Preference aggregation for couples

Abstract

We study the aggregation of a couple’s preferences over their respective jobs when the couple enters a centralized labor market jointly, such as the market for hospital residencies. In such markets couples usually need to submit a joint preference ordering over pairs of jobs and thus we are interested in preference aggregation rules which start with two individual preference orderings over single jobs and produce a preference ordering of pairs of jobs. We first study the Lexicographic and the Rank-Based Leximin aggregation rules, as well as a large class of preference aggregation rules which contains these two rules. Then we propose a smaller family of parametric aggregation rules, the k-Lexi-Pairing rules, which call for a systematic way of compromising between the two partners. The parameter k indicates the degree to which one partner is prioritized, with the most equitable Rank-Based Leximin rule at one extreme and the least equitable Lexicographic rule at the other extreme. Since couples care about geographic proximity, a parametric family of preference aggregation rules which build on the k-Lexi-Pairing rules and express the couple’s preference for togetherness is also identified. We provide axiomatic characterizations of the proposed preference aggregation rules.

Appendix

Proof of Theorem 1

Claim1.1 The Rank-Based Leximin rule satisfies Limited Equity.

Proof

Fix $(r_1,r_2) \in R \times R$ and let $P = \varphi (r_1, r_2)$, where $\varphi$ is the Rank-Based Leximin Rule. Let $( h_1, h_2 )$ and $( h'_1, h'_2 )$ be such that neither pair dominates the other and $g>g'$. We will show that then $(h'_1,h'_2)P(h_1, h_2)$.

Case 1: Max > Max$'$.

Then $(h'_1,h'_2)P(h_1, h_2)$.

Case 2: Max < Max$'$

Since $g>g'$, we must have Min < Min$'$. This implies that $( h_1, h_2 )$ dominates $( h'_1, h'_2 )$, which is a contradiction.

Case 3: Max $=$ Max$'$

Then if Min$' \ne$ Min there is a dominance relation between $( h_1, h_2 )$ and $( h'_1, h'_2 )$, and thus Min $=$ Min$'$. This implies that $g = g'$, which is a contradiction. $\square$

Claim 1.2 The Rank-Based Leximin rule satisfies Cross-Dominance.

Proof

Fix $(r_1,r_2) \in R \times R$ and let $P = \varphi (r_1, r_2)$, where $\varphi$ is the Rank-Based Leximin Rule. Let $(h_1, h_2)$ and $( h'_1, h'_2 )$ be such that $r_{1} ( h_1 )\geqslant r_{2} ( h'_2 )$ and $r_{2} ( h_2 )\geqslant r_{1} ( h'_1 )$ with at least one inequality. We will show that then $(h'_1, h'_2)P (h_1, h_2)$.

Case 1: Max > Max$'$

Then $(h'_1, h'_2)P (h_1, h_2)$.

Case 2: Max < Max$'$

Since Max $\geqslant r_1(h_1) \geqslant r_2(h'_2)$ and Max $\geqslant r_2(h_2) \geqslant r_1(h'_1)$, Max $\geqslant$ Max$'$, which is a contradiction.

Case 3: Max $=$ Max$'$

If Min > Min$'$ then $(h'_1, h'_2)P (h_1, h_2)$. Assume that Min $\leqslant$ Min$'$.

Subcase 3.1: If Min $= r_{1} ( h_1 )$ then $r_1( h_1) \leqslant r_2(h'_2)$ and $r_1( h_1) \leqslant r_1(h'_1)$. Then $r_1( h_1) = r_2(h'_2)$ and $r_2( h'_2) \leqslant r_1(h'_1)$. Therefore, Max$' = r_1(h'_1)$ and, since Max $= r_2(h_2)$, $r_2( h_2) = r_1(h'_1)$. This is a contradiction since $r_1( h_1) = r_2(h'_2)$ implies that $r_2( h_2) > r_1(h'_1)$.

Subcase 3.2: If Min $= r_{2} ( h_2 )$ then $r_2( h_2) \leqslant r_1(h'_1)$ and $r_2( h_2) \leqslant r_2(h'_2)$. Then $r_2( h_2) = r_1(h'_1)$ and $r_1( h'_1) \leqslant r_2(h'_2)$. Therefore, Max$' = r_2(h'_2)$ and, since Max $= r_1(h_1)$, $r_1( h_1) = r_2(h'_2)$. This is a contradiction since $r_2( h_2) = r_1(h'_1)$ implies that $r_1( h_1) > r_2(h'_2)$. $\square$

Claim 1.3 If a pairing rule satisfies Strong Pareto, Cross-Dominance and Limited Equity, then it is the Rank-Based Leximin rule.

Proof

Fix $(r_1,r_2) \in R \times R$ and let $P = \varphi (r_1, r_2)$, where $\varphi$ satisfies Strong Pareto, Cross-Dominance and Limited Equity. Let $( h_1, h_2 )$ and $( h'_1, h'_2 )$ be two distinct hospital pairs.

If Max $=$ Max$'$ and Min > Min$'$, $(h'_1, h'_2)$ dominates $(h_1, h_2)$. Then Dominance implies that $( h'_1, h'_2 )P ( h_1, h_2 )$.

If Max $=$ Max$'$ and Min $=$ Min$'$ then, given that $( h_1, h_2 ) \ne ( h'_1, h'_2 )$, we have symmetric opposites: $r_{1} ( h_1 ) = r_{2} ( h'_2 )$, $r_{2} ( h_2 ) = r_{1} ( h'_1 )$ and $r_{1} ( h_1) \ne r_{1} ( h'_1 )$. For symmetric opposites, $r_{1} ( h_1 ) > r_{1} ( h'_1 )$ implies $( h'_1, h'_2 )P ( h_1, h_2 )$ by convention.

If Max > Max$'$, we consider three cases regarding $\sum$ and $\sum '$. Note first that Max $=$ Min $+g=\sum -$ Min, and hence 2Max $=\sum +g$. Similarly, 2Max$'=\sum ' + g'$. Thus Max > Max$'$ if and only if $\sum +g>\sum '+g'$.

Case 1: $\sum = \sum '$

Given that $\sum = \sum ',$ $\sum + g > \sum ' + g'$ is equivalent to $g>g'$, and thus $( h'_1, h'_2 )P ( h_1, h_2 )$ by Equal-Sum Equity (a weaker version of Limited Equity).

Case 2: $\sum < \sum '$

Given that $\sum < \sum ',$ $( h'_1, h'_2 )$ does not dominate $( h_1, h_2 )$. Since Max > Max$'$, $( h_1, h_2 )$ does not dominate $( h'_1, h'_2 )$. Now note that $\sum + g > \sum ' + g'$ implies that $g > g'$. Thus, given that $( h_1, h_2 )$ does not dominate $( h'_1, h'_2 )$ and vice versa, Limited Equity implies that $( h'_1, h'_2 )P ( h_1, h_2 )$.

Case 3: $\sum > \sum '$

Let $( {\tilde{h}}_1, {\tilde{h}}_2 )$ be such that $\tilde{\text{ Max }} =$ Max$'+1$ and $\tilde{\text{ Min }}=\sum -$ Max$'-1$. Note that $\tilde{\text{ Min }}$ is feasible, since Min$' \geqslant 1$, $\sum ' \geqslant$ Max$'+1$ and thus $\sum > \sum '$ implies that $\sum >$ Max$' +1$, and hence $\tilde{\text{ Min }} > 0$. Observe that ${\tilde{\sum }} = \sum$. We will show next that ${\tilde{g}} \leqslant g$. Given that Max$'<$ Max, Max$'+1 \leqslant$ Max. Then 2Max$'+2 \leqslant 2$Max $=$ Max − Min $+$ Max $+$ Min. This means that 2Max$' + 2 - \sum \leqslant$ Max − Min, which is equivalent to ${\tilde{g}} \leqslant g$, since ${\tilde{g}} = \tilde{\text{ Max }} - \tilde{\text{ Min }} =$ Max$' + 1 - \sum +$ Max$' + 1 = 2$Max$' + 2 - \sum$.

Now note that Max$' < \tilde{\text{ Max }}$. We will show that Min$' \leqslant \tilde{\text{ Min }}$. Since $\sum > \sum '$, $\sum ' + 1 \leqslant \sum$. Thus, Min$' +$ Max$' +1 \leqslant \sum$ and Min$' \leqslant \sum -$ Max$' -1 = \tilde{\text{ Min }}$. Therefore, $( h'_1, h'_2 )$ dominates $( {\tilde{h}}_1, {\tilde{h}}_2 )$ and Dominance implies that $( h'_1, h'_2 )P( {\tilde{h}}_1, {\tilde{h}}_2 )$.

If ${\tilde{g}} < g$ then, given that ${\tilde{\sum }} = \sum$, Equal-Sum Equity implies $( {\tilde{h}}_1, {\tilde{h}}_2 )P( h_1, h_2 )$. If ${\tilde{g}} = g$, either $( {\tilde{h}}_1, {\tilde{h}}_2 ) = ( h_1, h_2 )$ or $( {\tilde{h}}_1, {\tilde{h}}_2 )$ and $( h_1, h_2 )$ are symmetric opposites. Since $( h'_1, h'_2 )$ dominates $( {\tilde{h}}_1, {\tilde{h}}_2 )$, it follows that in both cases $( h'_1, h'_2 )$ dominates $( h_1, h_2 )$, and therefore $( h'_1, h'_2 )P ( h_1, h_2 )$ by Dominance. $\square$

Finally, note that it follows from Claim 2.1 below that the Rank-Based Leximin rule satisfies Strong Pareto (the $k=0$ case in Claim 2.1). Together with this result, Claims 1.1, 1.2, and 1.3 prove Theorem 1.

Proof of Theorem 2

Claim 2.1 For all $k \in \{ 0,\ldots ,q-1 \}$, the k-Lexi-Pairing rule $\psi ^k$ satisfies Strong Pareto.

Proof

Let $k \in \{ 0,\ldots ,q-1 \}$. Fix $(r_1,r_2) \in R \times R$ and let $P^k= \psi ^k(r_1, r_2).$ Let $( h_1, h_2 )$ and $( h'_1, h'_2 )$ satisfy $r_1 ( h_1 )\geqslant r_1 ( {h'_1} )$ and $r_2 ( h_2 )\geqslant r_2 ( {h'_2} ),$ with at least one strict inequality. We need to show that $( h'_1, h'_2 )P^k ( h_1, h_2 )$.

Case 1. $r_1 ( h_1 ) > r_1 ( {h'_1} )$ and $r_2 ( h_2 ) > r_2 (h'_2)$

Subcase 1.1: If $r_1 (h_1)\geqslant r_2 ( {h_2} )- k$ then Max $=r_1 (h_1) >r_1 ( h'_1)$, and Max $>r_2 ( h'_2 )- k$. Thus, Max > Max$'$.

Subcase 1.2: If $r_1 ( {h_1} )< r_2 ( {h_2} )- k$ then Max $= r_2 (h_2 )- k> r_1(h_1) > r_1 ( h'_1 )$, and Max $>r_2 ( h'_2 )- k$. Thus, Max > Max$'$.

Case 2. $r_1 ( h_1 )= r_1 ( {h'_1} )$ and $r_2 ( h_2 ) > r_2 ( {h'_2} )$

Subcase 2.1: If $r_1 ( {h_1} )\geqslant r_2 (h_2)- k$, then Max $=r_1 ( {h_1} )=r_1 ( h'_1 )$ and Max $>r_2 (h'_2)-k$. Thus, Max$' = r_1 ( h'_1 )$ and Max $=$ Max$'$. Furthermore, Min $=r_2 ( {h_2} )- k$, Min $>r_2 ( h'_2 )- k$, and since Max$'=r_1 ( h'_1 )$, Min$'=r_2 ( h'_2 )- k$. Thus, Min > Min$'$.

Subcase 2.2: If $r_1 ( {h_1} )< r_2 ( {h_2} )- k$ then Max $=r_2 (h_2)-k$, Max $>r_2 ( h'_2 )-k > r_1(h_1) = r_1 ( h'_1 )$. Thus, Max > Max$'$.

Case 3. $r_1 (h_1)> r_1 (h'_1)$ and $r_2 ( h_2 )= r_2 (h'_2)$

Subcase 3.1: If $r_1 ( {h_1} )> r_2 ( {h_2} )- k$ then Max $=r_1 ( {h_1} ) >r_1 (h'_1)$, and Max $>r_2 ( h'_2 )-k$. Thus, Max > Max$'$.

Subcase 3.2: If $r_1 ( {h_1} )\leqslant r_2 ( {h_2} )- k$ then Max $= r_2(h_2)-k = r_2(h'_2) - k \geqslant r_1(h_1) > r_1(h'_1)$. Thus, Max $=$ Max$'$. Then Min $=r_1 ( {h_1} )$ and Min $>r_1 ( h'_1 )$, which implies that Min > Min$'$.

Therefore, $( h'_1, h'_2 )P^k ( h_1, h_2 )$ in each case. Since this holds for all $(r_1, r_2) \in R \times R,$ for all $k \in \{0, \ldots , q-1 \}$, the k-Lexi-Pairing rule satisfies Strong Pareto. $\square$

Claim 2.2 For all $k \in \{ 0,\ldots , q-1 \}$, the k-Lexi-Pairing rule satisfies k-Compromise.

Proof

Let $k \in \{ 0, \ldots , q-1 \}$. Fix $(r_1,r_2) \in R \times R$ and let $P^k = \psi ^k(r_1, r_2).$ Let $( h_1, h_2 )$ and $( h'_1, h'_2 )$ satisfy $r_1({h_1}) < r_1(h'_1)$ and $r_2(h_2) > r_2(h'_2)$.

Case 1: We will show that $r_1 (h'_1) <r_2 (h_2) - k$ implies $( h'_1, h'_2 )P^k ( h_1, h_2 )$.

Note first that $r_2 ( {h_2} )-k>r_1 ( h'_1 ) > r_1 (h_1)$ and thus Max $=r_2 ( {h_2} )-k$.

Subcase 2.1: If Max$'=r_1 ( h'_1 )$ then $r_2 ( {h_2} )-k>r_1 ( h'_1 )$ implies Max > Max$'$.

Subcase 2.2: If Max$'=r_2 ( h'_2 )-k$ then $r_2 ( {h_2} )>r_2 ( h'_2 )$ implies Max > Max$'$.

Hence, $( h'_1, h'_2 )P^k ( h_1, h_2 )$ in both subcases.

Case 2: We will show that $r_1(h'_1) > r_2 (h_2) - k$ implies $( h_1, h_2 )P^k ( h'_1, h'_2 )$.

Note first that $r_1 ( h'_1 ) + k> r_2(h_2) > r_2 (h'_2)$ and thus Max$'=r_1 ( h'_1 )$.

Subcase 1.1: If Max $=r_1 ( {h_1} )$ then $r_1 ( {h_1} )<r_1 ( h'_1 )$ implies Max < Max$'$.

Subcase 1.2: If Max $=r_2 ( {h_2} )-k$ then $r_2 ( {h_2} )-k<r_1 ( h'_1 )$ implies Max < Max$'$.

Hence, $(h_1, h_2 )P^k ( h'_1, h'_2 )$ in both subcases.

Therefore, k-Compromise is satisfied in each case. Since this holds for all $(r_1, r_2) \in R \times R,$ for all $k \in \{0, \ldots , q-1 \}$, the k-Lexi-Pairing rule satisfies k-Compromise. $\square$

Claim 2.3 For all $k \in \{0,\ldots ,q-1\}$, the k-Lexi Pairing rule satisfies k-Threshold-Consistency.

Proof

Let $k \in \{ 0,\ldots ,q-1 \}$. Fix $(r_1,r_2) \in R \times R$ and let $P^k= \psi ^k(r_1, r_2).$ Let $( h_1, h_2 )$ and $( h'_1, h'_2 )$ satisfy $r_1 ( {h_1} )< r_1 ( h'_1)$, $r_2 (h_2) > r_2 (h'_2)$ and $r_1 (h'_1)= r_2 ( h_2 )-k$.

Then $r_1 (h_1)< r_1 (h'_1) = r_2 (h_2)-k$ and thus Max = $r_2(h_2)-k$. Also, $r_1 (h'_1)= r_2 (h_2) - k > r_2 (h'_2)-k$ and thus Max$' = r_1(h'_1)$. Therefore, Max $=$ Max$'$. Moreover, Min $= r_1(h_1)$ and Min$' = r_2(h'_2)-k$.

Case 1: $r_1(h_1) > r_2 ( h'_2 )-k$

Then Min > Min$'$ and $(h'_1, h'_2)P^k (h_1, h_2)$.

Case 2: $r_1(h_1) < r_2 ( h'_2 )-k$

Then Min < Min$'$ and $(h_1, h_2)P^k (h'_1, h'_2)$.

Case 3: $r_1(h_1) = r_2 ( h'_2 )-k$

Then Min $=$ Min$'$ and, since $r_1(h_1) < r_1(h'_1)$, $(h_1, h_2)P^k (h'_1, h'_2)$.

Therefore, k-Threshold-Consistency is satisfied in each case. Since this holds for all $(r_1, r_2) \in R \times R,$ for all $k \in \{0, \ldots , q-1 \}$, the k-Lexi-Pairing rule satisfies k-Threshold-Consistency. $\square$

Claim 2.4 Let $k \in \{ 0,\ldots ,q-1 \}$. If a pairing rule satisfies Strong Pareto, k-Compromise, and k-Threshold-Consistency then it is the k-Lexi-Pairing rule.

Proof

Fix $k \in \{ 0, \ldots , q-1 \}$ and $(r_1,r_2) \in R \times R$. Let $( h_1, h_2 ), ( h'_1, h'_2) \in H \times H$ such that $( h_1, h_2 )\ne ( h'_1, h'_2 )$. Let Max, Min, Max$'$ and Min$'$ be defined as we defined them in the definition of k-Lexi-Pairing rules. We consider four scenarios (I-IV) depending on the values these take, and show that if P satisfies Strong Pareto, k-Compromise, and k-Threshold-Consistency then $P= \psi ^k(r_1, r_2)$.

I. Max ${\varvec{= r_1(h_1)}}$, Min ${\varvec{=r_2(h_2 )-k}}$, Max${\varvec{' = r_1 (h'_1)}}$, Min${\varvec{' =r_2 (h'_2)-k}}$

Case 1: Let Max > Max$'.$ Then $r_1 ( h_1 ) > r_1 ( {h'_1} )$.

Subcase 1.1: If $r_2 ( h_2) \geqslant r_2 ( {h'_2} )$ then $( h'_1, h'_2 )P ( h_1, h_2 )$ by Strong Pareto.

Subcase 1.2: If $r_2 ( h_2) < r_2 ( {h'_2} )$ then, since Max$' \geqslant$ Min$'$ implies $r_1 ( h'_1 ) \geqslant r_2 ( h'_2 )-k$, we have $r_1 ( h_1) > r_2 ( h'_2 ) - k$. Thus, $( h'_1, h'_2 )P ( h_1, h_2 )$ by k-Compromise.

Case 2: Let Max $=$ Max$'$ and Min > Min$'$. Then $r_1 ( h_1 )=r_1 ( {h'_1} )$ and $r_2 ( h_2 ) > r_2 (h'_2 )$. Then $( h'_1, h'_2 )P ( h_1, h_2 )$ by Strong Pareto.

Case 3: Let Max $=$ Max$'$ and Min $=$ Min$'$. Then $r_1 ( h_1 )=r_1 ( {h'_1} )$ and $r_2 ( h_2 )=r_2 ( {h'_2} )$. Thus, $( h_1, h_2 )= ( h'_1, h'_2 )$, which is ruled out.

II. Max ${\varvec{= r_2(h_2)-k}}$, Min ${\varvec{=r_1(h_1)}}$, Max${\varvec{' = r_2 (h'_2)-k}}$, Min${\varvec{' =r_1 (h'_1)}}$

Case 1: Let Max > Max$'.$ Then $r_2 ( h_2 ) > r_2 ( {h'_2} )$.

Subcase 1.1: If $r_1 ( h_1) \geqslant r_1 ( {h'_1} )$ then $( h'_1, h'_2 )P ( h_1, h_2 )$ by Strong Pareto.

Subcase 1.2: If $r_1 ( h_1) < r_1 ( {h'_1} )$ then, since Max$' \geqslant$ Min$'$ implies $r_2 ( h'_2 )-k \geqslant r_1 ( h'_1)$, which means $r_2 ( h'_2 ) \geqslant r_1 ( h'_1)+k$, we have $r_2 ( h_2) > r_1 ( h'_1 ) + k$. Thus, $r_1 ( h'_1) < r_2 ( h_2 ) - k$ and $( h'_1, h'_2 )P ( h_1, h_2 )$ by k-Compromise.

Case 2: Let Max $=$ Max$'$ and Min > Min$'$. Then $r_2 ( h_2 )=r_2 ( {h'_2} )$ and $r_1 ( h_1 ) > r_1 (h'_1 )$. Then $( h'_1, h'_2 )P ( h_1, h_2 )$ by Strong Pareto.

Case 3: Let Max $=$ Max$'$ and Min $=$ Min$'$. Then $r_2 ( h_2 )=r_2 ( {h'_2} )$ and $r_1 ( h_1 ) = r_1 (h'_1 )$. Thus, $( h_1, h_2 )= ( h'_1, h'_2 )$, which is ruled out.

III. Max ${\varvec{= r_2(h_2)-k}}$, Min ${\varvec{=r_1(h_1)}}$, Max${\varvec{' = r_1 (h'_1)}}$, Min${\varvec{' =r_2 (h'_2)-k}}$

Case 1: Let Max > Max$'.$ Then Max > Max$' \geqslant$ Min$'$ and thus $r_2 (h_2) > r_2 ( h'_2)$.

Subcase 1.1: If $r_1 ( h_1 ) \geqslant r_1 (h'_1)$ then $( h'_1, h'_2 )P ( h_1, h_2 )$ by Strong Pareto.

Subcase 1.2: If $r_1 ( h_1 ) < r_1 (h'_1)$ then, since Max > Max$'$, $r_2(h_2) - k > r_1 (h'_1)$ and thus $(h'_1, h'_2) P (h_1, h_2)$ by k-Compromise.

Case 2: Let Max $=$ Max$'$ and Min > Min$'$. Then Max $\geqslant$ Min >Min$'$ and thus $r_2 ( h_2 ) > r_2 (h'_2)$.

Subcase 2.1: If $r_1 ( h_1 ) \geqslant r_1 (h'_1)$ then $( h'_1, h'_2 )P ( h_1, h_2 )$ by Strong Pareto.

Subcase 2.2: If $r_1 ( h_1 )<r_1 (h'_1)$, Max = Max$'$ implies $r_2 ( {h_2} )-k = r_1 ( h'_1 )$, and Min > Min$'$ implies $r_1(h_1) > r_2 ( h'_2)-k$, and thus $( h'_1, h'_2 )P (h_1,h_2)$ by k-Threshold-Consistency.

Case 3: Let Max $=$ Max$'$ and Min $=$ Min$'$. Suppose that Max$' =$ Min$'$. Then $( h_1, h_2 )= ( h'_1, h'_2 )$, which is ruled out. Thus Max$' >$ Min$' =$ Min and $r_1(h'_1) > r_1(h_1)$. Also, Max > Min$'$ and thus $r_2 (h_2) > r_2(h'_2)$. Moreover, Max $=$ Max$'$ implies $r_1 (h'_1) = r_2(h_2)-k$ and Min $=$ Min$'$ implies $r_1 (h_1) = r_2(h'_2)-k$. Thus, $(h_1, h_2 ) P (h'_1, h'_2)$ by k-Threshold-Consistency.

IV. Max ${\varvec{= r_1(h_1)}}$, Min ${\varvec{=r_2(h_2)-k}}$, Max${\varvec{' = r_2 (h'_2)-k}}$, Min${\varvec{' =r_1 (h'_1)}}$

Case 1: Let Max > Max$'.$ Then Max > Min$'$ and thus $r_1 (h_1) > r_1 ( h'_1)$.

Subcase 1.1: If $r_2 ( h_2) \geqslant r_2 (h'_2)$ then $( h'_1, h'_2 )P ( h_1, h_2 )$ by Strong Pareto.

Subcase 1.2: If $r_2 ( h_2 ) < r_2 (h'_2)$ then, since Max > Max$'$, $r_1(h_1)> r_2 (h'_2)-k$ and thus $(h'_1, h'_2) P (h_1, h_2)$ by k-Compromise.

Case 2: Let Max $=$ Max$'$ and Min > Min$'$. Then Max > Min$'$ and thus $r_1(h_1) > r_1 (h'_1)$.

Subcase 2.1: If $r_2 ( h_2 ) \geqslant r_2 (h'_2)$ then $( h'_1, h'_2 )P ( h_1, h_2 )$ by Strong Pareto.

Subcase 2.2: If $r_2 ( h_2 ) < r_2(h'_2)$, Max = Max$'$ implies $r_1 (h_1) = r_2 ( h'_2 )-k$, and Min > Min$'$ implies $r_2(h_2) -k > r_1 ( h'_1)$, and thus $( h'_1, h'_2 )P ( h_1, h_2 )$ by k-Threshold-Consistency.

Case 3: Let Max $=$ Max$'$ and Min $=$ Min$'$. Suppose that Max$' =$ Min$'$. Then $( h_1, h_2 )= ( h'_1, h'_2 )$, which is ruled out. Thus Max$' >$ Min$' =$ Min and $r_2(h'_2) > r_2(h_2)$. Also, Max > Min$'$ and thus $r_1 (h_1) > r_1(h'_1)$. Moreover, Max $=$ Max$'$ implies $r_1 (h_1) = r_2(h'_2)-k$ and Min $=$ Min$'$ implies $r_1 (h'_1) = r_2(h_2)-k$. Thus, $(h'_1, h'_2 ) P (h_1, h_2)$ by k-Threshold-Consistency. $\square$

Claims 2.1, 2.2, 2.3 and 2.4 together prove Theorem 2.

Proof of Theorem 3

Claims 2.1, 2.2. and 2.3 in the proof of Theorem 2 readily generalize to prove that for all $k,t \in \{ 0,\ldots ,q-~1 \}$ the (k,t)-Lexi-Pairing rule satisfies t-Strong Pareto, (k,t)-Compromise and (k,t)-Threshold-Consistency, respectively. We now verify that for all $k,t \in \{ 0,\ldots ,q-1 \}$ the (k,t)-Lexi-Pairing rule satisfies t-Tie-Breaking. Fix $k,t \in \{0, \ldots , q-1\}$, and $(r_1,r_2)\in R \times R$. Suppose that the following holds for two hospital pairs $( h_1, h_2 )$ and $( h'_1, h'_2)$: ${\hat{r}}^t_1 (h_1,h_2) = {\hat{r}}^t_1 (h'_1,h'_2)$, ${\hat{r}}^t_2 (h_1,h_2) = {\hat{r}}^t_2 (h'_1,h'_2)$, $(h'_1, h'_2) \in G$ and $(h_1, h_2 )P^{(k,t)} ( h'_1, h'_2 )$, where $P^{(k,t)} = \psi ^{(k,t)} (r_1, r_2)$. Then Max $=$ Max$'$, Min $=$ Min$'$ and since $(h_1, h_2 )P^{(k,t)} ( h'_1, h'_2 )$, 4. in the definition of (k,t)-Lexi-Pairing rules implies that $(h_1,h_2) \in G$. This is a contradiction since ${\hat{r}}^t_1 (h_1,h_2) = {\hat{r}}^t_1 (h'_1,h'_2)$ and ${\hat{r}}^t_2 (h_1,h_2) = {\hat{r}}^t_2 (h'_1,h'_2)$ can only occur if one of the two hospital pairs is incompatible. Therefore, all (k,t)-Lexi-Pairing rules satisfy t-Tie-Breaking.

To prove the converse statement, namely that for all $k,t \in \{ 0, \ldots , q-1 \}$ if a pairing rule satisfies t-Strong Pareto, (k,t)-Compromise, (k,t)-Threshold-Consistency and t-Tie-Breaking then it is the (k,t)-Couple-Lexi-Pairing rule, we can use Claim 2.4 in the proof of Theorem 2 with the following modifications. In Case 3 of both I. and II., we have ${\hat{r}}^t_1 (h_1,h_2) = {\hat{r}}^t_1 (h'_1,h'_2)$ and ${\hat{r}}^t_2 (h_1,h_2) = {\hat{r}}^t_2 (h'_1,h'_2)$, which cannot be ruled out, unlike in the proof of Theorem 2, but we can infer in this case that one hospital pair is incompatible and the other one is compatible, since otherwise two distinct hospital pairs could not have identical rank numbers. Thus, without loss of generality we can let $(h_1,h_2) \notin G$ and $(h'_1,h'_2) \in G$. Then, given that P satisfies t-Tie-Breaking, $(h'_1, h'_2 )P ( h_1, h_2)$, which complies with 4. in the definition of (k,t)-Couple-Lexi-Pairing rules, since in Case 3 of both I. and II. we have Max $=$ Max$'$, Min $=$ Min$'$ and ${\hat{r}}_1 (h_1,h_2) = {\hat{r}}_1 (h'_1,h'_2)$. Furthermore, in Case 3 of both III. and IV. we can apply a similar argument to the above to show that if Max $=$ Max$'$, Min $=$ Min$'$ and Max$' =$ Min$'$ then t-Tie-Breaking implies that $(h'_1, h'_2 )P ( h_1, h_2)$, where $(h'_1, h'_2) \in G$, which complies with 4. in the definition of (k,t)-Couple-Lexi-Pairing rules. This leaves the case of Max $=$ Max$'$, Min $=$ Min$'$ and Max$'>$ Min$'$, and the proof can be concluded as shown in Case 3 of III. and IV., respectively, in the proof of Claim 2.4.