Appendix: Proof of Proposition 1
Condition 1 requires that n/2 − x > y − z. Substituting (n − y − z) into this expression in place of x, removing parentheses, and simplifying gives
$$n/2 - n + y + z > y - z,$$
which further simplifies to Condition 1U. Thus z > n/4 puts Z, rather than Y, into the runoff with X under B′.
Condition 2 requires that Z beat X in the runoff under B′. This implies that Z also beats X under B. Therefore, Condition 2U is necessary for UMF, but it needs to be shown that it is also sufficient.
In the event that y
x
≥ y − z, all the first-preference ballots that X must gain at Y’s expense to make Y the plurality loser under B′ can come from the y
x
ballots that would in any case transfer to X in a runoff with Z under B, so Z beats X by the same margin under B′ as under B. If y
x
< y − z, it is evidently more difficult for Z to beat X under B′ than under B because, to the extent that y − z exceeds y
x
, Z loses and X gains [(y − z) − y
x
] transferred ballots from Y in the runoff. Therefore, it must be that
$$z \, + \, y_{z} - \left[ {\left( {y - z} \right) - y_{x} > x \, + \, y_{x} + \left( {y - z} \right) - y_{x} } \right].$$
Suppose to the contrary that
$$z + y_{z} - \left[ {\left( {y - z} \right) - y_{x} } \right] \le x + y_{x} + \left[ {\left( {y - z} \right) - y_{x} } \right].$$
Removing parentheses and rearranging terms, we get
Substituting (n − z) for (x + y) and further simplifying, we get z ≤ n/4, contradicting Condition 1U. Thus, given that Condition 1U holds, it follows that, if Z beats X under B, Z also beats X under B′ and therefore implies UMF.