Answer: Polynomial-time solvable ($\mathrm{P}$) under the stated requirement of finite interim expectations: answer yes exactly when $q\le S_{\mathrm{eff}}$.
Theorem 4 — class: other
PDF text-layer excerpt — equations and symbol order may be approximateAssume that traders’ levels are not observable and that the population level distributions for buyers and sellers include all levels from L1 up to at least some K > 1. Assume further that 𝑓(∙), g(∙) ≠1 almost everywhere and that there is no x for which both 𝐹(𝑥) = 𝑥 and 𝐺(𝑥) = 𝑥.
statement extracted from the paper’s text layer; consult the paper for the exact wording
Given unit-mass buyer and seller populations with types \((v,k)\) and \((c,\ell)\), respectively, where values and costs lie in \([0,1]\), levels are independent of values, and each side has positive mass at every level from \(L1\) through \(L_K\), randomly match buyers and sellers for independent encounters. For rational polynomial value densities \(f\) and \(g\) satisfying Theorem 4’s generic conditions and rational level proportions, find a common budget-balanced direct mechanism \((p,x)\) maximizing aggregate surplus per encounter, subject to level-\(k\) incentive compatibility and interim individual rationality for every populated level; in particular, compute an optimal posted price \(\pi\), or an \(\varepsilon\)-optimal one.
For a finitely supported society, the input is the list of its \(\tau\) positive-mass types with binary-rational masses; \(\tau\) is the actual support size. The non-atomic societies below instead use an explicit density encoding, not a finite point-type support list.
Each side has total mass \(1\). Buyer types are \((V,k)\), seller types \((C,k)\), with independent values and levels, independent encounters, and unobservable levels \(k\in\{1,\ldots,K\}\), where \(K\ge2\); each side's level proportions are positive binary rationals summing to \(1\).
The value densities \(f,g\) are densely encoded rational polynomials, strictly positive on \([0,1]\), integrating to \(1\), and neither identically \(1\). Write \(F(t)=\int_0^t f(s)\,ds\) and \(G(t)=\int_0^t g(s)\,ds\); require no \(t\in(0,1)\) with \(F(t)=G(t)=t\). No additional unspecified genericity exclusion is imposed.
A common direct mechanism consists of measurable \(p:[0,1]^2\to[0,1]\) and \(x:[0,1]^2\to\mathbb R\), with all interim expectations finite: \(p\) is trade probability and \(x\) is the expected transfer from buyer to seller, ensuring budget balance at every report pair. Utilities are \(Vp(v,c)-x(v,c)\) and \(x(v,c)-Cp(v,c)\).
Level \(0\) reports uniformly on \([0,1]\), independently of value, and has zero population mass; level \(k\) best responds to the opposite side's level \(k-1\) reporting strategy under its true value density. Require truthful reporting to be a weak best response for every value and every populated level, selecting truth when indifferent, and require nonnegative truthful interim expected utility under that level's beliefs.
Surplus per encounter is \(S(p)=\int_0^1\int_0^1(V-C)p(V,C)f(V)g(C)\,dC\,dV\). A posted price \(\pi\in[0,1]\) gives \(p(v,c)=\mathbf1\{c\le\pi\le v\}\) and \(x(v,c)=\pi p(v,c)\).
Is deciding, for a binary-rational threshold \(q\), whether a feasible common mechanism achieves \(S(p)\ge q\) in \(\mathrm P\), or is it \(\mathrm{NP}\)-hard? State running times in the maximum density degree \(m\), \(K\), and total input bit length \(L\); the full type support is infinite.
The supplied anchor is a continuous structural result, not a discrete complexity theorem. Crawford (2021) claims generic optimality of a deterministic posted price, subject to further unnamed exceptional densities, and characterizes the feasible mechanisms through simultaneous uniform-belief and truthful-belief incentive constraints. [Source](https://econweb.ucsd.edu/~vcrawfor/LkBilateralTrading27January21.pdf)
That characterization supplies neither a complexity classification for this explicit input domain nor a specified promise excluding all exceptional densities.
The decision problem is in $\mathrm P$ as stated. Its exact criterion is
$$
q\le S_{\mathrm{eff}}
:=
\int_0^1\int_0^v(v-c)f(v)g(c)\,dc\,dv.
$$
The essential point is the transfer domain: finite interim expectations for every report do not require joint absolute integrability. Under the stated condition, efficient trade admits a common transfer satisfying all the incentive and participation constraints. I prove this below for the full nonuniform density domain.
This argument does not invoke the generic posted-price result in [Crawford (2021)](https://econweb.ucsd.edu/~vcrawfor/LkBilateralTrading27January21.pdf). Adding joint absolute integrability would change the problem and invalidate the construction used here.
Define the belief densities
$$
f_0(v)=g_0(c)=1,\qquad f_1(v)=f(v),\qquad g_1(c)=g(c).
$$
Level $1$ faces uniform reports. If truth is selected as a best response at that level, level $2$ faces the opposite side’s true value density. Inductively, every higher level faces that same truthful-report density. Independence of values and levels ensures that these densities are $f$ and $g$, regardless of level proportions.
Consequently, to construct a feasible mechanism, it suffices to make truth a best response and truthful utility nonnegative for buyers under both $g_0,g_1$ and for sellers under both $f_0,f_1$. The construction below makes truth the unique best response under each belief.
The main issue is realizing the required four interim transfer schedules with one common transfer.
I first prove a general lemma.
Lemma. Let $f,g$ be the densities in the task. Given any bounded measurable functions
$$
r_0,r_1,z_0,z_1:[0,1]\longrightarrow\mathbb R,
$$
there exists a measurable, finite-valued function $x$ such that, for every $v,c\in[0,1]$,
$$
\int_0^1x(v,c)g_j(c)\,dc=r_j(v),
\qquad j\in\{0,1\},
$$
and
$$
\int_0^1x(v,c)f_i(v)\,dv=z_i(c),
\qquad i\in\{0,1\}.
$$
Furthermore, every horizontal and vertical section of $x$ is absolutely integrable.
The lemma imposes no equality between the aggregate buyer and seller transfer integrals. That is possible because $x$ need not be jointly absolutely integrable.
Let $d$ be either $f$ or $g$. Since $d$ integrates to $1$ and is not identically $1$, it is a nonconstant polynomial. Its derivative is therefore not identically zero. Choose an interior point $a$ and $\varepsilon>0$ such that $d$ is strictly monotone on
$$
[a-\varepsilon,a+\varepsilon]\subset(0,1).
$$
For every integer $n\ge1$, define the open intervals
$$
I_n^-=(a-\varepsilon2^{-n},a-\varepsilon2^{-(n+1)}),
$$
$$
I_n^+=(a+\varepsilon2^{-(n+1)},a+\varepsilon2^{-n}).
$$
These intervals are pairwise disjoint, including across signs and indices.
Let $\phi_n^\pm$ be the normalized indicator of $I_n^\pm$, and let
$$
\mu_n^\pm=\int_0^1d(u)\phi_n^\pm(u)\,du.
$$
Strict monotonicity implies that $d(a)$ lies strictly between $\mu_n^-$ and $\mu_n^+$. Hence there exists $\theta_n\in(0,1)$ such that
$$
\theta_n\mu_n^-+(1-\theta_n)\mu_n^+=d(a).
$$
Set
$$
A_n=\theta_n\phi_n^-+(1-\theta_n)\phi_n^+.
$$
Each $A_n$ is nonnegative, bounded, and measurable, and
$$
\int_0^1A_n(u)\,du=1,
\qquad
\int_0^1d(u)A_n(u)\,du=d(a).
$$
The supports are pairwise disjoint. In particular, every point belongs to the support of at most one $A_n$.
Apply the preceding construction to $f$ at a center $a$, obtaining functions $A_n$, and to $g$ at a center $b$, obtaining functions $B_n$. Thus
$$
\int_0^1f_i(v)A_n(v)\,dv=f_i(a),
\qquad
\int_0^1g_j(c)B_n(c)\,dc=g_j(b).
$$
Define
$$
h_{a,b}(v,c)
=
\sum_{n=1}^{\infty}
A_n(v)\bigl(B_n(c)-B_{n+1}(c)\bigr).
$$
This function is measurable and finite-valued. For a fixed $v$, at most one summand is nonzero. For a fixed $c$, at most two summands are nonzero. Therefore every horizontal and vertical section is absolutely integrable. These statements include interval boundaries and accumulation points, where the relevant indicator functions vanish.
For every $v$ and each $j$,
$$
\int_0^1h_{a,b}(v,c)g_j(c)\,dc=0,
$$
because $B_n$ and $B_{n+1}$ have identical $g_j$ moments.
For a fixed $c$ in the support of $B_1$, the vertical section is $A_1(v)B_1(c)$. For $c$ in the support of $B_n$, with $n\ge2$, it is
$$
\bigl(A_n(v)-A_{n-1}(v)\bigr)B_n(c).
$$
The identical $f_i$ moments of the $A_n$ therefore give, for every $c$,
$$
\int_0^1h_{a,b}(v,c)f_i(v)\,dv=f_i(a)B_1(c).
$$
In particular,
$$
\int_0^1g_j(c)
\left[\int_0^1h_{a,b}(v,c)f_i(v)\,dv\right]dc
=f_i(a)g_j(b).
$$
There is no application of Fubini’s theorem here: each section is integrated separately. Indeed, under uniform weights, the two iterated integrals are $0$ and $1$. Thus $h_{a,b}$ is not jointly absolutely integrable.
Choose centers $a_0,a_1$ for $f$ such that
$$
f(a_0)\ne f(a_1),
$$
and centers $b_0,b_1$ for $g$ such that
$$
g(b_0)\ne g(b_1).
$$
Such choices exist within intervals where the respective derivatives have a fixed nonzero sign.
Define
$$
U=
\begin{pmatrix}
1&1\\
f(a_0)&f(a_1)
\end{pmatrix},
\qquad
V=
\begin{pmatrix}
1&1\\
g(b_0)&g(b_1)
\end{pmatrix}.
$$
Both matrices are invertible.
For the prescribed schedules, define
$$
M^R_{ij}=\int_0^1f_i(v)r_j(v)\,dv,
\qquad
M^Z_{ij}=\int_0^1g_j(c)z_i(c)\,dc,
$$
and let
$$
\Delta=M^Z-M^R.
$$
These are finite because the schedules and densities are bounded.
Set
$$
\Lambda=U^{-1}\Delta(V^{\mathsf T})^{-1},
\qquad
h=\sum_{s,t=0}^1\Lambda_{st}h_{a_s,b_t}.
$$
The function $h$ has absolutely integrable sections, and its buyer moments vanish for every report.
Let
$$
H_i(c)=\int_0^1h(v,c)f_i(v)\,dv.
$$
These functions are bounded: each is a finite linear combination of the bounded functions $B_1$ appearing in the construction. Moreover,
$$
\int_0^1g_j(c)H_i(c)\,dc
=
(U\Lambda V^{\mathsf T})_{ij}
=\Delta_{ij}.
$$
Consequently, the adjusted seller schedules
$$
\widetilde z_i=z_i-H_i
$$
satisfy the compatibility identities
$$
\int_0^1g_j(c)\widetilde z_i(c)\,dc
=
\int_0^1f_i(v)r_j(v)\,dv
=M^R_{ij}.
$$
The functions $f_0,f_1$ are linearly independent. Otherwise $f$ would be constant and, by normalization, identically $1$. Their Gram matrix
$$
\Gamma^f_{\ell i}=\int_0^1f_\ell(v)f_i(v)\,dv
$$
is positive definite: a vanishing quadratic form would make a nontrivial linear combination vanish almost everywhere, and continuity would then make it vanish everywhere.
Inverting this matrix gives bounded continuous functions $\alpha_0,\alpha_1$, each a linear combination of $f_0,f_1$, satisfying
$$
\int_0^1f_\ell(v)\alpha_i(v)\,dv
=
\begin{cases}
1,&\ell=i,\\
0,&\ell\ne i.
\end{cases}
$$
The same argument gives bounded functions $\beta_0,\beta_1$ with
$$
\int_0^1g_\ell(c)\beta_j(c)\,dc
=
\begin{cases}
1,&\ell=j,\\
0,&\ell\ne j.
\end{cases}
$$
Write $M_{ij}=M^R_{ij}$ and define
$$
J_i(c)=\sum_{j=0}^1M_{ij}\beta_j(c),
$$
$$
y(v,c)
=
\sum_{j=0}^1r_j(v)\beta_j(c)
+
\sum_{i=0}^1\alpha_i(v)
\bigl(\widetilde z_i(c)-J_i(c)\bigr).
$$
This function is bounded and measurable.
The compatibility identities imply
$$
\int_0^1g_j(c)
\bigl(\widetilde z_i(c)-J_i(c)\bigr)\,dc=0.
$$
Thus, for every $v$,
$$
\int_0^1y(v,c)g_j(c)\,dc=r_j(v).
$$
The dual identities for the $\alpha_i$ similarly give, for every $c$,
$$
\int_0^1y(v,c)f_i(v)\,dv
=
J_i(c)+\widetilde z_i(c)-J_i(c)
=\widetilde z_i(c).
$$
Finally, set
$$
x=y+h.
$$
Its buyer moments are $r_j$ and its seller moments are $\widetilde z_i+H_i=z_i$. Every section is absolutely integrable because $y$ is bounded and $h$ has absolutely integrable sections. This proves the lemma.
Take the efficient allocation
$$
p^*(v,c)=\mathbf1\{c\le v\}.
$$
Prescribe buyer transfer schedules
$$
r_j(v)=\int_0^v c\,g_j(c)\,dc
$$
and seller transfer schedules
$$
z_i(c)=\int_c^1v\,f_i(v)\,dv.
$$
All four schedules are bounded and measurable. The lemma supplies one common transfer $x$ realizing them, with absolutely integrable sections for every report.
Under belief $g_j$, a buyer of value $v$ reporting $r$ has expected utility
$$
u_j(v,r)
=
v\int_0^r g_j(c)\,dc-r_j(r)
=
\int_0^r(v-c)g_j(c)\,dc.
$$
If $r\le v$, then
$$
u_j(v,v)-u_j(v,r)
=
\int_r^v(v-c)g_j(c)\,dc\ge0.
$$
If $r\ge v$, then
$$
u_j(v,v)-u_j(v,r)
=
\int_v^r(c-v)g_j(c)\,dc\ge0.
$$
Since $g_j$ is strictly positive, either inequality is strict whenever $r\ne v$.
Truthful utility is
$$
u_j(v,v)=\int_0^v(v-c)g_j(c)\,dc\ge0.
$$
Thus buyer truthfulness and interim individual rationality hold under both beliefs, for every value.
Under belief $f_i$, a seller of cost $c$ reporting $r$ has expected utility
$$
w_i(c,r)
=
z_i(r)-c\int_r^1f_i(v)\,dv
=
\int_r^1(v-c)f_i(v)\,dv.
$$
If $r\le c$, then
$$
w_i(c,c)-w_i(c,r)
=
\int_r^c(c-v)f_i(v)\,dv\ge0.
$$
If $r\ge c$, then
$$
w_i(c,c)-w_i(c,r)
=
\int_c^r(v-c)f_i(v)\,dv\ge0.
$$
Strict positivity of $f_i$ again makes truth the unique best response.
Truthful utility is
$$
w_i(c,c)=\int_c^1(v-c)f_i(v)\,dv\ge0.
$$
Hence level $1$ reports truthfully under uniform beliefs, and induction gives truthful reporting at every subsequent level. All required interim expectations are finite. Using the single transfer $x$ with opposite signs in the two agents’ utilities supplies budget balance at every report pair.
This proves feasibility of $p^*$ for every admissible input.
Under uniform beliefs, the prescribed schedules are
$$
r_0(v)=\frac{v^2}{2},
\qquad
z_0(c)=\frac{1-c^2}{2}.
$$
Their aggregate integrals differ:
$$
\int_0^1r_0(v)\,dv=\frac16,
\qquad
\int_0^1z_0(c)\,dc=\frac13.
$$
A jointly absolutely integrable transfer would make these integrals equal by Fubini’s theorem. Thus the efficient mechanism constructed above necessarily uses the broader transfer class explicitly permitted in the task.
For every allocation probability $p(v,c)\in[0,1]$,
$$
(v-c)p(v,c)\le\max(v-c,0).
$$
For $v\ge c$, this follows from $p\le1$; for $v<c$, it follows from $p\ge0$.
Multiplication by the positive densities and integration therefore yield
$$
S(p)\le S_{\mathrm{eff}}.
$$
The feasible allocation $p^*$ attains equality. Consequently,
$$
\exists\text{ feasible common mechanism with }S(p)\ge q
\quad\Longleftrightarrow\quad
q\le S_{\mathrm{eff}}.
$$
In particular, the optimum is attained, so equality thresholds present no approximation issue.
Pad both density coefficient lists to degree $m$ and write
$$
f(v)=\sum_{i=0}^m a_i v^i,
\qquad
g(c)=\sum_{j=0}^m b_j c^j.
$$
For each $j$,
$$
\int_0^v(v-c)c^j\,dc
=
\frac{v^{j+2}}{(j+1)(j+2)}.
$$
Integrating the resulting polynomial gives
$$
S_{\mathrm{eff}}
=
\sum_{i=0}^m\sum_{j=0}^m
\frac{a_i b_j}
{(j+1)(j+2)(i+j+3)}.
$$
This is an explicitly computable rational number.
Let $Q_f$ and $Q_g$ be the products of the encoded coefficient denominators of $f$ and $g$. Their total bit length is $O(L)$. Write
$$
A_i=Q_fa_i,\qquad B_j=Q_gb_j,
$$
so that $A_i,B_j$ are integers.
Set
$$
T=(m+2)!(2m+3)!.
$$
Each denominator
$$
(j+1)(j+2)(i+j+3)
$$
divides $T$: the first two factors divide $(m+2)!$, and the third divides $(2m+3)!$. Hence
$$
S_{\mathrm{eff}}=\frac{N}{Q_fQ_gT},
$$
where
$$
N=
\sum_{i,j=0}^m
A_iB_j
\frac{T}{(j+1)(j+2)(i+j+3)}
$$
is an integer.
All intermediate integers have bit length
$$
B=O\bigl(L+(m+1)\log(m+2)\bigr).
$$
There are $O((m+1)^2)$ terms. Using ordinary integer arithmetic and exact cross multiplication to compare with $q$, a conservative bit-operation bound is
$$
O\!\left(
K+L+(m+1)^2
\bigl[L+(m+1)\log(m+2)\bigr]^2
\right).
$$
The level proportions affect only input reading; they do not affect the criterion. No finite point-type support or support-size parameter is used.
For example, the admissible nonuniform densities
$$
f(v)=\frac12+v,\qquad g(c)=\frac32-c
$$
satisfy
$$
F(t)-t=-\frac{t(1-t)}2,\qquad
G(t)-t=\frac{t(1-t)}2,
$$
so they meet the stated interior exclusion. The formula above gives their attained efficient surplus as
$$
S_{\mathrm{eff}}=\frac{31}{120}.
$$
⟨structured verdict parsed — rendered above⟩
The argument establishes polynomial-time solvability under the transfer condition stated in the task. The construction satisfies finite interim expectations for every report, despite lacking joint absolute integrability. I found one minor terminology issue concerning the word “support”; it does not affect the construction or conclusion.
1. Scope and interpretation of feasibility
The proof addresses the continuous density input directly. It does not use a finite support parameter or assume the unnamed genericity exclusions in the supplied anchor.
Its essential interpretation of the transfer requirement is justified by the task: a measurable, finite-valued transfer may have absolutely integrable horizontal and vertical sections without being jointly absolutely integrable. The task imposes finite interim expectations and pointwise budget balance; it does not impose joint absolute integrability or an aggregate expected-transfer identity.
Accordingly, the failure of Fubini’s theorem for the constructed transfer is not itself a feasibility defect. No cited structural theorem is needed for the argument.
2. Reduction to two belief distributions
This step is correct.
At level $1$, the opposite side’s level $0$ reports are uniform. Once truth is selected at level $1$, level $2$ faces the opposite side’s true value distribution. The same reasoning applies inductively at every higher level.
Independence of values and levels ensures that conditioning on a level does not change the value distribution. Thus the four interim constraints described in the proof suffice. Since the subsequent construction gives strict incentives for every unequal report, the required selection of truth when indifferent introduces no additional difficulty.
3. Construction of equal-moment functions
The polynomial assumptions justify the choices made in Section 2.1.
A normalized polynomial density that is not identically $1$ cannot be constant. Its derivative is therefore nonzero at some interior point. Continuity of the derivative gives an interior interval on which the derivative has a fixed nonzero sign.
The displayed intervals are disjoint as open sets. Their normalized indicators have integral $1$, and strict monotonicity places the two density averages strictly on opposite sides of $d(a)$. Consequently, the asserted coefficient $\theta_n\in(0,1)$ exists, and both moment identities for $A_n$ follow.
There is one minor terminology issue: under the usual topological definition, the supports of successive indicator functions share interval endpoints. Thus the statement “The supports are pairwise disjoint” is literally false under that convention. The sets where the functions are nonzero are pairwise disjoint, which is exactly what the proof uses. The explicit boundary discussion confirms this intended meaning.
Minimum clarification: replace “supports” with “sets where the functions are nonzero” wherever disjointness or membership counts are asserted.
4. The transfer with unequal iterated integrals
The construction in Section 2.2 is valid.
For each fixed $v$, only one index can contribute. For each fixed $c$, only two indices can contribute. Each contributing section is a finite linear combination of bounded indicator functions, so every section is absolutely integrable. This remains true at all endpoints and accumulation points because the indicators vanish there.
The horizontal moments vanish because $B_n$ and $B_{n+1}$ have identical moments under both seller belief densities.
For a point where $B_n(c)$ is nonzero, the stated vertical-section formulas correctly account for the positive contribution from index $n$ and, when $n\ge2$, the negative contribution from index $n-1$. Their equal buyer-density moments cancel for every $n\ge2$. Hence
$$
\int_0^1 h_{a,b}(v,c)f_i(v)\,dv=f_i(a)B_1(c)
$$
holds for every $c$, including points where all the indicators vanish.
The two uniform iterated integrals are indeed $0$ and $1$. This proves the claimed failure of joint absolute integrability: if joint absolute integrability held, Fubini’s theorem would make those iterated integrals equal.
5. Correction of the aggregate discrepancy matrix
Section 2.3 is correct.
The required pairs of centers exist because each density varies strictly on a suitable interior interval. The unequal density values make both displayed matrices invertible.
The orientation of the matrix multiplication is correct:
$$
(U\Lambda V^{\mathsf T})_{ij}
=
\sum_{s,t=0}^1
f_i(a_s)\Lambda_{st}g_j(b_t).
$$
This is exactly the aggregate seller moment generated by the finite linear combination of the constructed functions.
The functions $H_i$ are bounded because they are finite linear combinations of fixed bounded functions $B_1$. Subtracting them from the prescribed seller schedules therefore preserves boundedness and measurability. The displayed compatibility identities follow with the correct sign.
Overlaps between functions from different center choices cause no problem: only finitely many constructed functions are combined.
6. Realization of compatible schedules
Section 2.4 is correct.
The functions $1$ and $f$ are linearly independent, as are $1$ and $g$. Their Gram matrices are positive definite by the argument given. Inverting these matrices supplies the stated dual functions, which are bounded because they are linear combinations of continuous functions on a compact interval.
The formula for $y$ is bounded and measurable. Integrating horizontally eliminates its correction term by compatibility and recovers $r_j$. Integrating vertically uses the dual identities and recovers $\widetilde z_i$.
Thus $x=y+h$ has exactly the four prescribed interim transfer schedules and absolutely integrable sections for every report. The general lemma is established, subject only to the terminology clarification above.
7. Incentives, participation, and budget balance
The chosen allocation and transfer schedules yield the displayed buyer and seller utility formulas.
All four deviation calculations have the correct signs and integration limits. Strict positivity of the belief densities makes each deviation loss strictly positive whenever the report differs from the true value, including deviations involving boundary values.
Truthful utility is nonnegative for every buyer value and seller cost. The section-integrability result ensures finite interim expected transfers and utilities for every report under each required belief.
Pointwise budget balance follows from using the same finite transfer with opposite signs in the two utilities. The task places no further restriction on transfers at pairs where trade does not occur.
The uniform aggregate transfer calculations are also correct:
$$
\int_0^1r_0(v)\,dv=\frac16,
\qquad
\int_0^1z_0(c)\,dc=\frac13.
$$
They demonstrate why this particular construction cannot be jointly absolutely integrable, without contradicting the stated feasibility requirements.
8. Optimality and threshold decision
The pointwise upper bound
$$
(v-c)p(v,c)\le\max(v-c,0)
$$
is correct for both signs of $v-c$. Its integral exists because the allocation is bounded and the densities are continuous.
The feasible efficient allocation attains that upper bound. Therefore the exact threshold criterion follows, including equality thresholds. The additional exclusion involving $F$ and $G$ is unnecessary for this construction, which works on a larger domain containing every admissible input.
9. Exact arithmetic and running time
The integration formula is correct:
$$
\int_0^v(v-c)c^j\,dc
=
\frac{v^{j+2}}{(j+1)(j+2)}.
$$
The resulting expression for $S_{\mathrm{eff}}$ has the correct denominator $i+j+3$.
The common-denominator construction is valid. The distinct consecutive factors $j+1$ and $j+2$ occur in $(m+2)!$, while $i+j+3$ occurs in $(2m+3)!$. Thus their product divides the proposed $T$.
The products of encoded coefficient denominators have bit length $O(L)$. The factorial contribution has bit length $O((m+1)\log(m+2))$, and summation adds only logarithmic overhead. The asserted intermediate bit-length bound and conservative schoolbook arithmetic bound are therefore justified.
The level proportions require only input reading; no computation proportional to an infinite type support is used. The example satisfies the stated density restrictions, and its surplus evaluates correctly to $31/120$.
The output resolves the assigned question. Its only needed repair is a precise use of “support.”
⟨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.