On the aversion to incomplete preferences

Abstract

We propose an axiomatization of aversion to incomplete preferences. Some prevailing models of incomplete preferences rely on the hypothesis that incompleteness is temporary and that by keeping their opportunity set open individuals reveal a preference for flexibility. We consider that the maintenance of incomplete preference is also aversive. Our model allows us to show how incompleteness induces an aversive attitude in two different ways: intrinsic and instrumental. Intrinsic aversion holds when one instance of incomplete preference in the set suffices to decrease its utility. Instrumental aversion holds only insofar dominating options are affected by incompleteness. Given two partially overlapping sets of axioms on the binary relation over sets we formalize their consistency with the two types of aversion to incompleteness. Finally, we relate our model to the classical Sen’s distinction between tentative and assertive incompleteness. The spelling out of this distinction in the terms of our approach uncovers to what extent aversion to incompleteness may be compatible with a preference for flexibility.

Appendix

1.1 Proof of Theorem 4.1

In order to prove the sufficient part of the Theorem we have to prove that if $\succsim$ is P-intrinsic consistent with some asymmetric binary relation P then $\succsim$ satisfies RI, SC, MC, IA and CE.

• RI: $\{x,y\}\succ \{y\}$ implies, by the definition of P-intrinsic consistency, that xPy, which again by the definition of P-intrinsic consistency implies $\{x,y\}\sim \{y\}$, and vice versa.

• SC: By the definitions of P and $\otimes$, given any pair $x,y\in X$ either xPy, yPx or $x\otimes y$. By Definition 1, if xPy then $\{x,y\}\sim \{x\}$; if yPx then $\{x,y\}\succ \{x\}$; and if $x\otimes y$ then $\{x\}\succ \{x,y\}$.

• MC: Let $A\in \pi (X)$ and let $x\in X{\setminus } A$. Assume that $\{x,a\}\succ \{a\}$ for all $a\in A$. This implies, by Definition 1, that xPa for all $a\in A$, which by the same definition implies $A\cup \{x\} \succ A$.

• IA: Let $A\in \pi (X)$ and let $x\in X{\setminus } A$. Assume that $\{a,x\}\succsim \{a\}$ $\forall a\in A$. Thus, $\not \exists a\in A$ such that $\{a\}\succ \{a,x\}$, which by Definition 1 implies (i): $\not \exists a\in A$ s.t $a\otimes x$. Assume that, in addition, $\exists a^{*}\in A$ such that $\{a^{*},x\}\sim \{a^{*}\}$. This, by Definition 1, implies (ii): $a^{*}Px$. Given that (i) and (ii) hold, by Definition 1, $A\cup \{x\}\sim A$.

• CE: Assume that $A\succ A\cup \{x\}$. By Definition 1 this implies that there exists $a\in A$ such that $x\otimes a$. Therefore, if $A\subseteq B$ then there exists $b\in B$,($b=a$) such that $x\otimes b$, which by Definition 1 implies $B\succ B\cup \{x\}$.

In order to prove the necessary part of the theorem, assume that a binary relation $\succsim$ satisfies all the axioms of Theorem 4.1. Then there exists an asymmetric binary relation P defined on X with which $\succsim$ is P-intrinsic consistent.

We will show that, in particular, $\succsim$ is consistent with the relation P defined by: $P=\{(x,y)$ such that $\{x,y\}\succ \{y\}\}$.

First, we have to prove that P is asymmetric: Assume xPy, then by the definition of P, $\{x,y\}\succ \{y\}$. Assume that P is not asymmetric and therefore yPx. Then $\{x,y\}\succ \{x\}$, but this is in contradiction with RI.

Now, take $A\subseteq \pi (X)$ and $x\in X{\setminus } A$:

• (i) Assume that xPa for all $a\in A$. This, by the definition of P, implies $\{a,x\}\succ \{a\}$ $\forall a\in A$, which by MC implies $A\cup \{x\}\succ A$.

• (ii) Assume, now, that there exists a in A such that $x\otimes a$. Then, by the definition of $\otimes$, we know that $\lnot (aPx)$ and $\lnot (xPa)$. Given that $\lnot (aPx)$, by the definition of P, $\lnot (\{ax\}\succ \{x\})$, which by RI implies $\lnot (\{ax\}\sim \{a\})$. On the other hand, given that $\lnot (xPa)$, again by the definition of P, $\lnot (\{ax\}\succ \{a\})$. By SC, given that $\lnot (\{ax\}\sim \{a\})$ and $\lnot (\{ax\}\succ \{a\})$, we have that $\{a\}\succ \{ax\}$. Since $\{a\}\subseteq A$, by CE, $A\succ A\cup \{x\}$.

• (iii) Assume that none of the two assumptions in (i) and (ii) hold. Therefore, for all $a\in A$, either aPx or xPa and $\exists a^{*}\in A$ such that $a^{*}Px$. By the definition of P we have that $\{a^{*},x\}\succ \{x\}$ and by RI $\{a^{*},x\}\sim \{a^{*}\}$. Also, for all $a\in A$ either aPx, which as we have shown implies $\{a,x\}\sim \{a\}$, or xPa which by the definition of P implies $\{a,x\}\succ \{a\}$. Therefore, $\forall a\in A$ $\{a,x\}\succsim \{a\}$. Given that $\{a^{*},x\}\sim \{a^{*}\}$ and $\forall a\in A$ $\{a,x\}\succsim \{a\}$, by IA we have $A\cup \{x\} \sim A$.

To prove that the binary relation P with which $\succsim$ is P-consistent is unique, assume that there exists another binary relation $P'\ne P$ with which $\succsim$ is $P'$-consistent. $P\ne P'$ implies that there exist $x,y\in X$ such that $xP'y$ and $\lnot (\{x,y\}\succ \{y\})$. In this case, by the definition of $P'$-intrinsic consistency we have that $xP'y$ implies $\{x,y\}\succ \{y\}$ reaching a contradiction.

1.2 Proof of theorem 4.2

In order to prove the sufficient part of the Theorem we have to prove that if a reflexive binary relation $\succsim$ is consistent with some asymmetric and transitive binary relation P then $\succsim$ satisfies RI, SC, CC, IC, MR and CR.

• RI: $\{x,y\}\succ \{y\}$ implies, by Definition 2, that xPy, which again by the same definition implies $\{x,y\}\sim \{y\}$, and vice versa.

• SC: By the definition of binary relations P and $\otimes$, given any pair $x,y\in X$, either xPy, yPx or $x\otimes y$. By Definition 2, if xPy then $\{x,y\}\sim \{x\}$; if yPx then $\{x,y\}\succ \{x\}$; and if $x\otimes y$, then $\{x\}\succ \{x,y\}$.

• CC: By Definition 2$\{a\}\succ \{a,x\}$ implies $\lnot (aPx)$ and $\lnot (xPa)$. Therefore, if $\{a\}\succ \{a,x\}$ $\forall a\in A$ this means that $\not \exists a\in A$ s.t aPx or xPa, which again by Definition 2 implies $A\succ A\cup \{x\}$.

• IC: Let $x,y\in X$ and $A\in \pi (X)$ such that $y\notin A$, $\{x,y\}\sim \{x\}$ and $A\cup \{x\}\sim A$. By Definition 2, $\{x,y\}\sim \{x\}$ implies xPy. If $x\in A$, then, $\exists a\in A$ ($a=x$) such that aPy, and by Definition 2$A\cup \{y\}\sim A$. If $x\notin A$ then, by Definition 2, $A\cup \{x\}\sim A$ implies $\exists a\in A$ such that aPx and, by transitivity of P, aPy. Therefore, by Definition 2, $A\cup \{y\}\sim A$.

• MR: Let $A\in \pi (X)$ and $x,y\in X{\setminus } A$ such that $\lnot (\{y\}\sim \{x,y\})$. By Definition 2$A\cup \{x\} \succ A$ implies xPa for some $a\in \max _{P}(A)$. Also by Definition 2$A\succsim A\cup \{y\}$ implies that, for all $a\in \max _{P}(A)$, $\lnot (yPa)$. Therefore $\max _{P}(A)\subseteq \max _{P}(A\cup \{y\})$. This implies that xPa for some $a\in \max _{P}(A\cup \{y\})$, which by Definition 2 implies $A\cup \{x,y\}\succ A\cup \{y\}$.

• CR: Let $A\in \pi (X)$ and $x,y\in X{\setminus } A$. By Definition 2$A\succ A\cup \{x\}$ implies $x\otimes a$ for all $a\in \max _{P}(A)$. On the other hand, also by Definition 2, $A\cup \{y\}\sim A$ implies aPy for some $a\in A$. Therefore, given that P is transitive, $\max _{P}(A\cup \{y\})=\max _{P}(A)$, which implies $x\otimes a$ for all $a\in \max _{P}(A\cup \{y\})$, and by Definition 2 this implies $A\cup \{y\}\succ A\cup \{x,y\}$.

In order to prove the necessary part of the theorem we have to prove that, if a reflexive binary relation $\succsim$ satisfies all the axioms in Theorem 4.2, then there exists an asymmetric and transitive binary relation P defined on X with which $\succsim$ is P-instrumental consistent.

We will show that, in particular, $\succsim$ is P-consistent with the relation P defined by: $P=\{(x,y)$ such that $\{x,y\}\succ \{y\}\}$. For that purpose:

First, we have to show that P is asymmetric and transitive. Assume xPy, then, by the definition of P, $\{x,y\}\succ \{y\}$. Assume that P is not asymmetric and therefore yPx. Then $\{x,y\}\succ \{x\}$, but this is in contradiction with RI.

Now we prove the transitivity of P: Take $a,b,c\in X$ such that aPb and bPc. By the definition of P this implies $\{a,b\}\sim \{a\}$ and $\{b,c\}\sim \{b\}$. By applying IC (with a in the role of A; b in the role of x and c in the role of y), we have $\{a,c\}\sim \{a\}$, this by RI implies $\{a,c\}\succ \{c\}$, which by the definition of P implies aPc.

Now, take $A\subseteq \pi (X)$ and $x\in X{\setminus } A$:

(i) Let us assume that xPa for some $a\in \max _{P}(A)$:

Step 1. We claim that $\forall {\hat{A}}\subset \max _{P}(A)$, $\forall a^{*}\in \max _{P}(A){\setminus } {\hat{A}}$, ${\hat{A}}\succ {\hat{A}}\cup \{a^{*}\}$. Let $a^{*}\in {\hat{A}}$. Given that ${\hat{a}}, a^{*}\in \max _{P}(A)$ we know that ${\hat{a}}\otimes a^{*}$. This, by the definition of P, implies $\lnot (\{{\hat{a}},a^{*}\}\succ \{{\hat{a}}\})$ and also $\lnot (\{{\hat{a}},a^{*}\}\succ \{a^{*}\})$, which by RI implies $\lnot (\{{\hat{a}},a^{*}\}\sim \{{\hat{a}}\})$. Thus we have that $\lnot (\{{\hat{a}},a^{*}\}\succsim \{{\hat{a}}\})$, which by SC implies $\{{\hat{a}}\}\succ \{{\hat{a}},a^{*}\}$. Therefore, by CC, ${\hat{A}}\succ {\hat{A}}\cup \{a^{*}\}$.

Step 2. Let $\max _{P}(A)=\{a_{1}^{*},\ldots ,a_{n}^{*}\}$. Given that xPa for some $a\in \max _{P}(A)$ there exists $a_{i}^{*}\in \max _{P}(A)$ such that $xPa_{i}^{*}$. Assume, without loss of generality, that $a_{i}^{*}=a_{1}^{*}$. By the definition of P $\{x,a_{1}^{*}\}\succ \{a_{1}^{*}\}$.

Step 3. We claim that $\not \exists a\in A$ s.t $\{a\}\sim \{x,a\}$. Given that $xPa^{*}$ for some $a^{*}\in \max _{P}(A)$ and given that P is transitive, if there exists some $a\in A$ s.t aPx, then $aPa^{*}$ and therefore $a^{*}\notin \max _{P}(A)$, reaching a contradiction. Therefore, $\not \exists a\in A$ s.t aPx, which by the definition of P and RI implies that $\not \exists a\in A$ s.t $\{a\}\sim \{x,a\}$.

Step 4. Now we claim that $\max _{P}(A)\cup \{x\}\succ \max _{P}(A)$. If $\# \max _{P}(A)=1$, then the claim is proved. If $\max _{P}(A)=\{a^{*}_{1}, a^{*}_{2}\}$ then, given that $a^{*}_{1}\otimes a^{*}_{2}$, as we have shown in Step 1, $\{a^{*}_{1}\}\succ \{a^{*}_{1},a^{*}_{2}\}$. By Step 2 $\{a_{1}^{*},x\}\succ \{a_{1}^{*}\}$ and by Step 3 $\lnot \{a_{2}^{*}\}\sim \{a_{2}^{*},x\}$. Thus we can apply MR to obtain $\{a^{*}_{1},a_{2}^{*},x\}\succ \{a_{1}^{*},a_{2}^{*}\}$. If $\sharp max_P(A)=2$ Step 4 is proved, if not, By Step 1 $\{a_{1}^{*},a_{2}^{*}\}\succ \{a_{1}^{*},a_{2}^{*},a_{3}^{*}\}$ and by Step 3, $\lnot \{a_{3}^{*}\}\sim \{a_{3}^{*},x\}$, therefore, by MR, $\{a_{1}^{*},a_{2}^{*},a_{3}^{*},x\}\succ \{a_{1}^{*},a_{2}^{*},a_{3}^{*}\}$. By repeating the same reasoning until exhausting $\max _{P}(A)$ we reach $\max _{P}(A)\cup \{x\}\succ \max _{P}(A)$.

Step 5. We claim that, $\forall A'\subseteq A$ s.t $\max _{P} (A) \subseteq (A')$ $\forall a_{i}\in A{\setminus } \max _{P}(A)$, $A'\sim A'\cup \{a_i\}$. If $a_i\in A'$ then, by reflexivity, $A'\sim A'\cup \{a_i\}$. If $a_i\notin A'$ we know that, for all $a_{i}\in A{\setminus } \max _{P}(A)$ there exists $a^{*}_{j}\in \max _{P}(A)$ such that $a^{*}_{j}Pa_{i}$. Thus, by the definition of P and RI, $\{a_{j}^{*},a_{i}\}\sim \{a_{j}^{*}\}$. Since $\max _{P} (A) \subseteq (A')$ we know by reflexivity that $A'\sim A'\cup \{a^{*}_{j}\}$. Therefore, by IC, $A'\sim A'\cup \{a_i\}$.

Step 6. Now, let $A{\setminus } \max _{P}(A)=\{a_{1},\ldots ,a_{m}\}$. We know by Step 5 that $\max _{P}(A)\cup \{a_{1}\}\sim \max _{P}(A)$. By Step 4 we have that $\max _{P}(A)\cup \{x\}\succ \max _{P}(A)$ and by Step 3 $\lnot \{a_{1}^{*}\}\sim \{a_{1}^{*},x\}$. MR can be applied to obtain $\max _{P}(A)\cup \{a_{1},x\}\succ \max _{P}(A)\cup \{a_{1}\}$. If $\#(A{\setminus } \max _{P}(A))=1$ the proof is completed. If $\#(A{\setminus } \max _{P}(A))=2$ take $a_{2}$. We have $\max _{P}(A)\cup \{a_{1},x\}\succ \max _{P}(A)\cup \{a_{1}\}$. By Step 3 we have $\lnot \{a_{2}^{*}\}\sim \{a_{2}^{*},x\}$ and, given that $\max _{P}(A)\cup \{a_{1}\}\subset \max _{P}(A)$, we also have that $\max _{P}(A)\cup \{a_{1},a_{2}\}\sim \max _{P}(A)\cup \{a_{1}\}$. Again by MR we obtain $\max _{P}(A)\cup \{a_{1},a_{2}\}\cup \{x\}\succ \max _{P}(A)\cup \{a_{1},a_{2}\}$, completing the proof. If $\#(A{\setminus } \max _{P}(A))>2$, by repeating the same reasoning until exhausting $A{\setminus } \max _{P}(A)$ we reach $A\cup \{x\}\succ A$ as desired.

(ii) Now, assume that aPx for some $a\in A$. By the definition of P $\{a,x\}\succ \{x\}$. By RI $\{a,x\}\sim \{a\}$. By reflexivity of $\succsim$ we have $A\cup \{a\}\sim A$ and, given that $\{a,x\}\sim \{a\}$, by IC $A\cup \{x\}\sim A$.

(iii) Now, assume that assumptions (i) and (ii) do not hold, that is, $x\otimes a^{*}$ for all $a^{*}\in \max _{P}(A)$. Then, for all $a^{*}\in \max _{P}(A)$, by the definition of P we have that $\lnot (\{x,a^{*}\}\succ \{a^{*}\})$. We also have that $\lnot (\{x,a^{*}\}\succ \{x\})$, which by RI implies $\lnot (\{x,a^{*}\}\sim \{a^{*}\})$. Thus, by SC, $\{a^{*}\}\succ \{x,a^{*}\}$ for all $a^{*}\in \max _{P}(A)$. Therefore, by CC, $\max _{P}(A)\succ \max _{P}(A)\cup \{x\}$.

Now, we know that, for all $a\in A{\setminus } \max _{P}(A)$, there exists $a^{*}\in \max _{P}(A)$ such that $a^{*}Pa$ and, by the definition of P and RI, $\{a^{*},a\}\sim \{a^{*}\}$. We can apply the claim of Step 5 in part (i) of the proof to obtain $\max _{P}(A)\cup \{a\}\sim \max _{P}(A)$ and $\forall A'\in \pi (X)$ such that $\max _{P}(A)\subseteq A' \subset A$, $\forall a_{i}\in A{\setminus } A'$, $A'\cup \{a_{i}\}\sim A'$.

Thus, let $A{\setminus } \max _{P}(A)=\{a_{1},\ldots ,a_{m}\}$. We have already proved that $\max _{P}(A)\succ \max _{P}(A)\cup \{x\}$ and we also have that $\max _{P}(A)\sim \max _{P}(A)\cup \{a_{1}\}$. By CR this implies $\max _{P}(A)\cup \{a_{1}\}\succ \max _{P}(A)\cup \{a_{1}\}\cup \{x\}$. If $\#(A{\setminus } \max _{P}(A))=1$ then the proof is done. If $\#(A{\setminus } \max _{P}(A))=2$ we have that $\max _{P}(A)\cup \{a_{1}\}\succ \max _{P}(A)\cup \{a_{1}\}\cup \{x\}$ and that $\max _{P}(A)\cup \{a_{1}\}\sim \max _{P}(A)\cup \{a_{1}\}\cup \{a_{2}\}$. Again by CR we obtain $\max _{P}(A)\cup \{a_{1},a_{2}\}\succ \max _{P}(A)\cup \{a_{1},a_{2},x\}$, completing the proof. If $\#(A{\setminus } \max _{P}(A))>2$, then we repeat the same reasoning until exhausting set $A{\setminus } \max _{P}(A)$ to obtain $A\succ A\cup \{x\}$.

The proof of unicity of P is similar to the corresponding part of the proof for Theorem 4.1.

1.3 Proof of Corollary 3

It is easy to check that $\succsim$ is P-instrumental consistent (i.e. it satisfies the conditions of Definition 1). Next we prove that, if $\succsim$ is transitive and P-intrinsic consistent, then the three conditions in the corollary hold:

Let $B{\setminus } A=\{b_{1},b_{2},\ldots , b_{k}\}$.

• (i) Assume, w.l.o.g. that $b_{1}Pa$ for all $a\in A$. Given that $\succsim$ is P-intrinsic consistent, $A\cup \{b_{1}\}\succ A$. Given that $\otimes (A)=\otimes (B)$ we know that all the elements in $B{\setminus } A$ are pairwise related by means of P and that all of them are also related by means of P with the elements of A. That is, for any proper subset $B'$ of $\{b_{2},\ldots , b_{k}\}$, for any ${\hat{b}}\notin B'$, we know by P-intrinsic consistency of $\succsim$ that $A\cup \{b_{1}\}\cup B'\cup \{{\hat{b}}\}\succsim A\cup \{b_{1}\}\cup B'$. Therefore it is possible to construct a chain $B\succsim B{\setminus } \{b_{k}\} \succsim B{\setminus } \{b_{k},b_{k-1}\}\}\succsim \ldots \succsim A\cup \{b_{1}\} \succ A$, and by transitivity of $\succsim$ we obtain $B\succ A$.

• (ii) By the repeated application of P-intrinsic consistency we have $A\succ A\cup \{b_{1}\} \succ \ldots \succ B$. By transitivity $A\succ B$.

• (iii) $\otimes (A)=\otimes (B)$ implies that all the elements in $B{\setminus } A$ are related by means of P with those of A and also that all the elements in $B{\setminus } A$ are pairwise related by means of P. Given that $\max _{P}(A)=\max _{P}(B)$ and given that by Corollary 1 P is acyclic we know that there exists $b^*\in B{\setminus } A$ such that $aPb^*$ for some $a \in A$. Let, w.l.o.g. $b^*=b_{1}$. Then, by P-intrinsic consistency, $A\sim A\cup \{b_{1}\}$. The argument can be repeated as many times as elements are in $B{\setminus } A$ to obtain $A\sim A\cup \{b_{1}\}\sim \ldots \sim B$, and by transitivity of $\succsim$, $A\sim B$.

We also have to prove that the definition does not enter in contradiction with the assumption of transitivity of $\succcurlyeq$. That is, from the very conditions of Corollary 3, it is not possible to conclude that, for some $A,B,C\in \pi (X)$ that are pairwise related by set inclusion, $A\succcurlyeq B$, $B\succcurlyeq C$ and $C\succ A$.

Assume that from the conditions of Corollary 3 we state $A\succcurlyeq B$ and $B\succcurlyeq C$. Note that, if $A\succcurlyeq B$, by either conditions (i), (ii) or (iii), $\otimes (A)\le \otimes (B)$. Similarly $B\succcurlyeq C$ implies $\otimes (B)\le \otimes (C)$. Therefore $\otimes (A)\le \otimes (C)$. Thus, $C\succ A$ is only possible with $\otimes (A)= \otimes (C)$ because none of the conditions of Corollary 3 lead to $C\succ A$ and $\otimes (A)>\otimes (C)$. There are four possible situations under which $A\succcurlyeq B$, $B\succcurlyeq C$ and $\otimes (A)= \otimes (C)$.

• (1) $[B\subset A, \otimes (A)=\otimes (B)$ and $\exists a\in A{\setminus } B$ such that aPb for all $b\in B]$ and $[C\subset B, \otimes (B)=\otimes (C)$ and $\exists b\in B{\setminus } C$ such that bPc for all $c\in C]$. In this case, given that $C\subset B$ we know by hypothesis that $\exists a\in A{\setminus } C$ such that aPc for all $c\in C]$ and by the conditions of the corollary $A\succ C$.

• (2) $[B\subset A, \otimes (A)=\otimes (B)$ and $\exists a\in A{\setminus } B$ such that aPb for all $b\in B]$ and $[\max _{P}(B)=\max _{P}(C)$ and $\otimes (B)=\otimes (C)]$. We distinguish three cases:

  • $C\subset B$. Then $C\subset A$ and, by hypothesis, $\exists a\in A{\setminus } C$ such that aPc for all $c\in C]$. This, by the conditions of the corollary, implies $A\succ C$.

  • $B\subset C$ and $C\subset A$. Then $C\succ A$ can only be derived from the conditions in the Corollary if for all $a\in A{\setminus } C$ there exists $c\in C$ such that $a\otimes c$. This enters in contradiction with $\otimes (A)= \otimes (C)$.

  • $B\subset C$ and $A\subset C$. Then, for $C\succ A$ to be derived from the conditions in the Corollary it is necessary that there exists $c\in C{\setminus } A$ such that cPa for all $a\in A$. Note that any $c\in C$ such that cPa for some $a\in A$ cannot belong to $\max _{P}(C)$ because $\max _{P}(B)=\max _{P}(C)$ and by hypothesis aPb for all $a\in A{\setminus } B$. Therefore any $c^*\in C$ such that $c^*Pa$ should be such that $c^*\notin \max _{P}(C)$. This, given that P is acyclic by Corollary 1, implies that it is possible to find a chain $c_{1}Pc_{2}P\ldots Pc^{*}$ such that $c_{1}\in \max _{P}(C)$. We know by hypothesis that $aPc'$ for all $c'\in \max _{P}(C)$. Therefore we have $aPc_{1}P\ldots P c^*Pa$, which is in contradiction with the acyclicity of P.

• (3) $[\max _{P}(A)=\max _{P}(B)$ and $\otimes (A)=\otimes (B)]$ and $[C\subset B, \otimes (B)=\otimes (C)$ and $\exists b\in B{\setminus } C$ such that bPc for all $c\in C]$.

If $C\subset A$, $C\succ A$ can only be derived from the conditions in the Corollary if, for all $a\in A{\setminus } C$, there exists $c\in C$ such that $a\otimes c$. This enters in contradiction with $\otimes (A)= \otimes (C)$.

If $A\subset C$, we reach a situation that is incompatible with the hypothesis. Note that, given that $\max _{P}(A)=\max _{P}(B)$, $A\subset C$ implies $\max _{P}(B)\subset C$. Therefore it is not possible that, for some $b\in B{\setminus } C$, bPc for all $c\in C]$. 4) $\max _{P}(A)=\max _{P}(B)=\max _{P}(C)$ and $\otimes (A)=\otimes (B)=\otimes (C)$. In this case $A\sim C$.

1.4 Proof of Corollary 4

Firstly, notice that if $\succsim$ is P-instrumental consistent then, by Theorem 4.2, P is transititive. Secondly, it is easy to check that $\succsim$ is P-instrumental consistent (i.e. it satisfies the conditions of Definition 2). Next we prove that, if $\succsim$ is transitive and P-instrumental consistent, then the three conditions in the corollary hold:

Let $B{\setminus } A=\{b_{1},b_{2},\ldots , b_{k}\}$.

• (i) Assume, w.l.o.g. that $\max _P(B{\setminus } A)=\{b_1\}$. We know that there exists $a\in \max _P(A)$ s.t $b_1Pa$, which by Definition 2 implies $A\cup \{b_1\} \succ A$. Now, take $b_2$. Given that $b_{2}\notin \max _P(A)$ and P is transitive we know that $b_1Pb_2$, which again by the same definition implies $A\cup \{b_1,b_2\}\sim A\cup \{b_1\}$. Proceeding repeteadly we obtain $A\cup \{b_{1},b_{2},\ldots , b_{k}\} \sim \cdots \sim A\cup \{b_1,b_2\}\sim A\cup \{b_1\}\succ A$, which by the transitivity of $\succsim$ implies $B\succ A$.

• (ii) Take any $b\in \max _P(B{\setminus } A)$. Since b is incomparable with all the elements in $\max _P(A)$, we have by Definition 2 that $\max _P(A)\succ \max _P(A)\cup \{b\}$. By definition, all the alternatives contained in $\max _P(B{\setminus } A)$ are pairwise incomparable, and by hypothesis all of them are incomparable with the elements in $\max _P(A)$. Therefore we can apply repeatedly the definition of P-instrumental consistency to obtain $\max _P(A)\succ \max _P(A)\cup \{b\} \succ \cdots \succ \max _P(A)\cup \max _P(B)$. Also, by applying repeteadly the definition of P-instrumental consistency we know that $\max _P(A)\sim A$ and $\max _P(B)\sim B$. Thus, by transitivity of $\succsim$, $A\succ B$.

• (iii) Assume w.l.o.g. that $A\subset B$. Then, $\max _P(A)=\max _P(B)$ implies that $\forall b\in B{\setminus } A$ there exists $a\in A$ s.t aPb. Let $B{\setminus } A=\{b_1,\ldots ,b_k\}$ and take $b_1$. By P-instrumental consistency we have that $A\sim A\cup \{b_1\}\sim \ldots \sim B$. By transitivity of $\succsim$ we obtain $A\sim B$.

We also have to prove that the definition does not enter in contradiction with the assumption of transitivity of $\succcurlyeq$.

Assume that from the conditions of Corollary 4 we state $A\succcurlyeq B$ and $B\succcurlyeq C$.

$A\succsim B$ implies either (1) $B\subset A$ and $\max _P(A{\setminus } B)= \{a^*\}$ with $a^*Pb$ for some $b\in max_P(B)$; (2) $A\subset B$ and $a\otimes b$ $\forall a\in \max _P(A)$ and $\forall b\in \max _P(B{\setminus } A)$, or (3) $\max _P(A)= \max _P(B)$.

$B\succsim C$ implies either (4) $C\subset B$ and $\max _P(B{\setminus } C)= \{b^*\}$ with $b^*Pc$ for some $c\in max_P(C)$; (5) $B\subset C$ and $b\otimes c$ $\forall b\in \max _P(B)$ and $\forall c\in \max _P(C{\setminus } B)$, or (6) $\max _P(B)= \max _P(C)$.

Suppose (1) and (4). Then we have that $a^*Pb$ for some $b\in \max _P(B)$. If $b=b^*$, then, by transitivity of P, $a^*Pc$ for some $c\in \max _P(C)$ and the conditions of the corollary imply $A\succ C$. If not, then either $b^*\in \max _P(B)$ or $b^*\notin \max _P(B)$. If $b^*\in \max _P(B)$, then $b^*,b\in \max _P(B)$. By hypothesis this is only possible if $b\in C$. In this case $a^*Pc$ for some $c\in \max _P(C{\setminus } A)$ $(c=b)$ and the conditions of the corollary imply $A\succ C$. If $b^*\notin \max _P(B)$, then, by transitivity of P, $bPb^*$, and again by transitivity of P, $a^*PbPb^*Pc$ implies $a^*Pc$. Then, again by the conditions of the corollary, $A\succ C$.

Suppose (1) and (5). Notice that (5) implies that $\max _P(C)= \max _P(B)\cup \max _P(C)$. The reason is that $b\otimes c$ $\forall b\in \max _P(B)$ and $\forall c\in \max _P(C{\setminus } B)$ implies that if $c\in \max _P(C{\setminus } B)$ then $c\in \max _P(C)$. Otherwise $\exists b\in B$ such that bPc, which by hypothesis is only possible if $b\notin \max _P(B)$. Therefore, by transitivity of P, $\exists b^*\in B$ such that $b^*PbPc$, which implies, by transitivity of P, $b^*Pc$, which is in contradiction with the hypothesis (5). Thus, $\max _P(C)= \max _P(B)\cup \max _P(C)$. Therefore, by hypothesis, $a^*Pb$ for some $b\in \max _P(C)$, which by the conditions of the corollary implies $A\succ C$.

Suppose (1) and (6). Since $\max _P(B)= \max _P(C)$ we have $a^*Pb$ for some $b\in \max _P(C)$ which by the conditions of the corollary implies $A\succ C$.

Suppose (2) and (4). We distinguish two possibilities. (I) $C\subset A$ and (II) $A\subset C$

(I) $C\subset A$ implies $C\subset A\subset B$. In this case, for $C\succ A$ to be derived from the conditions in the corollary, it is necessary that $[ a\otimes c$ $\forall c\in \max _P(A{\setminus } C)$ and $\forall a\in \max _P(A)]$. By hypothesis $\exists b^*\in \max _P(B{\setminus } C)$ such that $b^*Pc$ for some $c\in \max _P(C)$. Notice that $(A{\setminus } C) \subset (B{\setminus } C)$. If $b^*\in \max _P(A{\setminus } C)$ then $C\succ A$ cannot be derived from the conditions in the corollary. If $b^*\notin \max _P(A{\setminus } C)$, then $a^*Pb^*$ for some $a^*\in \max _P(A{\setminus } C)$ and we have $a^*Pb^*Pc$. Therefore, by transitivity, $\exists a^*\in \max _P(A{\setminus } C)$ such that $a^*Pc$ and, again, $C\succ A$ cannot be derived from the conditions in the corollary.

(II) $A\subset C$ implies $A\subset C\subset B$. In this case, for $C\succ A$ to be derived from the conditions in the corollary, it is necessary that $\max _P(C{\setminus } A)=\{c^*\}$ and $c^*Pa$ for some $a\in max_P(A)$. By hypothesis we know that $\exists b^*\in \max _P(B{\setminus } C)$ such that $b^*Pc$ for some $c\in \max _P(C)$. Notice that if $c\in A$, given that $A\subset C$, $c\in \max _P(A)$ and then we reach a contradiction with the hypothesis. Therefore $c\in C{\setminus } A$. $c\in \max _P(C)$ implies $c\in \max _P(C{\setminus } A)$, and given that $\#\max _P(C{\setminus } A)=1$, $c=c^*$. Therefore we have $b^*Pc^*Pa$, which by transitivity implies $b^*Pa$, reaching again a contradiction. Therefore $C\succ A$ cannot be derived from the conditions in the corollary.

Suppose (2) and (5). By (2) we know that $\max _P(B)= \max _P(A)\cup \max _P(B)$. Therefore, $b\otimes c$ $\forall b\in \max _P(B)$ and $\forall c\in \max _P(C{\setminus } B)$ implies $a\otimes c$ $\forall a\in \max _P(A)$ and $\forall c\in \max _P(C{\setminus } A)$, which by the conditions of the corollary implies $A\succ C$.

Suppose (2) and (6). The conditions of the corollary lead to $C\succ A$ under two situations: (I) $A\subset C$ and $\max _P(C{\setminus } A)Pa$ for some $a\in \max _P(A)$ and (II) $C\subset A$ and $[ a\otimes c$ $\forall a\in \max _P(A{\setminus } C)$ $\forall c\in \max _P(C)]$. We prove that both cases lead to a contradiction.

(I) Let $c^*=\max _P(C{\setminus } A)$. Notice that if $A\subset C$ and $c^*Pa$ for some $a\in \max _P(A)$ then $c^*\in \max (C)$. Otherwise either $\exists c'\in C{\setminus } A$ such that $c'Pc^*$, in which case $c^*\notin \max _P(C{\setminus } A)$ or $\exists c'\in A$ such that $c'Pc^*$, in which case $c'Pc^*Pa$ and, by transitivity of P, $c'Pa$, which is in contradiction with $a\in \max _P(A)$. Given that $\max _P(B)=\max _P(C)$ we have that $\exists c^*\in \max _P(B)$ such that $c^*Pa$, which is only possible if $c^*\in B{\setminus } A$, but this is in contradiction with the hypothesis (2) that $a\otimes b$ $\forall a\in \max _P(A)$ and $\forall b\in \max _P(B{\setminus } A)$.

(II) In this case $C\subset A\subset B$. Given that $\max _P(B)=\max _P(C)$ we have that $\forall b\in B{\setminus } C$ there exists c such that cPb. Therefore, given that $C\subset A$, $\forall b\in B{\setminus } A$ there exists a such that aPb. Notice that $a\notin \max _P(A)$, otherwise we reach a contradiction with the hypothesis (2) $[ a\otimes b$ $\forall a\in \max _P(B{\setminus } A)$ $\forall a\in \max _P(A)]$. This implies that $\exists a^*\in \max _P(A)$ such that $a^*Pa$. By transitivity of P, $a^*PaPb$ impies $a^*Pb$, in which case we reach again a contradiction with the hypothesis $[ a\otimes b$ $\forall a\in \max _P(B{\setminus } A)$ $\forall a\in \max _P(A)]$.

Suppose (3) and (4). Let $b^*=\max _P(B{\setminus } C)$. If $b^*\in \max _P(B)$ then, given that $\max _P(A)= \max _P(B)$, $\exists b^*\in \max _P(A{\setminus } C)$ such that $b^*Pc^*$ for some $c^*\in \max _P(C)$ and, by the conditions of the corollary, $A\succ C$. If $b^*\notin \max _P(B)$, then $\exists {\hat{c}}\in C$ such that ${\hat{c}}Pb^*$. By transitivity of P ${\hat{c}}Pb^*Pc^*$ implies that $c^*\notin \max _P(C)$, reaching a contradiction.

Suppose (3) and (5). Then $C\succ A$ can only be derived from the conditions in the corollary if either $[A\subset C$, $\max _P(C{\setminus } A)=c^*$ and $c^*Pa$ for some $a\in \max _P(A)]$, or $[C\subset A$ and $a\otimes c$ $\forall a\in \max _P(A{\setminus } C)$ and $\forall c\in \max _P(C)]$.

In the first case, given that $\max _P(A)=\max _P(B)$, we know that $c^*Pb$ for some $b\in \max _P(B)$. This, by hypothesis, is only possible if $c^*\notin \max _P(C{\setminus } B)$. Therefore $\exists {\hat{c}}\in C{\setminus } B$ such that ${\hat{c}}Pc^*$. By transitivity this implies ${\hat{c}}Pa$ for some ${\hat{c}}\in C{\setminus } B$, reaching a contradiction with the hypothesis.

In the second case, we have that $B\subset C\subset A$. Given that $\max _P(A)=\max _P(B)$, we know that $\forall a\in A{\setminus } B$ there exists $b\in \max _P(B)$ such that bPa. Therefore $C\succ A$ in this second case is only possible if $b\notin \max _P(C)$, which implies that $\exists c\in C{\setminus } B$ such that cPb. This is again in contradiction with the hypothesis.

Suppose (3) and (6). Then $\max _P(A)= \max _P(B) = \max _P(C)$ which by the conditions in the corollary implies $A\sim B \sim C$.

1.5 Proof of Theorem 5.1

First, we prove that if $\succsim$ is consistent with a preference with a structured incompleteness $(P,\otimes _{1},\otimes _{2})$ then $\succsim$ satisfies WRI, SC, IA, MC and CE.

• WRI: Take any $x,y\in X$ such that $\{x,y\}\sim \{y\}$. This, by Definition 3, implies yPx, which again by the same definition implies $\{x,y\}\succ \{x\}$. Now, take any $x,y\in X$ such that $\{x,y\}\succ \{y\}$. By Definition 3 this implies either $y\otimes _{1}x$ or xPy. By the same definition, if $y\otimes _{1}x$ then $\{x,y\}\succ \{x\}$ and if xPy then $\{x,y\}\sim \{x\}$.

• SC: Take any $x,y\in X$ such that $\lnot (\{x,y\}\succsim \{x\})$. $\lnot (\{x,y\}\succ \{x\})$ implies by Definition 3$\lnot (x\otimes _{1}y)$ and $\lnot (yPx)$. Moreover, $\lnot (\{x,y\}\sim \{x\})$ implies by Definition 3 either $x\otimes _{2}y$ or $\lnot (xPy)$. In sum, $\lnot (\{x,y\}\succsim \{x\})$ implies necessarily $x\otimes _{2}y$, which by Definition 3 implies $\{x\}\succ \{x,y\}$.

• IA: Let $A\in \pi (X)$ and let $x\in X{\setminus } A$. Assume $\{a,x\}\succsim \{a\}$ $\forall a\in A$. Thus, $\not \exists a\in A$ such that $\{a\}\succ \{a,x\}$, which by Definition 3 implies (i) $\not \exists a\in A$ s.t $a\otimes _{2} x$. Assume that there exists at least one element $a^{*}\in A$ such that $\{a^{*},x\}\sim \{a^{*}\}$. This by Definition 3 implies (ii) $a^{*}Px$. Given (i) and (ii), by Definition 3, $A\cup \{x\}\sim A$

• MC: Let $A\in \pi (X)$ and let $x\in X{\setminus } A$. Assume that $\{x,a\}\succ \{a\}$ $\forall a\in A$. This implies, by Definition 3, that, for all $a\in A$, either $x\otimes _{1}a$ or xPa. Therefore, by Definition 3, $A\cup \{x\}\succ A$.

• CE: Let $A,B\in \pi (X)$ such that $A\subseteq B$ and let $x\in X{\setminus } B$. Assume that $A\succ A\cup \{x\}$. by Definition 3 this implies that $\exists a\in A$ such that $x\otimes _2 a$. Given that $A\subseteq B$, $\exists b\in B$,($b=a$) such that $x\otimes _2 b$, which by the Definition 3 implies $B\succ B\cup \{x\}$.

In order to prove the necessary part of the theorem, assume that a binary relation $\succsim$ satisfies all the axioms of Theorem 5.1. We have to prove that in this case there exists a preference with a structured incompleteness $(P,\otimes _{1},\otimes _{2})$ defined on X with which $\succsim$ is $(P,\otimes _{1},\otimes _{2})$-intrinsic consistent.

We will show that, in particular, $\succsim$ is consistent with the preference with a structured incompleteness $(P,\otimes _{1},\otimes _{2})$ where P, $\otimes _{1}$ and $\otimes _{2}$ are defined by: $P=\{(x,y)$ such that $\{x,y\}\sim \{x\}\}$; $\otimes _{1}= \{(x,y)$ such that $\{x,y\}\succ \{x\}$ and $\{x,y\}\succ \{y\}\}$ and $\otimes _{2}= \{(x,y)$ such that $\{x\}\succ \{x,y\}\}$.

First, we prove that $(P,\otimes _{1},\otimes _{2})$ is a preference with a structured incompleteness:

• P is asymmetric by WRI.

• $\otimes _{1}$ is symmetric by definition.

• $\otimes _{2}$ is symmetric: Assume that $\{x\}\succ \{x,y\}$ but $\lnot (\{y\}\succ \{x,y\})$. By SC $\lnot (\{y\}\succ \{x,y\})$ implies $\{x,y\}\succsim \{y\}$. By WRI this implies $\{x,y\}\succsim \{x\}$, reaching a contradiction.

• $\otimes _{1}\cap \otimes _{2}=\emptyset$ holds by definition.

• Now we prove that $X^{2}{\setminus } P= \otimes _{1}\cup \otimes _{2}$: $X^{2}{\setminus } P=\{(x,y)\in X$ such that $\lnot (xPy)$ and $\lnot (yPx)\}$. $\lnot (xPy)$ and $\lnot (yPx)$ implies by definition $\lnot (\{x,y\}\sim \{x\})$ and $\lnot (\{x,y\}\sim \{y\})$. By SC $\lnot (\{x,y\}\sim \{x\})$ implies either $\{x,y\}\succ \{x\}$ or $\{x\}\succ \{x,y\}$. By SC, if $\{x,y\}\succ \{x\}$ there are three possibilities: (i) $\{x,y\}\succ \{y\}$, in which case $x\otimes _{1}y$ by definition; (ii) $\{x,y\}\sim \{y\}$, which is a contradiction with the assumption that $\lnot (yPx)$, and (iii) $\{y\}\succ \{x,y\}$, which is in contradiction with WRI. If $\{x\}\succ \{x,y\}$ we have again three possibilities: (i) $\{x,y\}\succ \{y\}$, which is in contradiction with WRI; (ii) $\{x,y\}\sim \{y\}$, which also leads to a contradiction with the assumption that $\lnot (xPy)$ and (iii) $\{y\}\succ \{x,y\}$, which by definition implies $x\otimes _{2}y$.

Next we show that, if $\succsim$ satisfies all the axioms in Theorem 5.1, then it is $(P,\otimes _{1},\otimes _{2})$-consistent with $(P,\otimes _{1},\otimes _{2})$ defined as above:

• (i) Assume that for all $a\in A$ either xPa or $x\otimes _1 a$. Then, either by the definition of $\otimes _1$ or by the definition of P and WRI we know that $\{x,a\}\succ \{a\}$, therefore, by MC we have $A\cup \{x\}\succ A$.

• (ii) Now, assume that there exists a in A such that $x\otimes _2 a$. Then, by the definition of $\otimes _2$, $\{a\}\succ \{a,x\}$. Since $\{a\}\subseteq A$, by CE, $A\succ A\cup \{x\}$.

• (iii) Finally, assume that there exists $a^{*}\in A$ such that $a^{*}Px$ and $\not \exists a\in A$ s.t $a\otimes _2x$. By the definition of P we have that $\{a^{*},x\}\sim \{a^{*}\}$. Since $\not \exists a\in A$ s.t $a\otimes _2x$, we have that, for all $a\in A$, either aPx, which by P definition implies $\{a,x\}\sim \{a\}$, or xPa, or $x\otimes _1 a$. In these two last cases, as we have shown at (i), we have $\{a,x\}\succ \{a\}$. Therefore $\{a,x\}\succsim \{a\}$ $\forall a\in A$ and $\{a^{*},x\}\sim \{a^{*}\}$, which by IA implies $A\cup \{x\}\sim A$. $\square$

1.6 Proof of Theorem 5.2

Assume that a reflexive binary relation $\succsim$ satisfies all the axioms of Theorem 5.2. Then there exists a transitive preference with a structured incompleteness $(P,\otimes _{1},\otimes _{2})$ defined on X with which $\succsim$ is $(P,\otimes _{1},\otimes _{2})$-instrumental consistent. In particular, consider P, $\otimes _{1}$ and $\otimes _{2}$ defined as in the proof of Theorem 5.1. That is, $P=\{(x,y)$ such that $\{x,y\}\sim \{x\}\}$; $\otimes _{1}= \{(x,y)$ such that $\{x,y\}\succ \{x\}$ and $\{x,y\}\succ \{y\}\}$ and $\otimes _{2}= \{(x,y)$ such that $\{x\}\succ \{x,y\}\}$ . In the proof of Theorem 5.1 it is proved, by using axioms WRI and SC, that (P, $\otimes _{1}$, $\otimes _{2}$) is a preference with a structured incompleteness. Given that WRI and SC are also characterization axioms in Theorem 5.2 that part of the proof is identical. It only remains then to prove that P is, in addition, transitive: Take $a,b,c\in X$ such that aPb and bPc. By the definition of P this implies $\{a,b\}\sim \{a\}$ and $\{b,c\}\sim \{b\}$. By applying IC (with a in the role of A; b in the role of x and c in the role of y), we have $\{a,c\}\sim \{a\}$, which by the definition of P implies aPc.

Next we have to prove that, if a reflexive binary relation $\succsim$ satisfies all the axioms in Theorem 5.2, then it is $(P,\otimes _{1},\otimes _{2})$-instrumental consistent:

(i) Assume that $\exists a\in (A)$ s.t aPx. Then by the definition of P we know that $\{a,x\} \sim \{a\}$. By reflexivity of $\succsim$ we have $A\cup \{a\}\sim A$. Therefore, by IC, $A\cup \{x\}\sim A$.

(ii) Assume now that $\exists a_i\in \max _P (A)$ s.t $xPa_i$ or $x\otimes _1 a_i$; $\not \exists a\in \max _{P}(A)$ s.t aPx, and $\forall a^{*}\in \max _P(A)$ s.t $a^{*}\otimes _2 x$ $\not \exists a\in \max _P(A)$ s.t $a\otimes _1 a^{*}$.

Step 1. Let ${\hat{A}}=\{a\in \max _P(A): xPa$ or $x\otimes _1 a\}$. For all $a\in {\hat{A}}$, if xPa, by the definition of P, $\{x,a\}\sim \{x\}$, and by WRI $\{x,a\}\succ \{a\}$. If $x\otimes _1 a$, by the definition of $\otimes _1$, $\{x,a\}\succ \{a\}$. Thus, for all $a\in {\hat{A}}$, $\{x,a\}\succ \{a\}$. Then, by MC, we have that ${\hat{A}}\cup \{x\} \succ {\hat{A}}$.

Step 2. In this step we prove that $\{x\}\succ \{x,{\overline{a}}\}$ $\forall {\overline{a}}\in \max _P(A){\setminus } {{\hat{A}}}$. Given that ${\overline{a}}\in A{\setminus } {{\hat{A}}}$ we know that $\lnot (xP {\overline{a}})$ and $\lnot (x\otimes _1 {\overline{a}})$. By assumption $\not \exists a\in \max _{P}(A)$ s.t aPx, therefore $x\otimes _2 {\overline{a}}$ $\forall {\overline{a}}\in \max _{P}(A){\setminus } {{\hat{A}}}$, which by the definition of $\otimes _2$ implies $\{x\}\succ \{x,{\overline{a}}\}$ $\forall {\overline{a}}\in \max _{P}A{\setminus } {{\hat{A}}}$.

Step 3. Let $\max _P(A){\setminus } {{\hat{A}}}= \{{\overline{a}}_1,\ldots ,{\overline{a}}_n\}$. Take any ${\overline{a}}_j \in \max _P(A){\setminus } {{\hat{A}}}$. Given that ${\overline{a}}_j \in \max _P(A)$ we know that $\forall {\hat{a}} \in {\hat{A}}$ neither ${\overline{a}}_jP{\hat{a}}$ nor ${\hat{a}}P{\overline{a}}_j$. At Step 2 we have proved that $x\otimes _2{\overline{a}}_j$ $\forall {\overline{a}}_j\in \max _P(A){\setminus } {{\hat{A}}}$. Recall that, by assumption, $\forall a^{*}\in \max _P(A)$ s.t $a^{*}\otimes _2 x$ $\not \exists a\in \max _P(A)$ s.t $a\otimes _1 a^{*}$. Thus, for all $a \in \max _P(A)$, $\lnot ({\overline{a}}_j\otimes _1 a)$, which implies ${\overline{a}}_j\otimes _2 a$ $\forall a\in \max _P(A){\setminus } \{{\overline{a}}_j\}$. Take ${\overline{a}}_1$. By CC we know that ${\hat{A}}\succ {\hat{A}}\cup \{{\overline{a}}_1\}$; by Step 1 ${\hat{A}}\cup \{x\} \succ {\hat{A}}$, and by Step 3 $\{x\}\succ \{x,{\overline{a}}_1\}$. Therefore, we can apply MR to obtain ${\hat{A}}\cup \{x,{\overline{a}}_1\}\succ {\hat{A}}\cup \{{\overline{a}}_1\}$. If $\max _P(A){\setminus } {\hat{A}}=\{{\overline{a}}_1\}$ we have $\max _P(A)\cup \{x\}\succ \max _P(A)$. If not, take $\{{\overline{a}}_2\}$. We have ${\hat{A}}\cup \{x,{\overline{a}}_1\}\succ {\hat{A}}\cup \{{\overline{a}}_1\}$; by CC ${\hat{A}}\cup \{{\overline{a}}_1\} \succ {\hat{A}}\cup \{{\overline{a}}_1,{\overline{a}}_2\}$, and by Step 2 $\{x\}\succ \{x,{\overline{a}}_2\}$. Thus, we can apply MR to obtain ${\hat{A}}\cup \{x,{\overline{a}}_1,{\overline{a}}_2\} \succ {\hat{A}}\cup \{{\overline{a}}_1,{\overline{a}}_2\}$. Proceeding similarly until exhausting all elements of $\max _P(A){\setminus } {\hat{A}}$ we reach $\max _P(A)\cup \{x\}\succ \max _P(A)$.

Step 4. Let $A{\setminus } \max _P(A)=\{a_1,...,a_n\}$. Take $a_1$. We know that there exists $a^{*}\in \max _P(A)$ such that $a^{*}Pa_1$. Thus, by the definition of P, $\{a^{*},a_1\}\sim \{a^{*}\}$. By reflexivity of $\succsim$, $\max _{P}(A)\cup \{a^{*}\}\sim \max _{P}(A)$. Thus, we can apply IC to obtain $\max _P(A)\sim \max _P(A)\cup \{a_1\}$. By hypothesis $\not \exists a\in \max _{P}(A)$ such that aPx, thus, by transitivity of P, $\not \exists a\in A$ such that aPx, and by the definition of P, $\lnot (\{a_1\}\sim \{x,a_1\})$. By Step 3 $\max _P(A)\cup \{x\}\succ \max _P(A)$. Thus, we can apply MR to obtain $\max _P(A)\cup \{x,a_1\}\succ \max _P(A)\cup \{a_1\}$. If $A{\setminus } \max _P(A)=\{a_1\}$ Step 4 is proved. If not take $a_{2}\in A{\setminus } \max _P(A)=\{a_1,...,a_n\}$. Again, given that $a_{2}\notin \max _P(A)$ there exists $a^{*'} \in \max _P(A)\cup \{a_1\}$ such that $a^{*'}Pa_{2}$ and by applying similarly the definition of P, reflexivity of $\succsim$ and IC we obtain $\max _P(A)\cup \{a_1\}\sim \max _P(A)\cup \{a_1,a_2\}$. Again, by hypothesis, $\not \exists a\in \max _{P}(A)$ such that aPx, and by transitivity of P and the definition of P we obtain $\lnot (\{a_2\}\sim \{x,a_2\})$. Then, given that $\max _P(A)\cup \{x,a_1\}\succ \max _P(A)\cup \{a_1\}$, by MR we have $\max _P(A)\cup \{x,a_1,a_2\}\succ \max _P(A)\cup \{a_1,a_2\}$. Proceeding similarly until exhausting all elements of A we finally obtain $A\cup \{x\}\succ A$.

(iii) Now, assume that $x\otimes _2 a^{*}$ for all $a^{*}\in \max _{P}(A)$. By CC $\max _{P}(A)\succ \max _{P}(A)\cup \{x\}$. We can apply the definition of P, reflexivity of $\succsim$ and IC like in Step 4 above to prove that for all $a\in A{\setminus } \max _{P}(A)$, $\max _P(A)\sim \max _P(A)\cup \{a\}$. Thus, let $A{\setminus } \max _{P}(A)=\{a_{1},\ldots ,a_{k}\}$. We have that $\max _{P}(A)\succ \max _{P}(A)\cup \{x\}$ and $\max _{P}(A)\sim \max _{P}(A)\cup \{a_{1}\}$. By CR this implies $\max _{P}(A)\cup \{a_{1}\}\succ \max _{P}(A)\cup \{a_{1}\}\cup \{x\}$. If $A{\setminus } \max _{P}(A)=\{a_1\}$ then the proof is done. If $A{\setminus } \max _{P}(A)=\{a_1,a_2\}$ we have that $\max _{P}(A)\cup \{a_{1}\}\succ \max _{P}(A)\cup \{a_{1}\}\cup \{x\}$ and that $\max _{P}(A)\cup \{a_{1}\}\sim \max _{P}(A)\cup \{a_{1}\}\cup \{a_{2}\}$. Again by CR we obtain $\max _{P}(A)\cup \{a_{1},a_{2}\}\succ \max _{P}(A)\cup \{a_{1},a_{2},x\}$, completing the proof. If $\#(A{\setminus } \max _{P}(A))>2$, then we repeat the same reasoning until exhausting the set $A{\setminus } \max _{P}(A)$ to obtain $A\succ A\cup \{x\}$. $\square$

1.7 Proof of Theorem 5.3

• WRI: Take any $x,y\in X$ such that $\{x,y\}\sim \{y\}$. This, by Definition  4, implies yPx, which again by the same definition implies $\{x,y\}\succ \{x\}$. Now, take any $x,y\in X$ such that $\{x,y\}\succ \{y\}$. By Definition 4 this implies either xPy or $x\otimes _{1} y$. By Definition 4, if $y\otimes _{1}x$, then $\{x,y\}\succ \{x\}$ and if xPy, then $\{x,y\}\sim \{x\}$.

• SC: Take any $x,y\in X$ such that $\lnot (\{x,y\}\succsim \{x\})$. $\lnot (\{x,y\}\succ \{x\})$ implies by Definition 4$\lnot (x\otimes _{1}y)$ and $\lnot (yPx)$. Moreover, $\lnot (\{x,y\}\sim \{x\})$ implies by Definition 4$\lnot (xPy)$. Therefore, by the properties assumed on ($P,\otimes _{1},\otimes _{2}$), $\lnot (\{x,y\}\succsim \{x\})$ implies necessarily $x\otimes _{2}y$, which by Definition 4 implies $\{x\}\succ \{x,y\}$.

• CC: Let $\{a\}\succ \{a,x\}$ $\forall a\in A$. This, by Definition  4, implies $x\otimes _2 a$ $\forall a\in A$, therefore $x\otimes _2 a$ $\forall a\in max_P(A)$, which by Definition  4 implies $A\succ A\cup \{x\}$.

• IC: Let $x,y\in X$ and $A\in \pi (X)$ such that $x\notin A$, $\{x,y\}\sim \{x\}$ and $A\cup \{x\}\sim A$. By Definition 4$\{x,y\}\sim \{x\}$ implies xPy. We distinguish two possibilities: either $y\in A$ or $y\notin A$. If $y\in A$ then $A\cup \{y\} \sim A$ holds by reflexivity of $\succsim$. If $y\notin A$ then, by Definition 4, $\exists a\in A$ such that aPx. By the transitivity of P, aPy. Thus, by Definition 4, $A\cup \{y\} \sim A$.

• MR: Let $A\in \pi (X)$ and let $x,y\in X{\setminus } A$. Assume that $A\cup \{x\}\succ A$; $A\succsim A\cup \{y\}$ and $\lnot (\{y\}\sim \{x,y\})$.

  $A\cup \{x\}\succ A$ implies by Definition 4 that $\exists a_{i}\in \max _P(A)$ such that either $xPa_{i}$ or $a_{i}\otimes _{1} x$; $\not \exists a\in \max _{P}(A)$ s.t aPx, and $\forall a^{*}\in \max _P(A))$ s.t $x\otimes _2 a^{*}$ $\not \exists a\in \max _P(A)$ s.t $a\otimes _1 a^{*}$.

  Assume first that $A\sim A\cup \{y\}$. This implies, by Definition 4, that $\exists a\in A$ s.t aPy. Thus, given the transitivity of P, $\max _P(A)= \max _P(A\cup \{y\})$. Therefore: (i) $\exists a\in \max _P(A\cup \{y\})$ such that $xPa_{i}$ or $a_{i}\otimes _{1} x$ and (ii) $\forall a^{*}\in \max _P(A\cup \{y\})$ s.t $x\otimes _2 a^{*}$ $\not \exists a\in \max _P(A\cup \{y\})$ s.t $a\otimes _1 a^{*}$ and (iii) $\not \exists a\in A\cup \{y\}$ s.t aPx. That is, according to Definition 4 all the conditions ((i), (ii) and (iii)) for $A\cup \{x,y\} \succ A\cup \{y\}$ hold.

  Now, assume that $A\succ A\cup \{y\}$. By Definition  4 this implies $y\otimes _2 a$ $\forall a\in \max _P(A)$. Therefore $\max _P(A\cup \{y\})=\max _P(A)\cup \{y\}$, which implies (i) $\exists a\in \max _P(A\cup \{y\})$ such that $xPa_{i}$ or $a_{i}\otimes _{1} x$. $\lnot (\{y\}\sim \{x,y\})$ implies $\lnot (yPx)$, therefore, given that $\not \exists a\in \max _{P}(A)$ s.t aPx, we have that (ii) $\not \exists a\in \max (A\cup \{y\})$ s.t aPx. Given that $\lnot (yPx)$, we distinguish three possibilities: If it is the case that $x\otimes _2 y$, since $y\otimes _2 a$ $\forall a\in \max _P(A)$, we know that $\not \exists a\in \max _P (A)$ s.t $a\otimes _1 y$. We also know that $\forall a^{*}\in \max _P(A)$ s.t $x\otimes _2 a^{*}$ $\not \exists a\in \max _P (A)$ s.t $a\otimes _1 a^{*}$. Therefore, (iii) $\forall a^{*}\in \max _P(A\cup \{y\})$ s.t $x\otimes _2 a^{*}$ $\not \exists a\in \max _P(A\cup \{y\})$ s.t $a\otimes _1 a^{*}$. Thus, the three conditions for $A\cup \{x,y\} \succ A\cup \{y\}$ hold. If it is the case that either $x\otimes _{1}y$ or xPy, we already know that $\forall a^{*}\in \max _P(A)$ s.t $x\otimes _2 a^{*}$ $\not \exists a\in \max _P (A)$ s.t $a\otimes _1 a^{*}$. Given that $\lnot (x\otimes _{2}y)$, we have that (iii) $\forall a^{*}\in \max _P(A\cup \{y\})$ s.t $x\otimes _2 a^{*}$ $\not \exists a\in \max _P (A\cup \{y\}))$ s.t $a\otimes _1 a^{*}$. Thus, again, the three conditions for $A\cup \{x,y\} \succ A\cup \{y\}$ hold.

• MC: Assume that $\forall A\in \pi (X)$, $\forall x\in X{\setminus } A$, $\{x,a\}\succ \{a\}$ $\forall a\in A$. Then, by Definition  4, either xPa or $x\otimes _{1}a$ $\forall a\in A$. Therefore, by Definition  4, $A\cup \{x\}\succ A$.

• CR: Let $A\in \pi (X)$ and $x,y\in X{\setminus } A$. By Definition 4, $A\succ A\cup \{x\}$ implies $x\otimes _2 a$ for all $a\in \max _{P}(A)$. On the other hand, also by Definition 4, $A\cup \{y\}\sim A$ implies aPy for some $a\in A$. Therefore, given the transitivity of P, $\max _{P}(A\cup \{y\})=\max _{P}(A)$, which implies that $x\otimes _2 a$ for all $a\in \max _{P}(A\cup \{y\})$, and by Definition 4 this implies $A\cup \{y\}\succ A\cup \{x,y\}$.