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Unformatted text preview: x i y for all i G 1 and y i x for all i / G 1 , such that F ( 1 , . . . , n ) determines y to be at least as preferable as x . Therefore, by I , b c . Similarly, if G 2 is not almost decisive, then c a . Thus, by transitivity b a , but since G is decisive, a b , a contradiction. Thus, G 1 or G 2 is almost decisive. Proof of the Theorem: By Par , the set N is decisive. By the Group Contraction Lemma, every decisive set that includes more than one member has a proper subset that is decisive. Thus, there is a set { i } that is decisive, namely, F ( 1 , . . . , n ) i ....
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 Fall '10
 aswa

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