Voting by Committees under Constraints

Subcase 1.1.1: Assume B Ç D. Without loss of generality assume that
MB (B U {ж} , {ж}) ∩ TZ_f = {B U {ж} , {ж}} . Then we can generate, by an
additive preference with top on 0, the orderings D >-' {ж} >- ' B U {ж},
{ж} >■“ Dy^ CU {ж}, and {ж} >-³ В U {ж} >-³ D, by an additive preference
with top on B, the orderings D >’ BU {ж} >~⁴ {ж} , B U {ж} >-³ {ж} >-³ В,
and BU{x} >∙⁶ D >-⁶ {ж}. Moreover, by associating large negative values to
objects outside D U {ж}, we must be able to put these three alternatives at
the tops of the individual orderings. Therefore, we have a free-triple on the
elements of the range D, {ж}, and B U {ж}. Then the Gibbard-Satterthwaite
Theorem implies that there exists i ∈ N such that TV™ = W™ = {{i}}.

Subcase 1.1.2: Assume B = D. Because MB (D U {ж} , 0) ∩ TZ_f ≠ {{ж} , D}
then D U {ж} ∈ TZ_f. Then MB (D U {ж} , {ж}) ∩ TZ_f = {{^} , D U {ж}} ,
MB (D U {ж} , D) ∩ TZ_f = {D, D U {ж}} . Notice that MB (D, ⅛)Γ∖TZ_f =
{D}. Therefore, using an argument similar to the one already used in the
proof of Subcase 1.1.1, we have a free triple on elements of the range D.{x}
and D U {ж}, and again, the Gibbard-Satterthwaite Theorem implies that
there exists i ∈ N such that TV™ = TV™ = {{i}}.

Subcase 1.2: Assume MB (D U {ж} , 0) = {{ж} , D} .

Subcase 1.2.1: There exists C ∈ TZ_f, such that C ∩ (D U {ж}) ¢ {{ж} , D} .
Let C = C∏Di) {ж} and without loss of generality assume MB {C^,, C^,} ∩
TZ_f = C. Since MB {C, {ж}} ∩ TZ_f = {ж} and MB {C, D} Γ∖TZ_f = {D} we
have a free triple on elements of the range D, {ж} and C, implying that there
exists i ∈ N such that TV™ = TV™ = {{i}}, because у ∈ D.

Subcase 1.2.2: For all C ∈ TZ_f, C ∩ D U {ж} ∈ {{ж} , D} .

Claim 1 Assume that for all C ∈ TZ_f either {ж} Ç C or D Ç C. Then,
there exists A, B ∈ TZ_f and Z ∈ {{ж} , D} such that:

(1.1) MB (A, B) ∩ TZ_f = {A, B} .

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