Voting by Committees under Constraints

is impossible by the Gibbard-Satterthwaite Theorem. As for the case where
a section has two active components only, notice that we can divide the ob-
jects of this section into two sets, such that all the elements in one of the sets
obtains when those on the other don’t, and vice-versa. Our restriction that
the committees corresponding to these two sets of objects are complementary
guarantees that no vote can lead to choose at the same time objects from
these two active components. Otherwise, no further restriction is imposed
on our committees by strategy-proofness when only two outcomes arise.

Now, we state and prove that whenever a section in the minimal Cylindric
decomposition of R_f contains more than two active components, then we get
a dictator. This is achieved by showing that a free triple always exists in this
case.

Proposition 5 Assume that the following properties of R_f hold: (1) the
minimal cylindric decomposition of R_f has a unique section and (2) ffR_F ≥
3. Then, there exists i ∈ N such that for all к ∈ R_f, TV™ = {{i}}.

Proof of Proposition 5 By conditions (1) and (2) there exists Z ∈ 2^κ
such that Z R_f . Without loss of generality first assume that there exists
x such that either Z U {x} ∈ R_f or Z∖{rr} ∈ R_f. Moreover, by rotating the
hypercube to locate Z to its origin and redefining all coordinates accordingly,
assume that Z = 0 and {ж} ∈ R_f. Let у ∈ R_f∖{x} be arbitrary. We will
show that there exists i ∈ N such that W''' = TV™ = {{i}}. We will
distinguish between two cases.

Case 1: There exists D E R_f such that у E D and MB (D, 0) ∩ R_f = {D}.

Subcase 1.1: Assume M B (DU {ж} , 0) ≠ {{^} , D} . Since M B(D, ^)Γ∖R_f =
{D} there exists B such that 0 ≠ B Ç D, B U {ж} ≠ D, and B U {ж} ∈
MB (DU{^},0)∩‰.

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