Voting by Committees under Constraints

Lemma 2 Let Bi and B₂ be two sections of Rp. Then B = Bi U B₂ is
also a section of Rp.

Proof Let B = Bi U B₂ and assume that B₁ and B₂ are sections of Rp.
Let X, Y ∈ Rp be arbitrary. They can also be written as

X = (A ∩ B) U (A ∩ B^c)

and

Y = (У ∩ B) U (У ∩ B^c).

To show that B is a section, it is sufficient to show that (A ∩ B) U (У ∩ B^c) ∈
R_f . Rewrite X and Y as

X = (A ∩ (B₁∖B₂)) U (A ∩ (B₂∖B₁)) U (A ∩ (B₁ ∩ B₂)) U (A ∩ B^c)

and

Y = (У ∩ (B₁∖B₂)) U (У ∩ (B₂∖B₁)) U (У ∩ (B₁ ∩ B₂)) U (У ∩ B^c).

Since B₁ is a section, (У ∩ (B₁∖B₂)) U (У ∩ (B₁ ∩ B₂)) and (A ∩ (B₁∖B₂)) U
(A ∩ (B₁ ∩ B₂)) belong to AC(Bi), and (У ∩ (B₂∖B₁)) U (У ∩ B^c) ∈ Cj¹ ((У ∩
(B₁∖B₂)) ∩ (У ∩ (B₁ ∩ B₂))). Therefore,

(A ∩ (B₁∖B₂)) U (A ∩ (B₁ ∩ B₂)) U (У ∩ (B₂∖B₁)) U (У ∩ B^c) ∈ Rp.

By definition of the range complement of (У ∩ (B₂∖B₁)) U (X ∩ (B₁ ∩ B₂))
relative to B₂,

(A ∩ (B₁∖B₂)) U (У ∩ B^c) ∈ ф((У ∩ (B₂∖B₁)) U (A ∩ (B₁ ∩ B₂))). (2)

Also, since X and Y belong to Rp and B₂ is a section,

(A ∩ B₂) U (У ∩ Bf ∈ Rp.                  (3)

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