A Characterization of Profiles in Rich Pairwise-aligned Domains

PROOF: One implication follows from Lemmas 3 and 4. To prove the remaining implication, we show that the domain of all preference profiles that do not admit n-cycles, n= 2 3 satisfies R1 and R2. Fix a profile A that does not admit n-cycles, n= 2 3 To prove that R1 is satisfied, we construct an extension of the partial ordering constructed in the proof of Lemma 5 (the construction did not rely on the strictness of preferences assumed in the lemma). We recursively construct the extension so that all proper coalitions are comparable while maintaining transitivity and acyclicity. Let relation k⊂ C × C be a transitive and acyclic extension of . Take any proper coalitions C and C ′ that are not comparable under k (if there are no such coalitions, then the extension is complete). Let k+1 ⊂ C × C be the smallest transitive extension of k ∪{(C C ′)}. Relation k+1 is transitive by definition and is acyclic, as otherwise either k would violate acyclicity or C and C ′ would be comparable under . Since there is a finite number of proper coalitions, this process terminates, producing the postulated extension of . The extension satisfies relation (3) from the proof of Lemma 5 because all pairs of proper coalitions C and C ′ with nonempty intersection are comparable under . In the remainder of the proof, let us refer to the extension as .1 To prove R1, we take three different coalitions C0, C, and C1, and agent a such that C0 a C1, and construct a preference profile A whose existence is postulated in R1. At least one of the coalitions C0 or C1 is proper, and because of symmetry, we may assume that C1 = A. Define the preference profile A