by $\lambda $, is a $m^{\prime } \times n^{\prime }$ partial permutation belonging to ${\mathcal {L}}^1$, with $m^{\prime }=m_1+\cdots +m_{p^{\prime }}+p-p^{\prime }$. Consequently, by Eqs. (24)–(26), we derive that

$$\begin{aligned} \lambda \sim ^{m^{\prime }}_{n^{\prime }}I(\mu (\sigma )). \end{aligned}$$

(27)

Since it is clear that $(m^{\prime },n^{\prime })>(p,q)>(1,n)$, due to Corollary 4 we obtain that Eq. (23) implies $I(\mu (\sigma ))\sim ^{m^{\prime }}_{n^{\prime }}I(\tau )$, which combined with Eq. (27) leads to $\lambda \sim ^{m^{\prime }}_{n^{\prime }}I(\tau )$, showing that the searched partial permutation is just $\lambda $ and definitely closing the proof. $\square $

Proposition 6

Let $\preceq $ be a total preorder over ${\mathcal {F}}$ satisfying Axioms 1–3. Then there exists a strictly increasing function $f:{\mathbb {N}}^1\rightarrow {\mathbb {N}}^1$ fulfilling the following property:

(A1)

for any $\tau \in {\mathcal {F}}_{1,n}\cap {\mathcal {L}}^k$, with $k\ge 2$, there exists a $n^{\prime }>n$ and a partial permutation $\sigma \in {\mathcal {L}}^1$ such that $\sigma \sim ^{f(k)}_{n^{\prime }}I(\tau )$.

Proof

Let $\tau \in {\mathcal {F}}_{1,n}\cap {\mathcal {L}}^k$, with $k\ge 2$ and $n>1$. Now, set ${\mathcal {U}}(\tau ):=\{(m^{\prime },n^{\prime })>(1,n): \sigma \sim ^{m^{\prime }}_{n^{\prime }}I(\tau )\,$ for some $\sigma \in {\mathcal {L}}^1\}$. By Proposition 5, we know that ${\mathcal {U}}(\tau )$ is not void. Let $m^{*}=\min \{m^{\prime }>1: (m^{\prime },n^{\prime })\in {\mathcal {U}}(\tau )$ for some $n^{\prime }\in {\mathbb {N}}\}$. It is clear that, once we have fixed k, n and $\tau $, the index $m^{*}=m^{*}(k,n,\tau )$ is uniquely determined.

We assert that $m^{*}(k,n,\tau )=m^{*}(k,q,\lambda )$ for any $q>1$ and for any $\lambda \in {\mathcal {F}}_{1,q}\cap {\mathcal {L}}^k$. Suppose without loss of generality that $n\le q$. Since $\tau $ is also a $1\times q$ partial permutation, by Axiom 1 we get $\tau \sim ^{1}_{q}\lambda $ and consequently, taking into account Corollary 4, we find

$$\begin{aligned} I(\tau )\sim ^{m^{\prime }}_{n^{\prime }}I(\lambda ) \end{aligned}$$

(28)

for any $(m^{\prime },n^{\prime })>(1,q)$. This immediately leads to ${\mathcal {U}}(\lambda )\subseteq {\mathcal {U}}(\tau )$, which in turn implies that $m^{*}(k,n,\tau )\le m^{*}(k,q,\lambda )$. Set $m_1:=m^{*}(k,n,\tau )$ for the sake of convenience. Then, there exists at least a $n^{\prime }$ such that $\sigma \sim ^{m_1}_{n^{\prime }}I(\tau )$ for some $\sigma \in {\mathcal {L}}^1$. Thus, fixing any $n^{\prime \prime }>q$ and invoking Lemma 10, we obtain $\sigma \sim ^{m_1}_{n^{\prime \prime }}I(\tau )$, hence, by Eq. (28), we get $\sigma \sim ^{m_1}_{n^{\prime \prime }}I(\lambda )$, which definitely shows the assertion. As a direct consequence of the assertion, we derive that the index $m^{*}$ only depends upon k, hence we set $f(k):=m^{*}(k)$ for every $k\ge 2$. Evidently, f is a well-defined function whose domain and range are ${\mathbb {N}}^1$ (with regard to the range, note that $m^{*}\ge 2$ by definition). Further, condition A1 is clearly satisfied.

With respect to the strict increasing monotonicity of f, suppose by contradiction that $f(k+1)\le f(k)$ for some $k\ge 2$. Now, consider any $\sigma ,\tau \in {\mathcal {F}}_{1,q}$, with $\sigma \in {\mathcal {L}}^k$ and $\tau \in {\mathcal {L}}^{k+1}$. By Axiom 1 we have $\sigma \prec ^{1}_{n}\tau $ for any $n>q$, hence $\sigma $ and $\tau $ verify the assumptions of Corollary 5 with $m=1$ and consequently one gets

$$\begin{aligned} I(\sigma )\prec ^{f(k)}_{n}I(\tau ) \end{aligned}$$

(29)

for any $n>q$. Suppose first that $f(k+1)=f(k)$. Then, by A1, there exists a sufficiently large n such that $\lambda \sim ^{f(k)}_{n}I(\sigma )$ and $\mu \sim ^{f(k)}_{n}I(\tau )$ for some $\lambda ,\mu \in {\mathcal {L}}^1$. Thus, applying Corollary 3 to $\lambda $ and $\mu $, one finds that $I(\sigma )\sim ^{f(k)}_{n}I(\tau )$, so contradicting Eq. (29). Finally, suppose that $f(k+1)<f(k)$. In this case, by A1 there exists a $n^{\prime }>q$ and a $\mu \in {\mathcal {F}}_{f(k+1),n^{\prime }}\cap {\mathcal {L}}^1$ such that $\mu \sim ^{f(k+1)}_{n^{\prime }}I(\tau )$, hence by applying Corollary 4 to $I(\tau )$ and $\mu $ yields

$$\begin{aligned} I(\mu )\sim ^{f(k)}_{n}I(\tau ) \end{aligned}$$

(30)

for any $n>n^{\prime }$. Repeating the above argument, there exists a $\lambda \in {\mathcal {F}}_{f(k),n}\cap {\mathcal {L}}^1$ such that

$$\begin{aligned} \lambda \sim ^{f(k)}_{n}I(\sigma ) \end{aligned}$$

(31)

for a sufficiently large $n>n^{\prime }$. If we consecutively apply Eqs. (29)–(31), we find

$$\begin{aligned} I(\mu )\sim ^{f(k)}_{n} I(\tau )\succ ^{f(k)}_{n}I(\sigma )\sim ^{f(k)}_{n}\lambda . \end{aligned}$$

However, we can also apply Corollary 7 to $\lambda $ and $\mu $, with $m_1=f(k)$ and $m_2=f(k+1)$, so getting $\lambda \succ ^{f(k)}_{n}I(\mu )$, which contradicts the previous equation and concludes the proof. $\square $

Before stating the next results, we shall introduce some new notation. Given any $\sigma \in {\mathcal {F}}_{p,q}$ and any $f\in {\mathcal {V}}$, let $m(\sigma )=S_f(\sigma )+|N_{\sigma }|$. Moreover, let ${\mathcal {L}}^{0,1}={\mathcal {L}}^0\cup {\mathcal {L}}^1$.

It is quite easy to see that, for any $\sigma \in {\mathcal {F}}_{p,q}$ and any $f\in {\mathcal {V}}$, the following property holds:

(S1): $S_f(\sigma )=|D^{\prime }_{\sigma }|$ if, and only if, $\sigma \in {\mathcal {L}}^{0,1}$.

Proposition 7

Let $\preceq $ be a total preorder over ${\mathcal {F}}$ satisfying Axioms 1–3. Then there exists a $f\in {\mathcal {V}}$ fulfilling A1 and the following property:

(A2)

for any $\sigma \in {\mathcal {F}}_{p,q}$ not belonging to ${\mathcal {L}}^1$ and such that $N_{\sigma }=\emptyset $, there exists a partial permutation belonging to ${\mathcal {L}}^1$ which is equivalent to $I(\sigma )$ with respect to $\sim ^{S_f(\sigma )}_{n}$ for any sufficiently large n.

Proof

Observe that by Proposition 6 there exists a strictly increasing function $f:{\mathbb {N}}^1\rightarrow {\mathbb {N}}^1$ satisfying A1: our purpose is to show that the extension of such f to ${\mathcal {V}}$, denoted again as f with a little abuse of notation, verifies A2. Let $\sigma \in {\mathcal {F}}_{p,q}$ be such that $N_{\sigma }=\emptyset $ and suppose that $\sigma $ does not belong to ${\mathcal {L}}^1$: note in passing that by S1 we get $S_f(\sigma )>p$. By Remark 1 we have that $\sigma =\bigcup _{j=1}^{p} \sigma _{j}$, with $\sigma _{j}\in {\mathcal {F}}_{1,q}\cap {\mathcal {L}}^{k_j}$ for $j=1,\ldots ,p\,$ and some $k_1,\ldots ,k_p\in {\mathbb {N}}_0$. Note that the assumption $N_{\sigma }=\emptyset $ entails $k_j>0$ for all j: further, we have that $k_j>1$ at least for a j, as consequence of the fact that $\sigma \notin {\mathcal {L}}^1$. Without any loss of generality, suppose that $k_j>1$ if, and only if, $j\in {\mathbb {N}}_{p^{\prime }}$ for some $p^{\prime }$ such that $1\le p^{\prime }\le p$. By A1, for every $j\le p^{\prime }$ there exists a $n_j>q$ such that

$$\begin{aligned} \alpha _{j}\sim ^{f(k_j)}_{n_j}I(\sigma _{j}) \end{aligned}$$

(32)

for a suitable partial permutation $\alpha _{j}\in {\mathcal {L}}^1$. Consider now the set of partial permutations belonging to ${\mathcal {L}}^1$ given by $\{\alpha _j\in {\mathcal {F}}_{f(k_j),n_j}:j\in {\mathbb {N}}_{p^{\prime }}\} \cup \{\sigma _j\in {\mathcal {F}}_{1,q}:j=p^{\prime }+1,\ldots ,p\}$ and fix any sufficiently large n: based upon Lemma 15, there exists a set $\{\lambda _1,\ldots ,\lambda _p\}$ of mutually disjoint partial permutations belonging to ${\mathcal {L}}^1$ such that

$$\begin{aligned} \lambda _{j}\sim ^{f(k_j)}_{n}\alpha _j\quad {\text {for }} \ j=1,\ldots ,p^{\prime } \end{aligned}$$

(33)

and

$$\begin{aligned} \lambda _{j}\sim ^{1}_{n}\sigma _j\quad {\text {for }} \ j=p^{\prime }+1,\ldots ,p. \end{aligned}$$

(34)

If $p^{\prime }>1$, due to Remark 6, we may assume that the set $\{I(\sigma _{j})\in {\mathcal {F}}_{f(k_j),n}:j\in {\mathbb {N}}_{p^{\prime }}\}$ is made of mutually disjoint partial permutations. Therefore, whatever is $p^{\prime }$, it is trivial to check that also the set $T(\sigma )=\{ I(\sigma _{j})\in {\mathcal {F}}_{f(k_j),n}:j\in {\mathbb {N}}_{p^{\prime }} \}\cup \{ \sigma _j\in {\mathcal {F}}_{1,n}:j= p^{\prime }+1,\ldots ,p\}$ is made of mutually disjoint partial permutations. It is not difficult to see that the union of all mutually disjoint partial permutations of the set $T(\sigma )$ coincides with $I(\sigma )$. Analogously, the union of all mutually disjoint partial permutations of the set $\{ \lambda _1,\ldots ,\lambda _p\}$, hereafter denoted by $\lambda $, is a $m \times n$ partial permutation belonging to ${\mathcal {L}}^1$, with $m=\sum _{j=1}^{p^{\prime }} f(k_j)+p-p^{\prime }$. Consequently, by Eqs. (32)–(34), we derive that $\lambda \sim ^{m}_{n}I(\sigma )$. Eventually, the claim easily follows from the evident fact that $m=S_f(\sigma )$, by definition of ${\mathcal {L}}^k$. $\square $

Proposition 8

Let $\preceq $ be a total preorder over ${\mathcal {F}}$ satisfying Axioms 1–3. Then, for any $(p,q)\in {\mathcal {A}}$ and any $\sigma ,\tau \in {\mathcal {F}}_{p,q}\,\cap \,{\mathcal {L}}^{0,1}$, we have that $\sigma \preceq ^{p}_{q}\tau $ if, and only if, $D^{\prime }_{\sigma }\subseteq D^{\prime }_{\tau }$.

Proof

Let $\sigma ,\tau \in {\mathcal {F}}_{p,q}\,\cap \,{\mathcal {L}}^{0,1}$ and assume first that $\sigma \preceq ^{p}_{q}\tau $. Suppose ab absurdo that $D^{\prime }_{\tau }$ is a proper subset of $D^{\prime }_{\sigma }$. First of all, note that $D^{\prime }_{\tau }$ cannot be empty, otherwise we would find $\tau \in {\mathcal {L}}^0,\,\sigma \notin {\mathcal {L}}^0$ and $\sigma \preceq ^{p}_{q}\tau $, so contradicting Lemma 13. Thus, $|D^{\prime }_{\tau }|=r$, with $1\le r<p$ : consequently, there exists a proper subset $D^{*}$ of $D^{\prime }_{\sigma }$ of cardinality r. Consider now the compatible pair $(p(\sigma ),p(\tau ))$ given by the natural partitions of $\sigma $ and $\tau $ associated with $D^{*}$ and $D^{\prime }_{\tau }$, respectively. Since $\sigma _{1}$ and $\tau _{1}$ trivially belong to ${\mathcal {L}}^1$, by Corollary 3 we get

$$\begin{aligned} \sigma _{1}\sim ^{r}_{q}\tau _{1}. \end{aligned}$$

(35)

Furthermore, $D^{*}$ being a proper subset of $D^{\prime }_{\sigma }$, one finds that $\sigma _{2}\notin {\mathcal {L}}^0$, while $\tau _{2}\in {\mathcal {L}}^0$, thus, invoking Lemma 13, we obtain

$$\begin{aligned} \sigma _{2}\succ ^{p-r}_{q}\tau _{2}. \end{aligned}$$

(36)

Accordingly, owing to Axiom 2 and the fact that $\sigma \preceq ^{p}_{q}\tau $, Eqs. (35) and (36) would immediately lead to $\sigma \sim ^{p}_{q}\tau $, but this contradicts Lemma 11, so closing the first part of the proof. Conversely, if $D^{\prime }_{\sigma }\subseteq D^{\prime }_{\tau }$, it is really elementary to see that there exists an injection $s:D_{\sigma }\rightarrow D_{\tau }$ such that $\delta _{\sigma }(x)\le \delta _{\tau }(s(x))$ for every $x\in D_{\sigma }$, hence, by Lemma 9, we get the desired claim. $\square $

Corollary 8

Let $\preceq $ be a total preorder over ${\mathcal {F}}$ satisfying Axioms 1–3 and let $f\in {\mathcal {V}}$. Then, given any $(p,q)\in {\mathcal {A}}$ and any $\sigma ,\tau \in {\mathcal {F}}_{p,q}\,\cap \,{\mathcal {L}}^{0,1}$, we have that $\sigma \preceq ^{p}_{q}\tau $ if, and only if, $S_f(\sigma )\le S_f(\tau )$.

Proof

Let $\sigma ,\tau \in {\mathcal {F}}_{p,q}\,\cap \,{\mathcal {L}}^{0,1}$: due to the previous proposition, we derive that $\sigma \preceq ^{p}_{q}\tau $ if, and only if, $|D^{\prime }_{\sigma }|\le |D^{\prime }_{\tau }|$. Now, it suffices to apply S1 and the proof is concluded. $\square $

Lemma 16

Let $\sigma ,\tau \in {\mathcal {F}}_{p,q}$, with $p=m(\sigma )$. Suppose that $\tau \in {\mathcal {L}}^{0,1}$ and $|N_{\tau }|=|N_{\sigma }|$. Then $S_f(\tau )=S_f(\sigma )$ for any $f\in {\mathcal {V}}$.

Proof

By S1, we have that $S_f(\tau )=|D^{\prime }_{\tau }|$: further, since $D_{\tau }$, whose cardinality is equal to $p=m(\sigma )$, is the disjoint union of $D^{\prime }_{\tau }$ and $N_{\tau }$, it directly follows that $|D^{\prime }_{\tau }|= m(\sigma )-|N_{\tau }|$. Finally, by employing the assumption $|N_{\tau }|=|N_{\sigma }|$ yields $S_f(\tau )= m(\sigma )- |N_{\sigma }|=S_f(\sigma )$. $\square $

Corollary 9

Let $\preceq $ be a total preorder over ${\mathcal {F}}$ satisfying Axioms 1–3. Then, there exists a $f\in {\mathcal {V}}$ satisfying A1-A2 and the following property:

(A3)

for any $\sigma \in {\mathcal {F}}_{p,q}$ not belonging to ${\mathcal {L}}^{0,1}$, there exists a $\lambda \in {\mathcal {L}}^{0,1}$ such that $\lambda \sim ^{m(\sigma )}_{n} I(\sigma )$ for any sufficiently large n and $|N_{\lambda }|=|N_{\sigma }|$.

Proof

Observe that by Proposition 7 there holds a $f\in {\mathcal {V}}$ verifying A1-A2: our purpose is to show that such f also fulfills A3. Let $\sigma \in {\mathcal {F}}_{p,q}$ and suppose that $\sigma $ does not belong to ${\mathcal {L}}^{0,1}$. Remark that, in the trivial case $|N_{\sigma }|=0$, A3 reduces to A2, so we can suppose that $|N_{\sigma }|=r$, with $1\le r<p$, hence $|D^{\prime }_{\sigma }|=p-r>0$: note in passing that $|N_{\sigma }|>0$ implies $m(\sigma )>S_f(\sigma )$. Let $p(\sigma )$ be the natural partition associated with $D^{\prime }_{\sigma }$: obviously, $\sigma _{1}\notin {\mathcal {L}}^1$ and $|N_{\sigma _{1}}|=0$, thus Proposition 7 applies. Accordingly, by A2, observing that $S_f(\sigma )=S_f(\sigma _{1})$, there exists a $\mu \in {\mathcal {L}}^1$ such that

$$\begin{aligned} \mu \sim ^{S_f(\sigma )}_{n^{\prime }}I(\sigma _{1}), \end{aligned}$$

for a sufficiently large $n^{\prime }$. By Corollary 4, the previous equation goes to

$$\begin{aligned} I(\mu )\sim ^{m(\sigma )}_{n}I(\sigma _{1}) \end{aligned}$$

(37)

for any sufficiently large n. Consider now $I(\sigma ),I(\sigma _{1})\in {\mathcal {F}}_{m(\sigma ),n}$ and let $(p(I(\sigma )),p(I(\sigma _{1})))$ be the compatible pair given by the natural partitions of $I(\sigma )$ and $I(\sigma _{1})$ associated with $D^{\prime }_{\sigma }$ and $D_{\sigma _{1}}$. Since clearly $D^{\prime }_{\sigma }=D_{\sigma _{1}}$, we immediately get $I_1(\sigma )\sim ^{p-r}_{n}I_1(\sigma _{1})$, while $I_2(\sigma )\sim ^{m(\sigma )-p+r}_{n}I_2(\sigma _{1})$ follows from the fact that $I_2(\sigma ),I_2(\sigma _{1})\in {\mathcal {L}}^0$. Hence, by Axiom 2, we deduce that

$$\begin{aligned} I(\sigma _{1})\sim ^{m(\sigma )}_{n}I(\sigma ) \end{aligned}$$

for every sufficiently large n. If we set $\lambda :=I(\mu )\in {\mathcal {F}}_{m(\sigma ),n}$, it is trivial to see that $\lambda \in {\mathcal {L}}^{0,1}$ and $|N_{\lambda }|=|N_{\sigma }|$. Finally, the last equation, combined with Eq. (37), concludes the proof. $\square $

E.2 Proof of Theorem 1

Proof

We leave it to the reader to prove the sufficiency.

To show the necessity, recall that by Corollary 9 there is a $f\in {\mathcal {V}}$ satisfying conditions A1-A3: our purpose is to show that the restricted ordering $\preceq ^{p}_{q}$ is defined via an equation of type (1) with $M=S_f$ for every $(p,q)\in {\mathcal {A}}$. In other terms, we have to prove that by means of the function f assured by Corollary 9 the following equation is satisfied for any $(p,q)\in {\mathcal {A}}$:

$$\begin{aligned} \sigma \preceq \tau \quad {\text {if, and only if,}} \ S_f(\sigma )\le S_f(\tau ) \end{aligned}$$

(38)

for all $\sigma ,\tau \in {\mathcal {F}}_{p,q}$. Owing to Corollary 8, we can skip the case when both $\sigma $ and $\tau $ belong to ${\mathcal {L}}^{0,1}$. Thus, without any loss if generality, we can suppose that $\sigma ,\tau \notin {\mathcal {L}}^{0,1}$ (the remaining cases may be treated in complete analogy). Note that by Corollary 6 Eq. (38) is equivalent to

$$\begin{aligned} I(\sigma )\preceq ^{m}_{n} I(\tau )\quad {\text {if, and only if,}} \ S_f(\sigma )\le S_f(\tau ) \end{aligned}$$

(39)

for at least a $(m,n)>(p,q)$. Owing to A3, there exist $\lambda ,\mu \in {\mathcal {L}}^{0,1}$ and a sufficiently large n such that

$$\begin{aligned} \lambda \sim ^{m(\sigma )}_{n} I(\sigma ) \end{aligned}$$

(40)

and

$$\begin{aligned} \mu \sim ^{m(\tau )}_{n} I(\tau ), \end{aligned}$$

(41)

with $|N_{\lambda }|=|N_{\sigma }|$ and $|N_{\mu }|=|N_{\tau }|$. Remark that $S_f(\lambda )=S_f(\sigma )$ and $S_f(\mu )=S_f(\tau )$ due to Lemma 16. Assume first that $m(\sigma )=m(\tau )$: by Corollary 8 and the previous remark, we find

$$\begin{aligned} \lambda \preceq ^{m(\sigma )}_{n}\mu \quad {\text {if, and only if,}} \ S_f(\sigma )\le S_f(\tau ). \end{aligned}$$

This equation, together with Eqs. (40) and (41), directly leads to Eq. (39), with $m=m(\sigma )$, recalling that $m(\sigma )>p$ as a consequence of S1, so closing this first sub-case. Finally, assume without loss of generality that $m(\sigma )> m(\tau )$. Observe that, fixing any sufficiently large $n^{\prime }>n$, by Lemma 10 and Eq. (40) it follows that $\lambda \sim ^{m(\sigma )}_{n^{\prime }}I(\sigma )$. At the same time, from Corollary 4 and Eq. (41) it follows that $I(\mu )\sim ^{m(\sigma )}_{n^{\prime }}I(\tau )$. Now, since obviously $\lambda $ and $I(\mu )$ still belong to ${\mathcal {L}}^{0,1}$, we can proceed in the same way as in the first sub-case, by replacing $\mu $ with $I(\mu )$, so definitely concluding the proof. $\square $

E.3 Proof of Theorem 2

Proof

By Theorem 1, we know that $\preceq $ satisfies Axioms 1-3 if, and only if, there holds a $f\in {\mathcal {V}}$ such that Eq. (38) is satisfied for any $(p,q)\in {\mathcal {A}}$. Eventually, as a straightforward consequence of Proposition 2, it is immediate to see that Axiom 4 is verified if, and only if, f is strictly convex. $\square $

Appendix F: Proof of Theorem 3

Proof

We leave it to the reader to prove the sufficiency.

To show the necessity, recall that by Theorem 2 there is a strictly convex $f\in {\mathcal {V}}$ such that Eq. (38) is satisfied for any $(p,q)\in {\mathcal {A}}$. Our purpose now is to prove that the ordering relation $\preceq $ is defined via Eq. (1) with $M=M_f$ by means of the same f. Given any $(p,q),(p^{\prime },q^{\prime })\in {\mathcal {A}}$, let $\sigma \in {\mathcal {F}}_{p,q}$ and $\tau \in {\mathcal {F}}_{p^{\prime },q^{\prime }}$: by Axiom 5, we derive that $\sigma \preceq \tau $ if, and only if, there exists a pair $(p^{*},q^{*})>(\max \{p,p^{\prime }\},\max \{q,q^{\prime }\})$ and two suitable reproductions $\lambda (\sigma )$ and $\mu (\tau )$ such that

$$\begin{aligned} \lambda (\sigma )\preceq ^{p^{*}}_{q^{*}}\mu (\tau ). \end{aligned}$$

Owing to Eq. (38), the last equation may be rewritten as

$$\begin{aligned} S_f(\lambda (\sigma ))\le S_f(\mu (\tau )). \end{aligned}$$

Consequently, if we recall the definition of reproduction, we conclude that

$$\begin{aligned} \sigma \preceq \tau \quad {\text {if, and only if,}} \ k^{\prime }S_f(\sigma )\le k S_f(\tau )\quad {\text {for some }} \ k,k^{\prime }\in {\mathbb {N}}. \end{aligned}$$

(42)

Now, our purpose is to show that Eq. (42) leads to the fulfillment of Eq. (1) with $M=M_f$. Assume first that $M_f(\sigma )\le M_f(\tau )$ or, equivalently,

$$\begin{aligned} \frac{S_f(\sigma )}{S^M_f(p,q)}\le \frac{S_f(\tau )}{S^M_f(p^{\prime },q^{\prime })}. \end{aligned}$$

Then the right hand-side of Eq. (42) is clearly satisfied with $k^{\prime }=S^M_f(p^{\prime },q^{\prime })$ and $k=S^M_f(p,q)$, hence $\sigma \preceq \tau $. Conversely, assume that $\sigma \preceq \tau $: by Eq. (42), this is equivalent to saying that there exist $k,k^{\prime }\in {\mathbb {N}}$ such that

$$\begin{aligned} k^{\prime }S_f(\sigma )\le k S_f(\tau ). \end{aligned}$$

By an elementary algebraic manipulation, the previous equation is tantamount to

$$\begin{aligned} c^{\prime }\cdot S^M_f(p^{\prime },q^{\prime })S_f(\sigma )\le c\cdot S^M_f(p,q)S_f(\tau ), \end{aligned}$$

where $c^{\prime }:=k^{\prime }/S^M_f(p^{\prime },q^{\prime })$ and $c:=k/S^M_f(p,q)$. The proof is evidently finished if we show that $c^{\prime }=c$. Reasoning by contradiction, suppose for instance that $c^{\prime }<c$ (the remaining case is analogous): then, we would have

$$\begin{aligned} k^{\prime } S^M_f(p,q)< k S^M_f(p^{\prime },q^{\prime }). \end{aligned}$$

However, on account of Eq. (42), the previous equation would imply that $\sigma ^{*}_{p,q}\prec \sigma ^{*}_{p^{\prime },q^{\prime }}$ which clearly contradicts Axiom 6, so concluding the proof.

Appendix G: Proof of Lemma 1

Proof

As a consequence of Remark 5, we derive that the claim is equivalent to requiring that the following condition is satisfied for any $n\in {\mathbb {N}}^1$ and for any $\sigma ,\tau \in {\mathcal {P}}_n$:

$$\begin{aligned} |Q_{\sigma }|<|Q_{\tau }|\quad {\text {if, and only if,}} \ |Q_{\lambda (\sigma )}|=|Q_{I(\tau )}| \end{aligned}$$

for some outer extension $\lambda (\sigma )$. Suppose first that $|Q_{\sigma }|<|Q_{\tau }|$ and let $r=|Q_{\tau }|-|Q_{\sigma }|$. Let $\lambda (\sigma )\in {\mathcal {P}}_{2r}$ be the outer extension of $\sigma $ such that the restriction of $\lambda (\sigma )$ to ${\{n+i,n+i+1\}}$ coincides with $\nu (n+i,n+i+1)$ for every $i=1,3,5,\ldots ,2r-1$. Since $\lambda $ is an extension of $\sigma $ made of r inversions, from K1 and K2 it follows that $|Q_{\lambda (\sigma )}|=|Q_{\sigma }|+r$ and thus $|Q_{\lambda (\sigma )}|=|Q_{\tau }|$, so concluding the first part, seing evidently that $|Q_{\tau }|=|Q_{I(\tau )}|$. The converse is trivial since $|Q_{\lambda (\sigma )}|>|Q_{\sigma }|$ for every outer extension $\lambda (\sigma )$. $\square $

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Roberto Ghiselli Ricci
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1.Dipartimento di Economia e ManagementUniversitá di FerraraFerraraItaly

An axiomatic characterization of a class of rank mobility measures

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