Marginalism, egalitarianism and efficiency in multi-choice games

Abstract

The search for a compromise between marginalism and egalitarianism has given rise to many discussions. In the context of cooperative games, this compromise can be understood as a trade-off between the Shapley value and the Equal division value. We investigate this compromise in the context of multi-choice games in which players have several activity levels. To do so, we propose new extensions of the Shapley value and of the Equal division value to multi-choice games. Contrary to the existing solution concepts for multi-choice games, each one of these values satisfies a Core condition introduced by Grabisch and Xie (Math Methods Oper Res 66(3):491–512, 2007), namely Multi-Efficiency. We compromise between marginalism and egalitarianism by introducing the multi-choice Egalitarian Shapley values, computed as the convex combination of our extensions. To conduct this study, we introduce new axioms for multi-choice games. This allows us to provide an axiomatic foundation for each of these values.

Appendix

This section contains all the proofs of our results. In order to carry out these proofs, we introduce some definitions and remarks. For each $s\in {\mathcal {M}}$, denote by $s^{T}$ the highest activity level played in s, i.e., $s^{T}=\max _{i\in N}s_{i}$. Consider a multi-choice coalition $t\in {\mathcal {M}}$, $t\ne (0,\ldots , 0)$. The Dirac game $(m,\delta _{t})\in {\mathcal {G}}$, induced by t, is defined as

$$\begin{aligned} \forall s\in {\mathcal {M}},\quad \delta _{t}(s)={\left\{ \begin{array}{ll}1 &{}\text { if } s_{i}= t_{i} \text { for each }i\in N, \\ 0 &{}\text { otherwise}. \end{array}\right. } \end{aligned}$$

(19)

For each multi-choice game $(m,v)\in {\mathcal {G}}$, the characteristic function v admits a unique linear decomposition in terms of Dirac games as follows

$$\begin{aligned} v=\sum _{t\le m}v(t)\delta _{t}. \end{aligned}$$

(20)

Remark 5

For $t\in {\mathcal {M}}$, $t\ne {\mathbf {0}}$, we formulate two distinct remarks regarding minimal effort games. Take any $(m,u_{t})\in {\mathcal {G}}$. Each pair $(i,j)\in M^{+}$, such that $j> t_{i}$, is a null pair in $(m,u_{t})$. Let $(i,j),(i',j)\in M^{+}$ be two distinct pairs such that $j\le t_{i}$ and $j\le t_{i'}$. Both pairs are equal in $(m,u_{t})$.

Remark 6

For $t\in {\mathcal {M}}$, $t\ne {\mathbf {0}}$, we formulate two distinct remarks regarding Dirac games. Take any $(m,\delta _{t})\in {\mathcal {G}}$. Each pair $(i,j)\in M^{+}$, such that $j> t_{i}+1$, is a null pair in $(m,\delta _{t})$. If there exists two distinct players $i,i'\in N$ such that $t_{i}=t_{i'}$, then $(i,t_{i})$ and $(i',t_{i'})$ are equal in $(m,\delta _{t})$.

1.1 Proof of Proposition 3.2

We show that the multi-choice Shapley value admits an alternative expression given by (14). Observe that there are $|{\overline{O}}_{l}|=|Q(l)|!$ ways to order the pairs in $M^{+,l}$, for each $l\le \max _{k\in N}m_{k}$. Additionally, there are $\prod _{l<j}|Q(l)|!$ ways to order the pairs in $M^{+,1}$, then the pairs in $M^{+,2}$, and so on, until the pairs in $M^{+,l-1}$. Similarly, there are $\prod _{l>j}|Q(l)|!$ ways to order the pairs in $M^{+,j+1}$, then the pairs in $M^{+,j+2}$, and so on. Observe that, for each $\sigma \in {\overline{O}}$, there exists exactly one order $\sigma _{j}\in {\overline{O}}_{j}$ such that $s^{\sigma ,\sigma (i,j)}=s^{\sigma _{j},\sigma _{j}(i,j)}$. Additionally, for each $\sigma _{j}\in {\overline{O}}_{j}$, there are $\prod _{l<j}|Q(l)|!\times \prod _{l>j}|Q(l)|!$ orders $\sigma \in {\overline{O}}$ such that $s^{\sigma ,\sigma (i,j)}=s^{\sigma _{j},\sigma _{j}(i,j)}$. It follows that, for each $(m,v)\in {\mathcal {G}}$ and $(i,j)\in M^{+}$,

$$\begin{aligned} \varphi _{ij}(m,v)&=\frac{1}{\prod _{l\le \max _{k\in N}m_{k}}|Q(l)|!}\sum _{\sigma \in {\overline{O}}}\Big [v(s^{\sigma ,\sigma (i,j)})-v(s^{\sigma ,\sigma (i,j)-1})\Big ]\\&=\frac{(\prod _{l<j}|Q(l)|!)(\prod _{l>j}|Q(l)|!)}{\prod _{l\le \displaystyle \max _{k\in N}m_{k}}|Q(l)|!}\sum _{\sigma _{j}\in {\overline{O}}_{j} }\Big [v(s^{\sigma _{j},\sigma _{j}(i,j)})-v(s^{\sigma _{j},\sigma _{j}(i,j)-1})\Big ]. \end{aligned}$$

The first line comes from the definition of the multi-choice Shapley value, the second line follows from (13) and the fact that there are $\prod _{l<j}|Q(l)|!\times \prod _{l>j}|Q(l)|!$ orders $\sigma \in {\overline{O}}$ such that $s^{\sigma ,\sigma (i,j)}=s^{\sigma _{j},\sigma _{j}(i,j)}$, for each $\sigma _{j}\in {\overline{O}}_{j}$. Since $\prod _{l\le \max _{k\in N}m_{k}}|Q(l)|!=\prod _{l<j}|Q(l)|!\times |Q(j)|!\times \prod _{l>j}|Q(l)|!$, we obtain the desired result:

$$\begin{aligned} \forall (i,j)\in M^{+}, \quad \varphi _{ij}(m,v)=\frac{1}{|Q(j)|!}\sum _{\sigma _{j} \in {\overline{O}}_{j} }\Big [v(s^{\sigma _{j},\sigma _{j}(i,j)})-v(s^{\sigma _{j},\sigma _{j}(i,j)-1})\Big ]. \end{aligned}$$

$\square$

1.2 Proof of Proposition 2

Let $(m,v)\in {\mathcal {G}}$, $h\le \max _{k\in N}m_{k}$ and f a value satisfying (E) and (IH). Consider the sub-game $((h\wedge m_{k})_{k\in N},v)$. By (E), it holds that

$$\begin{aligned} \sum _{i\in N}\sum _{j=1}^{h\wedge m_{i}}f_{ij}((h\wedge m_{k})_{k\in N},v)&=v((h\wedge m_{k})_{k\in N}). \end{aligned}$$

(21)

By (IH), it holds that

$$\begin{aligned} \sum _{i\in N}\sum _{j=1}^{h\wedge m_{i}}f_{ij}((h\wedge m_{k})_{k\in N},v)&=\sum _{i\in N}\sum _{j=1}^{h\wedge m_{i}}f_{ij}(m,v). \end{aligned}$$

(22)

Combining (21) with (22), we obtain the desired result. $\square$

1.3 Proof of Theorem 1

The proof is divided in two steps.

Step 1: We show that $\varphi$ satisfies all the axioms of the statement of Theorem 1. For each $(m,v)\in {\mathcal {G}}$, it holds that

$$\begin{aligned} \sum _{i\in N}\sum _{j\in M^{+}_{i}}\varphi _ {ij}(m,v)&= \sum _{j\le \displaystyle \max _{k\in N}m_{k}}\sum _{i\in Q(j)} \varphi _ {ij}(m,v)\\&{\mathop {=}\limits ^{(14)}} \sum _{j\le \displaystyle \max _{k\in N}m_{k}}\frac{1}{|Q(j)|!} \sum _{\sigma _{j} \in {\overline{O}}_{j} } \sum _{i\in Q(j)} \Big [v(s^{\sigma _{j},\sigma _{j}(i,j)})-v(s^{\sigma _{j},\sigma _{j}(i,j)-1})\Big ]. \end{aligned}$$

Observe that, for each $\sigma _{j} \in {\overline{O}}_{j}$,

$$\begin{aligned}&\sum _{i\in Q(j)}\Big [ v(s^{\sigma _{j},\sigma _{j}(i,j)})-v(s^{\sigma _{j},\sigma _{j}(i,j)-1})\Big ] =v(s^{\sigma _{j},|Q(j)|})-v(s^{\sigma _{j},0}). \end{aligned}$$

By (13), for each $\sigma _{j}\in {\overline{O}}_{j}$, it holds that

$$\begin{aligned} s^{\sigma _{j},|Q(j)|}=(j\wedge m_{k})_{k\in N}, \quad \text {and} \quad s^{\sigma _{j},0}=((j-1)\wedge m_{k})_{k\in N}. \end{aligned}$$

It follows that

$$\begin{aligned} \sum _{i\in N}\sum _{j\in M^{+}_{i}}\varphi _ {ij}(m,v)&= \sum _{j \le \displaystyle \max _{k\in N}m_{k}}\frac{1}{|Q(j)|!} \sum _{\sigma _{j} \in {\overline{O}}_{j} }\Big [ v((j\wedge m_{k})_{k\in N} )-v(((j-1)\wedge m_{k})_{k\in N})\Big ]. \end{aligned}$$

(23)

Since the quantity $v((j\wedge m_{k})_{k\in N} )-v(((j-1)\wedge m_{k})_{k\in N})$ is independent from any order $\sigma _{j} \in {\overline{O}}_{j}$, it follows that it is summed as many times in (23) as there are orders in ${\overline{O}}_{j}$. Therefore, it holds that

$$\begin{aligned} \sum _{i\in N}\sum _{j\in M^{+}_{i}}\varphi _ {ij}(m,v)&=\sum _{j\le \displaystyle \max _{k\in N}m_{k}} \frac{1}{|Q(j)|!}Q(j)!\Big [v((j\wedge m_{k})_{k\in N})-v(((j-1)\wedge m_{k})_{k\in N}\Big ]\\&=\sum _{j\le \displaystyle \max _{k\in N}m_{k}}\Big [ v((j\wedge m_{k})_{k\in N})-v(((j-1)\wedge m_{k})_{k\in N})\Big ]\\&=v(m), \end{aligned}$$

which shows that the value satisfies (E). By the definition of the multi-choice Shapley value (see (14)), the payoff of a pair is independent of any activity level different from the activity level contained in this pair. Therefore, we have that $\varphi$ satisfies (IH). (L) follows directly from (14). By the definition of equal pairs (see (3)), $\varphi$ satisfies (ET), which implies that $\varphi$ satisfies (SS). By the definition of null pairs (see (2)), we have that $\varphi$ satisfies (N). This concludes Step 1.

Step 2: To complete the proof, it remains to show that there is at most one value satisfying all the axioms of the statement of Theorem 1. Take any f satisfying all the axioms of the statement of Theorem 1. Consider any $(m,v)\in {\mathcal {G}}$. We know that each multi-choice game admits a unique linear decomposition in terms of minimal effort games $\{u_{s}\}_{s\in {\mathcal {M}}}$. Consider $s\in {\mathcal {M}}$ such that $s\ne {\mathbf {0}}$. The set of top pairs T(s) (see (1)) can be re-written as

$$\begin{aligned} T(s)=\{(i,s^{T})\in M^{+,s^{T}}:s_{i}=s^{T}\}, \end{aligned}$$

where $s^{T}=\max _{i\in N}s_{i}$. Let us show that $f(m,u_{s})$ is uniquely determined. We divide this Step into several sub-steps.

Step 2.1. Let us show that, for each $(i,j)\in M^{+}$ such that $j\ne s^{T}$, $f_{ij}(m,u_{s})$ is uniquely determined.

Step 2.1.1. If $j<s^{T}$, then $(j\wedge m_{k})_{k\in N} \ngeq s$. It follows that $((j\wedge m_{k})_{k\in N},u_{s})$ is the null game since $u_{s}(t)=0$ for each $t\le (j\wedge m_{k})_{k\in N}$. Recall that each pair is a null pair (see (2)) in the null game. Combining these observations with (N), for each $(i,j)\in M^{+}$ such that $j< s^{T}$, we obtain

$$\begin{aligned} f_{ij}(m,u_{s}){\mathop {=}\limits ^{(IH)}}f_{ij}((j\wedge m_{k})_{k\in N},u_{s}) {\mathop {=}\limits ^{(N)}}0. \end{aligned}$$

Step 2.1.2. If $j> s^{T}$ then, by Remark 5, (i, j) is a null pair in $(m,u_{s})$. By (N), for each $(i,j)\in M^{+}$ such that $j> s^{T}$, it holds that

$$\begin{aligned} f_{ij}(m,u_{s}){\mathop {=}\limits ^{(N)}}0. \end{aligned}$$

We have shown that $f_{ij}(m,u_{s})=0$, and so is uniquely determined for each $(i,j)\in M^{+}$ such that $j\ne s^{T}$.

Step 2.2. Now, we show that, for each pair $(i,j)\in M^{+}$ such that $j=s^{T}$ i.e. each pair $(i,s^{T})\in M^{+,s^{T}}$, $f_{is^{T}}(m,u_{s})$ is uniquely determined. To that end, consider the game $(m,w)\in {\mathcal {G}}$ defined as

$$\begin{aligned} \forall t\le m, \quad w(t)=u_{s}(t) - \sum _{\begin{array}{c} (i,s^{T})\in T(s) \end{array}}\varphi _{is^{T}}(m,u_{s})u_{(0_{-i},s^{T})}(t). \end{aligned}$$

(24)

Step 2.2.1. We show that

$$\begin{aligned} \sum _{(i,s^{T})\in M^{+,s^{T}}}f_{is^{T}}(m,w)=0. \end{aligned}$$

We consider pairs in $M^{+,s^{T}}$. By definition of $M^{+,s^{T}}$, observe that

$$\begin{aligned} \sum _{i\in Q(s^{T})}f_{is^{T}}(m,w)=\sum _{(i,s^{T})\in M^{+,s^{T}}}f_{is^{T}}(m,w). \end{aligned}$$

We have that any pair $(i,s^{T})\in M^{+,s^{T}}$ is either in T(s) or not. Since f satisfies (E) and (IH), by Proposition 2, f also satisfies (ME). Therefore, it holds that

$$\begin{aligned} \sum _{(i,s^{T})\in M^{+,s^{T}}}f_{is^{T}}(m,w){\mathop {=}\limits ^{(ME)}}&w((s^{T}\wedge m_{k})_{k\in N})-w(((s^{T}-1)\wedge m_{k})_{k\in N}) \nonumber \\ {\mathop {=}\limits ^{(24)}}&u_{s}((s^{T}\wedge m_{k})_{k\in N}) - \sum _{\begin{array}{c} (i,s^{T})\in T(s) \end{array}}\varphi _{is^{T}}(m,u_{s})u_{(0_{-i},s^{T})}((s^{T}\wedge m_{k})_{k\in N}) \nonumber \\ &-u_{s}(((s^{T}-1)\wedge m_{k})_{k\in N})\nonumber \\ &+\sum _{\begin{array}{c} (i,s^{T})\in T(s) \end{array}}\varphi _{is^{T}}(m,u_{s})u_{(0_{-i},s^{T})}(((s^{T}-1)\wedge m_{k})_{k\in N}). \end{aligned}$$

(25)

Observe that $((s^{T}\wedge m_{k})_{k\in N})\ge s\ge ((0_{-i},s^{T})$, $(((s^{T}-1)\wedge m_{k})_{k\in N})\not \ge s$ and $(((s^{T}-1)\wedge m_{k})_{k\in N})\not \ge (0_{-i},s^{T})$, where $(i,s^{T})\in T(s)$. By definition of a minimal effort game (5), it holds that

$$\begin{aligned}&u_{s}((s^{T}\wedge m_{k})_{k\in N})=1,\quad&\text {and,}\quad&\forall (i,s^{T})\in M^{+,s^{T}}, \quad u_{(0_{-i},s^{T})}((s^{T}\wedge m_{k})_{k\in N})=1,\\&u_{s}(((s^{T}-1)\wedge m_{k})_{k\in N})=0,\quad&\text {and,}\quad&\forall (i,s^{T})\in M^{+,s^{T}},\quad u_{(0_{-i},s^{T})}(((s^{T}-1)\wedge m_{k})_{k\in N})=0. \end{aligned}$$

It follows that (25) becomes

$$\begin{aligned} \sum _{(i,s^{T})\in M^{+,s^{T}}}f_{is^{T}}(m,w)=&1-\sum _{(i,s^{T})\in T(s) }\varphi _{is^{T}}(m,u_{s})-0+0. \end{aligned}$$

(26)

Observe that, since $(i,s^{T})\notin T(s)$ if and only if $s^{T}>s_{i}$, then each $(i,s^{T})\notin T(s)$ is a null pair in $(m,u_{s})$. Since $\varphi$ satisfies (N), we have that $\varphi _{is^{T}}(m,u_{s})=0$ for each $(i,s^{T})\notin T(s)$. Since $\varphi$ satisfies (E), (IH), by Proposition 2 the value satisfies (ME). Therefore, it holds that

$$\begin{aligned} \sum _{\begin{array}{c} (i,s^{T})\in T(s) \end{array}}\varphi _{is^{T}}(m,u_{s}){\mathop {=}\limits ^{(N)}}&\sum _{(i,s^{T})\in T(s) }\varphi _{is^{T}}(m,u_{s})+\sum _{\begin{array}{c} (i,s^{T})\notin T(s) \end{array}}\varphi _{is^{T}}(m,u_{s})\\ =&\sum _{ (i,s^{T})\in M^{+,s^{T}} }\varphi _{is^{T}}(m,u_{s})\\ {\mathop {=}\limits ^{(ME)}}&u_{s}((s^{T}\wedge m_{k})_{k\in N})\\ =&1. \end{aligned}$$

Therefore, (26) becomes

$$\begin{aligned} \sum _{(i,s^{T})\in M^{+,s^{T}}}f_{is^{T}}(m,w)=&1-1=0, \end{aligned}$$

(27)

which concludes Step 2.2.1.

Step 2.2.2. We show that, for each $(i,s^{T})\in M^{+,s^{T}}$,

$$\begin{aligned} f_{is^{T}}(m,w)=0. \end{aligned}$$

We know that each pair $(i,s^{T})\notin T(s)$ is a null pair in $(m,u_{s})$. Moreover, each pair $(i,s^{T})\notin T(s)$ is a null pair in each $(m,u_{0_{-i'},s^{T}})$, $(i',s^{T})\in T(s)$. Indeed, in $(m,u_{0_{-i'},s^{T}})$, $(i',s^{T})$ is the only productive pair and all other pairs are null pairs. It follows that each pair $(i,s^{T})\notin T(s)$ is a null pair in (m, w). By (N), for each $(i,s^{T})\notin T(s)$, it holds that

$$\begin{aligned} f_{is^{T}}(m,w)=0. \end{aligned}$$

(28)

It follows that

$$\begin{aligned} \sum _{(i,s^{T})\in M^{+,s^{T}}}f_{is^{T}}(m,w)=&\sum _{ (i,s^{T})\in T(s) }f_{is^{T}}(m,w)+\sum _{(i,s^{T})\notin T(s) }f_{is^{T}}(m,w) \nonumber \\ {\mathop {=}\limits ^{(N)}}&\sum _{ (i,s^{T})\in T(s)}f_{is^{T}}(m,w)+0\nonumber \\ {\mathop {=}\limits ^{(27)}}&0. \end{aligned}$$

(29)

To complete the proof of Step 2.2.2, it remains to show that if there exist two distinct pairs $(i,s^{T}),(i',s^{T})\in T(s)$, then these pairs are equal. By Remark 5 two distinct pairs $(i,s^{T}),(i',s^{T})\in T(s)$ are equal in $(m,u_{s})$. Since $\varphi$ satisfies (ET), it follows that $\varphi _{is^{T}}(m,u_{s})=\varphi _{i's^{T}}(m,u_{s})$. By definition of a minimal effort game, for each $t\in {\mathcal {M}}$ such that $t_{i}=t_{i'}=s^{T}-1$, it holds that

$$\begin{aligned}&u_{(0_{-i},s^{T})}(t)=u_{(0_{-i'},s^{T})}(t)=0, \end{aligned}$$

(30)

$$\begin{aligned} \text {and}\quad&u_{(0_{-i},s^{T})}(t+e_{i})=u_{(0_{-i'},s^{T})}(t+e_{i'})=1. \end{aligned}$$

(31)

Therefore, for each $t\in {\mathcal {M}}$ such that $t_{i}=t_{i'}=s^{T}-1$, we have that

$$\begin{aligned} \sum _{\begin{array}{c} (k,s^{T})\in T(s) \end{array}}\varphi _{ks^{T}}(m,u_{s})u_{(0_{-k},s^{T})}(t+e_{i})&=\sum _{\begin{array}{c} (k,s^{T})\in T(s) \end{array}}\varphi _{ks^{T}}(m,u_{s})u_{(0_{-k},s^{T})}(t)+\varphi _{is^{T}}(m,u_{s})\\&=\sum _{\begin{array}{c} (k,s^{T})\in T(s) \end{array}}\varphi _{ks^{T}}(m,u_{s})u_{(0_{-k},s^{T})}(t)+\varphi _{i's^{T}}(m,u_{s})\\&=\sum _{\begin{array}{c} (k,s^{T})\in T(s) \end{array}}\varphi _{ks^{T}}(m,u_{s})u_{(0_{-k},s^{T})}(t+e_{i'}). \end{aligned}$$

Where the first equality and third equality follow from (30) and (31), and the second equality follows from $\varphi _{is^{T}}(m,u_{s})=\varphi _{i's^{T}}(m,u_{s})$, since $\varphi$ satisfies (ET). It follows that

$$\begin{aligned} w(t+e_{i})=w(t+e_{i'}), \end{aligned}$$

for each $t\in {\mathcal {M}}$ such that $t_{i}=t_{i'}=s^{T}-1$, showing that $(i,s^{T}),(i',s^{T})\in T(s)$ are equal pairs in (m, w). By (SS), we have that $sign(f_{is^{T}}(m,w))=sign(f_{i's^{T}}(m,w))$. It follows from (29) that, for each $(i,s^{T})\in T(s)$,

$$\begin{aligned} f_{is^{T}}(m,w)=0. \end{aligned}$$

(32)

Combining (28) with (32), the proof of Step 2.2.2 is complete.

Step 2.2.3. We show that, for each $(i,s^{T})\in M^{+,s^{T}}$,

$$\begin{aligned} f_{is^{T}}(m,u_{s})&=\varphi _{is^{T}}(m,u_{s}). \end{aligned}$$

By (24), (32) and (L), for each $(i,s^{T})\in M^{+,s^{T}}$, it holds that

$$\begin{aligned} f_{is^{T}}(m,w)&{\mathop {=}\limits ^{(24),(L)}}f_{is^{T}}(m,u_{s})-f_{is^{T}}\Big (m,\sum _{ (k,s^{T})\in T(s) }\varphi _{ks^{T}}(m,u_{s})u_{(0_{-k},s^{T})}\Big )\\ \iff f_{is^{T}}(m,u_{s})&{\mathop {=}\limits ^{(32)}}f_{is^{T}}\Big (m,\sum _{(k,s^{T})\in T(s) }\varphi _{ks^{T}}(m,u_{s})u_{(0_{-k},s^{T})}\Big )\\&{\mathop {=}\limits ^{(L)}}\sum _{ (k,s^{T})\in T(s) }\varphi _{ks^{T}}(m,u_{s})f_{is^{T}}\Big (m,u_{(0_{-k},s^{T})}\Big ). \end{aligned}$$

Additionally, by (N) and (ME), we have that

$$\begin{aligned} f_{is^{T}}(m,u_{(0_{-i},s^{T})})=1 \end{aligned}$$

since $(i,s^{T})$ is the only productive pair in $(m,u_{(0_{-i},s^{T})})$. Therefore, for each $(i,s^{T})\in M^{+,s^{T}}$, it holds that

$$\begin{aligned}&\varphi _{ks^{T}}(m,u_{s})f_{is^{T}}\Big (m,u_{(0_{-k},s^{T})}\Big )={\left\{ \begin{array}{ll} \varphi _{ks^{T}}(m,u_{s}) &{} \text {if }k=i,\\ 0 &{} \text {otherwise.} \end{array}\right. } \end{aligned}$$

It follows that, for each $(i,s^{T})\in M^{+,s^{T}}$,

$$\begin{aligned} f_{is^{T}}(m,u_{s})&=\varphi _{is^{T}}(m,u_{s}). \end{aligned}$$

Thus, $f_{is^{T}}(m,u_{s})$ is uniquely determined. This concludes Step 2.2.3.

From Step 2.1 and Step 2.2, we conclude that $f(m,u_{s})$ is uniquely determined. By (L), we have that f(m, v) is uniquely determined, which concludes the proof of Theorem 1. $\square$

Logical independence: The axioms invoked in Theorem 1 are logically independent, as shown by the following alternative solutions.

• The value f defined as: for each $(m,v)\in {\mathcal {G}}$,

  $$\begin{aligned} \forall (i,j)\in M^{+},\quad f_{ij}(m,v)=0, \end{aligned}$$

  satisfies all the axioms except (E).

• The value f defined as:

  $$\begin{aligned} f_{ij}(m,v)= \left\{ \begin{array}{cl} \varphi _{ij}(m,v) + \displaystyle \frac{\Delta _{v}((2\wedge m_{k})_{k\in N})}{|Q(1)|} &{}\text {if }j=1 \text { and } m^{T}>1,\\ &{} \\ \varphi _{ij}(m,v) - \displaystyle \frac{\Delta _{v}((2\wedge m_{k})_{k\in N})}{|Q(2)|} &{}\text {if }j=2 \text { and } m^{T}>1,\\ &{} \\ \varphi _{ij}(m,v) &{}\text {otherwise,} \end{array} \right. \end{aligned}$$

  satisfies all the axioms except (IH).

• The value f defined as: for each $(m,v)\in {\mathcal {G}}$,

  $$\begin{aligned} \forall (i,j)\in M^{+},\quad f_{ij}(m,v)= \sum _{\scriptstyle\begin{array}{c} s\le m \\ (i,j)\in T(s) \end{array}}\frac{(v((j-1\wedge m_{h})_{h\in N})+e_{i})^{2}+1}{\sum _{(k,s_{k})\in T(s)}v((j-1\wedge m_{h})_{h\in N})+e_{k})^{2})+1}\Delta _{v}(s), \end{aligned}$$

  satisfies all the axioms except (L).

• The multi-choice Equal division value $\xi$ satisfies all the axioms except (N).

• Take any $(m,v)\in {\mathcal {G}}$, and for each $(i,j)\in M^{+}$ fix any arbitrary $\beta _{ij}\in {\mathbb {R}}_{++}$. The value $f^{\beta }$ defined as: for each $(m,v)\in {\mathcal {G}}$,

  $$\begin{aligned} \forall (i,j)\in M^{+},\quad f^{\beta }_{ij}(m,v)=\sum _{\scriptstyle\begin{array}{c} s\le m\\ (i,j)\in T(s) \end{array}}\frac{\beta _{ij}}{\sum _{(k,l)\in T(s)}\beta _{kl}}\Delta _{v}(s), \end{aligned}$$

  satisfies all the axioms except (SS).

1.4 Proof of Corollary 2

By the the proof of Theorem 1, $\varphi$ satisfies (E), (IH), (L), (N) and (ET). Consider $(m,u_{s})\in {\mathcal {G}}$, $s\in {\mathcal {M}}$ such that $s\ne {\mathbf {0}}$. Similarly to (28), for each $(i,j)\notin T(s)$, (N) and (ME) imply

$$\begin{aligned} \varphi _{ij}(m,u_{s})=0. \end{aligned}$$

(33)

All pairs in T(s) pairs are equal pairs in $(m,u_{s})$. Thus, by (ET), it holds that

$$\begin{aligned} \varphi _{is^{T}}(m,u_{s})=\cdots =\varphi _{i's^{T}}(m,u_{s}). \end{aligned}$$

(34)

By (E) and (L), we obtain the desired result. $\square$

1.5 Proof of Theorem 2

From Theorem 1, we know that $\varphi$ satisfies (E), (IH) and (SS). By definition (see (14)), the multi-choice Shapley value satisfies (SM).

Next, we show that $\varphi$ is the unique value satisfying all the axioms of the statement of Theorem 2. Take any f satisfying all the axioms of the statement of and consider any $(m,v)\in {\mathcal {G}}$. Recall that $(m,v)\in {\mathcal {G}}$ can be rewritten as $(m,\sum _{t\in M}\Delta _{v}(t)u_{t})$. We define the set of coalitions for which the Harsanyi dividend is non null as

$$\begin{aligned} {\mathcal {T}}(v)=\{t\in {\mathcal {M}}\;|\; \Delta _{v}(t)\ne 0\}. \end{aligned}$$

By induction on the cardinality of ${\mathcal {T}}(v)$, we show that

$$\begin{aligned} f(m,v)=\varphi (m,v). \end{aligned}$$

Initialization: If $|{\mathcal {T}}(v)|=0$, then each dividend is null. The only game $(m,v)\in {\mathcal {G}}$ such that $|{\mathcal {T}}(v)|=0$ is the null game. Recall that $M^{+,j}=\{(i,j)\in M^{+}: i\in Q(j)\}$. Since f satisfies (E), (IH), by Proposition 2, it satisfies (ME). It follows that, for each $j\le \max _{k\in N}m_{k}$,

$$\begin{aligned} \sum _{(i,j)\in M^{+,j}}f_{ij}(m,v){\mathop {=}\limits ^{(ME)}}&v((j\wedge m_{k})_{k\in N})-v(((j-1)\wedge m_{k})_{k\in N})\nonumber \\ =\,&0. \end{aligned}$$

(35)

Recall that any two distinct pairs $(i,j)(i',j)\in M^{+,j}$ are equal pairs in the null game (m, v). Therefore, by (SS), we obtain

$$\begin{aligned} sign(f_{ij}(m,v))=sign(f_{i'j}(m,v)). \end{aligned}$$

(36)

Combining (35) and (36), for each $j\le \max _{k\in N}m_{k}$ and each $(i,j)\in M^{+,j}$, we obtain

$$\begin{aligned} f_{ij}(m,v)=0. \end{aligned}$$

Recall also that each pair is a null pair in the null game. Since $\varphi$ satisfies (N), for each $j\le \max _{k\in N}m_{k}$ and each $(i,j)\in M^{+,j}$, it holds that

$$\begin{aligned} \varphi _{ij}(m,v){\mathop {=}\limits ^{(N)}}0=f_{ij}(m,v). \end{aligned}$$

This concludes the initialization.

Hypothesis: Fix $r\in {\mathbb {N}}$ such that $r<|{\mathcal {M}}|-1$. We assume that, for each $(m,v)\in {\mathcal {G}}$ such that $|{\mathcal {T}}(v)|\le r$,

$$\begin{aligned} f(m,v)=\varphi (m,v). \end{aligned}$$

Induction: Consider any $(m,v)\in {\mathcal {G}}$ such that $|{\mathcal {T}}(v)|=r+1$. Let us show that

$$\begin{aligned} f(m,v)=\varphi (m,v). \end{aligned}$$

We define the minimum of the set ${\mathcal {T}}(v)$ as

$$\begin{aligned} p=\bigwedge _{t\in {\mathcal {T}}(v)}t. \end{aligned}$$

Two cases can be distinguished. First, assume that $p\ne {\mathbf {0}}$. Consider any pair $(i,j)\in M^{+}$ such that $j>p_{i}$. By definition of p, there exists a $t\in {\mathcal {T}}(v)$ such that $j>t_{i}$. For such t, consider the game $(m,v-\Delta _{v}(t)u_{t})$. By definition of a minimal effort game (5) and Remark 1, we have that (i, j) is a null pair in $(m,\Delta _{v}(t)u_{t})$ . Therefore (i, j) has the same marginal contributions in (m, v) and in $(m,v-\Delta _{v}(t)u_{t})$. Moreover, observe that $|{\mathcal {T}}(v)|>|{\mathcal {T}}(v-\Delta _{v}(t)u_{t})|$. Therefore, we have that $r \ge |{\mathcal {T}}(v-\Delta _{v}(t)u_{t})|$. By the induction hypothesis and (SM), for each $(i,j)\in M^{+}$ such that $j>p_{i}$, we obtain

$$\begin{aligned} f_{ij}(m,v){\mathop {=}\limits ^{SM}}f_{ij}(m,v-\Delta _{v}(t)u_{t}){\mathop {=}\limits ^{Hyp}}\varphi _{ij}(m,v-\Delta _{v}(t)u_{t}){\mathop {=}\limits ^{SM}}\varphi _{ij}(m,v). \end{aligned}$$

(37)

Next, we assume that $p={\mathbf {0}}$. For each $(i,j)\in M^{+}$, there exists a $t\in {\mathcal {T}}(v)$ such that $j>t_{i}$. In this case, (37) holds for each $(i,j)\in M^{+}$ and the proof is complete.

It remains to show that, if $p\ne {\mathbf {0}}$, then for each $(i,j)\in M^{+}$ such that $j\le p_{i}$,

$$\begin{aligned} f_{ij}(m,v)=\varphi _{ij}(m,v). \end{aligned}$$

We proceed in two steps.

Step 1. We define the game $(m,w)\in {\mathcal {G}}$ as

$$\begin{aligned} \displaystyle w=v-\sum _{\scriptstyle\begin{array}{c} (i,j)\in M^{+}\\ j\le p_{i} \end{array}}\varphi _{ij}(m,v)u_{(0_{-i},j)}, \end{aligned}$$

(38)

and we show that, for each $(i,l)\in M^{+}$ such that $l\le p_{i}$,

$$\begin{aligned} f_{il}(m,w)=0. \end{aligned}$$

(39)

Step 1.1. To that end, we show that

$$\begin{aligned} \sum _{\scriptstyle\begin{array}{c} (i,l)\in M^{+,l}\\ l\le p_{i} \end{array}}f_{il}(m,w)=0. \end{aligned}$$

By Proposition 2, f satisfies (ME). By (ME) and (38), for each $l\le \max _{k\in N}m_{k}$, it holds that

$$\begin{aligned} \sum _{\begin{array}{c} (i,l)\in M^{+,l} \end{array}}f_{il}(m,w){\mathop {=}\limits ^{ME}}&w((l\wedge m_{k})_{k\in N})-w((l-1\wedge m_{k})_{k\in N} \nonumber \\ \iff \sum _{\begin{array}{c} (i,l)\in M^{+,l}\\ l\le p_{i} \end{array}}f_{il}(m,w)=&w((l\wedge m_{k})_{k\in N})-w((l-1\wedge m_{k})_{k\in N}-\sum _{\begin{array}{c} (i,l)\in M^{+,l}\\ l> p_{i} \end{array}}f_{il}(m,w)\nonumber \\ {\mathop {=}\limits ^{(38)}}&v((l\wedge m_{k})_{k\in N})-v((l-1\wedge m_{k})_{k\in N} -\sum _{\begin{array}{c} (i,j)\in M^{+}\\ j\le p_{i} \end{array}}\varphi _{ij}(m,v)u_{(0_{-i},j)}((l\wedge m_{k})_{k\in N})\nonumber \\&+\sum _{\begin{array}{c} (i,j)\in M^{+}\\ j\le p_{i} \end{array}}\varphi _{ij}(m,v)u_{(0_{-i},j)}((l-1\wedge m_{k})_{k\in N})-\sum _{\begin{array}{c} (i,l)\in M^{+,l}\\ l> p_{i} \end{array}}f_{il}(m,w). \end{aligned}$$

(40)

Before proceeding further into the computation of (40), observe that

$$\begin{aligned}&-\sum _{\begin{array}{c} (i,j)\in M^{+}\\ j\le p_{i} \end{array}}\varphi _{ij}(m,v)u_{(0_{-i},j)}((l\wedge m_{k})_{k\in N})+\sum _{\begin{array}{c} (i,j)\in M^{+}\\ j\le p_{i} \end{array}}\varphi _{ij}(m,v)u_{(0_{-i},j)}((l-1\wedge m_{k})_{k\in N})\\&\quad = -\sum _{\begin{array}{c} (i,j)\in M^{+}\\ j\le p_{i}\\ j<l \end{array}}\varphi _{ij}(m,v)u_{(0_{-i},j)}((l\wedge m_{k})_{k\in N})-\sum _{\begin{array}{c} (i,j)\in M^{+}\\ j\le p_{i}\\ j=l \end{array}}\varphi _{ij}(m,v)u_{(0_{-i},j)}((l\wedge m_{k})_{k\in N})\\&\qquad-\sum _{\begin{array}{c} (i,j)\in M^{+}\\ j\le p_{i}\\ j>l \end{array}}\varphi _{ij}(m,v)u_{(0_{-i},j)}((l\wedge m_{k})_{k\in N})\\&\qquad+\sum _{\begin{array}{c} (i,j)\in M^{+}\\ j\le p_{i}\\ j<l \end{array}}\varphi _{ij}(m,v)u_{(0_{-i},j)}((l-1\wedge m_{k})_{k\in N})+\sum _{\begin{array}{c} (i,j)\in M^{+}\\ j\le p_{i}\\ j\ge l \end{array}}\varphi _{ij}(m,v)u_{(0_{-i},j)}((l-1\wedge m_{k})_{k\in N}) \end{aligned}$$

By definition, $p_{i}\le m_{i}$, for each $i\in N$. For each $i\in N$ and $j\le p_{i}\le m_{i}$, it holds that

$$\begin{aligned} u_{(0_{-i},j)}((l\wedge m_{k})_{k\in N}) {\left\{ \begin{array}{ll} 1 &{} \text {if }j\le (l\wedge m_{i}),\\ 0 &{} \text {otherwise.} \end{array}\right. } \end{aligned}$$

It follows that

$$\begin{aligned}&-\sum _{\scriptstyle\begin{array}{c} (i,j)\in M^{+}\\ j\le p_{i} \end{array}}\varphi _{ij}(m,v)u_{(0_{-i},j)}((l\wedge m_{k})_{k\in N})+\sum _{\scriptstyle\begin{array}{c} (i,j)\in M^{+}\\ j\le p_{i} \end{array}}\varphi _{ij}(m,v)u_{(0_{-i},j)}((l-1\wedge m_{k})_{k\in N})\nonumber \\&\quad =-\sum _{\scriptstyle\begin{array}{c} (i,j)\in M^{+}\\ j\le p_{i}\\ j<l \end{array}}\varphi _{ij}(m,v)-\sum _{\scriptstyle\begin{array}{c} (i,j)\in M^{+}\\ j\le p_{i}\\ j=l \end{array}}\varphi _{ij}(m,v)+\sum _{\scriptstyle\begin{array}{c} (i,j)\in M^{+}\\ j\le p_{i}\\ j<l \end{array}}\varphi _{ij}(m,v)\nonumber \\&= \quad -\sum _{\scriptstyle\begin{array}{c} (i,j)\in M^{+}\\ j\le p_{i}\\ j=l \end{array}}\varphi _{ij}(m,v)=-\sum _{\scriptstyle\begin{array}{c} (i,l)\in M^{+,l}\\ l\le p_{i} \end{array}}\varphi _{il}(m,v). \end{aligned}$$

(41)

By (41), (40) becomes

$$\begin{aligned} \sum _{\scriptstyle\begin{array}{c} (i,l)\in M^{+,l}\\ l\le p_{i} \end{array}}f_{il}(m,w)&=v((l\wedge m_{k})_{k\in N})-v((l-1\wedge m_{k})_{k\in N} -\sum _{\scriptstyle\begin{array}{c} (i,l)\in M^{+,l}\\ l\le p_{i} \end{array}}\varphi _{il}(m,v)-\sum _{\scriptstyle\begin{array}{c} (i,l)\in M^{+,l}\\ l> p_{i} \end{array}}f_{il}(m,w). \end{aligned}$$

(42)

By (37), for each $(i,l)\in M^{+,l}$ such that $l> p_{i}$, it holds that

$$\begin{aligned} f_{il}(m,w)=\varphi _{il}(m,w). \end{aligned}$$

(43)

Combining (42) and (43), we obtain

$$\begin{aligned} \sum _{\scriptstyle\begin{array}{c} (i,l)\in M^{+,l}\\ l\le p_{i} \end{array}}f_{il}(m,w)&=v((l\wedge m_{k})_{k\in N})-v((l-1\wedge m_{k})_{k\in N} -\sum _{\scriptstyle\begin{array}{c} (i,l)\in M^{+,l}\\ l\le p_{i} \end{array}}\varphi _{il}(m,v)-\sum _{\scriptstyle\begin{array}{c} (i,l)\in M^{+,l}\\ l> p_{i} \end{array}}\varphi _{il}(m,w). \end{aligned}$$

(44)

Moreover, each $(i,l)\in M^{+,l}$, such that $l> p_{i}$, is a null pair in $(m,u_{0_{-i},j})$, for each $(i,j)\in M^{+}$ such that $j\le p_{i}$. By definition of (m, w) (see (38)), it follows that each pair (i, l), such that $l> p_{i}$, has the same marginal contributions in (m, w) and in (m, v). Since $\varphi$ satisfies (SM), for each $(i,l)\in M^{+,l}$ such that $l> p_{i}$, it holds that

$$\begin{aligned} \varphi _{il}(m,w)=\varphi _{il}(m,v). \end{aligned}$$

(45)

Combining (44), (45) and the fact that $\varphi$ satisfies (ME), we obtain

$$\begin{aligned} \sum _{\scriptstyle\begin{array}{c} (i,l)\in M^{+,l}\\ l\le p_{i} \end{array}}f_{il}(m,w)=&v((l\wedge m_{k})_{k\in N})-v((l-1\wedge m_{k})_{k\in N}-\sum _{\scriptstyle\begin{array}{c} (i,l)\in M^{+,l} \end{array}}\varphi _{il}(m,v)\nonumber \\ {\mathop {=}\limits ^{(ME)}}&0. \end{aligned}$$

(46)

This concludes Step 1.1.

Step 1.2. We show that all the pairs $(i,l)\in M^{+,l}$, such that $l\le p_{i}$, are equal in (m, w).

By definition of $M^{+,l}$, we have that $l\ge 1$. Consider two pairs $(i,l),(i',l)\in M^{+,l}$ such that $l\le p_{i}$ and $l\le p_{i'}$. Since $p=\bigwedge _{t\in {\mathcal {T}}(v)}t$, we have that each $t\in {\mathcal {T}}(v)$ verifies $t_{i}\ge l$ and $t_{i'}\ge l$. In other words, for each $s\in {\mathcal {M}}$, such that $s_{i}< l$ or $s_{i'}<l$, we have that $\Delta _{v}(s)=0$. Therefore, for each $s\in {\mathcal {M}}$ such that $s_{i}=s_{i'}=l-1$, we obtain

$$\begin{aligned} v(s+e_{i})=v(s+e_{i'})=0. \end{aligned}$$

(47)

Therefore, (i, l) and $(i',l)$ are equal in (m, v). Since $\varphi$ satisfies (ET) and by (38), for each $s\in {\mathcal {M}}$ such that $s_{i}=s_{i'}=l-1$, it holds that

$$\begin{aligned}w(s+e_{i})&{\mathop {=}\limits ^{(38)}}v(s+e_{i})-\sum _{\scriptstyle\begin{array}{c} (h,j)\in M^{+}\\ j\le p_{h} \end{array}}\varphi _{hj}(m,v)u_{(0_{-h},j)}(s+e_{i})\\ &\quad=v(s+e_{i})-\sum _{\scriptstyle\begin{array}{c} (h,j)\in M^{+}\\ j\le p_{h}\\ h\ne i,i' \end{array}}\varphi _{hj}(m,v)u_{(0_{-h},j)}(s+e_{i})-\sum _{\scriptstyle\begin{array}{c} j\le p_{i'}\\ j\le l-1 \end{array}}\varphi _{i'j}(m,v)u_{(0_{-i'},j)}(s+e_{i})\\&\qquad -\sum _{\scriptstyle\begin{array}{c} j\le p_{i}\\ j\le l-1 \end{array}}\varphi _{ij}(m,v)u_{(0_{-i},j)}(s+e_{i})-\varphi _{il}(m,v)u_{(0_{-i},l)}(s+e_{i})\\&\quad=v(s+e_{i})-\sum _{\scriptstyle\begin{array}{c} (h,j)\in M^{+}\\ j\le p_{h}\\ h\ne i,i' \end{array}}\varphi _{hj}(m,v)u_{(0_{-h},j)}(s+e_{i})-\sum _{\scriptstyle\begin{array}{c} j\le p_{i'}\\ j\le l-1 \end{array}}\varphi _{i'j}(m,v)u_{(0_{-i'},j)}(s+e_{i})\\&\qquad -\sum _{\scriptstyle\begin{array}{c} j\le p_{i}\\ j\le l-1 \end{array}}\varphi _{ij}(m,v)u_{(0_{-i},j)}(s+e_{i})-\varphi _{il}(m,v)\\&{\mathop {=}\limits ^{(ET),(47)}}v(s+e_{i'})-\sum _{\scriptstyle\begin{array}{c} (h,j)\in M^{+}\\ j\le p_{h}\\ h\ne i,i' \end{array}}\varphi _{hj}(m,v)u_{(0_{-h},j)}(s+e_{i'})-\sum _{\scriptstyle\begin{array}{c} j\le p_{i'}\\ j\le l-1 \end{array}}\varphi _{i'j}(m,v)u_{(0_{-i'},j)}(s+e_{i'})\\&\qquad -\sum _{\scriptstyle\begin{array}{c} j\le p_{i}\\ j\le l-1 \end{array}}\varphi _{ij}(m,v)u_{(0_{-i},j)}(s+e_{i'})-\varphi _{i'l}(m,v)\\&{\mathop {=}\limits ^{(38)}}w(s+e_{i'}). \end{aligned}$$

Therefore, two pairs (i, l) and $(i',l)$, such that $l\le p_{i}$ and $l\le p_{i'}$, are equal in (m, w). This concludes Step 1.2.

By (SS), it holds that

$$\begin{aligned} sign(f_{il}(m,w))=sign(f_{i'l}(m,w)). \end{aligned}$$

(48)

Combining (46) and (48), for each $(i,l)\in M^{+}$ such that $l \le p_{i}$, we obtain

$$\begin{aligned} f_{il}(m,w)=0, \end{aligned}$$

which concludes Step 1.

Step 2. For each $(i,j)\in M^{+}$, such that $0<j\le p_{i}$, we define the game $(m,w^{ij})\in {\mathcal {G}}$ as

$$\begin{aligned} w^{ij}=v-\varphi _{ij}(m,v)u_{(0_{-i},j)}. \end{aligned}$$

(49)

In this step, we first show that, for each $(i,j)\in M^{+}$ such that $j\le p_{i}$,

$$\begin{aligned} \varphi _{ij}(m,v)=f_{ij}(m,v)-f_{ij}(m,w^{ij}). \end{aligned}$$

The game $(m,w^{ij})$ is defined in such a way that the pair (i, j) has the same marginal contribution in (m, w) as in $(m,w^{ij})$. Indeed, observe that the pair (i, j) has null marginal contributions to coalition in each game $(m,u_{(0_{-i'},j')})$ such that $i'\ne i$ or $i'=i$ and $j'\ne j$. Therefore, by (SM), for each $(i,j)\in M^{+}$ such that $j\le p_{i}$, it holds that

$$\begin{aligned} f_{ij}(m,w)=f_{ij}(m,w^{ij}). \end{aligned}$$

(50)

Additionally, by (ME), (49) and the definition of a minimal effort game (see (5)), it holds that

$$\begin{aligned} \sum _{(k,j)\in M^{+,j}}f_{kj}(m,w^{ij}){\mathop {=}\limits ^{ME}}&w^{ij}((j\wedge m_{k})_{k\in N})-w^{ij}(((j-1)\wedge m_{k})_{k\in N}) \nonumber \\ {\mathop {=}\limits ^{(49)}}&v((j\wedge m_{k})_{k\in N})-v(((j-1)\wedge m_{k})_{k\in N}) -\varphi _{ij}(m,v)u_{(0_{-i},j)}((j\wedge m_{k})_{k\in N})\nonumber \\&+\varphi _{ij}(m,v)u_{(0_{-i},j)}(((j-1)\wedge m_{k})_{k\in N})\nonumber \\ {\mathop {=}\limits ^{(5)}}&v((j\wedge m_{k})_{k\in N})-v(((j-1)\wedge m_{k})_{k\in N})-\varphi _{ij}(m,v). \end{aligned}$$

(51)

Each pair in $M^{+,j}\setminus \{(i,j)\}$ is dummy in $(m,u_{0_{-i},j})$. Therefore, by (49), each pair in $M^{+,j}\setminus \{(i,j)\}$ has the same marginal contribution in $(m,w^{ij})$ and in (m, v). It follows that, by (SM), each pair in $M^{+,j}\setminus \{(i,j)\}$ receives the same payoff in $(m,w^{ij})$ and in (m, v). Then, we obtain

$$\begin{aligned} \sum _{(k,j)\in M^{+,j}}f_{kj}(m,w^{ij})=&\sum _{\scriptstyle\begin{array}{c} (k,j)\in M^{+,j}\\ k\ne i \end{array}}f_{kj}(m,w^{ij}) + f_{ij}(m,w^{ij})\nonumber \\ {\mathop {=}\limits ^{SM}}&\sum _{\scriptstyle\begin{array}{c} (k,j)\in M^{+,j}\\ k\ne i \end{array}}f_{kj}(m,v) + f_{ij}(m,w^{ij})\nonumber \\ {\mathop {=}\limits ^{ME}}&v((j\wedge m_{k})_{k\in N})-v(((j-1)\wedge m_{k})_{k\in N}-f_{ij}(m,v)+f_{ij}(m,w^{ij}). \end{aligned}$$

(52)

Combining (51) and (52), for each $(i,j)\in M^{+}$ such that $j\le p_{i}$, we obtain

$$\begin{aligned} \varphi _{ij}(m,v)=f_{ij}(m,v)-f_{ij}(m,w^{ij}), \end{aligned}$$

(53)

which concludes Step 2.

We have the material to conclude the proof of the Induction step. By (50), we have that $f_{ij}(m,w^{ij})=f_{ij}(m,w)$ and by (39) we have that $f_{ij}(m,w)=0$, for each $(i,j)\in M^{+}$ such that $j\le p_{i}$. By (53), for each $(i,j)\in M^{+}$ such that $j\le p_{i}$, it holds that

$$\begin{aligned} f_{ij}(m,v)=\varphi _{ij}(m,v). \end{aligned}$$

Therefore, for each $(m,v)\in {\mathcal {G}}$ and each $(i,j)\in M^{+}$, it holds that $f_{ij}(m,v)=\varphi _{ij}(m,v).$ The proof of the theorem is complete. $\square$

Logical independence: The axioms invoked in Theorem 2 are logically independent, as shown by the following alternative solutions.

• The value f defined as: for each $(m,v)\in {\mathcal {G}}$,

  $$\begin{aligned} \forall (i,j)\in M^{+},\quad f_{ij}(m,v)=0, \end{aligned}$$

  satisfies all the axioms except (E).

• The value f defined as: for each $(m,v)\in {\mathcal {G}}$,

  $$\begin{aligned} \forall (i,j)\in M^{+},\quad f_{ij}(m,v)=\left\{ \begin{array}{ll} \varphi _{ij}(m,v)+ m^{T} &{} \text { if } (i,j)=(1,1) \text { and } \varphi _{ij}(m,v)>m^{T},\\ \varphi _{ij}(m,v)- m^{T} &{} \text { if } (i,j)=(n,1) \text { and } \varphi _{ij}(m,v)>m^{T},\\ \varphi _{ij}(m,v) &{} \text { otherwise,} \end{array} \right. \end{aligned}$$

  satisfies all the axioms except (IH).

• The multi-choice Equal division value $\xi$ satisfies all the axioms except Strong monotonicity.

• The value f defined as: for each $(m,v)\in {\mathcal {G}}$,

  $$\begin{aligned} \forall (i,j)\in M^{+}, \quad f_{ij}(m,v)=\left\{ \begin{array}{ll} \varphi _{ij}(m,v)+ 1 &{} \text { if } (i,j)=(1,1),\\ \varphi _{ij}(m,v)- 1 &{} \text { if } (i,j)=(n,1),\\ \varphi _{ij}(m,v) &{} \text { otherwise,} \end{array} \right. \end{aligned}$$

  satisfies all the axioms except (SS).

1.6 Proof of Theorem 3

The proof is divided in two steps.

Step 1: We show that $\xi$ satisfies all the axioms of the statement of Theorem 3. For each $(m,v)\in {\mathcal {G}}$, it holds that

$$\begin{aligned} \sum _{i\in N}\sum _{j\in M^{+}_{i}}\xi _{ij}(m,v)&=\sum _{j \le \displaystyle \max _{k\in N}m_{k}}\sum _{i\in Q(j)}\frac{1}{|Q(j)|}\biggl [v((j\wedge m_{k})_{k\in N})-v(((j-1)\wedge m_{k})_{k\in N})\biggl ]\\&=\sum _{j\le \displaystyle \max _{k\in N}m_{k}}v((j\wedge m_{k})_{k\in N})-v(((j-1)\wedge m_{k})_{k\in N})\\&=v(m). \end{aligned}$$

This shows that the value satisfies (E). By definition of the Equal division value (see (15)), the payoff of a pair does not depend on activity levels different from the one contained in this pair. Therefore, $\xi$ satisfies (IH). By definition of $\xi$, it is straightforward to see that $\xi$ satisfies (L), (SP) and (ET). This concludes Step 1.

Step 2: We show the uniqueness part of the theorem. Let f be a value satisfying all the axioms of the statement of Theorem 3. We know that each characteristic function v admits a linear decomposition in terms of Dirac games (see 19). By (L), for each $(m,v)\in {\mathcal {G}}$, it holds that

$$\begin{aligned} f(m,v)=\sum _{s\le m}v(s)f(m, \delta _{s}). \end{aligned}$$

For each $s\in {\mathcal {M}}$, we show that

$$\begin{aligned} f(m, \delta _{s})=\xi _{ij}(m,\delta _{s}). \end{aligned}$$

We consider two cases.

Case 1. Suppose that $s\in {\mathcal {M}}$ is not a synchronized coalition, that is $s\ne ((l\wedge m_{k})_{k\in N})$, for each $l\le \max _{k\in N}m_{k}$. Since f satisfies (E) and (IH), by Proposition 2 it satisfies (ME). Therefore, by (ME), for each $j\le \max _{k\in N}m_{k}$,

$$\begin{aligned} \sum _{(i,j)\in M^{+,j}}f_{ij}(m,\delta _{s})&=\delta _{s}((j\wedge m_{k})_{k\in N})-\delta _{s}(((j-1)\wedge m_{k})_{k\in N}). \end{aligned}$$

Since $s\ne ((j\wedge m_{k})_{k\in N})$ and $s\ne (((j-1)\wedge m_{k})_{k\in N})$, by definition of a Dirac game,

$$\begin{aligned} \sum _{(i,j)\in M^{+,j}}f_{ij}(m,\delta _{s})&=0. \end{aligned}$$

(54)

Since $\delta _{s}((j\wedge m_{k})_{k\in N})-\delta _{s}(((j-1)\wedge m_{k})_{k\in N})\ge 0$, by (SP) and (54), for each $(i,j)\in M^{+,j}$, it holds that

$$\begin{aligned} f_{ij}(m,\delta _{s})\ge 0. \end{aligned}$$

(55)

Combining (54) and (55), for each $(i,j)\in M^{+,j}$, we obtain

$$\begin{aligned} f_{ij}(m,\delta _{s})=0=\xi _{ij}(m,\delta _{s}). \end{aligned}$$

Case 2. Suppose that $s\in {\mathcal {M}}$ is a synchronized coalition, that is $s= (l\wedge m_{k})_{k\in N}$, where $l\le \max _{k\in N}m_{k}$.

Case 2.1. Take any activity level j such that $j<l$. By (ME), it holds that

$$\begin{aligned} \sum _{(i,j)\in M^{+,j}}f_{ij}(m,\delta _{s})&=\delta _{s}((j\wedge m_{k})_{k\in N})-\delta _{s}(((j-1)\wedge m_{k})_{k\in N}). \end{aligned}$$

Since $s\ne ((j\wedge m_{k})_{k\in N})$ and $s\ne (((j-1)\wedge m_{k})_{k\in N})$, by definition of a Dirac game, we have that

$$\begin{aligned} \sum _{(i,j)\in M^{+,j}}f_{ij}(m,\delta _{s})&=0. \end{aligned}$$

Observe that $\delta _{s}((j\wedge m_{k})_{k\in N})-\delta _{s}(((j-1)\wedge m_{k})_{k\in N})= 0$. Similarly to Case 1, by (SP) and (54), for each pair $(i,j)\in M^{+,j}$ such that $j<l$,

$$\begin{aligned} f_{ij}(m,\delta _{s})&=0=\xi _{ij}(m,\delta _{s}). \end{aligned}$$

Case 2.2. Take any activity level $j>l+1$. Similarly to Case 2.1, for each $(i,j)\in M^{+}$ such that $j>l+1$

$$\begin{aligned} f_{ij}(m,\delta _{s})=0=\xi _{ij}(m,\delta _{s}). \end{aligned}$$

Case 2.3. Consider the pairs $(i,j)\in M^{+}$ such that $j=l$, that is the pairs in $M^{+,l}$. By (ME) and the definition of a Dirac game, it holds that

$$\begin{aligned} \sum _{(i,l)\in M^{+,l}}f_{il}(m,\delta _{s})&=\delta _{s}((l\wedge m_{k})_{k\in N})-\delta _{s}((l-1\wedge m_{k})_{k\in N})\nonumber \\&=1. \end{aligned}$$

(56)

Two distinct pairs $(i,l),(i',l)\in M^{+,l}$ are equal in $(m,\delta _{s})$. By (ET), one obtains

$$\begin{aligned} f_{il}(m,\delta _{s})=f_{i'l}(m,\delta _{s}). \end{aligned}$$

(57)

From (56) and (57), it follows that

$$\begin{aligned} f_{il}(m,\delta _{s})&=\frac{1}{|Q(l)|}=\xi _{ij}(m,\delta _{s}). \end{aligned}$$

Case 2.4. Consider the pairs $(i,l+1)\in M^{+,l+1}$. By (ME) and the definition of a Dirac game, it holds that

$$\begin{aligned} \sum _{(i,l+1)\in M^{+,l+1}}f_{i(l+1)}(m,\delta _{s})&=\delta _{s}(((l+1)\wedge m_{k})_{k\in N})-\delta _{s}((l\wedge m_{k})_{k\in N})=0-1=-1. \end{aligned}$$

Similarly to Case 2.3, for each $(i,l+1)\in M^{+,l+1}$,

$$\begin{aligned} f_{i(l+1)}(m,\delta _{s})=-\frac{1}{|Q(l+1)|}=\xi _{i(l+1)}(m,\delta _{s}). \end{aligned}$$

Therefore, for each $s\in {\mathcal {M}}$, we have that $f(m,\delta _{s})=\xi (m,\delta _{s})$. By (L), we conclude the proof of Theorem 3. $\square$

Logical independence: The axioms invoked in Theorem 3 are logically independent, as shown by the following alternative solutions.

• The value f defined as: for each $(m,v)\in {\mathcal {G}}$,

  $$\begin{aligned} \forall (i,j)\in M^{+},\quad f_{ij}(m,v)=0, \end{aligned}$$

  satisfies all the axioms except (E).

• The value f defined as: for each $(m,v)\in {\mathcal {G}}$,

  $$\begin{aligned} \forall (i,j)\in M^{+}, \quad f_{ij}(m,v)=\frac{1}{|Q(j)|}\sum _{k\ge j} \frac{v((k\wedge m_{h})_{h\in N})-v(((k-1)\wedge m_{h})_{h\in N})}{k} \end{aligned}$$

  satisfies all the axioms except (IH).

• The value f defined as: for each $(m,v)\in {\mathcal {G}}$,

  $$\begin{aligned} \forall (i,j)\in M^{+},\quad f_{ij}(m,v)&=\frac{(v(((j-1)\wedge m_{h})_{h\in N}+e_{i})^{2})+1}{\displaystyle \sum _{k\in Q(k)}((v(((j-1)\wedge m_{h})_{h\in N}+e_{k})^{2})+1}\\&\quad \times \Big [v((j\wedge m_{k})_{k\in N})-v(((j-1)\wedge m_{k})_{k\in N}\Big ], \end{aligned}$$

  satisfies all the axioms except (L).

• The multi-choice Shapley value $\varphi$ satisfies all the axioms except (SP).

• Take any $(m,v)\in {\mathcal {G}}$ and for each $(i,j)\in M^{+}$ fix any arbitrary integer $\beta ^{ij}\in \{1,2\}$. The value $f^{\beta }$ defined as: for each $(m,v)\in {\mathcal {G}}$,

  $$\begin{aligned} \forall (i,j)\in M^{+},\quad f^{\beta }_{ij}(m,v)=\frac{\beta ^{ij}}{\sum _{k\in Q(j)}\beta ^{kj}}\Big [v((j\wedge m_{k})_{k\in N})-v(((j-1)\wedge m_{k})_{k\in N})\Big ], \end{aligned}$$

  satisfies all the axioms except (ET).

1.7 Proof of Theorem 4

Before starting the proof, which is divided in two steps, we provide a useful remark.

Remark 7

By definition, (LD) implies (ET). If $(m,v)\in {\mathcal {G}}$ is the null game, then (ME) and (LD) imply $f_{ij}(m,v)=0$ for each $(i,j)\in M^{+}$.

Observe that $\varphi$ and $\xi$ both satisfy (LD) and (WM). Consider any parameter system $\alpha$. Since that multi-choice Egalitarian Shapley values are convex combinations of the multi-choice Shapley value and the multi-choice Equal division value (see (16)), $\chi ^{\alpha }$, inherits all the axioms of the statement of Theorem 4.

Next, we show that the multi-choice Egalitarian Shapley values are the only values satisfying all the axioms of the statement of Theorem 4. Consider a value f satisfying all the axioms of the statement of Theorem 4. To prove the uniqueness part, we show that, for each $(m,v)\in {\mathcal {G}}$, there exists a parameter system $\alpha$ such that

$$\begin{aligned} f(m,v)=\chi ^{\alpha }(m,v). \end{aligned}$$

By (L), for each $(m,v)\in {\mathcal {G}}$,

$$\begin{aligned} f(m,v)=\sum _{t\le m} \Delta _{t}(v)f(m,u_{t}). \end{aligned}$$

Recall that $t^{T}=\max _{i\in N}t_{i}$ denotes the highest activity level played in coalition $t\in {\mathcal {M}}$. Take any $1\le l \le m^{T}$. We show that, for each $(m,u_{t})$ such that $t^{T}=l$, f can be written as

$$\begin{aligned}&\forall (i,j)\in M^{+}, \\&f_{ij}(m,u_{t})= {\left\{ \begin{array}{ll} \alpha ^{l}\varphi _{il}(m,u_{t})+ (1-\alpha ^{l})\xi _{il}(m,u_{t}) &{} \quad\text {if } j= l,\\ 0 &{}\quad \text {otherwise,} \end{array}\right. } \end{aligned}$$

for some $0\le \alpha ^{l}\le 1$. To that end, consider all pairs $(i,j)\in M^{+}$ such that $j< l$. By (IH),

$$\begin{aligned} f_{ij}(m,u_{t})=f_{ij}((j\wedge m_{k})_{k\in N},u_{t}). \end{aligned}$$

Since $((j\wedge m_{k})_{k\in N},u_{t})$ is the null game, by Remark 7, for each $(i,j)\in M^{+}$ such that $j< l$,

$$\begin{aligned} f_{ij}(m,u_{t})=0. \end{aligned}$$

(58)

Consider all pairs $(i,j)\in M^{+}$ such that $j>l$. Observe that these pairs are null pairs in $(m,u_{t})$ and thus are equal. From Remark 7 and (ME), for each $(i,j)\in M^{+}$ such that $j>l$,

$$\begin{aligned} f_{ij}(m,u_{t})=0. \end{aligned}$$

(59)

Now, consider any pair $(i,l)\in M^{+}$. We show that f can be written as

$$\begin{aligned} f_{il}(m,u_{t})=\alpha ^{l}\varphi _{il}(m,u_{t})+ (1-\alpha ^{l})\xi _{il}(m,u_{t}), \end{aligned}$$

for some $0\le \alpha ^{l}\le 1$. We proceed by induction on $q^{t}(l)$ the number of players that play l in coalition t.

Initialization: Pick any minimal effort game $(m,u_{t})\in {\mathcal {G}}$ such that $t^{T}=l$ and $q^{t}(l)=1$. In such game, there is exactly one player, say $k\in N$, that plays the activity level l in t.

Before proceeding further into the initialization step, we prove the following claim.

Claim: Pick any minimal effort game $(m,u_{t})\in {\mathcal {G}}$, where $t^{T}=l$ and $q^{t}(l)=1$. Then, there exists $c^{l}\in {\mathbb {R}}$ such that

$$\begin{aligned} \forall (i,l)\in M^{+},i\ne k, \quad f_{il}(m,u_{t})=c^{l}, \end{aligned}$$

(60)

where k refers to the only player that plays the activity level l in t. In other words, the payoff assigned to the pair $(i,l)\in M^{+},i\ne k$, does not depend on the player k that plays $t^{T}=l$ in $(m,u_{t})$. To prove this claim, we distinguish three cases.

Case 1. If $|Q(m^{T})|\ge 3$, then there is at least three players in Q(l). Consider any three distinct players in Q(l) denoted by k, i and $i'$. Consider the three coalitions $t,t',t''\in {\mathcal {M}}$ defined as

$$\begin{aligned}&t_{k}=l, \text{ and } \forall h\in N\setminus \{k\}, \quad t_{h}<l,\\&t'_{i}=l, \text{ and } \forall h\in N\setminus \{i\}, \quad t'_{h}<l,\\&t''_{i'}=l, \text{ and } \forall h\in N\setminus \{i'\}, \quad t''_{h}<l. \end{aligned}$$

Consider the minimal effort games $(m,u_{t})$, $(m,u_{t'})$ and $(m,u_{t''})$. Observe that:

• $(i,l)\in M^{+}$ is a null pair in $(m,u_{t})$ and $(m,u_{t''})$, and (i, l) is the only non null pair in $(m,u_{t'})$;

• $(i',l)\in M^{+}$ is a null pair in $(m,u_{t})$ and $(m,u_{t'})$, and $(i',l)$ is the only non null pair in $(m,u_{t''})$;

• $(k,l)\in M^{+}$ is a null pair in $(m,u_{t'})$ and $(m,u_{t''})$, and (k, l) is the only non null pair in $(m,u_{t})$;

• $(i,l),(i',l)\in M^{+}$ are equal pairs in $(m,u_{t})$;

• $(i,l),(k,l)\in M^{+}$ are equal pairs in $(m,u_{t''})$;

• $(i',l),(k,l)\in M^{+}$ are equal pairs in $(m,u_{t'})$.

Moreover, observe that

$$\begin{aligned} u_{t}((l\wedge m_{h})_{h\in N})&=u_{t'}((l\wedge m_{h})_{h\in N}) \nonumber \\&=u_{t''}((l\wedge m_{h})_{h\in N})\nonumber \\ \text {and}\quad u_{t}(((l-1)\wedge m_{h})_{h\in N})&=u_{t'}(((l-1)\wedge m_{h})_{h\in N})\nonumber \\&=u_{t''}(((l-1)\wedge m_{h})_{h\in N}). \end{aligned}$$

(61)

By (61), the fact that $(i,l)\in M^{+}$ is a null pair in $(m,u_{t})$ and $(m,u_{t''})$, and (WM), one obtains

$$\begin{aligned} f_{il}(m,u_{t})&=f_{il}(m,u_{t''}). \end{aligned}$$

(62)

Similarly,

$$\begin{aligned} f_{i'l}(m,u_{t})=f_{i'l}(m,u_{t'})\quad \text {and}\quad f_{kl}(m,u_{t'})=f_{kl}(m,u_{t''}). \end{aligned}$$

(63)

By the fact that $(i,l),(i',l)\in M^{+}$ are equal pairs in $(m,u_{t})$ and (LD), one obtains

$$\begin{aligned} f_{il}(m,u_{t})=f_{i'l}(m,u_{t}). \end{aligned}$$

(64)

Similarly,

$$\begin{aligned} f_{i'l}(m,u_{t'})=f_{kl}(m,u_{t'})\quad \text {and}\quad f_{il}(m,u_{t''})=f_{kl}(m,u_{t''}). \end{aligned}$$

(65)

Combining (62), (63), (64) and (65), one obtains

$$\begin{aligned} f_{il}(m,u_{t}){\mathop {=}\limits ^{(62)}}f_{il}(m,u_{t''}) {\mathop {=}\limits ^{(65)}}f_{kl}(m,u_{t''}) {\mathop {=}\limits ^{(63)}}f_{kl}(m,u_{t'}) {\mathop {=}\limits ^{(65)}}f_{i'l}(m,u_{t'}) {\mathop {=}\limits ^{(63)}}f_{i'l}(m,u_{t}) =c^{l}, \end{aligned}$$

for some $c^{l}\in {\mathbb {R}}$.

We have shown that there exists a unique $c^{l}\in {\mathbb {R}}$ such that, for any minimal effort game $(m,u_{t})\in {\mathcal {G}}$, $|Q(m^{T})|\ge 3$, $t^{T}=l$, $q^{t}(l)=1$, (60) holds.

Case 2. If $|Q(m^{T})|=2$, then there is at least two players in Q(l). If there are at least three players in Q(l), then the proof is identical to the one in Case 1. Therefore, let us assume that $Q(l)=\{i,k\}$. Consider the two coalitions $t,t'\in {\mathcal {M}}$ defined as

$$\begin{aligned}&t_{i}=l, \quad \text {and} \quad \forall h\in N\setminus \{i\}, \quad t_{h}< l,\\&t'_{k}=l, \quad \text {and} \quad \forall h\in N\setminus \{k\}, \quad t'_{h}< l. \end{aligned}$$

Consider the minimal games $(m,u_{t})\in {\mathcal {G}}$, $(m,u_{t'})\in {\mathcal {G}}$, and $(m,u_{t}+u_{t'})\in {\mathcal {G}}$. Observe that $(i,l),(k,l)\in M^{+}$ are equal pairs in $(m,u_{t}+u_{t'})$. Therefore, by (LD), it holds that

$$\begin{aligned} f_{il}(m,u_{t}+u_{t'})=f_{kl}(m,u_{t}+u_{t'}). \end{aligned}$$

(66)

By (L), (66) becomes

$$\begin{aligned}&f_{il}(m,u_{t})+ f_{il}(m,u_{t'}) = f_{kl}(m,u_{t})+ f_{kl}(m,u_{t'})\nonumber \\ \iff \quad&f_{il}(m,u_{t}) = f_{kl}(m,u_{t})+ f_{kl}(m,u_{t'}) -f_{il}(m,u_{t'}) . \end{aligned}$$

(67)

Since f satisfies (E) and (IH), by Proposition 2, f satisfies (ME). Since i and k are the only two players in Q(l), by (ME), it holds that

$$\begin{aligned} f_{il}(m,u_{t})+f_{kl}(m,u_{t})=1 \quad \text {and}\quad f_{il}(m,u_{t'})+f_{kl}(m,u_{t'})=1. \end{aligned}$$

It follows that

$$\begin{aligned} f_{il}(m,u_{t})+f_{kl}(m,u_{t})= f_{il}(m,u_{t'})+f_{kl}(m,u_{t'}) \end{aligned}$$

(68)

Combining (67) and (68), one obtains

$$\begin{aligned}&f_{kl}(m,u_{t})+ f_{kl}(m,u_{t'}) -f_{il}(m,u_{t'})+f_{kl}(m,u_{t})\\&= f_{il}(m,u_{t'})+f_{kl}(m,u_{t'})\\ \iff \quad&f_{kl}(m,u_{t}) -f_{il}(m,u_{t'})+f_{kl}(m,u_{t})= f_{il}(m,u_{t'})\\ \iff \quad&f_{kl}(m,u_{t})=f_{il}(m,u_{t'})=c^{l}, \end{aligned}$$

for some $c^{l}\in {\mathbb {R}}$.

We have shown that there exists a unique $c^{l}\in {\mathbb {R}}$ such that, for any minimal effort game $(m,u_{t})\in {\mathcal {G}}$, $|Q(m^{T})|=2$, $t^{T}=l$, $q^{t}(l)=1$, (60) holds.

Case 3. Finally, assume that $|Q(m^{T})|=1$. If there is only one player in Q(l), then there is nothing to show. If there are two players in Q(l), then (60) holds according to Case 2. If there three players or more in Q(l), then (60) holds according to Case 1.

We have shown that there exists a unique $c^{l}\in {\mathbb {R}}$ such that, for any minimal effort game $(m,u_{t})\in {\mathcal {G}}$, $t^{T}=l$ and $q^{t}(l)=1$, (60) holds. This concludes the proof of the claim.

Next, pick any minimal effort game $(m,u_{t})\in {\mathcal {G}}$ such that $t^{T}=l$ and $q^{t}(l)=1$. Observe that the pair (k, l) is the only pair featuring the activity level l which is non null in $(m,u_{t})$, and $u_{t}(m)\ge 0$. By (60), the fact that null pairs featuring the activity levels l are equal pairs in $(m,u_{t})$ and (ME), it holds that

$$\begin{aligned} f_{kl}(m,u_{t})&=1-(|Q(l)|-1)c^{l}. \end{aligned}$$

Define $\alpha ^{l}=1-c^{l}|Q(l)|$ so that we obtain

$$\begin{aligned} c^{l}=\frac{1-\alpha ^{l}}{|Q(l)|}. \end{aligned}$$

Now, we show that $\alpha ^{l}\le 1$.

By Remark 7, each pair receives a zero payoff in the null game. Observe that each pair in $M^{+}$ has better marginal contributions to coalitions in $(m,u_{t})$ than in the null game. Moreover, it holds that

$$\begin{aligned} u_{t}((l\wedge m_{k})_{k\in N})- u_{t}(((l-1)\wedge m_{k})_{k\in N}) \ge 0. \end{aligned}$$

Thus, by (WM), $f_{il}(m,u_{t})\ge 0$ for each $(i,l)\in M^{+}$. It follows that

$$\begin{aligned} c^{l}&=\frac{1-\alpha ^{l}}{|Q(l)|}\ge 0 \implies \alpha ^{l}\le 1. \end{aligned}$$

Therefore, for each $(i,l)\in M^{+}$,

$$\begin{aligned} f_{il}(m,u_{t})={\left\{ \begin{array}{ll} \displaystyle \frac{1-\alpha ^{l} }{|Q(l)|} &{} \text {if }j=l \text { and }i\ne k, \\ \text {}\\ \displaystyle \frac{1-\alpha ^{l}}{|Q(l)|} + \alpha ^{l} &{} \text {if }j=l \text { and }i= k, \end{array}\right. } \end{aligned}$$

(69)

for some $0\le \alpha ^{l}\le 1$. Observe that, for each $(i,l)\in M^{+}$,

$$\begin{aligned} \xi _{il}(m,u_{t})= \displaystyle \frac{1}{|Q(l)|},&\quad \varphi _{il}(m,u_{t})={\left\{ \begin{array}{ll} 0 \quad &{}\text {if }i\ne k, \\ 1 \quad &{}\text {if }i= k. \end{array}\right. } \end{aligned}$$

Comparing $\xi _{il}(m,u_{t})$ and $\varphi _{il}(m,u_{t})$ with (69), one obtains

$$\begin{aligned} \forall (i,l)\in M^{+} \quad \text {and} \quad f_{il}(m,u_{t})=\alpha ^{l}\varphi _{il}(m,u_{t})+ (1-\alpha ^{l})\xi _{il}(m,u_{t}). \end{aligned}$$

for some $0\le \alpha ^{l}\le 1$. This concludes the initialization.

Hypothesis: Consider $r\in {\mathbb {N}}$ such that $1\le r <|Q(l)|$. Consider any t such that $q^{t}(l)=r$. In this case, there are r players that play l in t. Assume that

$$\begin{aligned} \forall (i,l)\in M^{+}, \quad f_{il}(m,u_{t})=\alpha ^{l}\varphi _{il}(m,u_{t})+ (1-\alpha ^{l})\xi _{il}(m,u_{t}). \end{aligned}$$

Induction: Consider any t such that $q^{t}(l)=r+1$. Let $s=t-e_{h}$, for some $h\in N$ such that $t_{h}=l$. Obviously, it holds that $q^{s}(l)=r$. Recall that $(i,l)\notin T(t)$ if $t_{i}<l$. Observe that if $(i,l)\notin T(t)$ then $(i,l) \notin T(s)$. If $(i,l)\notin T(t)$ then (i, l) is a null pair in $(m,u_{t})$ and is also a null pair in $(m,u_{s})$. Therefore, each $(i,l)\notin T(t)$ has the same marginal contributions in both games $(m,u_{t})$ and $(m,u_{s})$. Moreover, it holds that

$$\begin{aligned} u_{t}((l\wedge m_{h})_{h\in N})- u_{t}(((l-1)\wedge m_{h})_{h\in N}) = u_{s}((l\wedge m_{h})_{h\in N})- u_{s}(((l-1)\wedge m_{h})_{h\in N}). \end{aligned}$$

Thus, by double application of (WM), the induction hypothesis and the definitions of $\varphi$ and $\xi$, for each $(i,l)\notin T(t)$ it holds that

$$\begin{aligned} f_{il}(m,u_{t})&=f_{il}(m,u_{s})\nonumber \\&{\mathop {=}\limits ^{Hyp}}\alpha ^{l}\varphi _{il}(m,u_{s})+ (1-\alpha ^{l})\xi _{il}(m,u_{s})\nonumber \\&=\frac{(1-\alpha ^{l})}{|Q(l)|}. \end{aligned}$$

(70)

By (ME), (70) and the definition of a minimal effort game,

$$\begin{aligned} \sum _{(i,l)\in T(t)}f_{il}(m,u_{t})=&u_{t}((l\wedge m_{h})_{h\in N})-u_{t}(((l-1)\wedge m_{h})_{h\in N})\nonumber \\&- \sum _{(i,l)\notin T(t)}f_{il}(m,u_{t})\nonumber \\ =&1 -0- (|Q(l)|-|T(t)|)\frac{1-\alpha ^{l}}{|Q(l)|}. \end{aligned}$$

(71)

Additionally, any two distinct pairs $(i,l),(i',l)\in M^{+}$ such that $(i,l),(i',l) \in T(t)$, are equal in $(m,u_{t})$. By Remark 7 and by (LD), for each $(i,l)\in T(t)$,

$$\begin{aligned} f_{il}(m,u_{t})=c', \end{aligned}$$

for some $c'\in {\mathbb {R}}$. It follows that

$$\begin{aligned} \sum _{ (i,l)\in T(t)}f_{il}(m,u_{t})=|T(t)|c'. \end{aligned}$$

(72)

Therefore, combining (71) and (72), for each $(i,l)\in T(t)$, we obtain

$$\begin{aligned} c'= \displaystyle \frac{1 - (|Q(l)|-|T(t)|)\displaystyle \frac{1-\alpha ^{l}}{|Q(l)|}}{|T(t)|}. \end{aligned}$$

It follows that, for each $(i,l)\in T(t)$,

$$\begin{aligned} f_{il}(m,u_{t})&=\frac{1 - (|Q(l)|-|T(t)|)\displaystyle \frac{1-\alpha ^{l}}{|Q(l)|}}{|T(t)|}\nonumber \\&= \frac{\alpha ^{l}}{|T(t)|} + \frac{1-\alpha ^{l}}{|Q(l)|}\nonumber \\&=\alpha ^{l}\varphi _{il}(m,u_{t})+ (1-\alpha ^{l})\xi _{il}(m,u_{t}). \end{aligned}$$

(73)

Combining (70) and (73), if $t^{T}=l$ then, for each $(i,l)\in M^{+}$,

$$\begin{aligned} f_{il}(m,u_{t})=\alpha ^{l}\varphi _{il}(m,u_{t})+ (1-\alpha ^{l})\xi _{il}(m,u_{t}). \end{aligned}$$

This concludes the induction step.

We have shown that there exists a parameter system $\alpha$ such that f can be written, for each $(m,u_{t})$ such that $t^{T}=l$, as

$$\begin{aligned} \forall (i,j)\in M^{+}, \quad f_{ij}(m,u_{t})= {\left\{ \begin{array}{ll} \alpha ^{l}\varphi _{il}(m,u_{t})+ (1-\alpha ^{l})\xi _{il}(m,u_{t}) &{} \text {if } j= l,\\ 0 &{} \text {otherwise.} \end{array}\right. } \end{aligned}$$

By definition of multi-choice Egalitarian Shapley values (see (16)), for such parameter system $\alpha$, there is a $\chi ^{\alpha }$ such that, for each $(m,u_{t})$

$$\begin{aligned} f(m,u_{t})=\chi ^{\alpha }(m,u_{t}). \end{aligned}$$

We conclude by (L) that there exists a parameter system $\alpha$ such that, for each $(m,v)\in {\mathcal {G}}$

$$\begin{aligned} f(m,v)=\chi ^{\alpha }(m,v). \end{aligned}$$

This concludes the proof of the theorem. $\square$

Logical independence: The five axioms of the statement of Theorem 4 are logically independent, as shown by the following alternative solutions.

• The value f defined as: for each $(m,v)\in {\mathcal {G}}$,

  $$\begin{aligned} \forall (i,j)\in M^{+},\quad f_{ij}(m,v)=0, \end{aligned}$$

  satisfies all the axioms except (E).

• The value f defined as: for each $(m,v)\in {\mathcal {G}}$,

  $$\begin{aligned}&\forall (i,j)\in M^{+}, \\&f_{ij}(m,v)=\frac{\max _{k\in N}m_{k}}{\sum _{k\in N}m_{k}}\varphi (m,v) + \Big (1-\frac{\max _{k\in N}m_{k}}{\sum _{k\in N}m_{k}}\Big )\xi (m,v), \end{aligned}$$

  satisfies all the axioms except (IH).

• Take any $(m,v)\in {\mathcal {G}}$. Denote the difference in worth between the grand coalition and the $(m_{T}-1)$-synchronized coalition by

  $$\begin{aligned} V=v(m)-v(((m_{T}-1)\wedge m_{k})_{k\in N}. \end{aligned}$$

  Let f be the value defined as:

  • $f(m,v)=\varphi (m,v)$ if $|m^{T}|\ne 2$;

  • otherwise, for each $(i,j)\in M^{+}$ such that $j\ne m_{T}$, $f_{ij}(m,v)=\varphi _{ij}(m,v)$, and for the two remaining pairs $(i,m_{T}),(i',m_{T})\in M^{+}$,

    $$\begin{aligned}&(f_{im_{T}}(m,v),f_{i'm_{T}}(m,v))=\\&{\left\{ \begin{array}{ll} (\varphi _{im_{T}}(m,v),\varphi _{i'm_{T}}(m,v)) &{} \text {if } \varphi _{im_{T}}(m,v)\ge 0 \text { and } \varphi _{i'm_{T}}(m,v) \ge 0,\\ (0,V) &{} \text {if } \varphi _{im_{T}}(m,v)< 0, \varphi _{i'm_{T}}(m,v)> 0 \text { and } V \ge 0,\\ (V,0) &{} \text {if } \varphi _{im_{T}}(m,v)< 0, \varphi _{i'm_{T}}(m,v)> 0 \text { and } V< 0,\\ (\varphi _{im_{T}}(m,v),\varphi _{i'm_{T}}(m,v)) &{} \text {if } \varphi _{im_{T}}(m,v)\le 0 \text { and } \varphi _{i'm_{T}}(m,v) \le 0,\\ (0,V) &{} \text {if } \varphi _{im_{T}}(m,v)> 0, \varphi _{i'm_{T}}(m,v)< 0 \text { and } V \ge 0,\\ (V,0) &{} \text {if } \varphi _{im_{T}}(m,v)> 0, \varphi _{i'm_{T}}(m,v) < 0 \text { and } V > 0. \end{array}\right. } \end{aligned}$$

    The value f satisfies all the axioms except (L).

• The value f defined as: for each $(m,v)\in {\mathcal {G}}$,

  $$\begin{aligned} \forall (i,j)\in M^{+},&\\ f_{ij}(m,v)=&v(((j-1)\wedge m_{k})_{k\in N})+e_{i})\\ {}&+\frac{(v((j\wedge m_{k})_{k\in N})-v(((j-1)\wedge m_{k})_{k\in N}))}{|Q(j)|}\\&-\frac{\sum _{k\in Q(j)}v(((j-1)\wedge m_{k})_{k\in N})+e_{k})}{|Q(j)|}, \end{aligned}$$

  satisfies all the axioms except (WM). Observe that this value extends the Equal surplus division from TU-games to multi-choice games.

• Take any $(m,v)\in {\mathcal {G}}$. For each $(i,j)\in M^{+}$ fix an arbitrary integer $\beta ^{ij}\in \{1,2\}$. The value $f^{\beta }$ defined as: for each $(m,v)\in {\mathcal {G}}$,

  $$\begin{aligned}&\forall (i,j)\in M^{+},\\&f^{\beta }_{ij}(m,v)=\frac{\beta ^{ij}}{\sum _{k\in Q(j)}\beta ^{kj}}\Big [v((j\wedge m_{k})_{k\in N})-v(((j-1)\wedge m_{k})_{k\in N})\Big ], \end{aligned}$$

  satisfies all the axioms except (LD).

1.8 Proof of Proposition 3

Observe that, for each $i\in N$ and $j<q_{i}$, it holds that

$$\begin{aligned} \forall E\subseteq Q(j), \quad w^{(q,C)}_{j}(E)=w^{(q-e_{i},C)}_{j}(E). \end{aligned}$$

Therefore, we obtain

$$\begin{aligned} \forall i\in N, j< q_{i}, \quad Sh_{i}\Big (N, w^{(q,C)}_{j}\Big )=Sh_{i}\Big (N, w^{(q-e_{i},C)}_{j}\Big ). \end{aligned}$$

(74)

Additionally, recall that, for each $j\le \max _{k\in N}m_{k}$, the set of orders ${\overline{O}}_{j}$ over $M^{+,j}$ can be interpreted as the set of orders over the set of players in Q(j). An order over Q(j) is a bijection $\sigma _{j}^{N}:Q(j)\rightarrow \{1,\ldots , |Q(j)|\}$. We denote by ${\overline{Q}}(j)$ the set of of orders over Q(j). Consider an order $\sigma _{j}^{N} \in {\overline{Q}}_{j}$ and $h\in \{1,\ldots , |Q(j)|\}$. Recall that, for each $B\subseteq N$, the vector $e_{B}\in {\mathbb {R}}^{|N|}$ is defined by $(e_{B})_{i}=1$ if $i\in B$ and $(e_{B})_{i}=0$ otherwise. We denote by

$$\begin{aligned} ((j-1)\wedge q_{k})_{k\in N}+e_{E^{\sigma _{j}^{N},h}} \end{aligned}$$

the coalition in which each player in Q(j) ordered prior to step h with respect to $\sigma _{j}^{N}$, participates at its activity level j, whereas each player in Q(j) ordered after step h with respect to $\sigma _{j}^{N}$, participates at its activity level $j-1$. Each player not in Q(j) participates at its maximal activity level. Obviously, this coalition coincides with $s^{\sigma _{j},h}$, where $\sigma _{j}$ is the counterpart of $\sigma _{j}^{N}$ among the orders in ${\overline{O}}_{j}$. We use the convention $((j-1)\wedge q_{k})_{k\in N}+e_{E^{\sigma _{j}^{N},0}}=((j-1)\wedge q_{k})_{k\in N}$. Consider an order $\sigma _{j}^{N} \in {\overline{Q}}_{j}$. For each $i\in Q(j)$, we denote by

$$\begin{aligned} \mu ^{\sigma _{j}^{N}}_{i}(q,C)=C\big (((j-1)\wedge q_{k})_{k\in N}+e_{E^{\sigma _{j}^{N},\sigma _{j}^{N}(i)}}\big )-C\big (((j-1)\wedge q_{k})_{k\in N}+e_{E^{\sigma _{j}^{N},\sigma _{j}^{N}(i)-1}}\big ), \end{aligned}$$

(75)

the marginal contribution of player i for its activity level j with respect to the order $\sigma _{j}^{N}$. By (13), (14) and (75), for each $(q,C)\in \mathrm {C}$, the multi-choice Shapley value can be re-written as

$$\begin{aligned} \forall (i,j)\in M^{+,j}, \quad \varphi _{ij}(q,C)=\frac{1}{|Q(j)|!}\sum _{\begin{array}{c} \sigma _{j}^{N} \in {\overline{Q}}_{j} \end{array}}\mu ^{\sigma _{j}^{N}}_{i}(q,C). \end{aligned}$$

In addition, by definition of the Shapley value for TU-games (see Shapley (1953)), for each $j\le \max _{k\in N}m_{k}$, it holds that

$$\begin{aligned} \forall i\in Q(j), \quad Sh_{i}\Big (N,w^{(q,C)}_{j}\Big )=\frac{1}{|Q(j)|!}\sum _{\begin{array}{c} \sigma _{j}^{N} \in {\overline{Q}}_{j} \end{array}}\mu ^{\sigma _{j}^{N}}_{i}(q,C) =\varphi _{ij}(q,C). \end{aligned}$$

(76)

Therefore, for each $i\in N$, we obtain

$$\begin{aligned} SCS_{i}(q,C)-SCS_{i}(q-e_{i},C)&=\sum _{j=1}^{q_{i}}Sh_{i}\Big (N, w^{(q,C)}_{j}\Big )-\sum _{j=1}^{q_{i}-1}Sh_{i}\Big (N, w^{(q-e_{i},C)}_{j}\Big )\\&=Sh_{i}\Big (N, w^{(q,C)}_{q_{i}}\Big ) \\&=\varphi _{iq_{i}}(q,C), \end{aligned}$$

where the second equality follows from (18) and (74), and the third equality follows from (76). $\square$