Core of a simple game is convex hull of the veto players 
For $i,j \in N$ let the indicator vector $e^{\{i\}} \in \mathbb{R}^n$ be such that $e^{\{i\}}_j = 1$ if $i=j$ and $0$ if $i \neq j$. Let $\operatorname{veto}(v)$ be the set of veto players, defined by $\operatorname{veto}(v) = \bigcap \{S: \> v(S) = 1 \}$.
Let $v \in TU^N$ be a simple game and prove that $$C(v) = \text{Conv}(\{e^{\{i\}} | i \in \operatorname{veto}(v)\}) $$ where $C(v)$ is the core of $v$ and $\text{Conv}$ denotes a convex hull.

I already know that $C(v) \neq \emptyset \iff \operatorname{veto}(v) \neq 0.$ However I do not know how to prove this. Intuitively it is clear, as I've done examples by hand with 3 players, wherein I solved a system of inequalities to show the core is empty if there are no veto players. But I don't know how to go about extending this result to all $N$.
 A: To show equality, prove 
$$(i1) \quad C(v) \supseteq \operatorname{Conv}(\{e^{\{i\}} | i \in \operatorname{veto}(v)\})$$
and
$$(i) \quad C(v) \subseteq \operatorname{Conv}(\{e^{\{i\}} | i \in \operatorname{veto}(v)\}) $$
(i) Assume $i \in \operatorname{veto}(v)$. If $i \in S$, then $e^i(S) = 1 \ge v(S)$; if $i \not\in S$, then $e^i(S) = 0 = v(S)$. Then $e^i \in C(v)$. The $\supseteq$ inclusion follows because $C(v)$ is a convex set.
(ii) Assume $\mathbf{x}\in C(v)$ and show that $x_i = 0$ for any $i \not\in \operatorname{veto}(v)$. Suppose to the contrary that $x_i > 0$. Since $i$ is not a veto player, there exists a coalition $S$ with $v(S)=1$ and $i \not\in S$. Then we would have $x(S) = x(N) - x(S^c) = 1 - x_i < 1 = v(S)$, contradicting the assumption that $\mathbf{x}\in C(v)$.
See Theorem 16.12 in Peters (2015), Game Theory, second edition.
A: I checked the proof of Them 16.12(ii) in the Peters' Game Theory book, which is wrong. The proof runs through contraposition and not by contradiction. 
A proof by contraposition establishes $(\neg B \Rightarrow \neg A)$ which is equivalent to $(A \Rightarrow B)$. We need to establish that whenever $\mathbf{x} \not\in conv(\{\mathbf{e}^{i} \in \mathbb{R}^{n}\,\big\arrowvert\,i \in veto(v) \})$, then $\mathbf{x} \not\in C(v)$, which is equivalent to $\mathbf{x} \in C(v)$, then $\mathbf{x} \in conv(\{\mathbf{e}^{i} \in \mathbb{R}^{n}\,\big\arrowvert\,i \in veto(v) \})$ Hence, there is no need to assume that $\mathbf{x}$ is in the core. 
This means, we suppose that $i \not\in veto(v)$ with $x_{i} > 0$ for some non-veto player $i$ is satisfied, i.e., $\neg B$ holds. Take $S$ with $v(S) = 1$ and $i \not\in S$, then $x(S)=x(N)-x(N\backslash S) \le 1 - x_{i} < 1 = v(S)$. Thus, it holds $\mathbf{x} \not\in C(v)$. Therefore, $\neg A$ holds. This concludes the inclusion $C(v) \subseteq conv(\{\mathbf{e}^{i} \in \mathbb{R}^{n}\,\big\arrowvert\,i \in veto(v) \})$. 
