satisfying the conditions in part (ii) of Lemma 4 and prove that
$\boldsymbol{r}\in{\cal S}_{0}$. This in particular implies that

$\displaystyle\min_{j\notin{\cal C}}(u_{j}+r_{j})\geq\max_{i\in{\cal
C}}(u_{i}+r_{i})\geq 0.$ (53)

For any decisive coalition $\tilde{\cal C}\in{\cal D}^{S}$, consider the
partition of $\tilde{\cal C}$ defined by $\tilde{\cal C}={\cal C}_{1}\cup{\cal
C}_{2}\cup{\cal C}_{3}$. Members of the coalition ${\cal C}_{1}$ belong both
to ${\cal C}$ and $\tilde{\cal C}$, that is, ${\cal C}_{1}:={\cal
C}\cap\tilde{\cal C}$. The coalition ${\cal C}_{2}$ is a subset of
$\tilde{\cal C}\backslash{\cal C}$ such that when merged with the coalition
${\cal C}_{1}$, it forms a coalition with cardinality $\widehat{\kappa}$, that
is, $|{\cal C}_{1}|+|{\cal C}_{2}|=\widehat{\kappa}$. Finally, the coalition
${\cal C}_{3}$ formed by the residual members of $\tilde{\cal C}$ who do not
belong to ${\cal C}_{1}$ or ${\cal C}_{2}$, that is, ${\cal
C}_{3}:=\tilde{\cal C}\backslash({\cal C}_{1}\cup{\cal C}_{2})$. Note that
$u_{i}+r_{i}\leq u_{j}+r_{j}$ for all $i\in{\cal C}\backslash{\cal C}_{1}$ and
$j\in{\cal C}_{2}$ and that, $|{\cal C}\backslash{\cal C}_{1}|=|{\cal
C}_{2}|$. Thus, the inequality (53) implies

$\displaystyle\sum_{j\in{\cal C}_{2}}(u_{j}+r_{j})\geq\sum_{i\in{\cal
C}\backslash{\cal C}_{1}}(u_{i}+r_{i}).$ (54)

Since ${\cal C}_{3}\cap{\cal C}=\emptyset$, we have
$u_{k}+r_{k}\geq\max_{i\in{\cal C}}(u_{i}+r_{i})\geq 0$ for all $k\in{\cal
C}_{3}$, and hence $\sum_{k\in{\cal C}_{3}}(u_{k}+r_{k})\geq 0$. Using this
last inequality and (54) yields

$\displaystyle\sum_{i\in\tilde{\cal C}}(u_{i}+r_{i})$ $\displaystyle=$
$\displaystyle\sum_{i\in{\cal C}_{1}}(u_{i}+r_{i})+\sum_{j\in{\cal
C}_{2}}(u_{j}+r_{j})+\sum_{k\in{\cal C}_{3}}(u_{k}+r_{k})$ $\displaystyle\geq$
$\displaystyle\sum_{i\in{\cal C}_{1}}(u_{i}+r_{i})+\sum_{i\in{\cal
C}\backslash{\cal C}_{1}}(u_{i}+r_{i})+0=\sum_{i\in{\cal C}}(u_{i}+r_{i})\geq
0.$

Now it follows from Proposition 1 again that ${\boldsymbol{r}}\in{\cal
S}_{0}$. ∎

We now prove Proposition 5. We first show that ${\cal T}_{\bm{r}}\geq U^{S}$
for any stable promises profile $\boldsymbol{r}\in{\cal S}_{0}$. Indeed, since
$|{\cal C}^{R}|<\kappa$, we have ${\cal C}^{S}\in{\cal D}^{S}$ so Proposition
1 implies then that $\sum_{{\cal C}^{S}}(u_{i}+r_{i})\geq 0$. Therefore,

$\displaystyle\sum_{i\in{\cal C}^{S}}|r_{i}|\geq\sum_{i\in{\cal
C}^{S}}r_{i}\geq\sum_{i\in{\cal C}^{S}}(-u_{i})=U^{S}.$ (55)

Then it follows from Lemma 4 (i) that

$\displaystyle{\cal T}_{\bm{r}}\geq|\sum_{i\in{\cal C}^{S}}r_{i}|\geq
U^{S},\mbox{ for all }\boldsymbol{r}\in{\cal S}_{0}.$ (56)

Step $1$: We start by considering the case where $|{\cal C}^{R}|<\kappa-1$,
and cover first the ”if” direction ($\Leftarrow$), then the ”only if”
direction ($\Rightarrow$).

If a promises profile $\boldsymbol{r}$ satisfies conditions $1$ and $2.a$ from
Proposition 5, then it can be directly checked that $\boldsymbol{r}\in{\cal
P}$ and ${\cal T}_{\boldsymbol{r}}=U^{S}$. Moreover, by construction we see
that $u_{i}+r_{i}=0$ for $i\in{\cal C}^{S}$ and $u_{j}+r_{j}\geq 0$ for
$j\in{\cal C}^{R}$. Therefore, $\sum_{i\in{\cal C}}(u_{i}+r_{i})\geq 0$ for
all coalitions ${\cal C}\in{\cal D}^{S}$ and hence, by Proposition 1 the
promises profile $\boldsymbol{r}$ is stable, that is $\boldsymbol{r}\in{\cal
S}_{0}$. Since ${\cal T}_{\boldsymbol{r}}=U^{S}$, inequality (56) shows that
the promises profile $\boldsymbol{r}$ achieves the minimum total promises
transfer, and thus $\boldsymbol{r}\in{\cal E}$.181818In particular, we note
that the following promises profile $\boldsymbol{r}$ satisfies $1$ and $2.a$
from Proposition 5, and hence is an equilibrium with minimum total promises
transfer:  $\displaystyle r_{i}:=-u_{i},~{}i\in{\cal C}^{S};\qquad
r_{j}:=-{U^{S}\over U^{R}}u_{j},~{}j\in{\cal C}^{R}.$ (57)

We now prove the only if part. That is, we assume $\boldsymbol{r}\in{\cal E}$
and prove that conditions $1$ and $2.a$ from Proposition 5 hold. Note that

$\displaystyle\sum_{i\in{\cal C}^{R}}|r_{i}|\geq\sum_{i\in{\cal
C}^{R}}(-r_{i})=\sum_{i\in{\cal C}^{S}}r_{i}\geq U^{S},$ (58)

and combining (55) with (58) gives

$\displaystyle{\cal T}_{\bm{r}}=\frac{1}{2}\sum_{i\in{\cal
C}^{R}}|r_{i}|+\frac{1}{2}\sum_{i\in{\cal
C}^{S}}|r_{i}|\geq\frac{1}{2}\sum_{i\in{\cal
C}^{R}}(-r_{i})+\frac{1}{2}\sum_{i\in{\cal C}^{S}}(r_{i})\geq U^{S}.$ (59)

Since $\boldsymbol{r}\in{\cal E}$ has minimum total promises transfer, we must
have ${\cal T}_{\bm{r}}=U^{S}$. Therefore, inequalities (59) become
equalities, that is,

$\displaystyle\displaystyle U^{S}={\cal T}_{\bm{r}}=\frac{1}{2}\sum_{i\in{\cal
C}^{R}}|r_{i}|+\frac{1}{2}\sum_{i\in{\cal
C}^{S}}|r_{i}|=\frac{1}{2}\sum_{i\in{\cal
C}^{R}}(-r_{i})+\frac{1}{2}\sum_{i\in{\cal C}^{S}}(r_{i})=U^{S}.$

This in turn implies that inequalities (55) and (58) are also equalities:

$\displaystyle\sum_{i\in{\cal C}^{S}}|r_{i}|=\sum_{i\in{\cal
C}^{S}}r_{i}=U^{S},\quad\sum_{i\in{\cal C}^{R}}|r_{i}|=\sum_{i\in{\cal
C}^{R}}(-r_{i})=U^{S}.$

Then

$\displaystyle|r_{i}|=r_{i},~{}i\in{\cal
C}^{S};\quad|r_{i}|=-r_{i},~{}i\in{\cal
C}^{R};\quad\mbox{and}\quad\sum_{i\in{\cal C}^{S}}r_{i}=\sum_{i\in{\cal
C}^{R}}|r_{i}|=U^{S},$

and thus

$\displaystyle r_{i}\geq 0\geq r_{j},\quad\forall i\in{\cal C}^{S},j\in{\cal
C}^{R},\quad\mbox{and}\quad\sum_{i\in{\cal C}^{S}}r_{i}=U^{S}=\sum_{j\in{\cal
C}^{R}}(-r_{j}).$ (60)

Moreover, the equality $\sum_{i\in{\cal C}^{S}}r_{i}=U^{S}$ implies that
$\sum_{i\in{\cal C}^{S}}(u_{i}+r_{i})=0$.

Since $|{\cal C}^{S}|>\widehat{\kappa}$, then for any $i_{0}\in{\cal C}^{S}$,
${\cal C}^{S}\backslash\\{i_{0}\\}\in{\cal D}^{S}$, thus by Proposition 1 we
have $\sum_{i\in{\cal C}^{S}\backslash\\{i_{0}\\}}(u_{i}+r_{i})\geq 0$. This,
together with $\sum_{i\in{\cal C}^{S}}(u_{i}+r_{i})=0$, implies that
$u_{i_{0}}+r_{i_{0}}\leq 0$, for all $i_{0}\in{\cal C}^{S}$. Combining this
with the equality $\sum_{i\in{\cal C}^{S}}(u_{i}+r_{i})=0$, we must have
$u_{i}+r_{i}=0$ or $r_{i}=-u_{i}$ for all $i\in{\cal C}^{S}$.

Finally, for each $j\in{\cal C}^{R}$, since ${\cal C}^{S}\cup\\{j\\}\in{\cal
D}^{S}$, then by Proposition 1 we have $0\leq\sum_{i\in{\cal
C}^{S}\cup\\{j\\}}(u_{i}+r_{i})=u_{j}+r_{j}$. That is, $r_{j}\geq-u_{j}$ for
all $j\in{\cal C}^{R}$. To summarize, we have proven that: in addition to the
relations (60), $r_{i}=-u_{i}$ for all $i\in{\cal C}^{S}$ and $-u_{j}\leq
r_{j}\leq 0$ for all $j\in{\cal C}^{R}$. Thus, conditions $1$ and $2.a$ from
Proposition 5 hold. This concludes the proof for the case $|{\cal
C}^{R}|<\kappa-1$.

Step $2$: We now consider the case $|{\cal C}^{R}|=\kappa-1$. Notice that in
this case, $|{\cal C}^{S}|=\widehat{\kappa}$. First, if $\boldsymbol{r}$
satisfies conditions $1$ and $2.b$ of Proposition 5, then it can be checked as
in Step 1 that $\boldsymbol{r}\in{\cal P}$, ${\cal T}_{\bm{r}}=U^{S}$,
$\sum_{i\in{\cal C}^{S}}(u_{i}+r_{i})=0$, and $r_{j}\geq-u_{j}$ for all
$j\in{\cal C}^{R}$. Moroever, by condition $2.b$ of Proposition 5 and property
(ii) of Lemma 4, we see that $r\in{\cal S}_{0}$. Since ${\cal
T}_{\bm{r}}=U^{S}$, the promises profile $\boldsymbol{r}$ minimizes the total
promises transfer and hence $\boldsymbol{r}\in{\cal E}$.191919In particular,
we note that the promises profile $\boldsymbol{r}$ constructed in (57)
satisfies conditions $1$ and $2.b$ of of Proposition 5 and hence is in ${\cal
E}$. In contrast to the case where $|{\cal C}^{S}|<\kappa-1$ where all members
of the coalition ${\cal C}^{S}$ were ex post indifferent between $R$ and $S$
after receiving the equilibrium promises, we note that in the case $|{\cal
C}^{R}|=\kappa-1$ it is possible that $u_{i}+r_{i}>0$ for some $i\in{\cal
C}^{S}$.

We now prove the only if part. We assume $\boldsymbol{r}\in{\cal E}$ and show
that conditions $1$ and $2.b$ of Proposition 5 hold true. By the same
arguments in Step $1$ of this proof, we see that (60) still holds true.
Moreover, for any $i_{0}\in{\cal C}^{S}$ and $j_{0}\in{\cal C}^{R}$, note that
$({\cal C}^{S}\backslash\\{i_{0}\\})\cup\\{j_{0}\\}\in{\cal D}^{S}$, then by
Proposition 1 we have

$\displaystyle 0\leq\sum_{i\in({\cal
C}^{S}\backslash\\{i_{0}\\})\cup\\{j_{0}\\}}(u_{i}+r_{i})$ $\displaystyle=$
$\displaystyle\sum_{i\in{\cal
C}^{S}}(u_{i}+r_{i})-(u_{i_{0}}+r_{i_{0}})+(u_{j_{0}}+r_{j_{0}})$
$\displaystyle=$ $\displaystyle(u_{j_{0}}+r_{j_{0}})-(u_{i_{0}}+r_{i_{0}}).$

Thus, $u_{i_{0}}+r_{i_{0}}\leq u_{j_{0}}+r_{j_{0}}$ for all $i_{0}\in{\cal
C}^{S}$ and $j_{0}\in{\cal C}^{R}$ and hence, condition $2.(b).ii$ of
Proposition 5 is satisfied. The remaining properties of condition $2.b$ from
Proposition 5 are implied by (60). Similarly with the exception of the
inequality $-u_{j}\leq r_{j}$ for $j\in{\cal C}^{R}$, all other properties in
Condition $1$ from Proposition 5 are implied by (60). To prove this last
property, note that if $\sum_{i\in{\cal C}^{S}}(u_{i}+r_{i})=0$, then there
exists $i_{0}\in{\cal C}^{S}$ such that $u_{i_{0}}+r_{i_{0}}\geq 0$. Condition
$2.(b).ii$ from Proposition 5 implies that $u_{j}+r_{j}\geq
u_{i_{0}}+r_{i_{0}}\geq 0$ for any $j\in{\cal C}^{R}$, and thus $r_{j}\geq-
u_{j}$, for all $j\in{\cal C}^{R}$. ∎

### Proof of Proposition 6:

Note that in this case $\sum_{i=1}^{\widehat{\kappa}}u_{i}=-G^{S}\geq 0$. It
can be verified that the conditions in Lemma 4 (ii) for ${\cal
C}=\\{1,\cdots,\widehat{\kappa}\\}$ and $\boldsymbol{r}=\boldsymbol{0}$ hold.
Then $\boldsymbol{0}\in{\cal S}_{0}$, and this stable promises profile is both
unique and has zero total transfers. Hence ${\cal E}=\\{\boldsymbol{0}\\}$. ∎

### Proof of Proposition 7:

The proof proceeds in four steps. In the first step, we show that ${\cal
T}_{\bm{r}}\geq G^{S}$. In the second step, we prove the ”if” part of
Proposition 7. In the third step, we give an example of across the aisle
equilibrium promise that satisfies ${\cal T}_{\boldsymbol{r}}=G^{S}$. The
example is used in the fourth and last step where we prove the ”only if” part
of Proposition 7.

Step 1. We first show that ${\cal T}_{\bm{r}}\geq G^{S}$ for any
$\boldsymbol{r}\in{\cal S}_{0}$. Since ${\cal C}^{S}\cup\underline{{\cal
C}}^{R}\in{\cal D}^{S}$, by Proposition 1 and (18) we have

$\displaystyle 0\leq\sum_{i\in{\cal C}^{S}\cup\underline{{\cal
C}}^{R}}[u_{i}+r_{i}]=-G^{S}+\sum_{i\in{\cal C}^{S}\cup\underline{{\cal
C}}^{R}}r_{i}.$ (61)

Then by Lemma 4 (i) we have ${\cal T}_{\bm{r}}\geq|\sum_{i\in{\cal
C}^{S}\cup\underline{{\cal C}}^{R}}r_{i}|\geq\sum_{i\in{\cal
C}^{S}\cup\underline{{\cal C}}^{R}}r_{i}\geq G^{S}$.

Step 2. We next prove the if part: assume conditions (20), (21) and (22) hold
and prove that $\boldsymbol{r}\in{\cal E}$. Let $\boldsymbol{r}\in{\cal P}$
satisfy (20), note that $r_{k}\geq-u_{k}$ or $u_{k}+r_{k}\geq 0$ for all
$k\in{\cal C}^{R}\backslash\underline{{\cal C}}^{R}$, and by the calculation
in (61) we see that $\sum_{i\in{\cal C}^{S}\cup\underline{{\cal
C}}^{R}}v^{\boldsymbol{r}}_{i}\geq 0$. These observations together with
condition (21) show that all conditions required in Lemma 4 (ii) are
fulfilled, and therefore we have $\boldsymbol{r}\in{\cal S}_{0}$. Condition
(22) and Step $1$ of this proof show that in addition to being stable, the
promises profile $\boldsymbol{r}$ achieves the minimal total promises transfer
${\cal T}_{\bm{r}}=G^{S}$, and thus $\boldsymbol{r}\in{\cal E}$.

Step 3. In this step we construct an equilibrium $\boldsymbol{r}\in{\cal E}$.
Consider the following across the aisle promises profile $\boldsymbol{r}$:

$\displaystyle r_{i}:=-{G^{S}\over U^{S}}u_{i},~{}i\in{\cal C}^{S};\quad
r_{j}:=0,~{}j\in\underline{{\cal C}}^{R};\quad r_{k}:=-{G^{S}\over{\Delta
U_{\widehat{\kappa}}}}[u_{k}-u_{\widehat{\kappa}}],~{}k\in{\cal
C}^{R}\backslash\underline{{\cal C}}^{R}.$ (62)

We can show directly that it satisfies conditions (20), (21) and,
(22).202020To see this, first observe that $\boldsymbol{r}\in{\cal P}$ since,
using (4) and (19), it can be checked that $\sum_{\mathbb{I}}r_{i}=0$.
Noticing that $u_{\widehat{\kappa}}\geq 0$, $G^{S}\leq U^{S}$, and
$G^{S}\leq\Delta U_{\widehat{\kappa}}$, we have for any $i\in{\cal C}^{S}$,
$j\in\underline{{\cal C}}^{R}$, and $k\in{\cal
C}^{R}\backslash\underline{{\cal C}}^{R}$,  $\displaystyle r_{i}\geq 0,\quad
r_{j}=0,\quad r_{k}\leq 0;\qquad r_{k}\geq-{G_{s}\over
U_{\widehat{\kappa}}}u_{k}\geq-u_{k}\,,$ and therefore $\boldsymbol{r}$
satisfies condition (20). Moreover, we have,  $\displaystyle
v^{\bm{r}}_{i}=[1-{G^{S}\over U_{s}}]u_{i}<0\leq u_{\widehat{\kappa}},\quad
v^{\bm{r}}_{j}=u_{j}\leq u_{\widehat{\kappa}},$ $\displaystyle
v^{\bm{r}}_{k}=u_{k}-{G_{s}\over\Delta
U_{\widehat{\kappa}}}[u_{k}-u_{\widehat{\kappa}}]\geq
u_{k}-[u_{k}-u_{\widehat{\kappa}}]=u_{\widehat{\kappa}};$ and thus condition
(21) is satisfied. Finally,  $\displaystyle\sum_{i\in{\cal
C}^{S}\cup\underline{{\cal C}}^{R}}r_{i}={G^{S}\over U^{S}}\sum_{i\in{\cal
C}^{S}}[-u_{i}]=G^{S},$ $\displaystyle\sum_{k\in{\cal
C}^{R}\backslash\underline{{\cal C}}^{R}}r_{k}=-{G_{s}\over\Delta
U_{\widehat{\kappa}}}\sum_{k\in{\cal C}^{R}\backslash\underline{{\cal
C}}^{R}}[u_{k}-u_{\widehat{\kappa}}]=-{G_{s}\over\Delta
U_{\widehat{\kappa}}}\Delta U_{\widehat{\kappa}}=-G^{S};$ and hence condition
(22) is satisfied. To sum up the promises profile $\boldsymbol{r}$ satisfy
conditions (20), (21) and, (22), and therefore it is an equilibrium. Then,
using the result in Step $2$ of this proof shows that the promises profile
$\boldsymbol{r}$ is an equilibrium.

Step 4. We now prove the only if part. Let $\boldsymbol{r}\in{\cal E}$ and
prove that $\boldsymbol{r}$ satisfies conditions (20), (21) and, (22).
Applying Step $1$ of this proof shows that, since $\boldsymbol{r}\in{\cal
S}_{0}$, we have ${\cal T}_{\bm{r}}\geq G^{S}$. Recall (61) and note that

$\displaystyle\displaystyle\sum_{i\in{\cal C}^{S}\cup\underline{{\cal
C}}^{R}}|r_{i}|\geq\sum_{i\in{\cal C}^{S}\cup\underline{{\cal
C}}^{R}}r_{i}\geq G^{S};$ (63) $\displaystyle\displaystyle\sum_{k\in{\cal
C}^{R}\backslash\underline{{\cal C}}^{R}}|r_{k}|\geq\sum_{k\in{\cal
C}^{R}\backslash\underline{{\cal C}}^{R}}(-r_{k})=\sum_{i\in{\cal
C}^{S}\cup\underline{{\cal C}}^{R}}r_{i}\geq G^{S}.$ (64)

Combining (63) and (64) gives

$\displaystyle{\cal T}_{\bm{r}}=\frac{1}{2}\sum_{i\in{\cal
C}^{S}\cup\underline{{\cal C}}^{R}}|r_{i}|+\frac{1}{2}\sum_{k\in{\cal
C}^{R}\backslash\underline{{\cal C}}^{R}}|r_{k}|\geq\frac{1}{2}\sum_{i\in{\cal
C}^{S}\cup\underline{{\cal C}}^{R}}r_{i}+\frac{1}{2}\sum_{k\in{\cal
C}^{R}\backslash\underline{{\cal C}}^{R}}(-r_{k})\geq G^{S}.$ (65)

Since ${\cal T}_{\bm{r}}\geq G^{S}$ for any stable promises profile, and since
the equilibrium promises profile defined in (62) achieves the total promises
transfer $G^{S}$, it must be that ${\cal T}_{\bm{r}}=G^{S}$ for any
$\boldsymbol{r}\in{\cal E}$. When ${\cal T}_{\bm{r}}=G^{S}$, inequalities (65)
become equalities:

$\displaystyle G^{S}={\cal T}_{\bm{r}}=\frac{1}{2}\sum_{i\in{\cal
C}^{S}\cup\underline{{\cal C}}^{R}}|r_{i}|+\frac{1}{2}\sum_{k\in{\cal
C}^{R}\backslash\underline{{\cal C}}^{R}}|r_{k}|=\frac{1}{2}\sum_{i\in{\cal
C}^{S}\cup\underline{{\cal C}}^{R}}r_{i}+\frac{1}{2}\sum_{k\in{\cal
C}^{R}\backslash\underline{{\cal C}}^{R}}(-r_{k})=G^{S}.$

This in turn implies inequalities (63) and (64) are also equalities:

$\displaystyle\sum_{i\in{\cal C}^{S}\cup\underline{{\cal
C}}^{R}}|r_{i}|=\sum_{i\in{\cal C}^{S}\cup\underline{{\cal
C}}^{R}}r_{i}=G^{S},\quad\sum_{j\in{\cal C}^{R}\backslash\underline{{\cal
C}}^{R}}|r_{j}|=\sum_{j\in{\cal C}^{R}\backslash\underline{{\cal
C}}^{R}}(-r_{j})=G^{S}.$

Then similarly to the approach used to prove (60), we deduce that
$|r_{i}|=r_{i}$ for all $i\in{\cal C}^{S}\cup\underline{{\cal C}}^{R}$ and
$|r_{j}|=-r_{j}$ for all $j\in{\cal C}^{R}\backslash\underline{{\cal C}}^{R}$.
Hence

$\displaystyle r_{j}\leq 0\leq r_{i},\quad\forall i\in{\cal
C}^{S}\cup\underline{{\cal C}}^{R},j\in{\cal C}^{R}\backslash\underline{{\cal
C}}^{R}.$

Recalling that ${\cal C}^{S}\cup\underline{{\cal
C}}^{R}=\\{1,..,\widehat{\kappa}\\}$ and ${\cal
C}^{R}\backslash\underline{{\cal C}}^{R}=\\{\widehat{\kappa}+1,..,I\\}$, we
see that the last equation is equivalent to

$\displaystyle r_{j}\leq 0\leq r_{i},\quad\forall i\leq\hat{\kappa}<j.$

Moreover,

$\displaystyle{\cal T}_{\boldsymbol{r}}=\sum_{i\in{\cal
C}^{S}\cup\underline{{\cal
C}}^{R}}r_{i}\equiv\sum_{i=1}^{\widehat{\kappa}}r_{i}=\sum_{j\in{\cal
C}^{R}\backslash\underline{{\cal
C}}^{R}}(-r_{j})\equiv\sum_{j=\widehat{\kappa}+1}^{I}(-r_{j})=G^{S}.$

Using (18), observe further that

$\displaystyle\sum_{i\in{\cal C}^{S}\cup\underline{{\cal
C}}^{R}}(u_{i}+r_{i})=\sum_{i\in{\cal C}^{S}}u_{i}+\sum_{i\in\underline{{\cal
C}}^{R}}u_{i}+\sum_{i\in{\cal C}^{S}\cup\underline{{\cal
C}}^{R}}r_{i}=-U^{S}+\underline{U}^{R}+G^{S}=0.$

To verify the individual rationality constraint $-u_{j}\leq r_{j}$ for any
$j\in{\cal C}^{R}\backslash\underline{{\cal C}}^{R}$, observe that $({\cal
C}^{S}\cup\underline{{\cal C}}^{R})\cup\\{j\\}\in{\cal D}^{S}$, then by
Proposition 1 we have

$\displaystyle 0\leq\sum_{i\in({\cal C}^{S}\cup\underline{{\cal
C}}^{R})\cup\\{j\\}}(u_{i}+r_{i})=\sum_{i\in{\cal C}^{S}\cup\underline{{\cal
C}}^{R}}(u_{i}+r_{i})+(u_{j}+r_{j})=u_{j}+r_{j},$

which implies $r_{j}\geq-u_{j}$. So far, we have shown that the payment
promises profile $\boldsymbol{r}$ satisfies (20) and (22).

All that’s left to prove is that (21) also holds. For any $j\in{\cal
C}^{S}\cup\underline{{\cal C}}^{R}$ and $k\in{\cal
C}^{R}\backslash\underline{{\cal C}}^{R}$, noting that ${\cal
C}^{S}\cup\underline{{\cal C}}^{R}\cup\\{k\\}\backslash\\{j\\}\in{\cal D}^{S}$
because $|({\cal C}^{S}\cup\underline{{\cal
C}}^{R}\cup\\{k\\})\backslash\\{j\\}|=\widehat{\kappa}$, then by (12) from
Proposition (1) we have

$\displaystyle 0\leq\sum_{i\in({\cal C}^{S}\cup\underline{{\cal
C}}^{R}\cup\\{k\\})\backslash\\{j\\}}(u_{i}+r_{i})=\sum_{i\in{\cal
C}^{S}\cup\underline{{\cal
C}}^{R}}(u_{i}+r_{i})+(u_{k}+r_{k})-(u_{j}+r_{j})=v^{\bm{r}}_{k}-v^{\bm{r}}_{j}.$

Thus, $v^{\bm{r}}_{j}\leq v^{\bm{r}}_{k}$ for all $j\in{\cal
C}^{S}\cup\underline{{\cal C}}^{R}$ and $k\in{\cal
C}^{R}\backslash\underline{{\cal C}}^{R}$, and the promises profile
$\boldsymbol{r}$ satisfies (21). ∎

### Proof of Proposition 8:

We start the proof with two preliminary lemmas.

######  Lemma 5.

Consider a committee operating under the $\kappa$-majority rule with
$|\kappa|\geq 2$ and such that $G^{S}>\Delta U_{\widehat{\kappa}}$. The
following statements hold:

(i) Recalling the intensity $u_{*}$ defined in (29), we have

$\displaystyle u_{*}<u_{\widehat{\kappa}}.$ (66)

(ii) There exists a unique group member $k_{*}\in\underline{{\cal C}}^{R}$,
i.e. $n<\kappa_{*}\leq\widehat{\kappa}$, such that inequalities (30) hold,
that is, $u_{k_{*}-1}\leq u_{*}<u_{k_{*}}.$

(iii) The constant ${\cal T}_{*}$ defined in (34) satisfies the inequality

$\displaystyle{\cal T}_{*}>G^{S}.$ (67)

Proof. To prove that (i) holds, observe that

$\displaystyle G^{S}$ $\displaystyle=$ $\displaystyle
U^{S}-\underline{U}^{R}=-\sum_{i=1}^{\widehat{\kappa}}u_{i}=-\sum_{i=1}^{I}u_{i}+\sum_{i=\widehat{\kappa}+1}^{I}u_{i}$
$\displaystyle=$
$\displaystyle-\sum_{i=1}^{I}u_{i}+\sum_{i=\widehat{\kappa}+1}^{I}[u_{i}-u_{\widehat{\kappa}}]+(I-\widehat{\kappa})u_{\widehat{\kappa}}$
$\displaystyle=$ $\displaystyle-\sum_{i=1}^{I}u_{i}+\Delta
U_{\widehat{\kappa}}+(\kappa-1)u_{\widehat{\kappa}},$

where we have used the definition of $\Delta U_{\widehat{\kappa}}$ in (19) and
the relation $\widehat{\kappa}=I-\kappa+1$. Using the definition of $u_{*}$
given in (29) yields

$\displaystyle G^{S}-\Delta
U_{\widehat{\kappa}}=(\kappa-1)u_{\widehat{\kappa}}-\sum_{i=1}^{I}u_{i}=(\kappa-1)[u_{\widehat{\kappa}}-u_{*}].$
(68)

Since $G^{S}>\Delta U_{\hat{\kappa}}$, this establishes inequality (66) in
part (i) of Lemma 5.

By the relations (3), (29), and (66), we have
$u_{n}<0<u_{*}<u_{\widehat{\kappa}}$. Then by the ordering condition (1) there
exists a unique $k_{*}\in\underline{{\cal C}}^{R}$, i.e.
$n<k_{*}\leq\widehat{\kappa}$, such that the inequalities (30) hold212121For
example, pick $k_{*}=\min k$ such that $u_{k}>u_{*}$.. This proves part (ii)
of Lemma 5.

To prove part (iii) of Lemma 5, observe that the total promises transfer
${\cal T}_{*}$ defined in (34) satisfies

$\displaystyle{\cal T}_{*}$ $\displaystyle\equiv$
$\displaystyle\sum_{j=k_{*}}^{\widehat{\kappa}}[u_{j}-u_{*}]+\sum_{j=\widehat{\kappa}+1}^{I}[u_{j}-u_{*}]$
(69) $\displaystyle=$
$\displaystyle\sum_{j=k_{*}}^{\widehat{\kappa}}[u_{j}-u_{*}]+\sum_{j=\widehat{\kappa}+1}^{I}[u_{j}-u_{\widehat{\kappa}}]+\sum_{j=\widehat{\kappa}+1}^{I}[u_{\widehat{\kappa}}-u_{*}]$
$\displaystyle=$
$\displaystyle\sum_{j=k_{*}}^{\widehat{\kappa}}[u_{j}-u_{*}]+\Delta
U_{\widehat{\kappa}}+(\kappa-1)[u_{\widehat{\kappa}}-u_{*}]$ $\displaystyle=$
$\displaystyle\sum_{j=k_{*}}^{\widehat{\kappa}}[u_{j}-u_{*}]+G^{S}>G^{S},$

where we used equation (68) in the fourth equality of the above sequence of
equalities. This establishes inequality (67) in part (iii) of Lemma 5. ∎

######  Lemma 6.

Consider the across the aisle promises profile $\boldsymbol{r}$ defined by

$\displaystyle r_{i}:=-{{\cal T}_{*}\over U^{S}}u_{i},~{}\forall i\leq
n;\qquad r_{k}:=0,~{}\forall n<k<k_{*};\quad r_{j}:=-[u_{j}-u_{*}],~{}\forall
j\geq k_{*},$ (70)

where we recall that $k_{*}$ is defined in part (ii) of Lemma 5. Then the
promises profile $\boldsymbol{r}$ is zero sum, $\boldsymbol{r}\in{\cal P}$ and
satisfies the conditions (31)-(34) of Proposition 8. Moreover, the promises
profile $\boldsymbol{r}$ defined in (70) is stable.

Proof. First, $\boldsymbol{r}\in{\cal P}$ since

$\displaystyle\sum_{i\in\mathbb{I}}r_{i}={{\cal T}_{*}\over
U^{S}}\sum_{i=1}^{n}(-u_{i})-\sum_{j=k_{*}}^{I}[u_{j}-u_{*}]={\cal
T}_{*}-{\cal T}_{*}=0.$

Second, the statements in (31), (32) and (34) can be directly checked from the
definition of the promises profile $\boldsymbol{r}$ given in (70).

Next, to prove (33), we need first to prove the preliminary result that
$0<{\cal T}_{*}<U^{S}$. Using (69), observe that

$\displaystyle U^{S}-{\cal
T}_{*}=U^{S}-\sum_{j=k_{*}}^{\widehat{\kappa}}[u_{j}-u_{*}]-G^{S}.$

Using the definition of $G^{S}$ given in (18) yields

$\displaystyle U^{S}-{\cal
T}_{*}=\underline{U}^{R}-\sum_{j=k_{*}}^{\widehat{\kappa}}[u_{j}-u_{*}]=\underline{U}^{R}-\sum_{j=k_{*}}^{\widehat{\kappa}}u_{j}+(\widehat{\kappa}-k_{*}+1)u_{*}.$
(71)

Since $k_{*}\in\underline{{\cal C}}^{R}$, we have
$n<k_{*}\leq\widehat{\kappa}$ and hence
$\underline{U}^{R}-\sum_{j=k_{*}}^{\widehat{\kappa}}u_{j}\geq 0$. Moreover,
$n<k_{*}\leq\widehat{\kappa}$ also implies that
$(\widehat{\kappa}-k_{*}+1)u_{*}>0$. Thus, using (71) gives $0<{\cal
T}_{*}<U^{S}$.

Now we are prepared to prove (33). For $i\leq n$, $n<k<k_{*}$, and $j\geq
k_{*}$, we have

$\displaystyle v^{\boldsymbol{r}}_{i}=u_{i}+r_{i}=[1-{{\cal T}_{*}\over
U^{S}}]u_{i}\leq 0;\quad v^{\boldsymbol{r}}_{k}=u_{k}+r_{k}=u_{k}\geq 0;\quad
v^{\boldsymbol{r}}_{j}=u_{j}+r_{j}=u_{*}>0$

where the left inequality is implied by $0<{\cal T}_{*}<U^{S}$. Thus, we see
that the condition (33) is satisfied for the promises profile $\boldsymbol{r}$
defined in (70).

Finally, to show that $\boldsymbol{r}$ is stable, set ${\cal C}:={\cal
C}^{S}\cup\underline{{\cal C}}^{R}$ in Lemma 4 (ii) so that $|{\cal
C}|=\widehat{\kappa}$. Using condition (33) and observing that
$u_{j}+r_{j}=u_{*}$ for all $j\geq k_{*}$ shows that we have $u_{i}+r_{i}\leq
u_{j}+r_{j}$ for all $i\in{\cal C}$ and $j\notin{\cal C}$.

Moreover, since $\boldsymbol{r}\in{\cal P}$, by (66) we have

$\displaystyle\sum_{i\in{\cal C}}(u_{i}+r_{i})$ $\displaystyle=$
$\displaystyle\sum_{i\in{\cal C}^{S}\cup\underline{{\cal
C}}^{R}}u_{i}-\sum_{j\in{\cal C}^{R}\backslash\underline{{\cal
C}}^{R}}r_{j}=\sum_{i\in{\cal C}^{S}\cup\underline{{\cal
C}}^{R}}u_{i}-\sum_{j\in{\cal C}^{R}\backslash\underline{{\cal
C}}^{R}}[-u_{j}+u_{*}]$ $\displaystyle=$
$\displaystyle\sum_{i\in\mathbb{I}}u_{i}-(\kappa-1)u_{*}=0.$

Then by applying Lemma 4 (ii) we see that $\boldsymbol{r}\in{\cal S}_{0}$. ∎

Proof of Proposition 8: We proceed in two steps.

Step 1. In this step, we prove the only if part: we fix an equilibrium
promises profile $\boldsymbol{r}\in{\cal E}$, and show that $\boldsymbol{r}$
satisfies conditions (31)-(34) of Proposition 8. Notice that Proposition 3
shows that the set ${\cal E}$ is not empty, and thus it is possible to select
a promises profile from ${\cal E}$.

Recall Lemma 3 and let $\tilde{k}_{*}\in{\cal C}^{R}$ satisfy

$\displaystyle\underline{k}_{*}\equiv\max\\{i:r_{i}>0\\}<\tilde{k}_{*}\leq\min\\{i:r_{i}<0\\}\equiv\overline{k}_{*},$
(72)

and the requirements (16) (In this proof, we reserve the notation $k_{*}$ for
the requirement (30) instead of the requirements (16)), that is, $-u_{j}\leq
r_{j}\leq 0\leq r_{i}$ and $v^{\boldsymbol{r}}_{i}\leq v^{\boldsymbol{r}}_{j}$
for all $i<\tilde{k}_{*}\leq j$. Moreover, we assume $\tilde{k}_{*}$ is the
largest one satisfying the requirements, and thus

$\displaystyle\mbox{either}~{}r_{\tilde{k}_{*}}<0,\quad\mbox{or}~{}v^{\boldsymbol{r}}_{\tilde{k}_{*}}>\min_{j>\tilde{k}_{*}}v^{\boldsymbol{r}}_{j},$
(73)

because otherwise $\tilde{k}_{*}+1$ would also satisfies the desired
requirements222222To see this, assume the opposite of statement (73) holds,
that is, $r_{\tilde{k}_{*}}\geq 0$ and
$v^{\boldsymbol{r}}_{\tilde{k}_{*}}\leq\min_{j>\tilde{k}_{*}}v^{\boldsymbol{r}}_{j}$.
Condition (16) implies that $r_{\tilde{k}_{*}}\leq 0$ and hence the assumption
$r_{\tilde{k}_{*}}\geq 0$ implies that $r_{\tilde{k}_{*}}=0$. Thus, we have
$\underline{k}_{*}<\tilde{k}_{*}+1\leq\overline{k}_{*}$ and, $-u_{j}\leq
r_{j}\leq 0\leq r_{i}$ for all $i<\tilde{k}_{*}+1\leq j$. Finally, the
assumption
$v^{\boldsymbol{r}}_{\tilde{k}_{*}}\leq\min_{j>\tilde{k}_{*}}v^{\boldsymbol{r}}_{j}$
implies $v^{\boldsymbol{r}}_{i}\leq v^{\boldsymbol{r}}_{j}$ for all
$i<\tilde{k}_{*}+1\leq j$. Hence $\tilde{k}_{*}+1$ also satisfies the
requirements (72) and (16)..

Step 1.1. We first show that $\tilde{k}_{*}\leq k_{*}$. Since
$k_{*}\leq\widehat{\kappa}$, this implies $\tilde{k}_{*}\leq\widehat{\kappa}$.
Assume by contradiction that $\tilde{k}_{*}>k_{*}$. Then by (16) we have

$\displaystyle u_{j}+r_{j}\geq u_{k_{*}}+r_{k_{*}}\geq
u_{k_{*}},\quad\mbox{for all}\quad j\geq\tilde{k}_{*}.$

Moreover, in the case that $\tilde{k}_{*}>\widehat{\kappa}$, by (1) we also
have,

$\displaystyle u_{j}+r_{j}\geq u_{j}\geq u_{k_{*}},\quad\mbox{for
all}\quad\widehat{\kappa}\leq j<\tilde{k}_{*}.$

So in all the cases we have $u_{j}+r_{j}\geq u_{k_{*}}$ for all
$j\geq\widehat{\kappa}$. This in turn implies,

$\displaystyle\sum_{j\in{\cal C}^{R}\backslash\underline{{\cal
C}}^{R}}r_{j}\geq\sum_{j>\widehat{\kappa}}[u_{k_{*}}-u_{j}]=-\Delta
U_{k_{*}}>-G^{S}.$

Thus, since $\boldsymbol{r}\in{\cal P}$,

$\displaystyle\sum_{i\in{\cal C}^{S}\cup\underline{{\cal
C}}^{R}}[u_{i}+r_{i}]=\sum_{i\in{\cal C}^{S}\cup\underline{{\cal
C}}^{R}}u_{i}-\sum_{j\in{\cal C}^{R}\backslash\underline{{\cal
C}}^{R}}r_{j}<-G^{S}+G^{S}=0.$

This contradicts the inequality (12) which is required here because
$\boldsymbol{r}\in{\cal S}_{0}$ and ${\cal C}^{S}\cup\underline{{\cal
C}}^{R}\in{\cal D}^{S}$.

Step 1.2. In this step, we prove that the ex post intensities of the members
$j=\tilde{k}_{*},\cdots,I$ are equal:

$\displaystyle
v^{\boldsymbol{r}}_{\tilde{k}_{*}}=\cdots=v^{\boldsymbol{r}}_{I}.$ (74)

Consider for notational convenience, the order statistics of
$\\{v^{\boldsymbol{r}}_{j}\\}_{\tilde{k}_{*}\leq j\leq I}$:

$\displaystyle v^{\boldsymbol{r}}_{l_{\tilde{k}_{*}}}\leq\cdots\leq
v^{\boldsymbol{r}}_{l_{I}},$

where $\\{l_{\tilde{k}_{*}},\cdots,l_{I}\\}$ is a permutation of
$\\{\tilde{k}_{*},\cdots,I\\}$.

To simplify the exposition, we proceed in two substeps. First, we show that
$v^{\boldsymbol{r}}_{l_{\tilde{k}_{*}}}=\cdots=v^{\boldsymbol{r}}_{l_{\widehat{\kappa}+1}}$.
Second, we show that the equality
$v^{\boldsymbol{r}}_{l_{\tilde{k}_{*}}}=\cdots=v^{\boldsymbol{r}}_{I}$ is also
true.

Step 1.2.1. In this sub-step we show that

$\displaystyle
v^{r}_{l_{\tilde{k}_{*}}}=v^{r}_{l_{\widehat{\kappa}+1}}\quad\mbox{and
hence}\quad v^{r}_{l_{\tilde{k}_{*}}}=\cdots=v^{r}_{l_{\widehat{\kappa}+1}}.$
(75)

We emphasize that we are considering the term $l_{\widehat{\kappa}+1}$, rather
than $l_{\widehat{\kappa}}$.

Assume by contradiction that

$\displaystyle
v^{\boldsymbol{r}}_{l_{\tilde{k}_{*}}}<v^{\boldsymbol{r}}_{l_{\widehat{\kappa}+1}}.$
(76)

We claim that

$\displaystyle r_{l_{\tilde{k}_{*}}}<0.$ (77)

Indeed, recall (73). In the case $r_{\tilde{k}_{*}}<0$, by the definition of
order statistics and the ordering (1) we have

$\displaystyle u_{l_{\tilde{k}_{*}}}+r_{l_{\tilde{k}_{*}}}\leq
u_{\tilde{k}_{*}}+r_{\tilde{k}_{*}}<u_{\tilde{k}_{*}}\leq
u_{l_{\tilde{k}_{*}}},$

which implies (77). In the case $r_{\tilde{k}_{*}}=0$ and
$v^{\boldsymbol{r}}_{\tilde{k}_{*}}>\min_{j>\tilde{k}_{*}}v^{\boldsymbol{r}}_{j}$,
by (1) again we have

$\displaystyle
u_{\tilde{k}_{*}}=v^{\boldsymbol{r}}_{\tilde{k}_{*}}>\min_{j\geq\tilde{k}_{*}}v^{\boldsymbol{r}}_{j}=v^{\boldsymbol{r}}_{l_{\tilde{k}_{*}}}=u_{l_{\tilde{k}_{*}}}+r_{l_{\tilde{k}_{*}}}\geq
u_{\tilde{k}_{*}}+r_{l_{\tilde{k}_{*}}},$

implying (77) again.

Clearly $r_{i}>0$ for some $i<\tilde{k}_{*}$, and assume without loss of
generality that $r_{1}>0$. We now modify $\boldsymbol{r}$ as follows: for some
$\varepsilon>0$ small,

$\displaystyle\tilde{r}_{1}=r_{1}-\varepsilon>0,\quad\tilde{r}_{l_{\tilde{k}_{*}}}=r_{l_{\tilde{k}_{*}}}+\varepsilon<0,\quad\mbox{and}\quad\tilde{r}_{i}=r_{i}\quad\mbox{for
all}~{}i\neq 1,l_{\tilde{k}_{*}}.$

Set $\tilde{\cal
C}\\!=\\!\\{1,\cdots,\tilde{k}_{*}-1\\}\cup\\{l_{\tilde{k}_{*}},\cdots,l_{\widehat{\kappa}}\\}$
with $|\tilde{\cal C}|=\widehat{\kappa}$. Note that, for $\varepsilon>0$ small
enough,

$\displaystyle u_{1}+\tilde{r}_{1}<u_{1}+r_{1}\leq
v^{\boldsymbol{r}}_{l_{\widehat{\kappa}+1}};\qquad
u_{l_{\tilde{k}_{*}}}+\tilde{r}_{l_{\tilde{k}_{*}}}=v^{\boldsymbol{r}}_{l_{\tilde{k}_{*}}}+\varepsilon<v^{\boldsymbol{r}}_{l_{\widehat{\kappa}+1}};$
$\displaystyle u_{i}+\tilde{r}_{i}=u_{i}+r_{i}\leq
v^{\boldsymbol{r}}_{l_{\widehat{\kappa}+1}},\quad i\in\tilde{\cal
C}\backslash\\{1,l_{\tilde{k}_{*}}\\};\quad
u_{j}+\tilde{r}_{j}=u_{j}+r_{j}\geq
v^{\boldsymbol{r}}_{l_{\widehat{\kappa}+1}},\quad j\notin\tilde{\cal C}.$

So $u_{i}+\tilde{r}_{i}\leq v^{\boldsymbol{r}}_{l_{\widehat{\kappa}+1}}\leq
u_{j}+\tilde{r}_{j}$ for all $i\in\tilde{\cal C}$ and $j\notin\tilde{\cal C}$.
Note further that, since $\boldsymbol{r}\in{\cal E}\subset{\cal S}_{0}$,

$\displaystyle\sum_{i\in\tilde{\cal
C}}[u_{i}+\tilde{r}_{i}]=\sum_{i\in\tilde{\cal
C}}[u_{i}+r_{i}]+\Big{[}(\tilde{r}_{1}-r_{1})+(\tilde{r}_{l_{\tilde{k}_{*}}}-r_{l_{\tilde{k}_{*}}})\Big{]}=\sum_{i\in\tilde{\cal
C}}[u_{i}+r_{i}]\geq 0.$

Then by Lemma 4 (ii) we have $\boldsymbol{\tilde{r}}\in{\cal S}_{0}$. However,
as in the last part of the proof for Lemma 1, we have ${\cal
T}_{\tilde{\boldsymbol{r}}}={\cal T}_{\boldsymbol{r}}-\varepsilon<{\cal
T}_{\boldsymbol{r}}$. This contradicts the assumption that
$\boldsymbol{r}\in{\cal E}$ has the minimum total promises transfer.
Therefore, (75) holds true.

Step 1.2.2. Now we proceed to prove (74). Assume by contradiction that, for
the order statistics in the previous step,

$\displaystyle
v^{\boldsymbol{r}}_{l_{\tilde{k}_{*}}}=\cdots=v^{\boldsymbol{r}}_{l_{k_{2}}}<v^{\boldsymbol{r}}_{l_{k_{2}+1}}\leq\cdots\leq
v^{\boldsymbol{r}}_{l_{I}},\quad\mbox{for some}~{}\widehat{\kappa}+1\leq
k_{2}<I.$

First, by (77) we also have

$\displaystyle r_{l_{j}}<0,\quad\mbox{for all}\quad
j=\tilde{k}_{*},\cdots,k_{2}.$ (78)

Again assume $r_{1}>0$. We then modify $\boldsymbol{r}$ as follows: for
$\varepsilon>0$ small,

$\displaystyle\displaystyle\hat{r}_{1}=r_{1}-[\widehat{\kappa}-\tilde{k}_{*}+1]\varepsilon>0,\quad\hat{r}_{l_{j}}=r_{l_{j}}+\varepsilon<0,~{}j=\tilde{k}_{*},\cdots,k_{2};$
$\displaystyle\displaystyle\hat{r}_{l_{k_{2}+1}}=r_{l_{k_{2}+1}}-[k_{2}-\widehat{\kappa}]\varepsilon<0;\quad\hat{r}_{i}=r_{i}\quad\mbox{for
all other $i$}.$

One can check that $\hat{\boldsymbol{r}}\in{\cal P}$:

$\displaystyle\sum_{i\in\mathbb{I}}\hat{r}_{i}=\sum_{i\in\mathbb{I}}r_{i}-[\widehat{\kappa}-\tilde{k}_{*}+1]\varepsilon+\sum_{j=\tilde{k}_{*}}^{k_{2}}\varepsilon-[k_{2}-\widehat{\kappa}]\varepsilon=0$

Similarly to Step $1.2.1$, we see that, for all $i<\tilde{k}_{*}$ and
$j>k_{2}+1$,

$\displaystyle u_{i}+\hat{r}_{i}\leq
u_{l_{\tilde{k}_{*}}}+\hat{r}_{l_{\tilde{k}_{*}}}=\cdots=u_{l_{k_{2}}}+\hat{r}_{l_{k_{2}}}<u_{l_{k_{2}+1}}+\hat{r}_{l_{k_{2}+1}}<u_{l_{j}}+\hat{r}_{l_{j}},$

where the second inequality holds for $\varepsilon>0$ small enough. Now for
the same $\tilde{\cal
C}=\\{1,\cdots,\tilde{k}_{*}-1\\}\cup\\{l_{\tilde{k}_{*}},\cdots,l_{\widehat{\kappa}}\\}$
with $|\tilde{\cal C}|=\widehat{\kappa}$ as in Step $1.2.1$, we have
$u_{i}+\hat{r}_{i}\leq u_{l_{\tilde{k}_{*}}}+\hat{r}_{l_{\tilde{k}_{*}}}\leq
u_{j}+\hat{r}_{j}$ for all $i\in\tilde{\cal C}$ and $j\notin\tilde{\cal C}$,
and

$\displaystyle\sum_{i\in\tilde{\cal
C}}[u_{i}+\hat{r}_{i}]=\sum_{i\in\tilde{\cal
C}}[u_{i}+r_{i}]-[\widehat{\kappa}-\tilde{k}_{*}+1]\varepsilon+\sum_{j=\tilde{k}_{*}}^{\widehat{\kappa}}\varepsilon=\sum_{i\in\tilde{\cal
C}}[u_{i}+r_{i}]\geq 0.$

Then by Lemma 4 (ii) we see that $\hat{\boldsymbol{r}}\in{\cal S}_{0}$.
Moreover, note that

$\displaystyle{\cal T}_{\hat{\boldsymbol{r}}}-{\cal T}_{\boldsymbol{r}}$
$\displaystyle=$ $\displaystyle{1\over
2}\Big{[}|\hat{r}_{1}|-|r_{1}|+\sum_{j=\tilde{k}_{*}}^{k_{2}}[|\hat{r}_{l_{j}}|-|r_{l_{j}}|]+|\hat{r}_{l_{k_{2}+1}}|-|r_{l_{k_{2}+1}}|\Big{]}$
$\displaystyle=$ $\displaystyle{1\over
2}\Big{[}-[\widehat{\kappa}-\tilde{k}_{*}+1]\varepsilon+\sum_{j=\tilde{k}_{*}}^{k_{2}}(-\varepsilon)+[k_{2}-\widehat{\kappa}]\varepsilon\Big{]}=-[\widehat{\kappa}-\tilde{k}_{*}+1]\varepsilon<0,$

where the last inequality is due to $\tilde{k}_{*}\leq\widehat{\kappa}$ from
Step $1.1.$ This contradicts the assumption that $\boldsymbol{r}\in{\cal E}$
has the minimum total promises transfer, so (74) holds true.

Step 1.3. We now collect all the results from the intermediate steps to show
that $\boldsymbol{r}$ satisfies conditions (31)-(34) of Proposition 8.

Let $y_{*}$ denote the common value in (74). Then $r_{j}=y_{*}-u_{j}\leq 0$
for all $j\geq\tilde{k}_{*}$. On one hand, since $\tilde{k}_{*}\leq
k_{*}\leq\widehat{\kappa}$ by Step $1.1$,

$\displaystyle 0$ $\displaystyle\leq$
$\displaystyle\sum_{i\leq\widehat{\kappa}}[u_{i}+r_{i}]=\sum_{i\leq\widehat{\kappa}}u_{i}-\sum_{j>\widehat{\kappa}}r_{j}=\sum_{i\leq\widehat{\kappa}}u_{i}-\sum_{j>\widehat{\kappa}}[y_{*}-u_{j}]$
$\displaystyle=$
$\displaystyle\sum_{i\in\mathbb{I}}u_{i}-(\kappa-1)y_{*}=(\kappa-1)(u_{*}-y_{*}).$

Therefore, $y_{*}\leq u_{*}$. On the other hand, by (17),

$\displaystyle{\cal
T}_{\boldsymbol{r}}=\sum_{j\geq\tilde{k}_{*}}(-r_{j})\geq\sum_{j\geq
k_{*}}(-r_{j})=\sum_{j\geq k_{*}}[u_{j}-y_{*}]\geq\sum_{j\geq
k_{*}}[u_{j}-u_{*}]={\cal T}_{*}.$ (79)

Since $\boldsymbol{r}$ minimizes the total promises transfer and we already
constructed a stable promises profile in Lemma 6 with total promises transfer
${\cal T}_{*}$, then we must have ${\cal T}_{\boldsymbol{r}}={\cal T}_{*}$,
and thus all the inequalities in (79) are equalities. In particular, the
second inequality in (79) implies that $u_{*}=y_{*}$. Moreover, by (73) and
(74) we have $r_{\tilde{k}_{*}}<0$, so that the first inequality in (79)
implies that $\tilde{k}_{*}=k_{*}$. Now it can be directly checked that the
conditions (31)-(34) of Proposition 8 hold. This concludes the proof of the
only if part of Proposition 8.

Step 2. In this step we show the if part: we fix $\boldsymbol{r}\in{\cal P}$
that satisfies conditions (31)-(33) of Proposition 8 and show that
$\boldsymbol{r}\in{\cal E}$ and that (34) holds.

To show that $\boldsymbol{r}\in{\cal S}_{0}$, set ${\cal C}:={\cal
C}^{S}\cup\underline{{\cal C}}^{R}$ in Lemma 4 (ii) so that $|{\cal
C}|=\widehat{\kappa}$. Using condition (33) and observing that
$u_{j}+r_{j}=u_{*}$ for all $j\geq k_{*}$ (condition (32)) shows that we have
$u_{i}+r_{i}\leq u_{j}+r_{j}$ for all $i\in{\cal C}$ and $j\notin{\cal C}$.
Moreover, since $\boldsymbol{r}\in{\cal P}$ and $r_{j}=-u_{j}+u_{*}$ for
$j\geq\kappa_{*}$, we have

$\displaystyle\sum_{i\in{\cal C}}(u_{i}+r_{i})$ $\displaystyle=$
$\displaystyle\sum_{i\in{\cal C}^{S}\cup\underline{{\cal
C}}^{R}}u_{i}-\sum_{j\in{\cal C}^{R}\backslash\underline{{\cal
C}}^{R}}r_{j}=\sum_{i\in{\cal C}^{S}\cup\underline{{\cal
C}}^{R}}u_{i}-\sum_{j\in{\cal C}^{R}\backslash\underline{{\cal
C}}^{R}}[-u_{j}+u_{*}]$ $\displaystyle=$
$\displaystyle\sum_{i\in\mathbb{I}}u_{i}-(\kappa-1)u_{*}=0.$

Then by applying Lemma 4 (ii) we see that $\boldsymbol{r}\in{\cal S}_{0}$.
Moreover,

$\displaystyle{\cal T}_{\boldsymbol{r}}=\sum_{j\geq k_{*}}(-r_{j})=\sum_{j\geq
k_{*}}[u_{j}-u_{*}]={\cal T}_{*}.$

By Proposition 3 there exists $\boldsymbol{r}^{*}\in{\cal E}$. By Step $1$
(the only if direction), the promises profile $\boldsymbol{r}^{*}$ satisfies
conditions (31)-(34), in particular, ${\cal T}_{\boldsymbol{r}^{*}}={\cal
T}_{*}$, thus ${\cal T}_{*}$ is the minimum total promises transfer for all
$\boldsymbol{r}\in{\cal S}_{0}$. Finally, if $\boldsymbol{r}\in{\cal S}_{0}$
satisfies (31)-(33), we have ${\cal T}_{\boldsymbol{r}}={\cal T}_{*}$, so
$\boldsymbol{r}$ matches the minimum total promises transfer, and thus
$\boldsymbol{r}\in{\cal E}$. ∎

### Appendix B

In this appendix, we first provide necessary and sufficient conditions under
which the equilibrium promises are of the reaching across the aisle type.
Second, we propose an equilibrium selection based on an a finite sequence of
transfers from the strongest supporters of the reform to the weakest ones.

### Reaching across the aisle equilibria

The following proposition generalizes the discussions on whether the
equilibrium rules out circling the wagon transfers where some promises
recipients are reform supporters.

######  Proposition 9.

[Reaching across the aisle equilibria.] Consider a committee with a
$\kappa$-majority voting rule. When $\kappa\geq 2$, all the equilibrium
promises are of the reaching across the aisle type if and only if one of the
following conditions hold:

  1. 1.

Reform supporter lack voting power to enact the reform, $|{\cal
C}^{R}|<\kappa$, as in Proposition 5.

  2. 2.

Reform supporters have enough voting power, $|{\cal C}^{R}|\geq\kappa$, and
$G^{S}\leq\Delta U_{\widehat{\kappa}}$ as in Proposition 7, and the two
following conditions are satisfied

    1. (a)

The weakest reform supporters $\underline{{\cal
C}}^{R}=\\{n+1,\cdots,\widehat{\kappa}\\}$ have equal ex ante intensities:

$u_{n+1}=\cdots=u_{\hat{\kappa}}.$

    2. (b)

Either $G^{S}=\Delta U_{\hat{\kappa}}$, or $G^{S}<\Delta U_{\hat{\kappa}}$ and
the reform supporter $\widehat{\kappa}+1$ has an ex ante intensity that is
equal to that of the members $\\{n+1,\cdots,\widehat{\kappa}\\}$:

$G^{S}=\Delta U_{\hat{\kappa}},\quad\mbox{or}\quad G^{S}<\Delta
U_{\hat{\kappa}}\mbox{ and
}u_{n+1}=\cdots=u_{\hat{\kappa}}=u_{\hat{\kappa}+1}.$

  3. 3.

Reform supporters have enough voting power, $|{\cal C}^{R}|\geq\kappa$, and
$\Delta U_{\widehat{\kappa}}<G^{S}$ as in Proposition 8, and

$u_{n+1}=\cdots=u_{k_{*}-1}=u_{*}.$

Proposition 9 shows that in the case of a frustrated $\kappa-$minority covered
in Proposition 5, promises recipients are reform opponents in all equilibria.
Despite their multiplicity, equilibrium promises share the common feature of
being of the reaching across the aisle type. By contrast, in the cases covered
in Proposition 7 and Proposition 8 where there are enough supporters to enact
the reform to begin with ($|{\cal C}^{R}|\geq\kappa$), we show in Proposition
9 that for equilibrium promises to always be of the reaching across the aisle
type additional restrictions are required. To rule out circle the wagon type
transfers, we broadly need a stale distribution of intensities among a
specific subset of reform supporters with weakest intensities. In words, the
proposition shows that when the weakest reform supporters derive uniform
utility from the reform, then equilibrium requires that all promises
recipients are reform opponents. We now give the proof of Proposition 9.

### Proof of Proposition 9:

When $|{\cal C}^{R}|<\kappa$, it is clear from Proposition 5 that all promises
recipients are reform supporters, and thus statement $1$ in Proposition 9
holds.

We now prove statement $2$. We first show that properties $2.a.$ and $2.b$
collectively imply that any $\boldsymbol{r}\in{\cal E}$ belongs to the
reaching across the aisle type.

Consider now the first subcase where $u_{n+1}=\cdots=u_{\widehat{k}}$ and
$G^{S}=\Delta U_{\widehat{k}}$. Note that

$\displaystyle u_{j}+r_{j}=v^{\boldsymbol{r}}_{j}\geq
v^{\boldsymbol{r}}_{\hat{\kappa}}\geq u_{\hat{\kappa}},\quad\forall
j>\hat{\kappa}.$

Then

$\displaystyle{\cal
T}_{\boldsymbol{r}}=\sum_{j>\hat{\kappa}}(-r_{j})\leq\sum_{j>\hat{\kappa}}(u_{j}-u_{\hat{\kappa}})=\Delta
U_{\hat{\kappa}}=G^{S}.$

Since $\boldsymbol{r}\in{\cal E}$, then ${\cal T}_{\boldsymbol{r}}=G^{S}$ (
Proposition 7, condition 3), and thus equality holds above. This implies that
$-r_{j}=u_{j}-u_{\hat{\kappa}}$, and thus
$v^{\boldsymbol{r}}_{j}=u_{\hat{\kappa}}$ for all $j>\hat{\kappa}$. Note
further that $u_{i}\leq v^{\boldsymbol{r}}_{i}\leq
v^{\boldsymbol{r}}_{j}=u_{\hat{\kappa}}$ for all $n<i\leq\hat{\kappa}<j$. By
the assumption in this subcase, we see that $r_{i}=0$ for
$n<i\leq\hat{\kappa}$. Then the promises profile $\boldsymbol{r}$ is of the
reaching across the aisle type.

Consider the second subcase where
$u_{n+1}=\cdots=u_{\widehat{k}}=u_{\widehat{k}+1}$. By Proposition 7 Part 1,
we have $r_{i}\geq 0$ for $n<i\leq\hat{\kappa}$ and $r_{\hat{\kappa}+1}\leq
0$. Then $v^{\boldsymbol{r}}_{i}\geq u_{i}=u_{\hat{\kappa}+1}\geq
v^{\boldsymbol{r}}_{\hat{\kappa}+1}$. By Proposition 7 Part 2, we have
$v^{\boldsymbol{r}}_{\hat{\kappa}+1}\geq v^{\boldsymbol{r}}_{i}$, and thus we
must have $v^{\boldsymbol{r}}_{\hat{\kappa}+1}=v^{\boldsymbol{r}}_{i}$. Thus,
$r_{i}=0$ for $n<i\leq\hat{\kappa}$, and therefore $\boldsymbol{r}$ is of the
reaching across the aisle type promise.

We next prove the only if part of statement 2 in Proposition 5. To do so, we
assume that either statement $2.a$ or statement $2.b$ is false, and construct
an equilibrium promises profile $\boldsymbol{r}\in{\cal E}$ where some
promises recipients are reform supporters. Note that, when either statement
$2.a$ or statement $2.b$ is false, together with (1) and the assumption that
$0<G^{S}\leq\Delta U_{\hat{\kappa}}$, one of the following two statements must
be true:

$\displaystyle\displaystyle u_{n+1}<u_{\hat{\kappa}}\quad\mbox{ and }\quad
0<G^{S}\leq\Delta U_{\hat{\kappa}};$ (80) $\displaystyle\displaystyle
u_{n+1}=\cdots=u_{\hat{\kappa}}<u_{\hat{\kappa}+1}\quad\mbox{ and }\quad
0<G^{S}<\Delta U_{\hat{\kappa}}.$ (81)

Now let $0<\varepsilon<G^{S}$ and we modify the equilibrium promises profile
in (62) as follows:

$\displaystyle\left.\begin{array}[]{c}\displaystyle
r_{i}:=-{G^{S}-\varepsilon\over U^{S}}u_{i},~{}i\in{\cal C}^{S};\quad
r_{n+1}:=\varepsilon;\\\ \displaystyle r_{j}:=0,~{}n+1<j\leq\hat{\kappa};\quad
r_{k}:=-{G^{S}\over{\Delta
U_{\widehat{\kappa}}}}[u_{k}-u_{\widehat{\kappa}}],~{}k\in{\cal
C}^{R}\backslash\underline{{\cal C}}^{R}.\end{array}\right.$ (84)

It can be directly checked that $\boldsymbol{r}\in{\cal P}$ and satisfies
conditions (20) and (22) from Proposition 7, so to establish that
$\boldsymbol{r}$ is an equilibrium, we only need to prove (21).

In the sub-case (80), assume further that
$\varepsilon<u_{\hat{\kappa}}-u_{n+1}$. Then one can check that

$\displaystyle v^{\boldsymbol{r}}_{i}\leq 0\leq v^{\boldsymbol{r}}_{j}\leq
u_{\hat{\kappa}}\leq v^{\boldsymbol{r}}_{k},\quad\mbox{for all}~{}i\leq
n<j\leq\hat{\kappa}<k.$

In the sub-case (81), assume further that $\varepsilon<[1-{G^{S}\over\Delta
U_{\hat{\kappa}}}][u_{\hat{\kappa}+1}-u_{\hat{\kappa}}]$. Then using
$u_{n+1}=u_{\widehat{\kappa}}$ one can see

$\displaystyle v^{\boldsymbol{r}}_{i}\leq 0\leq v^{\boldsymbol{r}}_{j}\leq
u_{\hat{\kappa}}+\varepsilon\leq u_{\hat{\kappa}}+[1-{G^{S}\over\Delta
U_{\hat{\kappa}}}][u_{\hat{\kappa}+1}-u_{\hat{\kappa}}]\leq
v^{\boldsymbol{r}}_{k},\quad\mbox{for all}~{}i\leq n<j\leq\hat{\kappa}<k.$

To sum up, in all subcases, the promises profile $\boldsymbol{r}$ defined in
(84) also satisfies the condition (21) of Proposition 7 when $\varepsilon$ is
small enough, and as a result, $\boldsymbol{r}\in{\cal E}$. We conclude then
by observing that since $r_{n+1}>0$ the reform supporter $n+1$ is a promise
recipient, and therefore the equilibrium $\boldsymbol{r}$ has some circle the
wagon transfers.

We now prove statement $3$ in Proposition 9.

We first show the if part. Fix an arbitrary $\boldsymbol{r}\in{\cal E}$. By
Proposition 8, we have $r_{i}\geq 0$ and $u_{i}\leq v^{\boldsymbol{r}}_{i}\leq
u_{*}$ for $n+1\leq i\leq\hat{\kappa}$. However, since we assume
$u_{n+1}=\cdots=u_{k_{*}-1}=u_{*}$ here, we must have $r_{i}=0$ for $n+1\leq
i\leq\hat{\kappa}$. Since $r_{j}=-(u_{j}-u_{*})\geq 0$ for $j\geq k_{*}$, we
see that the promises profile $\boldsymbol{r}$ is of the reaching across the
aisle type.

We now prove the only if part using the contrapositive. Equivalently,
recalling (1), we assume $u_{n+1}<u_{*}$ and we shall construct an
$\boldsymbol{r}\in{\cal E}$ which has some circle the wagon type transfers.
Let $0<\varepsilon<{\cal T}_{*}$ and we modify the promises profile described
in (70) as follows:

$\displaystyle\left.\begin{array}[]{c}\displaystyle r_{i}:=-{{\cal
T}_{*}-\varepsilon\over U^{S}}u_{i},~{}i\leq n;\qquad r_{n+1}:=\varepsilon;\\\
\displaystyle r_{k}:=0,~{}n+1<k<k_{*};\quad r_{j}:=-[u_{k}-u_{*}],~{}j\geq
k_{*}.\end{array}\right.$ (87)

Assume further that $\varepsilon<u_{*}-u_{n+1}$, then one can check that

$\displaystyle v^{\boldsymbol{r}}_{i}\leq 0\leq v^{\boldsymbol{r}}_{j}\leq
u_{*}=v^{\boldsymbol{r}}_{k},\quad\mbox{for all}~{}i\leq
n<j\leq\hat{\kappa}<k.$

It can be checked that $\boldsymbol{r}\in{\cal P}$ and satisfies all the other
requirements in Proposition 8. We conclude by observing that $r_{n+1}>0$, and
therefore the equilibrium promises profile $\boldsymbol{r}\in{\cal E}$
constructed in (87) has some circle the wagon transfers. ∎

### An equilibrium selection algorithm

In this subsection, we propose an algorithm describing how to select a
sequence of incremental promises that reach an equilibrium after a finite
number of steps. Unlike the algorithm from Example 5, the algorithm described
in this section dictates in each step how to proceed for the next step and
does not leave room for multiple choices. Importantly, the algorithm achieves
an equilibrium for all distributions of ex ante intensities; this covers all
cases that we have discussed in this paper.

To describe the algorith, we introduce an operator $\Phi$ as follows. Given an
intensity profile $\boldsymbol{u}$ satisfying (1), with $u_{1}<u_{I}$, let
$k_{1}(\boldsymbol{u})\leq k_{2}(\boldsymbol{u})$ be such that, abbreviating
as $k_{1},k_{2}$ for notational simplicity,

$\displaystyle u_{1}=\cdots=u_{k_{1}}<u_{k_{1}+1},\quad
u_{k_{2}}<u_{k_{2}+1}=\cdots=u_{I}.$ (88)

We then define, for any $X>0$,

$\displaystyle\left.\begin{array}[]{c}\displaystyle\Phi(\boldsymbol{u},X):=\boldsymbol{v},\quad\mbox{where}\quad
v_{i}=u_{i},~{}k_{1}<i\leq k_{2},\quad\mbox{and}\\\ \displaystyle
v_{1}=\cdots=v_{k_{1}}:=u_{1}+{X\over k_{1}},\quad\quad
v_{k_{2}+1}=\cdots=v_{I}:=u_{I}-{X\over I-k_{2}}.\end{array}\right.$ (91)

We first establish the following simple lemma, whose proof is postponed to the
end of Appendix B.

######  Lemma 7.

Denote $v:=\Phi(\boldsymbol{u},X)$.

(i) For any $X>0$, $\boldsymbol{r}:=\boldsymbol{v}-\boldsymbol{u}\in{\cal P}$.

(ii) Assume
$X\leq\min\big{(}k_{1}(u_{k_{1}+1}-u_{k_{1}}),(I-k_{2})(u_{k_{2}+1}-u_{k_{2}})\big{)}$,
then $v_{1}\leq\cdots\leq v_{I}$.

(iii) For the $X$ in (ii), if $v_{1}<v_{I}$, then $k_{1}(\boldsymbol{u})\leq
k_{1}(\boldsymbol{v})\leq k_{2}(\boldsymbol{v})\leq k_{2}(\boldsymbol{u})$.

We now present the algorithm more formally:

Step 0 (Initialization). If
$\displaystyle\sum_{i=1}^{\widehat{\kappa}}u_{i}\geq 0$, $\boldsymbol{u}$ is
already stable, so $\boldsymbol{r}=(0,\cdots,0)$ is the unique equilibrium. We
thus assume $\displaystyle\sum_{i=1}^{\widehat{\kappa}}u_{i}<0$, and set
$\boldsymbol{v}^{0}:=\boldsymbol{u}$.

Step 1. Given $\boldsymbol{v}^{l}$ satisfying $v^{l}_{1}\leq\cdots\leq
v^{l}_{I}$ and $\displaystyle\sum_{i=1}^{\widehat{\kappa}}v^{l}_{i}<0$, we
define $\boldsymbol{v}^{l+1}$ recursively as follows until
$\displaystyle\sum_{i=1}^{\widehat{\kappa}}v^{l+1}_{i}\geq 0$: denoting
$k_{1}:=k_{1}(\boldsymbol{v}^{l})$, $k_{2}:=k_{2}(\boldsymbol{v}^{l})$ as in
(88),

$\displaystyle\boldsymbol{v}^{l+1}:=\Phi(\boldsymbol{v}^{l},X_{l+1})\quad\mbox{where}\quad
X_{l+1}:=\min\big{(}k_{1}(v^{l}_{k_{1}+1}-v^{l}_{k_{1}}),(I-k_{2})(v^{l}_{k_{2}+1}-v^{l}_{k_{2}})\big{)},$

We remark that by Lemma 7 (ii) $\boldsymbol{v}^{l+1}$ remains ordered.

Step 2. We now assume
$\displaystyle\sum_{i=1}^{\widehat{\kappa}}v^{l}_{i}<0\leq\sum_{i=1}^{\widehat{\kappa}}v^{l+1}_{i}$
for some $l$. Then we set

$\displaystyle\boldsymbol{v}^{*}:=\Phi(\boldsymbol{v}^{l},X^{*})\quad\mbox{where
$0<X^{*}\leq X_{l+1}$ is determined
by}~{}\sum_{i=1}^{\widehat{\kappa}}v^{*}_{i}=0,$

and stop the algorithm.

We note that, in each recursion of Step 1, we will have either
$v^{l+1}_{k_{1}+1}=v^{l+1}_{k_{1}}$ or $v^{l+1}_{k_{2}+1}=v^{l+1}_{k_{2}}$,
namely $\boldsymbol{v}^{l+1}$ will have at least one more tied component than
$\boldsymbol{v}^{l}$, so Step 1 will stop after finitely many recursions.
Moreover, clearly $\Phi(\boldsymbol{v}^{l},X)$ is continuous and strictly
monotone in $X$, so the transfer $X^{*}$ in Step 2 is unique. The following
proposition shows that the above algorithm implements an equilibrium.

######  Proposition 10.

The promises profile $\boldsymbol{r}^{*}:=\boldsymbol{v}^{*}-\boldsymbol{u}$
produced after a finite number of steps from the above algorithm is an
equilibrium and $v^{*}_{1}\leq\cdots\leq v^{*}_{I}$ with $v^{*}\equiv
v^{\boldsymbol{r}^{*}}$. Moreover, denoting by
$k_{1}^{*}=k_{1}(\boldsymbol{v}^{*})$ and
$k_{2}^{*}=k_{2}(\boldsymbol{v}^{*})$. the committee members defined as in
(88), the following properties hold true:

(i) Members of the coalition of promisers $\\{k_{2}^{*}+1,\cdots,I\\}$, make
individual promises $r_{i}\leq 0$ and get equalized ex post intensities
$v_{i}^{*}=\overline{v}>0$, for all $i>k_{2}^{*}$.

(ii) Members of the the coalition of promisees $\\{1,\cdots,k_{1}^{*}\\}$,
receive individual transfers $r_{i}\geq 0$ and get equalized ex post
intensities $v_{i}^{*}=\underline{v}\leq 0$, for all $i\leq k_{1}^{*}$.

(iii) Members of the coalition $\\{k_{1}^{*}+1,\cdots,k_{2}^{*}\\}$ make no
promises, that is, $r_{i}=0$ and $v_{i}^{*}=u_{i}$ for all $k_{1}^{*}+1\leq
i\leq k_{2}^{*}$.

(iv) Among all equilibria, the promises profile $\boldsymbol{r}^{*}$ produced
by the algorithm generates the ex post intensity $\boldsymbol{v}^{*}$ with the
smallest dispersion. That is, for any $\boldsymbol{r}\in{\cal E}$ such that
$\boldsymbol{r}\neq\boldsymbol{r}^{*}$, we have

$\displaystyle\min_{i}v^{\boldsymbol{r}}_{i}\leq
v^{*}_{1},\quad\max_{i}v^{\boldsymbol{r}}_{i}\geq
v^{*}_{I},\quad\mbox{and}\quad\max_{i}v^{\boldsymbol{r}}_{i}-\min_{i}v^{\boldsymbol{r}}_{i}>v^{*}_{I}-v^{*}_{1}.$

Proposition 10 shows that the algorithm that we describe in this subsection
converges after a finite number of steps to a unique equilibrium. The
algorithm has an egalitarian underpinning: in each step, the utility transfers
flow from the members with the highest intensities to the members with the
lowest intensities while preserving the ordering of the interim intensities.
Perhaps not surprisingly, property (iv) of Proposition 10 shows that the
resulting equilibrium is the one with lowest dispersion. The algorithm is in
fact an equilibrium selection that comes with a dynamic implementation based
on transfers between members of the two tails of the distribution of
intensities. There may be other algorithms based on other principles that will
select other types of equilibria.

We now give the proof of Lemma 7 and that of Proposition 10.

### Proof of Lemma 7:

(i) By (91), we have

$\displaystyle\sum_{i\in\mathbb{I}}r_{i}$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{k_{1}}r_{i}+\sum_{i=k_{1}+1}^{k_{2}}r_{i}+\sum_{i=k_{2}+1}^{I}r_{i}$
$\displaystyle=$ $\displaystyle\sum_{i=1}^{k_{1}}{X\over
k_{1}}+\sum_{i=k_{1}+1}^{k_{2}}0+\sum_{i=k_{2}+1}^{I}(-{X\over
I-k_{2}})=X-X=0.$

That is, $\boldsymbol{r}\in{\cal P}$.

(ii) Recall $u_{1}=u_{k_{1}}$. Since $X\leq k_{1}(u_{k_{1}+1}-u_{k_{1}})$, we
have

$\displaystyle v_{k_{1}}=u_{k_{1}}+{X\over k_{1}}\leq
u_{k_{1}}+(u_{k_{1}+1}-u_{k_{1}})=u_{k_{1}+1}=v_{k_{1}+1}.$ (92)

Similarly, recalling $u_{I}=u_{k_{2}+1}$, since
$X\leq(I-k_{2})(u_{k_{2}+1}-u_{k_{2}})$, then

$\displaystyle v_{k_{2}+1}=u_{k_{2}+1}-{X\over I-k_{2}}\geq
u_{k_{2}+1}-(u_{k_{2}+1}-u_{k_{2}})=u_{k_{2}}=v_{k_{2}}.$ (93)

Now since $v_{i}=u_{i}$ for $k_{1}<i\leq k_{2}$, by (1) we have
$v_{k_{1}+1}\leq\cdots\leq v_{k_{2}}$. Then we obtain

$\displaystyle v_{1}=\cdots=v_{k_{1}}\leq v_{k_{1}+1}\leq\cdots\leq
v_{k_{2}}\leq v_{k_{2}+1}=\cdots=v_{I}.$ (94)

(iii) Inequalities (94) imply $k_{1}(\boldsymbol{u})=k_{1}\leq
k_{1}(\boldsymbol{v})\leq k_{2}(\boldsymbol{v})\leq
k_{2}=k_{2}(\boldsymbol{u})$. ∎

### Proof of Proposition 10:

In the case $\sum_{i=1}^{\widehat{\kappa}}u_{i}\geq 0$ (there are no blocking
coalitions), the algorithm stops at Step 0. All the statements are obvious.

Henceforth we assume $\sum_{i=1}^{\widehat{\kappa}}u_{i}<0$ in this proof.
Note that $\sum_{i=1}^{I}u_{i}>0$, then we must have $u_{1}<0<u_{I}$, and thus
$k_{1}(\boldsymbol{u})$ and $k_{2}(\boldsymbol{u})$ are well defined.

For all $l$ in Step 1 such that
$\displaystyle\sum_{i=1}^{\widehat{\kappa}}v^{l}_{i}<0$, by Lemma 7 (iii) we
see that $k_{1}(\boldsymbol{v}^{l})$ is weakly increasing in $l$ and
$k_{2}(\boldsymbol{v}^{l})$ is weakly decreasing in $l$. Moreover, by (91) we
have

$\displaystyle v^{l+1}_{i}>v^{l}_{i},~{}i\leq k_{1}(\boldsymbol{v}^{l});\quad
v^{l+1}_{k}=v^{l}_{k},~{}k_{1}(\boldsymbol{v}^{l})<k\leq
k_{2}(\boldsymbol{v}^{l});\quad
v^{l+1}_{j}<v^{l}_{j},j>k_{2}(\boldsymbol{v}^{l}).$

Then, since $\boldsymbol{v}^{0}=\boldsymbol{u}$, by induction on $l$ we see
that

$\displaystyle v^{l+1}_{i}>u_{i},~{}i\leq k_{1}(\boldsymbol{v}^{l});\quad
v^{l+1}_{k}=u_{k},~{}k_{1}(\boldsymbol{v}^{l})<k\leq
k_{2}(\boldsymbol{v}^{l});\quad
v^{l+1}_{j}<u_{j},j>k_{2}(\boldsymbol{v}^{l}).$

In particular, for the $l$ in Step 2, we have $k_{1}(\boldsymbol{v}^{l})\leq
k_{1}^{*}\leq k_{2}^{*}\leq k_{2}(\boldsymbol{v}^{l})$, and

$\displaystyle v^{*}_{i}>u_{i},~{}i\leq k_{1}(\boldsymbol{v}^{l});\quad
v^{*}_{k}=u_{k},~{}k_{1}(\boldsymbol{v}^{l})<k\leq
k_{2}(\boldsymbol{v}^{l});\quad v^{*}_{j}<u_{j},j>k_{2}(\boldsymbol{v}^{l}).$

This implies that

$\displaystyle\displaystyle r_{i}>0,~{}i\leq k_{1}(\boldsymbol{v}^{l});\quad
r_{i}=0,~{}k_{1}(\boldsymbol{v}^{l})<i\leq k_{1}^{*};\quad
r_{k}=0,~{}k_{1}^{*}<k\leq k_{2}^{*};$ $\displaystyle\displaystyle
r_{j}=0,~{}k_{2}^{*}<j\leq k_{2}(\boldsymbol{v}^{l});\quad
r_{j}<0,~{}j>k_{2}(\boldsymbol{v}^{l}).$

This proves (iii), and by the definition of $k_{1}^{*},k_{2}^{*}$ in (88), we
have $v^{*}_{1}=\cdots=v^{*}_{k_{1}^{*}}$, and
$v^{*}_{k_{2}^{*}+1}=\cdots=v^{*}_{I}$. Note further that
$\sum_{i=1}^{\widehat{\kappa}}v^{*}_{i}=0<\sum_{i=1}^{I}v^{*}_{i}$ and by
Lemma 7 $\boldsymbol{v}^{*}$ is ordered, so we must have
$\boldsymbol{v}^{*}_{1}\leq 0<v^{*}_{I}$. This completes the proof for (i) and
(ii).

Moreover, again since $\sum_{i=1}^{\widehat{\kappa}}v^{*}_{i}=0$ and
$\boldsymbol{v}^{*}$ is ordered, by setting ${\cal
C}=\\{1,\cdots,\widehat{\kappa}\\}$ in Lemma 4 (ii) we see that
$\boldsymbol{r}^{*}\in{\cal S}_{0}$. Let $\underline{k}_{*},\overline{k}_{*}$
follow the definition in (49) from Lemma 3, using $\boldsymbol{r}^{*}$. By
(i)-(iii) it is clear that

$\displaystyle\underline{k}_{*}\leq k_{1}^{*}\leq
k_{2}^{*}<\overline{k}_{*}\quad\mbox{and}\quad\underline{k}_{*}\leq n\leq
k_{2}^{*}.$

We next prove $\boldsymbol{r}^{*}\in{\cal E}$ and (iv) in three cases.

Case 1: $|{\cal C}^{R}|<\kappa$, and thus $n=I-|{\cal
C}^{R}|\geq\widehat{\kappa}$. By (i) it is clear that $u_{i}\geq v^{*}_{i}>0$
for $i>k_{2}^{*}$, then $n\leq k_{2}^{*}$ and by (ii) and (iii) we have
$v^{*}_{i}\leq 0$ for all $i\leq n$. On the other hand, since
$\sum_{i=1}^{\widehat{\kappa}}v^{*}_{i}=0$ and $\widehat{\kappa}\leq n$, then
we must have $v^{*}_{i}=0$ for all $i\leq\widehat{\kappa}$. Since
$\boldsymbol{v}^{*}$ is ordered, we actually have $v^{*}_{i}=0$, and thus
$r_{i}>0$ for all $i\leq n$. Then $n\leq\underline{k}_{*}$ and in fact they
are equal. Then

$\displaystyle{\cal
T}_{\boldsymbol{r}^{*}}=\sum_{i=1}^{\underline{k}_{*}}r^{*}_{i}=\sum_{i=1}^{n}r^{*}_{i}=\sum_{i\in{\cal
C}^{S}}(-u_{i})=U^{S}.$

Now by Proposition 5 we see that $\boldsymbol{r}^{*}\in{\cal E}$.

Moreover, let $\boldsymbol{r}\in{\cal E}$ be an arbitrary equilibrium. By
Proposition 5 Part 1 we have

$\displaystyle\sum_{i>k_{2}^{*}}(u_{i}-v^{*}_{I})=U^{S}={\cal
T}_{\boldsymbol{r}}=\sum_{i>n}(-r_{i})\geq\sum_{i>k_{2}^{*}}(u_{i}-v^{\boldsymbol{r}}_{i}).$

Here we used the fact that $k_{2}^{*}\geq n$. This clearly implies that
$\max_{i\in\mathbb{I}}v^{r}_{i}\geq v^{*}_{I}$, and if equality holds, we must
have $v^{r}_{i}=v^{*}_{I}$ for all $i>k_{2}^{*}$ and $r_{i}=0$ for all
$n<i\leq k_{2}^{*}$. Similarly, by Proposition 5 Part 2, in both cases we have

$\displaystyle\sum_{i\in{\cal C}^{S}}(v^{*}_{1}-u_{i})=\sum_{i\in{\cal
C}^{S}}r^{*}_{i}=U^{S}={\cal T}_{r}=\sum_{i\in{\cal
C}^{S}}r_{i}=\sum_{i\in{\cal C}^{S}}(v^{r}_{i}-u_{i}).$

Then clearly $\min_{i\in\mathbb{I}}v^{r}_{i}\leq v^{*}_{1}$, and if equality
holds, we must have $v^{r}_{i}=v^{*}_{1}$ for all $i\in{\cal C}^{S}$. The same
logic applies to maximum of $v_{i}^{r}-u_{i}$, directly implying
$\max_{i\in\mathbb{I}}v^{\boldsymbol{r}}_{i}-\min_{i\in\mathbb{I}}v^{\boldsymbol{r}}_{i}\geq
v^{*}_{I}-v^{*}_{1}$. Now if the two sides are actually equal, then
$\max_{i\in\mathbb{I}}v^{r}_{i}=v^{*}_{I}$ and
$\min_{i\in\mathbb{I}}v^{r}_{i}=v^{*}_{1}$, thus $v^{r}_{i}=v^{*}_{1}$ for
$i\in{\cal C}^{S}$, $r_{i}=0$ for $n<i\leq k_{2}^{*}$, and
$v^{r}_{i}=v^{*}_{I}$ for $i>k_{2}^{*}$. This exactly means
$\boldsymbol{r}=\boldsymbol{r}^{*}$. So (iv) holds in this case.

Case 2: $\mid{\cal C}^{R}\mid\geq\kappa$ and $k_{2}^{*}\geq\widehat{\kappa}$.
This implies that $\underline{k}_{*}\leq n<\widehat{\kappa}\leq k_{2}^{*}$.
Note that $r_{i}=0$ for $\underline{k}_{*}<i\leq k_{2}^{*}$ and
$\displaystyle\sum_{i=1}^{\widehat{\kappa}}v^{*}_{i}=0$, then

$\displaystyle{\cal
T}_{\boldsymbol{r}^{*}}=\sum_{i=1}^{\underline{k}_{*}}r^{*}_{i}=\sum_{i=1}^{\widehat{\kappa}}r^{*}_{i}=\sum_{i=1}^{\widehat{\kappa}}(v^{*}_{i}-u_{i})=-\sum_{i=1}^{\widehat{\kappa}}u_{i}=G^{S}.$

Note that by Proposition 7 we see that the minimum total promises transfer is
$G^{S}$ in the case $0<G^{S}\leq\Delta U_{\widehat{\kappa}}$, and by
Proposition 8 it is larger than $G^{S}$ in the case $\Delta
U_{\widehat{\kappa}}<G^{S}$. So we must be in the case $0<G^{S}\leq\Delta
U_{\widehat{\kappa}}$ and $\boldsymbol{r}^{*}\in{\cal E}$.

Moreover, let $\boldsymbol{r}\in{\cal E}$ be an arbitrary equilibrium. By
Proposition 7 Part 1 we have

$\displaystyle\sum_{i>k_{2}^{*}}(u_{i}-v^{*}_{I})={\cal
T}_{\boldsymbol{r}^{*}}=G^{S}={\cal
T}_{r}=\sum_{i>\widehat{\kappa}}(-r_{i})\geq\sum_{i>k_{2}^{*}}(-r_{i})=\sum_{i>k_{2}^{*}}(u_{i}-v^{r}_{i}).$

This clearly implies that $\max_{i\in\mathbb{I}}v^{r}_{i}\geq v^{*}_{I}$, and
if equality holds, we must have $v^{r}_{i}=v^{*}_{I}$ for all $i>k_{2}^{*}$
and $r_{i}=0$ for all $\widehat{\kappa}<i\leq k_{2}^{*}$. Similarly, by
Proposition 7 Part 2 we have

$\displaystyle\sum_{i=1}^{k_{1}^{*}}(v^{*}_{1}-u_{i})=\sum_{i=1}^{k_{1}^{*}}r^{*}_{i}={\cal
T}_{r^{*}}=G^{S}={\cal
T}_{r}=\sum_{i=1}^{\widehat{\kappa}}r_{i}\geq\sum_{i=1}^{k_{1}^{*}}r_{i}=\sum_{i=1}^{k_{1}^{*}}(v^{r}_{i}-u_{i}).$

Then clearly $\min_{i\in\mathbb{I}}v^{r}_{i}\leq v^{*}_{1}$, and if equality
holds, we must have $v^{r}_{i}=v^{*}_{1}$ for all $i\leq k_{1}^{*}$ and
$r_{i}=0$ for all $k_{1}^{*}<i\leq\widehat{\kappa}$. (iv) holds in this case
by subtracting the minimum inequality from the maximum.

Case 3: $\mid{\cal C}^{R}\mid\geq\kappa$ and $k_{2}^{*}<\widehat{\kappa}$. By
the analysis in Case 2, we must have $\Delta U_{\widehat{\kappa}}<G^{S}$, and
thus we are in the situation of Proposition 8. Note that, since
$\sum_{i=1}^{\widehat{\kappa}}v^{*}_{i}=0$,

$\displaystyle\sum_{i>k_{2}^{*}}(u_{i}-v^{*}_{i})=\sum_{i>k_{2}^{*}}(-r^{*}_{i})={\cal
T}_{\boldsymbol{r}^{*}}=\sum_{i\leq k_{2}^{*}}r^{*}_{i}=\sum_{i\leq
k_{2}^{*}}(v^{*}_{i}-u_{i})=-\sum_{i=k_{2}^{*}+1}^{\widehat{\kappa}}v^{*}_{i}-\sum_{i\leq
k_{2}^{*}}u_{i}.$

This implies that, recalling (66),

$\displaystyle\sum_{i=1}^{I}u_{i}=\sum_{i=\widehat{\kappa}+1}^{I}v^{*}_{i}=(\kappa-1)v^{*}_{I},\quad\mbox{and
thus}\quad v^{*}_{I}={1\over\kappa-1}\sum_{i=1}^{I}u_{i}=u_{*}.$

This implies immediately that $k_{2}^{*}=k_{*}$ and ${\cal
T}_{\boldsymbol{r}^{*}}={\cal T}_{*}$, where $k_{*}$ and ${\cal T}_{*}$ are as
in Proposition 8. Therefore, $\boldsymbol{r}^{*}\in{\cal E}$.

Moreover, let $\boldsymbol{r}\in{\cal E}$ be an arbitrary equilibrium. By
Proposition 8 Part 1 we see that
$\max_{i\in\mathbb{I}}v^{r}_{i}=u_{*}=v^{*}_{I}$. On the other hand, recalling
again that $k_{*}=k_{2}^{*}\geq k_{1}^{*}$, by Proposition 8 Part 2 we have

$\displaystyle\sum_{i=1}^{k_{1}^{*}}(v^{*}_{1}-u_{i})=\sum_{i=1}^{k_{1}^{*}}r^{*}_{i}={\cal
T}_{\boldsymbol{r}^{*}}={\cal T}_{*}={\cal
T}_{\boldsymbol{r}}=\sum_{i=1}^{k^{*}}r_{i}\geq\sum_{i=1}^{k_{1}^{*}}r_{i}=\sum_{i=1}^{k_{1}^{*}}(v^{\boldsymbol{r}}_{i}-u_{i}).$

Then clearly $\min_{i\in\mathbb{I}}v^{\boldsymbol{r}}_{i}\leq v^{*}_{1}$, and
if equality holds, we must have $v^{\boldsymbol{r}}_{i}=v^{*}_{1}$ for all
$i\leq k_{1}^{*}$ and $r_{i}=0$ for all $k_{1}^{*}<i\leq k_{2}^{*}$. Putting
the maximum and minimum together shows that (iv) holds in this case as well. ∎