The Allocations for Dual Fuzzy Cooperative Games

Abstract

In our daily life, players may reduce their cooperation levels for various reasons. This paper studies how to evaluate and allocate the cooperation payoff when the players cooperate partially. Different from previous research that directly uses the cooperation level to calculate the payoffs of fuzzy coalitions, we introduce the concept of dual fuzzy cooperative games, where the payoffs of fuzzy coalitions equal the difference between the original payoff and the lost payoff caused by the decline of the cooperation level. Meanwhile, the dual Shapley value is introduced, and its axiomatic systems are studied. Further, a special kind of dual fuzzy cooperative game with Choquet integral form is presented. Finally, we provide an application of dual fuzzy cooperative games in the manufacturer–retailer supply chain to allocate the payoff for vertical co-op advertising.

Appendices

Appendix A

Proof of Theorem 2

Existence. Similar to Shapley [22], the detail proofs are omitted.

Uniqueness. Similar to traditional cooperative games, we have $v^{D} = \sum\limits_{\emptyset \ne T \subseteq U} {c_{T}^{D} u_{T} }$, where

$$ \begin{aligned} c_{T}^{D} = & \sum\limits_{S \subseteq T} {( - 1)^{{\left| {{\text{Supp}} T} \right| - \left| {{\text{Supp}} S} \right|}} \left( {v^{D} (S)} \right)} = \sum\limits_{S \subseteq T} {( - 1)^{{\left| {{\text{Supp}} T} \right| - \left| {{\text{Supp}} S} \right|}} \left( {v(1^{{{\text{Supp}} S}} ) - v(S^{D} )} \right)} \\ & = \sum\limits_{{{\text{Supp}} S \subseteq {\text{Supp}} T}} {( - 1)^{{\left| {{\text{Supp}} T} \right| - \left| {{\text{Supp}} S} \right|}} v(1^{{{\text{Supp}} S}} )} - \sum\limits_{{S^{D} \subseteq T^{D} }} {( - 1)^{{\left| {{\text{Supp}} T^{D} } \right| - \left| {{\text{Supp}} S^{D} } \right|}} v(S^{D} )} \\ \end{aligned} $$

and $u_{T} (S) = \left\{ \begin{gathered} 1\;\;\;\;\;S \subseteq T \hfill \\ 0\;\;\;\;{\text{otherwise}} \hfill \\ \end{gathered} \right.$.

Let $c_{{{\text{Supp}} T}} = \sum\limits_{{{\text{Supp}} S \subseteq {\text{Supp}} T}} {( - 1)^{{\left| {{\text{Supp}} T} \right| - \left| {{\text{Supp}} S} \right|}} v(1^{{{\text{Supp}} S}} )}$ and $c_{{T^{D} }} = \sum\limits_{{S^{D} \subseteq T^{D} }} {( - 1)^{{\left| {{\text{Supp}} T^{D} } \right| - \left| {{\text{Supp}} S^{D} } \right|}} v(S^{D} )}$ for any $T \subseteq U$, we have $v^{D} = \sum\limits_{\emptyset \ne T \subseteq U} {\left( {c_{{{\text{Supp}} T}} - c_{{T^{D} }} } \right)u_{T} }$. From DFA, we need to prove the conclusion on the fuzzy unanimity game $u_{T}$, i.e.,$f^{D} (U,u_{T} ) = Sh^{D} (U,u_{T} )$. By DFC and DFS, we have $f_{i}^{D} (U,u_{T} ) = \left\{ {\begin{array}{*{20}c} {\frac{1}{{\left| {{\text{Supp}} T} \right|}}} & {i \in {\text{Supp}} T} \\ 0 & {i \in {\text{Supp}} T} \\ \end{array} } \right.$.

On the other hand, we obtain $Sh_{i}^{D} (U,u_{T} ) = \left\{ {\begin{array}{*{20}c} {\frac{1}{{\left| {{\text{Supp}} T} \right|}}} & {i \in {\text{Supp}} T} \\ 0 & {i \in {\text{Supp}} T} \\ \end{array} } \right.$ by Eq. (8). Therefore,$Sh^{D} = f^{D}$.□

Appendix B

Proof of Theorem 3

Existence. $Sh^{D}$ obviously satisfies DFE. Next, we show that $Sh^{D}$ owns DFBC, namely

$$ \begin{gathered} Sh_{i} (1^{{{\text{Supp}} U}} ,v) - Sh_{i} (U^{D} ,v) - [Sh_{i} (1^{{{\text{Supp}} U\backslash j}} ,v) - Sh_{i} (U^{D} \backslash U^{D} (j),v)] \hfill \\ = Sh_{j} (1^{{{\text{Supp}} U}} ,v) - Sh_{j} (U^{D} ,v) - [Sh_{j} (1^{{{\text{Supp}} U\backslash i}} ,v) - Sh_{j} (U^{D} \backslash U^{D} (i),v)] \hfill \\ \end{gathered} $$

for all $i,j \in {\text{Supp}} U$ with $i \ne j$.

According to Theorem A in [6], we have

$$ Sh_{i} (1^{{{\text{Supp}} U}} ,v) - Sh_{i} (1^{{{\text{Supp}} U\backslash j}} ,v) = Sh_{j} (1^{{{\text{Supp}} U}} ,v) - Sh_{j} (1^{{{\text{Supp}} U\backslash i}} ,v) $$

for all $i,j \in {\text{Supp}} U$ with $i \ne j$.

Therefore, we need to show

$$ Sh_{i} (U^{D} ,v) - Sh_{i} (U^{D} \backslash U^{D} (j),v) = Sh_{j} (U^{D} ,v) - Sh_{j} (U^{D} \backslash U^{D} (i),v) $$

for all $i,j \in {\text{Supp}} U$ with $i \ne j$.

From Eq. (1), we know

$$ \begin{gathered} Sh_{i} (U^{D} ,v) - Sh_{i} (U^{D} \backslash U^{D} (j),v) \hfill \\ \quad \quad = \sum\limits_{{i \in {\text{Supp}} R^{D} \subseteq {\text{Supp}} U^{D} }} {\frac{{\left( {\left| {{\text{Supp}} R^{D} } \right| - 1} \right)!\left( {\left| {{\text{Supp}} U^{D} } \right| - \left| {{\text{Supp}} R^{D} } \right|} \right)!}}{{\left| {{\text{Supp}} U^{D} } \right|!}}} (v(R^{D} ) - v(R^{D} \backslash U^{D} (i))) \hfill \\ \quad \quad - \sum\limits_{{i \in {\text{Supp}} R^{D} \subseteq {\text{Supp}} (U^{D} \backslash U^{D} (j))}} {\frac{{\left( {\left| {{\text{Supp}} R^{D} } \right| - 1} \right)!\left( {\left| {{\text{Supp}} (U^{D} \backslash U^{D} (j))} \right| - \left| {{\text{Supp}} R^{D} } \right|} \right)!}}{{\left| {{\text{Supp}} (U^{D} \backslash U^{D} (j))} \right|!}}} (v(R^{D} ) - v(R^{D} \backslash U^{D} (i))) \hfill \\ \quad \quad = \sum\limits_{{i \in {\text{Supp}} R^{D} \subseteq {\text{Supp}} U^{D} }} {\frac{{\left( {\left| {{\text{Supp}} R^{D} } \right| - 1} \right)!\left( {\left| {{\text{Supp}} U^{D} } \right| - \left| {{\text{Supp}} R^{D} } \right|} \right)!}}{{\left| {{\text{Supp}} U^{D} } \right|!}}} (v(R^{D} ) - v(R^{D} \backslash U^{D} (i))) \hfill \\ \quad \quad - \sum\limits_{{i \in {\text{Supp}} R^{D} \subseteq {\text{Supp}} U^{D} }} {\frac{{\left( {\left| {{\text{Supp}} R^{D} } \right| - 1} \right)!(\left| {{\text{Supp}} U^{D} } \right| - \left| {{\text{Supp}} R^{D} } \right|)!}}{{\left| {{\text{Supp}} U^{D} } \right|!}}} (v((R^{D} \backslash U^{D} (j)) - v((R^{D} \backslash \{ U^{D} (j),U^{D} (i)\} ) \hfill \\ \quad \quad = \sum\limits_{{i \in {\text{Supp}} R^{D} \subseteq {\text{Supp}} U^{D} }} {\frac{{\left( {\left| {{\text{Supp}} R^{D} } \right| - 1} \right)!\left( {\left| {{\text{Supp}} U^{D} } \right| - \left| {{\text{Supp}} R^{D} } \right|} \right)!}}{{\left| {{\text{Supp}} U^{D} } \right|!}}} \left[ {v(R^{D} ) - v(R^{D} \backslash U^{D} (i)) - v(R^{D} \backslash U^{D} (j))} \right. \hfill \\ \left. {\quad \quad + v(R^{D} \backslash \{ U^{D} (i),U^{D} (j)\} )} \right] \hfill \\ \end{gathered} $$

Since the above expression is symmetric for i and j, we can similarly derive this expression for $Sh_{j} (U^{D} ,v) - Sh_{j} (U^{D} \backslash U^{D} (i),v)$. Therefore, DFBC holds.

Uniqueness. Let $U \in L(N)$ and $v^{D} \in DG(N)$. For any $U \in L(N)$, if $\left| {SuppU} \right| = 1$, without loss of generality, let U = {U(i)}. By DFE, we have.

$f^{D} (U,v^{D} ) = f_{i} (1^{{{\text{Supp}} U}} ,v) - f_{i} (U^{D} ,v) = v(1^{{{\text{Supp}} U(i)}} ) - v(U^{D} (i)) = Sh_{i} (1^{{{\text{Supp}} U(i)}} ,v) - Sh_{i} (U^{D} ,v)$.

Assume that $f^{D} = Sh^{D}$ when $\left| {{\text{Supp}} U} \right| \le n - 1$($n \ge 2$). Next, we prove that the conclusion is true when $\left| {{\text{Supp}} U} \right| = n$. By DFBC, we obtain

$$ \begin{gathered} f_{i} (1^{{{\text{Supp}} U}} ,v) - f_{i} (U^{D} ,v) - [f_{i} (1^{{{\text{Supp}} U\backslash j}} ,v) - f_{i} (U^{D} \backslash U^{D} (j),v)] \hfill \\ \quad \quad = f_{j} (1^{{{\text{Supp}} U}} ,v) - f_{j} (U^{D} ,v) - [f_{j} (1^{{{\text{Supp}} U\backslash i}} ,v) - f_{j} (U^{D} \backslash U^{D} (i),v)] \hfill \\ \end{gathered} $$

for all $i,j \in {\text{Supp}} U$ with$i \ne j$.

By assumption, we get

$$ \begin{aligned} f_{i} (1^{{{\text{Supp}} U}} ,v) - & f_{i} (U^{D} ,v) - [f_{j} (1^{{{\text{Supp}} U}} ,v) - f_{j} (U^{D} ,v)] \\ & = Sh_{i} (1^{{{\text{Supp}} U\backslash j}} ,v) - Sh_{i} (U^{D} \backslash U^{D} (j),v) - [Sh_{j} (1^{{{\text{Supp}} U\backslash i}} ,v) - Sh_{j} (U^{D} \backslash U^{D} (i),v)] \\ \end{aligned} $$

(a.1)

Because $Sh^{D}$ satisfies DFBC, from Eq. (a.1) we have namely

$$ \begin{aligned} f_{i} (1^{{{\text{Supp}} U}} ,v) - & f_{i} (U^{D} ,v) - [f_{j} (1^{{{\text{Supp}} U}} ,v) - f_{j} (U^{D} ,v)] \\ & = Sh_{i} (1^{{{\text{Supp}} U}} ,v) - Sh_{i} (U^{D} ,v) - [Sh_{j} (1^{{{\text{Supp}} U}} ,v) - Sh_{j} (U^{D} ,v)] \\ \end{aligned} $$

$$ \begin{aligned} f_{i} (1^{{{\text{Supp}} U}} ,v) - & f_{i} (U^{D} ,v) - [Sh_{i} (1^{{{\text{Supp}} U}} ,v) - Sh_{i} (U^{D} ,v)] \\ & = f_{j} (1^{{{\text{Supp}} U}} ,v) - f_{j} (U^{D} ,v) - [Sh_{j} (1^{{{\text{Supp}} U}} ,v) - Sh_{j} (U^{D} ,v)] \\ \end{aligned} $$

(a.2)

From Eq. (a.2), we derive

$$ f_{i}^{D} (U,v^{D} ) - [Sh_{i}^{D} (U,v^{D} )] = f_{j}^{D} (U,v^{D} ) - [Sh_{j}^{D} (U,v^{D} )] $$

Take the sum for $j \in {\text{Supp}} U$, we obtain

$$ \left| {{\text{Supp}} U} \right|[f_{i}^{D} (U,v^{D} ) - Sh_{i}^{D} (U,v^{D} )] = \sum\limits_{{j \in {\text{Supp}} U}} {f_{j}^{D} (U,v^{D} )} - \sum\limits_{{j \in {\text{Supp}} U}} {Sh_{j}^{D} (U,v^{D} )} = v^{D} (U) - v^{D} (U) = 0 $$

Therefore,$f_{i}^{D} (U,v^{D} ) = Sh_{i}^{D} (U,v^{D} )$ for all $i \in {\text{Supp}} U$, and the conclusion is true. □

Appendix C

Proof of Theorem 3

Existence. By Eq. (8), STPDFG holds. Next, let us show that $Sh_{i} (1^{{{\text{Supp}} U}} ,v) - Sh_{i} (U^{D} ,v) =$ $Sh_{i} (1^{{{\text{Supp}} R}} ,v^{{f,1^{{{\text{Supp}} R}} }} ) - Sh_{i} (R^{D} ,v^{{f,R^{D} }} )$ for any $R \subseteq U$ and any $i \in {\text{Supp}} R$. From Theorem B in [6], we know that $Sh_{i} (1^{{{\text{Supp}} U}} ,v) = Sh_{i} (1^{{{\text{Supp}} S}} ,v^{{f,1^{{{\text{Supp}} R}} }} )$. Therefore, we just need to show that $Sh_{i} (U^{D} ,v) = Sh_{i} (S^{D} ,v^{{f,S^{D} }} )$.

From the linearity of $Sh(U^{D} ,v)$ and $v = \sum\limits_{{\emptyset \ne T^{D} \subseteq U^{D} }} {c_{{T^{D} }} u_{{T^{D} }} }$, where $c_{{T^{D} }} = \sum\limits_{{S^{D} \subseteq T^{D} }} {( - 1)^{{\left| {{\text{Supp}} T^{D} } \right| - \left| {{\text{Supp}} S^{D} } \right|}} v(S^{D} )}$ and $u_{{T^{D} }} (S^{D} ) = \left\{ \begin{gathered} 1,\;\;\;\;\;S^{D} \subseteq T^{D} \hfill \\ 0\;\;\;\;\;{\text{otherwise}} \hfill \\ \end{gathered} \right.$. We need to prove the conclusion on the fuzzy unanimity game $u_{{T^{D} }}$. Take $u_{{T^{D} }}$ such that $T^{D} \subseteq U^{D}$ and $R^{D} \subseteq U^{D}$. By Definition 10, we know that

$$ u_{{T^{D} }}^{{Sh,R^{D} }} (Q^{D} ) = u_{{T^{D} }} (Q^{D} \vee (R^{D} )^{c} ) - \sum\limits_{{i \in {\text{Supp}} (R^{D} )^{c} }} {Sh_{i} (Q^{D} \vee (R^{D} )^{c} ,v)} $$

for any $Q^{D} \subseteq R^{D}$.

If ${\text{Supp}} (T^{D} \wedge R^{D} ) = \emptyset$, then $T^{D} \subseteq (R^{D} )^{c}$. For any $Q^{D} \subseteq R^{D}$, we have

$$ u_{{T^{D} }} (Q^{D} \vee (R^{D} )^{c} ) = \sum\limits_{{i \in {\text{Supp}} (R^{D} )^{c} }} {Sh_{i} (Q^{D} \vee (R^{D} )^{c} ,u_{{T^{D} }} )} = 1 $$

Thus,$u_{{T^{D} }}^{{Sh,R^{D} }} (Q^{D} ) \equiv 0$ for any $Q^{D} \subseteq R^{D}$. Further,$Sh_{i} (R^{D} ,u_{{T^{D} }}^{{Sh,R^{D} }} ) = Sh_{i} (U^{D} ,u_{{T^{D} }} ) = 0$ for any $i \in {\text{Supp}} R^{D}$.

If ${\text{Supp}} (T^{D} \wedge R^{D} ) \ne \emptyset$, then, for any $Q^{D} \subseteq R^{D}$ with $T^{D} \wedge R^{D} \subseteq Q^{D}$, we have

$$ \begin{aligned} u_{{T^{D} }}^{{Sh,R^{D} }} (Q^{D} ) = & u_{{T^{D} }} (Q^{D} \vee (R^{D} )^{c} ) \\ & - \sum\limits_{{i \in {\text{Supp}} (R^{D} )^{c} }} {Sh_{i} (Q^{D} \vee (R^{D} )^{c} ,u_{{T^{D} }} )} \\ & = 1 - (1 - \sum\limits_{{i \in {\text{Supp}} (R^{D} \wedge T^{D} )}} {Sh_{i} (Q^{D} \vee (R^{D} )^{c} ,u_{{T^{D} }} )} ) \\ & = 1 - \left( {1 - \frac{{\left| {{\text{Supp}} (R^{D} \wedge T^{D} )} \right|}}{{\left| {{\text{Supp}} T^{D} } \right|}}} \right) \\ & = \frac{{\left| {{\text{Supp}} (R^{D} \wedge T^{D} )} \right|}}{{\left| {{\text{Supp}} T^{D} } \right|}} \\ \end{aligned} $$

For any $Q^{D} \subseteq R^{D}$ with $T^{D} \wedge R^{D} \not\subset Q^{D}$, we get

$$ u_{{T^{D} }}^{{Sh,R^{D} }} (Q^{D} ) = u_{{T^{D} }} (Q^{D} \vee (R^{D} )^{c} ) - \sum\limits_{{i \in {\text{Supp}} (R^{D} )^{c} }} {Sh_{i} (Q^{D} \vee (R^{D} )^{c} ,u_{{T^{D} }} )} = 0 $$

Thus,

$$ u_{{T^{D} }}^{{Sh,R_{d} }} (Q^{D} ) = \left\{ \begin{gathered} \frac{{\left| {{\text{Supp}} (R^{D} \wedge T^{D} )} \right|}}{{\left| {{\text{Supp}} T^{D} } \right|}},\;\;\;\;\;\;\;\;R^{D} \wedge T^{D} \subseteq Q^{D} \hfill \\ 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;R^{D} \wedge T^{D} \not\subset Q^{D} \hfill \\ \end{gathered} \right. $$

Therefore, $Sh_{i} (R^{D} ,u_{{T^{D} }}^{{Sh,R^{D} }} ) = \left\{ \begin{gathered} \frac{1}{{\left| {{\text{Supp}} T^{D} } \right|}},\;\;\;\;i \in {\text{Supp}} (R^{D} \wedge T^{D} ) \hfill \\ 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;i \in {\text{Supp}} (R^{D} \backslash T^{D} )\;\; \hfill \\ \end{gathered} \right. = Sh_{i} (U^{D} ,u_{{T^{D} }} )$ for all $i \in {\text{Supp}} R^{D}$.

Uniqueness. From DFE, we obtain

$$ \sum\limits_{{i \in {\text{Supp}} U}} {f_{i}^{D} (U,v^{D} )} = \sum\limits_{{i \in {\text{Supp}} U}} {[f_{i} (1^{{{\text{Supp}} U}} ,v) - f_{i} (U^{D} ,v)]} = v(1^{{{\text{Supp}} U}} ) - v(U^{D} ) = v^{D} (U) $$

If $\left| {{\text{Supp}} U} \right| = 1$, we show that $f_{i}^{D} (U(i),v^{D} ) = f_{i} (1^{SuppU(i)} ,v) - f_{i} (U^{D} (i),v) = v(1^{{{\text{Supp}}U(i)}} ) - v(U^{D} (i)) = v^{D} (U(i))$. Let $v(1^{{{\text{Supp}} U(i)}} ) - v(U^{D} (i)) = \delta$. We consider the game $(\{ U(i),U(j)\} ,v^{D} )$, where $i \ne j$, such that $v^{D} (U(i)) = v^{D} (\{ U(i),U(j)\} ) = \delta$ and $v^{D} (U(j)) = 0$.

By STPDFG, we derive $f_{i}^{D} (\{ U(i),U(j)\} ,v^{D} ) = f_{i} (1^{{{\text{Supp}} \{ U(i),U(j)\} }} ,v) - f_{i} (\{ U^{D} (i),U^{D} (j)\} ,v) = \delta$ and $f_{j}^{D} (\{ U(i),U(j)\} ,v^{D} ) =$$f_{j} (1^{{{\text{Supp}} \{ U(i),U(j)\} }} ,v) - f_{j} (\{ U^{D} (i),U^{D} (j)\} ,v) = 0$. Hence,

$$ v_{{f^{D} ,U(i)}}^{D} (U(i)) = \delta - 0 = \delta = v^{D} (U(i)) $$

From the consistency, we have

$$ \begin{gathered} \delta = f_{i}^{D} (\{ U(i),U(j)\} ,v^{D} ) = f_{i} (1^{{{\text{Supp}} \{ U(i),U(j)\} }} ,v) - f_{i} (\{ U^{D} (i),U^{D} (j)\} ,v) \hfill \\ = f_{i} (1^{{{\text{Supp}} U(i)}} ,v^{{f,1^{{{\text{Supp}} U(i)}} }} ) - f_{i} (U^{D} (i),v^{{f,U^{D} (i)}} ) = f_{i} (1^{{{\text{Supp}} U(i)}} ,v) - f_{i} (U^{D} (i),v) = f_{i}^{D} (U(i),v^{D} ) \hfill \\ \end{gathered} $$

Thus, $f_{i}^{D} (U(i),v^{D} ) = v^{D} (U(i))$.

This indeed holds for $\left| {{\text{Supp}} U} \right| = 2$ by the property of STPDFG. Let $\left| {{\text{Supp}} U} \right| \ge 3$. Assume that

$$ \sum\limits_{{i \in {\text{Supp}} U}} {f_{i}^{D} (U,v^{D} )} = v^{D} (U) $$

for $v^{D}$ with less than $\left| {{\text{Supp}} U} \right|$ players, when there are $\left| {{\text{Supp}} U} \right|$ players. By DFC, we have

$$ \begin{gathered} \sum\limits_{{i \in {\text{Supp}} U}} {f_{i}^{D} (U,v^{D} )} = \sum\limits_{{i \in {\text{Supp}} U}} {[f_{i} (1^{{{\text{Supp}} U}} ,v) - f_{i} (U^{D} ,v)]} \hfill \\ = \sum\limits_{{i \in {\text{Supp}} U\backslash j}} {[f_{i} (1^{{{\text{Supp}} U\backslash j}} ,v^{{f,1^{{{\text{Supp}} U\backslash j}} }} ) - f_{i} (U^{D} \backslash U^{D} (j),v^{{f,U^{D} \backslash U^{D} (j)}} )]} + f_{j} (1^{{{\text{Supp}} U}} ,v) - f_{j} (U^{D} ,v) \hfill \\ \end{gathered} $$

By assumption,$f^{D}$ satisfies DFE when there are $\left| {{\text{Supp}} U} \right| - 1$ players.

Thus, the above equation can be written as

$$ \begin{aligned} \sum\limits_{{i \in {\text{Supp}} U}} {f_{i}^{D} (U,v^{D} )} = & \sum\limits_{{i \in {\text{Supp}} U}} {[f_{i} (1^{{{\text{Supp}} U}} ,v) - f_{i} (U^{D} ,v)]} \\ & = v^{{f,1^{{{\text{Supp}} U\backslash j}} }} (1^{{{\text{Supp}} U\backslash j}} ) - v^{{f,U^{D} \backslash U^{D} (j)}} (U^{D} \backslash U^{D} (j)) + f_{j} (1^{{{\text{Supp}} U}} ,v) - f_{j} (U^{D} ,v) \\ & = v_{{f^{D} ,R}}^{D} \left( {U\backslash U(j)} \right) + f_{j}^{D} (U,v^{D} ) \\ \end{aligned} $$

According to Definition 10, we obtain $\sum\limits_{{i \in {\text{Supp}} U}} {f_{i}^{D} (U,v^{D} )} = v^{D} \left( U \right)$. Thus, DFE holds for $f^{D}$.

If $\left| {{\text{Supp}} U} \right| = 2$, by STPDFG we derive

$$ \begin{aligned} f_{i}^{D} (U,v^{D} ) = & f_{i} (1^{{{\text{Supp}} U}} ,v) - f_{i} (U^{D} ,v) = \frac{1}{2}(v^{D} (U) + v^{D} (U(i)) - v^{D} (U(j))) \\ & = Sh_{i} (1^{{{\text{Supp}} U}} ,v) - Sh_{i} (U^{D} ,v) = Sh_{i}^{D} (U,v^{D} ) \\ \end{aligned} $$

(a.3)

and

$$ \begin{aligned} f_{j}^{D} (U,v^{D} ) = & f_{j} (1^{{{\text{Supp}} U}} ,v) - f_{j} (U^{D} ,v) = \frac{1}{2}(v^{D} (U) + v^{D} (U(j)) - v^{D} (U(i))) \\ & = Sh_{j} (1^{{{\text{Supp}} U}} ,v) - Sh_{j} (U^{D} ,v) = Sh_{j}^{D} (U,v^{D} ) \\ \end{aligned} $$

(a.4)

for all $i,j \in {\text{Supp}} U$ such that $i \ne j$.

From Eqs. (a.3) and (a.4), we get

$$ f_{i}^{D} (U,v^{D} ) - f_{j}^{D} (U,v^{D} ) = f_{i} (1^{{{\text{Supp}} U}} ,v) - f_{i} (U^{D} ,v) - [f_{j} (1^{{{\text{Supp}} U}} ,v) - f_{j} (U^{D} ,v)] $$

$$ = Sh_{i} (1^{{{\text{Supp}} U}} ,v) - Sh_{i} (U^{D} ,v) - [Sh_{j} (1^{{{\text{Supp}} U}} ,v) - Sh_{j} (U^{D} ,v)] $$

$$ = Sh_{i}^{D} (U,v^{D} ) - Sh_{j}^{D} (U,v^{D} ) $$

$$ = v^{D} (U(i)) - v^{D} (U(j)) $$

(a.5)

We assume that $f^{D} = Sh^{D}$ for v^D with $\left| {{\text{Supp}} U} \right| - 1$ players. Now, we show that $f^{D} = Sh^{D}$ for v^D with $\left| {{\text{Supp}} U} \right|$ players. For each fuzzy coalition $R \subseteq U$ with $\left| {{\text{Supp}} R} \right| = 2$, by $v_{{f^{D} ,R}}^{D}$, we denote the dual fuzzy reduced game on ${\text{Supp}} R = \{ i,j\}$ for $f^{D}$, and by $v_{{Sh^{D} ,R}}^{D}$, we express the reduced game for $Sh^{D}$. Since ${\text{Supp}} R = \{ i,j\}$,${\text{Supp}} Q = \{ i\}$, and ${\text{Supp}} (R^{c} \cup Q) = {\text{Supp}} U\backslash j$, $v_{{f^{D} ,R}}^{D} (U(i))$ and $v_{{Sh^{D} ,R}}^{D} (U(i))$ depend on $f^{D}$ and $Sh^{D}$ through dual fuzzy cooperative game v^D with $\left| {{\text{Supp}} U} \right| - 1$ players. By the induction assumption, we get

$$ v_{{f^{D} ,R}}^{D} (U(i)) = v_{{Sh^{D} ,R}}^{D} (U(i)) $$

(a.6)

and

$$ v_{{f^{D} ,R}}^{D} (U(j)) = v_{{Sh^{D} ,R}}^{D} (U(j)) $$

(a.7)

By DFC, we have

$$ f_{i}^{D} (R,v_{{f^{D} ,R}}^{D} ) - f_{j}^{D} (R,v_{{f^{D} ,R}}^{D} ) = f_{i}^{D} (U,v^{D} ) - f_{j}^{D} (U,v^{D} ) $$

(a.8)

and

$$ Sh_{i}^{D} (R,v_{{Sh^{D} ,R}}^{D} ) - Sh_{j}^{D} (R,v_{{Sh^{D} ,R}}^{D} ) = Sh_{i}^{D} (U,v^{D} ) - Sh_{j}^{D} (U,v^{D} ) $$

(a.9)

According to Eq. (a.5), the left sides of Eqs. (a.8) and (a.9) are equal to $v_{{f^{D} ,R}}^{D} (U(i)) - v_{{f^{D} ,R}}^{D} (U(j))$ and $v_{{Sh^{D} ,R}}^{D} (U(i)) - v_{{Sh^{D} ,R}}^{D} (U(j))$, respectively.

Equations (a.6)-(a.9) show that $f_{i}^{D} (U,v^{D} ) - f_{j}^{D} (U,v^{D} ) = Sh_{i}^{D} (U,v^{D} ) - Sh_{j}^{D} (U,v^{D} )$. As $\sum\limits_{{i \in {\text{Supp}} U}} {f_{i}^{D} (U,v^{D} )} = \sum\limits_{{i \in {\text{Supp}} U}} {Sh_{i}^{D} (U,v^{D} )}$, we derive $f_{i}^{D} (U,v^{D} ) = Sh_{i}^{D} (U,v^{D} )$ for any $i \in {\text{Supp}} U$. □

Appendix D

Proof of Theorem 7

By Eq. (10), we have

$$ \begin{aligned} v^{D,C} (S) = & v(1^{{{\text{Supp}} S}} ) - h_{1} (S^{D} )v(1^{{{\text{Supp}} S}} ) - \sum\limits_{l = 2}^{{q(S_{d} )}} {v(1^{{{\text{Supp}} S_{{h_{l} (S^{D} )}}^{D} }} )\left( {h_{l} (S^{D} ) - h_{l - 1} (S^{D} )} \right)} \\ & = \left( {1 - h_{1} (S^{D} )} \right)v(1^{{{\text{Supp}} S}} ) - \sum\limits_{l = 2}^{{q(S_{d} )}} {v(1^{{{\text{Supp}} S_{{h_{l} (S^{D} )}}^{D} }} )\left( {h_{l} (S^{D} ) - h_{l - 1} (S^{D} )} \right)} \\ & = h_{q(S)} (S)v(1^{{{\text{Supp}} S}} ) - \sum\limits_{l = 2}^{{q(S^{D} )}} {v(1^{{{\text{Supp}} S_{{h_{l} (S^{D} )}}^{D} }} )\left( {h_{l} (S^{D} ) - h_{l - 1} (S^{D} )} \right)} \\ & = \left( {1 - h_{{q(S^{D} )}} (S^{D} )} \right)v(1^{{{\text{Supp}} S}} ) + \left( {h_{q(S)} (S) - \left( {1 - h_{{q(S^{D} )}} (S^{D} )} \right)} \right)v(1^{{{\text{Supp}} S}} ) - \sum\limits_{l = 2}^{{q(S^{D} )}} {v(1^{{{\text{Supp}} S_{{h_{l} (S^{D} )}}^{D} }} )\left( {h_{l} (S^{D} ) - h_{l - 1} (S^{D} )} \right)} \\ & = \left( {1 - h_{{q(S^{D} )}} (S^{D} )} \right)v(1^{{{\text{Supp}} S}} ) + \left( {h_{q(S)} (S) - \left( {1 - \left( {1 - h_{1} (S)} \right)} \right)} \right)v(1^{{{\text{Supp}} S}} ) - \sum\limits_{l = 2}^{{q(S^{D} )}} {v(1^{{{\text{Supp}} S_{{h_{l} (S^{D} )}}^{D} }} )\left( {h_{l} (S^{D} ) - h_{l - 1} (S^{D} )} \right)} \\ & = \left( {1 - h_{{q(S^{D} )}} (S^{D} )} \right)v(1^{{{\text{Supp}} S}} ) + \left( {h_{q(S)} (S) - h_{1} (S)} \right)v(1^{{{\text{Supp}} S}} ) - \sum\limits_{l = 2}^{{q(S^{D} )}} {v(1^{{{\text{Supp}} S_{{h_{l} (S^{D} )}}^{D} }} )\left( {h_{l} (S^{D} ) - h_{l - 1} (S^{D} )} \right)} \\ & = h_{1} (S)v(1^{{{\text{Supp}} S}} ) + \sum\limits_{l = 2}^{q(S)} {\left( {v(1^{{{\text{Supp}} S}} ) - v(1^{{{\text{Supp}} S\backslash {\text{Supp}} S_{{h_{l} (S)}} }} )} \right)\left( {h_{l} (S) - h_{l - 1} (S)} \right)} \\ \end{aligned} $$

(a.10)

On the other hand,

$$ v^{C} (S) = \sum\limits_{l = 1}^{q(S)} {v(1^{{{\text{Supp}} S_{{h_{l} (S)}} }} )\left( {h_{l} (S) - h_{l - 1} (S)} \right)} = h_{1} (S)v(1^{{{\text{Supp}} S}} ) + \sum\limits_{l = 2}^{q(S)} {v(1^{{{\text{Supp}} S_{{h_{l} (S)}} }} )\left( {h_{l} (S) - h_{l - 1} (S)} \right)} $$

(a.11)

Since v is supper-additive, we have $v(1^{{{\text{Supp}} S}} ) - v(1^{{{\text{Supp}} S\backslash {\text{Supp}} S_{{h_{l} (S)}} }} ) \ge v(1^{{{\text{Supp}} S_{{h_{l} (S)}} }} )$ for any $l = 1,2, \ldots ,q(S)$. Thus, $v^{D,C} (S) \ge v^{C} (S)$ for any $S \subseteq U$.□

Appendix E

Proof of Theorem 9

Existence. For DFE, we derive.

$$ \begin{aligned} \sum\limits_{{i \in {\text{Supp}} U}} {Sh_{i}^{D,C} (U,v^{D,C} )} = & \sum\limits_{{i \in {\text{Supp}} U}} {Sh_{i} (1^{{{\text{Supp}} U}} ,v)} - \sum\limits_{{i \in {\text{Supp}} U}} {\sum\limits_{l = 1}^{{q(U^{D} )}} {Sh_{i} (1^{{{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} }} ,v)\left( {h_{l} (U^{D} ) - h_{l - 1} (U^{D} )} \right)} } \\ & = v(1^{{{\text{Supp}} U}} ) - \sum\limits_{l = 1}^{{q(U^{D} )}} {v(1^{{{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} }} )\left( {h_{l} (U^{D} ) - h_{l - 1} (U^{D} )} \right)} \\ & = v^{D,C} (U). \\ \end{aligned} $$

For DFBC: we need to show that

$$ \begin{gathered} \left[ {Sh_{i} (1^{{{\text{Supp}} U}} ,v) - \sum\limits_{l = 1}^{{q(U^{D} )}} {Sh_{i} (1^{{{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} }} ,v)\left( {h_{l} (U^{D} ) - h_{l - 1} (U^{D} )} \right)} } \right] - \left[ {Sh_{i} (1^{{{\text{Supp}} U\backslash j}} ,v)} \right. \hfill \\ \left. {\quad - \sum\limits_{l = 1}^{{q(U^{D} )}} {Sh_{i} (1^{{{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} \backslash j}} ,v)\left( {h_{l} (U^{D} \backslash U^{D} (j)) - h_{l - 1} (U^{D} \backslash U^{D} (j))} \right)} } \right] \hfill \\ \quad = \left[ {Sh_{j} (1^{{{\text{Supp}} U}} ,v) - \sum\limits_{l = 1}^{{q(U^{D} )}} {Sh_{j} (1^{{{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} }} ,v)\left( {h_{l} (U^{D} ) - h_{l - 1} (U^{D} )} \right)} } \right] - [Sh_{j} (1^{{{\text{Supp}} U\backslash i}} ,v) \hfill \\ \quad \left. { - \sum\limits_{l = 1}^{{q(U_{d} )}} {Sh_{j} \left( {1^{{{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} \backslash i}} ,v} \right)\left( {h_{l} (U^{D} \backslash U^{D} (i)) - h_{l - 1} (U^{D} \backslash U^{D} (i))} \right)} } \right] \hfill \\ \quad - \left[ {Sh_{i} (1^{{{\text{Supp}} U\backslash j}} ,v)} \right. \hfill \\ \end{gathered} $$

for any $S \subseteq U$ and all $i,j \in {\text{Supp}} S$ with $i \ne j$.

By Theorem A in [6], we have

$$ Sh_{i} (1^{{{\text{Supp}} U}} ,v) - Sh_{i} (1^{{{\text{Supp}} U\backslash j}} ,v) = Sh_{j} (1^{{{\text{Supp}} U}} ,v) - Sh_{j} (1^{{{\text{Supp}} U\backslash i}} ,v) $$

for any ${\text{Supp}} S \subseteq {\text{Supp}} U$ and all $i,j \in {\text{Supp}} S$ with $i \ne j$.

Therefore, we need to prove that

$$ \begin{gathered} \sum\limits_{l = 1}^{{q(U^{D} )}} {Sh_{i} (1^{{{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} }} ,v)\left( {h_{l} (U^{D} ) - h_{l - 1} (U^{D} )} \right)} - \sum\limits_{l = 1}^{{q(U^{D} )}} {Sh_{i} (1^{{{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} \backslash j}} ,v)\left( {h_{l} (U^{D} \backslash U^{D} (j)) - h_{l - 1} (U^{D} \backslash U^{D} (j))} \right)} \hfill \\ = \sum\limits_{l = 1}^{{q(U^{D} )}} {Sh_{j} (1^{{{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} }} ,v)\left( {h_{l} (U^{D} ) - h_{l - 1} (U^{D} )} \right)} - \sum\limits_{l = 1}^{{q(U^{D} )}} {Sh_{j} (1^{{{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} }} \backslash i,v)\left( {h_{l} (U^{D} \backslash U^{D} (i)) - h_{l - 1} (U^{D} \backslash U^{D} (i))} \right)} \hfill \\ \end{gathered} $$

for any $S \subseteq U$ and all $i,j \in {\text{Supp}} S$ with $i \ne j$.

From Eq. (2), we know that

$$ \begin{gathered} \sum\limits_{l = 1}^{{q(U^{D} )}} {Sh_{i} (1^{{{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} }} ,v)\left( {h_{l} (U^{D} ) - h_{l - 1} (U^{D} )} \right)} - \sum\limits_{l = 1}^{{q(U^{D} )}} {Sh_{i} (1^{{{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} \backslash j}} ,v)\left( {h_{l} (U^{D} \backslash U^{D} (j)) - h_{l - 1} (U^{D} \backslash U^{D} (j))} \right)} \hfill \\ \quad = \sum\limits_{l = 1}^{{q(U^{D} )}} {\sum\limits_{{i \in {\text{Supp}} R \subseteq {\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} }} {\frac{{(\left| {{\text{Supp}} R} \right| - 1)!(\left| {{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} } \right| - \left| {{\text{Supp}} R} \right|)!}}{{\left| {{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} } \right|!}}} (v(1^{{{\text{Supp}} R}} ) - v(1^{{{\text{Supp}} R\backslash i}} ))} \hfill \\ \quad - \sum\limits_{l = 1}^{{q(U^{D} )}} {\sum\limits_{{i \in {\text{Supp}} R \subseteq {\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} \backslash j}} {\frac{{(\left| {{\text{Supp}} R} \right| - 1)!(\left| {{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} \backslash j} \right| - \left| {{\text{Supp}} R} \right|)!}}{{\left| {{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} \backslash j} \right|!}}} (v(1^{{{\text{Supp}} R}} ) - v(1^{{{\text{Supp}} R\backslash i}} ))} \hfill \\ \quad = \sum\limits_{l = 1}^{{q(U^{D} )}} {\sum\limits_{{i \in {\text{Supp}} R \subseteq {\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} }} {\frac{{(\left| {{\text{Supp}} R} \right| - 1)!(\left| {{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} } \right| - \left| {{\text{Supp}} R} \right|)!}}{{\left| {{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} } \right|!}}} (v(1^{{{\text{Supp}} R}} ) - v(1^{{{\text{Supp}} R\backslash i}} ))} \hfill \\ \quad - \sum\limits_{l = 1}^{{q(U^{D} )}} {\sum\limits_{{i \in {\text{Supp}} R \subseteq {\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} }} {\frac{{(\left| {{\text{Supp}} R} \right| - 1)!(\left| {{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} } \right| - \left| {{\text{Supp}} R} \right|)!}}{{\left| {{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} } \right|!}}} (v(1^{{{\text{Supp}} R\backslash j}} ) - v(1^{{{\text{Supp}} R\backslash \{ i,j\} }} ))} \hfill \\ \quad = \sum\limits_{l = 1}^{{q(U^{D} )}} {\sum\limits_{{i \in {\text{Supp}} R \subseteq {\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} }} {\frac{{(\left| {{\text{Supp}} R} \right| - 1)!(\left| {{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} } \right| - \left| {{\text{Supp}} R} \right|)!}}{{\left| {{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} } \right|!}}} \left[ {v(1^{{{\text{Supp}} R}} ) - v(1^{{{\text{Supp}} R\backslash i}} )} \right.} \hfill \\ \quad - v(1^{{{\text{Supp}} R\backslash j}} ) + v(1^{{{\text{Supp}} R\backslash \{ i,j\} }} )] \hfill \\ \end{gathered} $$

As the above expression is symmetric for the players i and j, we can also derive the above conclusion for

$$ \sum\limits_{l = 1}^{{q(U^{D} )}} {Sh_{j} (1^{{{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} }} ,v)(h_{l} (U^{D} ) - } h_{l - 1} (U^{D} )) - \sum\limits_{l = 1}^{{q(U^{D} )}} {Sh_{j} (1^{{{\text{Supp}} U_{{h_{l} (U^{D} )}}^{D} \backslash i}} ,v)(h_{l} (U^{D} \backslash U^{D} (i)) - } h_{l - 1} (U^{D} \backslash U^{D} (i))) $$

Uniqueness. Let $v^{D,C} \in DG_{C} (N)$ and $U \in L(N)$. Assume that $f^{D,C}$ is another solution that satisfies DFE and DFBC. If $\left| {{\text{Supp}} U} \right|{ = }1$, by DFE, we have $f_{i}^{D,C} (U,v^{D,C} ) = U(i)v(1^{{{\text{Supp}} U(i)}} ) = Sh_{i}^{D,C} (U,v^{D,C} )$. Assume that $f^{D,C} = Sh^{D,C}$ with no more than $\left| {{\text{Supp}} U} \right| = n - 1$ players, where $n \ge 2$. Next, we prove $f^{D,C} = Sh^{D,C}$, where $\left| {{\text{Supp}} U} \right| = n$. By DFBC, we obtain

$$ f_{i}^{D,C} (U,v^{D,C} ) - f_{i}^{D,C} (U\backslash U(j),v^{D,C} ) = f_{j}^{D,C} (U,v^{D,C} ) - f_{j}^{D,C} (U\backslash U(i),v^{D,C} ) $$

for all $i,j \in {\text{Supp}} U$ with $i \ne j$.

By induction, we get

$$ f_{i}^{D,C} (U,v^{D,C} ) - f_{j}^{D,C} (U,v^{D,C} ) = Sh_{i}^{D,C} (U\backslash U(j),v^{D,C} ) - Sh_{j}^{D,C} (U\backslash U(i),v^{D,C} ) $$

Because $Sh^{D,C}$ satisfies DFBC, we have

$$ f_{i}^{D,C} (U,v^{D,C} ) - f_{j}^{D,C} (U,v^{D,C} ) = Sh_{i}^{D,C} (U,v^{D,C} ) - Sh_{j}^{D,C} (U,v^{D,C} ) $$

Thus,

$$ f_{i}^{D,C} (U,v^{D,C} ) - Sh_{i}^{D,C} (U,v^{D,C} ) = f_{j}^{D,C} (U,v^{D,C} ) - Sh_{j}^{D,C} (U,v^{D,C} ) $$

Sum $j \in {\text{Supp}} U$ for the above equation, we derive

$$ \begin{gathered} \left| {{\text{Supp}} U} \right|[f_{i}^{D,C} (U,v^{D,C} ) - Sh_{i}^{D,C} (U,v^{D,C} )] \hfill \\ \quad = \sum\limits_{{j \in {\text{Supp}} U}} {f_{j}^{D,C} (U,v^{D,C} )} - \sum\limits_{{j \in {\text{Supp}} U}} {Sh_{j}^{D,C} (U,v^{D,C} )} \hfill \\ \quad = v^{D,C} (U) - v^{D,C} (U) = 0 \hfill \\ \end{gathered} $$

Therefore, $f_{i}^{D,C} (U,v^{D,C} ) = Sh_{i}^{D,C} (U,v^{D,C} )$ for any $i \in {\text{Supp}} U$. □