Resignation monotonicity — a winner resigns; can the count throw out someone who stayed?¶

Level: 301 · deep dive

One line: the obvious demand is that re-running a count after a winner resigns must not unseat anybody else; the non-obvious finding is that plain Approval is the only rule that can promise it — every proportional rule tested here breaks it, including all three of this engine's score-based PR rules, and that is a theorem rather than an accident.

Not the "resignation" this repo already talks about. Two pages use the word as a metaphor for a weak mandate — winning a race everyone wanted to leave (preference vs support, weak mandate). That is about how much a win is worth. This page is about a winner literally quitting after the count, and the two have nothing to do with each other.

The axiom¶

A committee is elected. Before it sits, one of its members resigns — health, a new job, a grant recipient moving institutions, a workshop organiser who can no longer attend. The seat has to be filled from the candidates who remain, so the rule is re-run on the shortened ballot. The question is what happens to everyone else.

Definition (Oh & Peters). For an instance E = (N, C, (Aᵢ), k) and a set T ⊆ C, write E − T for the instance with T struck from the candidate set and from every ballot. A winning committee W ∈ R(E) is resignation monotone if for every T ⊆ W there is a replacement set T′ ⊆ C \ W with (W \ T) ∪ T′ ∈ R(E − T). A rule is resignation monotone when every committee it outputs is. — Candidate Resignation Monotonicity in Approval-Based Committee Elections, arXiv:2608.06156, 6 Aug 2026, §3.

The paper's own framing of why this is not negotiable is worth keeping: "A politician who won their seat would rightly object to being removed because a colleague resigned." Note what the axiom does not ask — it says nothing about who gets the vacant seat. Fill it with anyone. It only forbids evicting the survivors.

This is a different axiom from its two neighbours in this folder, and they are close enough to conflate:

	what changes	what must not happen
Committee monotonicity	a seat is added	a sitting winner is dropped
Support monotonicity	a ballot gains an approval	the approved winner is dropped
Resignation monotonicity	a candidate leaves	the other winners are dropped

And none of the three is the monotonicity that sinks IRV, which is a single-winner, ranked-ballot pathology.

AV passes, and it is the only one that does¶

Approval scores each candidate independently of who else is elected. Strike a winner and the remaining scores are unchanged, so the other winners are still the top of the list and the vacant seat goes to whoever was next. That argument is the whole proof — and the paper's Theorem 3.1 shows it is also the only such argument available: among Thiele methods and sequential Thiele methods, AV is the only resignation monotone rule. Footnote 1 sharpens it: every other Thiele method admits an instance where all its output committees fail, so no tie-breaking rescues them.

Witness — five rules, five voters, two seats¶

Example 3.2. Five voters, four candidates, two seats:

v1:{c1}   v2:{c1,c3}   v3:{c1,c4}   v4:{c2,c3}   v5:{c2,c4}

PAV, Chamberlin–Courant, Monroe, leximax-Phragmén and Minimax Approval Voting each uniquely elect {c1,c2}. Then c1 resigns, and each of them uniquely elects {c3,c4} — so c2, who did nothing, is off the committee:

Abridged for the lesson — not verbatim engine output

before          {c1,c2}
c1 resigns      {c3,c4}      <- c2 is gone, and there was no tie to blame

Witness — the representation trap¶

Example 3.3 is the one that shows why this is structural rather than a quirk of one rule. Five voters, seven candidates, five seats: two voters share c1 and split on a second name, three voters approve the same slate of four.

v1:{c1,c2}   v2:{c1,c3}   v3:{c4,c5,c6,c7}   v4:{c4,c5,c6,c7}   v5:{c4,c5,c6,c7}

PAV, seq-Phragmén and the Method of Equal Shares all elect {c1,c4,c5,c6,c7}. c1 resigns. One seat is free — and now justified representation demands a seat for c2 and a seat for c3, because each of those voters is now a group large enough to deserve one and has exactly one candidate left to want. Two demands, one seat. Every JR-satisfying rule must break something, and what it breaks is the slate: none of the three returns all four of c4…c7 under any tie-breaking.

That generalises to Theorem 3.4: no resignation monotone rule satisfies JR — the weakest proportionality axiom in the literature. Perfect representation goes the same way (Proposition 3.5): on v1:{c1,c3} v2:{c1,c4} v3:{c2,c3} v4:{c2,c4} both {c1,c2} and {c3,c4} are perfectly representative, and once c1 resigns only {c3,c4} is.

So the trade is not "some rules are careless". Representation is why a rule fails: it is the property that makes a winner's seat depend on who else is seated, which is exactly the dependency resignation monotonicity forbids.

The question this repo asked next: the score rules¶

The paper covers approval-based rules. This engine's proportional rules take score ballots — Allocated Score (the STAR-PR rule BetterVoting runs), Sequentially Spent Score, and Reweighted Range Voting. They are not in the paper, and the question is well-posed: an approval ballot is a score ballot that only uses 0 and 5, so the paper's whole domain sits inside theirs.

All three fail. Each has a witness found by exhaustive search over every approval profile of its size, and in each one the rule has exactly one reachable committee before and after, so no tie-breaking is doing the work.

Allocated Score — the lone voter loses her seat¶

Two seats. Four of the five voters are one bloc: all of them back Bruno, and they split evenly on a second name. The fifth voter backs Ana and nobody else.

(No ballot art for resign_star_pr_seated_c4_b5 — draw it with build_style_ballot_images.py --from-yaml 04_Approval/03_Criteria/cases/resign_star_pr_seated_c4_b5.yaml.)

Row 1 = candidate names; each later row is one voter's 0–5 scores (a N × prefix = N identical ballots).

Ana,Bruno,Cleo,Dev
5,0,0,0     # the lone voter — Ana only
0,5,5,0     # bloc voter, Cleo wing
0,5,5,0     # bloc voter, Cleo wing
0,5,0,5     # bloc voter, Dev wing
0,5,0,5     # bloc voter, Dev wing

Allocated Score does what proportional representation is supposed to do: one seat to the four-voter bloc (Bruno), one to the lone voter (Ana).

--- Allocated Score Voting Method (2 winners) ---

[Allocated Score Voting]
 Tabulating 5 ballots to fill 2 seats.
Count × Ana,Bruno,Cleo,Dev
    2 ×   0,    5,   5,  0
    2 ×   0,    5,   0,  5
    1 ×   5,    0,   0,  0

[Allocated Score Voting: Round 1]
 The highest-scoring candidate wins a seat.
   Bruno         -- 20 -- First place
   Cleo          -- 10
   Dev           -- 10
   Ana           --  5
 Bruno wins a seat.

[Allocated Score Voting: Round 1: Ballot allocation round]
 Allocating 2+1/2 ballots.

[Allocated Score Voting: Round 1: Ballot allocation round: Round 1]
 Allocating 4 ballots at score 5.
 This allocation overfills the quota.  Returning fractional surplus.
 Allocating only 62.50% of these ballots.
 Keeping these ballots, but multiplying their weights by 3/8.
 4 ballots reweighted from 1 to 3/8.

[Allocated Score Voting: Round 2]
 The highest-scoring candidate wins a seat.
   Ana           -- 5     -- First place
   Cleo          -- 3+3/4
   Dev           -- 3+3/4
 Ana wins a seat.

[Allocated Score Voting: Winners — Allocated Score Voting Method (2 winners)]
 Ana
 Bruno

Bruno resigns. Nobody re-marks a ballot; Ana's supporter never mentioned Bruno at all:

[Divergence from STAR]
  STAR    = Cleo
  RCV-IRV = Dev   (differs from STAR)
  Note: no ballots had tied scores, so RCV-IRV vs STAR here is a genuine
        method difference, not a tie-breaking artifact.
  Note: Ranked Robin (RCV-RR) agrees with STAR, so RCV-IRV is the lone
        outlier — the classic center-squeeze signature.
  Full round-by-round reports (generated for review):
  RCV-IRV rounds: cases_tabulated/resign_star_pr_after_bruno_c3_b5_RCV-IRV_tabulated.txt

--- Allocated Score Voting Method (2 winners) ---

[Allocated Score Voting]
 Tabulating 5 ballots to fill 2 seats.
Count × Ana,Cleo,Dev
    2 ×   0,   5,  0
    2 ×   0,   0,  5
    1 ×   5,   0,  0

[Allocated Score Voting: Round 1]
 The highest-scoring candidate wins a seat.
   Cleo          -- 10 -- Tied for first place
   Dev           -- 10 -- Tied for first place
   Ana           --  5
 There's a two-way tie for first.

*(Ties are resolved by choosing the tied candidate with the highest-priority official lot number.)*
    Lot-number priority order: ['Ana', 'Cleo', 'Dev']

[Tiebreaker: Lot Number Priority]
  Tie among: ['Cleo', 'Dev']
  Resolved: ['Cleo'] (selected by lot-number priority).

[Lot-decided tie — rare]
  ⚠ The ballots did not break this tie: Allocated Score Voting has one
    deterministic rung per seat — the round's weighted score total —
    and the tied candidates came back equal on it, so the pre-published
    LOT order chose among them — the result here was set by lot, not by
    the votes. No head-to-head or five-star rung runs on this path: a
    tie on the weighted total goes straight to the lot. Verify the tied
    candidates' totals in the round above.

[Allocated Score Voting: Round 1: Ballot allocation round]
 Allocating 2+1/2 ballots.

[Allocated Score Voting: Round 1: Ballot allocation round: Round 1]
 Allocating 2 ballots at score 5.

[Allocated Score Voting: Round 2]
 Tabulating 3 remaining ballots.
Count × Ana,Cleo,Dev
    2 ×   0,   5,  0
    2 ×   0,   0,  5
    1 ×   5,   0,  0

[Allocated Score Voting: Winners — Allocated Score Voting Method (2 winners)]
 Cleo
 Dev

Cleo and Dev — both wings of the bloc that just lost its own winner — take both seats, and Ana is out.

The mechanism is visible in the two counts and worth stating plainly, because it is the same mechanism as Example 3.3. Bruno was absorbing the bloc's voting power: electing him spent their four ballots down to a remainder of ⅜ each, which left Cleo and Dev at 3¾ against Ana's untouched 5. Take Bruno away and that power was never spent, so the bloc buys both seats outright. The thing that protected Ana's seat was the presence of a candidate she did not vote for.

Sequentially Spent Score fails on the identical profile, for the identical reason.

Reweighted Range Voting — a one-supporter winner evicted¶

Three seats, five voters. Fern and Gus each have a single supporter; Hana, Ivan and Juno share three overlapping ones.

(No ballot art for resign_rrv_seated_c5_b5 — draw it with build_style_ballot_images.py --from-yaml 04_Approval/03_Criteria/cases/resign_rrv_seated_c5_b5.yaml.)

Row 1 = candidate names; each later row is one voter's 0–5 scores (a N × prefix = N identical ballots).

Fern,Gus,Hana,Ivan,Juno
5,0,0,0,0     # Fern only
0,5,0,0,0     # Gus only
0,0,5,5,0     # Hana + Ivan
0,0,5,0,5     # Hana + Juno
5,0,5,5,5     # Fern + the whole Hana/Ivan/Juno slate

RRV elects Fern, Gus and Hana. Hana resigns, and it elects Fern, Ivan and Juno — Gus is evicted, and the vacated seat goes to the slate that lost its own winner. RRV also fails on the paper's own Example 3.3, where it agrees with PAV on the committee and then loses all four survivors.

RRV's failure is really the paper's theorem in disguise, and the other two are not. On 0/1 ballots RRV is sequential PAV (why), so Theorem 3.1 already covers it and the witness above is a re-derivation rather than news. Allocated Score and Sequentially Spent Score are a different animal — they follow the quota / STV lineage, spending ballots down as seats are filled, and are not Thiele methods at all. Nothing in the paper reaches them, which is why they needed their own witness.

rule	resignation monotone?	witness
Approval (AV / bloc)	✓ (theorem)	— Theorem 3.1
Allocated Score (STAR-PR)	✗	`resign_star_pr_seated_c4_b5.yaml` → `resign_star_pr_after_bruno_c3_b5.yaml`
Sequentially Spent Score	✗	the same profile
Reweighted Range Voting	✗	`resign_rrv_seated_c5_b5.yaml` → `resign_rrv_after_hana_c4_b5.yaml`
Bloc STAR	no violation found	exhaustive over 4 candidates, ≤6 voters — evidence, not a theorem

The tie-breaking trap, which is the reason this page exists at all¶

The score rules are sequential, so a tie inside the count is settled by a tie-breaking order, and running one of them once is not enough to establish anything.

On the paper's own Example 3.2, all three score rules look like they fail under this engine's default tie-break — and all three are perfectly well-behaved once the alternatives are enumerated, because after c1 resigns the three remaining candidates are tied at 10 apiece and {c2,c3} and {c2,c4} are both reachable. A first pass at this page recorded three violations there. There are none.

So every verdict above is taken over all tie-breaking orders (every permutation of the candidates, via starvote's predefined_permutation_tiebreaker), and a violation is reported only when the survivors appear in no reachable committee. That distinction matters beyond this page: a resolute engine answering a question about an irresolute axiom will manufacture failures.

What is established here, and what is not¶

Every ✗ is DEMONSTRATED. Run the witness, watch a seated winner get evicted. Proof by counterexample, and complete.
AV's ✓ is CITED, to Theorem 3.1 — not demonstrated, because "no profile anywhere violates this" is a universal claim no finite replay settles.
Bloc STAR's row is neither. It is an exhaustive sweep of every 4-candidate profile up to 6 voters (94,872 of them, 1,591 flagged by the default tie-break, none surviving the full sweep) plus a random search at larger sizes. That is real evidence and it is not a proof; if you want one, it needs an argument about why Bloc STAR's score does not depend on the rest of the committee.
The three ✗ score rows are this repo's finding, not the paper's. Oh & Peters do not discuss score rules. If the result matters to you, the witnesses are four runnable files and one script.

Reproduce it¶

.venv/bin/python 06_Other/abcvoting_tabulation_engine/resignation_check.py --verbose

That replays the paper's three witnesses through abcvoting and the four score witnesses through starvote, and exits non-zero if any stops reproducing. --search N hunts for fresh violations on random profiles; --sweep-bloc runs the exhaustive Bloc STAR sweep. Gated by tests/test_resignation_monotonicity.py, which also checks that each "after" case file really is its "before" file minus one column.

The Approval control is runnable too — resign_av_holds_c7_b5.yaml and resign_av_holds_after_kai_c6_b5.yaml are the paper's Example 3.3 counted by plain multi-winner Approval, which keeps all four survivors where PAV, seq-Phragmén and Equal Shares cannot.

What the paper does about it¶

Two escapes, both worth knowing before anyone proposes one here:

Fractional committees. If seats can be split, rules based on maximum flow in the election's payment network satisfy both resignation monotonicity and a proportionality axiom (GRP). Real boards do not have fractional members, so this is a theoretical route rather than a proposal.
The Maximum Payment Rule (MPR). An integral rule satisfying a relaxed form — after a resignation there is some way to refill the seats that guarantees PJR+. It is NP-hard in general, tractable in structured domains, and on party-list instances it reduces to the largest remainder method. Neither this engine nor BetterVoting implements it.

The paper's last section is a different worry with the same shape: a losing candidate recruiting weak extra candidates to change the result. Resignation monotone rules that are independent of losers are immune, as are many sequential rules — but PAV can be manipulated this way. That is the same complaint as this repo's spoiler pages, one committee up.

Related: the criteria table · committee monotonicity — a seat is added instead · support monotonicity — a ballot changes instead · proportionality for solid coalitions — the other place Allocated Score's quota-spending bites · the ABC rules themselves.