### Proof

To ease on notation, we will write \(x^*\) instead of \(x^*(v)\) throughout this proof.

[

**Sufficiency**.] Let

*x* be the nucleolus of

*v*. We know from Theorem

1 that

*x* satisfies

\(x_i \le x_j\wedge \frac{M^v(i)}{2}\), for any weak player

\(i=i_1,\ldots ,i_m\) and veto player

\(j=i_{m+1},\ldots , i_n\). Assuming that (

3) holds, that is

$$\begin{aligned} \varepsilon (S,x^*)\le -\max _{i=i_1,\ldots ,i_m} x_i^*,\quad \forall S \in N^*(v), \end{aligned}$$

we will now show that we must, in fact, have

\(x=x^*\), where

\(x^*\) is defined in (

2).

By way of contradiction, suppose that

\(x\ne x^*\). Then, we must have the following:

^{6}$$\begin{aligned} x_{i}< t^v \wedge \frac{M^v(i)}{2} = x^*_{i},\quad \text{ for } \text{ some } i=i_{1},\ldots ,i_m. \end{aligned}$$

(7)

Let then

\(k\in \{1,\ldots ,m\}\) be the

*lowest ranked* weak player satisfying (

7), that is

$$\begin{aligned} x_{i_k} <x^*_{i_k}\quad \text{ and }\quad \left[ l\in \{1,\ldots ,k-1\}\Rightarrow x_{i_l} \ge x^*_{i_l} \right] . \end{aligned}$$

(8)

We show that the conditions (i)–(ii) of Lemma

1 are satisfied for

\(\bar{S}=\{i_k\}\) and

\(x'=x^*\)—that is,

\(\theta (x^*)\) is lexicographically inferior to

\(\theta (x)\). Indeed, note from (

8) that we have

\(\varepsilon (i_k,x^*)=-x^*_{i_k}<-x_{i_k}=\varepsilon (i_k,x)\) and condition (i) is satisfied.

For condition (ii), suppose now that \(\varepsilon (S,x^*)>\varepsilon (S,x)\) and (by induction) \(\varepsilon (S',x^*)\le \varepsilon (S',x)\) for any \(S'\) s.t. \(\varepsilon (S',x^*)>\varepsilon (S,x^*)\). Let us first consider coalitions of the form \(S=\{j\}\), for some \(j\in N{\setminus } i_k\). In addition, suppose that \(\bar{T}(v)\ne \{j\}\), so that \(v(S)=0\). Since \(\varepsilon (S,x^*)=-x^*_j>-x_j=\varepsilon (S,x)\), it follows that \(x^*_j<x_j\). Let us now consider two cases. Case 1: Suppose that \(j\in \bar{T}(N,v)\). Then, we know from (3) that \(\varepsilon (S,x^*)=-x^*_j=-t^v\le -\left( t^v\wedge \frac{M^v(i_k)}{2} \right) =-x^*_{i_k}=\varepsilon (\bar{S},x^*)\); hence, condition (ii) is met. Case 2: Suppose now that \(j\notin \bar{T}(v)\), that is, \(j\in W(v)\). Then, we know from Theorem 1 that \(x_j \le \frac{M^v(i)}{2}\), that is to say, \(x^*_j=t^v\wedge \left( {\begin{array}{c}M^v(j)\\ 2\end{array}}\right) < x_j \le \frac{M^v(j)}{2}\), which implies that \(x^*_j=t^v\). One may thus write: \(\varepsilon (j,x^*)=-t^v\le -\left( t^v\wedge \frac{M^v(i_k)}{2}\right) =-x^*_{i_k}=\varepsilon (\bar{S},x^*) \). In this case too, condition (ii) is satisfied.

Consider, now, a coalition

*S*, such that

\(|S|\ge 2\) and

\(S\cap \bar{T}(v)\ne \bar{T}(v)\) (so that

\(v(S)=0\)). Then,

\(\varepsilon (S,x^*)>\varepsilon (S,x)\) means that

\(-x^*_{S}>-x_{S}\), that is,

\(x^*_{S}<x_{S}\). Therefore, there must exist

\(j\in S\) s.t.

\(x^*_j<x_j\). Thus, recalling from the preceding paragraph that

\(x^*_j=t^v\), one gets

$$\begin{aligned} \varepsilon (S,x^*)= & {} -x^*_j-x^*_{S{\setminus } j}=-t^v-x^*_{S{\setminus } j}\le -\underbrace{\left( t^v\wedge \frac{M^v(i_k)}{2}\right) }_{=x^*_{i_k}} \\&-\underbrace{x^*_{S{\setminus } j}}_{\ge 0}\le -x_{i_k}= \varepsilon (i_k,x^*). \end{aligned}$$

Hence, condition (ii) is satisfied for any

*S* s.t.

\(|S|\ge 2\) and

\(S\cap \bar{T}(v)\ne \bar{T}(v)\).

There only remains the case of *S* s.t. \(\bar{T}(v)=\{i_{m+1},\ldots , i_n\}\subseteq S\). We also discuss two cases here. Case 1: If \(S\in N^*(v)\), then remark that condition (ii) easily follows from (3)). Case 2: If instead \(S\notin N^*(v)\), then we have \(\varepsilon (S,x^*)=\varepsilon (N\backslash S,x^*)\) and we get from that \(\varepsilon (N\backslash S,x^*)<\varepsilon (N\backslash S,x)\). Given our assumption that \(\varepsilon (S',x^*)\le \varepsilon (S',x)\) for any \(S'\) s.t. \(\varepsilon (S',x^*)>\varepsilon (S,x^*)\), it comes that the nucleolus \(\theta (x)\) is lexicographically inferior to \(\theta (x)^*\), which contradicts the definition of the nucleolus *x* and, hence, shows that Case 2 will never occur.

In summary, we have shown that assuming \(x\ne x^*\) leads to a contradiction. Therefore, we must have \(x= x^*\) under the condition (3).

[

**Necessity**.] Suppose that (

3) is not satisfied. Then, there exist

\(i\in \{i_1,\ldots ,i_m\}\) and

\(S\in N^*(v)\) s.t.

\(\varepsilon (S,x^*)>-x^*_i\). Let then

\(\alpha \equiv \varepsilon (S,x^*)+x^*_i>0\); define

\(x'\equiv x^*-\frac{\alpha }{2} \mathrm{e}^i+\frac{\alpha }{2|\bar{T}(v)|}\mathrm{e}^{\bar{T}(v)}\). It is easy to see that the requirement (i) of Lemma

1 is satisfied for

\(S=\bar{T}(v)\), since

\(\varepsilon (\bar{T}(v),x')=\varepsilon (\bar{T}(v),x^*)-\frac{\alpha }{2}<\varepsilon (\bar{T}(v),x^*)\). Moreover, the only coalitions

*S*, such that

\(\varepsilon (S,x')>\varepsilon (S,x^*)\) are those satisfying [

\(i\in S\) and

\(|S\cap \bar{T}(v)|< n-m\)]. However, for any such

*S*, we have

\(v(S)=0\); one may thus write

$$\begin{aligned} \varepsilon (S,x^*)= -x^*_S=\underbrace{-x^*_{i}}_{<\varepsilon (S,x^*)}-\underbrace{x^*_{S{\setminus } i}}_{\ge 0} <\varepsilon (S,x^*). \end{aligned}$$

Hence, the requirement (ii) is also met; it follows from Lemma

1 that

\(x^*\) is not the nucleolus of

*v*.

\(\square \)