Construction of the inheritance trees based on f

Fixa ∈ A. We proceed to construct an inheritance treeΓ_a =hTa,ζ_a^{N I},ζ^EO_a ifora∈ A. LetTabe a rooted tree that satisfies Condition(i)of Definition6.1. Letζ^EO_a : E(Ta) → A\ {a}be an edges-to-objects function that satisfies Condition (iii) of Definition6.1. We will defineζ_a^{N I} : V(Ta) → N, a nodes-to-individuals function, in accordance with property Condition(ii)of Definition6.1based on f.

LetP_N⁰ ⊆ P_N be the set of all preference profilesPN such thatτ(Pi) =afor alli∈ N.

Lemma D.4. There exists k∈ N such that f_k(PN) =a for all PN ∈ P_N⁰.

Proof of LemmaD.4. By Remark 4.1, for every given PN ∈ P_N⁰, there exists an individualk ∈ N such that fk(PN) = a. It remains to show that this individual is unique for all preference profile inP⁰_N, that is, fk(PN) = fk(P_N^′ ) = afor allPN,P_N^′ ∈ P⁰_N. Assume for contradiction that fj(PN) = fj^′(P_N^′ ) = a for some PN,P_N^′ ∈ P_N⁰ andj,j^′ ∈ Nsuch thatj6= j^′.

Since fj(PN) = a,τ(Pj) = a, andaPkfk(PN)for allk 6= j, by moving the preferences of the individuals k 6= j one by one from Pk to P_k^′, and by applying top-envy-proofness condition every time, we obtain f_j(P_j,P₋^′ _j) = a. Moreover, since f_j^′(P_N^′ ) = a and j 6= j^′, we have f_j(P_N^′ ) 6= a. This, together with the fact τ(P_j^′) = a, implies aP_j^′fj(P_N^′ ). However, the facts fj(Pj,P₋^′ _j) = a andaP_j^′fj(P_N^′ ) together contradict

strategy-proofness of f. This completes the proof of LemmaD.4.

By LemmaD.4, there existsi1 ∈ Nsuch that fi₁(PN) = afor all PN ∈ P_N⁰. Defineζ_a^{N I}(v¹_a) = i1where v¹_a is the root-node of Ta. Let (v¹_a, . . . ,v^r_a)with r ≥ 2 be a path from v¹_a to v^r_a in Ta. We define ζ_a^{N I} on {v^s_a |1≤ s≤r}in a recursive manner.

Assume thatζ^{N I}_a is defined on{v^s_a |1≤ s≤r−1}. Letζ_a^{N I}(v^s_a) =isfor alls=1, . . . ,r−1. We proceed to defineζ^{N I}_a onv^r_a. LetP_N^r−1 ⊆ P_N be the set of all preference profiles PN such thatPis = P^{(ζEOa (vsa,vsa+1);a)} for alls =1, . . . ,r−1, andτ(Pi) = aotherwise. Note that for allPN ∈ P^r_N⁻¹and alls,s^′ ∈ {1, . . . ,r−1}, τ(Pis)6=τ(Pi_s′)ifs6=s^′.

Lemma D.5. There exists k∈ N\ {i1, . . . ,ir−1}such that fk(PN) =a for all PN ∈ P^r_N⁻¹.

Proof of LemmaD.5.We first prove two claims that we will use to complete the proof of LemmaD.5.

Claim D.2. Let S = {h1, . . . ,hm} ( N be a set of distinct individuals with m < |A|and let {b1, . . . ,bm} ∈ A\ {a}be a set of distinct objects. Consider the preference profile PN such thatτ(P_hu) =bufor all u = 1, . . . ,m andτ(Pi) =a for all i∈/S. Then, there exists j∈ N\S such that fj(PN) = a.

Proof of ClaimD.2. By Remark4.1, for allc∈ {a,b1, . . . ,bm}, there existsjc ∈ Nsuch that fjc(PN) = c. It remains to showja ∈/S. Assume for contradiction thatja ∈ S. Let{j1, . . . ,jt−1} ⊆Sandjt∈/Sbe such that j1 = ja, fjs+1(PN) = τ(Pjs)for all 1≤ s ≤ t−1. SinceSis finite, to show such a sequence must exist, it is sufficient to show thatj1, . . . ,jt−1are all distinct. We show this in what follows. Assume for contradiction that lis the first index in the ordering 1, . . . ,t−1 for which there existsl < _l^′ ≤ t−1 such thatjl = jl^′. Suppose l = 1. The facts l = 1, j_l = j_l^′, j1 = ja, fja(PN) = a and fj_l′(PN) = τ(Pj_{l′ −1})together imply τ(Pj_{l′ −1}) = a. This is a contradiction since jl^′−1 ∈ S, which in particular meansτ(Pj_{l′ −1}) ∈ {b1, . . . ,bm}. Now, supposel>_{1. Thenjl} = j_l^′, fj_l(PN) =τ(Pj_l−1)and fj_l′(PN) =τ(Pj_{l′ −1})together imply

τ(P_jl−1) =τ(P_{jl′ −1}). (D.8) However, by our assumption onl, jl−16= jl^′−1. Becausejl−1,jl^′−1 ∈Sandjl−1 6= jl^′−1, by the construction ofPN,τ(Pj_l−1)6=τ(Pj_{l′ −1}), a contradiction to (D.8). This shows thatj1, . . . ,jt−1are all distinct.

By the construction of {j1, . . . ,jt}, {fjs(PN) | s = 1, . . . ,t} = {τ(Pjs) | s = 1, . . . ,t}. Define the allocationµ such that µ(i) = τ(Pi) for alli ∈ {j1, . . . ,jt} andµ(i) = fi(PN)for all i ∈ N\ {j1, . . . ,jt}. ClearlyµPareto dominates f(PN)atPN, which violates Pareto efficiency of f atPN. This completes the

proof of ClaimD.2.

Claim D.3. For all PN ∈ P_N^r−1and all s=1, . . . ,r−1, we have fis(PN) =τ(Pis). Proof of ClaimD.3. FixPN ∈ P_N^r−1. We prove this in two steps.

Step 1. In this step, we show that fis(PN)Pisa for all s = 1, . . . ,r−1. Assume for contradiction that aRi_s∗fi_s∗(PN) for somes^∗ ∈ {1, . . . ,r−1}. Consider the preference profile ˜PN such that ˜Pit = Pit for all t =1, . . . ,s^∗−1 andτ(P^˜i) =a, otherwise. By the recursive definition ofζ^{N I}_a ,

fi_s∗(P^˜N) =a. (D.9)

Sinceτ(P^˜i) = afor alli∈ N\ {i1, . . . ,is^∗−1}, (D.9) implies that fi_s∗(P^˜N) = τ(P^˜i_s∗)and fi_s∗(P^˜N)P^˜ifi(P^˜N)for alli ∈ N\ {i1, . . . ,is^∗}. Therefore, by moving the preferences of all the individuals i ∈ N\ {i1, . . . ,is^∗} from ˜Pito Pi, and by applying top-envy-proofness condition every time, it follows from the construction of ˜PN that

fi_s∗(P^˜i_s∗,P−i_s∗) =a. (D.10) By strategy-proofness, (D.10) implies

fi_s∗(PN)Ri_s∗a. (D.11)

By Claim D.2, there exists j ∈ N\ {i1, . . . ,ir−1}such that fj(PN) = a. Since j ∈ N\ {i1, . . . ,ir−1}and fj(PN) = a, (D.11) implies fi_s∗(PN)Pi_s∗a, a contradiction to our assumption. This proves fis(PN)Pisafor all s =1, . . . ,r−1.

Step 2. In this step, we show that fis(PN) =τ(Pis)for alls = 1, . . . ,r−1. Assume for contradiction that fis1(PN)6= τ(Pis1)for somes1 ∈ {1, . . . ,r−1}. Lets1, . . . ,sube themaximalsequence of distinct elements such that{s1, . . . ,su} ⊆ {1, . . . ,r−1}and fi_st+1(PN) = τ(Pi_st)for allt = 1, . . . ,u−1. Let j ∈ Nbe such that fj(PN) = τ(Pi_su). By the maximality assumption ofs1, . . . ,su, either j ∈ N\ {i1, . . . ,ir−1}or j = is1. We distinguish the following two cases.

CASE1: Supposej∈ N\ {i1, . . . ,ir−1}.

By the construction of su, we have fi_su(PN) 6= τ(Pi_su). Also, since su ∈ {1, . . . ,r−1}, by Step 1, fi_su(PN)Pi_sua. Combining the facts fi_su(PN)6=τ(Pi_su)and fi_su(PN)Pi_sua, we have

τ(Pi_su)Pi_su fi_su(PN)Pi_su a. (D.12) Also, sincesu ∈ {1, . . . ,r−1}, by the construction ofPN, we have Pi_su = P^(τ(Pisu);a). This, together with (D.12), implies

τ(Pi_su)≺ fi_su(PN)≺a or a≺ fi_su(PN)≺ τ(Pi_su). (D.13) Since j ∈ N\ {i1, . . . ,ir−1}, by the construction of PN, we have τ(Pj) = a. This, together with (D.13), implies

a Pj fi_su(PN)Pjτ(Pi_su). (D.14) Since fj(PN) = τ(Pi_su), (D.12) implies fj(PN)Pi_sufi_su(PN). Furthermore, since fj(PN) = τ(Pi_su), (D.14) implies f_isu(PN)P_j f_j(PN). However, the factsf_j(PN)P_isuf_isu(PN)and f_isu(PN)P_j f_j(PN)together contradict Pareto efficiency of f atPN.

CASE2: Supposej=is1.

By the construction of{s1, . . . ,su}andj, we have{fi_st(PN) | t = 1, . . . ,u} = {τ(Pi_st) | t = 1, . . . ,u}. Let µbe the allocation such that µ(i) = τ(Pi)for all i ∈ {ist | t = 1, . . . ,u}andµ(i) = fi(PN)for all

i∈ N\ {ist |t =1, . . . ,u}. Clearly,µPareto dominates f(PN)atPN, which violates Pareto efficiency of f atPN.

Case 1 and Case 2 together complete Step 2, and Step 1 and Step 2 together complete the proof of Claim

D.3.

Now we complete the proof of LemmaD.5. By ClaimD.2, for every givenPN ∈ P_N^r−1, there exists an individualk∈ N\ {i1, . . . ,ir−1}such that fk(PN) =a. It remains to show that this individual is unique for all preference profile inP_N^r−1, that is, f_k(PN) = f_k(P^˜N) =afor allPN, ˜PN ∈ P_N^r−1. Assume for contradiction that fj(PN) = fj˜(P^˜N) =afor somePN, ˜PN ∈ P_N^r−1andj, ˜j∈ N\ {i1, . . . ,ir−1}such thatj6= j.^˜

Consider the preference profile(P^˜_i1,P−i1)∈ P_N^r−1. SincePN,(P^˜_i1,P−i1)∈ P_N^r−1, by ClaimD.3, we have fi1(PN) = fi1(P^˜i1,P−i1). Using non-bossiness, fi1(PN) = fi1(P^˜i1,P−i1)implies

f(PN) = f(P^˜i₁,P−i₁).

Continuing in this manner, we can move the preferences of all individuals is, s = 0, . . . ,r−1, from the preferencePis to ˜Pis one by one and obtain

f(PN) = f(P^˜i1, . . . , ˜Pir−1,P−{i1,...,ir−1}). (D.15) The fact fj(PN) = a, together with (D.15), implies fj(P^˜i1, . . . , ˜Pir−1,P−{i1,...,ir−1}) = a. Since j ∈ N\ {i1, . . . ir−1}andτ(P_i) =afor alli∈ N\ {i1, . . . ir−1}, it follows from the fact f_j(P^˜_i1, . . . , ˜P_ir−1,P_{−{i1,...,ir−1}}) =a that fj(P^˜i1, . . . , ˜Pir−1,P−{i1,...,ir−1}) = τ(Pj)and fj(P^˜i1, . . . , ˜Pir−1,P−{i1,...,ir−1})Pifi(P^˜i1, . . . , ˜Pir−1,P−{i1,...,ir−1})for alli∈ N\ {i1, . . . ir−1,j}. Therefore, by moving the preferences of all the individualsi∈ N\ {i1, . . . ir−1,j} fromPi to ˜Pi, and by applying top-envy-proofness condition every time, we obtain

fj(Pj, ˜P−j) =a. (D.16)

Since fj˜(P^˜N) = a and j 6= j, we have^˜ fj(P^˜N) 6= a. Moreover, j ∈ N\ {i1, . . . ir−1} implies τ(P^˜j) = a.

Combining the facts fj(P^˜N)6=aandτ(P^˜j) =a, we obtainaP˜jfj(P^˜N). However, this, together with (D.16), contradicts strategy-proofness of f. This completes the proof of LemmaD.5.

By Lemma D.5, there exists ir ∈ N\ {i1, . . .ir−1}such that fir(PN) = a for all PN ∈ P_N^r−1. Define ζ^{N I}_a (v^r_a) =ir. This completes the recursive definition ofζ_a^{N I}, and thereby completes the construction ofΓ_a. Similarly for each object, an inheritance tree is constructed. Thus, we have constructed a collection of inheritance treesΓ, based on the assignment rule f.

Now, we prove f(PN) = f^Γ(PN)for allPN ∈ PN, where f^Γis the hierarchical exchange rule associated withΓ.

D.2.2 f(PN) = f^Γ(PN)for allPN ∈ P_N

FixPN ∈ P_N. We show f(PN) = f^Γ(PN). We prove this by induction on the stages of f^ΓatPN. Base Case:Assignments in Stage 1.

(i) fi(PN) = f_i^Γ(PN)for alli∈W¹(PN), and

(ii) fi(P_N^′ ) = fi(PN)for alli ∈ W¹(PN), whereP_N^′ ∈ PN is such that for alli ∈ W¹(PN)eitherτ(P_i^′) = fi(PN)orP_i^′ =Pi.

Proof of the Base Case.First, we prove a claim that we use in the proof of the Base Case.

Claim D.4. Let i∈ N and let a∈ E1(i,PN). SupposeP˜N ∈ P_N is such thatτ(P^˜i) =a. Then fi(P^˜N) =a.

Proof of ClaimD.4.By the definition of f^Γ,a∈ E1(i,PN)impliesζ^{N I}_a (v¹_a) =iwherev¹_a is the root-node of Ta.¹⁹ By the construction ofΓ_a,ζ_a^{N I}(v¹_a) =iimplies that

fi(P^¯N) =afor all ¯PN ∈ PN withτ(P^¯j) =afor allj∈ N. (D.17) Now we show fi(P^˜N) = afor all ˜PN withτ(P^˜i) =a. Consider the preference profile(P^˜i, ˆP−i)such that τ(P^ˆj) =a for allj6= i. By (D.17), we have fi(P^˜i, ˆP−i) = a. Sinceτ(P^˜i) = a, fi(P^˜i, ˆP−i) = a, and τ(P^ˆj) = a for allj6=i, we have fi(P^˜i, ˆP−i) = τ(P^˜i)and fi(P^˜i, ˆP−i)P^ˆjfj(P^˜i, ˆP−i)for allj6=i. Therefore, by moving the preferences of all the individualsj6=ifrom ˆPj to ˜Pj, and by applying top-envy-proofness condition every time, we have fi(P^˜N) =a. This completes the proof of ClaimD.4.

Now, we proceed to prove the Base Case. First we show(i)of the Base Case. Fixi∈W¹(PN). We com-plete the proof for(i)of the Base Case by using another level of induction on the number of individuals inC1(i,PN).

Base Case (for(i)of the Base Case). Suppose|C1(i,PN)|= 1. It follows from the definition of f^Γthat T1(i,PN)∈ E1(i,PN)andT1(i,PN) =τ(Pi). Therefore, by ClaimD.4, we have

f_i(PN) =T1(i,PN). (D.18)

By the definition of f^Γ,|C1(i,PN)|=1 means

f_i^Γ(PN) =T1(i,PN). (D.19)

By (D.18) and (D.19), we have fi(PN) = f_i^Γ(PN). This completes the proof of Base Case (for(i)of the Base Case). Note that since PN ∈ PN andi ∈ W¹(PN)are chosen arbitrarily, using similar logic as above, we have fj(P^˜N) = f_j^Γ(P^˜N)for all ˜PN ∈ P_N and allj∈W¹(P^˜N)with|C1(j, ˜PN)|=1.

19Recall thatΓ_a=hT_a,ζ^{N I}_a ,ζ^EO_a i.

Induction Hypothesis (for(i) of the Base Case). Let u ≥ 2. Assume that fi(PN) = f_i^Γ(PN)for |C1(i, PN)|= u−1. Assume, furthermore, that for all ˜PN ∈ P_Nand allj∈W¹(P^˜N)such that|C1(j, ˜PN)|= u−1, we have fj(P^˜N) = f_j^Γ(P^˜N).

We show fi(PN) = f_i^Γ(PN)for|C1(i,PN)|= u. LetC1(i,PN) ={j1, . . . ,ju}such that for alll=1, . . . ,u, T1(jl,PN)∈ E1(jl+1,PN), wherei= j1. Assume for contradiction that fj1(PN)6= f_j^Γ₁(PN).

Take ˆPj₁ =Pju and ˆPju = Pj₁. By the construction of ˆPj₁ and the definition of f^Γ, it follows thatτ(P^ˆj₁)∈ E1(j1,PN). Sinceτ(P^ˆj1)∈ E1(j1,PN), by ClaimD.4, we have

fj1(P^ˆj1, ˆPju,P−j1,ju) = fj1(P^ˆj1,P−j1) =τ(P^ˆj1). (D.20) By the definition ofC1(i,PN)and the construction of ˆPju, it follows that|C1(ju,(P^ˆj1, ˆPju,P−j1,ju))|= |C1(ju, (P^ˆju,P−ju))|=u−1. Therefore, by Induction Hypothesis (for(i)of the Base Case), we have

fju(P^ˆj1, ˆPju,P−j1,ju) = f_j^Γ_u(P^ˆj1, ˆPju,P−j1,ju), and (D.21a)

fju(P^ˆju,P−ju) = f_j^Γ_u(P^ˆju,P−ju). (D.21b) By the definition of f^Γ, we have

f_j^Γ₁(PN) =τ(Pj₁), and (D.22a) f_j^Γ_u(P^ˆj₁, ˆPju,P−j₁,ju) = f_j^Γ_u(P^ˆju,P−ju) =τ(P^ˆju). (D.22b) Since ˆPju =Pj₁, combining (D.21) and (D.22b), we obtain

f_ju(P^ˆ_j1, ˆP_ju,P−j1,ju) = f_ju(P^ˆ_ju,P−ju) =τ(P_j1). (D.23)

Since fj1(PN)6= f_j^Γ₁(PN)by our assumption, (D.22a) and (D.23) together imply

fju(P^ˆj1, ˆPju,P−j1,ju)Pj1fj1(PN). (D.24) By (D.20) and (D.23), we have

fh(P^ˆj₁, ˆPju,P−j₁,ju) = fh(P^ˆh,P−h)for allh= j1,ju. (D.25) Since ˆPj₁ =Pju, by (D.20), we have fj₁(P^ˆj₁, ˆPju,P−j₁,ju) =τ(Pju), which in particular means

fj₁(P^ˆj₁, ˆPju,P−j₁,ju)Rjufju(PN). (D.26)

However, (D.24), (D.25) and (D.26) together contradict pairwise reallocation-proofness of f. This

com-pletes the proof of(i)of the Base Case. Note, furthermore, that sincePN ∈ P_Nandi∈W¹(PN)are chosen arbitrarily, using similar logic as above, we have

fj(P^˜N) = f_j^Γ(P^˜N)for all ˜PN ∈ PN and allj∈W¹(P^˜N). (D.27)

Now we show(ii)of the Base Case. FixP_N^′ ∈ PN such that for alli∈W¹(PN)eitherτ(P_i^′) = fi(PN)or P_i^′ = Pi. From(i)of the Base Case, we have fi(PN) = f_i^Γ(PN)for alli ∈W¹(PN). This, together with the definition of f^Γ, implies

fi(PN) =τ(Pi)for alli∈W¹(PN). (D.28) It follows from the construction of P_N^′ and (D.28) thatτ(P_i^′) = τ(Pi)for alli ∈ W¹(PN). This, together with the definition of f^Γ, implies

W¹(PN)⊆W¹(P_N^′ ), and (D.29a)

f_i^Γ(P_N^′ ) = f_i^Γ(PN)for alli∈W¹(PN). (D.29b) (D.29) and (D.27) together complete the proof of(ii)of the Base Case. This completes the proof of the Base

Case.

Now, we proceed to prove the induction step.

Induction Hypothesis: Fix a staget≥2. Assume that (i) f_i(PN) = f_i^Γ(PN)for alli∈W^t−1(PN), and

(ii) fi(P_N^′ ) = fi(PN)for all i ∈ W^t−1(PN), whereP_N^′ is such that for alli ∈ W^t−1(PN)eitherτ(P_i^′) = fi(PN)orP_i^′ =Pi.

We show

(i) fi(PN) = f_i^Γ(PN)for alli∈W^t(PN), and

(ii) fi(P_N^′ ) = fi(PN)for alli∈ W^t(PN), whereP_N^′ is such that for alli ∈ W^t(PN)eitherτ(P_i^′) = fi(PN) orP_i^′ =P_i.

First, we prove a claim.

Claim D.5. Let i ∈ N\W^t−1(PN) and let a ∈ Et(i,PN). Suppose P˜N ∈ P_N is such that P˜j = Pj for all j∈W^t−1(PN)andτ(P^˜i,A\F^t−1(PN)) =a. Then, fi(P^˜N) =a.

Proof of Claim D.5. Sincei ∈ N\W^t−1(PN)anda ∈ Et(i,PN), it follows from the definition of f^Γ that there exists r ≥ 1 such that there is a path (v¹_a, . . . ,v^r_a)in Ta from v¹_a (root-node of Ta) tov^r_a such that ζ^{N I}_a (v^r_a) = iand for alls = 1, . . . ,r−1, we have ζ_a^{N I}(v^s_a) ∈ W^t−1(PN)and f_ζ^ΓN I

a (v^s_a)(PN) = ζ^EO_a (v^s_a,v^s_a⁺¹). Note that for alls =1, . . . ,r−1, by(i)of the Induction Hypothesis, f_ζa^{N I}_(v^s_a)(PN) = f_ζ^ΓN I

a (v^s_a)(PN).

First, we show that fi(P^¯N) = a for all ¯PN ∈ P_N such that ¯Pj = Pj for all j ∈ W^t−1(PN)andτ(P^¯j) = a for all j ∈ N\W^t−1(PN). Fix ¯PN ∈ P_N such that ¯Pj = Pj for all j ∈ W^t−1(PN) andτ(P^¯j) = a for all j ∈ N\W^t−1(PN). Ifr = 1, thena ∈ E1(i,PN), and hence by ClaimD.4, we have fi(P^¯N) = a. Suppose r > _{1. Let S} = {ζ_a^{N I}(v^s_a) | s = 1, . . . ,r−1}. By construction, S ⊆ W^t−1(PN). Consider the preference profile ˆPN such that ˆPj = P^(fj(PN);a) for allj ∈ S, τ(P^ˆj) = afor all j ∈ W^t−1(PN)\S, and ˆPj = P^¯j for all j∈ N\W^t−1(PN). Since f_ζ^ΓN I

a (v^s_a)(PN) = ζ_a^EO(v^s_a,v^s_a⁺¹)and f_ζ^{N I}_{a (v}^s_a)(PN) = f_ζ^ΓN I

a (v^s_a)(PN), by the construction ofΓ_a, we have

fi(P^ˆN) =a. (D.30)

By the construction of ˆPN, τ(P^ˆj) = a for all j ∈ N\S. Since i ∈ N\W^t−1(PN), S ⊆ W^t−1(PN), and τ(P^ˆj) =afor allj∈ N\S, by (D.30), we have fi(P^ˆN) =τ(P^ˆi)and fi(P^ˆN)P^ˆjfj(P^ˆN)for allj∈W^t−1(PN)\S.

Therefore, by moving the preferences of all the individualsj∈W^t−1(PN)\Sfrom ˆPjtoPj, and by applying top-envy-proofness condition every time, we have

fi(P_N) =a, (D.31)

where P_j = P^ˆj for all j ∈/ W^t−1(PN)\SandP_j = Pj for all j ∈ W^t−1(PN)\S. By the construction ofP_N, for allj∈ W^t−1(PN), eitherτ(P_j) = f_j(PN)or P_j = P_j. Therefore, by(ii)of the Induction Hypothesis, we obtain

f_j(P_N) = f_j(PN)for allj∈W^t−1(PN). (D.32) Take j ∈ S. Consider the preference profileP_N^′′, where P_j^′′ = P_j andP_k^′′ = P_k for allk 6= j. Since for all k ∈ W^t−1(PN), eitherτ(P_k^′′) = fk(PN)or P_k = Pk, by (ii)of the Induction Hypothesis, fj(P_N^′′) = fj(PN). By (D.32), this means f_j(P_N^′′) = f_j(P_N). Since only individualjchanges her preference fromP_N toP_N^′′ and fj(P_N^′′) = fj(P_N), by non-bossiness, we have f(P_N^′′) = f(P_N). By moving the preferences of all individuals j∈ SfromP_jtoPjone by one and every time applying a similar logic, we conclude

f(P^¯N) = f(P_N). (D.33)

Combining (D.31) and (D.33), we have

fi(P^¯N) =a. (D.34)

Now we complete the proof of ClaimD.5. Take ˜PN such that ˜Pj = Pj for all j ∈ W^t−1(PN)andτ(P^˜i, A\F^t−1(PN)) = a. By (D.34) and the construction of ¯PN, we have fi(P^¯N) = τ(P^¯i)and fi(P^¯N)P^¯jfj(P^¯N)for all j ∈/ W^t−1(PN)∪ {i}. Therefore, by moving the preferences of all the individualsj ∈/ W^t−1(PN)∪ {i} from ¯Pjto ˜Pj, and by applying top-envy-proofness condition every time, we obtain

fi(P^¯i, ˜P−i) =a. (D.35)

Since f is strategy-proof, (D.35) implies

fi(P^˜N)R^˜ia. (D.36)

By the choice of ˜PN, we have ˜Pj =Pjfor allj∈W^t−1(PN). By(ii)of the Induction Hypothesis

f_j(P^˜N) = f_j(PN)for allj∈W^t−1(PN). (D.37)

Sinceτ(P^˜i,A\F^t−1(PN)) = a, (D.36) and (D.37) together imply fi(P^˜N) = a. This completes the proof of

ClaimD.5.

Now the proof of the induction step follows by using similar logic as for the proof of the Base Case with ClaimD.5in place of ClaimD.4.

Im Dokument Strategy-proofAllocationofIndivisibleGoodswhenPreferencesareSingle-peaked Mandal,PinakiandRoy,Souvik MunichPersonalRePEcArchive (Seite 24-32)