d4-conf-atal-2024-0012 (d4-conf-atal-2024-0012)

Given a rational finite-support society μ of unit-size transaction types and capacity shares \(0<κ<ρ<1\), let reports form a mass distribution ν; include the top ρ mass and confirm the top κ mass. Optimize over anonymous linear rebate rules \(R_C^K(ν)=a+\int Q^C_ν(q)\,dK(q)\) and \(R_U^K(ν)=a+\int Q^U_ν(q)\,dK(q)\), where \(K\) is a finite signed measure over included ranks and the quantiles use the leave-one-out, one-sided rank limits for confirmed and unconfirmed units. For every ν, require restricted truthfulness and user individual rationality, require \(κ R_C^K+(ρ-κ)R_U^K\le κ p_{\mathrm{in}}\), and maximize \(f\) subject to \(κ R_C^K\ge fκ p_{\mathrm{out}}\). Here \(p_{\mathrm{in}}\) is the lowest confirmed-side cutoff bid and \(p_{\mathrm{out}}\) the highest unconfirmed-side cutoff bid, allowing a gap between them. Return the optimal rule and its allocation and payments on μ.

A society is a nonatomic population of total mass \(1\), represented by its support: \(\tau\) nonnegative rational confirmation values and their positive binary-rational masses. Each transaction has unit demand and equal size; \(\tau\) is the support size, not the number of possible types. Let \(m=1\) be the type dimension and \(L\) the total input bit length, including rational capacities \(0<\kappa<\rho<1\).
Reports may be any finite-support probability distribution \(\nu\) on nonnegative real bids. Include the highest \(\rho\) mass and confirm the highest \(\kappa\) mass; split tied masses uniformly. Write \(B_\nu(q)\) for the decreasing bid quantile at absolute mass rank \(q\), and set \(p_{\mathrm{in}}=B_\nu(\kappa-)\), \(p_{\mathrm{out}}=B_\nu(\kappa+)\); these cutoffs may differ.
For an included infinitesimal unit at rank \(s\), define its leave-one-out rank limit on \([0,\rho)\) by \(Q_{\nu,s}(q)=B_\nu(q-)\) when \(q<s\), and \(Q_{\nu,s}(q)=B_\nu(q+)\) when \(q\ge s\), with \(B_\nu(0-)=B_\nu(0+)\). Ranks are not renormalized after deletion.
An anonymous linear rule consists of \(a\in\mathbb R\) and a finite signed Borel measure \(K\) on \([0,\rho)\), independent of reports, giving \(r_\nu(s)=a+\int Q_{\nu,s}(q)\,dK(q)\). Define \(Q^C_\nu(q)=\kappa^{-1}\int_0^\kappa Q_{\nu,s}(q)\,ds\) and \(Q^U_\nu(q)=(\rho-\kappa)^{-1}\int_\kappa^\rho Q_{\nu,s}(q)\,ds\); thus \(R_C^K=a+\int Q^C_\nu\,dK\) and \(R_U^K=a+\int Q^U_\nu\,dK\) are class-average rebates.
A confirmed unit pays \(p_{\mathrm{out}}-r_\nu(s)\); an included unconfirmed unit pays \(-r_\nu(s)\); excluded units pay zero and receive no rebate. Utility is confirmation value minus payment.
Restricted truthfulness means that, holding the included cohort fixed, every infinitesimal member weakly maximizes utility by reporting its value, against arbitrary reports of the other members and arbitrary nonnegative unilateral reports; confirmation and the member's rank are recomputed. Individual rationality requires nonnegative utility under truthful reporting.
Maximize \(f\ge0\) over \((a,K)\), subject to restricted truthfulness and individual rationality for every profile, and, for every \(\nu\), \(\kappa R_C^K+(\rho-\kappa)R_U^K\le\kappa p_{\mathrm{in}}\) and \(\kappa R_C^K\ge f\kappa p_{\mathrm{out}}\), including profiles with zero cutoffs. Return an optimal rule with a finite exact description and its allocation and payments on the truthful society \(\mu\).

Is this exact optimization problem solvable in polynomial time in \(m,\tau,L\), or is its decision version asking whether a feasible rule achieves \(f\ge t\), for an input rational \(t\), NP-hard?

For finite populations with \(n\ge k+2\), the analogous anonymous linear optimization has the unique optimum \(f^*=k/n\), with coefficient \(c_k=k/n\) and all other coefficients zero. [Verified discrete result](https://arxiv.org/pdf/2401.13262).
No result settling this precise continuous formulation was located in the searches.

Status and interpretation

I prove a stronger structural result than the one in the notes: every feasible rule has $f=0$, and its measure must be atomless, vanish on $[0,\kappa]$, and satisfy explicit cumulative bounds. These conditions are also sufficient under the right-cutoff allocation convention.

The remaining issue is the pointwise confirmation of a mass-zero bidder whose bid lies strictly between $p_{\mathrm{out}}$ and $p_{\mathrm{in}}$. Under strict rank confirmation, the feasible set is empty. Under right-cutoff confirmation, the optimization is polynomial-time solvable with optimum $0$.

I do not claim an unconditional resolution, because these conventions agree on aggregate highest-mass selection but give different answers to the stated individual incentive constraints.

Throughout, a unilateral nonatomic report leaves the aggregate distribution unchanged. Uniform tie splitting is evaluated in expected utility. A member of confirmation value $v$ obtains value $v$ when confirmed and value $0$ otherwise.

1. Proven characterization of potentially feasible measures

For $x\in[\kappa,\rho]$, define
$$ H(x)=K([0,x)). $$

Theorem. Every feasible rule necessarily satisfies
$$ a=0,\qquad f=0, $$
$$ K|_{[0,\kappa]}=0, \qquad K(\{x\})=0\quad\text{for every }x\in[0,\rho), $$
and
$$ 0\le H(x)\le \frac{\kappa}{\rho} \qquad\text{for every }x\in[\kappa,\rho]. $$

Conversely, these conditions imply feasibility with $f=0$ if confirmation follows the pointwise convention
$$ b>p_{\mathrm{out}}\Longrightarrow\text{confirmed}, \qquad b<p_{\mathrm{out}}\Longrightarrow\text{unconfirmed}, $$
with the prescribed uniform splitting at equality.

The necessity proof below does not require choosing whether a gap bidder at rank $\kappa$ is confirmed.

2. Quantile jumps and the constant term

For a finite-support report distribution, put
$$ B^+(q)=B_\nu(q+), \qquad \Delta(q)=B_\nu(q-)-B_\nu(q+). $$
By the convention at zero, $\Delta(0)=0$. At every other rank, the definition of the leave-one-out quantile gives
$$ Q_{\nu,s}(q) = B^+(q)+\Delta(q)\mathbf 1_{\{q<s\}}. $$
There are only finitely many nonzero jumps. Consequently,
$$ r_\nu(s) = a+\int_{[0,\rho)}B^+(q)\,dK(q) +\sum_{\substack{q\in(0,\rho)\\q<s}} \Delta(q)K(\{q\}). $$
All integrals exist because the quantiles are bounded and $K$ is a finite signed measure.

In the all-zero truthful society, every included member receives rebate $a$ and pays no confirmation price. Individual rationality gives $a\ge0$, while the budget inequality gives $\rho a\le0$. Thus
$$ a=0. $$

3. Truthfulness eliminates atoms away from $\kappa$

Atoms below $\kappa$

Fix $x\in(0,\kappa)$ and choose $y\in(x,\kappa)$. Let the other members report according to
$$ \nu=x\delta_2+(y-x)\delta_1+(1-y)\delta_0. $$
The bid-two and bid-one groups are entirely included and confirmed, and $p_{\mathrm{out}}=0$.

Uniform tie ranks lie in $(0,x)$ for bid two and in $(x,y)$ for bid one, almost surely. By the jump formula, the expected rebate for bid one exceeds that for bid two by exactly
$$ K(\{x\}). $$
The jump at $y$ is crossed by neither group.

For a member of true value $2$, reporting $1$ preserves confirmation and its price. Truthfulness therefore implies $K(\{x\})\le0$. For a member of true value $1$, reporting $2$ likewise preserves confirmation and its price, giving $K(\{x\})\ge0$. Hence
$$ K(\{x\})=0 \qquad (0<x<\kappa). $$

These tests respect the input restrictions: each can be realized in a society with one rational confirmation value and mass $1$. The real masses $x$ and $y-x$ describe arbitrary reports of the other members, which the task permits.

Atoms above $\kappa$

Fix $x\in(\kappa,\rho)$ and choose
$$ \kappa<z<x<y<\rho. $$
Use the report distribution
$$ \nu=z\delta_3+(x-z)\delta_2+(y-x)\delta_1+(1-y)\delta_0. $$
Here $p_{\mathrm{out}}=3$. Both the bid-two and bid-one groups are entirely included and unconfirmed.

Their expected rebates again differ by exactly $K(\{x\})$, because moving between these groups crosses only the quantile jump at $x$. Applying truthfulness first to a true-value-two member and then to a true-value-one member gives the two opposite inequalities. Therefore
$$ K(\{x\})=0 \qquad (\kappa<x<\rho). $$

Thus only a possible atom at $\kappa$, and the as-yet-unexamined endpoint zero, remain.

4. The measure vanishes below $\kappa$

Let $x\in(0,\kappa)$ be dyadic, and consider the truthful society
$$ \mu_x=x\delta_1+(1-x)\delta_0. $$
This society has positive binary-rational masses, as required. Both cutoffs are zero.

The only quantile jump is at $x$, whose $K$-atom was eliminated above. Hence every included member receives the same rebate
$$ T_x=K([0,x)). $$
There are included zero-value members because $x<\kappa<\rho$. Their individual rationality gives $T_x\ge0$. The budget inequality gives
$$ \rho T_x\le \kappa p_{\mathrm{in}}=0. $$
Thus
$$ K([0,x))=0 $$
for every dyadic $x\in(0,\kappa)$.

For any $z\in[0,\kappa)$, choose dyadics $x_n$ decreasing strictly to $z$, with $x_n<\kappa$. Then
$$ \bigcap_n[0,x_n)=[0,z]. $$
Continuity from above of finite signed measures yields
$$ K([0,z])=0. $$
This continuity follows by applying ordinary measure continuity separately to the positive and negative parts of $K$.

Taking $z$ increasingly to $\kappa$ also gives $K([0,\kappa))=0$. The intervals just considered determine a finite signed Borel measure on $[0,\kappa)$, so
$$ K|_{[0,\kappa)}=0. $$
In particular, $K(\{0\})=0$.

5. The cutoff atom and every positive guarantee are impossible

Write
$$ d=K(\{\kappa\}). $$

For any $x\in(\kappa,\rho)$, use
$$ \nu_x=x\delta_1+(1-x)\delta_0. $$
Every confirmed rank lies below $\kappa<x$, so its leave-one-out quantile equals one on $[0,x)$ and zero elsewhere. Therefore
$$ R_C^K=K([0,x)), \qquad p_{\mathrm{out}}=1. $$
The guarantee inequality implies
$$ K([0,x))\ge f. $$
Letting $x$ decrease to $\kappa$, continuity from above gives
$$ K([0,\kappa])\ge f. $$
Since the measure below $\kappa$ vanishes,
$$ d\ge f. $$

Suppose now that $d>0$. Choose a rational value
$$ 0<v<\min\{d,1\}. $$
Take the one-value society with confirmation value $v$ and mass $1$. Fix an included cohort of mass $\rho$, and let the other members report one on mass $\kappa$ and zero elsewhere. The aggregate report distribution is
$$ \nu=\kappa\delta_1+(1-\kappa)\delta_0, $$
with
$$ p_{\mathrm{in}}=1,\qquad p_{\mathrm{out}}=0. $$

When the distinguished member reports truthfully, its bid $v$ lies in the gap and its recomputed rank is exactly $\kappa$. The stipulated leave-one-out formula gives
$$ Q_{\nu,\kappa}=\mathbf 1_{[0,\kappa)}. $$
Its rebate is therefore zero. Whether its confirmation probability is zero, one, or intermediate, its truthful expected utility is at most $v$.

If it reports zero, uniform tie splitting within the fixed included cohort places its rank above $\kappa$ almost surely. Its leave-one-out quantile then equals one also at $q=\kappa$, giving rebate
$$ K([0,\kappa)) + K(\{\kappa\})=d. $$
The zero-bid group has confirmation probability zero because the bid-one group already fills confirmation capacity. The deviating utility is consequently $d>v$, contradicting truthfulness.

Thus $d\le0$. Combining this with $d\ge f\ge0$ proves
$$ d=0,\qquad f=0. $$

Together with the preceding sections, this also proves that $K$ is atomless everywhere on its domain and vanishes on $[0,\kappa]$.

6. Exact cumulative bounds and their sufficiency

Once all atoms are eliminated, the jump formula shows that the rebate is independent of rank:
$$ r_\nu(s)=R_\nu = \int_{[0,\rho)}B_\nu(q+)\,dK(q). $$
Both class-average rebates equal $R_\nu$.

Necessity of the bounds

For $x\in(\kappa,\rho]$, again take
$$ \nu_x=x\delta_1+(1-x)\delta_0. $$
Then
$$ R_\nu=H(x),\qquad p_{\mathrm{in}}=p_{\mathrm{out}}=1. $$
This remains true at $x=\rho$, since $\rho$ is outside the integration domain.

Because $f=0$, the guarantee inequality gives $H(x)\ge0$. The budget inequality gives
$$ \rho H(x)\le\kappa. $$
Also $H(\kappa)=0$. This proves all the stated cumulative bounds.

A bound valid for every report distribution

Conversely, assume that $K$ is atomless, vanishes on $[0,\kappa]$, and satisfies those bounds.

Let $J$ be the finite set of quantile jumps strictly between $\kappa$ and $\rho$, and set
$$ \beta=B_\nu(\rho-). $$
On $(\kappa,\rho)$, the decreasing step function has the decomposition
$$ B_\nu(q+) = \beta+\sum_{x\in J}\Delta(x)\mathbf 1_{\{q<x\}}. $$
All coefficients are nonnegative, and
$$ \beta+\sum_{x\in J}\Delta(x)=p_{\mathrm{out}}. $$
Integrating gives
$$ R_\nu = \beta H(\rho)+\sum_{x\in J}\Delta(x)H(x). $$
The cumulative bounds therefore imply
$$ 0\le R_\nu\le \frac{\kappa}{\rho}p_{\mathrm{out}}. $$
In particular,
$$ \rho R_\nu \le \kappa p_{\mathrm{out}} \le \kappa p_{\mathrm{in}}. $$
Thus both financial inequalities hold with $f=0$, including when either cutoff is zero.

Incentives under the right-cutoff convention

Fix the included cohort and the other reports, and write $P=p_{\mathrm{out}}$. A unilateral nonatomic report changes neither $P$ nor the common rebate $R_\nu$.

If $x(b)$ is the confirmation probability for report $b$, expected utility is
$$ u_v(b)=R_\nu+x(b)(v-P). $$
Under the right-cutoff convention:

• If $v>P$, truthful reporting gives $x(v)=1$, maximizing utility.
• If $v<P$, truthful reporting gives $x(v)=0$, maximizing utility.
• If $v=P$, utility is independent of the report.

This proves restricted truthfulness. Under truthful reporting, confirmed members have value at least $P$, and $R_\nu\ge0$, so individual rationality also holds. Excluded members obtain zero utility.

This completes the conditional sufficiency proof.

The characterization permits nonzero diffuse measures. For example, with $\operatorname{Leb}$ denoting Lebesgue measure,
$$ K(A)= \frac{\kappa}{\rho(\rho-\kappa)} \operatorname{Leb}\bigl(A\cap(\kappa,\rho)\bigr) $$
has
$$ H(x)= \frac{\kappa(x-\kappa)}{\rho(\rho-\kappa)}, \qquad \kappa\le x\le\rho. $$
It satisfies all the bounds and has a finite exact description. Thus the necessity of $f=0$ does not imply uniqueness of the zero-rebate rule.

7. Strict rank confirmation makes the problem infeasible

Assume instead that an included member is confirmed exactly when its recomputed rank satisfies
$$ s<\kappa. $$

Choose any rational $v\in(0,1)$ and use the same report profile
$$ \nu=\kappa\delta_1+(1-\kappa)\delta_0. $$
A truthful gap bidder of value $v$ has rank $\kappa$ and is unconfirmed. Reporting one places it uniformly within the bid-one group, hence at rank below $\kappa$ almost surely, making it confirmed.

For the truthful rank $\kappa$ and every deviating rank $s<\kappa$, the leave-one-out quantile is exactly
$$ Q_{\nu,s}=\mathbf 1_{[0,\kappa)}. $$
Therefore the rebate is identical under the two reports for every $a$ and every finite signed measure $K$. The confirmation payment is also zero because $p_{\mathrm{out}}=0$.

The deviation increases utility by $v>0$. Consequently, under strict rank confirmation, no anonymous linear rule is feasible, independently of the financial constraints.

8. Conditional exact algorithm and remaining question

Under right-cutoff confirmation, an optimal finite-description rule is
$$ a=0,\qquad K=0,\qquad f^*=0. $$

Sort the truthful support as
$$ v_1>\cdots>v_\tau\ge0, \qquad W_0=0,\qquad W_i=\sum_{j=1}^i w_j. $$
The included and confirmed masses of type $i$ are
$$ I_i=\min\{w_i,\max\{0,\rho-W_{i-1}\}\}, $$
$$ C_i=\min\{w_i,\max\{0,\kappa-W_{i-1}\}\}. $$
Uniform splitting gives inclusion probability $I_i/w_i$ and confirmation probability $C_i/w_i$.

Let $j$ be the first index with $W_j>\kappa$. Then
$$ P=p_{\mathrm{out}}=v_j. $$
This correctly selects the next lower support value when $\kappa$ equals a cumulative mass.

Every confirmed unit pays $P$; all other units pay zero. Type $i$ has expected payment per member
$$ P\frac{C_i}{w_i}. $$
Sorting and exact rational arithmetic take polynomial time in $\tau$ and $L$ for the stipulated $m=1$. The decision answer is “yes” exactly when $t\le0$.

Under strict rank confirmation, the decision answer is always “no”, and no optimal rule exists.

The unresolved question is which pointwise allocation the task requires. Aggregate highest-mass selection and uniform splitting of positive-mass ties do not by themselves specify the confirmation of a mass-zero bidder with a strict gap bid. The prescribed leave-one-out formula specifies that bidder’s rebate, but does not explicitly specify its confirmation indicator. Until that point is fixed, the two conditional conclusions cannot be combined into one unconditional answer.

⟨structured verdict parsed — rendered above⟩

Verdict

The output establishes substantial conditional results, but does not resolve the exact optimization problem as stated. The measure characterization and both conditional incentive arguments are mathematically sound under their stated assumptions. The blocking issue is the unverified relationship between the prover’s pointwise confirmation conventions and the allocation rule in USER_TASK.

No false central computation was found. The appropriate classification is partial.

Step-by-step verification

1. Interpretation and scope

The prover explicitly assumes that a unilateral nonatomic report leaves the aggregate distribution unchanged, evaluates uniform splitting through expected utility, and awards confirmation value only when confirmed. These are natural interpretations of the task.

The substantive additional assumption is the treatment of a bidder whose bid lies strictly between the two cutoffs. The prover distinguishes:

• Confirmation when the bid exceeds $p_{\mathrm{out}}$.
• Confirmation exactly when the recomputed rank satisfies $s<\kappa$.

These give different confirmation outcomes for the bidder used in Sections 5 and 7.

The task specifies highest-mass selection and recomputation of rank. Consequently, the right-cutoff convention cannot simply be assumed to implement the assigned mechanism. Conversely, the output does not establish that the task leaves the convention unspecified. Its two conditional analyses are valid, but their applicability to the original problem remains unverified.

Severity: major. This prevents accepting the proposed optimal rule and decision algorithm as answers to the original task.

2. Quantile-jump formula

The identity
$$ Q_{\nu,s}(q)=B^+(q)+\Delta(q)\mathbf 1_{\{q<s\}} $$
correctly follows from the prescribed definition, including the case $q=s$, where the right limit is used.

For a finite-support distribution, $\Delta$ is nonzero at finitely many ranks. Integrating therefore gives exactly
$$ r_\nu(s) = a+\int B^+(q)\,dK(q) +\sum_{\substack{q\in(0,\rho)\\q<s}}\Delta(q)K(\{q\}). $$
The atom at zero contributes no jump because $\Delta(0)=0$. Boundedness of the quantile and finiteness of the signed measure justify the integrals.

Verified.

3. Elimination of the constant term

In the all-zero truthful profile, every rebate equals $a$, confirmation payments vanish, and individual rationality requires $a\ge0$. The budget inequality becomes
$$ \rho a\le0. $$
Since $\rho>0$, this forces $a=0$.

Verified, including the zero-cutoff case.

4. Atoms strictly below the confirmation cutoff

For the distribution in Section 3, both positive-bid groups are fully confirmed, and $p_{\mathrm{out}}=0$.

Moving from the bid-two group to the bid-one group crosses the jump at $x$, whose magnitude is $1$, and does not cross the jump at $y$. Thus the rebate difference is exactly $K(\{x\})$.

For a true-value-two member, truthfulness gives
$$ K(\{x\})\le0. $$
For a true-value-one member, the reverse deviation gives
$$ K(\{x\})\ge0. $$
Confirmation and payment are unchanged in both comparisons.

The use of arbitrary real report masses is permitted by the task. The distinguished members’ true values are rational.

Verified.

5. Atoms strictly above the confirmation cutoff

Here $\kappa<z$, so the confirmation cutoff lies inside the bid-three group and
$$ p_{\mathrm{out}}=3. $$
The bid-two and bid-one groups are fully included and unconfirmed because $y<\rho$.

Their rebate difference is again exactly $K(\{x\})$: the jump at $z$ contributes equally to both groups, the jump at $x$ distinguishes them, and the jump at $y$ contributes to neither.

The two truthfulness comparisons therefore eliminate the atom at $x$.

Verified.

6. Vanishing of the measure below $\kappa$

For dyadic $x\in(0,\kappa)$, the proposed truthful society satisfies the input restrictions. Both cutoffs vanish. Since the atom at $x$ has already been eliminated, every included member receives
$$ T_x=K([0,x)). $$
Included zero-value members exist, so individual rationality gives $T_x\ge0$. The budget gives $\rho T_x\le0$. Hence $T_x=0$.

For dyadics decreasing strictly to $z<\kappa$,
$$ \bigcap_n[0,x_n)=[0,z]. $$
Continuity from above applies to a finite signed measure through its positive and negative parts. Consequently,
$$ K([0,z])=0. $$
These interval identities determine the restriction of the finite signed Borel measure to $[0,\kappa)$. They establish vanishing of that restriction, rather than merely vanishing of its total mass.

Verified, including elimination of the atom at zero.

7. The atom at $\kappa$ and the guarantee

For $x\in(\kappa,\rho)$, the confirmed-class rebate is
$$ R_C^K=K([0,x)), $$
and $p_{\mathrm{out}}=1$. Thus the guarantee implies
$$ K([0,x))\ge f. $$
Continuity as $x$ decreases to $\kappa$ gives
$$ d=K(\{\kappa\})\ge f, $$
using the previously established vanishing below $\kappa$.

The contradiction for $d>0$ is also valid. A rational $v$ satisfying
$$ 0<v<\min\{d,1\} $$
exists. At truthful rank $\kappa$, the prescribed leave-one-out quantile is
$$ Q_{\nu,\kappa}=\mathbf 1_{[0,\kappa)}, $$
so the truthful rebate is zero and truthful expected utility is at most $v$, regardless of the bidder’s confirmation probability.

Reporting zero gives rank above $\kappa$ almost surely within the fixed included cohort. The rebate becomes $d$, and the zero-bid group is unconfirmed because the bid-one group fills confirmation capacity. The deviation therefore yields utility $d>v$.

This proves
$$ d\le0. $$
Together with $d\ge f\ge0$, it establishes $d=f=0$.

Verified. This necessity argument genuinely avoids deciding the truthful gap bidder’s confirmation status.

8. Cumulative bounds

After eliminating all atoms, rebates are independent of rank.

For $x\in(\kappa,\rho]$, the indicator profile gives
$$ R_\nu=H(x),\qquad p_{\mathrm{in}}=p_{\mathrm{out}}=1. $$
At $x=\rho$, the integration domain still contains precisely the portion on which the quantile equals one.

The financial inequalities yield
$$ 0\le H(x)\le\frac{\kappa}{\rho}. $$
Also, $H(\kappa)=0$.

Verified.

9. Financial sufficiency

The finite step-function decomposition on $(\kappa,\rho)$ is correct:
$$ B_\nu(q+) = \beta+\sum_{x\in J}\Delta(x)\mathbf 1_{\{q<x\}}. $$
Its coefficients satisfy
$$ \beta\ge0,\qquad \Delta(x)\ge0,\qquad \beta+\sum_{x\in J}\Delta(x)=p_{\mathrm{out}}. $$
The atomless hypothesis makes endpoint values irrelevant to integration. Therefore
$$ R_\nu=\beta H(\rho)+\sum_{x\in J}\Delta(x)H(x), $$
and the asserted bound follows:
$$ 0\le R_\nu\le\frac{\kappa}{\rho}p_{\mathrm{out}}. $$
Since $p_{\mathrm{out}}\le p_{\mathrm{in}}$, both financial constraints hold with $f=0$.

Verified, including zero cutoffs.

10. Conditional truthfulness and individual rationality

Under the explicitly assumed right-cutoff convention, a unilateral report changes neither $P$ nor the common rebate. The utility formula
$$ u_v(b)=R_\nu+x(b)(v-P) $$
then gives the stated maximizing behavior for each of $v>P$, $v<P$, and $v=P$.

The individual-rationality argument is also valid under that convention: confirmed truthful members have value at least $P$, and all included members receive a nonnegative rebate.

Verified conditionally. This does not establish feasibility under the original allocation specification until compatibility with that specification is justified.

The displayed Lebesgue-measure example has the stated cumulative function and satisfies the bounds.

11. Conditional infeasibility under strict rank confirmation

Under the assumption that confirmation requires $s<\kappa$, the truthful gap bidder is unconfirmed and becomes confirmed by joining the bid-one group.

For both truthful rank $\kappa$ and deviating ranks below $\kappa$, the leave-one-out quantile is indeed
$$ \mathbf 1_{[0,\kappa)}. $$
Thus every allowed linear rule gives the same rebate under the two reports. The confirmation payment is zero, so the deviation increases utility by $v>0$.

Verified conditionally. The argument establishes infeasibility under this convention for every $a$ and $K$, without needing the financial constraints.

12. Conditional algorithm

The inclusion and confirmation mass formulas are correct for aggregate highest-mass allocation with uniform splitting.

The first index satisfying $W_j>\kappa$ correctly identifies the right cutoff, including when $\kappa$ equals a cumulative support mass. Such an index exists because total mass is $1>\kappa$.

For $a=0$ and $K=0$, the stated payments and expected payments follow. Sorting and the required exact rational arithmetic take polynomial time in $\tau$ and $L$.

The two conditional decision answers are correct:

• Under right-cutoff confirmation, feasibility at threshold $t$ holds exactly when $t\le0$.
• Under strict rank confirmation, the feasible set is empty.

Verified conditionally. Neither decision answer is established unconditionally for the task.

Minimum repair needed

The prover must connect its pointwise confirmation convention to the original allocation requirement and then commit to the resulting answer. In particular, it must justify the confirmation outcome for an included bidder with recomputed rank $\kappa$ and bid strictly between the cutoffs.

If the intended interpretation is strict rank confirmation, Section 7 already supplies the relevant conditional impossibility argument. If the prover intends to use right-cutoff confirmation, it must establish that this convention is admissible under USER_TASK. Merely observing agreement of aggregate allocations does not verify an individual incentive constraint.

No further repair to the displayed measure calculations is needed.

⟨structured verdict parsed — rendered above⟩