Preliminary remarks and lemmata

Remarks 5 and 6 identify circumstances under which, given any three different alter-natives x, y, z ∈ A, the set SSP() contains preferences with top on x that may or may not freely order y and z. The two remarks will be useful in some of the proofs that follow.

Remark 5 Let x, y, z ∈ A be three different alternatives and let be a semilattice overA. If x≻y, then

(i) there exists R^x_i ∈ SSP() such that zP_i^xy, (ii) there exists R^x_i ∈ SSP() such that yP_i^xz.

Proof Suppose x, y, z ∈ A are three different alternatives with x ≻ y. We first show that part (i) of the Remark holds. If z ∈ [y, x], then y /∈ [z, x] 6= ∅ and the result follows fromrichness. If z /∈[y, x], we will show that neither condition (i) nor condition (ii) in Remark 1 characterizing semilattice single-peakedness imply yR^x_iz.

First, notice that since x ≻ y, condition (i) in Remark 1 cannot be applied, and hence the ordering between y and z is unrestricted. Second, notice that although y x, condition (ii) in Remark 1 only implies x = sup{x, y}R^x_iy, and hence the

ordering betweeny and z is also unrestricted. Thus, there exists R_i^x∈ SSP() such that zP_i^xy. To show that part (ii) holds, notice that if z /∈ [y, x] the result follows fromrichness; whereas if z ∈[y, x] it follows that x≻z ≻y and parts (i) and (ii) of Remark 1 can not be applied, and hence the ordering betweenyandzis unrestricted.

Thus, there exists R^x_i ∈ SSP() such thatyP_i^xz.

Remark 6 Let x, y, z ∈ A be three different alternatives and let be a semilattice over A. If x ⊁ y, y ⊁ x and z /∈ [x,sup{x, y}], then there is R^x_i ∈ SSP() such that yP_i^xz.

Proof Let x, y, z ∈ A be three different alternatives and suppose x ⊁ y, y ⊁ x and z /∈ [x,sup{x, y}]. We will show that neither condition (i) nor condition (ii) in Remark 1 characterizing semilattice single-peakedness imply that, for each R^x_i ∈SSP(),zR_i^xy holds. First, notice that since x⊀y, condition (i) in Remark 1 cannot be applied. Second, notice that since x ⊀y, condition (ii) in Remark 1 only implies that sup{x, y}R^x_iy. But since z 6= sup{x, y}, there exists R^x_i ∈SSP()

such thatyP_i^xz.

To illustrate the content of Remarks 5 and 6 return again to Example 1 and observe, for instance, that

• x3 ≻x1 and there exist P_i^x3,Pb_i^x3 ∈ SSP() such thatx1P_i^x3x2 and x2Pb_i^x3x1.

• x₈ ⊁ x₉, x₉ ⊁ x₈, sup{x₈, x₉} = x₆ and x₄ ∈/ [x₈, x₆] and there is P_i^x8 ∈ SSP() such thatx9P_i^x8x4.

We next state and prove several lemmata that will be used in the proofs of the Propositions and Theorem 1.

Lemma 2 Let be a semilattice over A, let f :SSP()ⁿ →A be a strategy-proof and simple rule, and let R ∈ SSP()ⁿ. Assume x, y ∈ A are such that x ≻ y or y≻x and t(R) ={x, y}. Then, f(R)∈ {x, y}.

Proof Let the hypothesis of the Lemma hold. Without loss of generality, as-sume x ≻ y. The proof is by induction on the cardinality of N(R, x). Consider first N(R, x) = {i} (this implies N(R, y) =N \ {i}) and assume f(R) =z /∈ {x, y}. By Remark 5 (ii), there isR^x_i ∈ SSP() such thatyP_i^xz.Bytops-onlyness,f(R^x_i, R−i) = f(R) =z. Furthermore, by unanimity, f(R^y_i, R_−i) =y for any R^y_i ∈ SSP(). Then,

f(R^y, R−i) =yP^xz =f(R^x, R−i),

contradicting strategy-proofness. Hence, f(R)∈ {x, y}. Now, suppose R∈ SSP()ⁿ is such that, for all 1≤k ≤n−2,|N(R, x)|=k and f(R) ={x, y}. We want to see that ifR^′ ∈ SSP()ⁿ is such that|N(R^′, x)|=k+ 1, thenf(R^′)∈ {x, y}.To obtain a contradiction, suppose f(R^′) = z /∈ {x, y}. Let i ∈ N(R^′, x). By Remark 5 (ii), there is R^x_i ∈ SSP() such that yP_i^xz. By tops-onlyness, f(R_i^x, R^′_−i) = f(R^′) = z.

Furthermore, by the inductive hypothesis and tops-onlyness, f(R^y_i, R^′_−i)∈ {x, y} for any R^y_i ∈ SSP(). Then, in both case,

f(R_i^y, R_−i^′ )P_i^xz =f(R^x_i, R^′_−i),

contradicting strategy-proofness. Hence, f(R^′)∈ {x, y}.

Lemma 3 Let be a semilattice over A, let f :SSP()ⁿ →A be a strategy-proof andsimplerule and letk be such that1≤k < n. Assumex, y ∈Aare such that x≻y and there is R ∈ SSP()ⁿ such that t(R) = {x, y}, |N(R, y)| = k and f(R) = y.

Then, f(R) =e y for each Re∈ SSP()ⁿ such that |N(R, y)| ≥e k.

Proof Let the hypothesis of the Lemma hold. Let Re ∈ SSP()ⁿ be such that

|N(R, y)| ≥e k. We want to show that f(R) =e y. Let S ⊂ N be such that S ⊆ N(R, y) ande |S| = k. By anonymity, we can assume that N(R, y) = S. By tops-onlyness, f(ReS, R−S) = y. Let i ∈ N \ S. We claim that f(ReS∪{i}, R−(S∪{i})) = y. If not, f(Re_S∪{i}, R_−(S∪{i})) = w 6= y and, as x ≻ y, by Remark 5 (i) there is R^x_i ∈ SSP() such that wP_i^xy. Since N(R, y) = S, t(Ri) = x. By tops-onlyness, f(ReS, R^x_i, R−(S∪{i})) =f(ReS, R−S). Therefore,

f(ReS∪{i}, R−(S∪{i})) = wP_i^xy=f(ReS, R^x_i, R−(S∪{i})),

contradicting strategy-proofness. Hence, f(ReS∪{i}, R−(S∪{i})) = y. Continuing in the same way, we can successively change the preferences of each of the remaining agents

inN \S to finally obtain f(R) =e y.

Lemma 4 Let be a semilattice over A and let f : SSP()ⁿ → A be a strategy-proof andsimple rule. Then, for each R∈ SSP()ⁿ,

f(R)∈ [

i∈N

[t(Ri),supt(R)].

Proof Let the hypothesis of the Lemma hold. The proof is by induction on the cardinality of t(R). If R ∈ SSP()ⁿ is such that |t(R)| = 1, then the result is

trivially true by unanimity. Suppose that the result holds for any R ∈ SSP()ⁿ such that |t(R)| = k and 1 ≤ k < n. Consider now a profile R ∈ SSP()ⁿ such that t(R) = {x1, . . . , xk+1} and assume f(R) = y /∈ ∪^k+1_j=1[xj,supt(R)]. There are two cases to consider.

1. supt(R)∈ t(R).Without loss of generality, assume supt(R) =xk+1.Then,

∪^k+1_j=1[xj,supt(R)] =∪^k_j=1[xj, xk+1].

If |N(R, xk+1)| >1, take any i ∈N(R, xk+1) and consider R^′_i ∈ SSP() such that t(R^′_i) ∈ t(R)\ {x_k+1}. We claim that f(R_i^′, R−i) = z /∈ ∪^k_j=1[xj, xk+1]. If z =y this is obvious, by the contradiction hypothesis, so assume z 6=y and, to obtain a contradiction, supposez ∈ ∪^k_j=1[xj, xk+1].We now show that z6=xk+1, because otherwise,

f(R^′_i, R−i) =z =xk+1P_i^xk+1y=f(R^x_i^k+1, R−i),

contradicting strategy-proofness. Thus, xk+1 ≻ z. By Remark 5 (ii) there is R^x_i^k+1 ∈ SSP() such thatzP_i^xk+1y.Notice that, bytops-onlyness,f(R^x_i^k+1, R_−i) = f(R) =y. Therefore,

f(R_i^′, R_−i) =zP_i^xk+1y=f(R^x_i^k+1, R_−i),

contradicting strategy-proofness. Hence, f(R^′_i, R_−i) ∈ ∪/ ^k_j=1[xj, x_k+1]. If neces-sary (i.e., if there are still several agents with top atx_k+1), we repeat this process until we obtain a profile Re such that t(R) =e {x₁, . . . , xk+1}, |N(R, xe k+1)| = 1, and f(R) =e t /∈ ∪^k_j=1[xj, xk+1]. Hence, let N(R, xe k+1) = {i} and consider R^′_i ∈ SSP() such that t(R^′_i) ∈ t(R)e \ {xk+1}. Let f(R^′_i,Re−i) = w and e

x = sup{x1, . . . , xk}. Then, by the inductive hypothesis, w ∈ ∪^k_j=1[xj,x].e Therefore, x_k+1 ≻ ex w and, by Remark 5 (ii), there is R^x_i^k+1 ∈ SSP() such thatwP_i^xk+1t. Bytops-onlyness, f(R^x_i^k+1,Re_−i) =f(R) =e t. Therefore,

f(R^′_i,Re_−i) = wP_i^xk+1t=f(R^x_i^k+1,Re_−i), contradictingstrategy-proofness.

2. supt(R)∈/ t(R).Letex= supt(R).Consider anyx^⋆ ∈t(R) and let t^⋆(R) = {x ∈ t(R) : x x^⋆ or x^⋆ x}. Notice that, since supt(R) ∈/ t(R), 1 ≤

|t^⋆(R)|<|t(R)|. Let N^⋆ ={i∈N :t(Ri)∈t^⋆(R)} and leti^⋆ ∈N(R, x^⋆).

Claim: Leti ∈N^⋆ and letR^′_i ∈ SSP() be such thatt(R^′⋆_i (R).Then, f(R^′_i, R−i) ∈ ∪/ ^k+1_j=1[xj,x].e

Let i ∈ N^⋆, xi = t(Ri), and z = f(R^′_i, R_−i). If z = y this is obvious, by the contradiction hypothesis, so assume z 6= y and, to obtain a contradiction, suppose z ∈ ∪^k+1_j=1[xj,ex].Then, there is R^x_iⁱ ∈ SSP() such that

zP_i^xiy, (3)

where, remember, y = f(R). To see that (3) holds, there are two cases to consider. First, assume thatz xi.Asy /∈ ∪^k+1_j=1[xj,ex],it follows thaty /∈[xi, z]

and (3) is implied byrichness. Second, assume thatz xi.Ifxi z,(3) follows from Remark 5 (ii). If xi z, and since y /∈[xi,ex],(3) follows from Remark 6.

Bytops-onlyness,f(R^x_iⁱ, R−i) =f(R) =y. Therefore, f(R^′_i, R−i) =zP_i^xiy=f(R^x_iⁱ, R−i),

contradictingstrategy-proofness. Hence, f(R^′_i, R−i)∈ ∪/ ^k+1_j=1[xj,x] and the Claime is proved.

Using the Claim, we proceed by changing the preferences of all the members of N^⋆ except i^⋆. We obtain a new profile Re ∈ SSP() such that t(R)e ⊂t(R), N(R, xe ^⋆) ={i^⋆}, and f(R)e ∈ ∪/ ^k+1_j=1[xj,x].e

To finish the proof, consider R^′_i⋆ ∈ SSP() such that t(R^′_i⋆) ∈ t(R)\t^⋆(R).

Let t = f(R), we = f(R^′_i⋆,Re_−i^⋆) and xb = sup[t(R)\ t^⋆(R)]. Then, by the inductive hypothesis, w ∈ S

xj∈t(R)\t^⋆(R)[xj,bx]. Since sup{x^⋆, w} = xe and t /∈ [x^⋆,ex],by Remark 6, there isR^x_i^⋆⋆ ∈ SSP() such thatwP_i^x⋆⋆t.Bytops-onlyness, f(R^x_i^⋆⋆,Re−i) =f(R) =e t. Therefore,

f(R_i^′⋆,Re−i^⋆) =wP_i^x⋆⋆t =f(R^x_i^⋆⋆,Re−i^⋆), contradictingstrategy-proofness.

Hence, f(R)∈ ∪^k+1_j=1[xj,supt(R)].

Lemma 5 Let be a semilattice over A and let f : SSP()ⁿ → A be a strategy-proof and simple rule. Let R ∈ SSP()ⁿ, x ∈ A, i ∈ N and R^x_i ∈ SSP(). If x≻t(Ri), x≻f(R) and f(R)∈/ [t(Ri), x], then

f(R^x_i, R−i) =f(R).

Proof Let the hypothesis of the Lemma hold. First, notice that

f(R^x_i, R−i)∈ {f(R), x}. (4)

Otherwise, since x ≻ f(R), by Remark 5 (ii) there is Re^x_i ∈ SSP() such that f(R)Pe_i^xf(R^x_i, R_−i). Bytops-onlyness, f(R^x_i, R_−i) =f(Re^x_i, R_−i). Therefore,

f(R)Pe_i^xf(Re^x_i, R_−i),

contradicting strategy-proofness. Second, assume f(R^x_i, R−i) = x. Since f(R) ∈/ [t(Ri), x], by richness there is Rei ∈ SSP() such that t(Rei) = t(Ri) and xPeif(R).

Bytops-onlyness, f(R) =f(Rei, R−i). Therefore, f(R^x_i, R−i)Peif(Rei, R−i),

contradicting strategy-proofness. To conclude, as (4) holds and f(R^x_i, R_−i) 6= x, the

result follows.

Lemma 6 Let be a semilattice over A and let f : SSP()ⁿ → A be a strategy-proofandsimplerule. IfR∈ SSP()ⁿis such thatsupt(R)∈/ t(R)andsupt(R)≻ f(R),then there isRe ∈ SSP()ⁿsuch thatsupt(R) = supe t(R),supt(R)e ∈t(R),e and f(R) =e f(R).

Proof Let the hypothesis of the Lemma hold. Take i ∈ N such that f(R) ∈/ [t(Ri),supt(R)] (such i exists, because otherwise f(R) = supt(R)). Let R^′_i ∈ SSP() be such that t(R^′_i) = supt(R) and consider the profile Re = (R^′_i, R−i).

Then, supt(R) = supe t(R) ∈ t(R) and sincee f(R) ∈/ [t(Ri),supt(R)] holds, by

Lemma 5, f(R) =e f(R).

Lemma 7 Let be a semilattice over A, let f :SSP()ⁿ →A be a strategy-proof andsimplerule and letk be such that1≤k < n. Assumex, y ∈Aare such that x≻y and there is R ∈ SSP()ⁿ such that t(R) = {x, y}, |N(R, y)| = k and f(R) = y.

Then, f(R)e ∈ {y,supt(R)}e for all Re∈ SSP()ⁿ such that |N(R, y)|e < k.

Proof Let the hypothesis of the Lemma hold. Let Re ∈ SSP()ⁿ be such that

|N(R, y)|e = k −1. If k = 1 then t(R) =e {x}, and by unanimity, f(R) =e x = supt(R) and the statement holds trivially. Supposee k >1. Let i∈N(R, y) and set S ≡ N(R, y)\ {i}. By anonymity, we can assume S = N(R, y).e By tops-onlyness, f(ReS, R_−S) =f(R). Letj ∈N(R, x). First, we show that

f(ReS∪{j}, R−(S∪{j})) =y. (5)

Otherwise, since x≻y, by Remark 5 (i) there is R^x_j ∈ SSP() such that f(ReS∪{j}, R−(S∪{j}))P_j^xy.

Bytops-onlyness, f(ReS, R^x_j, R−(S∪{j})) =f(ReS, R−S) =y. Therefore, f(ReS∪{j}, R−(S∪{j}))P_j^x(ReS, R^x_j, R−(S∪{j})),

contradicting strategy-proofness. Thus, (5) holds. Continuing in the same way, we can change the preferences of each of the remaining agents in N(R, x) to obtain

f(Re_−i, Ri) =y. (6)

Next, we show that f(R)e ∈ {y,supt(R)}.e Let s ≡ supt(R) and, to get a con-e tradiction, suppose f(R)e ∈ {y, s}./ Then, as f(R)e ≺ s by Lemma 4, we can assume that Re is such that s ∈ t(R) by Lemma 6. Bye anonymity, assume i ∈ N(R, s).

Since s ≻ f(R),e by Remark 5 (i) there is R^s_i ∈ SSP() such that yP_i^sf(R).e By tops-onlyness,f(R) =e f(R_i^s,Re−i). Therefore, using (6),,

f(Ri,Re_−i)P_i^sf(R_i^s,Re_−i), contradicting strategy-proofness. Hence, f(R)e ∈ {y, s}.

We just proved the Lemma when Re is such that |N(R, y)|e = k − 1. In order to prove the statement for a profile Rb such that N(R, y) =b k −2 we use a similar reasoning and the fact that the result is true whenever there arek−1 tops in y. We successively apply the same reasoning to profiles in which the cardinality of the set

of agents with top in y is smaller.

Lemma 8 Let be a semilattice over A such that there is supA ≡ α and let f :SSP()ⁿ →A be a strategy-proof and simple rule. Assume f(R) =α whenever α∈t(R). Then, f(R) = supt(R) for each R ∈ SSP()ⁿ.

Proof Let the hypothesis of the Lemma hold. To obtain a contradiction, assume there isR ∈ SSP()ⁿsuch thatf(R)6= supt(R).Sincef(R)∈ ∪i∈N[t(Ri),supt(R)]

by Lemma 4,

supt(R)≻f(R). (7)

By Lemma 6, it is without loss of generality to assume that supt(R) ∈ t(R). Let i ∈ N(R,supt(R)). As α supt(R)≻ f(R), by richness, there is R^′_i ∈ SSP()

such that t(R^′_i) = supt(R) and αP_i^′f(R). Bytops-onlyness, f(R^′_i, R−i) =f(R). By the hypothesis, f(R^α_i, R−i) =α for any R_i^α∈ SSP(). Therefore,

f(R^α_i, R−i) =αP_i^′f(R) = f(R^′_i, R−i),

contradicting strategy-proofness. Hence, f(R) = supt(R).

Lemma 9 Let be a semilattice over A such that supA does not exist and let f :SSP()ⁿ →A be a strategy-proof and simple rule. Then, f(R) = supt(R) for each R ∈ SSP()ⁿ.

Proof Let the hypothesis of the Lemma hold. Notice first that, if |t(R)|= 1 then the result follows by unanimity. To obtain a contradiction, let R ∈ SSPⁿ, with

|t(R)| > 1, be such that f(R) 6= supt(R). Since f(R) ∈ ∪i∈N[t(Ri),supt(R)] by Lemma 4,

supt(R)≻f(R). (8)

By Lemma 6, it is without loss of generality to assume that supt(R)∈t(R).Let

s≡supt(R). (9)

First, notice that there is x ∈ A such that x≻ s. Since, by the hypothesis, there is no supA, there exists y∈A such that sy. If y≻s, take x=y; whereas if s y and ys, take x= sup{s, y}. We now proceed through several steps.

Claim 1: If S ⊆ N(R, s) and ReS ∈ SSP()^|S| is such that t(Rei) = x for each i ∈ S, then f(ReS, R−S) = f(R).

The proof is by induction on the cardinality of S. Suppose first that S = {i} and consider anyRei ∈ SSP() such thatt(Rei) =x. As f(R)∈/ [s, x], by Lemma 5,

f(Rei, R−i) =f(R). (10) Suppose now thatf(ReS, R−S) =f(R) and consideri∈N(R, s)\S.LetRei ∈ SSP() be such thatt(Rei) =x. As f(ReS, R_−S)∈/ [s, x], by Lemma 5,

f(Re_S∪{i}, R_−S∪{i}) = f(ReS, R_−S). (11) Thus, f(ReS∪{i}, R−S∪{i}) = f(R) and the Claim holds.

Claim 2: If S ⊇ N(R, s) and ReS ∈ SSP()^|S| is such that t(Rei) = x for each i ∈ S, then f(ReS, R_−S) = f(R).

The proof is by induction on the cardinality of S. Let Ne ≡ N(R, s), i /∈ Ne and consider Re_Ne ∈ SSP()^|N|e with t(Rej) = x for each j ∈ Ne and R^′_i ∈ SSP() with t(R^′_i) = s. Since s ≻ f(R) and f(R_Ne, R_−Ne) = f(R) by Claim 1, by Lemma 5 it follows that

f(R^′_i,Re_Ne, R_{−(N∪{i})e} ) =f(Re_Ne, R_−Ne).

Now consider Rei ∈ SSP() such that t(Rei) =x.Since x≻f(R) andf(Re_Ne, R_−Ne) = f(R) by Claim 1, by Lemma 5 it follows that

f(Re_N∪{i}e , R_{−(N∪{i})e} ) =f(Re_Ne, R_−Ne).

To finish the proof of Claim 2, suppose next that S ) N(R, s), ReS ∈ SSP()^|S|

is such that t(Rei) = x for each i ∈ S and f(ReS, R−S) = f(R). Let i /∈ S. First, let R^′_i ∈ SSP() such that t(R^′_i) = s. By an analogous reasoning as the one pre-sented to obtain (10), we can prove that f(R^′_i,ReS, R_−(S∪i)) = f(ReS, R_−S). Second, let Rei ∈ SSP() be such thatt(Rei) =x. By an analogous reasoning as the one pre-sented to obtain (11), we can prove that f(ReS∪i, R−(S∪i)) = f(ReS, R−S). Therefore, f(ReS∪i, R−(S∪i)) =f(R) and the inductive proof is complete.

Concluding. Let Re ∈ SSP()ⁿ be such that t(Rei) = x for each i ∈ N. By una-nimity, f(R) =e x. Applying Claim 2 when S = N we get f(R) =e f(R). Then, f(R) =x. But, as x ≻ s and s ≻ f(R), it follows by (8) and (9) thatf(R) ≻f(R), a contradiction. Hence, f(R) = supt(R) for each R ∈ SSP()ⁿ.