Characterization of monotonic rules in minimum cost spanning tree problems

Abstract

We provide, in minimum cost spanning tree problems, a general framework to identify the family of rules satisfying monotonicity over cost and population. We also prove that the set of allocations induced by the family coincides with the so-called irreducible core, that results from decreasing the cost of the arcs as much as possible, without reducing the minimal cost.

Appendix

We prove the results of the paper.

1.1 Proof of Theorem 3.1

Let $\left( N_{0},C\right)$ be a mcstp. We first prove that $IC\left( N_{0},C\right) \subset AM\left( N_{0},C\right)$.

Given $\pi \in \Pi _{U}$ we define the rule $f^{\pi }$ such that for each $mcstp$ $\left( N_{0}^{\prime },C^{\prime }\right)$ and each $i\in N^{\prime }$,

$$\begin{aligned} f_{i}^{\pi }\left( N_{0}^{\prime },C^{\prime }\right) =c_{C^{\prime *}}\left( Pre\left( i,\pi _{N^{\prime }}\right) \cup \left\{ i\right\} \right) -c_{C^{\prime *}}\left( Pre\left( i,\pi _{N^{\prime }}\right) \right) . \end{aligned}$$

This rule $f^{\pi }$ is well defined because

$$\begin{aligned} \sum _{i\in N^{\prime }}f_{i}^{\pi }\left( N_{0}^{\prime },C^{\prime }\right) =c_{C^{\prime *}}\left( N^{\prime }\right) =m\left( N_{0}^{\prime },C^{\prime *}\right) =m\left( N_{0}^{\prime },C^{\prime }\right) . \end{aligned}$$

We now prove that for each $\pi \in \Pi _{U}$, $f^{\pi }$ satisfies ${ PM}$ and $\textit{CM}$.

Given a mcstp $\left( N_{0}^{\prime },C^{\prime }\right)$ and $S\subset N^{\prime }$ the next expression appears, for instance, in Lemma 0 $\left( b\right)$ of Bergantiños and Gómez-Rúa (2010),

$$\begin{aligned} c_{C^{\prime *}}\left( S\cup \left\{ i\right\} \right) -c_{C^{\prime *}}\left( S\right) =\min _{k\in S\cup \left\{ 0\right\} }\left\{ c_{ik}^{\prime *}\right\} . \end{aligned}$$

Then,

$$\begin{aligned} f_{i}^{\pi }\left( N_{0}^{\prime },C^{\prime }\right) =\min _{k\in Pre\left( i,\pi _{N^{\prime }}\right) \cup \left\{ 0\right\} }\left\{ c_{ik}^{\prime *}\right\} \end{aligned}$$

Let $S\subset N^{\prime }$, and $i\in S$. Since $Pre\left( i,\pi _{S}\right) \subset Pre\left( i,\pi _{N^{\prime }}\right)$,

$$\begin{aligned} \min _{k\in Pre\left( i,\pi _{N^{\prime }}\right) \cup \left\{ 0\right\} }\left\{ c_{ik}^{\prime *}\right\} \le \min _{k\in Pre\left( i,\pi _{S}\right) \cup \left\{ 0\right\} }\left\{ c_{ik}^{\prime *}\right\} . \end{aligned}$$

Under (1), $C_{S}^{\prime *}\le \left( C_{S}^{\prime }\right) ^{*}$. Then,

$$\begin{aligned} \min _{k\in Pre\left( i,\pi _{S}\right) \cup \left\{ 0\right\} }\left\{ c_{ik}^{\prime *}\right\} \le \min _{k\in Pre\left( i,\pi _{S}\right) \cup \left\{ 0\right\} }\left\{ \left( C_{S}^{\prime }\right) _{ik}^{*}\right\} =f_{i}^{\pi }\left( S_{0},C_{S}^{\prime }\right) . \end{aligned}$$

Hence, $f^{\pi }$ satisfies ${ PM}$.

Let $\left( N_{0}^{\prime },C^{\prime }\right)$ and $\left( N_{0}^{\prime },C^{\prime \prime }\right)$ be such that $C^{\prime }\le C^{\prime \prime }$. Bergantiños and Vidal-Puga (2007) prove in Lemma 4.2 that $C^{\prime *}\le C^{\prime \prime *}$. Then, for each $i\in N^{\prime }$,

$$\begin{aligned} f_{i}^{\pi }\left( N_{0}^{\prime },C^{\prime }\right) =\min _{k\in Pre\left( i,\pi _{N^{\prime }}\right) \cup \left\{ 0\right\} }\left\{ c_{ik}^{\prime *}\right\} \le \min _{k\in Pre\left( i,\pi _{N^{\prime }}\right) \cup \left\{ 0\right\} }\left\{ c_{ik}^{\prime \prime *}\right\} =f_{i}^{\pi }\left( N_{0},C^{\prime \prime }\right) . \end{aligned}$$

Hence, $f^{\pi }$ satisfies $\textit{CM}$.

It is well known that $\left( N,c_{C^{*}}\right)$ is a concave game. See, for instance Proposition 3.3 $\left( c\right)$ in Bergantiños and Vidal-Puga (2007). Then, $core\left( N,c_{C^{*}}\right)$ is the convex hull of the family of vector of marginal contributions. Namely, given $x=\left( x_{i}\right) _{i\in N}\in \textit{IC}\left( N_{0},C\right)$, there exists $w=\left( w_{\pi }\right) _{\pi \in \Pi _{N}}\in \Delta \left( \Pi _{N}\right)$ such that for each $i\in N$,

$$\begin{aligned} x_{i}=\sum _{\pi \in \Pi _{N}}w_{\pi }\left[ c_{C^{*}}\left( Pre\left( i,\pi \right) \cup \left\{ i\right\} \right) -c_{C^{*}}\left( Pre\left( i,\pi \right) \right) \right] . \end{aligned}$$

Given $\pi \in \Pi _{N}$ we define $\pi ^{U}\in \Pi _{U}$ such that we first order agents in $N$ as in $\pi$ and later agents in $U\setminus N$ as in the identical order. Formally, let $\pi ^{id}\in \Pi _{U}$ be such that $\pi ^{id}\left( i\right) =i$ for all $i\in U$. We now take $\pi ^{U}=\left( \pi ,\pi _{U\setminus N}^{id}\right)$.

We define the rule $f^{w}=\sum _{\pi \in \Pi _{N}}w_{\pi }f^{\pi ^{U}}$. For each $i\in N$ ,

$$\begin{aligned} f_{i}^{w}\left( N_{0},C\right)&= \sum _{\pi \in \Pi _{N}}w_{\pi }f_{i}^{\pi ^{U}}\left( N_{0},C\right) \\&= \sum _{\pi \in \Pi _{N}}w_{\pi }\left[ c_{C^{*}}\left( Pre\left( i,\pi _{N}^{U}\right) \cup \left\{ i\right\} \right) -c_{C^{*}}\left( Pre\left( i,\pi _{N}^{U}\right) \right) \right] \\&= \sum _{\pi \in \Pi _{N}}w_{\pi }\Big [ c_{C^{*}}\left( Pre\left( i,\pi \right) \cup \left\{ i\right\} \right) -c_{C^{*}}\left( Pre\left( i,\pi \right) \right) \Big ] \\&= x_{i}. \end{aligned}$$

It only remains to prove that $f^{w}$ satisfies ${ PM}$ and $\textit{CM}$. Let $\left( N_{0}^{\prime },C^{\prime }\right)$ and $\left( N_{0}^{\prime },C^{\prime \prime }\right)$ be such that $C^{\prime }\le C^{\prime \prime }$. Since for each $\pi \in \Pi _{N}$, $f^{\pi ^{U}}$ satisfies $\textit{CM}$, we have that $f^{\pi ^{U}}\left( N_{0}^{\prime },C^{\prime }\right) \le f^{\pi ^{U}}\left( N_{0}^{\prime },C^{\prime \prime }\right)$. Now

$$\begin{aligned} f^{w}\left( N_{0}^{\prime },C^{\prime }\right) =\sum _{\pi \in \Pi _{N}}w_{\pi }f^{\pi ^{U}}\left( N_{0}^{\prime },C^{\prime }\right) \le \sum _{\pi \in \Pi _{N}}w_{\pi }f^{\pi ^{U}}\left( N_{0}^{\prime },C^{\prime \prime }\right) =f^{w}\left( N_{0}^{\prime },C^{\prime \prime }\right) . \end{aligned}$$

Similarly, we can prove that $f^{w}$ satisfies ${ PM}$. Then, $IC\left( N_{0},C\right) \subset AM\left( N_{0},C\right)$.

Notice that we have proved that any allocation in the irreducible core could be obtained as an allocation induced by the convex hull of the “order induced” rules $\left\{ f^{\pi }\right\} _{\pi \in \Pi _{U}}$.

We now prove that $AM\left( N_{0},C\right) \subset \textit{IC}\left( N_{0},C\right)$. Let $x\in AM\left( N_{0},C\right)$. There exists a rule $f$ satisfying $\textit{CM}$ and ${ PM}$ such that $x=f\left( N_{0},C\right)$. Since $\textit{CM}$ implies $\textit{RED}$

$$\begin{aligned} f\left( N_{0},C\right) =f\left( N_{0},C^{*}\right) . \end{aligned}$$

Bergantiños and Vidal-Puga (2007) prove that if $f$ satisfies ${ PM}$, then it satisfies core selection, namely $f\left( N_{0},C\right) \in core\left( N,c_{C}\right)$. Therefore,

$$\begin{aligned} f\left( N_{0},C^{*}\right) \in core\left( N,c_{C^{*}}\right) =\textit{IC}\left( N_{0},C\right) . \end{aligned}$$

$\square$

We now provide a rule $f$ that does not satisfy ${ PM}$ (a similar example could be provided for $\textit{CM}$). Let $\pi ^{id}\in \Pi _{U}$ as in the proof of Theorem 3.1. We define $\pi ^{\prime }\in \Pi _{U}$ such that $\pi ^{\prime }\left( i\right) =\left| U\right| -i$ for all $i\in U$. Let $f^{\pi ^{id}}$ and $f^{\pi ^{\prime }}$ be as in the proof of Theorem 3.1. We define the rule $f$ as follows:

$$\begin{aligned} f\left( N_{0},C\right) =\left\{ \begin{array}{l@{\quad }l} f^{\pi ^{id}}\left( N_{0},C\right) &{} \quad \hbox {if}\,\left| N\right| \,\hbox {is even}\\ f^{\pi ^{\prime }}\left( N_{0},C\right) &{} \quad \hbox {if}\,\left| N\right| \, \hbox {is odd}.\\ \end{array} \right. \end{aligned}$$

Let $\left( N_{0},C\right)$ be a mcstp where $\left| N\right|$ is even (the case where $\left| N\right|$ is odd is similar and we omit it). Then, $f\left( N_{0},C\right) =f^{\pi ^{id}}\left( N_{0},C\right)$. Since $f^{\pi ^{id}}$ satisfies ${ PM}$ and $\textit{CM}$ (we have proved it in the proof of Theorem 3.1), $f^{\pi ^{id}}\left( N_{0},C\right) \in \textit{IC}\left( N_{0},C\right)$. Nevertheless, $f$ does not satisfy ${ PM}$. Let $N=\left\{ 1,2,3\right\}$, $c_{01}=c_{02}=c_{03}=9$, and $c_{12}=c_{13}=c_{23}=5$. Then, $f_{3}\left( \left\{ 1,3\right\} _{0},C_{\left\{ 13\right\} }\right) =5$ but $f_{3}\left( N_{0},C\right) =9$.

1.2 Proof of Theorem 4.1

Before proving it we need some previous results.

Some comments about neighborhoods.

• 1. From the definition of the irreducible problem $\left( N_{0},C^{*}\right)$ we deduce that neighborhoods of $\left( N_{0},C\right)$ and $\left( N_{0},C^{*}\right)$ coincide.

• 2. In general, $\left( C^{*}\right) _{S}\ne \left( C_{S}\right) ^{*}$. Take for example $N=\left\{ 1,2,3\right\}$, $c_{12}=c_{13}=1$, $c_{23}=2$ and $S=\left\{ 2,3\right\}$. Then, $c_{23}^{*}=1$ and hence $C^{\prime }=\left( C^{*}\right) _{S}$ satisfies $c_{23}^{\prime }=1$ whereas $C^{\prime \prime }=\left( C_{S}\right) ^{*}$ satisfies $c_{23}^{\prime \prime }=2$. Later on (Lemma 7.1) we will prove that the equality holds when $S$ is a neighborhood.

The next lemma gives some results about neighborhoods.

Lemma 7.1

• 1. $S\subset N$ is a neighborhood in $\left( N_{0},C\right)$ if and only if $S$ is a neighborhood in $\left( N_{0},C^{*}\right)$. Besides, $\left( C_{S}\right) ^{*}=\left( C^{*}\right) _{S}$ and

  $$\begin{aligned} \delta _{S}=\min _{i\in S,j\in N_{0}\setminus S}c_{ij}^{*}-\max _{i,j\in S}c_{ij}^{*}. \end{aligned}$$

• 2. If $S$ is a neighborhood in $\left( N_{0},C\right)$ and $i\in S$, then

  $$\begin{aligned} S=\left\{ j\in N:c_{ij}^{*}<\min _{k\in S,l\in N_{0}\setminus S}c_{kl}^{*}\right\} . \end{aligned}$$

• 3. If $S,S^{\prime }$ are two neighborhoods in $\left( N_{0},C\right)$, $\left( N_{0},C\right)$ is an irreducible problem, and $S\cap S^{\prime }\ne \emptyset$, then either $S\subset S^{\prime }$ or $S^{\prime }\subset S$.

• 4. For each $i\in N$, there exists a unique family of subsets of $N$, $S_{1},S_{2},\dots ,S_{q}$ with $q\ge 0$ ¹ such that $\left\{ S_{1},\dots ,S_{q}\right\}$ is the set of neighborhoods in $\left( N_{0},C\right)$ that contain $i$, and $S_{1}\subset S_{2}\subset ...\subset S_{q}$.

• 5. There exist no neighborhood in $\left( N_{0},C\right)$ if and only if $\left\{ \left\{ 0,i\right\} \right\} _{i\in N}$ is a mcst in $\left( N_{0},C\right)$.

Under part 1 of Lemma 7.1, for each neighborhood $S\subset N$, we have $\left( C^{*}\right) _{S}=\left( C_{S}\right) ^{*}$. We denote this matrix as $C_{S}^{*}$.

Proof of Lemma 7.1 (1) Assume that $S$ is a neighborhood in $\left( N_{0},C\right)$. Because of the definition of the irreducible problem $\left( N_{0},C^{*}\right)$, we have that $\min _{i\in S,j\in N_{0}\setminus S}\{c_{ij}\}=\min _{i\in S,j\in N_{0}\setminus S}\{c_{ij}^{*}\}$. Let $C^{1}=\left( C_{S}\right) ^{*}$ and let $C^{2}=\left( C^{*}\right) _{S}$. Let $\tau \left( S\right)$ be a mcst in $\left( S,C_{S}\right)$. Since $S$ is a neighborhood in $\left( N_{0},C\right)$, $\tau \left( S\right)$ is also a mcst in $\left( S,C^{1}\right)$ and $\left( S,C^{2}\right)$. Given $i,j\in S$, let $\tau _{ij}$ be the unique simple path in $\tau \left( S\right)$ from $i$ to $j$. Then,

$$\begin{aligned} c_{ij}^{1}=\max _{\left\{ i_{l},i_{l+1}\right\} \subset \tau _{ij}}c_{i_{l}i_{l+1}}=c_{ij}^{2}. \end{aligned}$$

Because of the definition of $C^{*}$ we have that $\max _{\left( i,j\right) \in \tau \left( S\right) }\{c_{ij}\}=\max _{\left( i,j\right) \in \tau \left( S\right) }\{c_{ij}^{*}\}=\max _{\left( i,j\right) \in S}\{c_{ij}^{*}\}$. Now,

$$\begin{aligned} \delta _{S}^{*}&= \min _{i\in S,j\in N_{0}\setminus S}c_{ij}^{*}-\max _{\left\{ i,j\right\} \in \tau \left( S\right) }c_{ij}^{*}\\&= \min _{i\in S,j\in N_{0}\setminus S}c_{ij}-\max _{\left\{ i,j\right\} \in \tau \left( S\right) }c_{ij}=\delta _{S} \end{aligned}$$

which means that $S$ is a neighborhood in $\left( N_{0},C^{*}\right)$.

Similarly we can prove that if $S$ is a neighborhood in $\left( N_{0},C^{*}\right)$, then $S$ is a neighborhood in $\left( N_{0},C\right)$.

• (2) ”$\supset$ ” Let $j\in N$ be such that $c_{ij}^{*}<\min _{k\in S,l\in N_{0}\setminus S}\{c_{kl}^{*}\}$. If $j\notin S$, then $c_{ij}^{*}\ge \min _{k\in S,l\in N_{0}\setminus S}\{c_{kl}^{*}\}$, which is a contradiction. Hence, $j\in S$. ”$\subset$ ”: Let $j\in N$ be such that $c_{ij}^{*}\ge \min _{k\in S,l\in N_{0}\setminus S}\{c_{kl}^{*}\}$. If $j\in S$, then

  $$\begin{aligned} \delta _{S}=\min _{k\in S,l\in N_{0}\setminus S}c_{kl}^{*}-\max _{k,l\in S}c_{kl}^{*}\le c_{ij}^{*}-c_{ij}^{*}=0 \end{aligned}$$

  which cannot be true because $S$ is a neighborhood. Hence, $j\notin S$.

• (3) Let $i\in S\cap S^{\prime }$. If $\min _{k\in S,l\in N_{0}\setminus S}\{c_{kl}^{*}\}\le \min _{k\in S^{\prime },l\in N_{0}\setminus S^{\prime }}\{c_{kl}^{*}\}$ then it follows from part 2 of Lemma 7.1 that $S\subset S^{\prime }$. If $\min _{k\in S^{\prime },l\in N_{0}\setminus S^{\prime }}\{c_{kl}^{*}\}\le \min _{k\in S,l\in N_{0}\setminus S}\{c_{kl}^{*}\}$ then it follows from part 2 of Lemma 7.1 that $S^{\prime }\subset S$.

• (4) It follows from part 3 of Lemma 7.1.

• (5) Assume that $\left\{ \left( 0,i\right) \right\} _{i\in N}$ is not a mcst. Let $\left\{ k,l\right\} \subset N$ be such that $c_{kl}=\min _{i,j\in N}\{c_{ij}\}$. Thus, $c_{kl} <\min _{i\in N}\{c_{0i}\}$. Then, $S=\left\{ k\right\} \cup \Big \{ i\in N:\max _{\left\{ i_{l}i_{l+1}\right\} \subset \tau _{ik}}\{c_{i_{l}i_{l+1}}\}\le c_{kl}\Big \}$ is a neighborhood in $\left( N_{0},C\right)$.

Assume $\left\{ \left( 0,i\right) \right\} _{i\in N}$ is a mcst. Then, given any $S\subset N$, we have $\min _{i\in S,j\in N_{0}\setminus S}\{c_{ij}\}=\min _{i\in S}\{c_{0i}\}$ and $\max _{\left\{ i,j\right\} \in \tau \left( S\right) }\{c_{ij}\}\ge \min _{i\in S}\{c_{0i}\}$. Hence

$$\begin{aligned} \delta _{S}=\min _{i\in S,j\in N_{0}\setminus S}c_{ij}-\max _{\left\{ i,j\right\} \in \tau \left( S\right) }c_{ij}\le 0 \end{aligned}$$

and $S$ is not a neighborhood. $\square$

In Lemma 7.2 below we prove that each $f^{e}$ is a rule, namely, $\sum _{i\in N}f_{i}^{e}\left( N_{0},C\right) =m\left( N_{0},C\right)$.

Lemma 7.2

For each extra-costs function $e$, $f^{e}$ is a rule.

Proof of Lemma 7.2 Let $\left( N_{0},C\right)$ be a mcstp. Then,

$$\begin{aligned} \sum _{i\in N}f_{i}^{e}\left( N_{0},C\right)&= \sum _{i\in N}c_{0i}^{*}-\sum _{i\in N}\sum _{S\text { }\in Ne\left( N_{0},C\right) ,i\in S}\left( \delta _{S}-e_{i}\left( C_{S}^{*},\delta _{S}\right) \right) \\&= \sum _{i\in N}c_{0i}^{*}-\sum _{S\text { }\in Ne\left( N_{0},C\right) }\left( \sum _{i\in S}\left( \delta _{S}-e_{i}\left( C_{S}^{*},\delta _{S}\right) \right) \right) \\&= \sum _{i\in N}c_{0i}^{*}-\sum _{S\text { }\in Ne\left( N_{0},C\right) }\left( \left| S\right| -1\right) \delta _{S}. \end{aligned}$$

Thus, it is enough to prove that for each mcstp $\left( N_{0},C\right)$,

$$\begin{aligned} m\left( N_{0},C\right) +\sum _{S\text { }\in Ne\left( N_{0},C\right) }\left( \left| S\right| -1\right) \delta _{S}=\sum _{i\in N}c_{0i}^{*}. \end{aligned}$$

Assume first there exists no neighborhood. Under part 5 of Lemma 7.1, $\left\{ \left\{ 0,i\right\} \right\} _{i\in N}$ is a $mcst$ in $\left( N_{0},C\right)$. Hence, $\left\{ \left\{ 0,i\right\} \right\} _{i\in N}$ is also a mcst in $\left( N_{0},C^{*}\right)$ and the result is easily checked.

Assume now that there are exactly $k>0$ neighborhoods and the result is true when there exists less than $k$ neighborhoods. Let $S^{\prime }$ be a minimal neighborhood (there is no neighborhood $S$ such that $S\varsubsetneq S^{\prime }$). Let $\tau \left( S^{\prime }\right)$ denote a mcst in $S^{\prime }$. Since $S^{\prime }$ is minimal, there exists $\alpha \ge 0$ such that $c_{ij}=\alpha$ for all $\left( i,j\right) \in \tau \left( S^{\prime }\right)$.

Let $t$ be a mcst in $\left( N_{0},C\right)$. We define $C^{\prime }$ as $c_{ij}^{\prime }=\alpha +\delta _{S^{\prime }}$ if $\left\{ i,j\right\} \subset S^{\prime }$ and $c_{ij}^{\prime }=c_{ij}$ otherwise. Thus,

• $t$ is also a mcst in $\left( N_{0},C^{\prime }\right)$;

• $c_{0i}^{\prime *}=c_{0i}^{*}$ for all $i\in N;$

• $m\left( N_{0},C^{\prime }\right) =m\left( N_{0},C\right) +\left( \left| S^{\prime }\right| -1\right) \delta _{S^{\prime }}$; and

• $\left\{ S:S\text { is a neighborhood in }\left( N_{0},C^{\prime }\right) \right\}$ coincides with $\{ S:S\text { is a neighborhood in} ( N_{0},C) \} \setminus \left\{ S^{\prime }\right\}$.

Now, applying the induction hypothesis, we have

$$\begin{aligned}&\qquad m\left( N_{0},C\right) +\sum _{S\text { }\in Ne\left( N_{0},C\right) }\left( \left| S\right| -1\right) \delta _{S} \\&\quad =m\left( N_{0},C^{\prime }\right) -\left( \left| S^{\prime }\right| -1\right) \delta _{S^{\prime }}+\sum _{S\text { }\in Ne\left( N_{0},C\right) }\left( \left| S\right| -1\right) \delta _{S} \\&\quad =m\left( N_{0},C^{\prime }\right) +\sum _{S\text { }\in Ne\left( N_{0},C^{\prime }\right) }\left( \left| S\right| -1\right) \delta _{S} \\&\quad =\sum _{i\in N}c_{0i}^{\prime *}=\sum _{i\in N}c_{0i}^{*}. \end{aligned}$$

$\square$

We firs give the idea of the proof of Theorem 2. Later, a formal proof. It is easy to see that each $f^{e}$ satisfies separability and reductionism. Given a rule $f$ satisfying both properties, we need to find $e$ such that $f=f^{e}$. Given an irreducible problem $\left( N,C\right)$ and $x\in \mathbb {R}_{+}$ we define the mcstp $\left( N_{0},C^{\prime }\right)$ and $\left( N_{0},C^{\prime \prime }\right)$ where

$$\begin{aligned} c_{ij}^{\prime }&= \left\{ \begin{array}{l@{\quad }l} c_{ij} &{} \hbox {if }\,0\notin \left\{ i,j\right\} \\ \max \left( N,C\right) &{} \hbox {if }\,0\in \left\{ i,j\right\} \\ \end{array} \right. \quad \text { and } \\ c_{ij}^{\prime \prime }&= \left\{ \begin{array}{l@{\quad }l} c_{ij} &{}\hbox {if }\,0\notin \left\{ i,j\right\} \\ \max \left( N,C\right) +x &{} \hbox {if } 0\in \left\{ i,j\right\} .\\ \end{array} \right. \end{aligned}$$

Now, for each $i\in N$ we define $e_{i}\left( (N,C), x \right) =f_{i}\left( N_{0},C^{\prime \prime }\right) -f_{i}\left( N_{0},C^{\prime }\right)$. We then prove that $e$ is an extra-costs function. Finally, using an induction argument over the number of neighborhoods, we prove that $f=f^{e}$.

Let $e$ be any extra-costs function and $f^{e}$ be the associated rule. It is obvious that $f^{e}$ satisfies $\textit{RED}$.

In order to prove that $f^{e}$ also satisfies ${ SEP}$, let $S\subset N$ such that $m\left( N_{0},C\right) =m\left( S_{0},C_{S_{0}}\right) +m\left( \left( N\setminus S\right) _{0},C_{\left( N\setminus S\right) _{0}}\right)$. Then,

$$\begin{aligned} Ne\left( N_{0},C\right) =Ne\left( S_{0},C_{S_{0}}\right) \cup Ne\left( \left( N\setminus S\right) _{0},C_{\left( N\setminus S\right) _{0}}\right) . \end{aligned}$$

Hence, $f_{i}^{e}\left( N_{0},C\right) =f_{i}^{e}\left( S_{0},C_{S_{0}}\right)$ and this proves that $f$ satisfies ${ SEP}$.

We now prove that if $f$ satisfies ${ SEP}$ and $\textit{RED}$, then $f=f^{e}$ for some extra-costs function $e$. Let $f$ be such a rule.

Given $\left( N,C^{*}\right) \in \mathcal {C}^{*}$ and $a\in \mathbb {R}_{+}$, we define $\left( N_{0},C^{*\left( a\right) }\right)$ as the $mcstp$ given by $c_{ij}^{*\left( a\right) }=c_{ij}^{*}$ for all $i,j\in N$ and $c_{0i}^{*\left( a\right) }=a$ for all $i\in N$. Notice that $\left( N,C^{*\left( a\right) }\right)$ is an irreducible problem when $a\ge \max \left( N,C^{*}\right)$.

For all $\left( N,C^{*}\right) \in \mathcal {C}^{*}$, $x\in \mathbb {R} _{+}$, and $i\in N$ we define

$$\begin{aligned} e_{i}\left( C^{*},x\right) =f_{i}\left( N_{0},C^{*\left( \max \left( N,C^{*}\right) +x\right) }\right) -f_{i}\left( N_{0},C^{*\left( \max \left( N,C^{*}\right) \right) }\right) . \end{aligned}$$

Given $i\in U\setminus N$ we define $e_{i}\left( C^{*},x\right) =0$.

We first prove that $e$ is an extra-costs function. By definition, $e_{i}\left( C^{*},x\right) =0$ for all $\left( N,C^{*}\right) \in \mathcal {C}^{*}$, $x\in \mathbb {R}_{+}$, $i\in U\setminus N$. Besides,

$$\begin{aligned} \sum _{i\in N}e_{i}\left( C^{*},x\right)&= m\left( N_{0},C^{*\left( \max \left( N,C^{*}\right) +x\right) }\right) -m\left( N_{0},C^{*\left( \max \left( N,C^{*}\right) \right) }\right) \\&= m\left( N,C^{*}\right) +\max \left( N,C^{*}\right) +x-m\left( N,C^{*}\right) -\max \left( N,C^{*}\right) \\&= x. \end{aligned}$$

Hence, $e$ is an extra-costs function.

We need to prove that $f=f^{e}$. It is obvious that for any mcstp $\left( N_{0},C\right)$, $f^{e}\left( N_{0},C\right) =f^{e}\left( N_{0},C^{*}\right)$. Since $f$ satisfies $\textit{RED}$, $f\left( N_{0},C\right) =f\left( N_{0},C^{*}\right)$. Thus, it is enough to prove that $f^{e}\left( N_{0},C^{*}\right) =f\left( N_{0},C^{*}\right)$.

We proceed by induction on the number of neighborhoods $Ne\left( N_{0},C\right)$. Assume $\left| Ne\left( N_{0},C\right) \right| =0$.

Under part 5 of Lemma 7.1, $\left\{ \left( 0,i\right) \right\} _{i\in N}$ is a mcst in $\left( N_{0},C\right)$. Since $f$ satisfies ${ SEP}$, $f_{i}\left( N_{0},C\right) =f_{i}\left( \left\{ i\right\} _{0},C_{\left\{ i\right\} _{0}}\right) =c_{0i}$. Besides, since $\left\{ \left( 0,i\right) \right\} _{i\in N}$ is a mcst in $\left( N_{0},C\right)$ , we have $c_{0i}=c_{0i}^{*}$ for all $i\in N$ and hence $f^{e}\left( N_{0},C\right) =f\left( N_{0},C\right)$.

Assume now the result is true for mcstp with less than $\left| Ne\left( N_{0},C\right) \right|$ neighborhoods.

Assume first that $\max \left( N_{0},C^{*}\right) \ge \max _{i\in N}\{c_{0i}^{*}\}$. By Proposition 3.1 in Bergantiños and Vidal-Puga (2007) we can find a mcstp $t$ in $\left( N_{0},C^{*}\right)$ and $\left( i,j\right) \in t$ such that $c_{ij}^{*}=\max \left( N_{0},C^{*}\right)$ and $i$ is in the unique simple path in $t$ from $j$ to $0$. Let $S$ be the set of agents $k$ satisfying that $j$ is in the unique simple path in $t$ from $k$ to $0$. Then, $S\ne \varnothing$, $S\ne N$ and $m\left( N_{0},C^{*}\right) =m\left( S_{0},C_{S_{0}}^{*}\right) +m\left( \left( N\setminus S\right) _{0},C_{\left( N\setminus S\right) _{0}}\right)$. Under ${ SEP}$, $f_{i}\left( N_{0},C^{*}\right) =f_{i}\left( S_{0},C_{S_{0}}^{*}\right)$ for all $i\in S$ and $f_{i}\left( N_{0},C^{*}\right) =f_{i}\left( \left( N\setminus S\right) _{0},C_{\left( N\setminus S\right) _{0}}\right)$ for all $i\in N\setminus S$. Repeating this argument we can find a partition $\left\{ S_{1},\dots ,S_{p}\right\}$ of $N$ satisfying that for each $k=1,\ldots ,p$ $\max \left( S_{k},C_{S_{k}}^{*}\right) <\max _{i\in S_{k}}\{c_{0i}^{*}\}$ and $f_{i}\left( N_{0},C^{*}\right) =f_{i}\left( \left( S_{k}\right) _{0},C_{\left( S_{k}\right) _{0}}^{*}\right)$ for each $i\in S_{k}$.

Hence, we can assume that $\max \left( N_{0},C^{*}\right) <\max _{i\in N}\{c_{0i}^{*}\}$. Since $\left( N_{0},C^{*}\right)$ is irreducible, $\max _{i\in N}\{c_{0i}^{*}\}=c_{0i}^{*}$ for all $i\in N$. Hence, $N$ is a neighborhood in $\left( N_{0},C^{*}\right)$ and $\delta _{N}=\max _{i\in N}\{c_{0i}^{*}\}-\max \left( N_{0},C^{*}\right)$. Now, for each $i\in N$,

$$\begin{aligned} f_{i}\left( N_{0},C^{*}\right)&= f_{i}\left( N_{0},C^{*\left( \max \left( N_{0},C^{*}\right) +\delta _{N}\right) }\right) \\&= e_{i}\left( C^{*},\delta _{N}\right) +f_{i}\left( N_{0},C^{*\left( \max C^{*}\right) }\right) . \end{aligned}$$

We define $C^{\prime }=C^{*\left( \max \left( N_{0},C^{*}\right) \right) }$. Then $\left( N_{0},C^{\prime }\right)$ is an irreducible problem satisfying $Ne\left( N_{0},C^{*}\right) =Ne\left( N_{0},C^{\prime }\right) \cup \left\{ N\right\}$. For each $S\in Ne\left( N_{0},C^{\prime }\right)$, $\delta _{S}=\delta _{S}^{\prime }$, and $c_{0i}^{\prime *}=c_{0i}^{*}-\delta _{N}$. Hence, applying the induction hypothesis, for each $i\in N$,

$$\begin{aligned} f_{i}\left( N_{0},C^{*}\right)&= e_{i}\left( C^{*},\delta _{N}\right) +f_{i}\left( N_{0},C^{\prime }\right) \\&= e_{i}\left( C^{*},\delta _{N}\right) +c_{0i}^{\prime *}+\sum _{S\in Ne\left( N_{0},C^{\prime }\right) }\left( e_{i}\left( C_{S}^{*},\delta _{S}\right) -\delta _{S}\right) \\&= e_{i}\left( C^{*},\delta _{N}\right) +c_{0i}^{*}-\delta _{N}+\sum _{S\in Ne\left( N_{0},C^{\prime }\right) }\left( e_{i}\left( C_{S}^{*},\delta _{S}\right) -\delta _{S}\right) \\&= c_{0i}^{*}+\sum _{S\in Ne\left( N_{0},C^{*}\right) }\left( e_{i}\left( C_{S}^{*},\delta _{S}\right) -\delta _{S}\right) \\&= f_{i}^{e}\left( N_{0},C^{*}\right) . \end{aligned}$$

$\square$

1.3 Proof of Theorem 5.1

We start the proof with the following lemma.

Lemma 7.3

$\left( i\right)$ Given $\left( N^{\prime },C^{\prime }\right) ,\left( N^{\prime \prime },C^{\prime \prime }\right) \in \mathcal {C} ^{*}$ and $a\in \mathbb {R}_{+}$ with $N^{\prime }\cap N^{\prime \prime }=\emptyset$ and $a\ge \max \left( N^{\prime \prime },C^{\prime \prime }\right) -\max \left( N^{\prime },C^{\prime }\right)$, then $C^{\prime }\oplus _{a}C^{\prime \prime }\in \mathcal {C}^{*}$.

$\left( ii\right)$ Given a disjoint sequence $\left\{ \left( N^{\gamma },C^{\gamma }\right) \right\} _{\gamma =1}^{\Gamma }\subset \mathcal {C} ^{*}$, $\Gamma >1$, $a\in \mathbb {R}_{+}^{\Gamma }$ with $a_{\gamma }\ge \max \left( N^{\gamma +1},C^{\gamma +1}\right) -\max \left( N^{\gamma },C^{\gamma }\right)$ for all $\gamma =1,\dots ,\Gamma -1$, and $y\in \left[ 0,a_{2}\right]$, then $C^{\gamma }\left( a\right) \in \mathcal {C}^{*}$ and $C^{\gamma }\left( a^{\prime }\right) \in \mathcal {C}^{*}$ for all $\gamma =1,\dots ,\Gamma$, where $a^{\prime }=\left( a_{1}+y,a_{2}-y,a_{3},\dots ,a_{\Gamma }\right)$.

Proof of Lemma 7.3. $\left( i\right)$ Let $C=C^{\prime }\oplus _{a}C^{\prime \prime }$. Then, $a+\max C^{\prime }=\max C$. Hence, we can find a mcst $t$ in $\left( N^{\prime }\cup N^{\prime \prime },C\right)$ and $\left( N^{\prime }\cup N^{\prime \prime },C^{*}\right)$ such that $t=t^{1}\cup t^{2}\cup \left\{ \left( k^{1},k^{2}\right) \right\}$ where $t^{1}$ is a mcst in $\left( N^{\prime },C^{\prime }\right)$, $t^{2}$ is a mcst in $\left( N^{\prime \prime },C^{\prime \prime }\right)$, $k^{1}\in N^{1}$ and $k^{2}\in N^{2}$. Since $c_{k^{1}k^{2}}=\max C\ge c_{ij}$ for all $\left( i,j\right) \in t^{1}\cup t^{2}$ we can deduce, using the definition of irreducible matrix, that $C=C^{*}$.

$\left( ii\right)$ We assume $\gamma >1$, since the case $\gamma =1$ is trivial. We proceed by induction on $\Gamma$. For $\Gamma =2$, the result follows from $\left( i\right)$ because $a_{1}^{\prime }=a_{1}+y\ge a_{1}\ge \max C^{2}-\max C^{1}$. Assume the result is true for sequences with less than $\Gamma$ mcstp’s, $\Gamma \ge 3$. Under the induction hypothesis, we have $C^{\gamma }\left( b\right)$, $C^{\gamma }\left( b^{\prime }\right) \in \mathcal {C}^{*}$ where $\gamma =1,\dots ,\Gamma -1$, $b=\left( a_{1},\dots ,a_{\Gamma -1}\right)$ and $b^{\prime }=\left( a_{1}+y,a_{2}-y,a_{3},\dots ,a_{\Gamma -1}\right)$. Now, it is clear that $C^{\gamma }\left( a\right) =C^{\gamma }\left( b\right)$ and $C^{\gamma }\left( a^{\prime }\right) =C^{\gamma }\left( b^{\prime }\right)$ for all $\gamma =1,\dots ,\Gamma -1$. Hence, the result holds for any $\gamma <\Gamma$. Assume now $\gamma =\Gamma$. We have

$$\begin{aligned} C^{\Gamma }\left( a\right) \overset{\left( 2\right) }{=} C^{\Gamma -1}\left( a\right) \oplus _{a_{\Gamma -1}}C^{\Gamma }\left( a\right) \overset{\left( i\right) }{\in }\mathcal {C}^{*} \end{aligned}$$

and

$$\begin{aligned} C^{\Gamma }\left( a^{\prime }\right) \overset{\left( 2 \right) }{=}C^{\Gamma -1}\left( a^{\prime }\right) \oplus _{a_{\Gamma -1}^{\prime }}C^{\Gamma }\left( a^{\prime }\right) . \end{aligned}$$

In order to apply $\left( i\right)$ to this last expression (so that $C^{\Gamma }\left( a^{\prime }\right) \in \mathcal {C}^{*}$) we have to prove that

$$\begin{aligned} a_{\Gamma -1}^{\prime }\ge \max C^{\Gamma }\left( a^{\prime }\right) -\max C^{\Gamma -1}\left( a^{\prime }\right) . \end{aligned}$$

(4)

By definition, $\max C^{\gamma }\left( a^{\prime }\right) =\max C^{\gamma }\left( a\right)$ for all $\gamma \ne 2$, whereas $\max C^{2}\left( a^{\prime }\right) =\max C^{2}\left( a\right) +y$. Hence, for $\Gamma >3$,

$$\begin{aligned} \max C^{\Gamma }\left( a^{\prime }\right) -\max C^{\Gamma -1}\left( a^{\prime }\right) =\max C^{\Gamma }\left( a\right) -\max C^{\Gamma -1}\left( a\right) \le a_{\Gamma -1}=a_{\Gamma -1}^{\prime } \end{aligned}$$

and for $\Gamma =3$,

$$\begin{aligned} \max C^{3}\left( a^{\prime }\right) -\max C^{2}\left( a^{\prime }\right) =\max C^{3}\left( a\right) -\max C^{2}\left( a\right) -y\le a_{2}-y=a_{2}^{\prime }. \end{aligned}$$

$\square$

We now prove that if $f=f^{e}$ with $e$ satisfying $\textit{ANM}$, then $f$ satisfies $\textit{CM}$ and ${ PM}$.

Following Norde et al. (2004), for each $N_{0}$ we define the set $\Sigma _{N_{0}}$ of linear orders on the edges as the set of all bijections $\sigma :\left\{ 1,\dots ,\left( {\begin{array}{c}n+1\\ 2\end{array}}\right) \right\} \rightarrow \left\{ \left\{ i,j\right\} :i,j\in N_{0}\right\}$. For each mcstp $\left( N_{0},C\right)$, there exists at least one linear order $\sigma \in \Sigma _{N_{0}}$ such that $c_{\sigma \left( 1\right) }\le c_{\sigma \left( 2\right) }\le \cdots \le c_{\sigma \left( \left( {\begin{array}{c}n+1\\ 2\end{array}}\right) \right) }$. For any $\sigma \in \Sigma _{N_{0}}$, we define the set

$$\begin{aligned} K^{\sigma }=\left\{ \left( N_{0},C\right) :c_{\sigma \left( k\right) }\le c_{\sigma \left( k+1\right) }\ \text {for all }k=1,2,\ldots ,\left( {\begin{array}{c}n+1\\ 2\end{array}}\right) \right\} , \end{aligned}$$

which we call the Kruskal cone with respect to $\sigma$. One can easily see that $\bigcup _{\sigma \in \Sigma _{N_{0}}}K^{\sigma }$ coincides with the set of all mcstp where the set of agents is $N$.

We say that a non-empty set $S\subset N$ is a quasi-neighborhood in $\left( N_{0},C\right)$ if $\delta _{S}\ge 0$. Let $qNe\left( N_{0},C\right) =\left\{ S\subset N,S\ne \emptyset :\delta _{S}\ge 0\right\}$ denote the set of quasi-neighborhoods in $\left( N_{0},C\right)$ . Clearly, $Ne\left( N_{0},C\right) \subset qNe\left( N_{0},C\right)$.

We now prove that $f$ satisfies $\textit{CM}$. It is enough to prove that $f\left( N_{0},C\right) \le f\left( N_{0},C^{\prime }\right)$ when there exists $\left\{ k,l\right\} \subset N_{0}$ such that $c_{kl}^{\prime }>c_{kl}$ and $c_{ij}^{\prime }=c_{ij}$ otherwise. Let $\left( k,l\right)$, $\left( N_{0},C\right)$ and $\left( N_{0},C^{\prime }\right)$ be defined in this way.

For any $x\in \left[ 0,1\right]$, the mcstp $\left( N_{0},C^{x}\right)$ defined as $c_{ij}^{x}=\left( 1-x\right) c_{ij}+xc_{ij}^{\prime }$ satisfies $c_{kl}^{\prime }\ge c_{kl}^{x}\ge c_{kl}$ and $c_{ij}^{x}=c_{ij}$ otherwise. Since $\Sigma _{N_{0}}$ is a finite set, there exist a sequence $\left\{ x^{1},x^{2},...x^{p}\right\} \subset \left[ 0,1\right]$ with $x^{1}=0$ and $x^{p}=1$ such that, for all $r$, we have $x^{r}<x^{r+1}$ and $\left( N_{0},C^{x^{r}}\right)$ and $\left( N_{0},C^{x^{r+1}}\right)$ belong to the same Kruskal cone.

Hence, it is enough to prove that $f\left( N_{0},C\right) \le f\left( N_{0},C^{\prime }\right)$ when both $\left( N_{0},C\right)$ and $\left( N_{0},C^{\prime }\right)$ belong to the same Kruskal cone. An immediate consequence is that there exists a common mcst $t$ in both $\left( N_{0},C\right)$ and $\left( N_{0},C^{\prime }\right)$.

Since $f$ satisfies $\textit{RED}$, $f\left( N_{0},C\right) =f\left( N_{0},C^{*}\right)$. If $\left\{ k,l\right\} \notin t$, then $C^{*}=C^{\prime *}$. Thus

$$\begin{aligned} f\left( N_{0},C\right) =f\left( N_{0},C^{*}\right) =f\left( N_{0},C^{\prime *}\right) =f\left( N_{0},C^{\prime }\right) . \end{aligned}$$

Hence, we assume $\left\{ k,l\right\} \in t$. This implies $c_{kl}=c_{kl}^{*}$ and $c_{kl}^{\prime }=c_{kl}^{\prime *}$. Let $\alpha =c_{kl}^{\prime *}-c_{kl}^{*}>0$.

Another consequence of $\left( N_{0},C\right)$, $\left( N_{0},C^{\prime }\right)$ being in the same Kruskal cone is that, for any $S\subset N$, $\left| S\right| >1$, there exist a mcst $\tau \left( S\right)$ in $\left( N_{0},C\right)$ and $\left( N_{0},C^{\prime }\right)$, $i^{1},i^{2},j^{2}\in S$, $j^{1}\in N_{0}\setminus S$ with $\left\{ i^{2},j^{2}\right\} \in \tau \left( S\right)$ such that

$$\begin{aligned} \delta _{S}&= \min _{i^{\prime }\in S,j^{\prime }\in N_{0}\setminus S}c_{i^{\prime }j^{\prime }}-\max _{\left\{ i^{\prime },j^{\prime }\right\} \in \tau \left( S\right) }c_{i^{\prime }j^{\prime }}=c_{i^{1}j^{1}}-c_{i^{2}j^{2}}\quad \text { and } \\ \delta _{S}^{\prime }&= \min _{i^{\prime }\in S,j^{\prime }\in N_{0}\setminus S}c_{i^{\prime }j^{\prime }}^{\prime }-\max _{\left\{ i^{\prime },j^{\prime }\right\} \in \tau \left( S\right) }c_{i^{\prime }j^{\prime }}^{\prime }=c_{i^{1}j^{1}}^{\prime }-c_{i^{2}j^{2}}^{\prime }. \end{aligned}$$

Thus $\delta _{S}$ and $\delta _{S}^{\prime }$ cannot have opposite sign. Namely, $\delta _{S}>0$ implies $\delta _{S}^{\prime }\ge 0$. Thus, $Ne\left( N_{0},C\right) \subset qNe\left( N_{0},C^{\prime }\right)$ and $Ne\left( N_{0},C^{\prime }\right) \subset qNe\left( N_{0},C\right)$.

Given any $X\subset 2^{N}$ with $Ne\left( N_{0},C\right) \subset X\subset qNe\left( N_{0},C\right)$ and $i\in N$ we have

$$\begin{aligned} f_{i}\left( N_{0},C\right) =c_{0i}^{*}-\sum _{i\in S\in X}\left( \delta _{S}-e_{i}\left( C_{S}^{*},\delta _{S}\right) \right) . \end{aligned}$$

(5)

The reason is that for any $S\in qNe\left( N_{0},C\right) \setminus Ne\left( N_{0},C\right)$, $\delta _{S}=0$ and hence $\delta _{S}-e_{i}\left( C_{S}^{*},\delta _{S}\right) =0-e_{i}\left( C_{S}^{*},0\right) =0$.

We define $X=Ne\left( N_{0},C\right) \cup Ne\left( N_{0},C^{\prime }\right)$ . Clearly, $Ne\left( N_{0},C\right) \subset X\subset qNe\left( N_{0},C\right)$ and $Ne\left( N_{0},C^{\prime }\right) \subset X\subset qNe\left( N_{0},C^{\prime }\right)$.

Fix $i\in N$. We need to prove that $f_{i}\left( N_{0},C\right) \le f_{i}\left( N_{0},C^{\prime }\right)$. Under (5), we have

$$\begin{aligned} f_{i}\left( N_{0},C\right)&= c_{0i}^{*}-\sum _{i\in S\in X}\left( \delta _{S}-e_{i}\left( C_{S}^{*},\delta _{S}\right) \right) \\ f_{i}\left( N_{0},C^{\prime }\right)&= c_{0i}^{\prime *}-\sum _{i\in S\in X}\left( \delta _{S}^{\prime }-e_{i}\left( C_{S}^{\prime *},\delta _{S}^{\prime }\right) \right) . \end{aligned}$$

We have seen above that

$$\begin{aligned} \delta _{S}=c_{i^{1}j^{1}}-c_{i^{2}j^{2}}\quad \text { and }\quad \delta _{S}^{\prime }=c_{i^{1}j^{1}}^{\prime }-c_{i^{2}j^{2}}^{\prime } \end{aligned}$$

for some $i^{1},i^{2},j^{2}\in S$, $j^{1}\in N_{0}\setminus S$ with $\left\{ i^{2},j^{2}\right\} \in t_{S}$.

By hypothesis, $c_{jj^{\prime }}=c_{jj^{\prime }}^{\prime }$ for all $\left\{ j,j^{\prime }\right\} \ne \left\{ k,l\right\}$. Hence, $\delta _{S}=\delta _{S}^{\prime }$ unless $\left\{ i^{1},j^{1}\right\} =\left\{ k,l\right\}$ or $\left\{ i^{2},j^{2}\right\} =\left\{ k,l\right\}$.

Given $S\in X$ and $\delta _{S}\ne \delta _{S}^{\prime }$ we study both cases:

• 1. If $\left\{ i^{1},j^{1}\right\} =\left\{ k,l\right\}$, then $\delta _{S}^{\prime }=\delta _{S}+\alpha$. Besides, there can be at most two such $S$. One of them contains node $k$ (if any) and the other contains node $l$ (if any). Assume, on the contrary, that there exist two $S^{\prime }\in X,S\ne S^{\prime }$ with $k\in S\cap S^{\prime }$ (the case for $l\in S$ is analogous). Hence,

  $$\begin{aligned} c_{kl}^{\prime }=c_{kl}^{\prime *}=\min _{i^{\prime }\in S,j^{\prime }\in N_{0}\setminus S}c_{i^{\prime }j^{\prime }}^{\prime *}=\min _{i^{\prime }\in S^{\prime },j^{\prime }\in N_{0}\setminus S^{\prime }}c_{i^{\prime }j^{\prime }}^{\prime *}. \end{aligned}$$

  Since $k\in S\cap S^{\prime }$, under part 4 of Lemma 7.1, $S\varsubsetneq S^{\prime }$ or $S^{\prime }\varsubsetneq S$. Assume w.l.o.g. $S\varsubsetneq S^{\prime }$. Then,

  $$\begin{aligned} c_{kl}^{\prime *}&= \min _{i^{\prime }\in S,j^{\prime }\in N_{0}\setminus S}c_{i^{\prime }j^{\prime }}^{\prime *}\le \min _{i^{\prime }\in S,j^{\prime }\in S^{\prime }\setminus S}c_{i^{\prime }j^{\prime }}^{\prime *} \\&\le \max _{i^{\prime },j^{\prime }\in S^{\prime }}c_{i^{\prime }j^{\prime }}^{\prime *}\le \min _{i^{\prime }\in S^{\prime },j^{\prime }\in N_{0}\setminus S^{\prime }}c_{i^{\prime }j^{\prime }}^{\prime *}=c_{kl}^{\prime *} \end{aligned}$$

  which implies that no inequality is strict. In particular, $\max _{i^{\prime },j^{\prime }\in S^{\prime }}\{c_{i^{\prime }j^{\prime }}^{\prime *}\}=c_{kl}^{\prime *}$. Since $\left\{ k,l\right\} \nsubseteq S^{\prime }$, $\max _{i^{\prime },j^{\prime }\in S^{\prime }}\{c_{i^{\prime }j^{\prime }}^{\prime *}\}=\max _{i^{\prime },j^{\prime }\in S^{\prime }}\{c_{i^{\prime }j^{\prime }}^{*}\}$ and hence

  $$\begin{aligned} \delta _{S^{\prime }}=\min _{i^{\prime }\in S^{\prime },j^{\prime }\in N_{0}\setminus S^{\prime }}c_{i^{\prime }j^{\prime }}^{*}-\max _{i^{\prime },j^{\prime }\in S^{\prime }}c_{i^{\prime }j^{\prime }}^{*}=c_{kl}^{*}-c_{kl}^{\prime *}=-\alpha <0, \end{aligned}$$

  which is a contradiction.

• 2. If $\left\{ i^{2},j^{2}\right\} =\left\{ k,l\right\}$, then $\delta _{S}^{\prime }=\delta _{S}-\alpha$. Besides, there can be at most one such $S$. Assume, on the contrary, that there exists $S^{\prime }\in X$, $S\ne S^{\prime }$, $k,l\in S\cap S^{\prime }$, and

  $$\begin{aligned} c_{kl}=c_{kl}^{*}=\max _{i^{\prime },j^{\prime }\in S}c_{i^{\prime }j^{\prime }}^{*}=\max _{i^{\prime },j^{\prime }\in S^{\prime }}c_{i^{\prime }j^{\prime }}^{*}. \end{aligned}$$

  Since $k\in S\cap S^{\prime }$, under 4 of Lemma 7.1, $S\varsubsetneq S^{\prime }$ or $S^{\prime }\varsubsetneq S$. Assume w.l.o.g. $S\varsubsetneq S^{\prime }$. Then,

  $$\begin{aligned} c_{kl}^{*}=\max _{i^{\prime },j^{\prime }\in S}c_{i^{\prime }j^{\prime }}^{*}\le \min _{i^{\prime }\in S,j^{\prime }\in N_{0}\setminus S}c_{i^{\prime }j^{\prime }}^{*}\le \min _{i^{\prime }\in S,j^{\prime }\in S^{\prime }\setminus S}c_{i^{\prime }j^{\prime }}^{*}\le \max _{i^{\prime },j^{\prime }\in S^{\prime }}c_{i^{\prime }j^{\prime }}^{*}=c_{kl}^{*} \end{aligned}$$

  which implies that no inequality is strict. Thus, $\min _{i^{\prime }\in S,j^{\prime }\in N_{0}\setminus S}\{c_{i^{\prime }j^{\prime }}^{*}\}=c_{kl}^{*}$ and hence

  $$\begin{aligned} \delta _{S}=\min _{i^{\prime }\in S,j^{\prime }\in N_{0}\setminus S}c_{i^{\prime }j^{\prime }}^{*}-\max _{i^{\prime },j^{\prime }\in S}c_{i^{\prime }j^{\prime }}^{*}=c_{kl}^{*}-c_{kl}^{*}=0, \end{aligned}$$

  which implies $\delta _{S}^{\prime }=\delta _{S}-\alpha =-\alpha <0$, which is a contradiction.

Let $S^{k}=\left\{ j\in N_{0}:c_{kj}^{\prime *}<c_{kl}^{\prime *}\right\}$ and let $S^{l}=\left\{ j\in N_{0}:c_{kj}^{\prime *}<c_{kl}^{\prime *}\right\}$. Both $S^{k}$ and $S^{l}$ are non-empty (because $k\in S^{k}$ and $l\in S^{l}$) and disjoint (it follows from $\left\{ k,l\right\} \in t$). Since they are disjoint, we can assume w.l.o.g. $0\notin S^{k}$. Let $S_{1}=S^{k}$. If $\left| S_{1}\right| >1$, then

$$\begin{aligned}&\quad \quad \quad l \notin S_{1}, \\&c_{kl}^{\prime *} =\min _{i^{\prime }\in S_{1},j^{\prime }\in N_{0}\setminus S_{1}}c_{i^{\prime }j^{\prime }}^{\prime *}, \\&\delta _{S_{1}}^{\prime } = c_{kl}^{\prime *}-\max _{i^{\prime },j^{\prime }\in S}c_{i^{\prime }j^{\prime }}^{\prime *}>0 \end{aligned}$$

and hence either $S_{1}\in Ne\left( N_{0},C^{\prime }\right)$ or $S_{1}=\left\{ k\right\}$.

Assume that $S_{1}\in Ne\left( N_{0},C^{\prime }\right)$. Since $\left( N_{0},C\right)$ and $\left( N_{0},C^{\prime }\right)$ are in the same Kruskal cone, $\delta _{S_{1}}=c_{i^{1}j^{1}}^{*}-c_{i^{2}j^{2}}^{*}$ and $\delta _{S_{1}}^{\prime }=c_{i^{1}j^{1}}^{\prime *}-c_{i^{2}j^{2}}^{\prime *}$. Since $\delta _{S_{1}}^{\prime }>0$ we deduce that $\delta _{S_{1}}\ge 0$. Hence $S_{1}\in qNe\left( N_{0},C\right)$. Now, satisfies condition 1. Then $\delta _{S_{1}}^{\prime }=\delta _{S_{1}}+\alpha$ when $\left| S_{1}\right| >1$.

Let $S_{2}=\left\{ j\in N_{0}:c_{kj}^{*}\le c_{kl}^{*}\right\}$. Clearly, $\left\{ k,l\right\} \subset S_{2}$. Notice that if $0\in S_{2}$ then $S_{2}\notin X$. Now, if $0\notin S_{2}$ then $S_{2}\in X$. Besides $S_{1}\varsubsetneq S_{2}$ and there is no $S\in X$, $S\ne S_{1}$, such that $S_{1}\varsubsetneq S\varsubsetneq S_{2}$.

In case $0\notin S_{2}$, we have that $S_{2}$ satisfies condition 2. Hence $\delta _{S_{2}}^{\prime }=\delta _{S_{2}}-\alpha$.

Let $F=\left\{ S\in Ne\left( N_{0},C\right) :S_{1}\subset S,\delta _{S}=\delta _{S}^{\prime }\right\}$ and let $F^{\prime }=\Big \{ S\in Ne\left( N_{0},C^{\prime }\right) :S_{1}\subset S,\delta _{S}=\delta _{S}^{\prime }\Big \}$. Then, $F=F^{\prime }$ ($F=F^{\prime }=\emptyset$ is also possible) and $S_{1},S_{2}\notin F$. By part 3 of Lemma 7.1 we can assume that $F=\left\{ S_{3},S_{4},\dots ,S_{\Gamma }\right\}$ for some $\Gamma \ge 2$ ($\Gamma =2$ when $F=\emptyset$) and $S_{\gamma }\varsubsetneq S_{\gamma +1}$ for all $\gamma =3,\dots ,\Gamma -1$.

Let $G=\left\{ S\in X:S_{1}\subset S\right\}$. Clearly, either $G=\left\{ S_{1},\dots ,S_{\Gamma }\right\}$ (when $S_{1}\in Ne\left( N_{0},C^{\prime }\right)$) or $G=\left\{ S_{2},\dots ,S_{\Gamma }\right\}$ (when $S_{1}=\left\{ k\right\}$). Besides, $S_{\gamma }\varsubsetneq S_{\gamma +1}$ for all $\gamma =1,2,\dots ,\Gamma -1$.

If $i\notin S_{\Gamma }$, then $f_{i}\left( N_{0},C\right) =f_{i}\left( N_{0},C^{\prime }\right)$. We assume $i\in S_{\gamma }$ for some $\gamma \in \left\{ 1,\dots ,\Gamma \right\}$. Let $\gamma _{i}$ be the minimum of these $\gamma$’s. We have two cases:

Case 1 $\Gamma =1$. This means $S_{2}\notin X$. Since $\delta _{S_{2}}\ge 0$, we have $0\in S_{2}$, which implies $c_{0k}^{*}\le c_{kl}^{*}$ and also $c_{0k}^{\prime *}\le c_{kl}^{\prime *}$.

Case 1.1 $S_{1}=\left\{ k\right\} =\left\{ i\right\}$. This implies $X=\emptyset$ and hence

$$\begin{aligned} f_{i}\left( N_{0},C^{\prime }\right) -f_{i}\left( N_{0},C\right) =c_{0i}^{\prime *}-c_{0i}^{*}\ge 0. \end{aligned}$$

Case 1.2 $S_{1}\in X$. This implies $c_{0k}^{\prime *}\ge c_{kl}^{\prime *}$ and hence $c_{0k}^{\prime *}=c_{kl}^{\prime *}$. Thus $c_{0i}^{\prime *}-c_{0i}^{*}=\alpha$ and $C_{S_{1}}^{*}=C_{S_{1}}^{\prime *}$. Hence,

$$\begin{aligned}&\qquad f_{i}\left( N_{0},C^{\prime }\right) -f_{i}\left( N_{0},C\right) \\&\quad \quad \quad \quad =c_{0i}^{\prime *}-\left( \delta _{S_{1}}^{\prime }-e_{i}\left( C_{S_{1}}^{\prime *},\delta _{S_{1}}^{\prime }\right) \right) -c_{0i}^{*}+\left( \delta _{S_{1}}-e_{i}\left( C_{S_{1}}^{*},\delta _{S_{1}}\right) \right) \\&\quad \quad \quad \quad =c_{0i}^{\prime *}-c_{0i}^{*}-\left( \delta _{S_{1}}+\alpha -e_{i}\left( C_{S_{1}}^{*},\delta _{S_{1}}+\alpha \right) \right) +\left( \delta _{S_{1}}-e_{i}\left( C_{S_{1}}^{*},\delta _{S_{1}}\right) \right) \\&\quad \quad \quad \quad =e_{i}\left( C_{S_{1}}^{*},\delta _{S_{1}}+\alpha \right) -e_{i}\left( C_{S_{1}}^{*},\delta _{S_{1}}\right) \ge 0 \end{aligned}$$

where the last inequality comes from applying $\textit{ANM}$ to $\left\{ \left( S_{1},C_{S_{1}}^{*}\right) \right\}$ with $\Gamma =1$, $a_{1}=\delta _{S_{1}}$ and $y=\alpha$.

Case 2 $\Gamma >1$. This means that $S_{2}\in X$ and hence $0\notin S^{l}$. Thus we can take $S_{1}=S^{k}$ or $S_{1}=S^{l}$. We have that $S_{2}=S^{k}\cup S^{l}$. If $i\in S_{2}$ we choose $S_{1}$ such that $i\in S_{1}$. Thus, $\gamma _{i}\ne 2$ which implies $c_{0i}^{\prime *}=c_{0i}^{*}$.

In this case,

$$\begin{aligned}&f_{i}\left( N_{0},C^{\prime }\right) -f_{i}\left( N_{0},C\right) \\&\quad \quad \quad =c_{0i}^{\prime *}-c_{0i}^{*}-\sum _{i\in S\in X}\left( \delta _{S}^{\prime }-\delta _{S}-e_{i}\left( C_{S}^{\prime *},\delta _{S}^{\prime }\right) +e_{i}\left( C_{S}^{*},\delta _{S}\right) \right) . \end{aligned}$$

For any $S\in X\setminus G$ with $i\in S$, we have $C_{S}^{*}=C_{S}^{\prime *}$, which implies $\delta _{S}=\delta _{S}^{\prime }$. Hence,

$$\begin{aligned}&f_{i}\left( N_{0},C^{\prime }\right) -f_{i}\left( N_{0},C\right) \\&\quad =\sum _{\gamma =\gamma _{i}}^{\Gamma }\left( -\delta _{S_{\gamma }}^{\prime }+\delta _{S_{\gamma }}+e_{i}\left( C_{S_{\gamma }}^{\prime *},\delta _{S_{\gamma }}^{\prime }\right) -e_{i}\left( C_{S_{\gamma }}^{*},\delta _{S_{\gamma }}\right) \right) \\&\quad =\sum _{\gamma =\gamma _{i}}^{\Gamma }e_{i}\left( C_{S_{\gamma }}^{\prime *},\delta _{S_{\gamma }}^{\prime }\right) -\sum _{\gamma =\gamma _{i}}^{\Gamma }e_{i}\left( C_{S_{\gamma }}^{*},\delta _{S_{\gamma }}\right) -\sum _{\gamma =\gamma _{i}}^{\Gamma }\left( \delta _{s_{\gamma }}^{\prime }-\delta _{s_{\gamma }}\right) . \end{aligned}$$

The last term is zero, because $\delta _{S_{1}}^{\prime }=\delta _{S_{1}}+\alpha$, $\delta _{S_{2}}^{\prime }=\delta _{S_{2}}-\alpha$, and $\delta _{S_{\gamma }}^{\prime }=\delta _{S_{\gamma }}$ otherwise (remember that $\gamma _{i}\ne 2$). Hence,

$$\begin{aligned} f_{i}\left( N_{0},C^{\prime }\right) -f_{i}\left( N_{0},C\right) =\sum _{\gamma =\gamma _{i}}^{\Gamma }\left( e_{i}\left( C_{S_{\gamma }}^{\prime *},\delta _{S_{\gamma }}^{\prime }\right) \right) -\sum _{\gamma =\gamma _{i}}^{\Gamma }\left( e_{i}\left( C_{S_{\gamma }}^{*},\delta _{S_{\gamma }}\right) \right) . \end{aligned}$$

We now define $\left\{ \left( N^{\gamma },C^{\gamma }\right) \right\} _{\gamma =1}^{\Gamma }$, $a\in \mathbb {R}_{+}^{\Gamma }$ and $y\in \left[ 0,a_{2}\right]$ so that $e_{i}\left( C_{S_{\gamma }}^{\prime *},\delta _{S_{\gamma }}^{\prime }\right) =e_{i}\left( C^{\gamma }\left( a^{\prime }\right) ,a_{\gamma }^{\prime }\right)$ and $e_{i}\left( C_{S_{\gamma }}^{*},\delta _{S_{\gamma }}\right) =e_{i}\left( C^{\gamma }\left( a\right) ,a_{\gamma }\right)$ for all $\gamma$. Under $\textit{ANM}$, this will prove that the above expression is non-negative.

Let $N^{1}=S_{1}$, $C^{1}=C_{N^{1}}^{*}$, and $a_{1}=\delta _{S_{1}}$. In general, for any $\gamma =2,...,\Gamma$, $N^{\gamma }=S_{\gamma }\setminus S_{\gamma -1}$, $C^{\gamma }=\left( C^{*}\right) _{N^{\gamma }}$, and $a_{\gamma }=\delta _{S_{\gamma }}$. We also define $y=\alpha$. Since $c_{kl}^{\prime *}=c_{kl}^{*}+\alpha$, we have that $\alpha \le a_{2}$ and hence $y\in \left[ 0,a_{2}\right]$.

Clearly, $C_{S_{1}}^{\prime *}=C^{1}$. Now, we prove that $C_{S_{2}}^{\prime *}=C^{1}\oplus _{a_{1}+\alpha }C^{2}=C^{2}\left( a^{\prime }\right)$. Let $C^{\alpha }=C_{S_{2}}^{\prime *}$ and $C^{\beta }=C^{1}\oplus _{a_{1}+\alpha }C^{2}$. We have that $C^{\alpha }=\left( C_{S_{2}}+\alpha I_{kl}\right) ^{*}$. Then, $c_{ij}^{\alpha }=c_{ij}^{\beta }$ for all $i,j\in N^{1}$ and all $i,j\in N^{2}$.

Let $k^{1}\in N^{1}$ and $k^{2}\in N^{2}$. Then,

$$\begin{aligned} c_{k^{1}k^{2}}^{\beta }&= \max C^{1}+a_{1}+\alpha =\max C^{1}+\delta _{S_{1}}+\alpha =\min _{\begin{array}{c} i\in N^{1} \\ j\in N_{0}\setminus N^{1} \end{array}}c_{ij}+\alpha \\&= c_{kl}+\alpha =c_{k^{1}k^{2}}^{\alpha }. \end{aligned}$$

Analogously, $C_{S_{3}}^{\prime *}=\left( C_{S_{3}}+\alpha I_{kl}\right) ^{*}=\left( C^{1}\oplus _{a_{1}+\alpha }C^{2}\right) \oplus _{a_{2}-\alpha }C^{3}=C^{3}\left( a^{\prime }\right)$. In general, $C_{S_{\gamma }}^{\prime *}=\left( C_{S_{\gamma }}+\alpha I_{kl}\right) ^{*}=C^{1}\oplus _{a_{1}+\alpha }C^{2}\oplus _{a_{1}-\alpha }C^{3}\oplus _{a_{3}}...\oplus _{a_{\gamma -1}}C^{\gamma }=C^{\gamma }\left( a^{\prime }\right)$ for all $\gamma =3,\dots ,\Gamma$.

Similarly, we can prove that $C_{S_{\gamma }}^{*}=C^{\gamma }\left( a\right)$ for all $\gamma =1,\dots ,\Gamma$.

Hence, by applying $\textit{ANM}$, we have

$$\begin{aligned} f_{i}\left( N_{0},C^{\prime }\right) -f_{i}\left( N_{0},C\right) \ge 0. \end{aligned}$$

We now prove that $f$ satisfies ${ PM}$. Under Theorem 2, we know that $f$ satisfies ${ SEP}$. It is enough to prove that for each mcstp $\left( N_{0},C\right)$ and $j\in N$, $f_{i}\left( N_{0},C\right) \le f_{i}\left( N_{0}\setminus \left\{ j\right\} ,C_{N_{0}\setminus \left\{ j\right\} }\right)$ for all $i\in N\setminus \left\{ j\right\}$. Let $\left( N_{0},C^{\prime }\right)$ be defined as $c_{ii^{\prime }}^{\prime }=c_{ii^{\prime }}$ for all $i,i^{\prime }\in N\setminus \left\{ j\right\}$ and $c_{ij}^{\prime }=\max C_{N_{0}\setminus \left\{ j\right\} }$ for all $i\in N_{0}\setminus \left\{ j\right\}$. Clearly, $m\left( N_{0},C^{\prime }\right) =m\left( N_{0}\setminus \left\{ j\right\} ,C_{N_{0}\setminus \left\{ j\right\} }^{\prime }\right) +m\left( \left\{ j\right\} _{0},C_{\left\{ j\right\} _{0}}^{\prime }\right)$. Under ${ SEP}$, $f_{i}\left( N_{0},C^{\prime }\right) =f_{i}\left( N_{0}\setminus \left\{ j\right\} ,C_{N_{0}\setminus \left\{ j\right\} }^{\prime }\right)$ for all $i\in N\setminus \left\{ j\right\}$. Given $i\in N\setminus \left\{ j\right\}$, under $\textit{CM}$,

$$\begin{aligned} f_{i}\left( N_{0},C\right) \le f_{i}\left( N_{0},C^{\prime }\right) =f_{i}\left( N_{0}\setminus \left\{ j\right\} ,C_{N_{0}\setminus \left\{ j\right\} }^{\prime }\right) =f_{i}\left( N_{0}\setminus \left\{ j\right\} ,C_{N_{0}\setminus \left\{ j\right\} }\right) . \end{aligned}$$

We now prove that if $f$ satisfies $\textit{CM}$ and ${ PM}$, then $f=f^{e}$ for some $e$ satisfying $\textit{ANM}$.

We define $e$ as in the proof of Theorem 2. Namely, for all $C^{*}\in \mathcal {C}^{*}$, $x\in \mathbb {R}_{+}$, and $i\in N$,

$$\begin{aligned} e_{i}\left( C^{*},x\right) =f_{i}\left( N_{0},C^{*\left( \max C^{*}+x\right) }\right) -f_{i}\left( N_{0},C^{*\left( \max C^{*}\right) }\right) \end{aligned}$$

and $e_{i}\left( C^{*},x\right) =0$ for all $i\notin N$. We already proved (proof of Theorem 2) that $e$ is an extra-costs function and $f=f^{e}$ .

Hence, we only need to check that $e$ satisfies $\textit{ANM}$. Let $\left\{ \left( N^{\gamma },C^{\gamma }\right) \right\} _{\gamma =1}^{\Gamma }\subset C^{*}$ be a disjoint sequence with $\Gamma \ge 1$, $i\in N^{\gamma _{i}}$ with $\gamma _{i}\ne 2$, $a\in \mathbb {R}_{+}^{\Gamma }$ with $a_{\gamma }\ge \max C^{\gamma +1}-\max C^{\gamma }$ for all $\gamma =1,\dots ,\Gamma -1$ and $y\in \left[ 0,a_{2}\right]$ (or simply $y\ge 0$, when $\Gamma =1$).

Assume first that $\Gamma =1$. We need to prove

$$\begin{aligned} e_{i}\left( C^{1},a_{1}+y\right) -e_{i}\left( C^{1},a_{1}\right) \ge 0. \end{aligned}$$

Let $C=C^{1}$. By definition,

$$\begin{aligned}&e_{i}\left( C,a_{1}+y\right) -e_{i}\left( C,a_{1}\right) \\&\quad =f_{i}\left( N_{0},C^{*\left( \max C^{*}+a_{1}+y\right) }\right) -f_{i}\left( N_{0},C^{*\left( \max C^{*}\right) }\right) \\&\qquad -f_{i}\left( N_{0},C^{*\left( \max C^{*}+a_{1}\right) }\right) +f_{i}\left( N_{0},C^{*\left( \max C^{*}\right) }\right) \\&\quad =f_{i}\left( N_{0},C^{*\left( \max C^{*}+a_{1}+y\right) }\right) -f_{i}\left( N_{0},C^{*\left( \max C^{*}+a_{1}\right) }\right) \ge 0 \end{aligned}$$

where the last inequality comes from the fact that $C^{*\left( \max C^{*}+a_{1}+y\right) }\ge$ $C^{*\left( \max C^{*}+a_{1}\right) }$ and $f$ satisfy $\textit{CM}$.

Assume now that $\Gamma >1$. We need to prove

$$\begin{aligned} \sum _{\gamma =\gamma _{i}}^{\Gamma }e_{i}\left( C^{\gamma }\left( a^{\prime }\right) ,a_{\gamma }^{\prime }\right) -\sum _{\gamma =\gamma _{i}}^{\Gamma }e_{i}\left( C^{\gamma }\left( a\right) ,a_{\gamma }\right) \ge 0 \end{aligned}$$

where $a^{\prime }=\left( a_{1}+y,a_{2}-y,a_{3},\dots ,a_{\Gamma }\right)$ and $C^{\gamma }\left( b\right) =C^{1}\oplus _{b_{1}}C^{2}\oplus _{b_{2}}...\oplus _{b_{\gamma -1}}C^{\gamma }$ for all $\gamma =1,\dots ,\Gamma$ and all $b\in \mathbb {R}_{+}^{\Gamma }$.

By definition,

$$\begin{aligned} e_{i}\left( C^{*},x\right) =f_{i}\left( N_{0},C^{*}\oplus _{x}\left( \left\{ 0\right\} ,0\right) \right) -f_{i}\left( N_{0},C^{*}\oplus _{0}\left( \left\{ 0\right\} ,0\right) \right) . \end{aligned}$$

Under ${ SEP}$, we have that

$$\begin{aligned} f_{i}\left( \bigcup _{r=1}^{\gamma }N^{r},C^{\gamma }\left( b\right) \oplus _{0}\left( \left\{ 0\right\} ,0\right) \right) =f_{i}\left( \bigcup _{r=1}^{\gamma -1}N^{r},C^{\gamma -1}\left( b\right) \oplus _{b_{\gamma -1}}\left( \left\{ 0\right\} ,0\right) \right) \end{aligned}$$

for all $\gamma =\gamma _{i}+1,\dots ,\Gamma$ and all $b\in \mathbb {R} _{+}^{\Gamma }$. Now,

$$\begin{aligned}&\sum _{\gamma =\gamma _{i}}^{\Gamma }e_{i}\left( C^{\gamma }\left( a^{\prime }\right) ,a_{\gamma }^{\prime }\right) \\&\quad =\sum _{\gamma =\gamma _{i}}^{\Gamma }\left[ f_{i}\left( \bigcup _{r=1}^{\gamma }N^{r},C^{\gamma }\left( a^{\prime }\right) \oplus _{a_{\gamma }^{\prime }}\left( \left\{ 0\right\} ,0\right) \right) -f_{i}\left( \bigcup _{r=1}^{\gamma }N^{r},C^{\gamma }\left( a^{\prime }\right) \oplus _{0}\left( \left\{ 0\right\} ,0\right) \right) \right] \\&\quad =f_{i}\left( \bigcup _{r=1}^{\Gamma }N^{r},C^{\Gamma }\left( a^{\prime }\right) \oplus _{a_{\Gamma }^{\prime }}\left( \left\{ 0\right\} ,0\right) \right) -f_{i}\left( \bigcup _{r=1}^{\gamma _{i}}N^{r},C^{\gamma _{i}}\left( a^{\prime }\right) \oplus _{0}\left( \left\{ 0\right\} ,0\right) \right) \end{aligned}$$

and

$$\begin{aligned}&\sum _{\gamma =\gamma _{i}}^{\Gamma }e_{i}\left( C^{\gamma }\left( a\right) ,a_{\gamma }\right) \\&\quad =\sum _{\gamma =\gamma _{i}}^{\Gamma }\left[ f_{i}\left( \bigcup _{r=1}^{\gamma }N^{r},C^{\gamma }\left( a\right) \oplus _{a_{\gamma }}\left( \left\{ 0\right\} ,0\right) \right) -f_{i}\left( \bigcup _{r=1}^{\gamma }N^{r},C^{\gamma }\left( a\right) \oplus _{0}\left( \left\{ 0\right\} ,0\right) \right) \right] \\&\quad =f_{i}\left( \bigcup _{r=1}^{\Gamma }N^{r},C^{\Gamma }\left( a\right) \oplus _{a_{\Gamma }}\left( \left\{ 0\right\} ,0\right) \right) -f_{i}\left( \bigcup _{r=1}^{\gamma _{i}}N^{r},C^{\gamma _{i}}\left( a\right) \oplus _{0}\left( \left\{ 0\right\} ,0\right) \right) . \end{aligned}$$

Hence,

$$\begin{aligned}&\sum _{\gamma =\gamma _{i}}^{\Gamma }e_{i}\left( C^{\gamma }\left( a^{\prime }\right) ,a_{\gamma }^{\prime }\right) -\sum _{\gamma =\gamma _{i}}^{\Gamma }e_{i}\left( C^{\gamma }\left( a\right) ,a_{\gamma }\right) \\&\quad =f_{i}\left( \bigcup _{r=1}^{\gamma _{i}}N^{r},C^{\gamma _{i}}\left( a\right) \oplus _{0}\left( \left\{ 0\right\} ,0\right) \right) -f_{i}\left( \bigcup _{r=1}^{\gamma _{i}}N^{r},C^{\gamma _{i}}\left( a^{\prime }\right) \oplus _{0}\left( \left\{ 0\right\} ,0\right) \right) \\&\qquad +f_{i}\left( \bigcup _{r=1}^{\Gamma }N^{r},C^{\Gamma }\left( a^{\prime }\right) \oplus _{a_{\Gamma }^{\prime }}\left( \left\{ 0\right\} ,0\right) \right) -f_{i}\left( \bigcup _{r=1}^{\Gamma }N^{r},C^{\Gamma }\left( a\right) \oplus _{a_{\Gamma }}\left( \left\{ 0\right\} ,0\right) \right) . \end{aligned}$$

Under $\textit{CM}$,

$$\begin{aligned} f_{i}\left( \bigcup _{r=1}^{\Gamma }N^{r},C^{\Gamma }\left( a^{\prime }\right) \oplus _{a_{\Gamma }^{\prime }}\left( \left\{ 0\right\} ,0\right) \right) \ge f_{i}\left( \bigcup _{r=1}^{\Gamma }N^{r},C^{\Gamma }\left( a^{\prime }\right) \oplus _{a_{\Gamma }^{\prime }}\left( \left\{ 0\right\} ,0\right) \right) . \end{aligned}$$

We now prove that

$$\begin{aligned} f_{i}\left( \bigcup _{r=1}^{\gamma _{i}}N^{r},C^{\gamma _{i}}\left( a\right) \oplus _{0}\left( \left\{ 0\right\} ,0\right) \right) =f_{i}\left( \bigcup _{r=1}^{\gamma _{i}}N^{r},C^{\gamma _{i}}\left( a^{\prime }\right) \oplus _{0}\left( \left\{ 0\right\} ,0\right) \right) . \end{aligned}$$

For $\gamma _{i}=1$, $C^{1}\left( a\right) =C^{1}\left( a^{\prime }\right) =C^{1}$ and the result holds trivially. Assume $\gamma _{i}>2$. Then, $N^{1}\cup \ldots \cup N^{\gamma _{i}-1}$ and $N^{\gamma _{i}}$ are two separable components in both $\left( \bigcup _{r=1}^{\gamma _{i}}N^{r},C^{\gamma _{i}}\left( a\right) \oplus _{0}\left( \left\{ 0\right\} ,0\right) \right)$ and $\left( \bigcup _{r=1}^{\gamma _{i}}N^{r},C^{\gamma _{i}}\left( a^{\prime }\right) \oplus _{0}\left( \left\{ 0\right\} ,0\right) \right)$. Besides, the restriction of $C^{*}$ to $N^{\gamma _{i}}$ coincides in both mcstp. Under ${ SEP}$, we obtain the equality.

Hence,

$$\begin{aligned} \sum _{\gamma =\gamma _{i}}^{\Gamma }e_{i}\left( C^{\gamma }\left( a^{\prime }\right) ,a_{\gamma }^{\prime }\right) -\sum _{\gamma =\gamma _{i}}^{\Gamma }e_{i}\left( C^{\gamma }\left( a\right) ,a_{\gamma }\right) \ge 0. \end{aligned}$$

$\square$

1.4 Proof of Proposition 5.1

$\left( 1\right)$ Using an obligation function $o$ we can arrive at a cost allocation as follows. We compute a mcst following Kruskal’s algorithm (Kruskal 1956), which consists in to construct a tree by sequentially adding edges with the lowest cost and without introducing cycles. The cost of each edge selected by Kruskal’s algorithm is divided among the agents who benefit from adding this edge. Each of these agents pays the difference between her obligation to two groups, one in which she belonged before the edge was added and the one after. We now define an obligation rule, $f^{o}$, formally.

Given a network $g$ we define $P\left( g\right) =\left\{ T_{k}\left( g\right) \right\} _{k=1}^{n\left( g\right) }$ as the partition of $N_{0}$ in connected components induced by $g$. Namely, $P\left( g\right)$ is the only partition of $N_{0}$ satisfying the following two properties: Firstly, if $i,j\in T_{k}\left( g\right)$, $i$ and $j$ are connected in $g$. Secondly, if $i\in T_{k}$, $j\in T_{l}$, and $k\ne l$, then $i$ and $j$ are not connected in $g$. Given a network $g$, let $S\left( P\left( g\right) ,i\right)$ denote the element of $P\left( g\right)$ to which $i$ belongs to.

Given a mcstp $\left( N_{0},C\right)$, let $g^{\left| N\right| }$ be a tree obtained applying Kruskal’s algorithm to $\left( N_{0},C\right)$, and for each $p=1,\dots ,\left| N\right|$, $\left( i^{p},j^{p}\right)$ is the edge selected by Kruskal’s algorithm at Stage $p$, and $g^{p}$ the set of edges selected by Kruskal’s algorithm at stages $1,\dots ,p$. For each $i\in N$, we define the obligation rule associated with the obligation function $o$ as

$$\begin{aligned} f_{i}^{o}\left( N_{0},C\right) =\sum _{p=1}^{\left| N\right| }c_{i^{p}j^{p}}\left( o_{i}\left( S\left( P\left( g^{p-1}\right) ,i\right) \right) -o_{i}\left( S\left( P\left( g^{p}\right) ,i\right) \right) \right) \end{aligned}$$

where by convention, $o_{i}\left( T\right) =0$ if $0\in T$.

Tijs et al. (2006) prove that $f^{o}$ is well defined, namely, it is independent of the mcst obtained following Kruskal’s algorithm.

We prove that if $f^{o}$ is an obligation rule, then $f^{o}=f^{e}$ where $e\left( C^{*},x\right) =xo_{i}\left( N\right)$ for each $\left( N,C^{*}\right)$ and $x$.

We proceed by induction on the number of agents. If $\left| N\right| =1$ the result holds trivially. Assume that $f^{o}=f^{e}$ when $\left| N\right| <q$ and we prove it when $\left| N\right| =q$.

Let $\left( N_{0},C\right)$ be a $mcstp$. Since $f^{o}$ and $f^{e}$ satisfy $\textit{CM}$, it is enough to prove that $f^{o}\left( N_{0},C^{*}\right) =f^{e}\left( N_{0},C^{*}\right)$.

Proposition 3.1 in Bergantiños and Vidal-Puga (2007) says the following. $\left( N_{0},C^{*}\right)$ is an irreducible problem if and only if there exists a mcst $t$ in $\left( N_{0},C^{*}\right)$ that satisfies the following two conditions:

$\left( A1\right)$ $t=\left\{ \left( \pi _{s-1},\pi _{s}\right) \right\} _{s=1}^{n}$ where $\pi _{0}=0$ (the source).

$\left( A2\right)$ Given $\pi _{p},\pi _{q}\in N_{0}$ with $p<q$, $c_{\pi _{p}\pi _{q}}^{*}=\max _{s\mid p<s\le q}\left\{ c_{\pi _{s-1}\pi _{s}}^{*}\right\}$.

Let $t$ be such $mcst$. Without loss of generality we assume that $\pi _{s}=s$ for each $s=1,\dots ,\left| N\right|$. We consider two cases.

• 1. There exists $s>1$ such that $c_{s-1,s}^{*}\ge c_{r-1,r}^{*}$ for all $r=1,\dots ,\left| N\right|$. Let $S=\left\{ 1,\dots ,s-1\right\}$. By Proposition 3.1 in Bergantiños and Vidal-Puga (2007) we have that $\left\{ \left( r-1,r\right) \right\} _{r=1}^{s-1}$ can be obtained applying Kruskal’s algorithm to $\left( S_{0},C_{S_{0}}^{*}\right)$ and $\left\{ \left( 0,s\right) \right\} \cup \left\{ \left( r-1,r\right) \right\} _{r=s+1}^{\left| N\right| }$ can be obtained applying Kruskal’s algorithm to $\left( \left( N\setminus S\right) _{0},C_{\left( N\setminus S\right) _{0}}^{*}\right)$. Thus, $m\left( N_{0},C^{*}\right) =m\left( S_{0},C_{S_{0}}^{*}\right) +m\left( \left( N\setminus S\right) _{0},C_{\left( N\setminus S\right) _{0}}^{*}\right)$. Let $i\in S$. Since $f^{o}$ and $f^{e}$ satisfy ${ SEP}$, we deduce that

  $$\begin{aligned} f_{i}^{o}\left( N_{0},C^{*}\right) =f_{i}^{o}\left( S_{0},C_{S_{0}}^{*}\right) \quad \text { and }\quad f_{i}^{e}\left( N_{0},C^{*}\right) =f_{i}^{e}\left( S_{0},C_{S_{0}}^{*}\right) . \end{aligned}$$

  By induction hypothesis $f_{i}^{o}\left( S_{0},C_{S_{0}}^{*}\right) =f_{i}^{e}\left( S_{0},C_{S_{0}}^{*}\right)$. Hence, $f_{i}^{o}\left( N_{0},C^{*}\right) =f_{i}^{e}\left( N_{0},C^{*}\right)$. Similarly, we can prove that $f_{i}^{o}\left( N_{0},C^{*}\right) =f_{i}^{e}\left( N_{0},C^{*}\right)$ when $i\in N\setminus S$.

• 2. $c_{01}^{*}>c_{r-1,r}^{*}$ for all $r=2,\dots ,\left| N\right|$. Let $\alpha =c_{01}^{*}-\max _{r=2,\dots ,\left| N\right| }\left\{ c_{r-1,r}^{*}\right\}$. Let $\left( N_{0},C^{\prime }\right)$ be the irreducible problem associated with $t$ and $C^{\prime }$ where $c_{01}^{\prime }=c_{01}^{*}-\alpha$ and $c_{r-1,r}^{\prime }=c_{r-1,r}^{*}$ for all $r=2,\dots ,\left| N\right|$. Since $C^{\prime }$ is under the conditions of the previous case, we have that $f^{o}\left( N_{0},C^{\prime }\right) =f^{e}\left( N_{0},C^{\prime }\right)$. Thus, it is enough to prove that for all $i\in N$,

  $$\begin{aligned} f_{i}^{o}\left( N_{0},C^{*}\right) -f_{i}^{o}\left( N_{0},C^{\prime }\right) =f_{i}^{e}\left( N_{0},C^{*}\right) -f_{i}^{e}\left( N_{0},C^{\prime }\right) . \end{aligned}$$

  Fix $i\in N$. We first compute $f_{i}^{o}\left( N_{0},C^{*}\right) -f_{i}^{o}\left( N_{0},C^{\prime }\right)$. We can apply Kruskal’s algorithm to both $C^{*}$ and $C^{\prime }$ in such a way that:

  • The edge selected at each stage belongs to $t$. Namely, for each $p=1,\dots ,\left| N\right|$, $\left( i^{p}\left( C^{*}\right) ,j^{p}\left( C^{*}\right) \right) \in t$ and $\left( i^{p}\left( C^{\prime }\right) ,j^{p}\left( C^{\prime }\right) \right) \in t$.

  • The edge selected at each stage is the same in both problems. Namely, for each $p=1,\dots ,\left| N\right|$, $\left( i^{p}\left( C^{*}\right) , j^{p}\left( C^{*}\right) \right) =\left( i^{p}\left( C^{\prime }\right) ,j^{p}\left( C^{\prime }\right) \right)$.

  • The last edge selected is $\left( 0,1\right)$. Namely, $\left( i^{\left| N\right| }\left( C^{*}\right) ,j^{\left| N\right| }\left( C^{*}\right) \right) =\Big ( i^{\left| N\right| }\left( C^{\prime }\right) , j^{\left| N\right| }\left( C^{\prime }\right) \Big ) =\left( 0,1\right)$.

  Thus, $f_{i}^{o}\left( N_{0},C^{*}\right) -f_{i}^{o}\left( N_{0},C^{\prime }\right) =c_{01}^{*}o_{i}\left( N\right) -c_{01}^{\prime }o_{i}\left( N\right) =\alpha o_{i}\left( N\right)$. We now compute $f_{i}^{e}\left( N_{0},C^{*}\right) -f_{i}^{e}\left( N_{0},C^{\prime }\right)$. Notice that if $S$ is a neighborhood of $i$ in $\left( N_{0},C^{\prime }\right)$, then $S$ is also a neighborhood of $i$ in $\left( N_{0},C^{*}\right)$. Besides, $N$ is the unique neighborhood of $i$ in $\left( N_{0},C^{*}\right)$ which is not a neighborhood of $i$ in $\left( N_{0},C^{\prime }\right)$. Thus,

  $$\begin{aligned} f_{i}^{e}\left( N_{0},C^{*}\right) -f_{i}^{e}\left( N_{0},C^{\prime }\right) =c_{0i}^{*}-\left( \delta _{N}-e_{i}\left( C_{N}^{*},\delta _{N}\right) \right) -c_{0i}^{\prime *}. \end{aligned}$$

  Since $\delta _{N}=\alpha$,

  $$\begin{aligned} f_{i}^{e}\left( N_{0},C^{*}\right) -f_{i}^{e}\left( N_{0},C^{\prime }\right) =e_{i}\left( C_{N}^{*},\alpha \right) =\alpha o_{i}\left( N\right) . \end{aligned}$$

Using arguments similar to those used above we can prove that if $f^{e}$ is associated with some $e$ as in the statement, then $f^{e}=f^{o}$ where $o\left( N\right) =e\left( C^{*},1\right)$. Notice that, by hypothesis, $o\left( N\right)$ does not depend on $C^{*}$.

$\left( 2\right)$ I is a trivial consequence of part $\left( 1\right)$ and the definition of optimistic weighted Shapley rules.

$\left( 3\right)$ I is a trivial consequence of part $\left( 1\right)$ and the definition of the folk rule. $\square$

1.5 Proof of Proposition 5.2

We prove that the extra-costs function $e$ satisfies the $\textit{ANM}$ property, which implies, under Theorem 3, that $f^{e}$ satisfies $\textit{CM}$ and ${ PM}$.

Consider a disjoint sequence $\left\{ \left( N^{\gamma },C^{\gamma }\right) \right\} _{\gamma =1}^{\Gamma }\subset \mathcal {C}^{*}$, $i\in N^{\gamma _{i}}\subset N$ with $\gamma _{i}\ne 2$, $a\in \mathbb {R}_{+}^{\Gamma }$ with $a_{\gamma }\ge \max C^{\gamma +1}-\max C^{\gamma }$ for all $\gamma =1,\dots ,\Gamma -1$, and $y\in \left[ 0,a_{2}\right]$ ($y\ge 0$ when $\Gamma =1$). We will prove that

$$\begin{aligned} \sum _{\gamma =\gamma _{i}}^{\Gamma }e_{i}\left( C^{\gamma }\left( a^{\prime }\right) ,a_{\gamma }^{\prime }\right) \ge \sum _{\gamma =\gamma _{i}}^{\Gamma }e_{i}\left( C^{\gamma }\left( a\right) ,a_{\gamma }\right) . \end{aligned}$$

If $\Gamma =1$ the result is obvious. Assume now that $\Gamma >1$. Since $a_{\gamma }^{\prime }=a_{\gamma }$ when $\gamma \ge 3$,

$$\begin{aligned} e_{i}\left( C^{\gamma }\left( a^{\prime }\right) ,a_{\gamma }^{\prime }\right)&= \int _{0}^{a_{\gamma }^{\prime }}o_{i}^{x}\left( N^{1}\cup \ldots \cup N^{\gamma }\right) dx \\&= \int _{0}^{a_{\gamma }}o_{i}^{x}\left( N^{1}\cup ldots\cup N^{\gamma }\right) dx=e_{i}\left( C^{\gamma }\left( a\right) ,a_{\gamma }\right) \end{aligned}$$

for all $\gamma \ge 3$.

In particular, if $\gamma _{i}\ge 3$ the inequality holds. Hence, we assume $i\in N^{1}$. We know that $e_{i}\left( C^{\gamma }\left( a^{\prime }\right) ,a_{\gamma }^{\prime }\right) =e_{i}\left( C^{\gamma }\left( a\right) ,a_{\gamma }\right)$ for all $\gamma \ge 3$. Thus, it is enough to prove that

$$\begin{aligned} \sum _{\gamma =1}^{2}e_{i}\left( C^{\gamma }\left( a^{\prime }\right) ,a_{\gamma }^{\prime }\right) \ge \sum _{\gamma =1}^{2}e_{i}\left( C^{\gamma }\left( a\right) ,a_{\gamma }\right) . \end{aligned}$$

We make some computations:

$$\begin{aligned} e_{i}\left( C^{1}\left( a^{\prime }\right) ,a_{1}^{\prime }\right)&= \int _{0}^{a_{1}^{\prime }}o_{i}^{x}\left( N^{1}\right) dx=\int _{0}^{a_{1}+y}o_{i}^{x}\left( N^{1}\right) dx \\ e_{i}\left( C^{2}\left( a^{\prime }\right) ,a_{2}^{\prime }\right)&= \int _{0}^{a_{2}^{\prime }}o_{i}^{x}\left( N^{1}\cup N^{2}\right) dx=\int _{0}^{a_{2}-y}o_{i}^{x}\left( N^{1}\cup N^{2}\right) dx \\ e_{i}\left( C^{1}\left( a\right) ,a_{1}\right)&= \int _{0}^{a_{1}}o_{i}^{x}\left( N^{1}\right) dx, \quad \text {and} \\ e_{i}\left( C^{2}\left( a\right) ,a_{2}\right)&= \int _{0}^{a_{2}}o_{i}^{x}\left( N^{1}\cup N^{2}\right) dx. \end{aligned}$$

Thus, the inequality holds if and only if

$$\begin{aligned}&\int _{0}^{a_{1}+y}o_{i}^{x}\left( N^{1}\right) dx+\int _{0}^{a_{2}-y}o_{i}^{x}\left( N^{1}\cup N^{2}\right) dx\\&\quad \ge \int _{0}^{a_{1}}o_{i}^{x}\left( N^{1}\right) dx+\int _{0}^{a_{2}}o_{i}^{x}\left( N^{1}\cup N^{2}\right) dx. \end{aligned}$$

Equivalently,

$$\begin{aligned} \int _{a_{1}}^{a_{1}+y}o_{i}^{x}\left( N^{1}\right) dx \quad \ge \int _{a_{2}-y}^{a_{2}}o_{i}^{x}\left( N^{1}\cup N^{2}\right) dx \end{aligned}$$

which is a particular case of the condition given in (3).$\square$