Inductive Reasoning in Social Choice Theory

Abstract

The usual procedure in the theory of social choice consists in postulating some desirable properties which an aggregation procedure should verify and derive from them the features of a corresponding social choice function and the outcomes that arise at each possible profile of preferences. In this paper we invert this line of reasoning and try to infer, up from what we call social situations (each one consisting of a profile and the associated social ordering) the criteria verified in the implicit aggregation procedure. This inference process, which extracts intensional from extensional information can be seen as an exercise in “qualitative statistics”.

Appendix

Proof of Proposition 1

It is immediate from the definition of $\simeq _{{\mathcal {D}}({\mathcal {G}}_{S})}$ that for any situation ${\mathcal {S}}$, ${\mathcal {S}} \simeq _{{\mathcal {D}}({\mathcal {G}}_{S})}{\mathcal {S}}$, just taking the identity permutation $Id_S$, which belongs to ${\mathcal {G}}_{S}$ by (A2). That is, $\simeq _{{\mathcal {D}}({\mathcal {G}}_{S})}$ is reflexive. To see that it is symmetric, assume ${\mathcal {S}} \simeq _{{\mathcal {D}}({\mathcal {G}}_{S})}{\mathcal {S}^{'}}$ under a $\gamma \in {\mathcal {G}}_{S}$. Then, by (A3), $\gamma ^{-1} \in {\mathcal {G}}_{S}$, and thus ${\mathcal {S}}^{'} \simeq _{{\mathcal {D}}({\mathcal {G}}_{S})}{\mathcal {S}}$. On the other hand, when ${\mathcal {S}} \simeq _{{\mathcal {D}}({\mathcal {G}}_{S})}{\mathcal {S}^{'}}$ and ${\mathcal {S}}^{'} \simeq _{{\mathcal {D}}({\mathcal {G}}_{S})}{\mathcal {S}^{''}}$ we have that there exist permutations $\gamma , \gamma ^{'} \in {\mathcal {G}}_{S}$ such that for each $s,t \in S$

$$\begin{aligned}&D^{{\mathcal {S}}}(\{s,t\})=D^{{\mathcal {S}}^{'}}(\{\gamma (s),\gamma (t)\}) \ \hbox {and} \\&\quad \ D^{{\mathcal {S}}^{'}}(\{\gamma (s),\gamma (t)\}) \ = \ D^{{\mathcal {S}}^{''}}(\{\gamma ^{'} \circ \gamma (s),\gamma ^{'} \circ \gamma (t)\}) \end{aligned}$$

That is

$$\begin{aligned} D^{{\mathcal {S}}}(\{s,t\}) \ = \ D^{{\mathcal {S}}^{''}}(\{\gamma ^{'} \circ \gamma (s),\gamma ^{'} \circ \gamma (t)\}) \end{aligned}$$

Since, by (A1) $\gamma ^{'} \circ \gamma \in {\mathcal {G}}_{S}$, ${\mathcal {S}} \simeq _{{\mathcal {D}}({\mathcal {G}}_{S})}{\mathcal {S}^{''}}$. That is, $\simeq _{{\mathcal {D}}({\mathcal {G}}_{S})}$ is transitive. $\square$

Proof of Theorem 1

$\Rightarrow )$ Given $\bar{\mathcal {S}} \in \widehat{\mathcal {S}}/\simeq _{{\mathcal {D}}({\mathcal {G}}_{S})}$ consider the family of its decision sets, ${\mathcal {F}}_{{\mathcal {G}}_{S}}(\bar{\mathcal {S}})$ and define a function ${\mathbf {f}}_{\bar{\mathcal {S}}}$ as follows. For each profile $\langle R_{1},\ldots , R_{n}\rangle$ and for each pair $s,t \in S$, consider the decisive set $D(\{s,t\})\in {\mathcal {F}}_{{\mathcal {G}}_{S}}(\bar{\mathcal {S}})$. Then, for every $\gamma \in {\mathcal {G}}_{S}$:

$$\begin{aligned} {\mathbf {f}}\big (\hbox {prof}_{\gamma ({\mathcal {S}})}\big ) = R(\gamma (s),\gamma (t)) \ \hbox {iff for every} \ i \in D(\{s,t\}), R_{i}(\gamma (s),\gamma (t)). \end{aligned}$$

By the definition of ${\mathcal {F}}_{{\mathcal {G}}_{S}}(\bar{\mathcal {S}})$, ${\mathbf {f}}_{\bar{\mathcal {S}}} \in \mathbf{F}_{{\mathcal {G}}_{S}}$. Moreover, if for every ${\mathcal {S}}=\langle S, R_{1}, \ldots , R_{n};{\bar{R}}\rangle \notin \bar{\mathcal {S}}$ we define ${\mathbf {f}}_{\bar{\mathcal {S}}}\big (R_{1},\ldots ,R_{n}\big )= \emptyset$, it is immediate that ${\mathbf {f}}_{\bar{\mathcal {S}}}$ is prime.

$\Leftarrow )$ Consider the class of situations with nonempty social orders which can be seen as defined by an arbitrary prime aggregation function ${\mathbf {f}} \in \mathbf{F}_{{\mathcal {G}}_{S}}$, $\widehat{\mathcal {S}}^{{\mathbf {f}}}=\{{\mathcal {S}} \in \widehat{\mathcal {S}}: {\mathcal {S}}= \langle S, R_{1}, \ldots , R_{n};{\bar{R}}\rangle , {\mathbf {f}}(R_{1},\ldots ,R_{n})= {\bar{R}} \ne \emptyset \}$. Assume, by way of contradiction, that there exist two equivalence classes $\bar{\mathcal {S}}, \bar{\mathcal {S}}^{'} \in \widehat{\mathcal {S}}/\simeq _{{\mathcal {D}}({\mathcal {G}}_{S})}$ such that $\widehat{\mathcal {S}}^{{\mathbf {f}}}\cap \bar{\mathcal {S}} \ne \emptyset$ and $\widehat{\mathcal {S}}^{{\mathbf {f}}}\cap \bar{\mathcal {S}}^{'} \ne \emptyset$. Consider two situations ${\mathcal {S}} \in \widehat{\mathcal {S}}^{{\mathbf {f}}}\cap \bar{\mathcal {S}}$ and ${\mathcal {S}}^{'} \in \widehat{\mathcal {S}}^{{\mathbf {f}}}\cap \bar{\mathcal {S}}^{'}$. Then, ${\mathcal {S}}$ and ${\mathcal {S}}^{'}$ are not equivalent. This can happens for two reasons: for every permutations $\gamma$ there exists at least a pair $s,t \in S$ such that either

• $D^{\bar{\mathcal {S}}}(\{s,t\})\ne \big (D^{\bar{\mathcal {S}}^{'}}(\{\gamma ^{-1}(s),\gamma ^{-1}(t)\})\big )$

or

• either ${\bar{R}}(s,t)$ or ${\bar{R}}^{'}(\gamma (s),\gamma (t))$ is not defined.

Let us analyze the latter case. Without loss of generality, assume that ${\mathcal {S}}=\langle S, R_{1}, \ldots , R_{n};{\bar{R}}\rangle$ and ${\mathcal {S}}^{'}=\langle S, R_{1}^{'}, \ldots , R_{n}^{'};{\bar{R}}^{'}\rangle$ are such that $R_{i}\equiv R_{i}^{'}$ for $i=1,\ldots ,n$ while ${\bar{R}}(s,t)$ and not ${\bar{R}}^{'}(\gamma (s),\gamma (t))$ (nor ${\bar{R}}^{'}(\gamma (t),\gamma (s))$), in particular for $\gamma \equiv Id_S$. Since ${\mathcal {S}}, {\mathcal {S}}^{'} \in \widehat{\mathcal {S}}^{{\mathbf {f}}}$, ${\bar{R}}={\mathbf {f}}(R_{1},\ldots ,R_{n})$ and ${\bar{R}}^{'}={\mathbf {f}}(R_{1},\ldots ,R_{n})$. But then, since we assumed that ${\bar{R}} \not \equiv {\bar{R}}^{'}$ we have a contradiction because ${\mathbf {f}}$ is assumed to be a function.

Now suppose that $D^{\bar{\mathcal {S}}}(\{s,t\})\ne D^{\bar{\mathcal {S}}^{'}}(\{\gamma ^{-1}(s),\gamma ^{-1}(t)\})$ for at least a pair of alternatives, s and t. According to the $\Rightarrow$ part of this proof there exist two prime aggregation functions ${\mathbf {f}}^{'}$ and ${\mathbf {f}}^{''}$ in $\mathbf{F}_{{\mathcal {G}}_{S}}$ such that ${\mathbf {f}}^{'}(R_{1},\ldots ,R_{n})= {\bar{R}}$ and ${\mathbf {f}}^{''}(R_{1},\ldots ,R_{n})= {\bar{R}}^{'}$. On the other hand we have that ${\mathbf {f}}^{'}(R_{1},\ldots ,R_{n})\subseteq {\mathbf {f}}(R_{1},\ldots ,R_{n})$ and ${\mathbf {f}}^{''}(R_{1},\ldots ,R_{n})\subseteq {\mathbf {f}}(R_{1},\ldots ,R_{n})$, but this is absurd since ${\mathbf {f}}$ is prime. This shows that there exists an equivalence class $\bar{\mathcal {S}}\in \widehat{\mathcal {S}}/\simeq _{{\mathcal {D}}({\mathcal {G}}_{S})}$, such that $\widehat{\mathcal {S}}^{{\mathbf {f}}}\subseteq \bar{\mathcal {S}}$.

Now assume that there exists a situation ${\mathcal {S}} \in \bar{\mathcal {S}}$ and ${\mathcal {S}} \notin \widehat{\mathcal {S}}^{{\mathbf {f}}}$. This means that if ${\mathcal {S}}$ is $\langle S, R_{1}, \ldots , R_{n};{\bar{R}}\rangle$, with ${\bar{R}} \not \equiv {\mathbf {f}}(R_{1},\ldots ,R_{n})$, there must exist a situation ${\mathcal {S}}^{'}= \langle S, R_{1}, \ldots , R_{n};{\mathbf {f}}(R_{1},\ldots ,R_{n})\rangle \in \widehat{\mathcal {S}}^{{\mathbf {f}}}$. But then, for every permutation $\gamma$, there must exist at least a pair $s,t \in S$ such that $D^{{\mathcal {S}}}(\{s,t\})\ne D^{{\mathcal {S}}^{'}}(\{\gamma ^{-1}(s),\gamma ^{-1}(t)\})$. Absurd, since ${\mathcal {S}}$ and ${\mathcal {S}}^{'}$ are both in $\bar{\mathcal {S}}$. Therefore $\widehat{\mathcal {S}}^{{\mathbf {f}}}=\bar{\mathcal {S}}$. $\square$

Proof of Proposition 2

For every aggregation function ${\mathbf {f}}$ there exists a family $\{{\mathbf {f}}^{j}\}_{j \in J_{-0}}$ of prime aggregation functions such that each ${\mathbf {f}}^{j} \preceq {\mathbf {f}}$. Consider each ${\bar{R}} \in \hbox {Im}\big ({\mathbf {f}}\big )$. If ${\bar{R}} \ne \emptyset$ just consider the equivalence classes associated to each prime function according to Theorem 1: $\{\bar{\mathcal {S}}^{j}\}_{j \in J_{-0}} \subseteq \widehat{\mathcal {S}}/\simeq _{{\mathcal {D}}({\mathcal {G}}_{S})}$. It is clear that for each $\hbox {prof}_{\mathcal {S}}$ such that ${\mathcal {S}} \in \cup _{j \in J_{-0}}\bar{\mathcal {S}}^{j}$ we have that there exists at least a $j \in J_{-0}$ that verifies that ${\mathbf {f}}^{j}(\hbox {prof}_{\mathcal {S}})= {\bar{R}}$. On the other hand, if $R=\emptyset$ then consider $\bar{\mathcal {S}}^{0}$ the class of all the situations of the form $\langle S, R_{1},\ldots , R_{n}; \emptyset \rangle$. This class corresponds to the prime function ${\mathbf {f}}^{0}$ such that ${\mathbf {f}}^{0}(\hbox {prof}_{\mathcal {S}})= \emptyset$. Then, if we consider the class $J=J_{-0}\cup \{0\}$, we have that ${\mathbf {f}}(\hbox {prof}_{\mathcal {S}})= {\bar{R}}$ for every ${\mathcal {S}} \in \cup _{j \in J}\bar{\mathcal {S}}^{j}$. $\square$

Proof of Theorem 2

First of all, let us note that each chain ${\mathcal {C}}$ in $\langle \langle \mathbf{F}_{{\mathcal {G}}_{S}}, \preceq \rangle$ has a length, defined as its cardinality and denoted as $|{\mathcal {C}}|$. Consider first the case in which $|{\mathcal {C}}| \le \aleph _{0}$, i.e. that ${\mathcal {C}}$ is countable. If ${\mathcal {C}} = \langle {\mathbf {f}}^{0}, {\mathbf {f}}^{1}, \ldots , {\mathbf {f}}^{|{\mathcal {C}}|}\rangle$, ${\mathbf {f}}^{0}$ is the only prime aggregation function in ${\mathcal {C}}$. According to Theorem 1, for ${\mathbf {f}}^{0}$ there exists a $\bar{\mathcal {S}}^{0} \in \widehat{\mathcal {S}}/\simeq _{{\mathcal {D}}({\mathcal {G}}_{S})}$ such that ${\mathbf {f}}^{0}(\hbox {prof}({\mathcal {S}}))= {\bar{R}}$ for every ${\mathcal {S}}= \langle S, R_{1},\ldots , R_{n}; {\bar{R}}\rangle \in \bar{\mathcal {S}}^{0}$. We will prove by induction that for each ${\mathbf {f}}^{k} \in {\mathcal {C}}$ there exists an element $\pi ^{k}$ in a partition $\Pi ^{\alpha ^{k}}$ such that ${\mathbf {f}}^{k}(\hbox {prof}({\mathcal {S}}))= {\bar{R}}$ for every ${\mathcal {S}}= \langle S, R_{1},\ldots , R_{n}; {\bar{R}}\rangle \in \pi ^{k}$. Moreover, that $\Pi ^{\alpha ^{k}}$ is a refinement of $\Pi ^{\alpha ^{k+1}}$ for each k:

• Consider the case of ${\mathbf {f}}^{1}$. According to Proposition 2 we have that there exists $\{\bar{\mathcal {S}}^{j}\}_{j \in J} \subseteq \widehat{\mathcal {S}}/\simeq _{{\mathcal {D}}({\mathcal {G}}_{S})}$ such that ${\mathbf {f}}(\hbox {prof}_{\mathcal {S}})= {\bar{R}}$ for every ${\mathcal {S}}= \langle S, \hbox {prof}_{\mathcal {S}}; {\bar{R}}\rangle \in \cup _{j \in J}\bar{\mathcal {S}}^{j}$. Suppose that $\bar{\mathcal {S}}^{0} \ne \bar{\mathcal {S}}^{j}$ for every $j \in J$, but then this means that ${\mathbf {f}}^{0}(\hbox {prof}_{\mathcal {S}})\ne {\mathbf {f}}^{1}(\hbox {prof}_{\mathcal {S}})$ for every ${\mathcal {S}}\in \bar{\mathcal {S}}^{0}$. Absurd, since we assumed that ${\mathbf {f}}^{0}\preceq {\mathbf {f}}^{1}$. Then $\bar{\mathcal {S}}^{0} \in \{\bar{\mathcal {S}}^{j}\}_{j \in J}$. That is, while ${\mathbf {f}}^{0}$ is supported by $\bar{\mathcal {S}}^{0} \in \widehat{\mathcal {S}}/\simeq _{{\mathcal {D}}({\mathcal {G}}_{S})}$, ${\mathbf {f}}^{1}$ is supported by $\bar{\mathcal {S}}^{0} \cup _{j\in J,j\ne 0}\bar{\mathcal {S}}^{j}$. If we call $\bar{\mathcal {S}}^{0}$, $\pi ^{0}$ and $\widehat{\mathcal {S}}/\simeq _{{\mathcal {D}}({\mathcal {G}}_{S})}$, $\Pi ^{\alpha _{0}}$, we see that if we denote $\bar{\mathcal {S}}^{0} \cup _{j\in J,j\ne 0}\bar{\mathcal {S}}^{j}$ as $\pi ^{1}$, it is clear that $\pi _{0} \subseteq \pi _{1}$. On the other hand, $\pi _{1}\cup _{j^{'} \notin J}\bar{\mathcal {S}}^{j^{'}}= \widehat{\mathcal {S}}$ while $\pi _{1}\cap \bar{\mathcal {S}}^{j^{'}}= \emptyset$ for every $j^{'} \notin J$. That means that $\pi _{1}$ and $\{\bar{\mathcal {S}}^{j^{'}}\}_{j^{'} \notin J}$ constitute a partition of $\widehat{\mathcal {S}}$, which we call $\Pi ^{\alpha _{1}}$. Since for $\pi ^{0} \in \Pi ^{\alpha ^{0}}$, $\pi ^{0}\subseteq \pi ^{1}$ and for each $\bar{\mathcal {S}}^{j}\in \Pi ^{\alpha ^{0}}$ ($j \in J$), $\bar{\mathcal {S}}^{j}\subseteq \pi ^{1}$ while $\bar{\mathcal {S}}^{j^{'}}\in \Pi ^{\alpha ^{1}}$ for $j^{'} \notin J$, $\Pi ^{\alpha ^{1}}$ is a coarsening of $\Pi ^{\alpha ^{0}}$.

• Assume that the result is valid up to k. That is, ${\mathbf {f}}^{k}$ is supported by a $\pi ^{k} \in \Pi ^{\alpha ^{k}}$. Without loss of generality we assume that if ${\mathbf {f}}^{k}$ is supported, according to Proposition 2, by a family $\{\bar{\mathcal {S}}^{l}\}_{l \in L} \subseteq \widehat{\mathcal {S}}/\simeq _{{\mathcal {D}}( {\mathcal {G}}_{S})}$ then $\pi ^{k}= \cup _{l \in L}\bar{\mathcal {S}}^{l}$ and $\Pi ^{\alpha ^{k}}=\{\pi ^{k}, \{\bar{\mathcal {S}}^{l^{'}}\}_{l^{'} \notin L}\}$. Now consider the case of ${\mathbf {f}}^{k+1}$. Again, by Proposition 2, we have that there exists $\{\bar{\mathcal {S}}^{j}\}_{j \in J} \subseteq \widehat{\mathcal {S}}/\simeq _{{\mathcal {D}}({\mathcal {G}}_{n}, {\mathcal {G}}_{S})}$ such that ${\mathbf {f}}^{k+1}(\hbox {prof}_{\mathcal {S}})= {\bar{R}}$ for every ${\mathcal {S}} \in \cup _{j \in J}\bar{\mathcal {S}}^{j}$. Since ${\mathbf {f}}^{k}\preceq {\mathbf {f}}^{k+1}$ we have that ${\mathbf {f}}^{k}(\hbox {prof}_{\mathcal {S}})\subseteq {\mathbf {f}}^{k+1}(\hbox {prof}_{\mathcal {S}})$ for every ${\mathcal {S}} \in \pi ^{k}$. That is, $\pi ^{k} \in \{\bar{\mathcal {S}}^{j}\}_{j \in J}$. Let us call the latter $\pi ^{k+1}$. We have that $\pi ^{k+1}$ and $\{\bar{\mathcal {S}}^{j^{'}}\}_{j^{'} \notin J}$ is a partition of $\widehat{\mathcal {S}}$ which we call $\Pi ^{\alpha ^{k+1}}$. It is clear that $\pi ^{k} \subseteq \pi ^{k+1}$ and $\bar{\mathcal {S}}^{j} \subseteq \pi ^{k+1}$ for $j \in J$, while $\bar{\mathcal {S}}^{j^{'}}\in \Pi ^{\alpha ^{k+1}}$ for $j^{'} \notin J$. Therefore, $\Pi ^{\alpha ^{k+1}}$ is a coarsening of $\Pi ^{\alpha ^{k}}$.

Therefore, the claim is proved for every k.

Now consider the case in which $|{\mathcal {C}}| > \aleph _{0}$. Then, the set of indexes k in ${\mathcal {C}} = \langle {\mathbf {f}}^{0}, {\mathbf {f}}^{1}, \ldots , {\mathbf {f}}^{|{\mathcal {C}}|}\rangle$ can be decomposed in two classes, those of the functions with either successor or limit indexes. That is, those indexes $k^{'}$ that have the form $k^{'}=k+1$ (the successor indexes) and those which verify that $k^{'}= lim_{k<k^{'}}k$, the limit indexes. The proof for the countable length of ${\mathcal {C}}$ applies also for the successor indexes. The following is the proof for the limit indexes:

• Assume that the result is valid for each $k<k^{'}$ for a limit index $k^{'}$. That is, each ${\mathbf {f}}^{k}$ is supported by a $\pi ^{k} \in \Pi ^{\alpha ^{k}}$ and, as ${\mathbf {f}}^{k}$ is supported by a family $\{\bar{\mathcal {S}}^{l}\}_{l \in L} \subseteq \widehat{\mathcal {S}}/\simeq _{{\mathcal {D}}({\mathcal {G}}_{S})}$ then $\pi ^{k}= \cup _{l \in L}\bar{\mathcal {S}}^{l}$ and $\Pi ^{\alpha ^{k}}=\{\pi ^{k}, \{\bar{\mathcal {S}}^{l^{'}}\}_{l^{'} \notin L}\}$. Moreover, for each pair $k_{1}< k_{2}<k^{'}$, $\Pi ^{\alpha ^{k_{1}}}$ is a refinement of $\Pi ^{\alpha ^{k_{2}}}$. Let us now consider ${\mathbf {f}}^{k^{'}}$. By Proposition 2 we have that there exists $\{\bar{\mathcal {S}}^{j}\}_{j \in J} \subseteq \widehat{\mathcal {S}}/\simeq _{{\mathcal {D}}({\mathcal {G}}_{S})}$ such that ${\mathbf {f}}^{k^{'}}(\hbox {prof}_{\mathcal {S}})= {\bar{R}}$ for every ${\mathcal {S}} \in \cup _{j \in J}\bar{\mathcal {S}}^{j}$. For every ${\mathbf {f}}^{k}\preceq {\mathbf {f}}^{k+1}$ we have that ${\mathbf {f}}^{k}(\hbox {prof}_{\mathcal {S}})\subseteq {\mathbf {f}}^{k^{'}}(\hbox {prof}_{\mathcal {S}})$ for every ${\mathcal {S}} \in \pi ^{k}$. That is, each $\pi ^{k} \in \{\bar{\mathcal {S}}^{j}\}_{j \in J}$. Let us call the latter $\pi ^{k^{'}}$. We have that $\pi ^{k^{'}}$ and $\{\bar{\mathcal {S}}^{j^{'}}\}_{j^{'} \notin J}$ is a partition of $\widehat{\mathcal {S}}$ which we call $\Pi ^{\alpha ^{k^{'}}}$. It is clear that for each $k<k^{'}$, $\pi ^{k} \subseteq \pi ^{k^{'}}$ and $\bar{\mathcal {S}}^{j} \subseteq \pi ^{k^{'}}$ for $j \in J$, while $\bar{\mathcal {S}}^{j^{'}}\in \Pi ^{\alpha ^{k^{'}}}$ for $j^{'} \notin J$. Therefore, $\Pi ^{\alpha ^{k^{'}}}$ is a coarsening of $\{\Pi ^{\alpha ^{k}}\}_{k<k^{'}}$. $\square$

Proof of Proposition 3

According to Theorem 2, ${\mathcal {C}}$ generates a chain of partitions, and particularly of equivalence classes, a $\pi ^{k}$ for each ${\mathbf {f}}^{k} \in {\mathcal {C}}$. Since each $\pi ^{k}$ is fully described by ${\mathcal {DEC}}({\mathbf {f}}^{k})$, we have that ${\mathcal {C}}$ generates also a sequence of the form ${\mathcal {C}}^{{\mathcal {DEC}}}=\{{\mathcal {DEC}}({\mathbf {f}}^{k})\}_{k=0}^{|{\mathcal {C}}|}$. To see that ${\mathcal {C}}^{{\mathcal {DEC}}}$ is a chain under $\subseteq$, just consider the fact that for $k<k^{'}$, for every profile $\hbox {prof}_{\mathcal {S}}$ we have that ${\mathbf {f}}^{k}\big (\hbox {prof}_{\mathcal {S}}\big ) \subseteq {\mathbf {f}}^{k^{'}}\big (\hbox {prof}_{\mathcal {S}}\big )$. Since, in particular ${\mathbf {f}}^{k}\big (\hbox {prof}_{\mathcal {S}}\big )(s,t) \subseteq {\mathbf {f}}^{k^{'}}\big (\hbox {prof}_{\mathcal {S}}\big )(s,t)$ for each situation ${\mathcal {S}}$ and every pair $s,t \in S$, the class of agents that are decisive over s and t for the former function, $D^{k}(\{s,t\}) \in {\mathcal {F}}^{k}_{{\mathcal {G}}_{S}}(\bar{\mathcal {S}})\subseteq {\mathcal {DEC}}({\mathbf {f}}^{k})$ is also a class of decisive agents over s and t for the latter. That is, $D^{k}(\{s,t\}) \in {\mathcal {F}}^{k^{'}}_{{\mathcal {G}}_{S}}(\bar{\mathcal {S}})\subseteq {\mathcal {DEC}}({\mathbf {f}}^{k^{'}})$. In other words: ${\mathcal {DEC}}({\mathbf {f}}^{k}) \subseteq {\mathcal {DEC}}({\mathbf {f}}^{k^{'}})$. $\square$

Proof of Proposition 4

Trivial. Suppose ${\mathcal {S}}^{*}\in {\mathcal {SUPP}}({\mathbf {f}}^{j})$ for $j < j^{*}$. Then, there would exist ${\mathcal {S}}^{'}\in {\mathcal {SUPP}}({\mathbf {f}}^{j^{*}})$, such that for a given pair $s,t \in S$ and a permutation $\gamma$, $D^{{\mathcal {S}}^{*}}(\{s,t\})\subseteq$$D^{{\mathcal {S}}^{'}}(\{\gamma (s),\gamma (t)\})$. Then, either $D^{{\mathcal {S}}^{*}}(\{s,t\})=$$D^{{\mathcal {S}}^{'}}(\{\gamma (s),\gamma (t)\})$ (in which case ${\mathcal {S}}^{*}$ is decision-equivalent to ${\mathcal {S}}^{'}$ and therefore ${\mathcal {S}}^{*} \in {\mathcal {SUPP}}({\mathbf {f}}^{j^{*}})$) or $D^{{\mathcal {S}}^{*}}(\{s,t\})\subset$$D^{{\mathcal {S}}^{'}}(\{\gamma (s),\gamma (t)\})$. But then, since ${\mathcal {DEC}}({\mathbf {f}}^{j}) \subseteq {\mathcal {DEC}}({\mathbf {f}}^{j^{*}})$, there will exist a decisive set for a pair s and t, $D(\{s,t\})\in {\mathcal {DEC}}({\mathbf {f}}^{j^{*}})$ for which there does not exist a of permutation $\gamma$ such that $D^{{\mathcal {S}}^{*}}(\{s,t\})\subseteq$$D(\{\gamma (s),\gamma (t)\})$. Contradiction. $\square$

Proof of Proposition 7

By construction, each $\theta \in \bigcup {\mathcal {H}}({\mathbf {f}})$ is $\theta = \langle i, \mathcal {P}_{S}, \mathbf {G}_{n}, \mathbf {G}_{S} \rangle$, i.e. a single individual and a pair of states over which she is decisive(plus the permutations that preserve her decisiveness). Then $\theta \in \mathbf {H}$ for a $\mathbf {H} \in {\mathcal {H}}({\mathbf {f}})$. Define then $\Upsilon (\theta )$ as the corresponding $\mathbf {H}$. $\square$

Proof of Lemma 1

The first claim follows immediately from Definition 10 and Proposition 6. The characterization of the sentences in ${\mathcal {A}}^{{\mathbf {f}}}$ follows from the fact that sentences of the type $\phi ^{\forall \forall }_{\mathbf {H}}$, $\phi ^{\forall \exists }_{\mathbf {H}}$, $\phi ^{\exists \forall }_{\mathbf {H}}$, $\phi ^{\exists \exists }_{\mathbf {H}}$ or $\phi ^{\triangle }$ can verified in ${\mathcal {H}}^{{\mathbf {f}}}$ independently of S and n being finite or not. $\square$