∑ ∑ √pω
+ Σ ≈⅛)
p∈p⅛(p⅛)
+ Σ ≈⅛).
⅛∈P⅛(^sp,pfp)
P,≠P B , fB t в ,
xEF р (Pi P ,P_r )
< ∑ ∑ F'W
>,'≠p 1∈Λv°<⅛')
where the equality follows from condition (??) and the inequality follows from
condition (??). Therefore, F(P~i, Pi)PiF(P')∙, that is, F is not strategy-proof.
(b) Let {B1, ...,Bq} be a partition of K' Ç K and consider any P ∈A".
i ∈ N, and Pi ∈A. Since for all p = 1, ...,q the functions FBp are strategy-
proof, we have that FBp(PBp)Rfp FBp(Pfp, P^p); that is, for all p = 1, ...q,
∑ ≈⅛) > £ af"(j∕),
ι∈F⅞(P⅞) p∈P⅛(Pfp ,pff)
where ufp and ufp are any pair of functions on Bp representing Pfp and
FBp, respectively. Therefore, adding up,
Q Q
∑ ∑ <Tω ≥ ∑ ∑ ⅛⅛)∙
p=l xeFbp(pbp) p=l уерВр(£Вр,рвР)
Hence, F{P')RiF{Pi, P-i); that is, F is strategy-proof. That {Bi,..., Bq}
is a Cylindric decomposition of Rp = K' and Rp = Bi + ... + Bq follow
g
immediately from the fact that.F (F) = Q FBp (PbF} for all P EAn.
p=l
Our strategy of proof for necessity relies heavily on invoking the Gibbard-
Satterthwaite Theorem for the case where there are more than three active
components in a section Bp of the minimal Cylindric decomposition of the
range. This is done by proving that, then, there will be three feasible out-
comes which agents can rank as the three most-preferred, and in any relative
order (a “free triple”). But FBp must be strategy-proof if F is (Proposition
4). If Fbp was non-dictatorial, we could use it to construct a non-dictatorial
and strategy-proof social choice function over our free triple, which we know
23
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