Rewriting condition (??), we have
(У ∩ (B1∖B2)) U (A ∩ (B2∖B1)) U (A ∩ (B1 ∩ B2)) U (У ∩ Bc) ∈ Rf.
Therefore,
(A ∩ (B2∖B1)) U (A ∩ (B1 ∩ B2)) ∈ AC(B2). (4)
Then, by conditions (??) and (??), the fact again that B2 is a section, and
Remark 2,
(A ∩ (B2∖B1)) U (A ∩ (B1 ∩ B2)) U (A ∩ (B1∖B2)) U (У ∩ Bc) ∈ Rf.
This implies that (A ∩ B) U (У ∩ Bc) ∈ Rf Hence, B is a section of Rf. ■
Proposition 3 Any set Rf has a unique minimal cylindric decomposition.
Proof Assume not. Let {B^,...,B^1} and {B^,..., B^2} be two distinct
minimal cylindric decompositions of Rf. There exists at least one pair
such that Bpi ∩ B^2 ≠ 0. By Lemma 2, Bj(ι U B^2 is a section of Rf. By
Lemma 1, Bp1∖Bjζ is also a section of Rf implying, again by Lemma 1, that
{B^,..., B*1} was not minimal. ■
3.2 Additive Preferences
We can now state our first characterization.
Theorem 1 A social choice function F: An → 2κ is strategy-proof if and
only if it is voting by committees with the following properties:
(1) yVx and Vff are equal for all x and y in the same active component of any
section with two active components in Rff s minimal cylindric decomposition,
(2) Wx and Vff are complementary for all x and y in different active compo-
nents of the same section in Rffs minimal cylindrical decomposition, when
there are only two active components in this section, and
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