## Appendix
Then, \(F(P(\underline{t}))\subseteq A\subseteq P(\underline{t})\). From the path-independent property, \(S=F(A)=F(P(\underline{t}))\). Figure 8: Four-partition. Using Lemma 2 again, \(\mathcal{D}_{F}(P(\underline{t}))=\mathcal{D}_{F}(A)=E\setminus B\), \(E\setminus\mathcal{D}_{F}(P(\underline{t}))=B\) and \(P_{G}(E\setminus\mathcal{D}_{F}(P(\underline{t})))=P_{G}(B)=\underline{t}\); therefore, \(\underline{t}\) is a fixed point. Proof of Theorem 14.: We know from Theorems 5 and 6 that there exist a greatest and a least stable score vector. Let \(\underline{t}^{1}\) and \(\underline{t}^{2}\) be two arbitrary stable score vectors. We want to show that they have a join and a meet. Using Lemma 6, \(t^{min}=\min(\underline{t}^{1},\underline{t}^{2})\) is valid. Let us start the score-decreasing algorithm from \(t^{min}\); from the algorithm, we get a stable score vector, \(\underline{t}\). From Theorem 5, \(\underline{t}\) is the biggest among all the stable score vectors smaller than or equal to \(t^{min}\). Therefore, \(\underline{t}=\underline{t}^{1}\wedge\underline{t}^{2}\), because for every stable vector, such that \(\underline{t}^{\prime}\leq t^{1},t^{2}\), \(t^{\prime}\leq t^{min}\); therefore, \(\underline{t}^{\prime}\leq t\). We finish by showing the existence of \(\underline{t}^{1}\vee\underline{t}^{2}\). There exists a common upper bound of \(\underline{t}^{1}\) and \(\underline{t}^{2}\), for example \(\underline{s}_{C}\). Since the lattice is finite, there has to be at least one least common upper bound. Suppose there exist two least common upper bounds: \(\underline{a}\) and \(\underline{b}\). Since \(\underline{t}^{1}\) is a lower bound of \(\underline{a}\) and \(\underline{b}\), \(\underline{t}^{1}\leq\underline{a}\wedge\underline{b}\); similarly, \(\underline{t}^{2}\leq\underline{a}\wedge\underline{b}\). Therefore, we found a common upper bound of \(\underline{t}^{1}\) and \(\underline{t}^{2}\) smaller than \(a\); a contradiction. Proof of Theorem 15.: _Three-stable \(\Rightarrow\) dominating_  
There are \(A\) and \(B\), such that \(F(A)=S=G(B)\). From Lemma 1, \(\mathcal{D}_{F}(S)\supseteq A\setminus S\). The same goes for \(G(B)\); so, \(\mathcal{D}_{G}(S)\supseteq B\setminus S\). Their union is  
\(\mathcal{D}_{F}(S)\cup\mathcal{D}_{G}(S)\supseteq((A\setminus S)\cup(B \setminus S))=E\setminus S\). Additionally, \(S\) does not dominate itself; so, \(\mathcal{D}_{F}(S)\cup\mathcal{D}_{G}(S)=E\setminus S\). _Dominating \(\Rightarrow\) four-stable_  
We know that \(\mathcal{D}_{F}(S)\cup\mathcal{D}_{G}(S)=E\setminus S\). Let \(A=E\setminus\mathcal{D}_{G}(S)\) and \(B=E\setminus\mathcal{D}_{F}(S)\). \(A\subseteq S\cup\mathcal{D}_{F}(S)\), so \(F(A)=S\). From Lemma 2, \(\mathcal{D}_{F}(A)=\mathcal{D}_{F}(S)=E\setminus B\). Similarly, \(\mathcal{D}_{G}(B)=\mathcal{D}_{G}(S)=E\setminus A\). With this \((A,B)\) pair, \(S\) is four-stable. _Four-stable \(\Rightarrow\) three-stable_  
There exists subsets \(A\) and \(B\) of \(E\), such that \(F(A)=S=G(B)\) and \(A\cap B=S\), and \(\mathcal{D}_{F}(A)=E\setminus B,\mathcal{D}_{G}(B)=E\setminus A\). Let \(D=E\setminus(A\cup B)\) and \(A^{\prime}=A\cup D,B^{\prime}=B\). Now, \(A^{\prime}\cup B^{\prime}=E\), \(A^{\prime}\cap B^{\prime}=A\cap B=S\), and from Lemma 2, \(\mathcal{D}_{F}(S)=\mathcal{D}_{F}(A)=E\setminus B=(A\setminus S)\cup D=A^{ \prime}\setminus S\). From Lemma 1, \(F(A^{\prime})=S\), \(G(B^{\prime})=G(B)=S\); so, with the \((A^{\prime},B^{\prime})\) pair, \(S\) is three-stable. Proof of Theorem 16.: In the third part of the proof of Theorem 15, we did not use that \(G\) was path independent. Proof of Theorem 17.: Let \(S\subseteq E\) be the enrollment realized from stable score vector \(\underline{t}\). Define \(A\) as the set of contracts above score vector \(\underline{t}\). Let \(B\) be the union of \(S\) and the set of contracts under score limit \(\underline{t}\). From all the accepted contracts above score vector \(\underline{t}\), the applicants choose contract set \(S\); so, \(F(S)=S\).