Voting by Committees under Constraints

Let D be such that

MB(D,{y})∩‰ = {D}              (8)

and let B be such that

MB(B,0)∩7⅞ = {B}.                 (9)

If y ∈ B then we are back to Case 1. Therefore, assume that y B. For
each z ∈ B we can apply case 1 and obtain that there exists i ∈ N such that
w? = Ж» = {{≈}}.

Subcase 2.1: Assume that {x, y} ∈ TZ_f . We claim that MB ({y} , B) GTZ_f =
{B}. To see it, assume that there exists C≠B such that C ∈ MB({y}, B)∩
TZ_f. Ify ∈ CthenC ∈ MB(D, {y})∩7v⅛ Contradictingcondition (??). Ify
C then CGB, contradicting the fact that C≠B because M Bl B. Φ')∏TZ_f =
{B}. Moreover, since MB({y},D) GTZ_f = {D} and MB ({y} , {x, y}) ∩
TZ_f = {x, y} we can generate all orderings on D, B, {x, y} (with these three
subsets on the top); therefore, there exists i ∈ N such that TV™ = TV™ =
{≡∙

Subcase 2.2: Assume that {rr,y} TZ_f. First suppose that MB ({y} , B) ∩
TZ_f = {B}. Since MB ({y} , D) ∩ TZ_f = {D} and MB ({y} , {ж}) ∩ TZ_f =
{ж} (remember, by condition (??) we know that у ∈ TZ_f) we can gener-
ate all orderings on D,B, and {ж} (with these three subsets on the top);
therefore, there exists i ∈ N such that W™ = W™ = {{i}}. Suppose
that MB({y},B) ≠ {B}. We claim that D = BU {y} and therefore
MB({y},B) = {B,D}. To see it, let C ∈ MB({y},B). If у ∈ C then,
by condition (??), C = D and C = DU {y}. If у C then CGB and, by
condition (??), C = B. Now, if MB ({y},B) ∩ TZ_f = {B,D} we can also
generate all orderings on D, B, and {ж} with two preferences: one with top
on у (orderings D >- ' B >- ' {ж}, D >-² {ж} >-² B, and {ж} >- ³ D >- ³ B) and

27

<<   25   26   27   28   29   30   >>

More intriguing information