Answer: \(\mathsf{NP}\)-complete
Theorem 4.7 — class: hardness
Given \(C\), \(T=\mathcal{L}(C)\), rational \(\mu\in\mathbb{Q}_{\ge0}^{T}\) with \(\sum_t\mu_t=1\), rational \(\rho\le\mu\), and fixed STV tie-breaking, let \(x=\operatorname{STV}(\mu)\). Decide whether there exist \(y\ne x\) and rational \(\eta\ge0\) with \(\sum_t\eta_t=\sum_t\rho_t\), \(\eta_t>0\Rightarrow x\succ_t y\), and \(\operatorname{STV}(\mu-\rho+\eta)=y\).
Let \(C\) have \(m\ge 2\) candidates and let \(T=\mathcal L(C)\) be the strict complete rankings.
A society is given by its support: the list of the \(\tau\) types with positive mass, each with its mass as a binary rational; \(\tau\) is the support size, not \(m!\). The reported society \(\mu\) satisfies \(\sum_{t\in T}\mu_t=1\).
For each listed type, the input specifies a binary rational suspect mass \(0\le\rho_t\le\mu_t\); unlisted coordinates are zero. Put \(s=\sum_{t\in T}\rho_t\).
Use single-winner STV: each ranking contributes its entire mass to its highest remaining candidate; eliminate one candidate of minimum total mass each round until one remains. A fixed total order on candidate labels selects the first tied candidate for elimination, identically in every election.
Put \(x=\operatorname{STV}(\mu)\). The decision problem asks whether there exist \(y\in C\setminus\{x\}\) and \(\eta\in\mathbb Q_{\ge0}^{T}\) such that \(\sum_t\eta_t=s\), \(\eta_t>0\Rightarrow x\succ_t y\), and \(\operatorname{STV}(\mu-\rho+\eta)=y\). Replacement mass may be split arbitrarily among all rankings, including previously absent types.
State running times in \(m\), \(\tau\), and the total input bit length \(L\).
Is this continuous STV coalitional possible-manipulators decision problem in \(\mathsf P\), or is it \(\mathsf{NP}\)-hard?
For finite unweighted electorates, the corresponding problem is \(\mathsf{NP}\)-complete even with one specified suspect voter ([source](https://arxiv.org/pdf/1404.2367)).
No result settling exactly this divisible-mass version was identified in the literature search.
The problem is \(\mathsf{NP}\)-complete under the stated sparse-support encoding.
The hardness proof below handles arbitrary splitting among previously absent rankings. Its key property is stronger than needed: in the constructed election, every replacement profile of the prescribed mass has winner either \(x\) or \(y\), and \(y\) can win only if the source exact-cover instance is positive.
The construction adapts the selector-and-cover-check structure behind the discrete hardness result in [Dey, Misra, and Narahari, Theorem 7](https://arxiv.org/pdf/1404.2367). The numerical parameters and the fractional soundness argument are given explicitly below.
Write \(\nu=\mu-\rho\). Fix a proposed elimination order \(e_1,\ldots,e_m\), with proposed winner \(y=e_m\ne x\), and let \(x=e_h\).
For round \(i\), put \(R_i=\{e_i,\ldots,e_m\}\). Let \(b_i(c)\) be the tally contributed to \(c\in R_i\) by \(\nu\). These constants can be computed from the \(\tau\) listed rankings in \(\operatorname{poly}(m,\tau,L)\) bit operations.
Introduce replacement-tally variables \(z_i(c)\). Impose:
The last condition is necessary because a replacement ranking with \(x\succ y\) cannot contribute to \(y\) while \(x\) remains.
Let \(\triangleleft\) denote the fixed priority order for elimination. The required tally comparisons are
\[
\]
\(b_i(e_i)+z_i(e_i)\le b_i(c)+z_i(c)\qquad(c\in R_i\setminus\{e_i\}).\)
The comparison must be strict whenever \(c\triangleleft e_i\), since otherwise the tie-breaking rule would eliminate \(c\).
These constraints are also sufficient to realize the replacement tallies by complete rankings. Indeed, define
\(f_{i,c}=z_{i+1}(c)-z_i(c)\ge 0\qquad(c\in R_{i+1}).\)
The total-mass equations give
\(\sum_{c\in R_{i+1}}f_{i,c}=z_i(e_i).\)
Thus the initial allocations \(z_1(c)\), together with transfers \(f_{i,c}\), form a flow on the acyclic graph whose candidate arcs go from earlier to later positions in the elimination order. Decompose this flow into source-to-\(y\) paths.
For each path, rank its candidates in path order. If the path contains \(x\), it already ranks \(x\) above \(y\). If it omits \(x\), its predecessor of \(y\) must be eliminated after \(x\): transfers into \(y\) before \(x\)'s elimination are zero. Insert \(x\) immediately before \(y\). This insertion does not change the path's voting behavior, because \(x\) has already been eliminated when that path reaches \(y\). Append every unused candidate after \(y\).
Assign each resulting complete ranking its path-flow mass. This realizes all \(z_i(c)\) and satisfies \(x\succ y\).
To implement the strict comparisons, introduce \(0\le\epsilon\le 1\), replace each strict comparison by
\(b_i(e_i)+z_i(e_i)+\epsilon\le b_i(c)+z_i(c),\)
and maximize \(\epsilon\). The prescribed order is feasible exactly when this rational LP is feasible and its optimum is positive. A genuine election supplies a positive minimum over its finitely many required strict gaps; conversely, positive \(\epsilon\) enforces every required strict comparison.
There are \(O(m^2)\) variables and constraints, with polynomial encoding length. Exact rational linear programming therefore decides fixed-order feasibility in \(\operatorname{poly}(m,\tau,L)\) time. Its feasible region is bounded, so positive feasibility also has a polynomial-bit rational witness.
Consequently:
The flow decomposition uses \(O(m^2)\) paths, so a positive instance also admits a replacement profile supported on \(O(m^2)\) complete rankings with polynomial-bit masses.
Reduce from Exact Cover by 3-Sets. Let the universe be \(\{1,\ldots,3q\}\), and let its three-element subsets be \(S_1,\ldots,S_k\). An exact cover consists of \(q\) of these subsets covering the universe.
We may assume \(q\ge 1\) and \(k\ge q\). Instances with \(k<q\) are immediately negative and can be mapped to any fixed negative target instance.
Work first with unnormalized weights. Set
\(\delta=\frac14,\qquad H=20k,\qquad B_i=10k+4i.\)
The candidates are
\(C=\{x,y,d_0,d_1,\ldots,d_{3q}\}\cup\{a_i,\bar a_i,b_i,\bar b_i,g_i:1\le i\le k\}.\)
Thus \(m=5k+3q+3\).
The honest profile \(v\) consists of the following weighted rankings. Every displayed prefix is completed by appending the remaining candidates in a fixed order.
| Weight | Ranking prefix | Indices |
|---|---|---|
| \(H-1\) | \(x\succ y\) | |
| \(H\) | \(y\succ x\) | |
| \(H-2(k-q)\) | \(d_0\succ x\succ y\) | |
| \(H-2\) | \(d_j\succ x\succ y\) | \(1\le j\le 3q\) |
| \(H\) | \(g_i\succ x\succ y\) | \(1\le i\le k\) |
| \(B_i-5\) | \(b_i\succ\bar b_i\succ x\succ y\) | \(1\le i\le k\) |
| \(2\) | \(b_i\succ d_j\succ x\succ y\) | \(j\in S_i\) |
| \(B_i-1\) | \(\bar b_i\succ b_i\succ x\succ y\) | \(1\le i\le k\) |
| \(2\) | \(\bar b_i\succ d_0\succ x\succ y\) | \(1\le i\le k\) |
| \(B_i-3\) | \(a_i\succ g_i\succ x\succ y\) | \(1\le i\le k\) |
| \(1\) | \(a_i\succ b_i\succ g_i\succ x\succ y\) | \(1\le i\le k\) |
| \(2\) | \(a_i\succ\bar a_i\succ g_i\succ x\succ y\) | \(1\le i\le k\) |
| \(B_i-3\) | \(\bar a_i\succ g_i\succ x\succ y\) | \(1\le i\le k\) |
| \(1\) | \(\bar a_i\succ\bar b_i\succ g_i\succ x\succ y\) | \(1\le i\le k\) |
| \(2\) | \(\bar a_i\succ a_i\succ g_i\succ x\succ y\) | \(1\le i\le k\) |
All weights are positive.
We will analyze \(v+\zeta\) for every nonnegative replacement profile \(\zeta\) of total weight \(\delta\), without imposing any preference restriction. At every round, its contribution to any candidate lies between \(0\) and \(\delta\).
For each \(i\), the initial honest tallies are
\(v(a_i)=v(\bar a_i)=B_i,\qquad v(b_i)=v(\bar b_i)=B_i+1.\)
We prove that the first \(3k\) eliminations occur in \(k\) consecutive stages. Stage \(i\) eliminates:
At the start of stage \(i\), the pair \(a_i,\bar a_i\) has honest tally \(B_i\), whereas its \(b\)-pair has honest tally \(B_i+1\). Replacement weight cannot bridge this gap, since \(\delta<1\).
Untouched later selector pairs have honest tally at least \(B_i+4\). Every guard initially has honest tally at least \(H-2k=18k\), and guards only gain honest weight during these stages. The candidates \(x,y,g_j\), and earlier surviving \(b\)-candidates also have larger tallies. The relevant bound is
\(B_i+2+\delta\le 14k+2+\delta<18k.\)
Therefore the next elimination is one of \(a_i,\bar a_i\), regardless of replacement splitting or tie-breaking.
If \(a_i\) is eliminated, it transfers honest weight \(2\) to \(\bar a_i\) and \(1\) to \(b_i\). Hence \(\bar b_i\), with honest tally \(B_i+1\), is uniquely the next minimum: both \(\bar a_i\) and \(b_i\) have honest tally \(B_i+2\).
If \(\bar a_i\) is eliminated, the symmetric calculation makes \(b_i\) uniquely the next minimum.
After that \(b\)-candidate is eliminated, its counterpart has honest tally either \(2B_i-3\) or \(2B_i+1\), both exceeding \(H\). The remaining \(a\)-candidate has honest tally \(B_i+2\), so it is eliminated before the next stage begins, because
\(B_i+2+\delta<B_i+4.\)
This proves the stage structure inductively.
Define the selected index set
\(J=\{i:b_i\text{ is eliminated during stage }i\}.\)
Each stage therefore makes a discrete set choice, even though the replacement weight is arbitrarily divisible.
After the first \(3k\) eliminations, both \(a\)-candidates of every stage are gone, and precisely one \(b\)-candidate per stage remains.
Let
\(c_j=|\{i\in J:j\in S_i\}|.\)
The honest tallies at this checkpoint are:
| Remaining candidate | Honest tally |
|---|---|
| \(x\) | \(H-1\) |
| \(y\) | \(H\) |
| \(d_j\), \(1\le j\le 3q\) | \(H-2+2c_j\) |
| \(d_0\) | \(H+2(q-|J|)\) |
| \(g_i\) | \(H+2B_i-1\) |
| \(\bar b_i\), if \(i\in J\) | \(2B_i-3\) |
| \(b_i\), if \(i\notin J\) | \(2B_i+1\) |
For the guards, each eliminated \(b_i\) sends weight \(2\) to every element of \(S_i\), whereas each eliminated \(\bar b_i\) sends weight \(2\) to \(d_0\).
For a surviving \(b\)-candidate, add its initial \(B_i+1\), the weight \(1\) from the first eliminated \(a\)-candidate, and the eliminated counterpart's bulk weight, respectively \(B_i-5\) or \(B_i-1\).
For \(g_i\), the two \(a\)-candidates contribute their two bulk blocks, their two mutual-transfer blocks, and the weight \(1\) whose designated \(b\)-candidate has already disappeared. Their total contribution is
\(2(B_i-3)+4+1=2B_i-1.\)
These calculations also show that no honest weight has yet reached \(x\) or \(y\).
The useful bounds are:
A guard's honest tally is at least \(H\) exactly when its associated condition holds. Thus all guards have honest tally at least \(H\) if and only if
\(|J|\le q\quad\text{and}\quad c_j\ge 1\text{ for every }j.\)
Because each selected set has three elements, covering all \(3q\) elements requires \(|J|\ge q\). Consequently these conditions hold exactly when \(J\) is an exact cover.
If \(J\) is not an exact cover, at least one guard has honest tally at most \(H-2\).
Suppose \(J\) is not an exact cover.
A deficient guard has total tally at most \(H-2+\delta<H-1\), while \(x\)'s total tally is at least \(H-1\). The \(b\)- and \(g\)-candidates exceed \(H+\delta\). Hence the next eliminated candidate is a guard.
Its entire honest tally transfers to \(x\), giving \(x\) honest tally at least
\((H-1)+18k=38k-1.\)
Meanwhile \(y\)'s total tally is at most \(H+\delta\). Every surviving \(b\)- or \(g\)-candidate has honest tally strictly above \(H+\delta\). Therefore, until \(y\) is eliminated, only other guards can be eliminated first; each such elimination increases \(x\)'s honest tally. In particular, \(x\) cannot be eliminated before \(y\).
When \(y\) is eliminated, its honest weight \(H\) transfers to \(x\), giving \(x\) honest tally at least
\(58k-1.\)
Every other remaining candidate has total tally at most \(48k-1+\delta\), which is strictly smaller.
At this point every other remaining honest ballot transfers next to \(x\): both \(a\)-candidates and the other member of each \(b\)-pair are gone. Thus no other candidate can acquire further honest weight while \(x\) remains. Its replacement contribution is always at most \(\delta\). The strict dominance persists, and \(x\) wins.
This argument did not assume any restriction on the replacement rankings.
Suppose \(J\) is an exact cover.
Every guard has honest tally at least \(H\), whereas
\(\operatorname{tally}(x)\le H-1+\delta<H.\)
The \(b\)- and \(g\)-candidates have still larger honest tallies. Thus \(x\) is uniquely the next eliminated candidate.
Its honest weight \(H-1\) transfers to \(y\), giving \(y\) honest tally at least \(2H-1=40k-1\).
The next eliminated candidate is a guard or a surviving \(b\)-candidate. Indeed, a guard has total tally at most \(22k+\delta\), below both \(y\)'s tally and every \(g\)-candidate's tally. Every guard and surviving \(b\)-candidate has honest tally at least \(H\), and its honest ballots now transfer directly to \(y\).
Therefore, after this next elimination, \(y\)'s honest tally is at least
\(3H-1=60k-1>48k-1+\delta.\)
As in the preceding case, every remaining honest ballot transfers next to \(y\); other candidates cannot acquire further honest weight. The dominance persists, and \(y\) wins.
We have proved, for every replacement profile of weight \(\delta\),
\(\operatorname{STV}(v+\zeta)=y\quad\Longleftrightarrow\quad J\text{ is an exact cover},\)
and otherwise the winner is \(x\). These conclusions hold for any fixed candidate-label tie-breaking order.
Define the reported suspect ranking by the prefix
\(r_0:\bar a_1\succ\bar a_2\succ\cdots\succ\bar a_k\succ x\succ y.\)
Give this ranking weight \(\delta\).
During stage \(i\), this ballot protects \(\bar a_i\), so \(a_i\) is eliminated first and \(\bar b_i\) second. After \(\bar a_i\) is eliminated, the ballot moves to \(\bar a_{i+1}\). Thus its selected set is \(J=\varnothing\). Since \(q\ge 1\), the cover check fails, and the reported winner is \(x\).
Now suppose an exact cover \(J^\ast\) exists. Define
\(u_i=a_i\text{ if }i\in J^\ast,\qquad u_i=\bar a_i\text{ otherwise},\)
and use the replacement ranking with prefix
\(r^\ast:u_1\succ u_2\succ\cdots\succ u_k\succ g_1\succ x\succ y.\)
Give it weight \(\delta\). In each stage it protects \(u_i\), selecting precisely the indices in \(J^\ast\). Hence \(y\) wins. The ranking explicitly satisfies \(x\succ y\), as required.
Conversely, if the X3C instance has no exact cover, Section 5 shows that every replacement profile of weight \(\delta\) elects \(x\). This includes profiles with arbitrary support and profiles satisfying the preference restriction for any proposed alternative winner.
Finally, put
\(N=\sum_t v_t+\delta,\qquad \mu=\frac{v+\delta e_{r_0}}{N},\qquad \rho=\frac{\delta e_{r_0}}{N}.\)
Here \(e_{r_0}\) denotes unit mass on ranking \(r_0\). These are valid rational input profiles, with \(\rho\le\mu\), total reported mass \(1\), and suspect mass \(s=\delta/N\).
Multiplying all weights by the same positive number preserves every tally comparison and tie. Therefore normalized replacement profiles \(\eta\) correspond exactly to unnormalized profiles \(\zeta=N\eta\), and the reduction's equivalence is unchanged.
The reduction has
\(m=5k+3q+3,\qquad \tau\le 13k+3q+4.\)
Its unnormalized weights are integers of magnitude \(O(k)\), except for \(\delta=1/4\). After normalization, numerators and denominators have \(O(\log(k+q))\) bits. Writing every complete ranking gives total encoding length
\(L=O((k+q)^2\log(k+q)).\)
The reduction therefore takes polynomial time in the source-instance length.
For arbitrary target instances, fixed-order feasibility takes \(\operatorname{poly}(m,\tau,L)\) time, and exhaustive order enumeration takes \(m!\operatorname{poly}(m,\tau,L)\) time. The polynomial-time fixed-order verifier establishes membership in \(\mathsf{NP}\); the reduction establishes \(\mathsf{NP}\)-hardness.
Thus the divisible-mass problem is \(\mathsf{NP}\)-complete, including when the reported suspect mass is supported on just one ranking.
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The prover’s claim is verified: the argument establishes \(\mathsf{NP}\)-completeness for the stated sparse-support, divisible-mass problem. In particular, the hardness argument addresses arbitrary splitting among previously absent rankings. I found no mathematical error or gap requiring repair.
With remaining candidates \(R_i\), replacement tallies are nonnegative and sum to \(s\). A surviving candidate’s replacement tally cannot decrease: its existing ballots stay, and it may receive transfers from the eliminated candidate.
The condition \(z_i(y)=0\) for \(i\le h\) is necessary because \(x\) remains through round \(h\), and every permitted replacement ranking places \(x\) above \(y\).
The elimination comparisons also encode tie-breaking correctly. If \(c\triangleleft e_i\), equality would eliminate \(c\), so the comparison against \(c\) must be strict. For candidates later in the priority order, a weak comparison suffices.
The stated identity
\[
\sum_{c\in R_{i+1}}f_{i,c}=z_i(e_i)
\]
follows directly from the two consecutive mass equations. Monotonicity makes every transfer nonnegative.
These transfers conserve flow at each eliminated candidate: its initial allocation plus accumulated incoming transfers equals its tally at elimination, which equals its outgoing transfer mass. Candidate arcs go strictly forward in the elimination order, so the graph is acyclic and its terminal candidate is \(y\).
The ranking construction preserves the specified voting behavior. In particular:
Thus the construction realizes the tallies using complete rankings satisfying \(x\succ y\). When \(s=0\), the empty replacement profile handles the zero-flow case.
Introducing a common \(\epsilon\) correctly handles the finitely many strict comparisons. Any realization gives positive gaps for all required strict inequalities; a sufficiently small positive \(\epsilon\le1\) preserves them. Conversely, a positive feasible \(\epsilon\) enforces every required strict comparison.
If there are no strict comparisons, \(\epsilon\) is unconstrained except for its bounds and can equal \(1\), so this edge case also works.
The LP has \(O(m^2)\) variables and constraints. Its rational coefficients have polynomial bit length: each honest tally is a sum of input rationals, and the total denominator bit length remains polynomial in \(L\). All replacement variables lie between \(0\) and \(s\), and \(\epsilon\) lies between \(0\) and \(1\), so the feasible region is bounded.
The invoked rational-LP facts apply: feasibility and optimization are polynomial in the encoded LP size, and a positive optimum has a polynomial-bit rational solution. The stated polynomial-size flow decomposition consequently gives a polynomial-size replacement certificate.
Guessing an elimination order therefore establishes membership in \(\mathsf{NP}\). Enumerating the orders gives the stated \(m!\operatorname{poly}(m,\tau,L)\) bound.
Under \(1\le q\le k\), every displayed honest weight is positive. The candidate count is \(5k+3q+3\).
The initial selector tallies check:
\[
v(a_i)=v(\bar a_i)=B_i,\qquad
v(b_i)=v(\bar b_i)=B_i+1.
\]
Replacement mass contributes at most \(\delta=1/4\) to any candidate in any round. It therefore cannot reverse any of the unit-or-larger gaps used in the stage argument.
The bounds separating selector candidates from guards and other candidates hold even at \(k=1\). In particular,
\[
14k+2+\delta<18k.
\]
Both selector branches have the claimed transfers:
After the subsequent \(b\)-elimination, its surviving counterpart has honest tally \(2B_i-3\) or \(2B_i+1\), each exceeding \(H\). The remaining \(a\)-candidate has tally \(B_i+2\) and is eliminated before the next stage.
Earlier surviving \(b\)-candidates remain above subsequent selector tallies. The induction therefore establishes precisely the claimed first \(3k\) eliminations, independently of arbitrary replacement splitting and tie-breaking.
The guard tallies are correctly computed as
\[
v(d_j)=H-2+2c_j,\qquad
v(d_0)=H+2(q-|J|).
\]
The surviving \(b\)-candidate tallies include its initial mass, the transferred unit from the first eliminated selector, and its counterpart’s bulk transfer. They are respectively \(2B_i-3\) and \(2B_i+1\).
For each \(g_i\), the transferred honest mass is
\[
2(B_i-3)+4+1=2B_i-1,
\]
giving tally \(H+2B_i-1\). No honest weight has reached \(x\) or \(y\) at this checkpoint.
The stated bounds are valid:
All guards have tally at least \(H\) exactly when \(|J|\le q\) and every element is covered. Since selected sets each contain three elements, covering \(3q\) elements requires at least \(q\) selected sets. With exactly \(q\), covering every element forces an exact cover.
Conversely, failure of either guard condition gives an honest guard tally at most \(H-2\). The integer spacing here is essential and is correctly used.
Failed cover. A deficient guard has total tally below \(H-1\), so a guard is eliminated before \(x\). Its honest mass transfers entirely to \(x\), raising \(x\)’s honest tally to at least \(38k-1\).
While \(y\) remains, surviving \(b\)- and \(g\)-candidates have tallies above \(H+\delta\), whereas \(y\)’s tally is at most \(H+\delta\). Consequently only guards can precede \(y\)’s elimination, and their honest transfers further increase \(x\)’s tally.
When \(y\) is eliminated, \(x\)’s honest tally reaches at least \(58k-1\), exceeding every other remaining candidate’s total tally. All other remaining honest ballots next transfer to \(x\); hence another candidate cannot gain honest mass that would undo this dominance. Arbitrary replacement transfers remain bounded by \(\delta\).
Successful cover. Every guard has honest tally at least \(H\), while \(x\)’s total tally is below \(H\), so \(x\) is uniquely eliminated next. Its honest mass raises \(y\)’s tally to at least \(40k-1\).
A guard remains below both \(y\) and every \(g_i\), so the next elimination is a guard or a surviving \(b\)-candidate. Its honest mass is at least \(H\) and transfers directly to \(y\), raising \(y\)’s tally to at least \(60k-1\). This exceeds every remaining competitor’s total tally, and the stated transfer structure preserves dominance.
Thus every replacement profile elects either \(x\) or \(y\), with \(y\) winning precisely when the induced selection is an exact cover. No preference restriction was used in this soundness argument.
The suspect ranking \(r_0\) protects \(\bar a_i\) in each successive stage, inducing \(J=\varnothing\). Since \(q\ge1\), this gives reported winner \(x\).
Given an exact cover, \(r^\ast\) protects the appropriate selector in every stage and induces exactly that cover. It explicitly places \(x\) above \(y\), so it is a permitted replacement ranking.
If no exact cover exists, the unrestricted soundness result rules out every alternative winner, including alternatives other than the designated \(y\).
Normalization preserves tally comparisons and ties. The resulting input has total reported mass \(1\), nonnegative honest mass, \(\rho\le\mu\), and suspect support consisting of one ranking. Scaling replacement profiles by \(N\) gives the asserted correspondence.
Counting the displayed ranking blocks gives \(\tau\le13k+3q+4\). Their weights and the normalization denominator have the claimed polynomial magnitude and logarithmic bit length. Writing complete rankings gives the stated encoding bound \(L=O((k+q)^2\log(k+q))\).
The reduction from the explicitly represented X3C problem is polynomial. Restricting to \(q\ge1\) removes trivial source instances; the \(k<q\) branch concerns immediately negative instances and can use a fixed negative target instance.
The [cited Theorem 7](https://arxiv.org/pdf/1404.2367) concerns discrete STV possible manipulators. Here it supplies historical context; the divisible-mass hardness conclusion rests on the independently specified construction and fractional soundness argument verified above.
No additional clarification is needed to verify the mathematical conclusion.
⟨structured verdict parsed — rendered above⟩
Tier: machine-proved, adversarially reviewed by an independent model. No human has checked this proof. Generated from pipeline artifacts by tools/atlas_build.py.