V (Q, σ(Q, s2)). Hence, σ(Q, j1) < σ(Q, s2) = σ(Q, j2), contradicting (13). Since j1 = j2,
fs2 (j2) < fs2 (j1).
Let s = s1, s′ = s2, i = j1 , and i′ = j2. We have just shown that fs′ (i′) < fs′ (i). By
Step 4, fs(i) < fs(i’). Define Is := As(Q)∖i and Is’ := As’(Q)∖i'. By Step 1(b,c), Is and
Is are disjoint sets such that |Is| = qs — 1 and ∣Is' | = qs' — 1. Moreover, by definition of
As(Q) and i, Is ⊆ Uf(i). Similarly, Is’ ⊆ Uf'(is). Hence, schools s and s’ together with
students i and i’ constitute an X-cycle. □■
Proof of Theorem 7.2 Follows from Lemma B.12 and Proposition B.13. ■
C Appendix: Proofs of Results in Section 8
Proof of Lemma 8.1 Let φ := γ, τ. We will prove that φ(Pik,Q-i)(i) = φ(Q')(i') for
all Q—i ∈ Q(k)11∖ or φ(Pk,Q-i)Piφ(Qi,Q-i) for some Q-i ∈ Q(k)I∖i. (This obviously
completes the proof as it implies that no strategy k-dominates Pik.)
Suppose φ(Pik, Q-i)(i) = φ(Q')(i') for some Q-i ∈ Q(k)I∖i. We have to show that
for some Q-i ∈ Q(k)I∖i, φ(Pik,Q-i)Piφ(Qi,Q-i). Suppose that for some Q- ∈ Q1'∖^l,
φ(Qi,Q-i)Piφ(Pik, Q-i). Since φ(Pik,Q-i)(i')Rii, we have s := φ(Qi, Q-i)(i) ∈ S. From
Lemma A.1 (for γ) and Lemma B.5 (for τ), φ(s, Q-i)(i) = φ(Qi, Q-i)(i).
Suppose s is also listed in Pik. Then,
φ(Pk,C^-i)(i)Riφ(P ′k ,Q-i)(i = φ(s,Q-i}(i) = s, (14)
where P'k is the preference relation obtained from Pik by putting ' in the first position.
The first relation follows from Lemma 4.2. The second relation follows from the fact that
the assignment by the DA/TTC algorithm does not change if a student makes more schools
acceptable and puts them below the school he is assigned to. Clearly, (14) contradicts
φ(Qi, Q-i)Piφ(Pk, Q-i). Hence, ' is not listed in Pik.
Let S := {s ∈ S∖s : sQis }. The fact that ' is not listed in Pik together with the
definition of Pik implies that there is a school s ∈ S listed in Pik with sPis. Let s* be the
Pi-best school among the schools s ∈ S listed in Pik with sPis.
Suppose φ = γ. Since φ(Qi,Q-i) ∈ NW (Qi,Q-i), ∖φ(Qi, Q-i)(s)∣ = qs for all s ∈ S.
Clearly, for all s ∈ S, i ∈ φ(Qi,Q-i)(s'). Also, for all s,t ∈ S with s = t, φ(Qi,Q-i)(s') ∩
φ(Qi, Q-i)(t) = 0. So we can define for j ∈ I∖i,
Q’j:=
s if j ∈ φ(Qi, Q-i)(s) for some s ∈ S,
0 otherwise.
44
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