Facets of the cone of exact games

Abstract

The class of exact transferable utility coalitional games, introduced in 1972 by Schmeidler, has been studied both in the context of game theory and in the context of imprecise probabilities. We characterize the cone of exact games by describing the minimal set of linear inequalities defining this cone; these facet-defining inequalities for the exact cone appear to correspond to certain set systems (= systems of coalitions). We noticed that non-empty proper coalitions having non-zero coefficients in these facet-defining inequalities form set systems with particular properties. More specifically, we introduce the concept of a semi-balanced system of coalitions, which generalizes the classic concept of a balanced coalitional system in cooperative game theory. The semi-balanced coalitional systems provide valid inequalities for the exact cone and minimal semi-balanced systems (in the sense of inclusion of set systems) characterize this cone. We also introduce basic classification of minimal semi-balanced systems, their pictorial representatives and a substantial concept of an indecomposable (minimal) semi-balanced system of coalitions. The main result of the paper is that indecomposable semi-balanced systems are in one-to-one correspondence with facet-defining inequalities for the exact cone. The second relevant result is the rebuttal of a former conjecture claiming that a coalitional game is exact iff it is totally balanced and its anti-dual is also totally balanced. We additionally characterize those inequalities which are facet-defining both for the cone of exact games and for the cone of totally balanced games.

Appendices

Proof of Lemma 1

For reader’s convenience we recall the result.

Lemma 1    Given $|N|\ge 2$ and $\emptyset \ne {{\mathcal {S}}}\subseteq {\mathcal{P}(N)}{\setminus } \{\emptyset ,N\}$, let $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\cdot \chi _{S}=\rho$ be a non-zero semi-conic combination yielding a constant vector $\rho =[r,\ldots ,r]\in {\mathbb R}^{N}$. Then one has $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\ge r\ge 0$; moreover, $r>0$ in case of a conic combination.

In any case $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}>0$ and by a positive factor multiplication one gets an affine semi-conic combination $\sum _{S\in {{\mathcal {S}}}} {\tilde{\uplambda }}_{S}\cdot \chi _{S}$ yielding a constant vector ${\tilde{\rho }}=[{\tilde{r}},\ldots ,{\tilde{r}}]\in {\mathbb R}^{N}$ with ${\tilde{r}}\in [0,1]$.

Finally, if the considered linear combination is not conic, then ${{\mathcal {S}}}$ has to contain at least three different sets and the existence of such a set system forces $|N|\ge 3$.

Proof

The case of a conic combination is easy: choose $L\in {{\mathcal {S}}}$ with $\uplambda _{L}>0$, $i\in L$, and write $0<\uplambda _{L}\le \sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\cdot \chi _{S}(i)=r=\sum _{S\in {{\mathcal {S}}}:\,i\in S} \uplambda _{S}\le \sum _{S\in {{\mathcal {S}}}} \uplambda _{S}$. Thus, in the rest of the proof we assume that the combination is not conic.

Let $T\in {{\mathcal {S}}}$ be the set with $\uplambda _{T}<0$; note that $\emptyset \subset T\subset N$. The choice of an element $j\in N\setminus T$ gives $r=\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\cdot \chi _{S}(j)= \sum _{S\in {{\mathcal {S}}}:\,j\in S} \uplambda _{S}\ge 0$ because $\uplambda _{S}\ge 0$ whenever $S\in {{\mathcal {S}}}{\setminus }\{T\}$. Choose some $i\in T$ and observe implications for subsets of N: $i\not \in S\in {{\mathcal {S}}}~\Rightarrow ~ S\in {{\mathcal {S}}}{\setminus }\{T\}~\Rightarrow ~ \uplambda _{S}\ge 0$. This gives $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\ge \sum _{S\in {{\mathcal {S}}}:\,i\in S} \uplambda _{S}=\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\cdot \chi _{S}(i)=r$. For the verification of the claim $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}>0$ assume without loss of generality that $\uplambda _{S}\ne 0$ for any $S\in {{\mathcal {S}}}$, for otherwise one can replace ${{\mathcal {S}}}$ by ${{\mathcal {S}}}^{\prime }:=\{S\in {{\mathcal {S}}}\,:\ \uplambda _{S}\ne 0\}$. We distinguish two cases:

• In case [ $\exists \, L\in {{\mathcal {S}}}{\setminus }\{T\}$ with $T{\setminus } L\ne \emptyset$ ] we choose $k\in T{\setminus } L$. Then $\uplambda _{S}\ge 0$ whenever $k\not \in S\in {{\mathcal {S}}}$ and we get $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\ge \uplambda _{L}+ \sum _{S\in {{\mathcal {S}}}:\,k\in S} \uplambda _{S} > \sum _{S\in {{\mathcal {S}}}:\,k\in S} \uplambda _{S}= r\ge 0$.

• In case [ $\forall \, L\in {{\mathcal {S}}}{\setminus }\{T\}$ one has $T\subseteq L$ ] we first observe $\bigcup {{\mathcal {S}}}{\setminus } T\ne \emptyset$. Indeed, by contradiction: if $\bigcup {{\mathcal {S}}}\subseteq T$ then $\forall \, L\in {{\mathcal {S}}}{\setminus } \{T\}$ one has $L\subseteq T\subseteq L$, which means ${{\mathcal {S}}}=\{T\}$ contradicting $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\ge 0$. Thus, one can choose $k\in \bigcup {{\mathcal {S}}}{\setminus } T$, fix $K\in {{\mathcal {S}}}$ with $k\in K$ and write $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\ge r=\sum _{S\in {{\mathcal {S}}}:\,k\in S} \uplambda _{S}\ge \uplambda _{K}>0$.

Thus, in both cases we have $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}> 0$.

In particular, given a non-zero semi-conic combination $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\cdot \chi _{S}$ yielding a constant vector in ${\mathbb R}^{N}$ we put $\ell :=\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}>0$ and observe that $\sum _{S\in {{\mathcal {S}}}} (\ell ^{-1}\cdot \uplambda _{S})\cdot \chi _{S}$ is the required affine semi-conic combination. The inequality $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\ge r$ applied to an affine combination $\sum _{S\in {{\mathcal {S}}}} {\tilde{\uplambda }}_{S}\cdot \chi _{S}$ with $\sum _{S\in {{\mathcal {S}}}} {\tilde{\uplambda }}_{S}=1$ yields ${\tilde{r}}\in [0,1]$.

To verify the last claim take $T\in {{\mathcal {S}}}$ with $\uplambda _{T}<0$ and assume for a contradiction $|{{\mathcal {S}}}|\le 2$. If $|{{\mathcal {S}}}|=1$ take $j\in T$ and get a contradictory conclusion: $0\le r=\sum _{S\in {{\mathcal {S}}}:\,j\in S} \uplambda _{S}=\uplambda _{T}<0$. In case $|{{\mathcal {S}}}|=2$ one has ${{\mathcal {S}}}=\{T,L\}$ and contingent choice of $j\in T{\setminus } L$ leads to an analogous contradiction. Hence, one must have $T\subset L$ and the choice of $j\in T$ and $k\in L{\setminus } T$ leads to another contradiction: $r=\sum _{S\in {{\mathcal {S}}}:\,k\in S} \uplambda _{S}=\uplambda _{L}>\uplambda _{L}+\uplambda _{T}= \sum _{S\in {{\mathcal {S}}}:\,j\in S} \uplambda _{S}=r$. Thus, $|{{\mathcal {S}}}|\ge 3$; since $|N|=2 ~\Rightarrow ~ |{\mathcal{P}(N)}{\setminus } \{\emptyset ,N\}|=2$, the existence of ${{\mathcal {S}}}$ forces $|N|\ge 3$. $\square$

Proof of Lemma 2

For reader’s convenience we recall the result.

Lemma 2    Given $|N|\ge 2$, let $\emptyset \ne {{\mathcal {S}}}\subseteq {\mathcal{P}(N)}{\setminus } \{\emptyset ,N\}$ be a non-trivial set system on N. Then the following conditions on ${{\mathcal {S}}}$ are equivalent:

(a):

${{\mathcal {S}}}$ is a minimal set system such that there is a constant vector in ${\mathbb R}^{N}$ which can be written as a non-zero semi-conic combination of vectors $\{ \chi _{S}\,:\ S\in {{\mathcal {S}}}\}$,

(b):

${{\mathcal {S}}}$ is a minimal semi-balanced set system on N,

(c):

${{\mathcal {S}}}$ is semi-balanced on N, the vectors $\{ \chi _{S}\,:\ S\in {{\mathcal {S}}}\}$ are affinely independent and in case $\bigcup {{\mathcal {S}}}=N$ even linearly independent,

(d):

there is only one affine combination of vectors $\{ \chi _{S}\,:\ S\in {{\mathcal {S}}}\}$ yielding a constant vector in ${\mathbb R}^{N}$ and this unique combination is semi-conic and has all coefficients non-zero,

(e):

there is only one affine semi-conic combination of vectors $\{ \chi _{S}\,:\ S\in {{\mathcal {S}}}\}$ which is a constant vector in ${\mathbb R}^{N}$ and this unique combination has all coefficients non-zero.

Proof

To show (a) $\Rightarrow$(b) assume a non-zero semi-conic combination $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\cdot \chi _{S}$ which is a constant vector in ${\mathbb R}^{N}$ and put ${{\mathcal {S}}}^{\prime }:=\{ S\in {{\mathcal {S}}}\, :\ \uplambda _{S}\ne 0\}$. Because of minimality of ${{\mathcal {S}}}$ in (a) one has ${{\mathcal {S}}}^{\prime }={{\mathcal {S}}}$, which implies that ${{\mathcal {S}}}$ is semi-balanced. The rest is evident.

To show (b) $\Rightarrow$(a) it is enough to verify the minimality of ${{\mathcal {S}}}$ in (a). Assume for a contradiction that a set system ${{\mathcal {C}}}\subset {{\mathcal {S}}}$ exists with a non-zero semi-conic combination $\sum _{S\in {{\mathcal {C}}}} \mu _{S}\cdot \chi _{S}$ yielding a constant vector in ${\mathbb R}^{N}$ and put ${{\mathcal {C}}}^{\prime }:=\{ S\in {{\mathcal {C}}}\, :\ \mu _{S}\ne 0\}$. Then ${{\mathcal {C}}}^{\prime }$ is semi-balanced on N and ${{\mathcal {C}}}^{\prime }\subset {{\mathcal {S}}}$ contradicts the minimality of ${{\mathcal {S}}}$ in (b).

To show (b) $\Rightarrow$(c) let us fix a semi-conic combination $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\cdot \chi _{S}$ yielding a constant vector in ${\mathbb R}^{N}$ with $\uplambda _{S}\ne 0$ for $S\in {{\mathcal {S}}}$. We first consider the case when it is not a conic combination, that is, there is a unique set $T\in {{\mathcal {S}}}$ with $\uplambda _{T}<0$; hence, $\uplambda _{S}>0$ for $S\in {{\mathcal {S}}}{\setminus }\{T\}$.

We then verify that $\{ \chi _{S}\,:\ S\in {{\mathcal {S}}}\}$ are affinely independent. Assume for a contradiction that a non-zero linear combination $\sum _{S\in {{\mathcal {S}}}} \mu _{S}\cdot \chi _{S}=\underline{0}$ with $\sum _{S\in {{\mathcal {S}}}} \mu _{S}=0$ exists. One can assume without loss of generality $\mu _{T}\le 0$ for otherwise one can multiply the linear combination by $(-1)$. Moreover, $\sum _{S\in {{\mathcal {S}}}} \mu _{S}=0$ implies the existence of $L\in {{\mathcal {S}}}$ with $\mu _{L}<0$. In fact, there are at least two such sets: otherwise $\sum _{S\in {{\mathcal {S}}}} \mu _{S}\cdot \chi _{S}=\underline{0}$ is a semi-conic combination and, by Lemma 1, one has $\sum _{S\in {{\mathcal {S}}}} \mu _{S}>0$ which contradicts the assumption. In particular, there is $L\in {{\mathcal {S}}}{\setminus }\{T\}$ with $\mu _{L}<0$. For any $\varepsilon \ge 0$ and $S\in {{\mathcal {S}}}$ we put $\uplambda ^{\varepsilon }_{S}:= \uplambda _{S}+\varepsilon \cdot \mu _{S}$ and observe that $\sum _{S\in {{\mathcal {S}}}} \uplambda ^{\varepsilon }_{S}\cdot \chi _{S}$ yields the same constant vector, $\uplambda ^{\varepsilon }_{T}<0$, while $\uplambda ^{\varepsilon }_{S}>0$ for $S\in {{\mathcal {S}}}{\setminus }\{T\}$ and sufficiently small $\varepsilon$. Since $\uplambda ^{\varepsilon }_{L}$ tends to $-\infty$ with increasing $\varepsilon$ there is a maximal $\varepsilon ^{*}>0$ such that $\uplambda ^{\varepsilon ^{*}}_{S}\ge 0$ for all $S\in {{\mathcal {S}}}{\setminus }\{T\}$. There must be $K\in {{\mathcal {S}}}{\setminus }\{T\}$ with $\uplambda ^{\varepsilon ^{*}}_{K}=0$; then the system ${{\mathcal {C}}}:=\{S\in {{\mathcal {S}}}\,: \ \uplambda ^{\varepsilon ^{*}}_{S}\ne 0\}\not \ni K$ is semi-balanced on N, which contradicts the minimality of ${{\mathcal {S}}}$.

The second step is to show that if $\bigcup \mathcal{S}=N$ then the vectors $\{ \chi _{S}\,:\ S\in {{\mathcal {S}}}\}$ are linearly independent. Note that this case involves the case of a conic combination $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\cdot \chi _{S}=r\cdot \chi _{N}\in {\mathbb R}^{N}$ because then, by Lemma 1, $r>0$, which enforces if $\bigcup \mathcal{S}=N$ then the vectors. In the sequel, let T denote a contingent set $T\in {{\mathcal {S}}}$ with $\uplambda _{T}<0$, which, however, need not exist. Recall that $\uplambda _{S}>0$ for $S\in {{\mathcal {S}}}{\setminus }\{T\}$.

Assume for a contradiction that a non-zero linear combination $\sum _{S\in {{\mathcal {S}}}} \mu _{S}\cdot \chi _{S}=\underline{0}$ exists. We then show that there exists such a linear combination which, additionally, satisfies $\mu _{T}\le 0$ if T exists and $\mu _{L}<0$ for at least one $L\in {{\mathcal {S}}}{\setminus }\{T\}$. This is easy in case T does not exist or $\mu _{T}=0$ because possible multiplication of the linear combination by $(-1)$ reaches the goal. In case $\mu _{T}\ne 0$ possible multiplication ensures $\mu _{T}<0$. To show then the existence of $L\in {{\mathcal {S}}}{\setminus }\{T\}$ with $\mu _{L}<0$ assume for a contradiction the opposite, which means that $\sum _{S\in {{\mathcal {S}}}} \mu _{S}\cdot \chi _{S}=\underline{0}$ is a semi-conic combination and the system $\mathcal{C} \,:=\, \{S\in \mathcal{S}\ : \, \mu _{S}\ne 0 \}$ is semi-balanced. The minimality of ${{\mathcal {S}}}$ then implies that $\mathcal{C}=\mathcal{S}$’ and, thus, $\mu _{S}>0$ for $S\in {{\mathcal {S}}}{\setminus }\{T\}$. Since $T\subset N$ there exists $i\in N{\setminus } T$ and $\bigcup {{\mathcal {S}}}=N$ implies the existence of $K\in {{\mathcal {S}}}$ with $i\in K$. Since [ $i\in S\in {{\mathcal {S}}}~\Rightarrow ~ \mu _{S}>0$ ], this leads to a contradictory conclusion $0=\sum _{S\in {{\mathcal {S}}}} \mu _{S}\cdot \chi _{S}(i)= \sum _{S\in {{\mathcal {S}}}: i\in S} \mu _{S}\ge \mu _{K}>0$.

Finally, having a linear combination $\sum _{S\in {{\mathcal {S}}}} \mu _{S}\cdot \chi _{S}=\underline{0}$ with $\mu _{T}\le 0$ and $\mu _{L}<0$ for some $L\in {{\mathcal {S}}}{\setminus }\{T\}$, one can repeat the construction used in the previous case (of affine independence) to get a contradiction with the minimality of ${{\mathcal {S}}}$.

To show (c) $\Rightarrow$(d) use Lemma 1 to obtain an affine semi-conic combination $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\cdot \chi _{S}$ with all coefficients non-zero yielding a constant vector $\rho =[r,\ldots ,r]$ in ${\mathbb R}^{N}$ with $r\in [0,1]$. Let us fix this affine combination. Assume that $\sum _{S\in {{\mathcal {S}}}} \sigma _{S}\cdot \chi _{S}$ is an affine combination yielding a constant vector $\varsigma =[s,\ldots ,s]\in {\mathbb R}^{N}$. It is enough to show that these two combinations coincide. To this end we distinguish two cases.

In case $\bigcup {{\mathcal {S}}}\!\subset \! N$ we have $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\cdot \chi _{S}(i)\!=\!0\!=\!\sum _{S\in {{\mathcal {S}}}} \sigma _{S}\cdot \chi _{S}(i)$ for any $i\in N\setminus \bigcup \mathcal{S}$. This implies that both $\rho = \underline{0}$ and $\varrho = \underline{0}$. By subtracting we obtain the relation $\sum _{S\in {{\mathcal {S}}}} (\uplambda _{S}-\sigma _{S})\cdot \chi _{S}=\underline{0}$ with $\sum _{S\in {{\mathcal {S}}}} (\uplambda _{S}-\sigma _{S}) =0$ and by affine independence $\sigma _{S}=\uplambda _{S}$ for all $S\in {{\mathcal {S}}}$.

In case $\bigcup {{\mathcal {S}}}=N$ the linear independence of vectors $\{ \chi _{S}\,:\ S\in {{\mathcal {S}}}\}$ implies both $r\ne 0$ and $s\ne 0$ because solely their zero linear combination yields the vector $=\underline{0}$. Hence, we have $\sum _{S\in {{\mathcal {S}}}} (r^{-1}\cdot \uplambda _{S})\cdot \chi _{S}=\chi _{N}$ and $\sum _{S\in {{\mathcal {S}}}} (s^{-1}\cdot \sigma _{S})\cdot \chi _{S}=\chi _{N}$. By subtracting we get $\sum _{S\in {{\mathcal {S}}}} (r^{-1}\cdot \uplambda _{S}-s^{-1}\cdot \sigma _{S})\cdot \chi _{S}=\underline{0}$ and by linear independence $r^{-1}\cdot \uplambda _{S}=s^{-1}\cdot \sigma _{S}$ for all $S\in {{\mathcal {S}}}$. Thus, $\uplambda _{S}=r\cdot s^{-1}\cdot \sigma _{S}$ for $S\in {{\mathcal {S}}}$ and, because both combinations are affine, by summing over $S\in {{\mathcal {S}}}$ one derives $1=r\cdot s^{-1}$. Hence, $s=r$ and $\sigma _{S}=\uplambda _{S}$ for all $S\in {{\mathcal {S}}}$.

The implication (d) $\Rightarrow$(e) is evident.

To show (e) $\Rightarrow$(b) consider the affine semi-conic combination $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\cdot \chi _{S}$ with all coefficients non-zero yielding a constant vector in ${\mathbb R}^{N}$. Its existence implies that ${{\mathcal {S}}}$ is semi-balanced. To verify the minimality of ${{\mathcal {S}}}$ assume for a contradiction that ${{\mathcal {C}}}\subset {{\mathcal {S}}}$ exists which is semi-balanced on N. By Lemma 1 applied to ${{\mathcal {C}}}$ there exists an affine semi-conic combination $\sum _{S\in {{\mathcal {C}}}} \sigma _{S}\cdot \chi _{S}$ yielding a constant vector in ${\mathbb R}^{N}$. We extend it by putting $\sigma _{S}=0$ for $S\in {{\mathcal {S}}}{\setminus }{{\mathcal {C}}}$. Thus, we get two different affine semi-conic combinations of $\{ \chi _{S}\,:\ S\in {{\mathcal {S}}}\}$ yielding a constant vector in ${\mathbb R}^{N}$, which contradicts the assumption. $\square$

Proof of Lemma 7

For reader’s convenience we recall what is claimed; see Definition 5 for notation.

Lemma 7    Given $|N|\ge 2$, every set ${\tilde{\Theta }}^{N}_{D}$, where $\emptyset \ne D\subseteq N$, is a bounded polyhedron. Every vector $\theta \in \Theta ^{N}_{D}$ satisfies both $\theta (N)\ge 0$ and $\theta (\emptyset )\ge 0$ and every non-zero vector $\theta \in \Theta ^{N}_{D}$ satisfies $\theta (N)+\theta (\emptyset )>0$. Given $m\in {\mathbb R}^{\mathcal{P}(N)}$ with $m(\emptyset )=0$, one has

$$\begin{aligned} \qquad \qquad \quad m\in \mathcal{E}(N) ~~\Leftrightarrow ~~ \left[ \,\forall \, \theta \in \bigcup _{\emptyset \ne D\subseteq N}{\tilde{\Theta }}^{N}_{D}\qquad \langle \theta ,m\rangle \ge 0\right] \,. \qquad \qquad \qquad \qquad \quad (3) \end{aligned}$$

Proof

Note that $\Theta ^{N}_{N}\subseteq \Theta ^{N}_{D}$ if $D\subset N$; therefore, one can assume without loss of generality $\emptyset \ne D\subset N$. To show $\theta (N)\ge 0$ for $\theta \in \Theta ^{N}_{D}$ choose $i\in N{\setminus } D$ and write $\theta (N)=-\sum _{L\subset N:i\in L} \theta (L)\ge 0$. Note that, for any $j\in N$,

$$\begin{aligned} \sum _{S\subseteq N{\setminus }\{j\}} \theta (S)= \sum _{S\subseteq N} \theta (S)-\sum _{L\subseteq N:j\in L} \theta (L) = 0-0 =0. \end{aligned}$$

Hence, to show $\theta (\emptyset )\ge 0$ for $\theta \in \Theta ^{N}_{D}$ take $j\in D$ and write $\theta (\emptyset )=-\sum _{\emptyset \ne S\subseteq N{\setminus }\{j\}} \theta (S)\ge 0$.

Thus, we have observed that every $\theta \in \Theta ^{N}_{D}$ satisfies both $\theta (N)\ge 0$ and $\theta (\emptyset )\ge 0$. In particular, if $\theta \in {\tilde{\Theta }}^{N}_{D}$ then $\theta (N)+\theta (\emptyset )=1$ gives both $0\le \theta (\emptyset )\le 1$ and $0\le \theta (N)\le 1$.

Let us show that, for $\theta \in {\tilde{\Theta }}^{N}_{D}$, if $S\subset N$ and $S{\setminus } D\ne \emptyset$ then $0\ge \theta (S)\ge -1$. Indeed, because of $S\not \in \{\emptyset ,D,N\}$, the choice of $i\in S{\setminus } D$ gives

$$\begin{aligned} 0\ge \theta (S)\ge \sum _{L\subset N:i\in L} \theta (L)=-\theta (N)+ \underbrace{\sum _{L\subseteq N:i\in L} \theta (L)}_{=0} =-\theta (N)\ge -1\,. \end{aligned}$$

To observe that $0\ge \theta (S)\ge -1$ for $\theta \in {\tilde{\Theta }}^{N}_{D}$ whenever $\emptyset \ne S\subset N$ and $D{\setminus } S\ne \emptyset$ introduce a vector $\theta ^{\star }\in {\mathbb R}^{\mathcal{P}(N)}$ by $\theta ^{\star }(L):=\theta (N{\setminus } L)$ for $L\subseteq N$. It is easy to observe that $\theta ^{\star }\in {\tilde{\Theta }}^{N}_{N{\setminus } D}$: to this end write for any $i\in N$

$$\begin{aligned} \sum _{L\subseteq N:i\in L} \theta ^{\star }(L)=\sum _{L\subseteq N:i\in L} \theta (N{\setminus } L)=\sum _{S\subseteq N:i\not \in S} \theta (S) =\sum _{S\subseteq N} \theta (S)-\sum _{S\subseteq N:i\in S} \theta (S)=0-0=0. \end{aligned}$$

Thus, because of $\emptyset \ne D{\setminus } S= (N{\setminus } S){\setminus }(N{\setminus } D)$, one has $0\ge \theta ^{\star }(N{\setminus } S)\ge -1$ by the previous observation applied to $\theta ^{\star }$, which, however, means $0\ge \theta (S)\ge -1$.

Altogether, we have $0\ge \theta (S)\ge -1$ for $\theta \in {\tilde{\Theta }}^{N}_{D}$ and $S\in \mathcal{P}(N){\setminus } \{\emptyset ,D,N\}$, which implies $0\ge \sum _{S: S\not \in \{\emptyset ,D,N\}} \theta (S)\ge 3-2^{|N|}$. Taking into consideration that

$$\begin{aligned} \sum _{S: S\not \in \{\emptyset ,D,N\}} \theta (S)= -\theta (\emptyset )-\theta (D)-\theta (N)+\underbrace{\sum _{L\subseteq N} \theta (L)}_{=0}= -\theta (N)-\theta (\emptyset )-\theta (D)=-1-\theta (D) \end{aligned}$$

one gets $2^{|N|}-4\ge \theta (D)\ge -1$. In particular, ${\tilde{\Theta }}^{N}_{D}$ is bounded and so is ${\tilde{\Theta }}^{N}_{N}$.

The fact that, for any $\emptyset \ne D\subseteq N$, every non-zero vector $\theta \in \Theta ^{N}_{D}$ satisfies $\theta (N)+\theta (\emptyset )>0$ follows directly from (Kroupa and Studený 2019, Lemma 5.1). In particular, every non-zero $\theta \in \Theta ^{N}_{D}$ is a positive multiple of a vector ${\tilde{\theta }}\in {\tilde{\Theta }}^{N}_{D}$. Further important fact, which follows from (Kroupa and Studený 2019, Lemma 5.3), is that a game $m$ is exact, that is, $m\in \mathcal{E}(N)$, iff $[\,\forall \,\emptyset \ne D\subseteq N ~~\forall \, \theta \in \Theta ^{N}_{D}\quad \langle \theta ,m\rangle \ge 0\,]$. The combination of these two observations gives (3). $\square$

Proof of Lemma 8

For reader’s convenience we recall what is claimed; see Definition 5 for notation.

Lemma 8    Given $|N|\ge 2$ and $\emptyset \ne D\subseteq N$, every vertex of ${\tilde{\Theta }}^{N}_{D}$ has either the form $\theta _{{{\mathcal {B}}}}$, where ${{\mathcal {B}}}$ is a min-balanced set system on N, or the form $\theta _{{{\mathcal {S}}}}$, where ${{\mathcal {S}}}$ is a min-semi-balanced system on N having D as the exceptional set.

Conversely, in case $\emptyset \ne D\subset N$, every vector $\theta _{{{\mathcal {S}}}}$, where ${{\mathcal {S}}}$ is a min-semi-balanced system on N having D as the exceptional set, is a vertex of ${\tilde{\Theta }}^{N}_{D}$: $\theta _{{{\mathcal {S}}}}\in \hbox \mathrm{ext}\,({\tilde{\Theta }}^{N}_{D})$.

Proof

Given a vertex $\theta \in \hbox \mathrm{ext}\,({\tilde{\Theta }}^{N}_{D})$, we first observe that there exists a min-semi-balanced system ${{\mathcal {S}}}$ on N such that $\theta =\theta _{{{\mathcal {S}}}}$. To this end we put

$$ \begin{aligned} {{\mathcal {S}}}~:=~ \{S\subseteq N\,:\ \emptyset \ne S\subset N ~~ \& ~~ \theta (S)\ne 0\,\}\qquad \hbox {and}~~ \uplambda _{S} ~:=~ -\theta (S)\quad \hbox {for} \ S\in {{\mathcal {S}}}. \end{aligned}$$

Thus, $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\cdot \chi _{S}$ yields a constant vector $\theta (N)\cdot \chi _{N}$ and ${{\mathcal {S}}}$ is semi-balanced on N. To evidence that ${{\mathcal {S}}}$ is min-semi-balanced we use the condition (c) in Lemma 2. To verify affine independence of the vectors $\{ \chi _{S}\,:\ S\in {{\mathcal {S}}}\}$ assume for a contradiction that there is a non-zero linear combination $\sum _{S\in {{\mathcal {S}}}} \sigma _{S}\cdot \chi _{S} =\underline{0}$ with $\sum _{S\in {{\mathcal {S}}}} \sigma _{S}=0$. Then we put $\sigma _{L}:=0$ for remaining $L\subseteq N$ and $\theta ^{\varepsilon }(S):=\theta (S)+\varepsilon \cdot \sigma _{S}$ for any $S\subseteq N$ and $\varepsilon \in {\mathbb R}$. Then one has $\theta ^{\varepsilon }\in {\tilde{\Theta }}^{N}_{D}$ whenever $|\varepsilon |$ is small; thus, the relation $\theta =\frac{1}{2}\cdot \theta ^{\varepsilon }+\frac{1}{2}\cdot \theta ^{-\varepsilon }$ then contradicts the assumption of extremity of $\theta$ in ${\tilde{\Theta }}^{N}_{D}$, because of $\sigma _{L}\ne 0$ for some $L\in {{\mathcal {S}}}$.

To verify linear independence of the vectors $\{ \chi _{S}\,:\ S\in {{\mathcal {S}}}\}$ in case $\bigcup {{\mathcal {S}}}=N$ we first realize that one has $\theta (N)>0$ then. Indeed, if $D=N$ then $\theta (S)<0$ for any $S\in {{\mathcal {S}}}$ and if $D\ne N$ then we choose $i\in N{\setminus } D$ and have $\theta (S)<0$ for any $S\in {{\mathcal {S}}}$ with $i\in S$; this allows us to use $\sum _{L\subseteq N:\, i\in L} \theta (L)=0$ for $i\in N$ to derive $\theta (N)>0$.

Assume for a contradiction that a non-zero linear combination $\sum _{S\in \mathcal{S}} \sigma _{S}\cdot \chi _{S}=\underline{0}$ exists and put $\varsigma :=\sum _{S\in {{\mathcal {S}}}} \sigma _{S}$. The case $\varsigma =0$ leads to a contradiction as shown in the case of affine independence. Thus, consider $\varsigma \ne 0$, put $\sigma _{\emptyset }:=-\varsigma$, $\sigma _{L}:=0$ for remaining $L\subseteq N$, and $\theta ^{\varepsilon }(S):=(1-\varepsilon \cdot \varsigma )^{-1}\cdot (\theta (S)+\varepsilon \cdot \sigma _{S})$ for any $S\subseteq N$ and $\varepsilon \in {\mathbb R}$, $\varepsilon \ne \varsigma ^{-1}$. One has $\theta ^{\varepsilon }\in {\tilde{\Theta }}^{N}_{D}$ for small $|\varepsilon |$. Moreover, $\theta =\frac{1-\varepsilon \cdot \varsigma }{2}\cdot \theta ^{\varepsilon }+ \frac{1+\varepsilon \cdot \varsigma }{2}\cdot \theta ^{-\varepsilon }$. Because of $\varsigma \ne 0$ we have $\theta ^{\varepsilon }(N)=(1-\varepsilon \cdot \varsigma )^{-1}\cdot \theta (N)\ne \theta (N)$ if $\varepsilon \ne 0$. Hence, we get a contradiction with the assumption of extremity of $\theta$ in ${\tilde{\Theta }}^{N}_{D}$.

We have thus shown that the set system ${{\mathcal {S}}}$ is min-semi-balanced on N and one clearly has $\theta =\theta _{{{\mathcal {S}}}}$. If ${{\mathcal {S}}}={{\mathcal {B}}}$ is min-balanced then the first option $\theta =\theta _{{{\mathcal {B}}}}$ occurs. If ${{\mathcal {S}}}$ has an exceptional set $T\in {{\mathcal {S}}}$ then $\theta (T)=\theta _{{{\mathcal {S}}}}(T)>0$ which forces $T=D$. Thus, D is the exceptional set in ${{\mathcal {S}}}$ in this case.

To verify the second claim assume that ${{\mathcal {S}}}$ is a min-semi-balanced set system on N and that D is the exceptional set within ${{\mathcal {S}}}$. It is straightforward to evidence that $\theta _{{{\mathcal {S}}}}\in {\tilde{\Theta }}^{N}_{D}$. To show that $\theta _{{{\mathcal {S}}}}\in \hbox \mathrm{ext}\,({\tilde{\Theta }}^{N}_{D})$ assume for a contradiction that there is a non-trivial convex combination $\theta _{{{\mathcal {S}}}}=\alpha \cdot \theta ^{0}+ (1-\alpha )\cdot \theta ^{1}$ with $\theta ^{0},\theta ^{1}\in {\tilde{\Theta }}^{N}_{D}$, $\alpha \in (0,1)$, and $\theta ^{0}\ne \theta ^{1}$. We know that $\theta _{{{\mathcal {S}}}}(S)=0$ any $S\subseteq N$, $S\not \in \{\emptyset ,N\}\cup {{\mathcal {S}}}$ and observe that $\theta ^{0}(S)=0=\theta ^{1}(S)$ for any such $S\subseteq N$ as well. Indeed, the inclusion ${\tilde{\Theta }}^{N}_{D}\subseteq \Theta ^{N}_{D}$ implies $\theta ^{i}(L)\le 0$ for $L\subseteq N$, $L\not \in \{\emptyset ,D,N\}$, which forces $\theta ^{i}(S)=0$ for $S\subseteq N$, $S\not \in \{\emptyset ,N\}\cup {{\mathcal {S}}}$ .

For every $\varepsilon \ge 0$ we put $\theta ^{\varepsilon } := \theta ^{0} +\varepsilon \cdot (\theta ^{1}-\theta ^{0})$. Note that $\theta ^{\varepsilon }(S)=0$ for any $S\subseteq N$, $S\not \in \{\emptyset ,N\}\cup {{\mathcal {S}}}$. The fact that ${\tilde{\Theta }}^{N}_{D}$ is bounded (see Lemma 7) implies the existence of $\varepsilon ^{\diamond }:=\max \, \{\varepsilon \ge 1\,:\ \theta ^{\varepsilon }\in {\tilde{\Theta }}^{N}_{D}\,\}$. There exists $L\in {{\mathcal {S}}}$ with $\theta ^{\varepsilon ^{\diamond }}(L)=0$ because otherwise one gets a contradiction with maximality of $\varepsilon ^{\diamond }$. We put

$$\begin{aligned} {{\mathcal {C}}}:=\{ S\in {{\mathcal {S}}}\,:\ \theta ^{\varepsilon ^{\diamond }}(S)\ne 0\,\} \end{aligned}$$

and observe, using the fact $\theta ^{\varepsilon ^{\diamond }}\in {\tilde{\Theta }}^{N}_{D}$, that ${{\mathcal {C}}}$ is a semi-balanced system on N. Then the fact $L\in \mathcal{S}\setminus \mathcal{C}$ contradicts the minimality of ${{\mathcal {S}}}$. Thus, there is no non-trivial convex combination of $\theta ^{0},\theta ^{1}\in {\tilde{\Theta }}^{N}_{D}$ yielding $\theta _{{{\mathcal {S}}}}$, which means that $\theta _{{{\mathcal {S}}}}$ is a vertex of ${\tilde{\Theta }}^{N}_{D}$. $\square$

Proof of Lemma 12

For reader’s convenience we recall what is claimed.

Lemma 12    Given $|N|\ge 2$ and a min-semi-balanced system ${{\mathcal {S}}}$ on N, let $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\cdot \chi _{S}$ be the unique affine (and semi-conic) combination yielding a constant vector $r\cdot \chi _{N}$, $r\in [0,1]$.

One has then $r=0$ iff there exists a min-balanced system ${{\mathcal {B}}}$ on $M\subset N$, $|M|\ge 2$, such that ${{\mathcal {S}}}={{\mathcal {B}}}\cup \{M\}$; another equivalent condition is $\bigcup {{\mathcal {S}}}\subset N$. Moreover, a min-balanced system ${{\mathcal {B}}}$ on $M\subset N$, $|M|\ge 2$, yields a min-semi-balanced one, namely the system $\mathcal{S}:={{\mathcal {B}}}\cup \{M\}$ on N with $r=0$.

One has $r=1$ iff ${{\mathcal {S}}}$ is a complementary system to a system ${{\mathcal {S}}}^{\star }$ with $\bigcup {{\mathcal {S}}}^{\star }\subset N$; another equivalent condition is $\bigcap {{\mathcal {S}}}\ne \emptyset$.

On the other hand, every min-balanced system ${{\mathcal {S}}}$ on N satisfies $0<r<1$.

Proof

Recall that all coefficients $\uplambda _{S}$, $S\in {{\mathcal {S}}}$, are non-zero. If $r=0$ then there is $T\in {{\mathcal {S}}}$ with $\uplambda _{T}<0$, for otherwise $\uplambda _{S}>0$ for $S\in {{\mathcal {S}}}$ and ${{\mathcal {S}}}\ne \emptyset$ contradict $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\cdot \chi _{S} =\underline{0}$. Given $i\in N{\setminus } T$ one has $\uplambda _{S}>0$ whenever $i\in S\in {{\mathcal {S}}}$ and $0=\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\cdot \chi _{S}(i)=\sum _{S\in {{\mathcal {S}}}: i\in S} \uplambda _{S}$ implies that there is no $S\in {{\mathcal {S}}}$ with $i\in S$. Hence, $\bigcup {{\mathcal {S}}}=T$ and $\sum _{S\in {{\mathcal {S}}}{\setminus }\{T\}} \uplambda _{S}\cdot \chi _{S}=(-\uplambda _{T})\cdot \chi _{T}$. This implies ${{\mathcal {S}}}{\setminus }\{T\}\ne 0$ and, thus, forces $|T|\ge 2$ (because $\bigcup {{\mathcal {S}}}=T$). The latter equality also means that ${{\mathcal {S}}}{\setminus }\{T\}$ is balanced on T and one can put $M:=T$ and ${{\mathcal {B}}}:={{\mathcal {S}}}{\setminus }\{T\}$. To show that ${{\mathcal {B}}}$ is minimal assume for a contradiction that ${{\mathcal {D}}}\subset {{\mathcal {B}}}$ exists which is balanced on T, that is, there are $\sigma _{S}>0$, $S\in {{\mathcal {D}}}$, with $\sum _{S\in {{\mathcal {D}}}} \sigma _{S}\cdot \chi _{S}=\chi _{T}$. Hence, $\sum _{S\in {{\mathcal {D}}}} \sigma _{S}\cdot \chi _{S}+ (-1)\cdot \chi _{T} =\underline{0}$ and, by Lemma 1, one has $-1+\sum _{S\in {{\mathcal {D}}}} \sigma _{S}>0$ and one can multiply it to get an affine combination (in ${\mathbb R}^{N}$) different from $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\cdot \chi _{S} =\underline{0}$, which contradicts the minimality of ${{\mathcal {S}}}$ (use Lemma 2(d)). Thus, ${{\mathcal {B}}}$ has to be min-balanced on T.

The existence of such ${{\mathcal {B}}}$ then implies $\bigcup {{\mathcal {S}}}\subset N$. To complete the chain of implications realize that, if $\bigcup {{\mathcal {S}}}\subset N$ for a min-semi-balanced set system ${{\mathcal {S}}}$ on N then the choice of $i\in N{\setminus }\bigcup {{\mathcal {S}}}$ gives $r=\sum _{S\in {{\mathcal {S}}}:i\in S} \uplambda _{S}=0$.

To verify the additional claim assume that ${{\mathcal {B}}}$ is a min-balanced system on $M\subset N$, $|M|\ge 2$. By (Kroupa and Studený 2019, Lemma 2.1) $\{\chi _{S}\,:\ S\in \mathcal B\}$ is linearly independent (even in ${\mathbb R}^{N}$). Fix a combination $\sum _{S\in {{\mathcal {B}}}} \uplambda _{S}\cdot \chi _{S}=\chi _{M}$ with all coefficients strictly positive. Then $\sum _{S\in {{\mathcal {B}}}} \uplambda _{S}\cdot \chi _{S}+ (-1)\cdot \chi _{M} =\underline{0}$ is a semi-conic combination yielding a constant vector in ${\mathbb R}^{N}$ and ${{\mathcal {S}}}:={{\mathcal {B}}}\cup \{M\}$ is semi-balanced. We use the condition (d) in Lemma 2 to show that ${{\mathcal {S}}}$ is minimal. By Lemma 1 one has $k:=\sum _{S\in {{\mathcal {B}}}} \uplambda _{S}-1>0$ and $\sum _{S\in {{\mathcal {B}}}} k^{-1}\uplambda _{S}\cdot \chi _{S}+ (-k^{-1})\cdot \chi _{M} =\underline{0}$ is an affine semi-conic combination with all coefficients non-zero. We need to show that any affine combination $\sum _{S\in {{\mathcal {B}}}} \sigma _{S}\cdot \chi _{S}+ \sigma _{M}\cdot \chi _{M}=\rho$ yielding a constant vector in ${\mathbb R}^{N}$ coincides with the above mentioned affine combination. The substitution of some $i\in N{\setminus } M$ allows one to observe that $\rho = \underline{0}$.. One cannot have $\sigma _{M}=0$ because then $\sum _{S\in {{\mathcal {B}}}} \sigma _{S}\cdot \chi _{S} =\underline{0}$ contradicts linear independence of $\{\chi _{S}\,:\ S\in \mathcal B\}$. If $\sigma _{M}\ne 0$ then $\sum _{S\in {{\mathcal {B}}}} -\sigma _{M}^{-1}\sigma _{S}\cdot \chi _{S}=\chi _{M}$ and linear independence of $\{\chi _{S}\,:\ S\in \mathcal B\}$ gives $\uplambda _{S}=-\sigma _{M}^{-1}\sigma _{S}$ for all $S\in {{\mathcal {B}}}$. We substitute $\sigma _{S}=-\sigma _{M}\uplambda _{S}$ for $S\in {{\mathcal {B}}}$ to $\sum _{S\in {{\mathcal {B}}}} \sigma _{S}+ \sigma _{M}=1$ to get $\sigma _{M}=-k^{-1}$; hence, the considered affine combinations coincide.

As concerns the case $r=1$, by Lemma 11, ${{\mathcal {S}}}^{\star }$ is also min-semi-balanced and the respective unique affine combination for ${{\mathcal {S}}}^{\star }$ is $\sum _{L\in {{\mathcal {S}}}^{\star }} \uplambda _{N{\setminus } L}\cdot \chi _{L}=(1-r)\cdot \chi _{N}$. Thus, $r=1$ iff the previous case $r^{\star }=0$ occurs for ${{\mathcal {S}}}^{\star }$. The formula $\bigcap {{\mathcal {S}}}= N{\setminus } \bigcup {{\mathcal {S}}}^{\star }$ then gives the other equivalent condition.

The definition of a min-balanced system ${{\mathcal {B}}}$ on N implies $\bigcup {{\mathcal {B}}}=N$. By Lemma 11, the same is true for its complementary system: $\bigcup {{\mathcal {B}}}^{\star }=N$, which means $\bigcap {{\mathcal {B}}}=\emptyset$. Thus, one has both $\bigcup {{\mathcal {B}}}=N$ and $\bigcap {{\mathcal {B}}}=\emptyset$, and by previous claims, $r\ne 0$ and $r\ne 1$.

$\square$

Proof of Lemma 13

For reader’s convenience we recall what is claimed.

Lemma 13    Assume $|N|\ge 3$. If ${{\mathcal {B}}}$ is a min-balanced set system on N such that $|{{\mathcal {B}}}|\ge 3$ and $Z\in {{\mathcal {B}}}$ then $Y:=N{\setminus } Z$ is not in ${{\mathcal {B}}}$, the set system ${{\mathcal {S}}}:=({{\mathcal {B}}}{\setminus }\{Z\})\cup \{Y\}$ is purely min-semi-balanced on N and Y is the exceptional set within ${{\mathcal {S}}}$.

Conversely, if ${{\mathcal {S}}}$ is a (purely) min-semi-balanced system on N and $Y\in {{\mathcal {S}}}$ the exceptional set within ${{\mathcal {S}}}$ then $Z:=N{\setminus } Y$ is not in ${{\mathcal {S}}}$ and ${{\mathcal {B}}}:=({{\mathcal {S}}}{\setminus }\{Y\})\cup \{Z\}$ is a min-balanced system on N such that $|{{\mathcal {B}}}|\ge 3$.

Proof

I. :

Let ${{\mathcal {B}}}$ be a balanced system on N and $Z\in {{\mathcal {B}}}$ such that $Y:=N{\setminus } Z\not \in {{\mathcal {B}}}$. Then ${{\mathcal {S}}}:=({{\mathcal {B}}}{\setminus }\{Z\})\cup \{Y\}$ is semi-balanced on N. Indeed, there exists a conic combination $\sum _{S\in {{\mathcal {B}}}} \uplambda _{S}\cdot \chi _{S}=\chi _{N}$, $\uplambda _{S}>0$ for $S\in {{\mathcal {B}}}$. We add $-\uplambda _{Z}\cdot \chi _{N}$ to that and obtain a semi-conic combination $\sum _{S\in {{\mathcal {B}}}{\setminus }\{Z\}} \uplambda _{S}\cdot \chi _{S} + (-\uplambda _{Z})\cdot \chi _{Y}=(1-\uplambda _{Z})\cdot \chi _{N}$ yielding a constant vector.

II. :

Analogously, given a semi-balanced system ${{\mathcal {S}}}$ over N with an exceptional set $Y\in {{\mathcal {S}}}$ and $Z:=N{\setminus } Y\not \in \mathcal{S}$, the set system ${{\mathcal {B}}}:=({{\mathcal {S}}}{\setminus }\{Y\})\cup \{Z\}$ is balanced on N. Indeed, given a semi-conic combination $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\cdot \chi _{S}=r\cdot \chi _{N}$ with $r\in {\mathbb R}$, where $\uplambda _{Y}<0$ and $\uplambda _{S}>0$ for $S\in {{\mathcal {S}}}{\setminus }\{Y\}$ we add $-\uplambda _{Y}\cdot \chi _{N}$ to that and get $\sum _{S\in {{\mathcal {S}}}{\setminus }\{Y\}} \uplambda _{S}\cdot \chi _{S} + (-\uplambda _{Y})\cdot \chi _{Z}=(r-\uplambda _{Y})\cdot \chi _{N}$, which is a conic combination yielding a constant vector. Then use Lemma 1.

III. :

Let ${{\mathcal {B}}}$ be a min-balanced system on N with $|{{\mathcal {B}}}|\ge 3$ and $Z\in {{\mathcal {B}}}$. Then one has $Y:= N{\setminus } Z \not \in {{\mathcal {B}}}$ as otherwise ${{\mathcal {D}}}:=\{Y,Z\}\subset {{\mathcal {B}}}$ is a balanced system contradicting the minimality of ${{\mathcal {B}}}$. Step I. implies that ${{\mathcal {S}}}:=({{\mathcal {B}}}{\setminus }\{Z\})\cup \{Y\}$ is semi-balanced on N, specifically, that there exists a semi-conic combination $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\cdot \chi _{S}$ yielding a constant vector in ${\mathbb R}^{N}$ with $\uplambda _{Y}<0$. To show that ${{\mathcal {S}}}$ is minimal assume for a contradiction that a min-semi-balanced system ${{\mathcal {S}}}^{\prime }\subset {{\mathcal {S}}}$ exists. Then necessarily $Y\in {{\mathcal {S}}}^{\prime }$ as otherwise $\mathcal{S}^{\prime } \subset \mathcal{B}$ contradicts the minimality of ${{\mathcal {B}}}$ (use Corollary 3). Let $\sum _{S\in {{\mathcal {S}}}^{\prime }} \sigma _{S}\cdot \chi _{S}$ be the (unique) affine semi-conic combination yielding a constant vector in ${\mathbb R}^{N}$. Observe that $\sigma _{Y}<0$ as otherwise one can put $\sigma _{S}:=0$ for $S\in {{\mathcal {S}}}{\setminus }{{\mathcal {S}}}^{\prime }$, and then $\tau _{S}:=\alpha \cdot \sigma _{S}+(1-\alpha )\cdot \uplambda _{S}$ for $S\in {{\mathcal {S}}}$ with $\alpha :=-\uplambda _{Y}\cdot (\sigma _{Y}-\uplambda _{Y})^{-1}\in (0,1]$ to get a semi-conic combination $\sum _{S\in {{\mathcal {S}}}} \tau _{S}\cdot \chi _{S}$ yielding a constant vector in ${\mathbb R}^{N}$ where $\tau _{Y}=0$; this means that ${{\mathcal {E}}}:=\{S\in {{\mathcal {S}}}\,:\ \tau _{S}\ne 0\}\subset {{\mathcal {B}}}$ is a semi-balanced system which fact contradicts the minimality of ${{\mathcal {B}}}$, by Corollary 3. Thus, Y has to be an exceptional set within ${{\mathcal {S}}}^{\prime }$ and, by step II. applied to ${{\mathcal {S}}}^{\prime }$, the system ${{\mathcal {B}}}^{\prime }:=({{\mathcal {S}}}^{\prime }{\setminus }\{Y\})\cup \{Z\}\subset {{\mathcal {B}}}$ is a balanced system on N contradicting the minimality of ${{\mathcal {B}}}$. Thus, the first claim in Lemma 13 has been verified.

IV. :

Let ${{\mathcal {S}}}$ be a min-semi-balanced system on N with the exceptional set $Y\in {{\mathcal {S}}}$ within it. Then $Z:=N{\setminus } Y\not \in {{\mathcal {S}}}$ as otherwise ${{\mathcal {D}}}:=\{Y,Z\}\subset {{\mathcal {S}}}$ is a balanced system contradicting the minimality of ${{\mathcal {S}}}$ (note that $|{{\mathcal {S}}}|\ge 3$ by Lemma 1). Step II. implies that ${{\mathcal {B}}}:=({{\mathcal {S}}}{\setminus }\{Y\})\cup \{Z\}$ is balanced on N. To show that ${{\mathcal {B}}}$ is minimal assume for a contradiction that a balanced system ${{\mathcal {B}}}^{\prime }\subset {{\mathcal {B}}}$ on N exists. Then necessarily $Z\in {{\mathcal {B}}}^{\prime }$ as otherwise ${{\mathcal {B}}}^{\prime }\subset {{\mathcal {S}}}$ contradicts the minimality of ${{\mathcal {S}}}$. By step I. applied to ${{\mathcal {B}}}^{\prime }$, the set system ${{\mathcal {S}}}^{\prime }:=({{\mathcal {B}}}^{\prime }{\setminus }\{Z\})\cup \{Y\}\subset {{\mathcal {S}}}$ is semi-balanced on N, which contradicts the minimality of ${{\mathcal {S}}}$. Of course, $|{{\mathcal {B}}}|=|{{\mathcal {S}}}|\ge 3$, which concludes the proof of the second claim in Lemma 13. $\square$

Proof of Lemma 17

A key induction step in the proof is based on the following auxiliary observations.

Lemma 20

Given $|N|\ge 3$, let ${{\mathcal {S}}}$ be a purely min-semi-balanced system on N. Denote ${{\mathcal {W}}}:=\mathcal{P}(N){\setminus } (\{\emptyset ,N\}\cup {{\mathcal {S}}})$ and introduce the next special polytopes (see Definition 5):

$$\begin{aligned} \Sigma ({{\mathcal {S}}}):= & {} \hbox \mathrm{conv}\,\left( \bigcup _{D\in {{\mathcal {W}}}} {\tilde{\Theta }}^{N}_{D}\right) ,\\ \Delta ({{\mathcal {S}}}):= & {} \Sigma ({{\mathcal {S}}})\,\cap \, \{\theta \in {\mathbb R}^{\mathcal{P}(N)}\,:\ \theta (W)\ge 0 \quad \hbox {for any}\ W\in {{\mathcal {W}}}\,\}\,. \end{aligned}$$

Moreover, for any ${{\mathcal {Z}}}\subseteq {{\mathcal {W}}}$, we put:

$$\begin{aligned} \Sigma ^{{{\mathcal {Z}}}}({{\mathcal {S}}}):= & {} \Sigma ({{\mathcal {S}}})\,\cap \, \{\theta \in {\mathbb R}^{\mathcal{P}(N)}\,:\ \theta (Z)= 0 \quad \hbox {for any} \ Z\in {{\mathcal {Z}}}\,\}\,,\\ \Delta ^{{{\mathcal {Z}}}}({{\mathcal {S}}}):= & {} \Sigma ^{{{\mathcal {Z}}}}({{\mathcal {S}}})\,\cap \, \Delta ({{\mathcal {S}}})\,. \end{aligned}$$

Then, for any ${{\mathcal {Z}}}\subseteq {{\mathcal {W}}}$, one has:

(i):

$\Sigma ^{{{\mathcal {W}}}}({{\mathcal {S}}})=\emptyset$,

(ii):

if $\emptyset \ne \Delta ^{{{\mathcal {Z}}}}({{\mathcal {S}}})$ and $\Sigma ^{{{\mathcal {Z}}}}({{\mathcal {S}}}){\setminus } \Delta ^{{{\mathcal {Z}}}}({{\mathcal {S}}})\ne \emptyset$  then  $\emptyset \ne \Delta ^{{{\mathcal {Z}}}\cup \{D\}}({{\mathcal {S}}})$ for some set $D\in \mathcal{W}\setminus \mathcal{Z}$

(iii):

if $\theta \in \hbox \mathrm{ext}\,(\Sigma ^{{{\mathcal {Z}}}}({{\mathcal {S}}}))$ then $\theta \in \bigcup _{D\in {{\mathcal {W}}}} {\tilde{\Theta }}^{N}_{D}$.

Proof

To show (i) assume for a contradiction that $\theta \in \Sigma ^{{{\mathcal {W}}}}({{\mathcal {S}}})$ exists. Thus, $\theta (W)=0$ for any $W\in {{\mathcal {W}}}$ forces ${{\mathcal {B}}}:=\{ S\subset N\,:\ S\ne \emptyset ~ \& ~ \theta (S)\ne 0\,\}\subseteq {{\mathcal {S}}}$. Since, however, every $\eta \in \bigcup _{D\in {{\mathcal {W}}}} {\tilde{\Theta }}^{N}_{D}$ satisfies $\eta (S)\le 0$ for any $S\in {{\mathcal {S}}}$, one has $\theta ^{\prime }(S)\le 0$ for any $S\in {{\mathcal {S}}}$ and $\theta ^{\prime }\in \Sigma ({{\mathcal {S}}})$. Hence, $\theta (S)\le 0$ for any $S\in {{\mathcal {S}}}$ and for our vector $\theta \in \Sigma ^{{{\mathcal {W}}}}({{\mathcal {S}}})$. The fact that $\theta \in \Sigma ({{\mathcal {S}}})$ implies the equality constraints $\theta (N)+\theta (\emptyset )=0$, $\sum _{L\subseteq N} \theta (L)=0$, and $\sum _{L\subseteq N:\,i\in L} \theta (L)=0$ for $i\in N$ (use Definition 5). The relations imply $\sum _{S\in {{\mathcal {S}}}} \theta (S)\cdot \chi _{S}+\theta (N)\cdot \chi _{N} =\underline{0}$, and, using $\sum _{L:\emptyset \ne L\subset N} \theta (L)=-1$, also $\theta (N)>0$. Therefore $\theta (N)\cdot \chi _{N}=\sum _{S\in {{\mathcal {S}}}} -\theta (S)\cdot \chi _{S}$ is a conic combination, which allows one to observe that ${{\mathcal {B}}}$ is balanced on N. The minimality of ${{\mathcal {S}}}$ then implies that ${{\mathcal {B}}}={{\mathcal {S}}}$, contradicting the assumption that ${{\mathcal {S}}}$ is not balanced on N.

To show (ii) assume the existence of $\eta \in \Delta ^{{{\mathcal {Z}}}}({{\mathcal {S}}})$. By (i) one has ${{\mathcal {W}}}{\setminus }{{\mathcal {Z}}}\ne \emptyset$. If there is $D\in {{\mathcal {W}}}{\setminus }{{\mathcal {Z}}}$ with $\eta (D)=0$ then $\eta \in \Delta ^{{{\mathcal {Z}}}\cup \{D\}}({{\mathcal {S}}})$ and we are done. Thus, assume that $\eta (W)\ne 0$ for any $W\in {{\mathcal {W}}}{\setminus }{{\mathcal {Z}}}$, which implies, by definition of $\Delta ({{\mathcal {S}}})$, that $\eta (W)> 0$ for any $W\in {{\mathcal {W}}}{\setminus }{{\mathcal {Z}}}$. The second assumption in (ii) means that there exists $\theta \in \Sigma ^{{{\mathcal {Z}}}}({{\mathcal {S}}}){\setminus } \Delta ^{{{\mathcal {Z}}}}({{\mathcal {S}}})$; thus, necessarily $\eta \ne \theta$. Put $\theta ^{\alpha }:=(1-\alpha )\cdot \eta +\alpha \cdot \theta$ for $0\le \alpha \le 1$; the convexity of $\Sigma ^{{{\mathcal {Z}}}}({{\mathcal {S}}})$ gives $\theta ^{\alpha }\in \Sigma ^{{{\mathcal {Z}}}}({{\mathcal {S}}})$. Take $\beta :=\max \, \{\,\alpha \ge 0\,:\ \theta ^{\alpha }\in \Delta ^{{{\mathcal {Z}}}}({{\mathcal {S}}})\,\}$ and $\theta ^{\prime }:=\theta ^{\beta }$. Note that there exists $D\in {{\mathcal {W}}}{\setminus }{{\mathcal {Z}}}$ with $\theta ^{\prime }(D)=\theta ^{\beta }(D)=0$ as otherwise one has $\theta ^{\beta }(W)>0$ for any $W\in {{\mathcal {W}}}{\setminus }{{\mathcal {Z}}}$, which contradicts the maximality of $\beta$. Thus, the facts $\theta ^{\prime }\in \Delta ^{{{\mathcal {Z}}}}({{\mathcal {S}}})$ and $\theta ^{\prime }(D)=0$ imply together that $\theta ^{\prime }\in \Delta ^{{{\mathcal {Z}}}\cup \{D\}}({{\mathcal {S}}})$.

The condition (iii) can be verified by induction on $|{{\mathcal {Z}}}|$. In case ${{\mathcal {Z}}}=\emptyset$ this follows directly from the definition of $\Sigma ({{\mathcal {S}}})$ using basic facts from polyhedral geometry recalled in Sect. 2.2. To verify the induction step assume that the claim is true for some ${{\mathcal {Z}}}\subset {{\mathcal {W}}}$, take $Z\in {{\mathcal {W}}}{\setminus }{{\mathcal {Z}}}$ and evidence the claim for ${{\mathcal {Z}}}\cup \{Z\}$. Note that the polytope $\mathsf{Q}:=\Sigma ^{{{\mathcal {Z}}}\cup \{Z\}}({{\mathcal {S}}})$ is the intersection of the polytope $\mathsf{P}:=\Sigma ^{{{\mathcal {Z}}}}({{\mathcal {S}}})$ with the hyperplane $\mathsf{H}:=\{\,\theta \,:\ \theta (Z)=0\}$. The characterization of vertices of $\mathsf{Q}=\mathsf{P}\cap \mathsf{H}$ recalled in Sect. 2.2 says that any vertex $\theta \in \hbox \mathrm{ext}\,(\mathsf{Q})$ is either a vertex of $\mathsf{P}$, in which case one has $\theta \in \bigcup _{D\in {{\mathcal {W}}}} {\tilde{\Theta }}^{N}_{D}$ by the induction assumption, or there is an edge $[\eta ,\sigma ]$ of $\mathsf{P}$ such that $\theta \in \,]\eta ,\sigma [$ and $[\eta ,\sigma ]\cap \mathsf{H}=\{\theta \}$. Since $\eta ,\sigma \in \hbox \mathrm{ext}\,(\mathsf{P})$, by the induction hypothesis, one has $\eta ,\sigma \in \bigcup _{D\in {{\mathcal {W}}}} {\tilde{\Theta }}^{N}_{D}$. As $[\eta ,\sigma ]\cap \mathsf{H}=\{\theta \}$ one has $\eta (Z)\ne 0\ne \sigma (Z)$ and can assume without loss of generality that $\eta (Z)>0$ and $\sigma (Z)<0$. This forces that $\eta \in {\tilde{\Theta }}^{N}_{Z}$; assume that $\sigma \in {\tilde{\Theta }}^{N}_{E}$ for some $E\in {{\mathcal {W}}}$ (possibly $E=Z$). It makes no problem to observe that these facts imply that $\theta \in {\tilde{\Theta }}^{N}_{E}$ (see Definition 5). Hence, $\theta \in \bigcup _{D\in {{\mathcal {W}}}} {\tilde{\Theta }}^{N}_{D}$ and the induction step has been verified. $\square$

For reader’s convenience we recall what is claimed.

Lemma 17    Let ${{\mathcal {S}}}$ be a purely min-semi-balanced system on N, $|N|\ge 3$, with exceptional set $T\in {{\mathcal {S}}}$ and ${{\mathcal {W}}}:= \mathcal{P}(N){\setminus } (\{\emptyset ,N\}\cup {{\mathcal {S}}})$. Then the following conditions are equivalent:

(a):

$\theta _{{{\mathcal {S}}}}\not \in \hbox \mathrm{ext}\,(\Delta )$, (see Definition 5)

(b):

there exists a convex combination $\theta _{{{\mathcal {S}}}}=\sum _{D\in {{\mathcal {W}}}\cup \{T\}} \alpha _{D}\cdot \theta ^{D}$ where $\alpha _{T}<1$ and $\theta ^{D}\in {\tilde{\Theta }}^{N}_{D}$ whenever $\alpha _{D}>0$,

(c):

the set $\Delta ({{\mathcal {S}}}):=\hbox \mathrm{conv}\,(\bigcup _{D\in {{\mathcal {W}}}} {\tilde{\Theta }}^{N}_{D}) \cap \{\,\theta \in {\mathbb R}^{\mathcal{P}(N)}\,:\ \theta (W)\ge 0 ~\hbox {for }W\in {{\mathcal {W}}}\,\}$ is non-empty,

(d):

there exists $E\in {{\mathcal {W}}}$ such that E is exceptional in ${{\mathcal {S}}}\cup \{E\}$,

(e):

there exists $E\in {{\mathcal {W}}}$ such that E is exceptional in $({{\mathcal {S}}}{\setminus }\{T\})\cup \{E\}$,

(f):

a min-semi-balanced system ${{\mathcal {D}}}$ on N exists such that ${{\mathcal {D}}}{\setminus }{{\mathcal {S}}}=\{E\}$ for some $E\in {{\mathcal {W}}}$, the set E is exceptional within ${{\mathcal {D}}}$, and $T\not \in {{\mathcal {D}}}$,

(g):

a min-semi-balanced system ${{\mathcal {D}}}$ on N exists with an exceptional set E such that $\mathcal{D}\setminus \mathcal{S}= \{E\}$.

Proof

To show (a)$\Rightarrow$(b) realize that $\theta _{{{\mathcal {S}}}}\in {\tilde{\Theta }}^{N}_{T}\subseteq \Delta$ and assume that $\theta _{{{\mathcal {S}}}}\in \,]\eta ,\sigma [$ for $\eta ,\sigma \in \Delta$, $\eta \ne \theta _{{{\mathcal {S}}}}\ne \sigma$. By the definition of $\Delta$ a finite convex combination yielding $\eta$ exists: $\eta =\sum _{j\in J} \beta _{j}\cdot \eta _{j}$, where $\beta _{j}>0$ and $\eta _{j}\in \bigcup _{\emptyset \ne D\subseteq N} {\tilde{\Theta }}^{N}_{D}$ for $j\in J$. Since $\eta \ne \theta _{{{\mathcal {S}}}}$, there exists $j\in J$ with $\eta _{j}\ne \theta _{{{\mathcal {S}}}}$. An analogous convex combination exists for $\sigma$ and by combining them observe that a finite convex combination yielding $\theta _{{{\mathcal {S}}}}$ exists:

$$\begin{aligned} \theta _{{{\mathcal {S}}}}=\sum _{i\in I} \alpha _{i}\cdot \theta _{i},\qquad \hbox {where }\alpha _{i}>0\; \hbox {and}\; \theta _{i}\in \bigcup _{\emptyset \ne D\subseteq N} {\tilde{\Theta }}^{N}_{D}\; \hbox {for all }\;i\in I, \end{aligned}$$

and, moreover, $\theta _{i}\ne \theta _{{{\mathcal {S}}}}$ for at least one $i\in I$. Assume without loss of generality that $\theta _{i}$, $i\in I$, differ from each other and, if there is $i\in I$ with $\theta _{i}=\theta _{{{\mathcal {S}}}}$ then, by easy modification, one gets such a convex combination yielding $\theta _{{{\mathcal {S}}}}$ where $\theta _{i}\ne \theta _{{{\mathcal {S}}}}$ for all $i\in I$.

Observe that one can even assume without loss of generality that $\theta _{i}(S)\le 0$ for any $i\in I$ and $S\in {{\mathcal {S}}}{\setminus }\{T\}$. Indeed, if there exists $j\in I$ and $S\in {{\mathcal {S}}}{\setminus }\{T\}$ with $\theta _{j}(S)>0$ then the convex combination can be replaced by another convex combination $\theta _{{{\mathcal {S}}}}={\hat{\alpha }}_{j}\cdot {\hat{\theta }}_{j}+\sum _{i\in I{\setminus } \{j\}} {\hat{\alpha }}_{i}\cdot \theta _{i}$, where ${\hat{\theta }}_{j}(S)=0$. To observe that realize that $\alpha _{j}<1$, as otherwise $\theta _{j}=\theta _{{{\mathcal {S}}}}$. Thus, one can put $\alpha := 1-\alpha _{j}=\sum _{i\in I{\setminus }\{j\}} \alpha _{i}>0$ and

$$\begin{aligned} \theta ~:=~ \sum _{i\in I{\setminus }\{j\}} \frac{\alpha _{i}}{\alpha }\cdot \theta _{i}\,,\quad \hbox {which gives}\; \theta _{{{\mathcal {S}}}}=\alpha _{j}\cdot \theta _{j}+ \alpha \cdot \theta , \hbox {that is}, \theta _{{{\mathcal {S}}}}\in \,]\theta _{j},\theta [. \end{aligned}$$

On the other hand, since $\theta _{j}(S)>0$ and $\theta _{{{\mathcal {S}}}}(S)<0$ there exists unique ${\hat{\theta }}_{j}\in \,]\theta _{j},\theta _{{{\mathcal {S}}}}[$ such that ${\hat{\theta }}_{j}(S)=0$. Note that necessarily $\theta _{j}\in {\tilde{\Theta }}^{N}_{S}$ while $\theta _{{{\mathcal {S}}}}\in {\tilde{\Theta }}^{N}_{T}$ which facts together allow one to observe that ${\hat{\theta }}_{j}\in {\tilde{\Theta }}^{N}_{T}$. The vectors $\theta _{j},{\hat{\theta }}_{j},\theta _{{{\mathcal {S}}}}$ and $\theta$ are on the same line, which implies that $\theta _{{{\mathcal {S}}}}\in \,]{\hat{\theta }}_{j},\theta [$. Thus, $0<\gamma <1$ exists with $\theta _{{{\mathcal {S}}}}=\gamma \cdot {\hat{\theta }}_{j}+ (1-\gamma )\cdot \theta$ and the substitution gives

$$\begin{aligned} \theta _{{{\mathcal {S}}}} ~=~ \gamma \cdot {\hat{\theta }}_{j} + \sum _{i\in I{\setminus }\{j\}} \underbrace{\frac{(1-\gamma )\cdot \alpha _{i}}{\alpha }}_{{\hat{\alpha }}_{i}}\,\cdot \, \theta _{i}\,,\qquad \hbox {where it suffices to put}\; {\hat{\alpha }}_{j}:=\gamma . \end{aligned}$$

For any $i\in I$, one has $\theta _{i}\in {\tilde{\Theta }}^{N}_{E}$ for some $\emptyset \ne E\subseteq N$. In case $E\in {{\mathcal {S}}}{\setminus }\{T\}$ the above inequalities $\theta _{i}(S)\le 0$ for $S\in {{\mathcal {S}}}{\setminus }\{T\}$ give $\theta _{i}\in {\tilde{\Theta }}^{N}_{N}$ and (see Definition 5) the inclusion ${\tilde{\Theta }}^{N}_{N}\subseteq {\tilde{\Theta }}^{N}_{T}$ allows one to conclude that $\theta _{i}\in {\tilde{\Theta }}^{N}_{D}$ for some $D\in {{\mathcal {W}}}\cup \{T\}$. To summarize that: there exists a convex combination $\theta _{{{\mathcal {S}}}}=\sum _{i\in I} \alpha _{i}\cdot \theta _{i}$, where $\alpha _{i}>0$, $\theta _{i}\ne \theta _{{{\mathcal {S}}}}$ and $\theta _{i}\in \bigcup _{D\in {{\mathcal {W}}}\cup \{T\}} {\tilde{\Theta }}^{N}_{D}$ for all $i\in I$.

Further observation is that there exists $j\in I$ and $W\in {{\mathcal {W}}}$ such that $\theta _{j}(W)>0$. To show that assume for a contradiction the converse, that is, $\theta _{i}(W)\le 0$ for any $i\in I$ and $W\in {{\mathcal {W}}}$. That basically means, that, for any $i\in I$, the condition $\theta _{i}\in {\tilde{\Theta }}^{N}_{W}$ implies $\theta _{i}\in {\tilde{\Theta }}^{N}_{N}\subseteq {\tilde{\Theta }}^{N}_{T}$. In particular, one would have $\theta _{i}\in {\tilde{\Theta }}^{N}_{T}$, $\theta _{i}\ne \theta _{{{\mathcal {S}}}}$ for any $i\in I$, which contradicts the fact $\theta _{{{\mathcal {S}}}}\in \hbox \mathrm{ext}\,({\tilde{\Theta }}^{N}_{T})$ claimed by (the second claim in) Lemma 8.

Thus, any $i\in I$ can be assigned to some $D\in {{\mathcal {W}}}\cup \{T\}$ such that $\theta _{i}\in {\tilde{\Theta }}^{N}_{D}$; let us fix that choice and write $i\mapsto D$ to denote that. We already know that there exists $j\in I$ and $W\in {{\mathcal {W}}}$ with $\theta _{j}(W)>0$, which necessitates $j\mapsto W$. For any $D\in {{\mathcal {W}}}\cup \{T\}$ we put

$$\begin{aligned} \alpha _{D} ~:=~ \sum _{i\in I:\, i\,\mapsto D} \alpha _{i} \qquad \hbox {and, if }\alpha _{D}>0,\quad \theta ^{D} ~:=~ \sum _{i\in I:\, i\,\mapsto D} \frac{\alpha _{i}}{\alpha _{D}}\cdot \theta _{i}\in {\tilde{\Theta }}^{N}_{D}\,. \end{aligned}$$

Thus, one has $\theta _{{{\mathcal {S}}}}=\sum _{D\in {{\mathcal {W}}}\cup \{T\}} \alpha _{D}\cdot \theta ^{D}$, where $\theta ^{D}$ can be chosen arbitrarily in case $\alpha _{D}=0$. It is a convex combination and the existence of $j\in I$ and $W\in {{\mathcal {W}}}$ with $j\mapsto W$ implies $\alpha _{W}>0$. Hence, $\alpha _{T}<1$, which gives the condition (b).

To show (b)$\Rightarrow$(c) consider a convex combination $\theta _{{{\mathcal {S}}}}=\sum _{D\in {{\mathcal {W}}}\cup \{T\}} \alpha _{D}\cdot \theta ^{D}$ where $\alpha _{T}<1$ and $\theta ^{D}\in {\tilde{\Theta }}^{N}_{D}$ whenever $\alpha _{D}>0$. The fact $\theta _{{{\mathcal {S}}}}(T)>0$ forces both $\alpha _{T}>0$ and $\theta ^{T}(T)>0$ because $\eta (T)\le 0$ for any $\eta \in \bigcup _{D\in {{\mathcal {W}}}} {\tilde{\Theta }}^{N}_{D}$. Thus, one has some $\theta ^{T}\in {\tilde{\Theta }}^{N}_{T}$. Let us put

$$\begin{aligned} \theta ~:=~ \sum _{D\in {{\mathcal {W}}}}\,\, \frac{\alpha _{D}}{1-\alpha _{T}}\,\cdot \, \theta ^{D}\qquad \hbox {and observe that}\,\, \theta _{{{\mathcal {S}}}}=(1-\alpha _{T})\cdot \theta + \alpha _{T}\cdot \theta ^{T}. \end{aligned}$$

By definition, $\theta \in \hbox \mathrm{conv}\,(\bigcup _{D\in {{\mathcal {W}}}} {\tilde{\Theta }}^{N}_{D})$. The fact $\theta _{{{\mathcal {S}}}}\in \,]\theta ,\theta ^{T}[$ together with $\theta _{{{\mathcal {S}}}}(W)=0$ and $\theta ^{T}(W)\le 0$ for any $W\in {{\mathcal {W}}}$ forces $\theta (W)\ge 0$ for any such W. In particular, $\theta \in \Delta ({{\mathcal {S}}})$ and the condition (c) has been verified.

To show (c)$\Rightarrow$(d) we use Lemma 20; the condition (c) means, by notation from Lemma 20, that $\Delta ^{{{\mathcal {Z}}}}({{\mathcal {S}}})\ne \emptyset$ for empty ${{\mathcal {Z}}}=\emptyset$. By inductive application of Lemma 20(ii) we find ${{\mathcal {Z}}}\subseteq {{\mathcal {W}}}$ with $\Sigma ^{{{\mathcal {Z}}}}({{\mathcal {S}}})=\Delta ^{{{\mathcal {Z}}}}({{\mathcal {S}}})\ne \emptyset$; indeed, by Lemma 20(i) one has $\Sigma ^{{{\mathcal {W}}}}({{\mathcal {S}}})=\emptyset$, which ensures that inductive enlarging of ${{\mathcal {Z}}}$ has to finish with some desired ${{\mathcal {Z}}}\subset {{\mathcal {W}}}$. Thus, $\emptyset \ne \hbox \mathrm{ext}\,(\Sigma ^{{{\mathcal {Z}}}}({{\mathcal {S}}}))\subseteq \Delta ^{{{\mathcal {Z}}}}({{\mathcal {S}}})$ for some ${{\mathcal {Z}}}\subset {{\mathcal {W}}}$. One can take some $\theta \in \hbox \mathrm{ext}\,(\Sigma ^{{{\mathcal {Z}}}}({{\mathcal {S}}}))$ and, by Lemma 20(iii), there exists $E\in {{\mathcal {W}}}$ such that $\theta \in {\tilde{\Theta }}^{N}_{E}$. Hence, $\theta (W)\le 0$ for any $W\in {{\mathcal {W}}}{\setminus }\{E\}$ while $\theta \in \Delta ({{\mathcal {S}}})$ says $\theta (W)\ge 0$ for any $W\in {{\mathcal {W}}}$. That together means $\theta (W)=0$ for any $W\in {{\mathcal {W}}}{\setminus }\{E\}$ and $\theta (E)\ge 0$. One cannot have $\theta (E)=0$ for otherwise $\theta \in \Sigma ^{{{\mathcal {W}}}}({{\mathcal {S}}})$ contradicts the claim in Lemma 20(i). Thus, necessarily $\theta (E)>0$; note also that $\theta (S)\le 0$ for any $S\in {{\mathcal {S}}}$. The equality constraints $\sum _{L\subseteq N:\,i\in N} \theta (L)=0$ for any $i\in N$ and $\theta \in {\tilde{\Theta }}^{N}_{E}$ allow one to conclude that $\sum _{S\in {{\mathcal {S}}}\cup \{E\}} -\theta (S)\cdot \chi _{S}=\theta (N)\cdot \chi _{N}$ is a semi-conic combination yielding a constant vector in ${\mathbb R}^{N}$. Thus, by definition, E is exceptional in ${{\mathcal {S}}}\cup \{E\}$ and the condition (d) has been verified.

To show (d)$\Rightarrow$(e) assume that $E\in {{\mathcal {W}}}$ is exceptional within ${{\mathcal {S}}}\cup \{E\}$. This means that there exists a linear combination $\sum _{S\in {{\mathcal {S}}}\cup \{E\}} \nu _{S}\cdot \chi _{S}$ yielding a constant vector in ${\mathbb R}^{N}$ where $\nu _{E}<0$ and $\nu _{S}\ge 0$ for $S\in {{\mathcal {S}}}$. Observe that E is exceptional within a smaller set system $({{\mathcal {S}}}{\setminus }\{T\})\cup \{E\}$. In case $\nu _{T}=0$ we are done. In case $\nu _{T}>0$ we apply Lemma 2(d) to ${{\mathcal {S}}}$ which says that there exists (unique) affine combination $\sum _{S\in {{\mathcal {S}}}} \uplambda _{S}\cdot \chi _{S}$ yielding a constant vector in ${\mathbb R}^{N}$ with both $\uplambda _{T}<0$ and $\uplambda _{S}> 0$ for $S\in {{\mathcal {S}}}{\setminus }\{T\}$. Then we put $\uplambda _{E}:=0$ and

$$\begin{aligned} \kappa _{S} ~:=~ \frac{\nu _{T}}{\nu _{T}-\uplambda _{T}}\cdot \uplambda _{S} + \frac{-\uplambda _{T}}{\nu _{T}-\uplambda _{T}}\cdot \nu _{S} \qquad \hbox {for}\; S\in {{\mathcal {S}}}\cup \{E\}. \end{aligned}$$

Hence, $\sum _{S\in {{\mathcal {S}}}\cup \{E\}} \kappa _{S}\cdot \chi _{S}$ yields a constant vector in ${\mathbb R}^{N}$, $\kappa _{E}<0$, and $\kappa _{T}=0$. Thus, $({{\mathcal {S}}}{\setminus }\{T\})\cup \{E\}$ is a semi-balanced system on N and E is an exceptional set within it.

To show (e)$\Rightarrow$(f) we fix a semi-conic combination $\sum _{S\in {{\mathcal {S}}}\cup \{E\}} \kappa _{S}\cdot \chi _{S}$ yielding a constant vector in ${\mathbb R}^{N}$ with $\kappa _{E}<0$ and $\kappa _{T}=0$. In particular, one can choose a min-semi-balanced system ${{\mathcal {D}}}\subseteq ({{\mathcal {S}}}{\setminus }\{T\})\cup \{E\}$. One has $E\in {{\mathcal {D}}}$ as otherwise ${{\mathcal {D}}}\subset {{\mathcal {S}}}$ contradicts the minimality of ${{\mathcal {S}}}$. By Lemma 2(d), there exists (unique) affine combination $\sum _{S\in {{\mathcal {D}}}} \sigma _{S}\cdot \chi _{S}$ yielding a constant vector in ${\mathbb R}^{N}$ which is semi-conic and has all coefficients non-zero. Assume for a contradiction that $\sigma _{E}>0$. Then we put $\sigma _{S}:=0$ for $S\in {{\mathcal {S}}}{\setminus }{{\mathcal {D}}}$ and

$$\begin{aligned} \mu _{S} ~:=~ \frac{\sigma _{E}}{\sigma _{E}-\kappa _{E}}\cdot \kappa _{S} + \frac{-\kappa _{E}}{\sigma _{E}-\kappa _{E}}\cdot \sigma _{S} \qquad \hbox {for}~ S\in {{\mathcal {S}}}\cup \{E\}. \end{aligned}$$

Hence, $\sum _{S\in {{\mathcal {S}}}\cup \{E\}} \mu _{S}\cdot \chi _{S}$ yields a constant vector in ${\mathbb R}^{N}$ and $\mu _{E}=0=\mu _{T}$. Since it is a semi-conic combination ${{\mathcal {T}}}:=\{ S\in {{\mathcal {S}}}\,:\ \mu _{S}\ne 0\}\subset {{\mathcal {S}}}$ is a semi-balanced system on N, which contradicts the minimality of ${{\mathcal {S}}}$. As $\sigma _{E}\ne 0$, one necessarily has $\sigma _{E}<0$ and the set E is exceptional within ${{\mathcal {D}}}$. Thus, the condition (f) has been verified.

The implication (f)$\Rightarrow$(g) is evident.

To show (g)$\Rightarrow$(a) assume that ${{\mathcal {D}}}$ is the min-semi-balanced system with an exceptional set $E\in {{\mathcal {W}}}$ such that ${{\mathcal {D}}}{\setminus }{{\mathcal {S}}}=\{E\}$. Note that, by (1), one has both $\theta _{{{\mathcal {D}}}}(E)>0$ and $\theta _{{{\mathcal {D}}}}(W)=0$ for any $W\in {{\mathcal {W}}}{\setminus }\{E\}$. Then we put

$$\begin{aligned} \theta ^{\varepsilon } ~:=~ (1+\varepsilon )\cdot \theta _{{{\mathcal {S}}}} + (-\varepsilon )\cdot \theta _{{{\mathcal {D}}}} ~=~ \theta _{{{\mathcal {S}}}} +\varepsilon \cdot (\theta _{{{\mathcal {S}}}}-\theta _{{{\mathcal {D}}}})\qquad \hbox {for every }\varepsilon \ge 0. \end{aligned}$$

It makes no problem to observe that, for small $\varepsilon >0$, one has $\theta ^{\varepsilon }\in {\tilde{\Theta }}^{N}_{T}\subseteq \Delta$. Because of $\theta _{{{\mathcal {S}}}}\in \, ]\theta _{{{\mathcal {D}}}},\theta ^{\varepsilon }[$ and $\theta _{{{\mathcal {D}}}}\in {\tilde{\Theta }}^{N}_{E}\subseteq \Delta$ one gets $\theta _{{{\mathcal {S}}}}\not \in \hbox \mathrm{ext}\,(\Delta )$ and (a) has been verified. $\square$