Oops, what I said in my previous post was mostly wrong, because I was only considering A>W voters trying to establish A ~> W, not W>A voters trying to break it. Here's what I think is correct: Any preference data from any subelection containing both A and W can be used, *as long as each voter's ballot is truncated somewhere no further down than the highest ranked candidate of A and W, inclusive.* Every safe sub-election must contain both of A and W. Strictly speaking, others are safe as long as the truncation is informed by where the voter ranks A and W even if these are eliminated from the sub-election. Thus we lose nothing by saying "A and W must be included". So, for instance, a voter who votes Y>B>C>A>D>W>X has safe truncated ballots Y, Y>B, Y>B>C, and Y>B>C>A. A voter who votes Y>B>C>W>X has safe truncated ballots Y, Y>B, Y>B>C, and Y>B>C>W. As a bonus, this properly handles chains (A ~> B, B ~> C) too. A C>A voter who seeks to break B ~> C by burying A can't have any effect on B ~> C because C is ranked ahead of A on his honest ballot, and thus in the B ~> C calculation, his ballot will be truncated no further down than C, which removes all information about A. (Note that the case "first preferences in an election containing A and W" is a special case, because every ballot is implicitly truncated below the first rank since we only make use of first preference info.) Thus my "pairwise tiebreak" idea of the previous post, that A indirectly disqualifies B in one step, becomes: A disqualifies B in one step if there exists an A2 so that 1. A disqualifies B in every subelection not containing A2 (initial safe concept), and 2. A is ranked ahead of A2, implicitly or explicitly, on half or more of the ballots that make a distinction between the two, after truncating just below A or B, whichever is higher (corrected concept). Some details about the latter calculation: The possible ballots are: B > A, A2 (in some order) Makes no difference because the B>A voters are truncated just below B. A > B > A2 Truncated at A, hence implicitly A>A2 A2 > B > A Truncated at B, hence implicitly A2>A A > A2 > B Truncated at A, hence implicitly A>A2 A2 > A > B Truncated at A, hence explicitly A2>A ("implicitly" means "by treatment of unranked candidates as equal-last") So the second condition can be simplified to: in the sub-election {A, A2, B}, A's first preferences >= A2's first preferences. The presence of B prevents B-first voters from having an influence on the count. -km