Redistribution to the less productive: parallel characterizations of the egalitarian Shapley and consensus values

Abstract

In cooperative game theory with transferable utilities (TU games), there are two well-established ways of redistributing Shapley value payoffs: using egalitarian Shapley values, and using consensus values. We present parallel characterizations of these classes of solutions. Together with the (weaker) axioms that characterize the original Shapley value, those that specify the redistribution methods characterize the two classes of values. For the class of egalitarian Shapley values, we focus on redistributions in one-person unanimity games from two perspectives: allowing the worth of coalitions to vary, while keeping the player set fixed; and allowing the player set to change, while keeping the worth of coalitions fixed. This class of values is characterized by efficiency, the balanced contributions property for equal contributors, weak covariance, a proportionately decreasing redistribution in one-person unanimity games, desirability, and null players in unanimity games. For the class of consensus values, we concentrate on redistributions in $(n-1)$-person unanimity games from the same two perspectives. This class of values is characterized by efficiency, the balanced contributions property for equal contributors to social surplus, complement weak covariance, a proportionately decreasing redistribution in $(n-1)$-person unanimity games, desirability, and null players in unanimity games.

Appendix

1.1 A.0. A modification of $\hbox {C}^-$ in Yokote et al. (2018)

As we have discussed in footnote 9, we add the supposition “Suppose $f(N,{\mathbf {0}})=(0,\dots ,0) \in {\mathbb {R}}^N$” to the definition of C⁻ in Yokote et al. (2018). This modification makes C⁻ weaker than C.

Lemma 1

C⁻ is weaker than C.

Proof

Let f satisfy C. We prove that f satisfies C⁻. Suppose $f(N, {\mathbf {0}})=(0,\dots ,0)$. Fix $i\in N$. Then,

$$\begin{aligned} f_j(N,u_i)&=f_j(N,{\mathbf {0}}+ u_i) \nonumber \\&={\left\{ \begin{array}{ll} f_j(N,{\mathbf {0}})+1 &{} \text { if } j=i, \\ f_j(N,{\mathbf {0}}) &{} \text { if } j\ne i. \end{array}\right. }\nonumber \\&={\left\{ \begin{array}{ll} 1 &{} \text { if } j=i, \\ 0 &{} \text { if } j\ne i, \end{array}\right. } \end{aligned}$$

(1)

where the second equality follows from C and the third equality follows from $f(N, {\mathbf {0}})=(0,\dots ,0)$. It follows that, for any $(N, v) \in G$ and $\lambda \in {\mathbb {R}}$,

$$\begin{aligned} f_j\left( N, v+\lambda u_{i}\right)&={\left\{ \begin{array}{ll}f_j(N,v)+\lambda &{} \text { if } j=i, \\ f_j(N,v) &{} \text { if } j\ne i, \end{array}\right. } \\&=f_j(N,v)+\lambda f_j(N, u_i) \text { for all } j\in N, \end{aligned}$$

where the first equality follows from C and the second equality follows from (1). $\square$

Furthermore, this modification does not matter for results in Yokote et al. (2018). This is because Yokote et al. (2018) use $\mathbf{C}^-$ together with E and BCEC. If we consider null games only, BCEC (and BCESS also) is equivalent to BC. Then, because E and BC characterize the Shapley value (Myerson 1980) and because the Shapley value satisfies NG, we obtain NG which guarantees the supposition part we add to the new definition of $\mathbf{C}^-$.

Remark 1

Let f satisfy E and BCEC (or BCESS). Then, f satisfies NG.

1.2 A.1. Proof of Theorem 1

$\alpha Sh+(1-\alpha )ED$ clearly satisfies ${\bf{PDR}}^{1}$, for any $\alpha \in {\mathbb {R}}$. It also satisfies D and NU if $\alpha \in [0,1]$. Together with Remark 1 above, Yokote et al. (2018, Theorem 1) show that f satisfies E, BCEC, and $\mathbf{C}^-$ if and only if there exists a sequence ${\mathbf {r}}\in \bigl \{\{r_{\ell }\}_{{\ell }=1}^{{\bar{n}}} : r_{\ell } \in {\mathbb {R}} \text { for all } {\ell } \in \{1, \dots , {\bar{n}}\}\bigr \}$, such that $f_i(N,v)=\frac{v(N)-v^{\mathbf {r}}(N)}{n}+Sh_i(N, v^{\mathbf {r}}) \text { for all } (N,v)\in G$ and $i \in N$, where $v^{\mathbf {r}}(S)=r_s v(S) \text { for all } S\subseteq N, S\ne \emptyset$, and any $(N,v) \in G$. Furthermore, the sequence ${\mathbf {r}}$ is recursively determined, as follows:¹³

• $1-r_1=2f_j(\{i,j\},u_i)$.

• For $\ell \in \{2, \dots , {\bar{n}}-1\}$, by choosing $N \subseteq {\mathcal {U}}$ with $n=\ell$, $i,j\in N$ with $i\ne j$, and $k\in {\mathcal {U}}\backslash N$:

  $$\begin{aligned} 1-r_{{\ell }}=\sum _{m=1}^{{\ell }}(1-r_m)-\sum _{m=1}^{{\ell }-1}(1-r_m)={\ell }({\ell }+1)f_j(N\cup k, u_i)-{\ell }({\ell }-1)f_j(N, u_i). \end{aligned}$$

• $r_{{\bar{n}}}$ is an arbitrary value.

By ${\bf{PDR}}^1$, $(\ell +1)f_j(N \cup k,u_i)=\ell f_j(N,u_i)$, for $\ell \in \{2,3,\dots , {\bar{n}}-1\}$ and, thus, $1-r_{\ell }=(\ell -(\ell -1))\ell f_j(N,u_i)=\ell f_j(N,u_i)=2f_j(\{i,j\},u_i)$. By letting $\alpha =1-2f_j(\{i,j\},u_i)$, $\alpha \in {\mathbb {R}}$ satisfies $r_{\ell }=r_{\ell +1}=\alpha$, for any $1 \le \ell \le {\bar{n}}-1$. By the linearity¹⁴ of Sh, $f_i(N,v)=(1-\alpha )\frac{v(N)}{n}+Sh_i(N,\alpha v)=\alpha Sh_i(N,v)+(1-\alpha )ED_i(N,v)$.

It suffices to show that D and NU imply $\alpha \in [0,1]$. We prove this by contradiction. Consider $(\{i,j\},u_i) \in G$. Now, $f_i(\{i,j\},u_i)=\alpha + \frac{1-\alpha }{2}$ and $f_j(\{i,j\},u_i)=\frac{1-\alpha }{2}$. Suppose $\alpha >1$. Then, $f_j(\{i,j\},u_i)<0$, which violates NU. Suppose $\alpha < 0$. Then, $f_i(\{i,j\},u_i)<f_j(\{i,j\},u_i)$, which violates D. Therefore, $\alpha \in [0,1]$.

1.3 A.2. Proof of Theorem 2

First, we prepare for the proof. In the proof, we use the following two axioms. Given $(N,v) \in G$, $i \in N$ and $j \in N {\setminus } i$ are symmetric in (N, v) if, for any $S \subseteq N {\setminus } \{i,j\}$, $v(S \cup i)=v(S \cup j)$.

Symmetry (S)::

For any $(N,v) \in G$, if $i \in N$ and $j \in N {\setminus } i$ are symmetric in (N, v), then $f_i(N,v)=f_j(N,v)$.

Anonymity (A)::

For any $(N,v) \in G$ and $\pi : N \rightarrow {\mathbb {N}}$, $f_i(N,v)=f_{\pi (i)}(\pi (N),\pi v)$, where $\pi v(S)=v(\pi ^{-1}(S))$, for any $S \subseteq \pi (N)$.

We divide the proof of Theorem 2 into 11 lemmas and remarks. The role of each lemma/remark is summarized as follows. Lemma 2 shows the existence of the value satisfying the six axioms. Remark 2 and Lemma 3 show the uniqueness of the value for $n=1$ and $n=2$, respectively. Remark 3 shows A of the value for $n=2$. Remark 4 shows S of the value for $n=t$ if the value satisfies A, for $n=t-1$. Lemmas 4 to 6 together show the uniqueness of the value for $n=3$ , with respect to a coefficient $\alpha _3 \in {\mathbb {R}}$. Remark 5 shows A of the value for $n=3$. Lemma 7 shows the uniqueness of the value for $n=t+1$ , with respect to a coefficient $\alpha _{t+1} \in {\mathbb {R}}$, if the value is unique for $n=t$ $(\ge 3)$ , with respect to a coefficient $\alpha _{t} \in {\mathbb {R}}$. Lemma 8 shows the coincidence of the coefficients for any game. Lastly, we narrow the range of coefficients to [0, 1].

In the following, the proofs of remarks are apparent, and hence, we omit them. Further, each of R3, R4, L5, $\dots$ denote Remark 3, Remark 4, Lemma 5, $\dots$, respectively.

Lemma 2

For any $\alpha \in {\mathbb {R}}$, $\alpha Sh+(1-\alpha )ESD$ satisfies E, BCESS, CC⁻ and PDR$^{n-1}$. Furthermore, for any $\alpha \in [0,1]$, $\alpha Sh+(1-\alpha )ESD$ satisfies D and NU.

Proof

E , D and NU are obvious. CC⁻ is from the linearity of Sh and ESD. For ${\bf{PDR}}^{n-1}$, if $n \ge 3$, $\alpha Sh_i(N,u_{N {\setminus } i})+(1-\alpha )ESD_i(N,u_{N {\setminus } i})=\frac{1-\alpha }{n}$. For BCESS, for any $(N,v) \in G$ and $i,j \in N$,

$$\begin{aligned} ESD_i(N,v)-ESD_j(N,v)=v(i)-v(j) \end{aligned}$$

(2)

and

$$\begin{aligned}&ESD_i(N {\setminus } j,v)-ESD_j(N {\setminus } i,v)\nonumber \\&\quad =\frac{v(N {\setminus } j)-\sum _{k \in N {\setminus } j}v(k)}{n-1}+v(i)-\frac{v(N {\setminus } i)-\sum _{k \in N {\setminus } i}v(k)}{n-1}-v(j) \nonumber \\&\quad =\frac{v(N {\setminus } j)-v(N {\setminus } i)}{n-1}+\frac{n-2}{n-1}(v(i)-v(j)). \end{aligned}$$

(3)

Here, $v(i)+v(N {\setminus } i)=v(j)+v(N {\setminus } j)$ implies that (2)=(3). Because Sh satisfies BC, and Sh and ESD both satisfy linearity, the desired result is obtained. $\square$

Remark 2

Let f satisfy E. Then, for any game $(N,v) \in G$, with $n=1$, $f_i(N,v)=v(i)$ (i.e., f is uniquely determined).

Lemma 3

¹⁵Let f satisfy E and BCESS. For any game $(N,v) \in G$, with $n=2$, $f_i(N,v)=\frac{v(N)+v(i)-v(j)}{2}$ (i.e., f is uniquely determined).

Proof

$n=2$ implies that $N {\setminus } j=i$ and $N {\setminus } j=i$, for $i,j \in N$. Hence, BCESS and Remark 2 together imply that $f_i(N,v)-f_j(N,v)=v(i)-v(j)$. Further, E implies that $f_i(N,v)+f_j(N,v)=v(N)$. These together imply the desired result. $\square$

Remark 3

Let f satisfy E and BCESS. Then, f satisfies A on the class of games with two players.

Remark 4

¹⁶ Let f satisfy E and BCESS. Let $t \in {\mathbb {N}}$, with $t \ge 3$. If f satisfies A on the class of games with $t-1$ players, then f satisfies S on the class of games with t players.

From here, we prepare for the induction with respect to the number of players by showing the uniqueness of the value with respect to a real number $\alpha _3$ for games with three players. The induction is shown in Lemma 7.

Lemma 4

Let ${|N|}=3$, and let f satisfy E, BCESS, and CC⁻. Then, there exists $\alpha _N$, such that $f_i(N,u_{N {\setminus } i})=\frac{1-\alpha _N}{ {3}}$, for any $i \in N$.

Proof

Players $k,\ell \in N {\setminus } i$ are symmetric in $(N,u_{N {\setminus } i})$. Hence,

$$\begin{aligned} f_k(N,u_{N {\setminus } i})-f_{\ell }(N,u_{N {\setminus } i}) \overset{\mathbf { {R3,R4}}}{=}0. \end{aligned}$$

For any $k \ne i$, let

$$\begin{aligned} \alpha _{N,i}= {6}f_k(N,u_{N {\setminus } i})- {2} \iff f_k(N,u_{N {\setminus } i})=\frac{ {2}+\alpha _{N,i}}{ {6}}. \end{aligned}$$

(4)

Then,

$$\begin{aligned} f_i(N,u_{N {\setminus } i})\overset{\mathrm {\mathbf{E}}}{=}u_{N{\setminus } i}(N)- {2}f_k(N,u_{N {\setminus } i})\overset{{\mathbf {(4)}}}{=}\frac{1-\alpha _{N,i}}{ {3}}. \end{aligned}$$

(5)

Similarly, consider the game $(N,u_{N {\setminus } j}) \in G$. For any $\ell \ne j$, by letting

$$\begin{aligned} \alpha _{N,j} = {6}f_{\ell }(N,u_{N {\setminus } j})- {2} \iff f_{\ell }(N,u_{N {\setminus } j})=\frac{ {2}+\alpha _{N,j}}{ {6}}, \end{aligned}$$

(6)

we obtain that

$$\begin{aligned} f_j(N,u_{N {\setminus } j})\overset{\mathrm {\mathbf{E}}}{=}u_{N{\setminus } j}(N)- {2}f_{\ell }(N,u_{N {\setminus } j})\overset{{\mathbf {(6)}}}{=}\frac{1-\alpha _{N,j}}{ {3}}. \end{aligned}$$

(7)

Now, consider a game $(N,u_{N {\setminus } i}+u_{N {\setminus } j}) \in G$. Because i and j are symmetric in $(N,u_{N {\setminus } i}+u_{N {\setminus } j})$,

$$\begin{aligned} f_i(N,u_{N {\setminus } i}+u_{N {\setminus } j})-f_j(N,u_{N {\setminus } i}+u_{N {\setminus } j})\overset{{\mathbf {{R3,R4}}}}{=}0. \end{aligned}$$

(8)

At the same time,

$$\begin{aligned}&f_i(N,u_{N {\setminus } i}+u_{N {\setminus } j})-f_j(N,u_{N {\setminus } i}+u_{N {\setminus } j})\nonumber \\&\quad \overset{\mathbf {{CC}^-}}{=}f_i(N,u_{N {\setminus } i})+f_i(N,u_{N {\setminus } j})-f_j(N,u_{N {\setminus } i})-f_j(N,u_{N {\setminus } j})\nonumber \\&\quad \overset{\mathbf {(5),(6),(4),(7)}}{=}\frac{-\alpha _{N,i}+\alpha _{N,j}}{{2}}. \end{aligned}$$

(9)

By (8) and (9), $\alpha _{N,i}=\alpha _{N,j}$, for any $i,j \in N$. By letting $\alpha _N$ denote the equal value, (5) establishes the desired result. $\square$

Lemma 5

Let ${|N|}=3$ and let f satisfy E, BCESS, CC⁻, and PDR$^{n-1}$. For any $(N,u_{N {\setminus } i}),(M,u_{M {\setminus } j}) \in G$, with $i \in N$, $j \in M$, and $|N|=|M|= {3}$, there exists $\alpha _{ {3}} \in {\mathbb {R}}$, such that $f_i(N,u_{N {\setminus } i})=f_j(M,u_{ {M} {\setminus } j})=\frac{1-\alpha _{ {3}}}{ {3}}$.

Proof

By Lemma 4,

$$\begin{aligned} \alpha _N\overset{\mathrm {(5)}}{=} 1- {3}f_i(N,u_{N {\setminus } i})\overset{\mathrm {\mathbf{PDR}^{n-1}}}{=} 1- {3}f_j(M,u_{ {M} {\setminus } j})\overset{\mathrm {(5)}}{=}\alpha _{M}. \end{aligned}$$

Thus, $\alpha _N=\alpha _M=\alpha _{ {3}}$. $\square$

Lemma 6

Let ${|N|}=3$, and let f satisfy E, BCESS, CC⁻, and ${\bf{PDR}}^{n-1}$. For any game with three players, there exists $\alpha _{ {3}} \in {\mathbb {R}}$, such that $f=\alpha _{ {3}} Sh+(1-\alpha _{ {3}})ESD$.

Proof

By Lemma 2, $f=\alpha _{ {3}} Sh+(1-\alpha _{ {3}})ESD$ satisfies the four axioms with respect to $\alpha _{ {3}} \in {\mathbb {R}}$.

In the following, we show the uniqueness of f with respect to $\alpha _{ {3}}$. Given $(N,v) \in G$, with $|N|={ {3}}$, let $i,j \in N$, and let $\lambda _{i,j}=v(j)+v(N {\setminus } j)-v(i)-v(N {\setminus } i)$. Consider the game $(N,v+\lambda _{i,j}u_{N {\setminus } i})$.¹⁷ In this game,

• $(v+\lambda _{i,j}u_{N {\setminus } i})(i)=v(i)$,

• $(v+\lambda _{i,j}u_{N {\setminus } i})(N {\setminus } i)=v(j)+v(N {\setminus } j)-v(i)$,

• $(v+\lambda _{i,j}u_{N {\setminus } i})(j)=v(j)$, and

• $(v+\lambda _{i,j}u_{N {\setminus } i})(N {\setminus } j)=v(N {\setminus } j)$.

Thus, $(v+\lambda _{i,j}u_{N {\setminus } i})(i)+(v+\lambda _{i,j}u_{N {\setminus } i})(N {\setminus } i)=(v+\lambda _{i,j}u_{N {\setminus } i})(j)+(v+\lambda _{i,j}u_{N {\setminus } i})(N {\setminus } j)$. BCESS implies that

$$\begin{aligned} f_i(N,v+\lambda _{i,j}u_{N {\setminus } i})-f_j(N,v+\lambda _{i,j}u_{N {\setminus } i})=f_i(N {\setminus } j,v)-f_j(N {\setminus } i,v+\lambda _{i,j}u_{N {\setminus } i}). \end{aligned}$$

(10)

Now,

$$\begin{aligned}&f_i(N,v+\lambda _{i,j}u_{N {\setminus } i})-f_j(N,v+\lambda _{i,j}u_{N {\setminus } i})\nonumber \\&\quad \overset{\mathbf {CC}^-}{=}f_i(N,v)+\lambda _{i,j}f_i(N,u_{N {\setminus } i})-f_j(N,v)-\lambda _{i,j}f_j(N,u_{N {\setminus } i})\nonumber \\&\quad \overset{\mathbf { {L5}},(5),(4)}{=}f_i(N,v)+\lambda _{i,j}\frac{1-\alpha _{ {3}}}{ {3}}-f_j(N,v)-\lambda _{i,j}\frac{ {2}+\alpha _{ {3}}}{ {6}}. \end{aligned}$$

(11)

Then,

$$\begin{aligned} f_i(N,v)-f_j(N,v)\overset{\mathbf {(10),(11)}}{=}f_i(N {\setminus } j,v)-f_j(N {\setminus } i,v+\lambda _{i,j}u_{N {\setminus } i})+\lambda _{i,j}\frac{\alpha _{ {3}}}{ {2}}. \end{aligned}$$

(12)

By Lemma 3, $f_i(N {\setminus } j,v)$ and $f_j(N {\setminus } i,v+\lambda _{i,j}u_{N {\setminus } i})$ are uniquely determined. Hence, $f_i(N,v)-f_j(N,v)$ is uniquely determined with respect to $\alpha _3$.

Taking the above argument for i and any $k \ne i,j$ we obtain that

$$\begin{aligned} f_i(N,v)-f_k(N,v)=f_i(N {\setminus } k,v)-f_k(N {\setminus } i,v+\lambda _{i,k}u_{N {\setminus } i})+\lambda _{i,k}\frac{\alpha _{ {3}}}{ {2}}. \end{aligned}$$

(13)

Furthermore, by E

$$\begin{aligned} \sum _{\ell \in N}f_{\ell }(N,v)=v(N). \end{aligned}$$

(14)

(12), (13), and (14) are linearly independent of each other.¹⁸ Thus, given $\alpha _{ {3}} \in {\mathbb {R}}$, they constitute mutually independent linear equations with three variables, with a unique solution equal to $\alpha _{ {3}} Sh+(1-\alpha _{ {3}}) ESD$. $\square$

Remark 5

Let f satisfy E, BCESS, CC⁻, and ${\bf{PDR}}^{n-1}$. f satisfies A on the class of games with three players.

Now, we can complete the induction.

Lemma 7

Let f satisfy E, BCESS, CC⁻, and ${\bf{PDR}}^{n-1}$, and let $t \ge 3$.(i) If there exists $\alpha _t \in {\mathbb {R}}$, such that $f=\alpha _t Sh+(1-\alpha _t)ESD$ for any game with t players, then there exists $\alpha _{t+1} \in {\mathbb {R}}$, such that $f=\alpha _{t+1} Sh+(1-\alpha _{t+1})ESD$ for any game with $t+1$ players. (ii) f satisfies S for any games with $t+2$ players.

Proof

(i) From Lemmas 4–6, there exists $\alpha _{ {3}} \in {\mathbb {R}}$, such that $f=\alpha _{ {3}} Sh+(1-\alpha _{ {3}})ESD$ for any game with three players. Now, suppose that there exists $\alpha _t \in {\mathbb {R}}$, such that $f=\alpha _t Sh+(1-\alpha _t)ESD$ for any game with t players, and consider any game (N, v) with $t+1$ players, where $t \ge 3$. We divide the proof into three steps.

• Step 1 (Corresponds to Lemma 4): There exists $\alpha _N$ such that $f_i(N,u_{N {\setminus } i})=\frac{1-\alpha _N}{t+1}$ for any $i \in N$. $\square$

Proof

Let $i,j \in N$. Players in $N {\setminus } i$ and $N {\setminus } j$ are symmetric in $(N ,u_{N {\setminus } i})$ and $(N,u_{N {\setminus } j})$, respectively. Hence, by letting $\alpha _{N,i}=t(t+1)f_k(N,u_{N {\setminus } i})-t$ and $\alpha _{N.j}=t(t+1)f_k(N,u_{N {\setminus } j})-t$ with $k \in N {\setminus } \{i,j\}$, E implies that $f_i(N,u_{N {\setminus } i})=\frac{1-\alpha _{N,i}}{t+1}$ and that $f_j(N,u_{N {\setminus } j})=\frac{1-\alpha _{N,j}}{t+1}$. Similar to the proof of Lemma 4, by considering a game $(N,u_{N {\setminus } i}+u_{N {\setminus } j})$, S and CC⁻ together imply $\alpha _{N,i}=\alpha _{N,j}$, and we let $\alpha _N=\alpha _{N,i}$. $\square$

• Step 2 (Corresponds to Lemma 5): There exists $\alpha _{t+1}$ such that $f_i(N,u_{N {\setminus } i})=f_j(M,u_{M {\setminus } j})=\frac{1-\alpha _{t+1}}{t+1}$ for any player set N and M with $|N|=|M|=t+1$.

Proof

Similar to Lemma 5, Step 1 and $\mathbf{PDR} ^{n-1}$ together imply the desired result. $\square$

• Step 3 (Corresponds to Lemma 6): $f=\alpha _{t+1}Sh+(1-\alpha _{t+1})ESD$ for any game with $t+1$ players.

Proof

Similar to Lemma 6, given a game (N, v) with $t+1$ players, given $i \in N$ and any $j \in N {\setminus } i$, we consider a new game $(N,v+\lambda _{i,j}u_{N {\setminus } i})$ with $\lambda =v(j)+v(N {\setminus } j)-v(i)-v(N {\setminus } i),$ and apply BCESS to it. Then, by inductive hypothesis, $\mathbf {CC}^-$ and Steps 1 and 2, $f_i(N,v)-f_j(N,v)$ is uniquely determined with respect to $\alpha _{t+1}$. E implies $\sum _{\ell \in N}f_{\ell }(N,v)=v(N)$. Hence, with respect to $\alpha _{t+1}$, f is uniquely determined for games with $t+1$ players. $\square$

(ii) Apparent from (i) and Remark 4.

By (i) and (ii), an induction with respect to $t \ge 3$ implies the desired result. $\square$

Lemma 8

Let f satisfy E, BCESS, CC⁻, and ${\bf{PDR}}^{n-1}$. Then, there exists $\alpha \in {\mathbb {R}}$, such that $f=\alpha Sh+(1-\alpha )ESD$.

Proof

Lemma 7 and ${\bf{PDR}}^{n-1}$ together imply that $\alpha _t=\alpha _{t+1}$, for any $t \ge 3$. For any game with one or two players, $Sh=ESD$. Therefore, the desired result is obtained. $\square$

Lastly, it suffices to show that D and NU imply $\alpha \in [0,1]$. We prove the fact by contradiction. Consider $(\{i,j,k\},u_{\{i,j\}}) \in G$. Now, $f_i(\{i,j,k\},u_{\{i,j\}})=\frac{\alpha }{2} + \frac{1-\alpha }{3}$ and $f_k(\{i,j,k\},u_{\{i,j\}})=\frac{1-\alpha }{3}$. Suppose $\alpha >1$. Then, $f_k(\{i,j,k\},u_{\{i,j\}})<0$, which violates NU. Suppose $\alpha < 0$. Then, $f_i(\{i,j,k\},u_{\{i,j\}})<f_k(\{i,j,k\},u_{\{i,j\}})$, which violates D.