Dilemma with approval and disapproval votes

Abstract

This paper looks at the issue of selecting candidates when the votes cast in ballots enable voters to approve or disapprove each candidate. More precisely, three options are offered: voters can approve, disapprove or remain neutral in regard to each candidate. We define a large family of rules that satisfy desirable properties and prove that solving a dilemma is sufficient to characterize any rule which belongs to this family. In this context a dilemma appears when candidates with only neutral votes face candidates with both supporters and opponents. On the basis of this result, we provide comparable axiomatizations of four rules including some proposed in the literature.

Appendix

Proof of Lemma 1

We construct the function $f:\mathcal {B}\rightarrow \mathcal {B}$ satisfying points 1 to 3 as follows. For each voting profile $B^V_C\in \mathcal {B}$, denote by $\{c_1, c_2, \ldots , c_m\}$ the set of candidates in C and by $\{1, \ldots , n\}$ the set of voters in V. From $B^V_C$, construct the voting profile $\bar{B}_{C}^{V}\in \mathcal {B}$ such that, for each candidate $c_p \in C$, the $n^A_p(B^V_C)$ first voters approve $c_p$, and the $n^D_p(B^V_C)$ last voters disapprove $c_p$. By Anonymity for a given candidate, we conclude that, for each ${\mathcal {W}}\in \tilde{\mathcal {R}}$, ${\mathcal {W}}(B^V_C) = {\mathcal {W}}(\bar{B}_{C}^{V})$. From this observation, we deduce that candidate $c_p \in C$ is Pareto dominated, with respect to $\succsim _{\bar{B}_{C}^{V}}$, by candidate $c_q\in C$ if and only if $n^A_q(\bar{B} _{C}^{V})\ge n^A_p(\bar{B}_{C}^{V})$ and $n^D_q(\bar{B}_{C}^{V}) \le n^D_p(\bar{B}_{C}^{V})$ with at least one strict inequality. Now, denote by $\bar{C} \not = \emptyset$ the subset of candidates in C that are not Pareto dominated in $\bar{B}_{C}^{V}$. By Independence of Pareto dominated candidates, for each ${\mathcal {W}}\in \tilde{\mathcal {R}}$, ${\mathcal {W}}(\bar{B}_{C}^{V}) ={\mathcal {W}}(\bar{B}_{\bar{C}}^{V})$. At this step, we distinguish two exclusive cases.

Case 1. If $\bar{C}$ is a singleton, then set $f(B^V_C)=\bar{B}_{\bar{C}}^{V}$. Because each rule ${\mathcal {W}}$ is nonempty-valued by definition, ${\mathcal {W}}(\bar{B}_{\bar{C}}^{V})=\bar{C}$, so that $f(B^V_C)\in \mathcal {T}$, ${\mathcal {W}}(f(B^V_C))={\mathcal {W}}(B^V_C)$ and $\bar{C}\subseteq C$. In particular, this shows that $\mathcal {T}$ is non empty.

Case 2. If $\bar{C}$ contains at least two elements, pick any two distinct candidates $c_p$ and $c_q$ in $\bar{C}$. Three exclusive subcases arise:

(a) $n^A_p(\bar{B}_{\bar{C}}^{V}) = n^A_q(\bar{B}_{\bar{C}}^{V})$ and $n^D_p(\bar{B}_{\bar{C}}^{V}) = n^D_q(\bar{B}_{\bar{C}}^{V})$;

(b) $n^A_p(\bar{B}_{\bar{C}}^{V}) > n^A_q(\bar{B}_{\bar{C}}^{V})$ and $n^D_p(\bar{B}_{\bar{C}}^{V}) > n^D_q(\bar{B}_{\bar{C}}^{V})$;

(c) $n^A_q(\bar{B}_{\bar{C}}^{V}) > n^A_p(\bar{B}_{\bar{C}}^{V})$ and $n^D_q(\bar{B}_{\bar{C}}^{V}) > n^D_p(\bar{B}_{\bar{C}}^{V})$.

From (a), (b) and (c), we can endow $\bar{C}$ with a weak order (i.e. a complete and transitive relation) $R_{\bar{C}}$ defined as follows: $\forall c_p, c_q \in \bar{C}$,

$$\begin{aligned} c_p R_{\bar{C}} c_q \text{ if } \text{[either } (a) \text{ or } (b) \text{ holds], } \text{ and } c_q R_{\bar{C}} c_p \text{ if } [\text{ either } (a) \text{ or } (c) \text{ holds] }. \end{aligned}$$

Therefore, the strict part of $R_{\bar{C}}$, denoted by $P_{\bar{C}}$, is as follows:

$$\begin{aligned} \forall c_p, c_q \in \bar{C}, \quad c_p P_{\bar{C}} c_q \text{ if } [(b) \text{ holds], } \text{ and } c_q P_{\bar{C}} c_p \text{ if } [(c) \text{ holds }]. \end{aligned}$$

And the indifference part of $R_{\bar{C}}$, denoted by $I_{\bar{C}}$, is as follows:

$$\begin{aligned} \forall c_p, c_q \in \bar{C}, \quad c_p I_{\bar{C}} c_q \quad \text{ if } \quad [(a) \text{ holds }]. \end{aligned}$$

Denote by $\bar{C}/ I_{\bar{C}}$ the set of indifference classes of $(\bar{C} , R_{\bar{C}})$, and denote by $k \ge 1$ the cardinality of $\bar{C}/ I_{\bar{C}}$. We can observe that the maximal elements of $(\bar{C}, R_{ \bar{C}})$ belongs to the same indifference class denoted by [1]. Similarly, the maximal elements of $(\bar{C}{\setminus } [1], R_{\bar{C} {\setminus } [1]})$, where $R_{\bar{C} {\setminus } [1]}$ refers to the relation $R_{\bar{C}}$ restricted to $\bar{C} {\setminus } [1]$, are in the same indifference class denoted by [2]. Proceeding in this way until the class [k], we deduce that the indifference classes can be ordered as follows:

$$\begin{aligned} {[}1] R^*_{\bar{C}} [2] R^*_{\bar{C}} \cdots R^*_{\bar{C}} [k -1] R^*_{\bar{C} } [k] \end{aligned}$$

(1)

where $R_{\bar{C}}^*$ is the induced quotient relation (complete, transitive and antisymmetric) on $\bar{C}/ I_{\bar{C}}$.

To complete the construction of f, we proceed by induction on the number of indifference classes. If $k = 1$, then, by Neutrality and the fact that each ${\mathcal {W}}\in \tilde{\mathcal {R}}$ is nonempty-valued, ${\mathcal {W}}(\bar{B}_{\bar{C}}^{V}) = [1]$ and so $\bar{B}_{\bar{C}}^{V}\in \mathcal {T}$. Set $f(B^V_C)=\bar{B}_{\bar{C}}^{V}$. Once again, we have $f(B^V_C)\in \mathcal {T}$, ${\mathcal {W}}(f(B^V_C))={\mathcal {W}}(B^V_C)$ and $\bar{C}\subseteq C$. Next, assume that f is defined for each $\bar{B}_{\bar{C}}^{V} \in \mathcal {B}$ such that $(\bar{C}, R_{\bar{C}})$ contains k indifference classes where $1 \le k < r$. Pick any $\bar{B}_{\bar{C}}^{V} \in \mathcal {B}$ such that $(\bar{C}, R_{\bar{C}})$ contains $k =r$ indifference classes. By construction, for each $c_{\ell } \in [k]$, each $p\in \{1,\ldots ,k\}$ and each $c_q\in [p]$, we have $n^A_{\ell } (\bar{B}_{\bar{C}}^{V})\le n^A_q( \bar{B}_{\bar{C}}^{V})$ and $n^D_{\ell }(\bar{B}_{\bar{C}}^{V})\le n^D_q( \bar{B}_{\bar{C}}^{V})$. Because the elements of each column of $\bar{B} _{\bar{C}}^{V}$ are ranked in nondecreasing order, we can deduce that the first $n^A_{\ell }(\bar{B}_{\bar{C}}^{V})$ voters are unconcerned voters and approve each candidate, while the last $n^D_{\ell }(\bar{B} _{\bar{C}}^{V})$ voters are unconcerned voters who disapprove each candidate. Denote by $V^{\prime }\subsetneq V$ this set of unconcerned voters in $\bar{B}_{\bar{C}}^{V}$. By Unconcerned voters, we deduce that, for each ${\mathcal {W}}\in \tilde{\mathcal {R}}$, ${\mathcal {W}}(\bar{B}_{\bar{C}}^{V})={\mathcal {W}}(\bar{B}_{\bar{C}}^{V{\setminus } V^{\prime }})$. Because $\bar{C}/ I_{\bar{C}}$ contains at least two indifference classes, one can observe that each candidate $c_{\ell }$ in [k] receive the same positive number of neutral votes N. Hence, we deduce that $\mathcal {C}_N(\bar{B}_{\bar{C}}^{V{\setminus } V^{\prime }})$ is nonempty and equal to [k], and for each other class $[p]\ne [k]$ and each $c_{\ell }\in [p]$, $c_{\ell }\in \mathcal {C}_E(\bar{B}_{\bar{C}}^{V{\setminus } V^{\prime }})$. This means that $\bar{B}_{\bar{C}}^{V{\setminus } V^{\prime }}\in \mathcal {D}$. Set $f(B^V_C)=\bar{B}_{\bar{C}}^{V{\setminus } V^{\prime }}$. Obviously, we have $f(B^V_C)\in \mathcal {D}$, ${\mathcal {W}}(f(B^V_C))={\mathcal {W}}(B^V_C)$ and $\bar{C}\subseteq C$. Thus, we have constructed a function $f:\mathcal {B}\rightarrow \mathcal {B}$ which satisfies points 1 to 3 of the statement of Lemma 1, as desired. $\square$

Proof of Theorem 1

By definition of a dilemma resolving property $\mathbb {A}$, there is at least one rule ${\mathcal {W}}\in \tilde{\mathcal {R}}$ satisfying $\mathbb {A}$. Consider two rules ${\mathcal {W}}$ and ${\mathcal {W}}'$ in $\tilde{\mathcal {R}}$ satisfying the dilemma resolving property $\mathbb {A}$. Let $B^V_C\in \mathcal {B}$ be an arbitrary voting profile. By Lemma 1, there exists $B_{\bar{C}}^{\bar{V}}=f(B^V_C)$ such that $B_{\bar{C}}^{\bar{V}}\in \mathcal {D}\cup \mathcal {T}$, ${\mathcal {W}}(B^V_C)={\mathcal {W}}(B_{\bar{C}}^{\bar{V}})$ and ${\mathcal {W}}'(B^V_C)={\mathcal {W}}'(B_{\bar{C}}^{\bar{V}})$. If $B_{\bar{C}}^{\bar{V}}\in \mathcal {T}$ then, by definition of $\mathcal {T}$, ${\mathcal {W}}(B^V_C)={\mathcal {W}}'(B^V_C)$; if $B_{\bar{C}}^{\bar{V}}\in \mathcal {D}$, then ${\mathcal {W}}(B^V_C)={\mathcal {W}}'(B^V_C)$ follows from the fact that both rules satisfy property $\mathbb {A}$. Thus, we have shown that there is exactly one rule belonging to $\tilde{\mathcal {R}}$ and satisfying property $\mathbb {A}$. $\square$

Proof of Proposition 1

Let ${\mathcal {W}}$ and ${\mathcal {W}}'$ be two rules in $\tilde{\mathcal {R}}$. We have to show that ${\mathcal {W}}$ and ${\mathcal {W}}'$ coincide on $\mathcal {D}$ whenever they satisfy one of the following properties.

• 1. Choosing neutral candidates. If ${\mathcal {W}}$ and ${\mathcal {W}}'$ satisfy Choosing neutral candidates, then, for each dilemma $B_{\bar{C}}^{\bar{V}}$ in $\mathcal {D}$,

  $$\begin{aligned}{\mathcal {W}}(B_{\bar{C}}^{\bar{V}})=\mathcal {C}_E(B_{\bar{C}}^{\bar{V}}) \quad \text{ and } \quad {\mathcal {W}}'(B_{\bar{C}}^{\bar{V}})=\mathcal {C}_E(B_{\bar{C}}^{\bar{V}}).\end{aligned}$$

  Hence, for each dilemma $B_{\bar{C}}^{\bar{V}}$ in $\mathcal {D}$, ${\mathcal {W}}(B_{\bar{C}}^{\bar{V}})={\mathcal {W}}'(B_{\bar{C}}^{\bar{V}})$, which proves that Choosing neutral candidates is dilemma resolving.

• 2. Independence to dropping neutral candidates. Assume that ${\mathcal {W}}$ and ${\mathcal {W}}'$ satisfy Independence to dropping neutral candidates, and consider any dilemma $B_{\bar{C}}^{\bar{V}}$ in $\mathcal {D}$. We construct a finite sequence $(B_{C_\ell }^{V_{\ell }})_{\ell \in \{0,\dots ,\ell ^*\}}$ of voting profiles from $\mathcal {D}\cup \mathcal {T}$ such that:

  • (a) $B_{C_0}^{V_{0}}=B_{\bar{C}}^{\bar{V}}$;

  • (b) $B_{C_{\ell ^*}}^{V_{\ell ^*}}\in \mathcal {T}$;

  • (c) for each $\ell \in \{0,\ldots ,\ell ^*\}$, ${\mathcal {W}}(B_{C_\ell }^{V_{\ell }})={\mathcal {W}}(B_{C_0}^{V_{0}})$ and ${\mathcal {W}}'(B_{C_\ell }^{V_{\ell }})={\mathcal {W}}'(B_{C_0}^{V_{0}})$.

  Observe that if such sequence exists, then, by the point (b),

  $$\begin{aligned} {\mathcal {W}}(B_{C_{\ell ^*}}^{V_{\ell ^*}})={\mathcal {W}}'(B_{C_{\ell ^*}}^{V_{\ell ^*}}), \end{aligned}$$

  and, by points (a) and (c),

  $$\begin{aligned} {\mathcal {W}}\left( B_{C_{\ell ^*}}^{V_{\ell ^*}}\right) ={\mathcal {W}}\left( B_{\bar{C}}^{\bar{V}}\right) \quad \text{ and } \quad {\mathcal {W}}'\left( B_{C_{\ell ^*}}^{V_{\ell ^*}}\right) ={\mathcal {W}}'\left( B_{\bar{C}}^{\bar{V}}\right) . \end{aligned}$$

  This will prove that ${\mathcal {W}}$ and ${\mathcal {W}}'$ coincide on $\mathcal {D}$. It remains to prove that such a sequence $(B_{C_\ell }^{V_{\ell }})_{\ell \in \{0,\dots ,\ell ^*\}}$ exists. Because ${\mathcal {W}}$ and ${\mathcal {W}}'$ satisfy Independence to dropping neutral candidates and $B_{C_0}^{V_{0}}=B_{\bar{C}}^{\bar{V}}$ belongs to $\mathcal {D}$, we have

  $$\begin{aligned} {\mathcal {W}}\left( B_{C_0}^{V_{0}}\right) ={\mathcal {W}}\left( B_{\mathcal {C}_E\left( B_{C_0}^{V_{0}}\right) }^{\bar{V}}\right) \quad \text{ and } \quad {\mathcal {W}}'\left( B_{C_0}^{V_{0}}\right) ={\mathcal {W}}'\left( B_{\mathcal {C}_E\left( B_{C_0}^{V_{0}}\right) }^{\bar{V}}\right) . \end{aligned}$$

  (2)

  Let $\ell \ge 0$ be a natural number such that $B_{C_\ell }^{V_{\ell }}\in \mathcal {D}$. Such a natural number $\ell$ exists because $B_{C_0}^{V_{0}} \in \mathcal {D}$. From $B_{C_\ell }^{V_{\ell }} \in \mathcal {D}$, we have $C_{\ell }=\mathcal {C}_E(B_{C_\ell }^{V_{\ell }})\cup \mathcal {C}_N(B_{C_\ell }^{V_{\ell }})$ and both $\mathcal {C}_E(B_{C_\ell }^{V_{\ell }})$ and $\mathcal {C}_N(B_{C_\ell }^{V_{\ell }})$ are nonempty. It follows trivially that $\mathcal {C}_E(B_{C_\ell }^{V_{\ell }})\subsetneq C_{\ell }$. By Lemma 1, there exists $B_{C_{\ell +1}}^{V_{\ell +1}}=f(B_{\mathcal {C}_E(B_{C_\ell }^{V_{\ell }})}^{V_{\ell }})$ such that

  • (a) $B_{C_{\ell +1}}^{V_{\ell +1}}\in \mathcal {D}\cup \mathcal {T}$;

  • (b) ${\mathcal {W}}(B_{\mathcal {C}_E(B_{C_\ell }^{V_{\ell }})}^{V_{\ell }})={\mathcal {W}}(B_{C_{\ell +1}}^{V_{\ell +1}})$ and ${\mathcal {W}}'(B_{\mathcal {C}_E(B_{C_\ell }^{V_{\ell }})}^{V_{\ell }})={\mathcal {W}}'(B_{C_\ell }^{V_{\ell }})$;

  • (c) $C_{\ell +1}\subseteq \mathcal {C}_E(B_{C_\ell }^{V_{\ell }})\subsetneq C_{\ell }$.

  By Eq. (2), a successive application of Independence to dropping neutral candidates and point (b), we obtain:

  $$\begin{aligned} {\mathcal {W}}\left( B_{C_0}^{V_{0}}\right) ={\mathcal {W}}\left( B_{C_{\ell +1}}^{V_{\ell +1}}\right) \quad \text{ and }\quad {\mathcal {W}}'\left( B_{C_0}^{V_{0}}\right) ={\mathcal {W}}'\left( B_{C_{\ell +1}}^{V_{\ell +1}}\right) . \end{aligned}$$

  Two cases arise. If $B_{C_{\ell +1}}^{V_{\ell +1}}\in \mathcal {T}$, then it suffices to set $\ell ^*=\ell +1$ to complete the proof. Otherwise, $B_{C_{\ell +1}}^{V_{\ell +1}}$ belongs to $\mathcal {D}$ and we repeat the procedure until to reach a voting profile in $\mathcal {T}$. This procedure terminates after a finite number of steps because $C_0$ is finite, $C_{\ell +1}\subsetneq C_{\ell }$ for each $\ell \ge 0$ such that $B_{C_\ell }^{V_{\ell }}\in \mathcal {D}$, and $B_{C_{\ell }}^{V_{\ell }}\in \mathcal {T}$ as soon as $|C_{\ell }|=1$.

• 3. Compromise. The proof is very much like that of the previous case and so left to the reader.

• 4. Choosing neutral if preferred by a majority. Assume that ${\mathcal {W}}$ and ${\mathcal {W}}'$ satisfy Choosing neutral if preferred by a majority, and consider any dilemma $B_{\bar{C}}^{\bar{V}}$ in $\mathcal {D}$. Two cases arise. In the first case, $\mathcal {C}_O(B_{\bar{C}}^{\bar{V}})\cap \mathcal {C}_N(B_{\bar{C}}^{\bar{V}})\ne \emptyset$, then, by definition of Choosing neutral if preferred by a majority, ${\mathcal {W}}(B_{\bar{C}}^{\bar{V}}) = \mathcal {C}_O(B_{\bar{C}}^{\bar{V}})$. If the first case does not hold, then, by Choosing neutral if preferred by a majority, ${\mathcal {W}}(B_{\bar{C}}^{\bar{V}})={\mathcal {W}}(B _{\mathcal {C}_{E}(B_{\bar{C}}^{\bar{V}})}^{\bar{V}})$. Then, we continue the proof in a similar way as for the proof of Independence to dropping neutral candidates.

$\square$

Proof of Theorem 2

Existence part.${\mathcal {W}}^S$, ${\mathcal {W}}^{min}$, ${\mathcal {W}}^{max}$ and ${\mathcal {W}}^S$ belong to $\tilde{\mathcal {R}}$ is straightforward and so it is left to the reader.

Compromise. We now show that ${\mathcal {W}}^S$ satisfies Compromise. Pick any voting profile $B^V_C$ in $\mathcal {D}$. To show:

$$\begin{aligned} {\mathcal {W}}^S\left( B^V_C\right) =\mathcal {C}_{N}\left( B _{C}^{V}\right) \cup {\mathcal {W}}^S\left( B _{\mathcal {C}_{E}\left( B _{C}^{V}\right) }^{V}\right) . \end{aligned}$$

Consider a voting profile $B^V_C$ in $\mathcal {D}$. Call a candidate mixed if it belongs to $\mathcal {C}_{E}(B _{C}^{V})$. We know that mixed candidates cannot be compared to a neutral candidate by the Suppes dominance relation. Therefore, the set of non-dominated candidates in $B^V_C$ with respect to the Suppes dominance relation are exactly the neutral candidates and the non-dominated candidates among the mixed ones, as asserted.

Choosing neutral candidates. We show that ${\mathcal {W}}^{min}$ satisfies Choosing neutral candidates on $\mathcal {B}$. Consider a voting profile $B^V_C\in \mathcal {D}$. Because the number of disapproval votes received by a candidate in $\mathcal {C}_{E}(B^V_C)$ is always positive, we can conclude that $\arg \min _{c_p\in C}n^D(B^V_C)=\mathcal {C}_N(B^V_C)$. By definition of the Leximin rule, we easily conclude that ${\mathcal {W}}^{min} (B^V_C)=\mathcal {C}_N(B^V_C)$, as desired.

Independence to dropping candidates. To show that ${\mathcal {W}}^{max}$ satisfies Independence to dropping candidates on $\mathcal {B}$, it suffices to proceed in a similar way as above for ${\mathcal {W}}^{min}$. The details are left to the reader.

Choosing neutral if preferred by a majority. Pick any voting profile $B^V_C$ in $\mathcal {D}$ and any candidate $c_p\in \mathcal {C}_{N}(B^V_C)$. Observe that $n_{p}^A(B^V_C)-n_{p}^D(B^V_C)=0$; and, for each candidate $c_q\in \mathcal {C}_E(B^V_C)$, $m_{pq}(B^V_C)=n_{q}^{D}(B^V_C)$ and $m_{qp}(B^V_C)=n_{q}^A(B^V_C)$. It follows that

$$\begin{aligned} m_{pq}\left( B^V_C\right) -m_{qp}\left( B^V_C\right) =n_{q}^D\left( B^V_C\right) -n_{q}^A\left( B^V_C\right) . \end{aligned}$$

Two cases arise:

• if $\mathcal {C}_O(B^V_C)\cap \mathcal {C}_N(B^V_C)\ne \emptyset$, then, for each $c_{\ell }\in C$, we have $n_{\ell }^A(B^V_C)-n_{\ell }^D(B^V_C)\le 0$, and, for each $c_{\ell }\in \mathcal {C}_O(B^V_C)$, we have $n_{\ell }^A(B^V_C)-n_{\ell }^D(B^V_C)=0$. It follows that ${\mathcal {W}}^{D \& A}(B^V_C)=\mathcal {C}_O(B^V_C)$;

• if $\mathcal {C}_O(B^V_C)\cap \mathcal {C}_N(B^V_C)= \emptyset$, then there exists $c_q\in \mathcal {C}_E(B^V_C)$ such that $n_{q}^A(B^V_C)-n_{q}^D(B^V_C)>0$. By definition of the Dis & approval rule, ${\mathcal {W}}^{D \& A}(B^V_C)={\mathcal {W}}^{D \& A}(B _{\mathcal {C}_{E}(B^V_C)}^{V})$.

Uniqueness part. The uniqueness part is a direct consequence of Theorem 1, Proposition 3 together with the above existence part. $\square$

Proof of Proposition 2

Let V and $V'$ be two disjoint sets of voters such that, for each $c_p\in C$, $n_p^A(B^V_C)=n_p^A(B_{C}^{V'})$ and $n_p^D(B^V_C)=n_p^D(B_{C}^{V'})$. Let ${\mathcal {W}}$ be a rule that satisfies Compensation and Consistency with respect to balanced population. To show: ${\mathcal {W}}(B^V_C)={\mathcal {W}}(B_{C}^{V'})$. Consider a set of voters $\bar{V}$ disjoint from both V and $V'$, choose a voting profile $B_{C}^{\bar{V}} \in \mathcal {B}$ such that:

$$\begin{aligned} \forall c_p \in C, \quad n_p^A(B_{C}^{\bar{V}}) = \max _{c_q \in C} n_q^A (B_{C}^{V})- n_p^A (B_{C}^{V}) + \max _{c_q \in C} n_q^D (B_{C}^{V}) \end{aligned}$$

and

$$\begin{aligned} n_p^D\left( B_{C}^{\bar{V}}\right) = \max _{c_q \in C} n_q^D \left( B_{C}^{V}\right) - n_p^D \left( B_{C}^{V}\right) + \max _{c_q \in C} n_q^A \left( B_{C}^{V}\right) . \end{aligned}$$

By construction, $B_{C}^{(V\cup \bar{V})}$ and $B_{C}^{(V'\cup \bar{V})}$ are two balanced voting profiles. By Compensation, we have:

$$\begin{aligned} {\mathcal {W}}\left( B_{C}^{\left( V'\cup \bar{V}\right) }\right) ={\mathcal {W}}\left( B_{C}^{\left( V\cup \bar{V}\right) }\right) =C \end{aligned}$$

(3)

Taking into account (3) and applying Consistency with respect to a balanced profile, we obtain:

$$\begin{aligned} {\mathcal {W}}\left( B_{C}^{V'\cup \left( \bar{V}\cup V\right) }\right) = {\mathcal {W}}\left( B_{C}^{V'}\right) \quad \text{ and } \quad {\mathcal {W}}\left( B_{C}^{V'\cup \left( \bar{V}\cup V\right) }\right) = {\mathcal {W}}\left( B_{C}^{V}\right) , \end{aligned}$$

and so ${\mathcal {W}}(B^V_C)={\mathcal {W}}(B_{C}^{V'})$, as desired. $\square$

Proof of Proposition 3

By Theorem 2, ${\mathcal {W}}^S$ satisfies these two properties. By the proof of Lemma 1, if a rule satisfies Anonymity for a given candidate and Independence of Pareto dominated candidates, then, for each $B^V_C\in \mathcal {B}$, this rule selects the nonempty subset $\bar{C}$ of candidates in C that are not Pareto dominated in $\bar{B}_{C}^{V}$. Recall from the proof of Lemma 1 that $\bar{B}_{C}^{V}$ is the voting profile constructed from $B^V_C$ in the following way: for each candidate $c_p \in C$, the $n^A_p(B^V_C)$ first voters approve $c_p$, and the $n^D_p(B^V_C)$ last voters disapprove $c_p$. From the definition of the Suppes relation $\sqsupseteq _{B^V_C}$, we know that $\sqsupseteq _{B^V_C}$ coincides with $\succsim _{\bar{B}_C^V}$. Consequently, $\bar{C}$ is precisely the subset of candidates selected by ${\mathcal {W}}^S(B^V_C)$. Conclude that ${\mathcal {W}}^S$ is the largest rule satisfying Anonymity for a given candidate and Independence of Pareto dominated candidates. $\square$

Proof of Proposition 4

By way of contradiction, assume there is $B^V_C\in \mathcal {B}$ such that

$$\begin{aligned} {\mathcal {W}}^{ACE}(B^V_C) \cap \text{ Par } (B^V_C) = \emptyset . \end{aligned}$$

This means that for each $c_p \in {\mathcal {W}}^{ACE}(B^V_C)$, there is $c_{q(p)} \in \text{ Par } (B^V_C)$ such that $c_{q(p)} \succ _{B^V_C} c_p$. Consider any $c_p \in {\mathcal {W}}^{ACE}(B^V_C)$. First $c_p$ is not a Condorcet winner. Otherwise, ${\mathcal {W}}^{ACE}(B^V_C) = \{c_p\}$ and $c_{q(p)}$ would be also a Condorcet winner, a contradiction. Second, $c_{q(p)}$ is not a Condorcet loser. Otherwise, $c_p$, which is Pareto dominated by $c_{q(p)}$, would be also a Condorcet loser, a contradiction with the fact that $c_p \in {\mathcal {W}}^{ACE}(B^V_C)$. Third, assume that $c_{q(p)} \in \arg \max _{c_r \in C}n_{r}^{D}(B^V_C)$. Since $c_{q(p)} \succ _{B^V_C} c_p$, we also have $c_p \in \arg \max _{c_r \in C}n_{r}^{D}(B^V_C)$, a contradiction with the fact that $c_p \in {\mathcal {W}}^{ACE}(B^V_C)$. From this, we deduce that $c_{q(p)}$ will not be removed at any step of the procedure defining ${\mathcal {W}}^{ACE}(B^V_C)$. Therefore, $c_{q(p)} \in {\mathcal {W}}^{ACE}(B^V_C)$ which contradicts the initial assumption. $\square$

Proof of Proposition 5

By Proposition 3, it holds that:

$$ \begin{aligned} \forall B^V_C\in \mathcal {B}, \quad {\mathcal {W}}^{S} (B^V_C) \supseteq {\mathcal {W}}^{D \& A} (B^V_C) \end{aligned}$$

(4)

In case

$$\begin{aligned} P(B^V_C) = \left\{ c_p \in C: n_p^{A} > n_p^{D} \right\} \not = \emptyset , \end{aligned}$$

it must be clear that

$$ \begin{aligned} {\mathcal {W}}^{D \& A} (B^V_C) \subseteq {\mathcal {W}}^{P} (B^V_C). \end{aligned}$$

(5)

From (4) and (5) we conclude that

$$\begin{aligned} {\mathcal {W}}^S(B^V_C)\cap {\mathcal {W}}^{P}(B^V_C) \ne \emptyset . \end{aligned}$$

In case $P(B^V_C) = \emptyset$, ${\mathcal {W}}^{P}(B^V_C) = C$, and so

$$\begin{aligned} {\mathcal {W}}^S(B^V_C)\cap {\mathcal {W}}^{P}(B^V_C) \ne \emptyset \end{aligned}$$

holds trivially.

Next, assume first that

$$\begin{aligned} \max _{ c_p \in C} m_p^n (B^V_C) = 1. \end{aligned}$$

By definition, ${\mathcal {W}}^{M}(B^V_C)$ selects the subset of candidates $c_p \in C$ such that $m_p^n (B^V_C) = 1$. Among the candidate(s) in ${\mathcal {W}}^{M}(B^V_C)$, consider those with the greatest number of approval votes. These candidates cannot be dominated with respect to the Suppes rule, which implies that they belong to ${\mathcal {W}}^{S}(B^V_C)$ as well. Conclude that

$$\begin{aligned} {\mathcal {W}}^S(B^V_C)\cap {\mathcal {W}}^{M}(B^V_C) \ne \emptyset . \end{aligned}$$

For

$$\begin{aligned} \max _{ c_p \in C} m_p^n (B^V_C) \in \{- 1, 0\}, \end{aligned}$$

we proceed in a similar way to obtain the desired conclusion. $\square$