Markus is right. One way of retaining monotonicity, I think, is to replace the Sets with objects that record the number of times that a A has beaten B.
Then for the pair ordering A>C, B>C C>D D>A, D>B, A>B Affirming A>C and B> C A(W):A(W) B(W):B(W) C(L):A(W),B(W),C(L) D(W):D(W) Affirming C>D A(W):A(W) B(W):B(W) C(L):A(W),B(W),C(L) D(L):A(W), B(W), C(L), D(L) Affirming D>A, D>B, A>B A(W):A(W) [A(W), B(W), C(L), D(L)] B(W):B(W) [A(W), B(W), C(L), D(L)]A(W) C(L):A(W),B(W),C(L)[A(W), B(W), C(L), D(L)][A(W), B(W), C(L), D(L)] D(L):A(W), B(W), C(L), D(L)[A(W), B(W), C(L), D(L)][A(W), B(W), C(L), D(L)] Now remove equal numbers of A from B and B from A. A(W):A(W) [A(W), C(L), D(L)] B(W):B(W) [A(W), B(W), C(L), D(L)] C(L):A(W),B(W),C(L)[A(W), B(W), C(L), D(L)][A(W), B(W), C(L), D(L)] D(L):A(W), B(W), C(L), D(L)[A(W), B(W), C(L), D(L)][A(W), B(W), C(L), D(L)] B is reclassified a Loser A(W):A(W) [A(W), C(L), D(L)] B(L):B(L) [A(W), B(L), C(L), D(L)] C(L):A(W),B(L),C(L)[A(W), B(L), C(L), D(L)][A(W), B(L), C(L), D(L)] D(L):A(W), B(L), C(L), D(L)[A(W), B(L), C(L), D(L)][A(W), B(L), C(L), D(L)] A wins
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