1. Since every {z i , z i+1 } an (A, b(z i ))-acceptable set by itself, as choice functions are simple, intensity function w must strictly decrease along circuit Q, which is impossible. Case 2: Suppose that for every z ? Z, {z} ? Z is A-acceptable. Suppose that {z 1 } is (A, s(z 1 ))-acceptable. From Lemma B.8 we can find a trail {z 2 , z 3 . . . z k } ? Z such that for every z i , either {z i , z i+1 } is a (A, b(z i ))-essential pair, (therefore w(z i ) > w(z i+1 )) or there are some y 1 . . . y l such that b(y j ) = s(z i ) for all 1 ? j ? l and {z i , y 1 . . . y l } is (A, b(z i ))-acceptable. trail terminates at the first occasion when {z i } is (A, b(z i ))-acceptable. Since the intensity strictly decreases, we cannot get back to a contract used earlier in the trail, so the trail must terminate;Suppose that choice functions satisfy full substitutability, simplicity, and IRC. Then an outcome is stable if and only if it is weakly trail-stable. Under simplicity all properties of trail-stable outcomes (including existence!) apply to stable outcomes
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