Unformatted text preview: x ± ² i y for all i ∈ G 1 and y ± ² i x for all i / ∈ G 1 , such that F ( ± ² 1 , . . . , ± ² 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
 Empty set, g2, Group Contraction Lemma

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