A derivation of a 3 candidate 1 winner

cvd_robert_d_richiepreferential voting method follows. Axioms that allow the conclusion to be made are stated.

[1] An axiom shall be that the method returns the right number of winners.

[2] An axiom of the method is P2. P2 is defined so that the set of winners is not altered when the following papers are added:

P2(0): {t:(A), t:(B), t:(C), t:(D),…};
P2(1): {-(N-1)t:(A), t:(AB), t:(AC), t:(AD),…} [a different ‘t’];
P2(2): {-(N-2)t:(AB), t:(ABC), t:(ABD),…};
P2(3): {-(N-3)t:(ABC), t:(ABCD),…}; and so on.

[3] An axiom is that there is an ‘aim of proportionality’. This axiom removes degrees of freedom remaining after the other axioms have imposed their strict constraints. Proportionality (here) is the idea that for each candidate, there is a internal total that has added to it the weight (or quantity) of every paper naming that candidate. If nw winners needed to be selected, then this proportionality axiom would make the nw winners be the nw candidates that have the most positive totals. This proportionality aim would ignore the order of the preferences. However the P1 axiom (introduced below) is strict and applied first and it makes the order of the preferences be not ignored.

[4] The proportionality aim is defined to select the First Past the Post winner when 1 winner is to be found and the papers have no 2nd preferences. So the winner of the 1 winner election with the 3 papers {a:(A),b:(B),c:(C)} is candidate A if and only if (b<a)&(c<a). [The notation “x:(AB+{C+)” means that there are x (A=1st,B=2nd) papers and the intersection of the set of winners with {A,B,C} is {B,C}. When there is no “+” then the winners might not be being noted.]

[5] The method is truncation resistant. Truncation resistance is defined to require that for each candidate, e.g. B, alterations of preferences after any preferences for candidate B, do not affect whether candidate B wins or loses. (“After” means preferences on the “right” side distant from the 1st preference.)

[6] An axiom of the method is that it satisfies P1. A definition of P1 is this: for each candidate, e.g. B, alterations of preferences after any preferences for candidate B, and optionally including preference B, never change candidate B from a loser into a winner. For papers where candidate B has the 1st preference, some (positive) part of its weight may be discarded. The definition is very similar to the definition of truncation resistance except that the preference itself may be altered.

[7] Using [3] and [4], in the 1 winner election, with papers {a:(AB),b:(B),c:(C)}, candidate A wins if and only if (b<a)&(c<a). The region where candidate A wins, is shown in Figure 1.

[8] Property P1 is about the same as both truncation resistance and monotonicity. Monotonicity is (here) defined to allow preferences to be shifted to the right (i.e. away from the first) with that never resulting in the candidate changing from a loser into a winner. Truncation resistance allows the preferences on the right to be reordered and added to and removed from, while the preferences of the candidate under consideration are being shifted to the right.

[9] Figure 2 illustrates axiom P1 casting a B-must-lose shadow, of the A wins region. Axiom P1 requires that candidate B must not change from a loser into a winner when (B) papers are altered in any way at all. This alteration includes making the weight more negative but no votes may be gained.

Note: P1 can’t be replaced with monotonicity and truncation resistance for the (Figure 2) alteration (B)-(C{B+), since (BC) papers have been excluded. If it were not for that, this argument could have been made:
(a) Must have the same win-lose state for B: (B)–>(BC); by truncation resistance.
(b) B won’t change from a loser into a winner for the alteration (BC)–>(C), by monotoncity.

[10] Using the axiom that the number of winners is correct, candidate C wins in the triangle (a<c)&(1/3<a) since both A and B lose there. Temporarily assume that the weights are normalised so that 1=(a+b+c