Congruence relations on a choice space

Abstract

A choice space is a finite set of alternatives endowed with a map associating to each menu a nonempty subset of selected items. A congruence on a choice space is an equivalence relation that preserves its structure. Intuitively, two alternatives are congruent if the agent is indifferent between them, and, in addition, her choice is influenced by them in exactly the same way. We give an axiomatic characterization of the notion of congruence in terms of three natural conditions: binary fungibility, common destiny, and repetition irrelevance. Further, we show that any congruence satisfies the following desirable properties: (hereditariness) it induces a well-defined choice on the quotient set of equivalence classes; (reflectivity) the primitive behavior can be always retrieved from the quotient choice, regardless of any feature of rationality; (consistency) all basic axioms of choice consistency are preserved back and forth by passing to the quotient. We also prove that the family of all congruences on a choice space forms a lattice under set-inclusion, having equality as a minimum, and a unique maximum, called revealed indiscernibility. The latter relation can be seen as a limit form of revealed similarity as the agent’s rationality increases.

Appendices

Appendix A: Mathematics of congruences

We discuss some properties of congruence relations, which have been left out of the main body of the paper not to disrupt the presentation of the main results. Albeit rather technical, the analysis undertaken in this section provides further insight in the notion of congruence. This section also contains the proofs of all the results mentioned in it. (The proofs of all the results stated in the main body of the paper are in Appendix B.)

1.1 Relations among BF, CD, and RI.

Theorem 1 states a first characterization of congruences, which appeals to the three properties of Binary Fungibility (BF), Common Destiny (CD), and Repetition Irrelevance (RI). As a matter of fact, these three properties are not independent, since we have:²⁵

Lemma 4

Let $\equiv$ be an equivalence relation on a choice space. If RI holds for $\equiv$, then

$$\begin{aligned} \equiv \hbox { satisfies }\textsf {BF} \qquad \Longleftrightarrow \qquad \equiv \hbox { satisfies }\textsf {CD}. \end{aligned}$$

At a first sight, Lemma 4 may appear rather unexpected, insofar as it states the equivalence of two properties that describe a similarity of behavior from very different perspectives. Indeed, BF says that two equivalent alternatives are interchangeable whenever they appear alone, whereas CD requires that they always share the same destiny every time they are together in a menu. However, as soon as we consider that property RI operates at a higher level of complexity rather than properties BF and CD (since the former property compares menus of different sizes), the equivalence stated in Lemma 4 becomes less surprising. Furthermore, since the rationalizability of a choice implies that binary observations fully describe a choice behavior (thus flattening all levels of complexity to one, namely, the level of menus with two items only), also the following fact appears not so surprising:

Lemma 5

For any equivalence relation on a rationalizable choice space, if BF holds, then so does RI.

On the other hand, not even quasi-transitive rationalizability gives further leverage to obtain additional implications of the type described by Lemmas 4 and 5. In fact, the next two examples show that there is an equivalence relation $\equiv$ on a ground set X of size three and two quasi-rationalizable choices $c_1$ and $c_2$ on X such that (i) in the space $(X,c_1)$, $\equiv$ satisfies CD but neither BF nor RI, and (ii) in the space $(X,c_2)$, $\equiv$ satisfies RI but neither BF nor CD.

Example 7

Let $c_1$ be the choice correspondence on $X = \{x,y,z\}$ defined by

$$\begin{aligned} \underline{x}\,\underline{y}, \quad \underline{x}\,\underline{z}, \quad \underline{y}\,z, \quad \underline{x}\,\underline{y}\,z. \end{aligned}$$

It is immediate to check that $c_1$ is rationalizable by means of the quasi-transitive revealed preference $\succsim _{c_1}$ defined by $x \sim _{c_1} y$, $x \sim _{c_1} z$, and $y \succ _{c_1} z$. Consider the equivalence relation $\equiv$ on X having $\{\{x,y\},\{z\}\}$ as associated partition. Now $\equiv$ obviously satisfies CD, since in every menu containing x and y, both items are always chosen. However, $\equiv$ satisfies neither BF nor RI. Indeed, BF fails for the menu $A= \{z\}$, since $c_1 (A \cup \{x\}) {\setminus } \{x\} = \{z\} \ne \emptyset = c_1 (A \cup \{y\}) {\setminus } \{y\}$. Further, RI fails for the full menu $X= \{x,y,z\}$, since $c_1(X {\setminus } \{y\}) = \{x,z\} \ne \{x\} = c_1(X) {\setminus } \{y\}$. (Notice that the failure of RI readily follows from Lemma 4.)

Example 8

Let $c_2$ be the choice correspondence on $X = \{x,y,z\}$ defined by

$$\begin{aligned} \underline{x}\,\underline{y}, \quad \underline{x}\,\underline{z}, \quad y\,\underline{z}, \quad \underline{x}\,y\,\underline{z}. \end{aligned}$$

It is immediate to check that $c_2$ is rationalizable by means of the quasi-transitive revealed preference $\succsim _{c_2}$ defined by $x \sim _{c_2} y$, $x \sim _{c_2} z$, and $z \succ _{c_2} y$. Consider the equivalence relation $\equiv$ on X having $\{\{x,y\},\{z\}\}$ as associated partition. Now $\equiv$ satisfies RI, since in the full menu X (the other cases are trivial), we have

$$\begin{aligned} c_2(X {\setminus } \{y\}) = \{x,z\} = c_2(X) {\setminus } \{y\} \qquad \hbox {and}\qquad c_2(X {\setminus } \{x\}) = \{z\} = c_2(X) {\setminus } \{x\}. \end{aligned}$$

However, $\equiv$ satisfies neither BF nor CD. Indeed, BF fails for the menu $A= \{z\}$, since $x \in c_2(A \cup \{x\})$ but $y \notin c_2(A \cup \{y\})$. Further, CD fails for the full menu $X= \{x,y,z\}$, since $x \in c_2(X)$ whereas $y \notin c_2(X)$.

Two observations about Examples 7 and 8 are in order. First, notice that the two choices $c_1$ and $c_2$ are dual to each other, since they are simply obtained by reversing the arrow in the revealed preference. Second, in both cases, the equivalence relation $\equiv$ is contained in the associated relation of revealed indifference, and is maximal (but not maximum) in it.

Finally, we rule out the last possibility to improve the results of Lemmas 4 and 5, showing that the implication “BF$\Rightarrow$CD” does not hold in general, even under very strong assumptions of rational behavior. In fact, the counterexample given below uses a choice that is rationalizable by a linear order.

Example 9

Let $c_3$ be the (single-valued) choice function on $X = \{x,y,z\}$ defined by

$$\begin{aligned} \underline{x}\,y, \quad \underline{x}\,z, \quad \underline{y}\,z, \quad \underline{x}\,y\,z. \end{aligned}$$

It is immediate to check that $c_3$ is rationalizable by means of the linear revealed preference $\succsim _{c_3}$ defined by $x \succ _{c_3} y \succ _{c_3} z$. Consider again the equivalence relation $\equiv$ on X having $\{\{x,y\},\{z\}\}$ as associated partition. Now it is easy to check that $\equiv$ satisfies BF, but CD fails to hold for it.

It is worth noticing that, however, the equivalence relation used in Example 9 to disprove the implication “BF$\Rightarrow$CD” is rather awkward, insofar as it fails to be contained in the associated relation of revealed indifference (which, in this special case, is equality). This observation makes it possible to further study the relationships among the three axioms BF, CD, and RI (and possibly other similar properties) for the cases of equivalence relations that are coherent with the given observed behavior, in the sense that they are contained in the associated relation of revealed indifference. We leave this topic for future research.

1.2 Alternative forms of congruency

Here we describe some special cases of the condition of Full Congruency (Co), introduced in Definition 3. To start, we list a few natural properties of the saturation operator associated to an equivalence relation (see Definition 2).

Lemma 6

Let $\equiv$ be an equivalence relation on a choice space (X, c). For each $A,B \in \varOmega$, we have:

• (expansivity) $A \subseteq \textsf {sat}(A)\,$;

• (monotonicity) $A \subseteq B \;\; \Longrightarrow \;\; \textsf {sat}(A) \subseteq \textsf {sat}(B)\,$;

• (idempotency) $\textsf {sat}\big (\textsf {sat}(A)\big ) = \textsf {sat}(A)\,$;

• (closure under union) $\textsf {sat}(A \cup B) = \textsf {sat}(A) \cup \textsf {sat}(B)\,$;

• (closure under saturated intersection) $\textsf {sat}(A \cap B) = \textsf {sat}(A) \cap \textsf {sat}(B)$ for saturated A.

Observe that Lemma 6 holds for an arbitrary set X, regardless of whether X is finite or is endowed with a choice correspondence c.

Remark 2

If we naturally extend the definition of the saturation operator to the whole powerset of X, then we get a map $\widehat{\textsf {sat}} :2^X \rightarrow 2^X$ such that $\widehat{\textsf {sat}} \upharpoonright \varOmega = \textsf {sat}$ and $\widehat{\textsf {sat}}(\emptyset ) = \emptyset$. Lemma 6 yields that $\widehat{\textsf {sat}}$ is a closure operator on X in the sense of Ward (1942), being expansive, monotone, and idempotent.²⁶ In fact, being also closed with respect to finite unions, $\widehat{\textsf {sat}}$ is a topological closure operator in the sense of Kuratowski, hence it induces a topology on X (in which a set is closed if and only if it is either a saturated menu or the empty set).

Recall that, if $\equiv$ is an equivalence relation on X, then a transversal set (for$\equiv$) is any nonempty set $T \subseteq X$ that intersects each equivalence class in at most one element, that is, $\vert T \cap x' \vert \le 1$ for each $x \in X$ (where $x'$ is, as usual, the $\equiv$-equivalence class of x). Given $A \in \varOmega$, a transversal set T is maximal inA if $T \subseteq A$, and there is no transversal set $T_1$ such that $T \subsetneq T_1 \subseteq A$.

Next, we rewrite property Co in three limit cases: fixed one menu $A \in \varOmega$, we let the other menu $B \in \varOmega$ having the same saturation as A be either minimal, or equal to A, or maximal.

$$\begin{aligned} (\textsf {Co}_{\mathrm {tsv}}: \textit{Transversal Congruency}) \quad c(A)= & {} \textsf {sat}\big (c(T)\big ) \cap A, \\ (\textsf {Co}_{\mathrm {loc}}: \textit{Local Congruency}) \quad c(A)= & {} \textsf {sat}\big (c(A)\big ) \cap A, \\ (\textsf {Co}_{\mathrm {sat}}: \textit{Saturated Congruency}) \quad c(A)= & {} \textsf {sat}\big (c\big (\textsf {sat}(A)\big )\big ) \cap A, \end{aligned}$$

where $T \subseteq A$ is any maximal transversal set.

Fig. 2

Computing c(A) from a maximal transversal set $T_0 \subseteq A$

Transversal Congruency is already illuminating of the nature of congruence relations, since it yields the following simple algorithm to compute the choice set c(A) of a menu $A \in \varOmega$, given a congruence relation $\equiv$ on (X, c):

• (1) partition A into congruence classes;

• (2) select one item per class, and collect them into a transversal set $T_0 \subseteq A$;

• (3) take the union of the classes intersecting $c(T_0)$;

• (4) intersect this union with A.

Since Co$_{\mathrm {tsv}}$ holds for any transversal set T that is maximal in A, the above algorithm is sound, being invariant of the selection of $T_0$ (see steps (2) and (3)).

Figure 2 describes the four steps (1)–(4) of the algorithm. In all pictures, the ground set X is partitioned in cells (squares), and each cell is a congruence class of $\equiv \,$.²⁷ In the first picture, a menu $A \in \varOmega$ is represented by the shaded area. The saturated menu $\textsf {sat}(A)$ is given in the second picture, where the items of a maximal transversal set $T_0 \subseteq A$ are identified by dots, and the items of $T_0$ selected by c are identified by circled dots. The third picture represents (in dark gray) the congruence classes of the items in the choice set $c(T_0)$; the union of these classes is the saturated menu $\textsf {sat}(c(T_0))$. In the last picture the choice set c(A) is obtained as the trace of $\textsf {sat}(c(T_0))$ in the original menu A.

Contrary to Transversal Congruency Co$_{\mathrm {tsv}}$, which involves two menus A and T, the other two derived forms of congruency Co$_{\mathrm {loc}}$ and Co$_{\mathrm {sat}}$ only refer to a single menu A. Both of them can be interpreted as suitable “tracing properties” of choice sets. Maybe unexpectedly, we will later show that, although Local Congruency Co$_{\mathrm {loc}}$ is definitively weaker than Full Congruency Co, Saturated Congruency Co$_{\mathrm {sat}}$ is instead equivalent to Full Congruency: see Theorem 7(vi) later in this appendix. In view of the mentioned equivalence, property Co$_{\mathrm {sat}}$ allows us to view any congruence as an equivalence relation that makes the two operators of saturation and choice commute with each other: in fact, the choice set of any menu is obtained by (1) taking saturation, (2) applying the choice operator to this saturation, (3) taking again saturation, and (4) intersecting with the original menu.

1.3 Commutativity and some other necessary conditions

The following property plays a relevant role in characterizing congruence relations (Theorem 7(ii)–(v)):

Definition 13

Let $\equiv$ be an equivalence relation on a choice space (X, c), and $\textsf {sat}$ its saturation operator. We say that $\equiv$ is commutative if the following property holds:

• (Com: Commutativity)    $c \circ \textsf {sat}= \textsf {sat}\circ c$.

Then, we have:

Lemma 7

Any congruence relation on a choice space is commutative.

The converse of Lemma 7 does not hold, as the next example shows.

Example 10

Let $c :\varOmega \rightarrow \varOmega$ be the choice on $X = \{x,y,z,w\}$ considered in Example 6, that is,

$$\begin{aligned} \underline{x}\,\underline{y}, \; \underline{x}\,\underline{z}, \; \underline{x}\,w, \; \underline{y}\,\underline{z}, \; \underline{y}\,w, \; \underline{z}\,w, \qquad \underline{x}\,\underline{y}\,\underline{z},\; \underline{x}\,\underline{y}\,w \; \underline{x}\,z\,w,\; \underline{y}\,z\,w, \qquad \underline{x}\,\underline{y}\,\underline{z}\,w. \end{aligned}$$

The relation $\sim _c$ of revealed indifference is an equivalence relation, with associated partition $X' = \{\{x,y,z\},\{w\}\}$. An easy computation shows that the equality $c \circ \textsf {sat}= \textsf {sat}\circ c$ holds. On the other hand, property Co fails, since for $A = \{y,z,w\}$, we have $\textsf {sat}(c(A)) \cap A = \{y,z\} \, \ne \, \{y\} = c(A).$

Remark 3

For any commutative equivalence relation on a choice space (X, c), saturated menus are mapped by c onto saturated menus: indeed, if $A \in \varOmega$ is saturated, then $c(A) = c\big (\textsf {sat}(A)\big ) = \textsf {sat}\big (c(A)\big )$. In particular, for a congruence relation, a choice always maps saturated menus on saturated menus.

Fig. 3

A commutative equivalence relation on a choice space

Figure 3 describes the meaning (and the shortcomings) of Commutativity for an arbitrary equivalence relation on a choice space. Similarly to Fig. 2, the ground set X is partitioned in squares, representing equivalence classes. In the top picture, a menu $A \in \varOmega$ is depicted by the shaded area. The two middle pictures describe the effect on A of, respectively, the choice correspondence c (on the left, where c(A) is the area in dark gray) and the saturation operator $\textsf {sat}$ (on the right). Finally, the bottom picture simultaneously represents (in dark gray) the two sets $\textsf {sat}(c(A))$ and $c(\textsf {sat}(A))$, which are equal by Commutativity.

It is worth noticing that the picture on the left depicts the choice set c(A) as intersecting some equivalence classes without filling them up: in fact, this is the main shortcoming of a commutative equivalence relation that fails to be congruent with the choice structure. The relation of revealed indifference described in Example 10 exhibits an instance of this kind: indeed, the $\sim _c$-equivalence class $\{x,y,z\}$ contains both an item selected in $A = \{y,z,w\}$ (namely, y) and an item not selected in A (namely, z).

The phenomenon described above never occurs for a congruence relation:

Lemma 8

Let $\equiv$ be a congruence on a choice space (X, c), and $X'$ the quotient set. For all menus $A \in \varOmega$, we have:

• (ZO: Zero-One)    for each $E \in X'$, $c(A) \cap E$ is either empty or equal to $A \cap E$.

The converse of Lemma 8 is false: take the equivalence relation $\equiv$ on the choice space $(X,c_2)$ considered in Example 4. Another feature of a congruence relation is the following:

Lemma 9

Let $\equiv$ be a congruence on a choice space (X, c), and $X'$ the quotient set. All equivalence classes are fixed points of c, that is,

• (FP: Fixed Point)    $c(E) = E$    for each $E \in X'$.

Notice that the equivalence relation of Example 4 shows that also the converse of Lemma 9 is false. As a matter of fact, it is hardly surprising that Fixed Point holds for any congruence class in a choice space. Indeed, the very notion of a congruence relation describes a similar status of the items in each equivalence class: either select them all, or reject them all. Since the choice set of a menu is nonempty, it follows that the items of a congruence class ought to be selected all together.

We summarize all necessary conditions obtained so far in the next result:

Corollary 2

A congruence relation on a choice space satisfies the following properties: BF, CD, RI, Com, ZO, and FP. None of these properties (singularly taken) suffices to make an equivalence relation on a choice space into a congruence.

We conclude this section by mentioning some additional results, which emphasize even more the role played by Commutativity in the setting of this paper.

Lemma 10

For any commutative equivalence relation on a choice space, the following properties are equivalent: (1) Transversal Congruency, (2) Local Congruency, (3) Saturated Congruency, (4) Common Destiny, and (5) Zero-One.

However, we also have:

Lemma 11

Saturated Congruency implies Commutativity.

Therefore, as a consequence of Lemmas 10 and 11, Saturated Congruency implies both Transversal Congruency and Local Congruency. In fact, it will turn out that Saturated Congruency is equivalent to Full Congruency, and these two properties are stronger than Transversal Congruency and Local Congruency: see Theorem 7 below.

1.4 Menu indiscernibility

The notion of menu indiscernibility has been used in Sect. 4.1 to define the notion of revealed indiscernibility (see Definition 10). The key result that allows one to link menu indiscernibility and revealed indiscernibility is Lemma 1, which states that the ground set of a choice space can be partitioned into maximal indiscernible menus. The proof of Lemma 1 is long and nontrivial. Here we provide some results that are helpful in deriving this key fact. To start, we list few simple properties of menu indiscernibility.

Lemma 12

Let (X, c) be a choice space, and $\textsf {Ind}(c)$ the family of all indiscernible menus.

• (a) Every singleton is indiscernible.

• (b) All elements of $\textsf {Ind}(c)$ are fixed points of c.

• (c) For all $\mathcal {A}\subseteq \textsf {Ind}(c)$ and $E \in \mathcal {A}$, if $E \cap c\big (\bigcup \mathcal {A}\big ) \ne \emptyset$, then $E \subseteq c\big (\bigcup \mathcal {A}\big )$.

Next, we provide a useful characterization of indiscernible menus, which also suggests the existence of a link with the notion of congruence.

Lemma 13

A menu E is indiscernible if and only if the following two conditions hold:

• (Zero-One)   $\textsf {Probe}(E)$ is extremal;

• (Tracing)   $c(P) = c(E \cup P) \cap P$ for each $P \in \textsf {Probe}(E)$.

The discussion following Definition 9 (see Sect. 4.1) shows that Zero-One is a weakening of Inner Indiscernibility property.²⁸ Tracing can be interpreted as follows: the items selected from a menu P intersecting an indiscernible menu E are exactly those obtained by taking the trace over P of the items selected from $E \cup P$. The next technical example shows a case of failure of menu indiscernibility.

Example 11

Let $c_2 :\varOmega \rightarrow \varOmega$ be the choice on $X = \{x,y,z\}$ defined in Example 1. Set $E = \{x,y\}$ and $Z =\{z\} \subseteq X {\setminus } E$, whence $\textsf {Probe}(E) = \varOmega {\setminus } \{\{z\}\}$, and $\textsf {Probe}(E|Z) = \{\{x,z\},\{y,z\},X\}$. One can readily check that E satisfies Zero-One. On the other hand, none of the other indiscernibility conditions holds for E. Indeed, Inner Indiscernibility fails, because $E \cap c_2(X) = \emptyset$ and $E \cap c_2(\{x,z\}) = \{x\}$, hence $\textsf {Probe}(E|Z)$ is neither nonintersecting nor fully intersecting. Outer Indiscernibility fails as well, since $c_2(\{x,z\}) {\setminus } E = \emptyset$ and $c_2(X) {\setminus } E = \{z\}$, and so $|\{c_2(P) {\setminus } E : P \in \textsf {Probe}(E | Z)\}| > 1$. Finally, $c_2$ does not satisfy Tracing, because for $P = \{x,z\} \in \textsf {Probe}(E)$, we have $c_2(P) = \{x\} \ne \{z\} = c_2(E \cup P) \cap P$.

Lemma 1 in Sect. 4.1 is one of the most technical results of this paper. In order to prove it, the following two closure properties of the family $\textsf {Ind}(c)$ of indiscernible menus come handy: (a) $\textsf {Ind}(c)$ is closed by set-inclusion; and (b) $\textsf {Ind}(c)$ is closed by non-disjoint union (hence we can maximally expand an indiscernible menu by adjoining all other indiscernible menus that intersect it). Formally:

Lemma 14

• (a) A submenu of an indiscernible menu is indiscernible.

• (b) The union of two intersecting indiscernible menus is indiscernible.

We close this section by mentioning two results, which show that menu indiscernibility and revealed indiscernibility are indeed two sides of the same coin:

Lemma 15

A menu is indiscernible if and only if any two of its items are indiscernible.

Lemma 16

Two items x, y in a choice space (X, c) are indiscernible if and only if the following two properties hold for each menu $A \in \varOmega$:

• (Binary Tracing)    if $\{x,y\} \cap A \ne \emptyset$, then $c(A) = c(A \cup \{x,y\}) \cap A$;

• (Common Destiny)    if $x,y \in A$, then $x \in c(A) \Leftrightarrow y \in c(A)$.

1.5 A multiple characterization of congruences

Here we link most of the notions introduced so far by providing a multiple characterization of congruence relations, which extends Theorems 1, 2 and 4 in different directions.

Theorem 7

The following statements are equivalent for an equivalence relation on a choice space:

• (i) it is a congruence relation;

• (ii) it satisfies Transversal Congruency and Commutativity;

• (iii) it satisfies Local Congruency and Commutativity;

• (iv) it satisfies Common Destiny and Commutativity;

• (v) it satisfies Zero-One and Commutativity;

• (vi) it satisfies Saturated Congruency;

• (vii) it satisfies Binary Fungibility, Common Destiny, and Repetition Irrelevance;

• (viii) it is reflective;

• (ix) it is contained in the relation of revealed indiscernibility.

Theorem 7 points out that a commutative equivalence relation on a choice space is a congruence whenever it satisfies some weak form of structural coherence (cf. parts (ii), (iii), (iv), and (v)). On the other hand, part (vi) says that the property of saturated congruency suffices by itself to characterize a congruence. The equivalence between (i) and (vii) is known (see Theorem 1). The equivalence between (i) and (viii) sharpens Theorem 2 by saying that only congruences allow one to retrieve the primitive choice from a quotient choice. Finally, since revealed indiscernibility is a congruence by Theorem 4, the equivalence between (i) and (ix) readily yields the following consequence:

Corollary 3

Revealed indiscernibility is the maximum congruence relation on a choice space.

1.6 Proofs of the results in Appendix A

Proof of Lemma 4²⁹    Let $\equiv$ be an equivalence relation on a choice space (X, c), and assume that RI holds for $\equiv$. We shall show that $\equiv$ satisfies BF if and only if it satisfies CD.³⁰

$(\Rightarrow )\;$ Suppose BF holds for $\equiv$. Let $A \in \varOmega$ and $x,y \in X$ be such that $x \ne y$, $x,y \in A$, and $x\equiv y$. Assume that $x \in c(A)$. Since $x \ne y$, property RI yields $x \in c(A) {\setminus } \{y\} = c(A {\setminus } \{y\}) = c((A {\setminus } \{x\})_{y \leftarrow x})$. By properties BF and RI, we obtain $y \in c(A {\setminus } \{x\}) = c(A) {\setminus } \{x\}$, whence $y \in c(A)$ since $y \ne x$. This proves that $x \in c(A)$ implies $y \in c(A)$. The reverse implication is proved by reversing the role of x and y.

$(\Leftarrow )\;$ Suppose CD holds for $\equiv$. Let $A \in \varOmega$ and $x,y \in X$ be such that $x \in A$, $y \notin A$, and $x\equiv y$. Assume that $x \in c(A)$. Set $B := A \cup \{y\} \in \varOmega$, hence $A = B {\setminus } \{y\}$. An application of property RI yields $x \in c(A) = c(B {\setminus } \{y\}) = c(B) {\setminus } \{y\}$, hence $x \in c(B)$. By property CD, we get $y \in c(B)$. Now another application of property RI entails $y \in c(B) {\setminus } \{x\} = c(B {\setminus } \{x\}) = c(A_{x \leftarrow y})$. To complete the proof of BF, it remains to show that the equality $c(A) {\setminus } \{x\} = c(A_{x \leftarrow y}) {\setminus } \{y\}$ holds. Thus, suppose $z \in c(A) {\setminus } \{x\}$, i.e., $z \in c(A)$ and $z \ne x$. By property RI, we get $z \in c(B)$, hence $z \in c(B) {\setminus } \{x\} = c(B {\setminus } \{x\}) = c(A_{x \leftarrow y})$ by another application of RI. Since $z \ne y$, we can conclude that $z \in c(A_{x \leftarrow y}) {\setminus } \{y\}$. This proves the inclusion $c(A) {\setminus } \{x\} \subseteq c(A_{x \leftarrow y}) {\setminus } \{y\}$. The reverse inclusion is proved similarly. $\square$

Proof of Lemma 5    Let $\equiv$ be an equivalence relation on a choice space (X, c) such that BF holds for $\equiv$. Assume that c is rationalizable. To show that RI holds for $\equiv$, let $A \in \varOmega$ and $x,y \in X$ be such that $x,y \in A$ and $x \equiv y$; we shall prove $c(A) {\setminus } \{y\} = c(A {\setminus } \{y\})$. Let $z \in c(A) {\setminus } \{y\}$, hence by rationalizability $z \succsim _c w$ for any $w \in A {\setminus } \{y\}$. Again by rationalizability, we obtain $z \in c(A {\setminus } \{y\})$. This proves the inclusion $c(A) {\setminus } \{y\} \subseteq c(A {\setminus } \{y\})$. (Notice that BF is not needed at this stage.) For the reverse inclusion, let $z \in c(A {\setminus } \{y\})$. By rationalizability, we have $z \succsim _c w$ for any $w \in A {\setminus } \{y\}$. If $z \notin c(A)$, then we must have $y \succ _c z$, again by rationalizability. Since $c(\{y,z\}) = \{y\}$, we get $c(\{x,z\}) = \{x\}$ by BF, hence $x \succ _c z$ by rationalizability. However, the latter fact contradicts $z \in c(A {\setminus } \{y\})$. $\square$

Proof of Lemma 6    We only show that the saturation operator is closed under saturated intersection, since the other properties are immediate. Let $A,B \in \varOmega$ be such that A is saturated with respect to the equivalence relation $\equiv$ on X. The inclusion $\textsf {sat}(A \cap B) \subseteq \textsf {sat}(A) \cap \textsf {sat}(B)$ readily follows from monotonicity. Conversely, assume that $x \in \textsf {sat}(A) \cap \textsf {sat}(B) = A \cap \textsf {sat}(B)$. Then the equivalence class of x must intersect B, since otherwise $x \notin \textsf {sat}(B)$, a contradiction. Thus, there exists $z \in \textsf {sat}(A) \cap B = A \cap B$ such that $x \equiv z$. We can conclude that $x \in \textsf {sat}(A \cap B)$, and the proof is complete. $\square$

Proof of Lemma 7    Assume that $\equiv$ is a congruence relation on (X, c), and let $A \in \varOmega$. Since $\textsf {sat}$ is monotone and idempotent by Lemma 6, property Co (applied to the menus $\textsf {sat}(A)$ and A) yields

$$\begin{aligned} c\big (\textsf {sat}(A)\big ) \; = \; \textsf {sat}\big (c(A)\big ) \cap \textsf {sat}(A) \; = \; \textsf {sat}\big (c(A)\big ). \end{aligned}$$

This proves that the commutativity property holds. $\square$

Proof of Lemma 8    Assume that $\equiv$ is a congruence relation on (X, c). Let A be an arbitrary menu, and E a congruence class. Suppose $x \in c(A) \cap E \ne \emptyset$. If $y \in A \cap E$, then $x \equiv y$, hence $y \in c(A)$ by Common Destiny (which holds by Theorem 1). It follows that $c(A) \cap E = A \cap E$, proving the claim. $\square$

Proof of Lemma 9    Let E be a congruence class. An application of Lemma 8 for $A := E$ yields that either $c(E) \cap E = \emptyset$ or $c(E) \cap E = E$ holds. Since the former case cannot happen because $c(E) \ne \emptyset$, it follows that the equality $c(E) = E$ holds, as claimed. $\square$

Proof of Lemma 10    Let $\equiv$ be a commutative equivalence relation on a choice space (X, c). For each menu $A \in \varOmega$ and maximal transversal set $T \subseteq A$, we have:

$$\begin{aligned} \textsf {sat}\big (c(A)\big )&= \; \textsf {sat}\big (\textsf {sat}\big (c(A)\big )\big ) \quad \text {(by Lemma}~6\text {)}\\&= \; \textsf {sat}\big (c\big (\textsf {sat}(A)\big ) \big ) \quad \text {(by Commutativity)}\\&= \; \textsf {sat}\big (c\big (\textsf {sat}(T)\big )\big ) \quad \text {(by the maximality of }T\text { in }A)\\&= \; \textsf {sat}\big (\textsf {sat}\big (c(T)\big )\big ) \quad \text {(by Commutativity)}\\&= \; \textsf {sat}\big (c(T)\big ) \quad \text {(by Lemma}~6\text {)}. \end{aligned}$$

In particular, we obtain $\textsf {sat}\big (c(A)\big ) \cap A = \textsf {sat}\big (c(\textsf {sat}(A))\big ) \cap A = \textsf {sat}\big (c(T)\big ) \cap A$, which implies that the three properties Co$_{\mathrm {loc}}$, Co$_{\mathrm {sat}}$, and Co$_{\mathrm {tsv}}$ are equivalent for a commutative equivalence relation. To complete the proof, we show that Local Congruency, Common Destiny, and Zero-One property are equivalent. Actually, the proof given below shows that the equivalence of these three properties also holds for generic (possibly non commutative) equivalence relations.

Assume that Local Congruency holds, and let $x,y \in X$ and $A \in \varOmega$ be such that $x,y \in A$ and $x \equiv y$. If $x \in c(A) = \textsf {sat}(c(A)) \cap A$, then $y \in \textsf {sat}(c(A)) \cap A$, hence $y \in c(A)$. Similarly, $y \in c(A)$ implies $x \in c(A)$. Conversely, assume that Common Destiny holds for $\equiv$, and let $A \in \varOmega$ be arbitrary. The inclusion $c(A) \subseteq \textsf {sat}(c(A)) \cap A$ is obvious. For the reverse inclusion, let $x \in \textsf {sat}(c(A)) \cap A$. Thus, there exists $y \in c(A)$ such that $x \equiv y$. Since $x \in A$, Common Destiny yields $x \in c(A)$. This shows that Local Congruency and Common Destiny are equivalent.

Finally, we prove that Local Congruency and Zero-One are equivalent, without the assumption of Commutativity. Let $A \in \varOmega$. The equality $c(A) = \textsf {sat}(c(A)) \cap A$ is equivalent to the statement that, for each equivalence class E, $c(A) \cap E = \textsf {sat}(c(A)) \cap A \cap E$. If $c(A) \cap E = \emptyset$, then $\textsf {sat}(c(A)) \cap E = \emptyset$, and both sides are empty. If $c(A) \cap E \ne \emptyset$, then $\textsf {sat}(c(A)) \cap E = E$, hence $c(A) \cap E= \textsf {sat}(c(A)) \cap A \cap E$ is equivalent to $c(A) \cap E= A \cap E$. It follows that $c(A) = \textsf {sat}(c(A)) \cap A$ is equivalent to the statement that, for each equivalence class E, $c(A) \cap E \ne \emptyset$ implies $c(A) \cap E= A \cap E$. That is Zero-One. $\square$

Proof of Lemma 11    Let $\equiv$ be an equivalence relation on a choice space (X, c). Assume that Saturated Congruency Co$_{\mathrm {sat}}$ holds for $\equiv \,$. Observe preliminarily that c maps saturated menus on saturated menus. Indeed, for each $B \in \varOmega$, property Co$_{\mathrm {sat}}$ yields

$$\begin{aligned} c\big (\textsf {sat}(B)\big ) = \textsf {sat}\big (c(\textsf {sat}(\textsf {sat}(B))) \big ) \cap \textsf {sat}(B) \end{aligned}$$

and the set on the right hand side is saturated being the intersection of saturated menus. Now let $A \in \varOmega$. Since we have

$$\begin{aligned} \textsf {sat}\big (c(A)\big )&= \; \textsf {sat}\big (\textsf {sat}(c(\textsf {sat}(A))) \cap A \big ) \quad \text {(by }\textsf {Co}_{\mathrm {sat}})\\&= \; \textsf {sat}\big (\textsf {sat}(c(\textsf {sat}(A))) \big ) \cap \textsf {sat}(A) \quad \text {(by Lemma}~6\text {)}\\&= \; c\big (\textsf {sat}(A) \big ) \cap \textsf {sat}(A) \quad \text {(since }c\big (\textsf {sat}(A)\big )\text { is saturated)}\\&= \; c\big (\textsf {sat}(A) \big ), \end{aligned}$$

it follows that Commutativity holds. $\square$

Proof of Lemma 12    Part (a) is immediate, since each set of the type $\textsf {Probe}(\{x\} | Y)$ has exactly one element, hence Inner Indiscernibility and Outer Indiscernibility hold trivially for every singleton $\{x\}$. Part (b) is immediate. Concerning (c), let $\mathcal {A}$ be a nonempty collection of indiscernible subsets of X, and let $E \in \mathcal {A}$ be such that $E \cap c(\bigcup \mathcal {A}) \ne \emptyset$. Inner Indiscernibility yields $E \cap c(\bigcup \mathcal {A}) = E \cap \bigcup \mathcal {A}= E$, which implies $E \subseteq c(\bigcup \mathcal {A})$. This completes the proof of the lemma. $\square$

Proof of Lemma 13    Assume that $E \in \varOmega$ is indiscernible. Since Inner Indiscernibility implies Zero-One, to prove necessity it suffices to show that Tracing holds for E. To that end, let $P \in \textsf {Probe}(E) = \bigcup _{Y \subseteq X {\setminus } E} \textsf {Probe}(E | Y)$. Thus there exists (a unique) $Y \subseteq X {\setminus } E$ such that $P = F \cup Y$, with $\emptyset \ne F \subseteq E$. Then we have:

$$\begin{aligned} c(P) \;&= \; c(F \cup Y)\\&= \; c(F \cup Y) \cap (F \cup Y)\\&= \; \big ( c(F \cup Y) \cap F \big ) \, \cup \, \big (c(F \cup Y) \cap Y \big ) \\&= \; \big ( c(E \cup Y) \cap F \big ) \, \cup \, \big (c(F \cup Y) \cap Y \big ) \quad \text {(by Inner Indiscernibility)} \\&= \; \big ( c(E \cup Y) \cap F \big ) \, \cup \, \big (c(E \cup Y) \cap Y \big ) \quad \text {(by Outer Indiscernibility)} \\&= \; c(E \cup Y) \cap (F \cup Y) \\&= \; c(E \cup P) \cap P. \end{aligned}$$

This proves the claim.

Conversely, assume that $E \in \varOmega$ satisfies Zero-One and Tracing, and let $Y \subseteq X {\setminus } E$. Tracing yields

$$\begin{aligned} c(F \cup Y) \; = \; c(E \cup (F \cup Y)) \cap (F \cup Y) \; = \; c(E \cup Y) \cap (F \cup Y) \end{aligned}$$

for every $\emptyset \ne F \subseteq E$, and so

$$\begin{aligned} c(F \cup Y) \cap E= & {} \big (c(E \cup Y) \cap E\big ) \cap F, \end{aligned}$$

(3)

$$\begin{aligned} c(F \cup Y) {\setminus } E= & {} c(E \cup Y) \cap Y. \end{aligned}$$

(4)

Since $c(E \cup Y) \cap E \in \{\emptyset ,E\}$ by Zero-One, from (3) it follows that $\textsf {Probe}(E|Y)$ is either nonintersecting or fully intersecting. This proves that E satisfies Inner Indiscernibility. Moreover, E satisfies Outer Indiscernibility by (4). $\square$

Proof of Lemma 14    (a) Assume that $E\in \varOmega$ is an indiscernible menu, and let F be a submenu of E. By Lemma 13, to prove the indiscernibility of F, it suffices to show that Zero-One and Tracing hold for it. Fix $P \in \textsf {Probe}(F)$. To prove the Zero-One property, assume that $F \cap c(P) \ne \emptyset$. Since $P \in \textsf {Probe}(E)$, the hypothesis yields the equality $E \cap c(P) = E \cap P$, hence

$$\begin{aligned} F \cap c(P) \; = \; (F \cap E) \cap c(P) \; = \; F \cap (E \cap P) \; = \; F \cap P. \end{aligned}$$

This proves the claim. Concerning Tracing, observe that $P, (F \cup P) \in \textsf {Probe}(E)$, hence we get the following chain of equalities:

$$\begin{aligned} c(F \cup P) \cap P \;&= \; c(E \cup (F \cup P)) \cap (F \cup P) \cap P \quad \text {(by Tracing for }E\text {)}\\&= \; c(E \cup P) \cap P\\&= \; c(P) \quad \text {(by Tracing for }E\text {)}. \end{aligned}$$

It follows that Tracing holds for F as well, and the proof of part (a) is complete.

(b) We first prove the following technical fact, which will be repeatedly used below:

(*):

Let$E,F,P \in \varOmega$be such that$\emptyset \ne E \cap F \subseteq P$. IfEis indiscernible and$(E \cap F) \cap c(P)$is empty, then$E \cap c(P)$is empty as well.

Let $E,F,P \in \varOmega$ be such that $\emptyset \ne E \cap F \subseteq P$. To prove that (*) holds, it suffices to show that if E satisfies Zero-One, then $E \cap c(P) \ne \emptyset$ implies $E \cap F \subseteq c(P)$. To that end, assume that $\textsf {Probe}(E)$ is extremal, and let $E \cap c(P) \ne \emptyset$. Since $P \in \textsf {Probe}(E)$, we obtain $E \cap c(P) = E \cap P \supseteq E \cap F$. Thus $E \cap F \subseteq c(P)$, and so (*) holds.

Now, let $E_0, E_1 \in \varOmega$ be two indiscernible menus such that $E_0 \cap E_1 \ne \emptyset$. By Lemma 13, to prove the indiscernibility of $E_0 \cup E_1$, it suffices to show that Zero-One and Tracing hold for it. Fix $P \in \textsf {Probe}(E_0 \cup E_1)$. For Tracing, assume without loss of generality that $E_0 \cap P \ne \emptyset$, whence $P \in \textsf {Probe}(E_0)$ and $(E_0 \cup P) \in \textsf {Probe}(E_1)$. Then we have:

$$\begin{aligned} c(P) \;&= \; c(E_{0} \cup P) \cap P \quad \text {(by Tracing for }E_{0}\text {)}\\&= \; c(E_1 \cup (E_{0} \cup P)) \cap (E_{0} \cup P) \cap P \quad \text {(by Tracing for }E_{1}\text {)}\\&= \; c((E_{0} \cup E_{1}) \cup P) \cap P. \end{aligned}$$

This proves the claim.

Concerning Zero-One, we distinguish the following two cases: (i) $E_{0} \cap P = \emptyset$ or $E_{1} \cap P = \emptyset$; (ii) $E_{0} \cap P \ne \emptyset \ne E_{1} \cap P$. For case (i), assume without loss of generality that $E_{0} \cap P = \emptyset$. Since $E_{1} \cap P = (E _{0} \cup E_{1}) \cap P \ne \emptyset$, the Zero-One property of $E_{1}$ yields

$$\begin{aligned} (E _{0} \cup E_{1}) \cap c(P) \; = \; E_{1} \cap c(P) \, \in \, \{\emptyset , E_{1} \cap P\} \; = \; \{\emptyset , (E _{0} \cup E_{1}) \cap P\} \end{aligned}$$

thus proving that $E_0 \cup E_1$ satisfies the claim. For case (ii), start by observing that, for each $j,h,k \in \{0,1\}$ such that $h \ne k$, we have

$$\begin{aligned} P,\,E_j \cup P \in \textsf {Probe}\big (E_j \, | \, (P {\setminus } E_j)\big ) \quad \hbox {and}\quad E_h \cup P,\, E_k \cup E_h \cup P \in \textsf {Probe}\big (E_k \, | \, (E_h \cup P) {\setminus } E_k \big ). \end{aligned}$$

Then, the following chain of implications holds:

$$\begin{aligned} E_{0} \cap c(P) = \emptyset \;&\; \Longrightarrow \;\; E_0 \cap c(E_0 \cup P) = \emptyset \quad \text {(by Inner Indiscernibility for }E_{0}\text {)}\\&\; \Longrightarrow \;\; (E_{0} \cap E_{1}) \cap c(E_{0} \cup P) = \emptyset \\&\; \Longrightarrow \;\; E_1 \cap c(E_0 \cup P) = \emptyset \quad \text {(by property (*))}\\&\; \Longrightarrow \;\; E_{1} \cap c(E_{0} \cup E_{1} \cup P) = \emptyset \quad \text {(by Inner Indiscernibility for }E_{1}\text {)}\\&\; \Longrightarrow \;\; (E_{0} \cap E_{1}) \cap c(E_{0} \cup E_{1} \cup P) = \emptyset \\&\; \Longrightarrow \;\; E_{0} \cap c(E_{0} \cup E_{1} \cup P) = \emptyset \quad \text {(by property (*))}\\&\; \Longrightarrow \;\; E_{0} \cap c(E_1 \cup P) = \emptyset \quad \text {(by Inner Indiscernibility for }E_{0}\text {)}\\&\; \Longrightarrow \;\; (E_{0} \cap E_{1}) \cap c(E_1 \cup P) = \emptyset \\&\; \Longrightarrow \;\; E_1 \cap c(E_1 \cup P) = \emptyset \quad \text {(by property (*))}\\&\; \Longrightarrow \;\; E_1 \cap c(P) = \emptyset \quad \text {(by Inner Indiscernibility for }E_{1}\text {).} \end{aligned}$$

Likewise, $E_1 \cap c(P) = \emptyset$ implies $E_0 \cap c(P) = \emptyset$. Therefore, if $E_{0} \cap c(P) = \emptyset$, then

$$\begin{aligned} \emptyset \; = \; (E_0 \cap c(P)) \cup (E_1 \cap c(P)) \; = \; (E_{0} \cup E_{1}) \cap c(P). \end{aligned}$$

On the other hand, if $E_{0} \cap c(P) \ne \emptyset$, then we also have $E_1 \cap c(P) \ne \emptyset$, so that by the Zero-One property of $E_{0}$ and $E_{1}$, we obtain $E_0 \cap c(P) = E_0 \cap P$ and $E_1 \cap c(P) = E_1 \cap P$, respectively. It follows that the equality $(E_{0} \cup E_{1}) \cap c(P) = (E_{0} \cup E_{1}) \cap P$ holds in this case. This completes the proof that $E_{0} \cup E_{1}$ satisfies Zero-One, and so (b) holds as well. $\square$

Proof of Lemma 15    Necessity is an immediate consequence of Lemma 14(a). Sufficiency readily follows from Lemma 14(b). $\square$

Proof of Lemma 16    For necessity, let $x,y \in X$ be indiscernible items, and A an arbitrary menu. Common Destiny is an immediate consequence of Zero-One applied to the indiscernible menu $\{x,y\}$. Binary Tracing readily follows from Tracing applied to the indiscernible menu $\{x,y\}$ and its probe A.

For sufficiency, assume that Binary Tracing and Common Destiny hold for each menu $A \in \varOmega$. We prove that the menu $\{x,y\}$ is indiscernible. By Lemma 13, it suffices to show that Zero-One and Tracing hold for $\{x,y\}$. The latter property is immediate, so we prove the Zero-One property. Let $A \in \textsf {Probe}(\{x,y\})$. If $x,y \in A$, then the hypothesis readily yields that $\{x,y\} \cap c(A)$ is either empty or equal to $\{x,y\}$. On the other hand, if $\{x,y\} \cap A = \{x\}$ (or, dually, $\{x,y\} \cap A = \{y\}$), then we can only have either $\{x,y\} \cap c(A) = \emptyset$ or $\{x,y\} \cap c(A) = \{x\}$ (or $\{x,y\} \cap c(A) = \{y\}$ in the dual case). In any of these possible cases, the set $\{x,y\} \cap c(A)$ is either empty, or equal to $\{x,y\} \cap A$. $\square$

Proof of Theorem 7   Let $\equiv$ be an equivalence relation on a choice space (X, c).

(ii) $\Longleftrightarrow$ (iii) $\Longleftrightarrow$ (iv) $\Longleftrightarrow$ (v): by Lemma 10.

(i) $\Longrightarrow$ (ii): by Definition 3 and Lemma 7.

(i) $\Longrightarrow$ (vi): by Definition 3.

(i) $\Longrightarrow$ (vii): by Theorem 1.

(i) $\Longrightarrow$ (viii): by Theorem 2.

(i) $\Longrightarrow$ (ix): Suppose $\equiv$ is a congruence relation. Toward a contradiction, assume that there are $x,y \in X$ such that $x \equiv y$ but . By Lemma 16, without loss of generality there exists a menu $A \in \varOmega$ such that one of the following cases happens:

(I) A contains both x and y, $x \in c(A)$, and $y \notin c(A)$; or

(II) A contains x but not y, and $c(A) \ne c(A \cup \{y\}) \cap A$.

Case (II) breaks down in two subcases, namely:

(II.a) $x \in A$ and $z \in c(A) {\setminus } c(A \cup \{y\})$ for some $z \in A$; or

(II.b) $x \in A$ and $z \in c(A \cup \{y\}) {\setminus } c(A)$ for some $z \in A$.

We show that Co yields a contradiction in all cases.

• Case (I): Property Co implies $y \not \in c(A) = \textsf {sat}_\equiv (c(A)) \cap A \ni y$.

• Case (II.a): The hypothesis $x \equiv y$ yields $\textsf {sat}_\equiv (A) = \textsf {sat}_\equiv (A \cup \{y\})$. Now property Co implies $z \not \in c(A \cup \{y\}) = \textsf {sat}_\equiv (c(A)) \cap (A \cup \{y\}) \ni z$.

• Case (II.b): Property Co implies $z \in c(A \cup \{y\}) = \textsf {sat}_\equiv (c(A)) \cap (A \cup \{y\})$. It follows that $z \not \in c(A) = \textsf {sat}_\equiv (c(A)) \cap A \ni z$, again by Co.

(iii) $\Longrightarrow$ (i): Assume that (iii) holds, and let $A,B \in \varOmega$ be such that $\textsf {sat}(A) = \textsf {sat}(B)$. Since

$$\begin{aligned} c(A) \;&= \; \textsf {sat}\big (c(A) \big ) \cap A \quad \text {(by Local Congruency)}\\&= \; c\big (\textsf {sat}(A)\big ) \cap A \quad \text {(by Commutativity)}\\&= \; c\big (\textsf {sat}(B)\big ) \cap A \quad \text {(by hypothesis)}\\&= \; \textsf {sat}\big (c(B)\big ) \cap A \quad \text {(by Commutativity)}, \end{aligned}$$

it follows that (i) holds.

(vi) $\Longrightarrow$ (ii): by Lemmas 10 and 11.

(vi) $\Longleftrightarrow$ (viii): Let $A \in \varOmega$. Using the definitions of quotient choice, canonical projection, and saturation operator, it is easy to check that the following chain of equalities hold:

$$\begin{aligned} \bigcup c' \big (\pi (A)\big ) \; = \; \bigcup \pi \big (c \big ( \bigcup \pi (A) \big ) \big ) \; = \; \textsf {sat}\big ( c (\textsf {sat}(A) )\big ). \end{aligned}$$

It follows that

$$\begin{aligned} \bigcup c' \big (\pi (A)\big ) \cap A \; = \; \textsf {sat}\big ( c (\textsf {sat}(A) )\big ) \cap A. \end{aligned}$$

Thus (vi) and (viii) are equivalent.

(vii) $\Longrightarrow$ (ix): Assume that c satisfies the three properties BF, CD, and RI. We prove that $\equiv$ is contained in $\Bumpeq _c$. Thus, let $x,y \in X$ be such that $x \equiv y$. By Lemma 16, to show that $x \Bumpeq _c y$ holds, it suffices to prove the two properties of Binary Tracing and Common Destiny hold. Since c satisfies Common Destiny by hypothesis, it remains to show that the following implication holds for each menu $A \in \varOmega$:

$$\begin{aligned} \{x,y\} \cap A \ne \emptyset \quad \Longrightarrow \quad c(A) = c(A \cup \{x,y\}) \cap A. \end{aligned}$$

(5)

By symmetry, it suffices to examine the following two cases: (I) $x,y \in A$; (I) $x \in A$ and $y \notin A$. The claim holds trivially in case (I). In case (II), the consequent of (5) becomes $c(A) = c(A \cup \{y\}) \cap A$. Since $x, y\in A \cup \{y\}$ and $x \equiv y$ by hypothesis, an application of RI to the menu $A \cup \{y\}$ yields

$$\begin{aligned} c(A) \; = \; c\big ((A \cup \{y\}) {\setminus } \{y\}\big ) \; = \; c(A \cup \{y \}) {\setminus } \{y\} \; = \; c(A \cup \{y \}) \cap A, \end{aligned}$$

as required.

(ix) $\Longrightarrow$ (i):³¹ Suppose $\equiv$ is contained in $\Bumpeq _c\,$, and let $A, B \in \varOmega$ be such that $\textsf {sat}_{\equiv }(A) = \textsf {sat}_{\equiv }(B)$. To prove the claim, we show that the equality

$$\begin{aligned} c(A) \; = \; \textsf {sat}_\equiv \big (c(B)\big ) \cap A \end{aligned}$$

(6)

holds. The hypothesis yields $\textsf {sat}_{\Bumpeq _c}(A) = \textsf {sat}_{\Bumpeq _c}(B)$, whence $c(A) = \textsf {sat}_{\Bumpeq _c}(c(A)) \cap A$ since $\Bumpeq _{c}$ is a congruence relation by Theorem 4. Therefore, to prove (7), it suffices to show that the equality

$$\begin{aligned} \textsf {sat}_\equiv \big (c(B)\big ) \cap A \; = \; \textsf {sat}_{\Bumpeq _c}\big (c(B)\big ) \cap A \end{aligned}$$

(7)

holds. The inclusion “$\subseteq$” is obvious by hypothesis. To prove the reverse inclusion, assume that $x \in \textsf {sat}_{\Bumpeq _c}(c(B)) \cap A$. Since $\textsf {sat}_{\equiv }(A) = \textsf {sat}_{\equiv }(B)$ and $x \in A$, there is $y \in B$ such that $x \equiv y$, hence $x \Bumpeq _c y$ by hypothesis. Furthermore, since $x \in \textsf {sat}_{\Bumpeq _c}(c(B))$, there is $z \in c(B)$ such that $x \Bumpeq _c z$. Since $\Bumpeq _c$ is transitive, it follows that $y \Bumpeq _c z$, and so $y \in c(B)$ by Common Destiny. Thus, $x \in \textsf {sat}_{\equiv }(c(B)) \cap A$, as required.

This completes the proof of Theorem 7. $\square$

Appendix B: Proofs of main results

Here we prove all the results stated in the main body of the paper. To begin with, we recall three important results, which connect the rationalizability of a choice with the satisfaction of suitable axioms of choice consistency.³²

Theorem 8

(Sen 1971) A finite choice is rationalizable if and only if $(\alpha )$ and $(\gamma )$ hold.

Theorem 9

(Arrow 1959; Sen 1971) A finite choice is transitively rationalizable if and only if WARP holds if and only if $(\alpha )$ and $(\beta )$ hold.

Theorem 10

(Eliaz and Ok 2006; Cantone et al. 2016) A finite choice is quasi-transitively rationalizable if and only if WARNI holds³³ if and only if $(\alpha )$, $(\gamma )$, $(\rho )$ hold.³⁴

We now present the proofs of the main results of this paper, organized by sections.

1.1 Sect. 2 (Congruence relations)

Proof of Theorem 1    By Lemma 4 (independently proved later on), it suffices to show that (i) implies (iii), and (iv) implies (i).

(i) $\Rightarrow$ (iii). Assume that $\equiv$ is a congruence on the choice space (X, c). We show that the two properties of Common Destiny and Repetition Irrelevance hold. Let $x,y \in X$ be such that $x \equiv y$. For CD, assume that $A \in \varOmega$ is such that $x,y \in A$ and $x \in c(A)$.

Since $x \equiv y$, we get $y \in \textsf {sat}\big (c(A)\big )$, hence $y \in \textsf {sat}\big (c(A)\big ) \cap A = c(A)$ using property Co$_{\mathrm {loc}}$. This proves $y \in c(A)$, and so the claim holds.

Next, we show that c satisfies RI as well. Assume that $x,y \in A$. We shall show that $c(A {\setminus } \{y\}) = c(A) {\setminus } \{y\}$. Since $x \equiv y$, the equality $\textsf {sat}(A) = \textsf {sat}(A {\setminus } \{y\})$ holds, hence full congruency applies to A and $A {\setminus } \{y\}$ (in both directions). Thus, we obtain:

$$\begin{aligned} c(A{\setminus } \{y\})&= \textsf {sat}(c(A)) \cap (A {\setminus } \{y\}) \quad \text {(by }\textsf {Co}\text { applied to }A {\setminus } \{y\}\text { and }A\text {)}\\&= \textsf {sat}(c(A {\setminus } \{y\})) \cap (A {\setminus } \{y\}) \quad \text {(since }x \equiv y\text { and }x,y \in A\text {)}\\&= \big (\textsf {sat}(c(A {\setminus } \{y\})) \cap A \big ) {\setminus } \{y\} \\&= c(A) {\setminus } \{y\} \quad \text {(by }\textsf {Co}\text { applied to }A \text { and }A {\setminus } \{y\}\text {).} \end{aligned}$$

This proves the claim.

(iv) $\Rightarrow$ (i). This follows from the chain of implications (vii) $\Rightarrow$ (ix) $\Rightarrow$ (i) in Theorem 7 (see its proof). $\square$

1.2 Sect. 3 (Congruences are reflective and consistent)

Proof of Theorem 2    Let $\equiv$ be a congruence relation on a choice space (X, c), and $(X',c')$ the quotient space induced by $\equiv \,$. To prove that $\equiv$ is reflective, we first prove the following version of the commutativity property for the quotient choice:

Lemma 17

If $\equiv$ is a congruence relation, then $c' \circ \pi = \pi \circ c$.

Proof of Lemma 17 For each menu $A \in \varOmega$, we have:

$$\begin{aligned} c' \big (\pi (A)\big ) \;&= \, c' \big ( \{a' : a \in A \} \big ) \quad \text {(by the definition of }\pi \text {)} \\&= \; \pi \big (c \big (\bigcup \,\{a' : a \in A \} \big ) \big ) \quad \text {(by the definition of }c'\text {)}\\&= \; \pi \big (c (\textsf {sat}(A))\big ) \quad \text {(by the definition of }\textsf {sat}\text {)}\\&= \; \pi \big ( \textsf {sat}(c(A))\big ) \quad \text {(by Lemma}~7\text {)} \\&= \; \pi \big ( c(A) \big ) \quad \text {(by a property of }\pi \text {)}. \end{aligned}$$

This proves Lemma 17. $\square$

Now let $A \in \varOmega$. Since we have

$$\begin{aligned} c(A) \;&= \; \textsf {sat}\big (c(A)\big ) \cap A \quad \text {(by }\textsf {Co}_{\mathrm {loc}}\text {)}\\&= \; \bigcup \pi \big (c(A)\big ) \cap A \quad \text {(by the definition of }\textsf {sat}\text {)}\\&= \; \bigcup c' \big (\pi (A)\big ) \cap A \quad \text {(by Lemma}~17\text {)}, \end{aligned}$$

the equality (2) in Definition 6 holds, and so $c'$ is a reflection of c. This completes the proof of the theorem. $\square$

Proof of Theorem 3    This is a special case of a general result, which states that certain types of second-order logic formulae are preserved—back and forth—in the passage to the quotient with respect to a congruence relation Cantone et al. (2017a).

For the sake of completeness, here we sketch a direct proof.³⁵

To provide the reader with a transparent argument, we begin by recalling the basic notation. Let (X, c) be a choice space, $\equiv$ a congruence relation on (X, c), and $\pi :X \rightarrow X'$ the canonical projection defined by $x \mapsto x' = [x]_\equiv$. Further, let $(X',c')$ be the quotient space induced by $\equiv \,$, where the quotient choice $c'$ is defined by $c'(\mathcal {A}) = \pi (c(\pi ^{-1}(\mathcal {A})))$ for each $\mathcal {A}\in \varOmega '$ (see Definition 5).

The proof of Theorem 3 consists of six steps.

• Step 1 (Reduction) It suffices to only show that the four axioms $(\alpha )$, $(\gamma )$, $(\beta )$, and $(\rho )$ are reflected by the quotient, since the reflection of WARP and WARNI is guaranteed by Theorems 9 and 10, respectively. For simplicity, we shall henceforth refer to the four properties $(\alpha )$, $(\gamma )$, $(\beta )$, and $(\rho )$ as the basic axioms (of choice consistency). To prove the reflectivity of the basic axioms, we shall only use the property of commutativity and local congruency: the first holds by Lemma 7, and the second is a special case of full congruency.

• Step 2 (Identity) The following consequence of Local Congruency is needed (its simple proof is left to the reader):

Lemma 18

For each $\mathcal {A}\in \varOmega '$, we have $c\big (\pi ^{-1}(\mathcal {A})\big ) \; = \; \textsf {sat}\big (c\big (\pi ^{-1}(\mathcal {A})\big )\big )$.

• Step 3 (Reformulation) It is convenient to rewrite the four basic axioms omitting any reference to items, thus only quantifying over menus:

$(\alpha )$ -axiom :

$(\forall A,B \in \varOmega )$$\Big [ A \subseteq B \;\; \Longrightarrow \;\; c(B) \cap A \subseteq c(A) \Big ]$;

$(\gamma )$ -axiom :

$(\forall A,B \in \varOmega )$$\Big [ c(A) \cap c(B) \subseteq c(A \cup B) \Big ]$;

$(\beta )$ -axiom :

$(\forall A,B \in \varOmega )$$\Big [ \big (A \subseteq B \; \wedge \; c(A) \cap c(B) \ne \emptyset \big ) \;\; \Longrightarrow \;\; c(A) \subseteq c(B) \Big ]$;

$(\rho )$ -axiom :

$(\forall A,B \in \varOmega )$$\Big [c(A) {\setminus } c(A \cup B) \ne \emptyset \;\; \Longrightarrow \;\; B \cap c(A \cup B) \ne \emptyset \Big ]$.

The proof that the above formulations are equivalent to the original ones is not difficult, and is again left to the reader.³⁶

• Step 4 (Lifting) This is the main step of the proof: it lifts the validity of the four basic axioms from the family of saturated menus to the whole choice domain. (See Cantone et al. (2017b) for a detailed analysis of the lifting property, with applications to the decision problem for fragments of set theory in presence of a choice correspondence.)

Lemma 19

If a basic axiom holds for saturated menus, then it also holds for all menus.

Proof of Lemma 19 Assume that the basic axioms of consistency hold for saturated menus in the choice space (X, c). Let A and B be arbitrary menus in (X, c).

$(\alpha )$: Assume that $A \subseteq B$, hence $\textsf {sat}(A) \subseteq \textsf {sat}(B)$. Since $(\alpha )$ applies to saturated menus, we get

$$\begin{aligned} c(B) \cap A \;&\subseteq \; \textsf {sat}\big (c(B)\big ) \cap \textsf {sat}(A) \cap A \quad \text {(by Lemma}~6\text {)} \\&= \; c\big (\textsf {sat}(B)\big ) \cap \textsf {sat}(A) \cap A \quad \text {(by Commutativity)} \\&\subseteq \; c\big (\textsf {sat}(A)\big ) \cap A \quad \text {(by }(\alpha )\text { for saturated menus)} \\&= \; \textsf {sat}\big (c(A)\big ) \cap A \quad \text {(by Commutativity)} \\&= \; c(A) \quad \text {(by Local Congruency),} \end{aligned}$$

which proves the claim.

$(\gamma )$: Since we have

$$\begin{aligned} c(A) \cap c(B) \;&= \; \big (\textsf {sat}(c(A)) \cap A\big ) \cap \big (\textsf {sat}(c(B)) \cap B \big ) \quad \text {(by Local Congruency)}\\&= \; \big (c(\textsf {sat}(A)) \cap A\big ) \cap \big (c(\textsf {sat}(B)) \cap B \big ) \quad \text {(by Commutativity)}\\&\subseteq \; c\big (\textsf {sat}(A) \cup \textsf {sat}(B)\big ) \cap (A \cap B) \quad \text {(by }(\gamma )\text { for saturated menus)} \\&= \; c\big (\textsf {sat}(A \cup B) \big ) \cap (A \cap B) \quad \text {(by Lemma}~6\text {)} \\&\subseteq \; c\big (\textsf {sat}(A \cup B)\big ) \cap (A \cup B) \\&= \; \textsf {sat}\big (c(A \cup B)\big ) \cap (A \cup B) \quad \text {(by Commutativity)} \\&= \; c(A \cup B) \quad \text {(by Local Congruency)}, \end{aligned}$$

property $(\gamma )$ holds for arbitrary menus.

$(\beta )$: Assume that $A \subseteq B$ and $c(A) \cap c(B) \ne \emptyset$. Lemma 6 and Commutativity yield

$$\begin{aligned} \textsf {sat}(A) \subseteq \textsf {sat}(B) \qquad \hbox {and}\qquad c\big (\textsf {sat}(A)\big ) \cap c\big (\textsf {sat}(B)\big ) \, = \, \textsf {sat}\big (c(A)\big ) \cap \textsf {sat}\big (c(B)\big ) \, \ne \, \emptyset . \end{aligned}$$

Thus, property $(\beta )$ applied to $\textsf {sat}(A)$ and $\textsf {sat}(B)$ entails $c\big (\textsf {sat}(A)\big ) \, \subseteq \, c\big (\textsf {sat}(B)\big )$. From the last inclusion, Commutativity, and Local Congruency, we conclude

$$\begin{aligned} c(A)= & {} \textsf {sat}\big (c(A)\big ) \cap A \, = \, c\big (\textsf {sat}(A)\big ) \cap A \, \subseteq \, c\big (\textsf {sat}(B)\big ) \cap B \\= & {} \textsf {sat}\big (c(B)\big ) \cap B \, = \, c(B). \end{aligned}$$

This proves the claim.

$(\rho )$: Assume that $c(A) {\setminus } c(A \cup B) \ne \emptyset$.

By Lemma 6, Commutativity, and Local Congruency, we obtain

$$\begin{aligned} c\big (\textsf {sat}(A)\big ) {\setminus } c\big (\textsf {sat}(A) \cup \textsf {sat}(B)\big ) \;&= \; c\big (\textsf {sat}(A)\big ) {\setminus } c\big (\textsf {sat}(A \cup B)\big ) \\&\supseteq \; \Big ( c\big (\textsf {sat}(A)\big ) {\setminus } c\big (\textsf {sat}(A \cup B)\big ) \Big ) \cap A \\&\supseteq \; \Big ( c\big (\textsf {sat}(A)\big ) \cap A \Big ) {\setminus } \Big ( c\big (\textsf {sat}(A \cup B)\big ) \cap (A \cup B) \Big ) \\&= \; \Big ( \textsf {sat}\big (c(A)\big ) \cap A \Big ) {\setminus } \Big ( \textsf {sat}\big (c(A \cup B)\big ) \cap (A \cup B) \Big ) \\&= \; c(A) {\setminus } c(A \cup B) \; \ne \; \emptyset . \end{aligned}$$

Property $(\rho )$ (applied to $\textsf {sat}(A)$ and $\textsf {sat}(B)$), Lemma 6, and Commutativity entail

$$\begin{aligned} \emptyset \, \ne \, \textsf {sat}(B) \cap c\big (\textsf {sat}(A) \cup \textsf {sat}(B)\big )= & {} \textsf {sat}(B) \cap c\big (\textsf {sat}(A \cup B)\big ) \\= & {} \textsf {sat}(B) \cap \textsf {sat}\big (c(A \cup B)\big ). \end{aligned}$$

Choose $x \in \textsf {sat}(B) \cap \textsf {sat}\big (c(A \cup B)\big )$. Thus, there are $b \in B$ and $a \in c(A \cup B)$ such that $b \equiv x \equiv a$. Now transitivity implies $a \equiv b$, and Common Destiny yields $b \in B \cap c(A \cup B)$. This proves $B \cap c(A \cup B) \ne \emptyset$, as claimed. $\square$

• Step 5 (Equivalence) The key equivalence in the proof of the reflection of all basic axioms is that a congruence class is selected by the quotient choice if and only if its representative is selected by the primitive choice in the corresponding saturated menu. The formal statement, which is an easy consequence of Lemma 18 and the definition of quotient choice, is the following:

Lemma 20

For each $x \in X$ and $\mathcal {A}\in \varOmega '$, $x' \in c'(\mathcal {A})$ if and only if $x \in c(\pi ^{-1}(\mathcal {A}))$.

• Step 6 (Conclusion) We prove that axiom $(\alpha )$ holds in (X, c) if and only if it holds in $(X',c')$; likewise, one can show that the same is true of the other three basic axioms.³⁷ The hard part is sufficiency, for which we need the lifting result (Lemma 19).

(Necessity) Assume that $(\alpha )$ holds in (X, c). Let $x \in X$ and $\mathcal {A},\mathcal {B} \in \varOmega '$ be such that $x' = \pi (x) \in \mathcal {A}\subseteq \mathcal {B}$ and $x' \in c'(\mathcal {B})$. Apply Lemma 20 and axiom $(\alpha )$ to get $x \in c(\pi ^{-1}(\mathcal {A}))$. Now apply Lemma 20 again to get $x' \in c'(\mathcal {A})$. This proves that $(\alpha )$ holds for $(X',c')$.

(Sufficiency) Assume that $(\alpha )$ holds in $(X',c')$. By Lemma 19, it suffices to show that $(\alpha )$ holds in (X, c) for saturated menus. Let $x \in X$ and $A,B \in \varOmega$ be such that $x \in A \subseteq B$ and $x \in c(B)$, where A and B are saturated. The saturation of A and B means that there are $\mathcal {A},\mathcal {B} \in \varOmega '$ such that $A = \pi ^{-1}(\mathcal {A})$ and $B = \pi ^{-1}(\mathcal {B})$. Now the proof proceeds as in the necessary part. Indeed, since $x \in \pi ^{-1} (\mathcal {A}) \subseteq \pi ^{-1}(\mathcal {B})$ and $x \in \pi ^{-1}(\mathcal {B})$, we can apply Lemma 20 and axiom $(\alpha )$ to get $\pi (x) = x' \in c'(\mathcal {A})$. Another application of Lemma 20 delivers $x \in c(\pi ^{-1}(\mathcal {A})) = c(A)$, as claimed.

This completes the proof of Theorem 3. $\square$

Proof of Corollary 1    The claim readily follows from Theorems 8, 9, 10, and 3. $\square$

1.3 Sect. 4 (Revealed indiscernibility and the lattice of congruences)

Proof of Theorem 4    Let (X, c) be a choice space, and $\Bumpeq _c$ the relation of revealed indiscernibility associated to it. First, we show that $\Bumpeq _c$ is an equivalence relation on X. To that end, it suffices to show that, for all $x,y \in X$, we have $x \Bumpeq _c y$ if and only if x and y belong to the same maximal indiscernible menu. Sufficiency readily follows from Lemma 14(a), whereas necessity is a direct consequence of Lemma 14(b).

Thus, to complete the proof, it remains to show that property Co holds. We need two preliminary results. (In what follows, $\textsf {sat}$ denotes the saturation operation with respect to revealed indiscernibility.)

Lemma 21

For each family $\mathcal {E}= \{E_{1},\ldots ,E_{n}\}$ of indiscernible menus, and menu $A \in \varOmega$ such that $A \cap E_i \ne \emptyset$ for all $i \in \{1,\ldots ,n\}$, we have:

• (a) $c(A) = c\big (\bigcup \mathcal {E}\cup A \big ) \cap A\,$;

• (b) if $A \subseteq \bigcup \mathcal {E}\,$, then $c(A) = c\big (\bigcup \mathcal {E}\big ) \cap A\,$;

• (c) $c(A) = c \big (\textsf {sat}(A)\big ) \cap A$.

Proof of Lemma 21    It suffices to prove (a) only, because (b) readily follows from (a), and (c) is a particular case of (b) (since $A \subseteq \textsf {sat}(A)$ by Lemma 6). We proceed by induction on n. For the base case $n=1$, observe that the claim is equivalent to say that an indiscernible menu $E_1$ satisfies Tracing, hence it is an immediate consequence of Lemma 13. For the inductive step, assume that the results holds for $n-1 \ge 1$. The hypothesis yields that $\bigcup _{i = 1}^{n-1} E_i \cup A$ is a probe for the indiscernible menu $E_n$. Therefore, we have:

$$\begin{aligned} c(A) \;&= \; c\big (\textstyle \bigcup _{i=1}^{n-1} E_{i} \cup A\big ) \cap A \quad \text {(by the induction hypothesis)}\\&= \; c\big (\textstyle \bigcup _{i=1}^{n} E_{i} \cup A\big ) \cap \big (\textstyle \bigcup _{i=1}^{n-1} E_{i} \cup A\big ) \cap A\quad \text {(by Tracing for }E_n\text {)}\\&= \; c\big (\textstyle \bigcup _{i=1}^{n} E_{i} \cup A\big ) \cap A. \end{aligned}$$

This proves Lemma 21. $\square$

Lemma 22

The saturation operator associated to revealed indiscernibility commutes with the primitive choice, that is, $\textsf {sat}\big (c(A)\big ) = c\big (\textsf {sat}(A)\big )$ for each $A \in \varOmega$.

Proof of Lemma 22    Let $A \in \varOmega$. Lemma 21(c) readily yields

$$\begin{aligned} \textsf {sat}\big (c(A)\big ) \; =\; \textsf {sat}\Big (c\big (\textsf {sat}(A) \big ) \cap A \Big ). \end{aligned}$$

Thus, to prove that $c(\textsf {sat}(A)) = \textsf {sat}(c(A))$, it suffices to show that

$$\begin{aligned} c\big (\textsf {sat}(A)\big ) \; = \; \textsf {sat}\Big (c\big (\textsf {sat}(A) \big ) \cap A \Big ). \end{aligned}$$

(8)

Assume that $x \in c\big (\textsf {sat}(A)\big )$. If $x \in A$, then $x \in \textsf {sat}\big (c\big (\textsf {sat}(A) \big ) \cap A \big )$. Next, let $x \notin A$. Since $x \in \textsf {sat}(A)$, there exists $a \in A$ such that $x \Bumpeq _c a$. Since $x \in c\big (\textsf {sat}(A)\big )$, the property of Common Destiny (Lemma 16) yields $a \in c\big (\textsf {sat}(A)\big ) \cap A$. It follows that $x \in \textsf {sat}\big (c\big (\textsf {sat}(A) \big ) \cap A \big )$. This proves one inclusion in (8).

For the reverse inclusion, assume that $x \in \textsf {sat}\big (c\big (\textsf {sat}(A) \big ) \cap A \big )$. Thus, there is $a \in c\big (\textsf {sat}(A)\big ) \cap A$ such that $x \Bumpeq _c a$. Since $x \in \textsf {sat}(A)$ by the monotonicity of the saturation operator (Lemma 6), Common Destiny entails $x \in c\big (\textsf {sat}(A)\big )$. This proves Lemma 22. $\square$

Now we resume the proof of Theorem 4. Assume that $A,B \in \varOmega$ are such that $\textsf {sat}(A) = \textsf {sat}(B)$. Since

$$\begin{aligned} c(A) \;&= c \big (\textsf {sat}(A) \big ) \cap A \quad \text {(by Lemma}~21\text {(c))}\\&= c \big (\textsf {sat}(B) \big ) \cap A \quad \text {(by hypothesis)}\\&= \textsf {sat}\big (c(B)\big ) \cap A \quad \text {(by Lemma}~22\text {)}, \end{aligned}$$

it follows that property Co holds. This completes the proof of Theorem 4. $\square$

Proof of Lemma 3³⁸    The two inclusions $\Bumpeq _c \; \subseteq \; \sim _c^\textsf {U}$ and $\sim _c^\textsf {B}\; \subseteq \; \sim _c\,$ are an immediate consequence of the definitions of the four relations of revealed similarity. Thus, it suffices to show that the inclusion $\sim _c^\textsf {U}\; \subseteq \; \sim _c^\textsf {B}\,$ holds as well. Let $x,y \in X$ be such that $x \sim _c^\textsf {U}y$. To prove $x \sim _c^\textsf {B}y$, we need to show that the two equalities $x^{\uparrow ,\succsim _c} = y^{\uparrow ,\succsim _c}$ and $x^{\downarrow ,\succsim _c} = y^{\downarrow ,\succsim _c}$ hold, where $x^{\uparrow ,\succsim _c} = \{z \in X : z \succsim _c x\}$ is the weak upper section of x, and $x^{\downarrow ,\succsim _c} = \{z \in X : x \succsim _c z\}$ is the weak lower section of x. By symmetry, it suffices to prove the equality of the two weak lower sections. To that end, assume by contradiction that there is $z \in X$ such that $z \in x^{\downarrow ,\succsim _c}$ but $z \notin y^{\downarrow ,\succsim _c}$. Thus, we have $x \succsim _c z$ and $z \succ _c y$ (where the strict preference of z over y holds by the completeness of $\succsim _c$). By the very definition of $\succsim _c$, it follows that there is a menu $A \in \varOmega$ containing both x and z such that $x \in c(A)$, and y is never chosen in any menu containing both y and z. By Common Destiny, the menu A cannot contain y. Now Binary Fungibility yields that $y \in c(A_{x \leftarrow y})$. However, the latter fact is impossible, since z belongs to $A_{x \leftarrow y}$. This proves that the inclusion $x^{\downarrow ,\succsim _c} \subseteq y^{\downarrow ,\succsim _c}$ holds. The reverse inclusion is proved similarly. Thus, we have $x^{\downarrow ,\succsim _c} = y^{\downarrow ,\succsim _c}$, as claimed. $\square$

Proof of Theorem 5³⁹    Let (X, c) be a choice space.

(a) Assume that c is rationalizable, i.e., $c(A) = \max (A,\succsim _c)$ for all $A \in \varOmega$.

First, we show that revealed behavioral indifference $\sim _c^\textsf {B}$ and revealed universal indifference $\sim _c^\textsf {U}$ coincide. By Lemma 3, it suffices to show that the inclusion $\sim _c^\textsf {B}\, \subseteq \, \sim _c^\textsf {U}$ holds. Thus, let $x,y \in X$ be such that $x \sim _c^\textsf {B}y$, i.e., $x^{\uparrow ,\succsim _c} = y^{\uparrow ,\succsim _c}$ and $x^{\downarrow ,\succsim _c} = y^{\downarrow ,\succsim _c}$. In what follows, we show that the properties of Binary Fungibility and Common Destiny hold for x and y: this will prove that $x \sim _c^\textsf {U}y$, as claimed. Toward a contradiction, assume that BF fails for x and y, i.e., without loss of generality, there exists a menu $A \in \varOmega$ such that $x \in c(A)$, $y \notin A$, and $y \notin c(A_{x \leftarrow y})$. By the hypothesis of rationalizability, we obtain $x \in \max (A,\succsim _c)$ and $y \notin \max (A_{x \leftarrow y},\succsim _c)$. Thus, there exists $z \in A_{x \leftarrow y}$ such that $z \succ _c y$. Since $z \ne x$, it follows that $z \in A$, and so $x \succsim _c z$. However, this is impossible, since we would have $z \in x^{\downarrow ,\succsim _c}$ and $z \notin y^{\downarrow ,\succsim _c}$, contradicting the hypothesis. This proves that BF holds. To prove that CD holds as well, again we argue by contradiction. Assume that there exists a menu $A \in \varOmega$ such that $x,y \in A$, $x \in c(A)$, but $y \notin c(A)$. By rationalizability, there is $z \in A$ such that $x \succsim _c z$ and $z \succ _c y$. However, this contradicts the equality $x^{\downarrow ,\succsim _c} = y^{\downarrow ,\succsim _c}$.

To conclude the proof of part (a), we show that revealed universal indifference $\sim _c^\textsf {U}$ and revealed indiscernibility $\Bumpeq _c$ coincide. By Lemma 3 and Corollary 3, it suffices to show that if x and y are revealed to be universally indifferent, then they also satisfy Repetition Irrelevance. This is an immediate consequence of Lemma 5.

(b) Assume that c is transitively rationalizable, i.e., $c(A) = \max (A,\succsim _c)$ for all $A \in \varOmega$, and $\succsim _c$ is a total preorder (reflexive, transitive, and complete). By part (a), it suffices to show that revealed existential indifference $\sim _c$ and revealed behavioral indifference $\sim _c^\textsf {B}$ coincide. By Lemma 3, it suffices to show that $\sim _c$ is contained in $\sim _c^\textsf {B}$. Thus, let $x,y \in X$ be such that $x \sim _c y$. By symmetry, to prove that $x \sim _c^\textsf {B}$ holds, it suffices to show that the weak lower sections of x and y are equal. Let $z \in x^{\downarrow ,\succsim _c}$, i.e., $x \succsim _c z$. If $z \succ _c y$ were to hold, then the transitivity of $\succsim _c$ would yield $x \succ _c y$, a contradiction. It follows that $y \succsim _c z$. This proves that $x^{\downarrow ,\succsim _c}$ is contained into $y^{\downarrow ,\succsim _c}$. The reverse inclusion is proved similarly.⁴⁰$\square$

Proof of Theorem 6    Let (X, c) be a choice space, and $\textsf {Congr}(X,c)$ the set of all congruence relations on (X, c). For each $R,S \in \textsf {Congr}(X,c)$, define

$$\begin{aligned} R \wedge S \; := \; R \cap S \qquad \hbox {and}\qquad R \vee S \; := \; \textsf {tc}(R \cup S), \end{aligned}$$

where “tc” stands for transitive closure. We claim that $R \wedge S$ and $R \vee S$ are, respectively, the maximal congruence relation on (X, c) contained in R and S, and the minimal congruence relation in (X, c) containing R and S: this will prove that $\textsf {Congr}(X,c)$ is a lattice under set-containment.

To start, observe that $R \wedge S$ is the maximal equivalence relation contained in both R and S. Dually, $R \vee S$ is the minimal equivalence relation on X containing both R and S. Therefore, the claim holds as long as we show that both $R \wedge S$ and $R \vee S$ are congruence relations on (X, c). The latter facts are easy consequences of the characterization of a congruence relation given in Theorem 7. Indeed, since R and S are congruence relations on (X, c), both of them are contained in the relation of revealed indiscernibility. In particular, $R \wedge S$ in contained in $\Bumpeq _c$, hence (being an equivalence relation) it is a congruence relation on (X, c) by Theorem 7. Furthermore, the minimality of the transitive closure and the fact that revealed indiscernibility is an equivalence relation yield that $R \vee S$ is contained in $\Bumpeq _c\,$, which in turn implies that $R \vee S$ is a congruence relation on (X, c), again using Theorem 7. This proves that any pair R and S of elements in $\textsf {Congr}(X,c)$ has both a meet $R \wedge S$ and a join $R \vee S$, as claimed.

Finally, Corollary 3 yields that revealed indiscernibility is the maximum element of $\textsf {Congr}(X,c)$, whereas equality is obviously the minimum element of $\textsf {Congr}(X,c)$. $\square$