Choosing two finalists and the winner

Abstract

We study a class of boundedly rational choice functions which operate as follows. The decision maker uses two criteria in two stages to make a choice. First, she shortlists the top two alternatives, i.e. two finalists, according to one criterion. Next, she chooses the winner in this binary shortlist using the second criterion. The criteria are linear orders that rank the alternatives. Only the winner is observable. We study the behavior exhibited by this choice procedure and provide an axiomatic characterization of it. We leave as an open question the characterization of a generalization to larger shortlists.

Appendices

Appendix 1: Independence of axioms 1–4

The following four examples show that the axiomatic system A1–A4 is tight, in other words, no three among A1–A4 imply the fourth. We will use the following notation: for any choice function $c$, $C(x)=\{A:|A|>1$ and $c(A)=x\}$.

Example 3

A2, A3 and A4 do not imply A1. Let $X=xyzw$. Consider the following choice function: $C(x)=\{xy$, $xz$, $wx$, $xyz\}$, $C(y)=\{yz$, $wy$, $wxy\}$, $C(z)=\{wz$, $wxz$, $wyz$, $X\}$ and $C(w)=\varnothing$. Note that $\Sigma _{*}=\{wxy$, $wxz$, $wyz$, $X\}$, $R(X)=R(wyz)=R(wxz)=wz$ and $R(wxy)=wy$. Hence $c$ satisfies A3 and A4. Moreover the unique set with three or more elements and which does not belong to $\Sigma _{*}$ is $xyz$, where $x$ is the choice and $z$ is the hidden choice. Since $c(xz)=x$, $c$ also satisfies A2. A1 fails because $R(X)=wz$ but $c(X\backslash z)=y\not =x=c(X\backslash w)$. $\blacktriangle$

Example 4

A1, A3 and A4 do not imply A2. Let $X=xyzw$. Consider the following choice function: $C(x)=\{xy$, $xz$, $wx$, $wxz\}$, $C(y)=\{yz$, $wy$, $xyz$, $wxy$, $wyz$, $X\}$, $C(z)=\{wz\}$ and $C(w)=\varnothing$. Note that $\Sigma _{*}=\{xyz,wxy\}$, $R(xyz)=yz$ and $R(wxy)=wy$. Hence $c$ satisfies A3 and A4. Furthermore A1 follows since $c(xyz\backslash y)=c(xyz\backslash z)=x$ and $c(wxy\backslash w)=c(wxy\backslash y)=x$. However A2 fails because $X\not \in \Sigma _{*}$ but $c(c(X)h(X))=c(yx)=x=h(X)$. $\blacktriangle$

Example 5

A1, A2 and A4 do not imply A3. Let $X=vwxyz$ and consider the following choice function: $C(v)=\{vz\}$, $C(w)=\{vw$, $wx$, $wy$, $wz$, $vwy$, $vwz$, $wxy$, $wxz\}$, $C(x)=\{vx$, $xy$, $xz$, $vwx$, $vxy$, $vxz$, $vwxy$, $vwxz\}$, $C(y)=\{vy$, $yz$, $vyz$, $wyz$, $xyz$, $vwyz$, $vxyz$, $wxyz$, $X\}$, and $C(z)=\varnothing$. Note that $\Sigma _{*}$ contains $X$, all subsets of $X$ with four elements, and the following three-element sets: $vwx$, $wyz$, $xyz$. One can check that there is a unique non-trivial reverser in all sets in $\Sigma _{*}$, which is $v$ in $vwx$, $vwxy$ and $vwxz$, and $z$ otherwise. Thus $c$ satisfies A4. One can also check that the removal of the choice and the removal of the non-trivial reverser leads to the choice of the hidden choice in all sets with choice reversal, which is $x$ in $X$ and in all sets that do not contain $w$, and $w$ otherwise. This gives A1. To check A2, consider the following table listing all sets with three or more elements outside $\Sigma _{*}$, their choices and hidden choices. Now note that the choice is always chosen over the hidden choice in binary comparison. Hence $c$ satisfies A2 as well. A3 fails because $z\in R(X)$, $vwxz\in \Sigma _{*}$ but $R(vwxz)=vx$.

$A$	$vwy$	$vwz$	$wxy$	$wxz$	$vxy$	$vxz$	$vyz$
$c(A)$	$w$	$w$	$w$	$w$	$x$	$x$	$y$
$h(A)$	$y$	$v$	$x$	$x$	$y$	$v$	$v$

Example 6

A1, A2 and A3 do not imply A4. Let $X=vwxyz$ and $\succ _{1}$ and $\succ _{2}$ be linear orders given by $x\succ _{1}y\succ _{1}z\succ _{1}w\succ _{1}v$ and $v\succ _{2}w\succ _{2}x\succ _{2}y\succ _{2}z$. Consider the following extension of two-stage choosers: $c(A)=\max _{\succ _{2}}(T_{\succ _{1}}^{3}(A))$ for every $A$. Hence $c$ shortlists the top 3 alternatives in $\succ _{1}$ chooses from the shortlist using $\succ _{1}$. Clearly A4 fails as all sets with choice reversal have three choice reversers: $R(X)=R(wxyz)=R(vxyz)=xyz$, $R(vwyz)=wyz$, $R(vwxz)=wxz$ and $R(vwxy)=wxy$. It is straightforward to check that $c$ satisfies A1, A2 and A3. $\blacktriangle$

Appendix 2: Proofs of Lemmas 1–5

In this appendix, we prove Lemmas 1–5.

Proof of Lemma 1

Fix a choice function $c$ satisfying axioms A1–A4. We will first show B1. Suppose that $c(xy)=x$, $c(yz)=y$ and $c(xz)=z$ for distinct alternatives $x$, $y$ and $z$ so that a binary cyle exists and without loss of generality let $c(xyz)=x$. Note $R(xyz)=xy$, however $c(xyz\backslash x)\not =c(xyz\backslash y)$ and A1 fails. Hence $c$ satisfies B1. To see B2, B3 and B4 take $A\in \Sigma _{*}$ and let $x,y$ and $z$ be the choice, the non-trivial choice reverser and the hidden choice in $A$ respectively. A3 and A4 imply that $x$ and $y$ are the only potential reversers in any subset of $A$ where they both belong and therefore if $xy\subseteq A^{\prime }\subset A$, then $c(A^{\prime })=x$, giving B3. Similarly $x$ and $z$ are the only potential reversers in any subset of $A\backslash y$ which contains them. Hence $c(A\backslash y)=c(xz)=z$. Furthermore $c(xy)=x$ by B3 and $c(yz)=z$ by B1, which establishes B2. Now suppose that $xyz\subseteq A^{\prime }\subset A$. By B3 $c(A^{\prime })=x$. Let $A\backslash A^{\prime }=a_{1}...a_{K}$, $A_{k}=A^{\prime }\cup a_{1},,,a_{k}$ for every $k=1,...,K$ and $A_{0}=A^{\prime }$. We must have $c(A\backslash x)=z=c(A_{k}\backslash x)$ for every $k$, since otherwise there exists a set $A_{k}$ such that $y,z\in R(A_{k}\backslash x)$ by A3 and $a_{k}\in R(A_{k}\backslash x)$ by construction, a contradiction to A4. Hence $h(A^{\prime })=z$ as well. Finally, since $c(xz)=z$, $A^{\prime }\in \Sigma _{*}$ by A2 and $R(A^{\prime })=xy$ by A3 and A4. This gives B4. $\square$

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 3

Take $a,b,x,A$ and $B$ such that $c(A)=a$, $c(B)=b\not =a$, $a\in B\subset A$ and $x\in B\backslash a$. Now there must exist a set $A^{\prime }$ and an alternative $y\in A^{\prime }\backslash B\subset A\backslash B$ such that $B\subset A^{\prime }\subseteq A$, $A^{\prime }\in \Sigma _{*}$ and $y\in R(A^{\prime })$ since otherwise $c(A)=c(B)$. Note that $y\not =x$ and, by A3 $a\in R(A^{\prime })$ as well. Now A4 implies $R(A^{\prime })=ay$. Since $x\in B\backslash a$, $x\in A^{\prime }\backslash R(A^{\prime })$ and consequently $aP_{0}x$ and $yP_{0}x$. $\square$

Proof of Lemma 4

Fix a choice function $c$ which satisfies A1-A4 and distinct alternatives $x,y$ and $z$ satisfying 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 . 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. $\square$

Proof of Lemma 5

Fix a choice function $c$ which satisfies A1-A4 and distinct alternatives $x,y$ and $z$ such that $xP_{0}y$ and $yP_{0}z$. We will show that $xP_{0}z$. There exists, by definition, $S\in \Sigma _{*}$ such that $x\in R(S)$ and $y\in S\backslash R(S)$. Using A4, write $R(S)=xw$ where $w\in S\backslash xy$ and note that $wP_{0}y$ as well. If $z\in S$, then $z\not \in R(S)$, since otherwise $zP_{0}y$. It follows that $xP_{0}z$, as we needed. Suppose now that $z\not \in S$. Let $S^{\prime }=S\cup z$ and note that $c(S^{\prime })\in xwz$ by A3. We have $yP_{0}z$ by hypothesis and by asymmetry of $P_{0}$ as established in Lemma 2. Now Lemma 4 implies that $z\not =c(xyz)$. If $c(S^{\prime })=z$, then $zP_{0}y$ by Lemma 3, a contradiction. Hence $c(S^{\prime })\in xw=R(S)$. The rest of the proof proceeds exactly as the proof of Case 2 of Lemma 6 in the main text: it follows that $S^{\prime }\in \Sigma _{*}$, $R(S^{\prime })=xw$. Consequently $xP_{0}z$. $\square$