In this appendix, we prove Lemmas 1–5.

### *Proof of Lemma 2*

Take a choice function \(c\) satisfying A1-A4 and alternatives \(x\not =y\). Suppose towards a contradiction that \(xP_{0}y\) and \(yP_{0}x\). There must then exist sets \(S,T\in \Sigma _{*}\) such that \(x,y\in S\cap T\), \(x\in R(S)\), \(y\in R(T)\), \(x\not \in R(T)\) and \(y\not \in R(S)\). Without loss of generality, let \(c\left( xy\right) =y\). Note that \(x\not =h(T)\) as otherwise B2 would imply \(x=c(xy)\). We will analyze two exhaustive cases identified by the role of \(y\) in \(S\).

**Case 1** Suppose to begin with that \(y=h(S)\). Let \(R(S)=x\alpha \), \( R(T)=y\beta \) and \(h(T)=\gamma \). By B4, \(R(xy\alpha )=x\alpha \), \( y=h(xy\alpha )\), \(R(xy\beta \gamma )=y\beta \) and \(\gamma =h(xy\beta \gamma ) \). Clearly \(\beta \not =\gamma \), as hidden choice can not be a choice reverser by A1. If \(\alpha =\beta \) or \(\alpha =\gamma \), then A3 fails: \( xy\alpha \subset xy\beta \gamma \), \(y\in R(xy\beta \gamma )\), \(xy\alpha \in \Sigma _{*}\) however \(y\not \in R(xy\alpha )\). Hence we need only consider the case where \(\alpha ,\beta \) and \(\gamma \) are distinct alternatives.

We will begin by showing that \(xy\alpha \beta \gamma \not \in \Sigma _{*}\). By A3, \(c(xy\alpha \beta \gamma )\in \alpha \beta \) as each one of the other feasible alternatives does not belong to either \(R(xy\alpha )\) or \( R(xy\beta \gamma )\). If \(xy\alpha \beta \gamma \in \Sigma _{*}\) then \( R(xy\alpha \beta \gamma )=\alpha \beta \) by A3, giving \(c(xy\beta \gamma )=c(xy\alpha \gamma )=y\), where the first equality is by A1 and the second follows since \(c(xy\alpha \gamma )\in R(xy\beta \gamma )=y\beta \) however \( \beta \not \in xy\alpha \gamma \). Now using A3 again, we get \(y\in R(xy\alpha )\), a contradiction. We conclude that \(xy\alpha \beta \gamma \not \in \Sigma _{*}\).

Now we will show that \(c(xy\alpha \beta \gamma )=\alpha \). If \(c(xy\alpha \beta \gamma )=\beta \) then \(c(x\alpha \beta \gamma )=\beta \) as well by A3. We also have \(c(x\beta \gamma )=\gamma \) since \(\gamma =h(xy\beta \gamma )\) and \(y\in R(xy\beta \gamma )\). Now consider the set \(xy\alpha \gamma \). Note that \(c(xy\alpha \gamma )\not =\gamma \) by A2 since \(xy\alpha \beta \gamma \not \in \Sigma _{*}\), \(c(xy\alpha \beta \gamma )=\beta \) and \(c(\beta \gamma )=\gamma \). Furthermore \(c(xy\alpha \gamma )\not =y\) by A3 as \(xy\alpha \in \Sigma _{*}\) and \(y\not \in R(xy\alpha )\). Thus \(c(xy\alpha \gamma )\in x\alpha \) and \(y\not \in R(xy\alpha \gamma )\) by A3. It follows that \(c(xy\alpha \gamma )=c(x\alpha \gamma )\in x\alpha \). This gives a violation of A1 as \(c(x\alpha \beta \gamma )=\beta \), \(c(x\beta \gamma )=\gamma \) and \(c(x\alpha \gamma )\in x\alpha \). We conclude that \( c(xy\alpha \beta \gamma )=\alpha \).

To finish, note that \(h(xy\alpha \beta \gamma )=c(xy\beta \gamma )=c(y\beta ) \). Applying B2 in \(xy\alpha \) gives \(c(\alpha y)=y\). Then \(h(xy\alpha \beta \gamma )\not =y\) by A2 and \(h(xy\alpha \beta \gamma )=c(xy\beta \gamma )=c(y\beta )=\beta \). Applying A2 again in \(xy\alpha \beta \gamma \), we get \( c(\alpha \beta )=\alpha \). Thus the choices out of \(\alpha y\), \(y\beta \) and \(\alpha \beta \) constitute a binary cycle and a violation of B1.

**Case 2** Now suppose that \(y\not =h(S)\). Let \(R(S)=x\alpha \), \( h(S)=\beta \), \(R(T)=y\gamma \) and \(h(T)=\delta \). Note that \(x\ne c\left( S\right) \), since otherwise \(x=c\left( xy\alpha \right) \) by B3 and since \( y=c\left( xy\right) \), it follows that \(xy\alpha \in \Sigma _{*}\), \( R\left( xy\alpha \right) =xy\) and \(h\left( xy\alpha \right) =y\) leading us to the same contradiction as in Case 1 with \(T=xy\alpha \). Now, using B4, we get \(R(xy\alpha \beta )=x\alpha \), \(c(xy\alpha \beta )=\alpha \), \(h(xy\alpha \beta )=\beta \), \(R(xy\gamma \delta )=y\gamma \) and \(h(xy\gamma \delta )=\delta \). We will first show that if any two alternatives among \(\alpha ,\beta ,\gamma \) and \(\delta \) are the same, we arrive at a contradiction. Clearly if \(\alpha =\beta \) or \(\gamma =\delta \), A1 fails. If \(\alpha =\gamma \) then \(c(xy\alpha \beta \delta )=\alpha \) and \(xy\alpha \beta \delta \not \in \Sigma _{*}\) by A3. Since \(c(xy\beta )=\beta \) and \( c(xy\delta )=\delta \), \(h(xy\alpha \beta \delta )=c(xy\beta \delta )\in \beta \delta \) by A1. Applying B2 on \(xy\alpha \beta \) yields \(c(\alpha \beta )=\beta \), and on \(xy\gamma \delta \) yields \(c(\gamma \delta )=c(\alpha \delta )=\delta \). Hence both candidates for \(h(xy\alpha \beta \delta )\) are chosen over \(\alpha =c(xy\alpha \beta \delta )\) in binary sets. Now B2 implies \(xy\alpha \beta \delta \in \Sigma _{*}\), a contradiction. Similar contradictions obtain if \(\alpha =\delta \) or \(\beta =\gamma \) or \(\beta =\delta \) using almost exactly the same argument. We will skip the details and consider in the rest of the proof the case where \( \alpha ,\beta ,\gamma \) and \(\delta \) are distinct alternatives.

We will now show that \(xy\alpha \beta \gamma \delta \not \in \Sigma _{*}\) . If \(xy\alpha \beta \gamma \delta \in \Sigma _{*}\) then \(R(xy\alpha \beta \gamma \delta )=\alpha \gamma \) by A3 and \(c(xy\beta \gamma \delta )=c(xy\alpha \beta \delta )=h(xy\alpha \beta \gamma \delta )\) by A1. Clearly \(h(xy\alpha \beta \gamma \delta )\not \in \alpha \gamma \) by A1 and \( h(xy\alpha \beta \gamma \delta )\not \in xy\) by A3. Now if \(h(xy\alpha \beta \gamma \delta )=\beta \) then \(c(xy\alpha \beta \delta )=\beta \), and \(\delta \in R(xy\alpha \beta \delta )\) since \(c(xy\alpha \beta )=\alpha \). It follows that \(xy\alpha \beta \delta \in \Sigma _{*}\) and \(h(xy\alpha \beta \delta )=c(xy\alpha \beta )=\alpha \) by A1. Now applying B2 on \( xy\alpha \beta \), we find \(c(\beta x)=c(\beta \alpha )=\beta \). Hence \(\beta =c(xy\alpha \beta \delta )\) is chosen over both candidates for \(h(xy\alpha \beta \delta )\), giving \(xy\alpha \beta \delta \not \in \Sigma _{*}\) by B2, a contradiction. Similarly if \(h(xy\alpha \beta \gamma \delta )=\delta \) then \(xy\beta \gamma \delta \in \Sigma _{*}\) as \(c(xy\gamma \delta )\in y\gamma \). However B2 gives \(c(\delta y)=c(\delta \gamma )=\delta \) and \( xy\beta \gamma \delta \in \Sigma _{*}\) by B2, an analogous contradiction. We conclude that \(xy\alpha \beta \gamma \delta \not \in \Sigma _{*}\).

Now there are two subcases to consider depending on the roles played by \(y\) and \(\gamma \) in \(xy\gamma \delta \), and we will consider these cases seperately. We have established above that \(c(xy\alpha \beta \gamma \delta )\in \alpha \gamma \). In all subcases that follow we will assume that \( c(xy\alpha \beta \gamma \delta )=\alpha \). This is without loss of generality and the arguments can easily be adopted to work if \(c(xy\alpha \beta \gamma \delta )=\gamma \).

**Case 2a** Suppose that \(\alpha =c(xy\alpha \beta )\) and \(y=c(xy\gamma \delta )\). To begin, note that \(c(y\alpha \beta )=\beta \) since \(x\in R(xy\alpha \beta )\) and \(\beta =h(xy\alpha \beta )\). We will now build up to a contradiction to this finding. We have \(c(xy\beta )=\beta \) and \( c(xy\gamma )=y\), giving \(c(xy\beta \gamma )\in \beta y\) by A1. Now consider the set \(xy\beta \gamma \delta \). A3 implies \(c(xy\beta \gamma \delta )\not \in x\delta \), as neither \(x\) nor \(\delta \) reverse choice in \(xy\gamma \delta \). Furthermore A2 implies \(c(xy\beta \gamma \delta )\not =\beta \) as \(c(\alpha \beta )=\beta \), \(\alpha =c(xy\alpha \beta \gamma \delta )\) and \( xy\alpha \beta \gamma \delta \not \in \Sigma _{*}\). If \(c(xy\beta \gamma \delta )=\gamma \), then \(\delta \) and \(\beta \) are two distinct non-trivial reversers in \(xy\beta \gamma \delta \) and A4 fails. Hence we must have \( c(xy\beta \gamma \delta )=y=h(xy\alpha \beta \gamma \delta )\), \(c(\alpha y)=\alpha \) by A2 and \(c(y\beta \gamma \delta )=y\) as \(x\) cannot be a reverser in \(xy\beta \gamma \delta \) by A3. Using A3 again, we conclude that \(\delta \) cannot be a reverser in \(y\beta \gamma \delta \) as it is not a reverser in \(y\gamma \delta \). Hence \(c(y\beta \gamma )=y\). On the other hand, \(c(xy\alpha \beta \gamma )=\alpha \) as \(xy\alpha \beta \gamma \delta \not \in \Sigma _{*}\) and \(c(xy\beta \gamma )=y\) by A3 which dictates that \(\delta \) can not be a reverser in \(xy\beta \gamma \delta \) either. Note that \(c(xy\alpha \beta \gamma )\) is chosen over \(h(xy\alpha \beta \gamma )=c(xy\beta \gamma )\) in binary comparison as we established \( c(\alpha y)=y\) above, giving \(xy\alpha \beta \gamma \not \in \Sigma _{*}\) by B2. This implies \(c(y\alpha \beta \gamma )=\alpha \). Now \(c(y\alpha \beta \gamma )\) is chosen over \(h(y\alpha \beta \gamma )=\) \(c(y\beta \gamma )=y\) in binary comparison, giving \(y\alpha \beta \gamma \not \in \Sigma _{*}\) by B2. It follows that \(c(y\alpha \beta )=\alpha \), which is exactly the contradiction we were aiming for.

**Case 2b** Finally suppose that \(\alpha =c(xy\alpha \beta )\) and \(\gamma =c(xy\gamma \delta )\) making \(x\) and \(y\) non-trivial reversers of \(xy\alpha \beta \) and \(xy\gamma \delta \) respectively. We will build up to a contradiction to A1. Now \(c(y\alpha \beta \gamma \delta )=\alpha \) as \( xy\alpha \beta \gamma \delta \not \in \Sigma _{*}\). Since, by B4, \(\delta \) is not a reverser in \(y\gamma \delta \), it can not reverse choice in \( y\alpha \beta \gamma \delta \) either by A3. Hence \(c(y\alpha \beta \gamma )=\alpha \). Next we will consider the set \(xy\beta \gamma \delta \). By A3 \( c(xy\beta \gamma \delta )\not \in x\delta \). Furthermore by A2, \(c(xy\beta \gamma \delta )\not =\beta \) either as \(c(\beta \alpha )=\beta \) and \( xy\alpha \beta \gamma \delta \not \in \Sigma _{*}\). Thus \(c(xy\beta \gamma \delta )\in y\gamma \). Note that \(c(xy\beta \delta )\in \beta \delta \) by A1 as \(c(xy\beta )=\beta \) and \(c(xy\delta )=\delta \). Furthermore \( c(xy\gamma \delta )=\gamma \), and \(xy\beta \delta , xy\gamma \delta \subseteq xy\beta \gamma \delta \). Hence \(c(xy\beta \gamma \delta )\not =y\) by A1 and A4. We conclude that \(c(xy\beta \gamma \delta )=\gamma \). Note also that \( x\not \in R(xy\beta \gamma \delta )\) and \(\delta \not \in R(y\beta \gamma \delta )\) by A3, giving \(c(y\beta \gamma )=\gamma \). Finally note that \( c(y\alpha \beta )=\beta \) as \(x\in R(xy\alpha \beta )\) and \(\beta =h(xy\alpha \beta )\). Hence we have \(c(y\alpha \beta \gamma )=\alpha \), \( c(y\beta \gamma )=\gamma \) and \(c(y\alpha \beta )=\beta \) contradicting A1. \(\square \)

### *Proof of Lemma 4*

Fix a choice function \(c\) which satisfies A1-A4 and distinct alternatives \( x,y\) and \(z\) satisfying Open image in new window and \(zP_{0}x\). Since \(zP_{0}x\), there exists a set \(S\in \Sigma _{*}\) such that \(z\in R(S)\) and \(x\in S\backslash R(S)\). Suppose, to begin, that \(y\in S\). It follows that \(y\in R(S)\) since otherwise \(zP_{0}y\), a contradiction. Hence, by A4, \(R(S)=yz\). Now by B3, \(c(S)=c(xyz)\) and since \(c(S)\in R(S)\), \(c(xyz)\not =x\).

Now suppose \(y\not \in S\) and using A4 let \(R(S)=zw\) for some \(w\in S\backslash zx\). Note in particular that \(zP_{0}s\) for every \(s\in S\backslash zw\) by definition. Denote \(S^{\prime }=S\cup y\) and \(S^{\prime \prime }=S^{\prime }\backslash w\). We *claim* here that \(c(S^{\prime \prime })\in yz\ \). Before proving this *claim*, let us show that if it is correct, the proof is complete. Again, there are two possibilities. If \(c(S^{\prime \prime })=y\) and \(c(xyz)=x\), then by Lemma 2 there exists some \( s\in S^{\prime \prime }\backslash xyz\) such that \(sP_{0}z\). However such \(s\) necessarily belongs to \(S\backslash zw\), and therefore \(zP_{0}s\) as well, contradicting the asymmetry of \(P_{0}\). If \(c(S^{\prime \prime })=z\) and \( c(xyz)=x\), on the other hand, \(zP_{0}y\) by Lemma 2, an impossibility. Hence we must have \(c(xyz)\not =x\)

All that remains to be done is to prove the *claim* that \(c(S^{\prime \prime })\in yz\). By A3 \(R(S^{\prime })\subset wyz\), as no alternative in \( S\backslash wz\) could belong to \(R(S^{\prime })\). Hence there are three possibilities regarding the choice in \(S^{\prime }\). (1) Suppose first that \( c(S^{\prime })=y\). If \(S^{\prime }\not \in \Sigma _{*}\) then \(c(S^{\prime \prime })=y\). If, on the other hand, \(S^{\prime }\in \Sigma _{*}\), then either \(w\not \in R(S^{\prime })\) and \(c(S^{\prime \prime })=c(S)=y\), or \(w\in R(S^{\prime })\) and by A1 \(c(S)=c(S^{\prime \prime })\). In the latter case \(c(S)=c(S^{\prime \prime })=z\) since \(c(S)\in zw\) and \(w\not \in S^{\prime \prime }\). Hence if \(c(S^{\prime })=y\), then \(c(S^{\prime \prime })\in yz\). (2) Now suppose that \(c(S^{\prime })=z\). If \(S^{\prime }\not \in \Sigma _{*}\), then \(c(S^{\prime \prime })=z\). If, on the other hand, \( S^{\prime }\in \Sigma _{*}\), then \(R(S^{\prime })=yz\) since otherwise \( zP_{0}y\), a contradiction. Hence \(c(S^{\prime \prime })=c(S^{\prime })=z\). Consequently, if \(c(S^{\prime })=z\), then \(c(S^{\prime \prime })=z\). (3) Finally suppose that \(c(S^{\prime })=w\). We will first consider the case \( S^{\prime }\not \in \Sigma _{*}\), where we will exploit the fact that \( S\in \Sigma _{*}\). Let \(\alpha =h(S)\) and \(\beta =h(S^{\prime })\). Hence \(\alpha =c(S\backslash z)=c(S\backslash w)\) by A1, and \(\beta =c(S^{\prime \prime })\) by definition. Note that B2 implies \(c(\alpha w)=\alpha \) and A2 implies \(c(\beta w)=w\). We conclude \(\alpha \not =\beta \). If \(\beta =y\), then there is nothing to show. If \(\beta \not =y\), then \(\beta \in S\backslash w\). Since \(c(S^{\prime \prime }\backslash y)=c(S\backslash w)=h(S)=\alpha \not =\beta =c(S^{\prime \prime })\), \(y\in R(S^{\prime \prime })\), and, by A4, \(R(S^{\prime \prime })=\beta y\). Now if \(\beta \not =z\), \( z\in S^{\prime \prime }\backslash R(S^{\prime \prime })\) and there exists \( s\in S\) such that \(sP_{0}z\), contradicting Lemma 1. Hence \(\beta =z\) and \( c(S^{\prime \prime })\in yz\), as we wanted to show. Now suppose that \( S^{\prime }\in \Sigma _{*}\). Clearly \(w=c(S^{\prime })\in R(S^{\prime })\) and by A4 there exists a unique non-trivial reverser in \(S^{\prime }\). It follows that \(z\not \in R(S^{\prime })\) since Open image in new window . If there exists some \(s\in S^{\prime }\backslash wyz\) such that \(R(S^{\prime })=sw\), then \(sP_{0}z\), a contradiction to Lemma 1, as \(s\in S^{\prime }\backslash wyz\subset S\backslash R(S)\). Hence \(R(S^{\prime })=wy\). Now by A1 \( c(S^{\prime }\backslash w)=c(S^{\prime }\backslash y)\) and since \(S^{\prime }\backslash y=S\), \(c(S^{\prime }\backslash w)=c(S^{\prime }\backslash y)\in R(S)=wz\), giving \(c(S^{\prime \prime })=c(S^{\prime }\backslash w)=z\). We conclude that if \(c(S^{\prime })=w\), then \(c(S^{\prime \prime })\in yz\), and this establishes the *claim*. \(\square \)