, (? ) ?) ? U i (a e i ?) ? U i (f (?) e * i (? ) ? ) ? U i

, Combining (10) and (11) yields ?i a e i : U i (f (?) e * i (? ) ? ) ? U i (a e i ? ) ? U i (f (?) e * i (? ) ? ) ? U i

, Since f (? ) = f (?), the preceding line is equivalent to ? R? . That f is evidencemonotonic now implies that f (? ) = f (? ) and so f (?) = f (? ). Finally, observe that since ? R?

, We show thatf is Maskin-monotonic on the extended outcome space. To prove this, fix any ?, (a e) ?f (?), and ? . We must show that if ?i b e : U i (a e i ?) ? U i (b e i ?) ? U i

, Condition (8) implies that (12) holds when a = f (?). Thus, Maskin monotonicity of (f e * * ) implies that (f (?) e) ?f (? ), which implies that f (?) = f (? ) and e ? e * * (? ), as required. Proof of Proposition 2. Consider the hard-evidence setting, i.e., separable preferences and for all i, ?, and e i , either c i (e i ?) = 0 or c i (e i ?) > sup a u i (a ?) ? inf a u i (a ?). First we prove sufficiency. For this, it suffices to show that (2) ? (*) and (**). So assume (2), hence Lemma 1 implies that f (? ) = f (?) and e ? e * * (? ), which together imply that (a e) ?f (? )

. Fix, For any e i ? E i (? ) \ E i (?), it is straightforward to check that the antecedent within (2) is always satisfied for any a (because c i (e i ?) > sup a u i (a ?) ? inf a u i (a ?)). Hence, for any a, the consequent of (2) must be true

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