Voting by Committees under Constraints

∑   ∑ √^pω

P^,≠P        B , fB _t в ,

xEF р (P_i 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, P_i)P_iF(P')∙, that is, F is not strategy-proof.

(b) Let {B₁, ...,B_q} be a partition of K' Ç K and consider any P ∈A".
i ∈ N, and P_i ∈A. Since for all p = 1, ...,q the functions F^Bp are strategy-
proof, we have that F^Bp(P^Bp)Rf^p F^Bp(Pf^p, P^^p); that is, for all p = 1, ...q,

∑ ≈⅛) >    £   af"(j∕),

ι∈F⅞(P⅞)           p∈P⅛(Pf^p ,pff)

where uf^p and uf^p are any pair of functions on B_p representing Pf^p and
F^Bp, respectively. Therefore, adding up,

Q                                          Q

∑ ∑ <Tω ≥ ∑   ∑  ⅛⅛)∙

p=l   _xeFb_p(pb_p)               p=l   _уерВр(£В_р,_рв_Р)

Hence, F{P')R_iF{P_i, P-_i); that is, F is strategy-proof. That {B_i,..., B_q}
is a Cylindric decomposition of Rp = K' and Rp = Bi + ... + B_q follow
g

immediately from the fact that.F (F) = Q F^Bp (P^bF} 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 B_p 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 F^Bp must be strategy-proof if F is (Proposition
4). If F^bp 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

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