Answer: —
Theorem 9 — class: hardness
Given a finite set \(\Theta\) of complete worker cost types, a rational mass vector \(\mu\in\Delta(\Theta)\), finite category-labelled chore templates, and horizon \(H\), with \(c_\theta(B)=|B|-r_\theta(B)\) for a partition-matroid rank \(r_\theta\), an adversary reveals a per-capita stream of chore copies online. Each copy is assigned immediately and wholly to one agent, with no batch coordination; the fluid state is the mass \(z_{\theta,B}\) of type-\(\theta\) agents holding integral bundle \(B\). Decide whether a nonanticipating deterministic policy can be complete and \(\alpha\)-EF1 for \(\mu\)-almost every ordered pair at every prefix, or minimize \(\alpha\ge1\).
An atomless society has total mass \(1\) and is given by its support: \(\tau\) complete worker types \(\theta\), each with positive binary-rational mass \(\mu_\theta\), summing to \(1\). Here \(\tau\) is the support size, not the number of possible types.
There are \(m\ge1\) chore templates. Each type specifies a partition of the templates into categories, with capacity \(1\) per category; distinct copies of a template share its category. For an integral multiset bundle \(B\), let \(r_\theta(B)\) count its occupied categories and \(c_\theta(B)=|B|-r_\theta(B)\in\mathbb Z_{\ge0}\).
The types, templates and positive binary-rational horizon \(H\) are known initially. Cumulative arrival volume \(s\) is measured in copies per capita. An adversary reveals only the current template \(a(s)\), using finitely many constant segments with rational endpoints, and may stop at any \(s\le H\).
Initially \(z_{\theta,\varnothing}=\mu_\theta\). A deterministic nonanticipating policy selects measurable nonnegative transfer rates \(u_{\theta,B}(s)\), moving worker mass from \(B\) to \(B+\mathbf e_{a(s)}\); every affected worker receives one whole copy immediately and irrevocably, without storage or reallocation.
Precisely, bundle-state coordinates are absolutely continuous and satisfy \(\dot z_{\theta,B}=\mathbf1_{\{B_{a(s)}>0\}}u_{\theta,B-\mathbf e_{a(s)}}-u_{\theta,B}\), with \(u_{\theta,B}=0\) wherever \(z_{\theta,B}=0\). Completeness requires \(\sum_{\theta,B}u_{\theta,B}=1\) almost everywhere, \(z\ge0\), \(\sum_Bz_{\theta,B}=\mu_\theta\), and \(\sum_{\theta,B}|B|z_{\theta,B}=s\).
At every prefix, \(\alpha\)-EF1 means that for every pair of positive-mass states \((\theta,B),(\phi,C)\) with \(B\ne\varnothing\), some copy \(e\in B\) satisfies \(c_\theta(B\setminus\{e\})\le\alpha c_\theta(C)\). This is fairness for almost every ordered worker pair under the product population measure.
Running times are measured in \(m,\tau,H,L\) and the number of revealed segments, where \(L\) is the binary encoding length of the instance and revealed stream.
Does some finite universal factor \(\alpha\ge1\) permit a complete deterministic nonanticipating policy maintaining \(\alpha\)-EF1 for every such society and adversarial stream, or is every finite universal factor impossible?
For finite societies, every finite EF1 factor is impossible for complete deterministic online allocation, already with two workers and capacity-one partition costs ([Wang–Wei](https://arxiv.org/html/2505.24321v2#S4.SS2.SSS2)). Offline, binary supermodular costs admit polynomial-time EF1 and Pareto-efficient allocation ([Barman–Narayan–Verma](https://arxiv.org/abs/2302.11530)).
No result settling this exact continuous online model was located in the literature search.
The universal-factor question remains unresolved here. The main new result is an exact criterion for a narrower question: from a given fair state, can one template be repeated arbitrarily long while maintaining some fixed finite factor?
A directed graph of zero-cost witnesses answers that question. A cycle prevents arbitrarily long continuation for every finite factor; an acyclic graph permits such continuation with a factor depending on the initial state. I also construct reachable deadlocks with strictly positive comparison costs, even when the two partitions are ordered by refinement.
These results do not prove that an adversary can force every policy into a deadlock.
Write $\mathbf e_a$ for one copy of template $a$. For every type $\theta$, define
$$
d_\theta(B)=\max\{c_\theta(B)-1,0\}.
$$
For every nonempty bundle $B$,
$$
\min_{e\in B}c_\theta(B\setminus\{e\})=d_\theta(B).
$$
Indeed, deletion decreases cardinality by one and decreases the number of occupied categories by either zero or one. Thus it decreases cost by at most one. If the cost is positive, some category contains at least two copies, and deleting a copy from that category decreases cost by exactly one. If the cost is zero, deletion leaves cost zero.
For a state $z$, let
$$
S=\{(\phi,C):z_{\phi,C}>0\},
\qquad
h_\theta=\min_{(\phi,C)\in S}c_\theta(C).
$$
The minimum exists because it is the minimum of a nonempty set of nonnegative integers.
Consequently, $\alpha$-EF1 is equivalent to
$$
d_\theta(B)\le \alpha h_\theta
\qquad
\text{for every }(\theta,B)\in S.
$$
In particular,
$$
h_\theta=0
\quad\Longrightarrow\quad
c_\theta(B)\le 1
\quad\text{for every supported bundle owned by }\theta.
$$
This implication holds for every finite factor.
Let $k_\theta$ be the number of categories of type $\theta$. At a complete state of volume $s$, some supported bundle $C$ satisfies
$$
|C|\le \lfloor s\rfloor.
$$
Otherwise the population mean cardinality would exceed $s$.
For every evaluator $\theta$,
$$
c_\theta(C)\le \max\{0,\lfloor s\rfloor-1\}.
$$
Fairness therefore implies, for every supported own-bundle $B$,
$$
|B|
\le
k_\theta+1+
\left\lfloor
\alpha\max\{0,\lfloor s\rfloor-1\}
\right\rfloor.
$$
There are finitely many multisets under this bound, so every finite-factor fair state has finite support.
Adding a copy never decreases any evaluator’s cost. This implies that each $h_\theta(s)$ is nondecreasing along every complete trajectory.
Here is a direct justification using the stipulated dynamics. Fix a time $s_0$ and put $h=h_\theta(s_0)$. If $h=0$, the assertion is immediate. Otherwise the set
$$
\mathcal D=\{C:c_\theta(C)<h\}
$$
is finite, because
$$
|C|=c_\theta(C)+r_\theta(C)\le h-1+k_\theta.
$$
No transfer can enter $\mathcal D$ from outside it: that would decrease $\theta$-cost. Summing the differential equations over all types and all bundles in $\mathcal D$ shows that their total mass is nonincreasing. It starts at zero and remains nonnegative, so it stays zero. Hence $h_\theta(s)\ge h$ for all later times.
In particular, once a type has a positive evaluator minimum, it never again has a zero-cost comparison bundle.
Fix a finite-factor fair state $z^0$ of volume $s_0$, and fix a template $a$. All definitions in this section refer to this initial state.
Let
$$
Z=\{\theta:h_\theta=0\}.
$$
For an evaluator $\theta$, define
$$
\varepsilon_\theta(C,a)=
\begin{cases}
1,&\text{if the category containing }a\text{ is unoccupied in }C,\\
0,&\text{otherwise}.
\end{cases}
$$
For every integer $j\ge 1$,
$$
c_\theta(C+j\mathbf e_a)
=
c_\theta(C)+j-\varepsilon_\theta(C,a).
$$
If $\phi\in Z$ owns a supported bundle $C$, then $c_\phi(C)\le 1$. Define
$$
q_\phi(C,a)=1-c_\phi(C)+\varepsilon_\phi(C,a).
$$
Thus $q_\phi(C,a)\in\{0,1,2\}$, and it is exactly the largest number of additional copies of $a$ that this worker can receive while its own cost remains at most one.
If $c_\theta(C)=0$, define
$$
t_\theta(C,a)=\varepsilon_\theta(C,a).
$$
This is exactly the largest number of additional copies of $a$ for which the bundle still has zero $\theta$-cost.
Define a directed graph $G_a(z^0)$ on the vertex set $Z$. Put an edge
$$
\theta\longrightarrow\phi
$$
if some supported bundle $C$ owned by $\phi$ satisfies
$$
c_\theta(C)=0,
\qquad
q_\phi(C,a)\le t_\theta(C,a).
$$
The edge means that $C$ remains a zero-cost witness for $\theta$ even after its owner $\phi$ has received every additional copy allowed while $h_\phi=0$.
There are no self-loops. Indeed, when $c_\theta(C)=0$,
$$
q_\theta(C,a)
=
1+\varepsilon_\theta(C,a)
>
t_\theta(C,a).
$$
Theorem. The following are equivalent:
More precisely:
$$ \beta=M+2, \qquad M=\max_{(\phi,C)\in S}c_\phi(C). $$
The factor in this theorem depends on the state. It is not a universal factor.
Choose a directed cycle with vertex set $T$. For each $\theta\in T$, choose its cycle edge $\theta\longrightarrow\phi$ and a supported bundle $C_\theta$ witnessing that edge. Write
$$
q_\theta=q_\phi(C_\theta,a).
$$
Every bundle in the chain
$$
C_\theta,\ C_\theta+\mathbf e_a,\ \ldots,\
C_\theta+q_\theta\mathbf e_a
$$
has zero $\theta$-cost.
Let $W_\theta(s)$ be the total mass in these chain states, owned by $\phi$. Initially,
$$
W_\theta(s_0)>0.
$$
Suppose all $W_\theta$ are positive on an interval. Then $h_\theta=0$ for every $\theta\in T$. In particular, the owner $\phi$ of each selected chain has $h_\phi=0$.
The final chain state has own cost one, and adding another $a$ would give own cost two. Such a target state must have zero mass while $h_\phi=0$, under every finite fairness factor. Its outgoing rate is zero by the zero-mass restriction. Its differential equation consequently forces the transfer rate out of the final chain state to be zero almost everywhere.
Summing the equations along the chain now gives
$$
\dot W_\theta\ge 0
$$
almost everywhere: there may be incoming mass at the beginning, but there is no outgoing mass at the end.
These inequalities show that
$$
W_\theta(s)\ge W_\theta(s_0)>0
$$
throughout any fair continuation. To make the persistence argument explicit, take the maximal interval starting at $s_0$ on which all the $W_\theta$ remain positive. On this interval they are nondecreasing. Continuity then prevents any of them from reaching zero at a finite endpoint, so the interval extends through the entire continuation.
Thus every selected witness persists.
Fix one selected chain and put
$$
D=|C_\theta|+q_\theta.
$$
At every time, some supported bundle in this chain has cardinality at most $D$. Hence, for every evaluator $\psi$,
$$
h_\psi(s)\le \max\{0,D-1\}.
$$
If the maintained factor is $\beta<\infty$, fairness implies
$$
|B|
\le
k_\psi+1+
\left\lfloor\beta\max\{0,D-1\}\right\rfloor
$$
for every supported bundle owned by $\psi$. Taking population means yields the absolute volume bound
$$
s\le
\sum_\psi\mu_\psi(k_\psi+1)
+
\left\lfloor\beta\max\{0,D-1\}\right\rfloor.
$$
Therefore the continuation cannot have arbitrarily large length.
Set $\beta=M+2$. The initial state is $\beta$-EF1: types with zero evaluator minimum have own cost at most one; types with positive evaluator minimum have own deletion-cost at most $M$, while their minimum is at least one.
For bookkeeping, distinguish workers by their initial type-state. These labels only split coordinates internally. Summing labelled transfer rates produces valid rates in the original model. A positive labelled source has a positive original source coordinate, so the zero-mass restriction is respected.
First perform the following transfers:
Process these finite chains in a fixed deterministic order.
Every worker owned by a type in $Z$ keeps own cost at most one. Every other owner has evaluator minimum at least one throughout, by Section 1.3. Such workers receive at most two additional copies, so their own cost is at most $M+2$, and their deletion-cost is at most $M+1\le\beta$. Thus every intermediate state is $\beta$-EF1.
At the end of this phase, an evaluator $\theta\in Z$ has a zero-cost comparison bundle exactly when an edge
$$
\theta\longrightarrow\phi
$$
remains to an owner $\phi\in Z$. Indeed:
$$ q_\phi(C,a)\le t_\theta(C,a). $$
Since the graph is acyclic, it has a vertex $\theta$ with no outgoing edge. Its evaluator minimum is therefore positive. Top up every worker of this type to exactly two added copies of $a$. These transfers remain fair by the same $M+2$ bound. They remove all zero-cost witnesses owned by this type.
Delete that vertex and repeat. A finite acyclic graph always has a vertex with no outgoing edge, so eventually every worker has received exactly two additional copies of $a$. The total additional volume used is exactly two, and every evaluator minimum is now positive.
Continue in full population passes, giving one further copy of $a$ to every worker in each pass. During a pass after $n$ completed additional copies per worker, every worker has received either $n$ or $n+1$ copies, where $n\ge 2$. Therefore
$$
h_\theta\ge n-1,
\qquad
d_\theta(B)\le M+n.
$$
The required inequality follows from
$$
(M+2)(n-1)-(M+n)
=
(M+1)(n-2)\ge 0.
$$
All transfers use finitely many labelled chains per pass. Their rates can be set to zero at depletion instants. The construction is deterministic and can be interrupted at any prefix; it does not require knowledge of the stopping time. This proves the theorem.
The graph criterion gives a stronger obstruction than an immediate deadlock: every current recipient may initially be admissible, yet a sufficiently long constant continuation can still be impossible.
Take types $A,B$, each of mass $1/2$, with partitions
$$
A:\ \{\{a,b\},\{c\}\},
\qquad
B:\ \{\{a,c\},\{b\}\}.
$$
At volume one, consider
$$
z_{A,\{b\}}
=
z_{A,\{c\}}
=
z_{B,\{b\}}
=
z_{B,\{c\}}
=
\frac14.
$$
Every bundle is a singleton, so all own costs and deletion-costs are zero. In particular, every recipient can initially accept $a$ fairly.
The bundle $\{c\}$ owned by $B$ witnesses $A\longrightarrow B$:
Similarly, $\{b\}$ owned by $A$ witnesses $B\longrightarrow A$.
Thus both evaluator minima remain zero throughout every finite-factor fair constant-$a$ continuation. All own costs must consequently remain at most one.
The four initial groups can receive at most the following numbers of additional copies:
$$
\begin{array}{c|cccc}
\text{initial type-state}
&(A,\{b\})&(A,\{c\})&(B,\{b\})&(B,\{c\})\\
\hline
\text{maximum added copies of }a
&1&2&2&1
\end{array}
$$
For example, adding $a$ to $(A,\{b\})$ immediately incurs cost one; a second added copy would incur cost two. For $(A,\{c\})$, the first copy occupies a new category and the second incurs cost one.
During a constant-$a$ continuation, the initial counts of $b$ and $c$ are preserved. The dynamics therefore retain mass $1/4$ in each of these four groups of descendants. The total additional volume is at most
$$
\frac14(1+2+2+1)=\frac32.
$$
This bound is attainable. First give one $a$ to every worker, using volume one. Then give another $a$ to the two groups whose first added copy occupied a new own category, using volume $1/2$. Own costs never exceed one, so all prefixes are EF1.
The resulting state, at volume $5/2$, is an immediate deadlock: every own cost is one, both evaluator minima are zero, and adding $a$ to any recipient raises its own cost to two.
The initial singleton state is reached by revealing:
Every prefix has only empty bundles and singletons, so every prefix is EF1.
This proves that identical bundle distributions across types, combined with initially admissible choices, can lead to a state from which a long constant continuation is impossible. It does not prove that every policy must reach this state.
Say that partition $P_\phi$ refines $P_\theta$ if each category of $P_\phi$ is contained in a category of $P_\theta$. Then
$$
c_\phi(C)\le c_\theta(C)
$$
for every bundle $C$.
Corollary. If all type partitions are totally ordered by refinement, then $G_a(z^0)$ is acyclic for every template $a$ and every finite-factor fair state.
Proof. Suppose there is an edge $\theta\longrightarrow\phi$ witnessed by $C$. Assume, for contradiction, that $P_\phi$ refines $P_\theta$.
Since $c_\theta(C)=0$, refinement gives $c_\phi(C)=0$, and hence
$$
q_\phi(C,a)=1+\varepsilon_\phi(C,a).
$$
If $\varepsilon_\theta(C,a)=0$, then
$$
q_\phi(C,a)\ge 1>t_\theta(C,a)=0.
$$
If $\varepsilon_\theta(C,a)=1$, refinement implies $\varepsilon_\phi(C,a)=1$: occupying the finer category of $a$ would also occupy its coarser category. Therefore
$$
q_\phi(C,a)=2>t_\theta(C,a)=1.
$$
Both cases contradict the edge condition.
Thus every edge points from a strictly finer partition to a coarser partition. A directed cycle is impossible. ∎
This corollary concerns zero-cost obstructions. The next construction shows that positive-cost obstructions remain possible.
Fix any finite $\alpha\ge 1$, and define integers
$$
p=1+\lfloor\alpha\rfloor,
\qquad
q=1+\lfloor\alpha p\rfloor.
$$
Then
$$
p-1\le\alpha<p,
\qquad
q-1\le\alpha p<q,
\qquad
q\ge p+1.
$$
Use $q$ templates
$$
a,b_2,\ldots,b_q
$$
and two types of mass $1/2$:
These partitions are ordered by refinement.
Consider the state
$$
z_{F,B}=\frac12,
\qquad
z_{G,C}=\frac12,
$$
where
$$
B=(p+1)\mathbf e_a,
\qquad
C=2\mathbf e_a+\sum_{j=2}^{q}\mathbf e_{b_j}.
$$
Its volume is
$$
s_0=\frac{p+q+2}{2}.
$$
The costs are
$$
c_F(B)=p,\qquad c_F(C)=1,
$$
$$
c_G(B)=p,\qquad c_G(C)=q.
$$
Thus
$$
h_F=1,\qquad h_G=p.
$$
The state is $\alpha$-EF1 because
$$
d_F(B)=p-1=\lfloor\alpha\rfloor\le\alpha h_F,
$$
$$
d_G(C)=q-1=\lfloor\alpha p\rfloor\le\alpha h_G.
$$
However, adding $a$ gives
$$
d_F(B+\mathbf e_a)=p>\alpha h_F,
$$
$$
d_G(C+\mathbf e_a)=q>\alpha h_G.
$$
To verify that these inequalities forbid every positive continuation, note that every coordinate is $1$-Lipschitz: its incoming and outgoing rates are nonnegative and their sum is at most the total rate one. Therefore both initial states remain positive for an additional interval of length less than $1/4$.
On that interval, the two initial comparison bundles force both target coordinates to remain zero by fairness. Their outgoing rates are zero, and their differential equations force the outgoing rates from both initial sources to be zero almost everywhere.
Every other coordinate starts at zero. Induction on cardinality shows that it cannot acquire mass or send mass: during a constant-$a$ segment, its only possible predecessor has smaller cardinality and has zero outgoing rate. Since there are countably many coordinates, all outgoing rates are simultaneously zero almost everywhere. This contradicts completeness.
Thus this is an immediate deadlock, although both evaluator minima are strictly positive.
The deadlock state has a complete, deterministic, nonanticipating fair reaching trajectory.
First reveal $a$ for volume two, giving two copies to every worker in two population passes. All bundles have cardinality at most two during this phase, so deleting one copy leaves cost zero under every type.
Continue revealing $a$ and give each type-$F$ worker another $p-1$ copies. Throughout this phase:
$$ p-1\le\alpha. $$
Every prefix remains $\alpha$-EF1.
Finally reveal each $b_j$ for volume $1/2$, giving one copy to every type-$G$ worker. These templates occupy new categories under $F$, so every type-$G$ bundle continues to have $F$-cost one.
Suppose the current $G$-cost before one such population pass is $\ell$. During the pass, the supported $G$-costs are $\ell$ and $\ell+1$, while the type-$F$ bundle has $G$-cost $p$. It is enough to check
$$
\ell\le\alpha\min\{p,\ell\}.
$$
If $\ell\le p$, this follows from $\alpha\ge 1$. If $\ell>p$, it follows from
$$
\ell\le q-1=\lfloor\alpha p\rfloor\le\alpha p.
$$
Thus every prefix is fair, and the stated deadlock is reached.
All segment endpoints are rational: every population pass has length $1$ or $1/2$, and $p,q$ are integers.
Again, this is fair reachability by particular choices, not an adversary defeating every policy.
The exact constant-template criterion separates two issues:
Nevertheless, positive-cost deadlocks exist for every finite factor, even with nested partitions. Therefore preventing zero-cost cycles alone does not furnish a universal policy.
The missing global argument is still one of the following: a fixed finite-factor policy that avoids all relevant obstructions on every stream, or an adversary that forces an obstruction against every deterministic nonanticipating policy.
⟨structured verdict parsed — rendered above⟩
Overall assessment: valid partial progress; the universal-factor question remains unresolved. The auxiliary criterion and both reachable-deadlock constructions are sound. They do not establish either a universal policy or an adversary defeating every policy. The prover accurately acknowledges this limitation.
1. Basic consequences of the model
The deletion-cost identity is correct:
$$
\min_{e\in B}c_\theta(B\setminus\{e\})
=
\max\{c_\theta(B)-1,0\}.
$$
Positive cost guarantees a category containing at least two copies; deleting one of those copies reduces cost by exactly one. At zero cost, deletion preserves zero cost.
The evaluator-minimum reformulation is also correct. The support is nonempty because total population mass is one, and the minimum exists because evaluator costs are nonnegative integers. Empty own-bundles impose no additional constraint: their defined deletion-cost is zero.
The finite-support argument is valid. If every supported cardinality exceeded $\lfloor s\rfloor$, the mean cardinality would exceed $s$. For a supported comparison bundle of cardinality at most $\lfloor s\rfloor$,
$$
c_\theta(C)\le \max\{0,\lfloor s\rfloor-1\}.
$$
Combining fairness with $r_\theta(B)\le k_\theta$ gives the stated bound on own-bundle cardinality. Finite template number then gives finite support.
The monotonicity proof for $h_\theta$ is rigorous. The set of bundles with evaluator cost below its initial minimum is finite and cannot receive transfers from outside that set. Summing its coordinate equations therefore proves that its initially zero mass remains zero. This argument applies across changes of the arriving template.
2. Constant-template witness graph
The repeated-copy formula is correct for every integer $j\ge1$:
$$
c_\theta(C+j\mathbf e_a)
=
c_\theta(C)+j-\varepsilon_\theta(C,a).
$$
The first added copy can occupy one new category; subsequent copies cannot occupy another category.
For owners with zero evaluator minimum, fairness gives own cost at most one. The formulas for $q_\phi(C,a)$ and $t_\theta(C,a)$ therefore correctly measure, respectively, the maximum permitted additions while own cost stays at most one and the maximum additions preserving zero evaluator cost.
The edge interpretation is correct, including the boundary case $q_\phi(C,a)=0$. The exclusion of self-loops follows from the displayed strict inequality.
3. Cycle obstruction
The persistence argument is valid.
While all selected witness-chain masses are positive, every cycle evaluator has minimum zero. Each selected chain’s owner is another cycle vertex, so its own supported cost must remain at most one. The last chain state has own cost one, and its successor has own cost two.
That successor must consequently have zero mass throughout the interval. The zero-mass rate restriction and its differential equation force the transfer out of the last chain state to vanish almost everywhere. Summing along each chain gives nondecreasing chain mass.
The continuity argument closes the apparent circularity: none of the initially positive chain masses can reach zero at a finite endpoint of the interval on which all remain positive.
The resulting volume bound is correct. A surviving chain supplies a comparison bundle of cardinality at most $D$ for every evaluator. Hence each supported own-bundle satisfies the claimed cardinality bound, and taking population means gives
$$
s\le
\sum_\psi\mu_\psi(k_\psi+1)
+
\left\lfloor\beta\max\{0,D-1\}\right\rfloor.
$$
This rules out arbitrarily long fair constant-template continuation for any fixed finite factor.
4. Acyclic continuation
The construction with $\beta=M+2$ is sound.
The initial state is $\beta$-EF1: zero-minimum types have own cost at most one, while positive-minimum types have minimum at least one and own deletion-cost at most $M$.
The labelled bookkeeping is compatible with the original dynamics. A positive labelled source gives a positive original source, and summing labelled flows preserves the coordinate equations and completeness. Processing finitely many labelled chains sequentially realizes the specified transfers; depletion instants do not affect almost-everywhere completeness.
During the first phase, owners initially in $Z$ retain own cost at most one. Other owners retain positive evaluator minima and have own deletion-cost at most $M+1$.
At the end of this phase, the characterization of remaining zero witnesses by graph edges is exact. Two added copies give positive cost under every evaluator, initially positive costs remain positive, and the remaining cases satisfy precisely the edge inequality.
Removing a vertex with no outgoing edge is justified: its evaluator minimum is positive, so topping up its workers is fair and removes every zero witness owned by that type. Repeating this operation terminates because the graph is finite and acyclic.
For later population passes, the stated bounds
$$
h_\theta\ge n-1,
\qquad
d_\theta(B)\le M+n
$$
hold. The critical arithmetic checks:
$$
(M+2)(n-1)-(M+n)=(M+1)(n-2)\ge0
$$
for $n\ge2$.
There is one minor scope clarification: the original model has a fixed finite horizon $H$, whereas the theorem discusses every additional length. The theorem should explicitly quantify over arbitrarily large finite horizons, or state that it uses an auxiliary unbounded-horizon extension. This does not affect its mathematical argument.
5. The two-type zero-witness example
The partitions, singleton state, and both graph edges are correct. The cycle proof applies and keeps both evaluator minima zero throughout any finite-factor fair continuation.
The four maximum addition counts are correctly computed as
$$
1,\ 2,\ 2,\ 1.
$$
Constant-$a$ transfers preserve the initial counts of $b$ and $c$. Thus each initial group retains its mass among its descendants, and the additional-volume bound is
$$
\frac14(1+2+2+1)=\frac32.
$$
The proposed schedule attains this bound while every own cost remains at most one. At its endpoint, every recipient already occupies its own category of $a$ and has own cost one, so another copy produces a forbidden own cost two.
The reaching stream has rational segment endpoints and is fair at every prefix. This proves reachability under the specified allocation choices, not inevitability under arbitrary policies.
6. Refinement corollary
The refinement inequality
$$
c_\phi(C)\le c_\theta(C)
$$
has the correct direction when $P_\phi$ refines $P_\theta$.
Both cases in the edge-exclusion argument are correct. If the coarser category of $a$ is unoccupied, its finer category is also unoccupied. Consequently, an edge cannot point toward an equally fine or finer partition. Under total ordering by refinement, edges strictly decrease refinement and cannot form a cycle.
7. Positive-cost deadlocks
The parameter inequalities are correct, including $\alpha=1$:
$$
p-1\le\alpha<p,
\qquad
q-1\le\alpha p<q,
\qquad
q\ge p+1.
$$
All four cost computations and both evaluator minima are correct. They establish initial $\alpha$-EF1 and show that each possible first target violates fairness while the corresponding initial comparison bundle remains supported.
The immediate-deadlock argument handles simultaneous transfers correctly. Every coordinate is $1$-Lipschitz because its incoming and outgoing rates are distinct nonnegative terms in the total rate one. Therefore both initial coordinates remain positive on the stated short interval.
Fairness forces both first target coordinates to remain zero there. Their equations then force both initial outgoing rates to vanish almost everywhere. Induction on cardinality excludes mass or outgoing rates at every other coordinate. Countability permits a common full-measure set on which all rates vanish, contradicting completeness.
The reaching trajectory is also valid. The first two passes have deletion-cost zero. The additional copies assigned to type $F$ respect its bound against comparison cost one. During each later type-$G$ pass, its maximum deletion-cost is $\ell$, and its evaluator minimum is at least $\min\{p,\ell\}$. The displayed inequality verifies fairness throughout the pass. The endpoints are rational.
Again, this establishes a deadlock reachable by particular fair choices, not one forced against every policy.
8. What prevents resolution of the assigned task
The unresolved universal quantifier is a major missing component relative to USER_TASK, although it is not an error in the prover’s stated partial claim.
Neither reachability example shows that every deterministic nonanticipating policy must enter the exhibited state or another obstruction. Conversely, the acyclic continuation theorem uses a state-dependent factor and addresses only constant-template continuations. It does not provide a fixed universal factor for arbitrary streams.
The minimum substantive addition needed is exactly the global argument identified by the prover: either a universal policy with a complete fairness guarantee, or an adversary forcing failure against every such policy. No complexity classification of the continuous problem has been established.
⟨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.