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Econ-12 · economics · Economics

Grand-coalition arrival in the patient limit

LLM status Solved unverified

Statement

ID: GT-092. Category: Dynamic coalition formation. Status: P for the patient-limit question; the catalog's counterexample claim requires a version correction.

Definitions and mathematical statement

Fix a finite player set $N$, $|N|\geq2$. Let $\mathcal X$ be the set of laminar families of subsets of $N$ of size at least two: any two members are disjoint or one contains the other. For $x\in\mathcal X$, let $P(x)$ comprise its maximal members and the singleton players outside their union. Put $\mathcal G=\{x:N\in x\}$. Payoffs are $\pi_i:\mathcal X\to\mathbb R$ and $V(x)=\sum_i\pi_i(x)$. Require the source's Assumptions 2.8 and 2.9: \[ V(h)=V^*\ (h\in\mathcal G),\quad \gamma=\min_{x\notin\mathcal G}(V^*-V(x))>0, \] and, for each $K\in x\cap P(x)$, positive shares $s_i(x,K)$ summing to one over $i\in K$, with \[ \Delta(x,K)=\sum_{j\in K}\left(\pi_j(x)-\pi_j(x\setminus\{K\})\right),\qquad \pi_i(x)=\pi_i(x\setminus\{K\})+s_i(x,K)\Delta(x,K). \]

A move has a target $\tau(m)$ and a nonempty relevant player set $R(m)$. A merge chooses at least two blocks $\mathcal A\subseteq P(x)$, sets $K=\bigcup\mathcal A$, and has target $x\cup\{K\}$ and $R(m)=K$. A unilateral termination chooses $K\in x$ and $i\in K$, and has target $x\setminus\{L\in x:K\subseteq L\}$ and $R(m)=\{i\}$. A stay by $i$ has target $x$ and $R(m)=\{i\}$. Write $M(x)$ for moving moves and $M^+(x)$ for all these moves.

For $\delta\in(0,1)$ and a stochastic transition matrix $p$ on $\mathcal X$, define \[ \ell_i=(1-\delta)(I-\delta p)^{-1}\pi_i,\qquad \ell_i^+(x)=\sum_y p_{xy}\ell_i(y). \] At $x$, rank targets by $\ell_i$, breaking equal values in favour of $x$; denote this total preorder by $\succeq_i^x$ and its strict part by $\succ_i^x$. A move $m$ is profitable if $\ell_i(\tau(m))\geq\ell_i^+(x)$ for every $i\in R(m)$. It is dominated if some $m'\in M^+(x)$ satisfies $R(m')\subseteq R(m)$ and $\tau(m')\succ_i^x\tau(m)$ for every $i\in R(m')$. Let $C(x)$ be the profitable undominated moves, and \[ F(x)=\{m\in C(x):\exists i\in N\ \forall m'\in C(x),\ \tau(m)\succeq_i^x\tau(m')\}. \] The matrix $p$ is an equilibrium if, at every $x$, \begin{align*} p_{xy}>0, y\neq x&\ \Longrightarrow\ \exists m\in C(x):\tau(m)=y,\\ C(x)\cap M(x)\neq\varnothing&\ \Longrightarrow\ p_{xx}<1,\\ m\in F(x)&\ \Longrightarrow\ p_{x,\tau(m)}>0. \end{align*} For the resulting Markov chain let $T_{\mathcal G}=\inf\{t\geq0:X_t\in\mathcal G\}$.

Does there exist one fixed payoff model satisfying the assumptions, a sequence $\delta_k\uparrow1$, equilibrium matrices $p_k$, and states $x_k$ such that \[ \mathbb P_{x_k}^{p_k}(T_{\mathcal G}=\infty)>0\quad\text{for every }k? \] Equivalently, is it true that every fixed model has a threshold $\bar\delta<1$ beyond which every equilibrium arrives at $\mathcal G$ almost surely from every initial state?

Short English statement

Can a fixed coalition-formation model sustain nonarrival for equilibria with discount factors arbitrarily close to one?

Variants and scope boundaries

The fixed-model quantifier is stronger than allowing payoffs or the number of players to change with $k$. A model-dependent patience threshold is weaker than a universal threshold. The question uses unilateral termination and sharing of per-period surplus. Sharing discounted surplus, unanimous termination, or deleting only a nested agreement while preserving its ancestors changes the problem.

Source relationship and status

The definitions follow the inspected v2's Sections 2.1--2.4, and the patient-limit target follows its Section 8. Remark 6.7 and the heading of Proposition 6.8 explicitly state that the earlier deterministic-cycle construction is not an equilibrium under revised Definition 2.14, under either termination rule. Consequently the catalog's claim of established failure for $\delta\leq3/4$ cannot be transferred to these definitions. The formulation above fixes the primitives, making one precise patient-limit question; the source's broader phrase does not itself distinguish all uniformity variants.

References

[S1] Jobst Heitzig, Does the grand coalition form? Persistence, arrival, and the role of the sharing rule in a dynamic process of nested binding agreements (2026), arXiv:2608.17766v2. Inspected locators: Definitions 2.1--2.4, Assumptions 2.8--2.9, Definitions 2.14--2.15, Remark 6.7 and Proposition 6.8, Section 8, Open problems (2). https://arxiv.org/html/2608.17766v2

Common conventions: Source mathematical conventions

Finite games. For a finite set $A$, $\Delta(A)$ is its probability simplex. Players choose independent mixed strategies in Nash equilibrium; correlation is allowed only when explicitly part of the solution concept. In a game with payoff functions $u_i$ and strategy profile $x$, an additive $\varepsilon$ Nash equilibrium satisfies \[ u_i(x)\geq u_i(x_i',x_{-i})-\varepsilon \quad\text{for every player $i$ and every admissible deviation $x_i'$}. \] For numerical approximation, payoffs are normalized as locally specified. An $\varepsilon$ payoff residual is distinct from distance at most $\varepsilon$ to the set of exact equilibria. Point convergence, convergence of empirical play, and convergence in distance to an equilibrium set are also distinct.

Computational representations. Unless stated otherwise, a complexity question takes rational data with binary encoding; polynomial time is measured in total bit length. A fixed-accuracy polynomial algorithm is not necessarily polynomial in $1/\varepsilon$, and neither is necessarily polynomial in $\log(1/\varepsilon)$. Succinct games and value, demand, or sampling oracles must declare their interfaces. Exact real solutions require an explicit representation; an unspecified list of arbitrary real numbers is not a finite computational output. A claimed algorithmic barrier is meaningful only for its specified model and reductions.

Dynamic games. Histories, information, observation, and allowable randomization are part of the model. A stationary strategy depends only on the current state; a positional strategy in a deterministic graph game chooses one action at each controlled vertex. Nash equilibrium at an initial state and equilibrium after every history are different requirements. An expectation of a pathwise $\liminf$ payoff, a $\liminf$ of expected averages, a limiting expected average, and a uniform finite-horizon equilibrium are different criteria. None is silently substituted for another.

Allocation and mechanisms. A complete allocation assigns every item exactly once unless otherwise stated. Zero-valued goods, negative values, capacity constraints, feasibility, lotteries, and copies of goods affect fairness definitions and are specified locally. Dominant-strategy incentive compatibility, Bayesian incentive compatibility, universal truthfulness, and truthfulness in expectation impose different inequalities. Whenever an objective ratio has zero denominator, its convention is stated or those instances are excluded.

Infinite-dimensional models. Expectations, integrals, stochastic equations, and optimization objectives require their local regularity, integrability, filtrations, and admissible domains. A backward--forward mean-field system is a boundary-value problem; it is not an initial-value semiflow without an additional construction. Long-time, large-population, and vanishing-noise limits require an order of limits and a convergence topology. If a minimum need not be attained, the objective is its infimum or supremum and the approximation requirement is stated separately.

Statement licence: CC-BY-4.0 · Source file

Contributions 3

Econ-12-A3 · Full solution claimed · LLM-generated · gpt 6 astra pro ·

GPT 6 Astra Pro attempt on Econ-12, version 2

Grand-coalition arrival in the patient limit

C71-12: research attempt and adversarial verification

9 October 2026

Problem ID: C71-12  Model: GPT 6 Astra Pro Version: 2

Revision provenance. This version incorporates the supplied Pro audit's logical correction and independently reproduced rational checks. The complete construction, arithmetic table, theorems and proofs from version 1 are retained. The audit is an unpublished verification record, not a proof-assistant formalization or a claim of peer review. Tools. Version 1 used web browsing and exact rational computation. This revision reran the supplied standard-library exact verifier; the literature record below is carried forward from version 1.

FORMAL RESTATEMENT

Work with one fixed finite set $N$, a fixed real payoff table, and fixed positive sharing weights. These primitives may not depend on the discount factor. The state space $\mathcal X$ consists of all laminar families of subsets of $N$ of size at least two; a family is laminar when two members are disjoint or one contains the other. Its top-level partition $P(x)$ consists of maximal agreements and uncovered singleton players. Write $\mathcal G=\{x:N\in x\}$.

Require $V(x)=\sum_i\pi_i(x)$, $V(h)=V^*$ for $h\in\mathcal G$, and $\gamma=\min_{x\notin\mathcal G}(V^*-V(x))>0$. For each top-level agreement $K\in x\cap P(x)$ require \[ \pi_i(x)=\pi_i(x\setminus\{K\})+s_i(x,K) \sum_{j\in K}[\pi_j(x)-\pi_j(x\setminus\{K\})],\quad i\in K, \] where every $s_i(x,K)>0$ and $\sum_{i\in K}s_i(x,K)=1$.

A merge joins at least two blocks of $P(x)$, adds their union $K$, and needs the relevant player set $R=K$. Terminating $K\in x$ by $i\in K$ removes every agreement containing $K$ and has $R=\{i\}$. Player $i$ may also stay, with target $x$ and $R=\{i\}$. These are all moves. Let $M^+(x)$ include stays, and let $M(x)$ exclude them.

For a stochastic matrix $p$ and $0<\delta<1$, set \[ \ell_i=(1-\delta)(I-\delta p)^{-1}\pi_i, \qquad \ell_i^+(x)=\sum_y p_{xy}\ell_i(y). \] The inverse exists: $\sum_{t\ge0}\delta^tp^t$ converges entrywise and in the maximum-row-sum matrix norm. A player ranks targets by $\ell_i$, with the current state strictly preferred to every different target having equal value; all other value ties remain ties. A move $m$ is profitable when $\ell_i(\tau(m))\ge\ell_i^+(x)$ for every $i\in R(m)$. It is dominated when some $m'\in M^+(x)$ has nonempty $R(m')\subseteq R(m)$ and every member of $R(m')$ strictly prefers its target. Let $C(x)$ be the profitable undominated moves. Let $F(x)$ consist of moves in $C(x)$ whose target is weakly best in $C(x)$ for at least one player, whether or not that player is relevant to the move.

An equilibrium satisfies precisely \begin{align*} p_{xy}>0, \ y\ne x&\ \Longrightarrow\ \exists m\in C(x):\tau(m)=y;\tag{E1}\\ C(x)\cap M(x)\ne\varnothing&\ \Longrightarrow\ p_{xx}<1;\tag{E2}\\ m\in F(x)&\ \Longrightarrow\ p_{x,\tau(m)}>0.\tag{E3} \end{align*} No positive lower bound on a nonzero transition probability is imposed. In particular, (E1) does not require a positive diagonal entry to be justified by a profitable stay.

Literal question. Do there exist such fixed primitives, $\delta_k\uparrow1$, equilibrium matrices $p_k$, and initial states $x_k$ such that $\Pr_{x_k}^{p_k}(T_{\mathcal G}=\infty)>0$ for every $k$? Here the hitting time includes time zero. The answer below is affirmative, with $x_k=\varnothing$ for every $k$ and probability equal to one.

Logical correction. The attachment's "equivalently, is it true that every fixed model has a threshold" is the negation of the preceding existential assertion, not the same assertion. The two formulations describe opposite sides of the same decision problem. We prove the existential assertion and disprove the threshold assertion.

More precisely, let $M$ range over fixed admissible payoff models. The universal eventual-arrival assertion is \[ \begin{aligned} \mathcal A:\quad &\forall M\ \exists\bar\delta_M\in[0,1)\ \forall\delta\in(\bar\delta_M,1)\\ &\forall p\in\operatorname{Eq}_M(\delta)\ \forall x\in\mathcal X_M,\quad \Pr_x^p(T_{\mathcal G}<\infty)=1. \end{aligned} \] Its negation is the existence of one fixed $M$ with failed almost-sure arrival at discount factors arbitrarily close to one. Because $T_{\mathcal G}$ is finite or infinite, failure means $\Pr_x^p(T_{\mathcal G}=\infty)>0$. Choosing these discount factors successively above both the preceding choice and $1-1/k$ gives the sequence in the literal question. Conversely, such a sequence violates every proposed threshold for that $M$. Thus the catalogue's equivalence must read: "Equivalently, is the following universal arrival assertion false: every fixed model has a threshold ...?" Theorem Theorem (Fixed-model nonarrival arbitrarily close to patience) answers the existential question affirmatively and $\mathcal A$ negatively.

QUICK LITERATURE/CONTEXT CHECK

Heitzig's version 2 explicitly withdraws the equilibrium status of the old deterministic cycle under the revised rules: see Remark 6.7 and Proposition 6.8. Some later passages still refer to that superseded construction without the qualification. We use neither those claims nor any equilibrium-existence theorem from that source. We reuse its concrete block-value primitives, then give a different stochastic kernel and verify the stated axioms independently.

Checked primary source. Jobst Heitzig, Does the grand coalition form? Persistence, arrival, and the role of the sharing rule in a dynamic process of nested binding agreements, arXiv:2608.17766v2, especially Definitions 2.14--2.15 and Remark 6.7. https://arxiv.org/html/2608.17766v2. This is a preprint; no novelty or peer-review claim is made for the construction below.

ATTACK PLAN

Proof track. Try to turn the welfare gap into a drift or supermartingale argument and examine closed communicating classes. The obstruction is that (E2)--(E3) require positivity but do not bound transition rates away from zero.

Construction track, selected. Introduce waiting while preserving discounted values by a time-change identity. Specify every off-cycle row, not merely a recurrent cycle. Enumerate all moves using exact fractions, then replace numerical evidence by a rational Bellman certificate and a uniform strict-margin bound.

Tools and their roles. Finite Markov-chain closure proves nonarrival; Bellman equations certify discounted values; diagonal time change separates patience from effective transition frequency; laminar-family enumeration checks the entire domain; subset domination incorporates unilateral deviations and stays; rational gap bounds make the result uniform in $\delta$; independent exact arithmetic detects ties and excludes rounding errors.

Tiny cases. With two players there are only $e=\varnothing$ and $h=\{N\}$. Sharing implies $\pi_i(h)>\pi_i(e)$ for both players. If $a=p_{eh}$ and $b=p_{he}$, subtraction of the two Bellman equations gives \[ \ell_i(h)-\ell_i(e) =\frac{(1-\delta)[\pi_i(h)-\pi_i(e)]} {1-\delta+\delta(a+b)}>0. \] At $h$, every termination is dominated by its player's stay, hence (E1) gives $b=0$. At $e$, merge-all is profitable and undominated; it is best for both players among the candidates. Thus (E3) gives $a>0$, and arrival occurs almost surely. The counterexample uses four players.

WORK

A fixed payoff model

Let $N=\{A,B,C,D\}$. Concatenated letters denote sets, not new players. Define a block-value function $v(K;P)$ to be zero except for \[ \begin{array}{c|r} (K;P)&v(K;P)\\\hline (N;\{N\})&256\\ (AB;AB|C|D)&240\\ (CD;AB|CD)&240\\ (C;A|B|C|D)&112\\ (D;A|B|C|D)&128. \end{array} \] Define $\pi(x)$ recursively on the number of agreements. For a singleton block $\{i\}\in P(x)$ set $\pi_i(x)=v(\{i\};P(x))$. For a nonsingleton block $K\in P(x)$ put $x^-=x\setminus\{K\}$ and set \begin{equation}\label{wsref-0-65-71-3a-72-65-63-75-72-73-69-76-65-2d-70-61-79-6f-66-66} \pi_i(x)=\pi_i(x^-)+\frac{v(K;P(x))-\sum_{j\in K}\pi_j(x^-)}{|K|},\qquad i\in K. \end{equation} Every player belongs to exactly one top-level block, so this defines a single payoff for each player. The recursive calls have fewer agreements, so they terminate. Summing over $K$ shows that its players' total payoff equals $v(K;P(x))$. Therefore $V(x)=\sum_{K\in P(x)}v(K;P(x))$. For grand states this is 256; for nongrand states it is either 240 or 0. Hence $V^*=256$ and $\gamma=16$. Equation (eq:recursive-payoff) is exactly the required sharing identity with $s_i(x,K)=1/|K|>0$. Negative individual payoffs are allowed by the problem.

An explicit auxiliary kernel on every state

Define \[ x_0=\varnothing,\quad x_1=\{AB\},\quad x_2=\{AB,CD\},\quad x_3=\{CD\},\qquad C_0=\{x_0,x_1,x_2,x_3\}. \] Let $Q$ be the following stochastic matrix. All unspecified entries are zero, and the cases are applied in the displayed order.
  1. For $x=x_j\in C_0$, set $Q_{xx}=2/3$ and $Q_{x,x_{j+1\bmod4}}=1/3$.
  2. For $x=\{AB,CD,N\}$, give probabilities $3/8,3/8,1/4$ to $x,x_3,x_2$, respectively.
  3. For $x=\{AC,BD\}$ or $x=\{BC,AD\}$, give probability $1/4$ to each of $x$, $x\cup\{N\}$, and the two states containing just one of the two agreements in $x$.
  4. For either grand state obtained in (iii), set $Q_{xx}=1$.
  5. In every other state give probability $1/2$ to $x$ and $1/2$ to $\{AB\}$ when $AB\in x$, or to $\varnothing$ when $AB\notin x$.
These cases exhaust all states and do not assign the same destination twice within a row. All diagonal entries are positive. Every off-diagonal destination is a legal moving target. To check the last assertion in case (v), a laminar family on four players is a chain unless it has two disjoint pairs. The three possible two-pair families were already handled. Removing the innermost agreement empties a chain; if $AB$ is present, terminating the next larger agreement leaves exactly $\{AB\}$. The remaining special cases are direct merges or chain-rule terminations.

Define the auxiliary value table \begin{equation}\label{wsref-0-65-71-3a-4c} L=\frac14(I-\tfrac34 Q)^{-1}\pi, \qquad H=QL. \end{equation} $Q$ is auxiliary: it is not asserted to be an equilibrium at discount $3/4$.

Time change and the role of undominated moves

For a state $x$, let $U(x)$ be all undominated moves, without imposing profitability, using the target values $L$ and the specified status-quo tie rule.
Lemma (Strict improvement for relevant players): If a moving move $m$ is undominated, then $L_i(\tau(m))>L_i(x)$ for every $i\in R(m)$.
Proof: If $L_i(\tau(m))\le L_i(x)$, staying at $x$ by $i$ is strictly preferred to the different target $\tau(m)$, including when values tie. Its relevant set $\{i\}$ is contained in $R(m)$. This stay therefore dominates $m$. Domination is tested against all moves, not only profitable moves, so no profitability hypothesis on the stay is needed. ∎
Lemma (A sufficient exact certificate): Suppose a finite payoff model and a stochastic matrix $Q$ with positive diagonal have the values in (eq:L). Suppose, at every state, \begin{equation}\label{wsref-0-65-71-3a-73-75-70-70-6f-72-74} \{\tau(m):m\in U(x),\ \tau(m)\ne x\} =\{y\ne x:Q_{xy}>0\}. \end{equation} Let $\eta>0$ be the minimum of $L_i(\tau(m))-L_i(x)$ over all moving $m\in U(x)$ and $i\in R(m)$, and let $B=\max_{x,i}|H_i(x)-L_i(x)|$. When the moving set is nonempty and $B>0$, put $\alpha_0=\min(1,\eta/(2B))$. For every $0<\alpha\le\alpha_0$, the matrix \[ p_\alpha=(1-\alpha)I+\alpha Q \] is an equilibrium at $\delta=3/(3+\alpha)$.
Proof: Lemma Lemma (Strict improvement for relevant players) makes the finite minimum positive. The identity $L=\pi/4+3QL/4$ gives $\pi=4L-3QL$. Substitution shows \[ (1-\delta)\pi+\delta[(1-\alpha)L+\alpha QL]=L \quad\text{when }\alpha=3(1-\delta)/\delta. \] The unique discounted Bellman solution for $p_\alpha$ is consequently $\ell=L$. Its continuation benchmark is $\ell^+(x)=L(x)+\alpha(H(x)-L(x))$. For every relevant player of an undominated moving move, \[ L_i(\tau(m))-\ell_i^+(x) \ge\eta-\alpha B\ge\eta/2>0. \] Thus all undominated moving moves are profitable. Some undominated stays can be unprofitable; this causes no problem because all diagonal entries of $p_\alpha$ are positive.

Every positive off-diagonal transition has an undominated moving witness by (eq:support); that witness is profitable, proving (E1). If $C(x)$ has a moving member, (eq:support) provides a positive off-diagonal entry, proving (E2). Every profitable undominated move has either a supported moving target by (eq:support) or the supported current state. In particular all moves in $F(x)$ have supported targets, proving (E3). This establishes the lemma without an appeal to equilibrium existence. ∎

Proposition (Finite rational certificate for the displayed model): For the payoff model and $Q$ above, (eq:support) holds at all 52 states. The minimum and maximum in Lemma Lemma (A sufficient exact certificate) are \[ \eta=\frac{24}{115},\qquad B=\frac{608}{25},\qquad \alpha_0=\frac{15}{3496}. \]
Proof: The complete rational table below specifies $L$ and the set of undominated moving targets at every state. The displayed values satisfy $L=\pi/4+3QL/4$ row by row. To test domination, enumerate the legal merges, every termination by every eligible player, and each stay, and compare their relevant sets and target values. This finite calculation checks 558 moves. The complete standard-library verifier in Section 5 independently regenerates the laminar families, payoffs and moves and checks every assertion using rational arithmetic. It does not invoke a linear-programming or equilibrium solver. It also checks the two extrema rather than merely testing a sample of discount factors. Together with the table, this is an explicit finite arithmetic certificate. ∎

Complete arithmetic certificate

State identifiers are local to this table. A coalition is encoded by the bit mask $A=1,B=2,C=4,D=8$; for example, $3=AB$ and $15=N$. The last column lists all undominated moving target states, not relevant player sets.

IDAgreements$\pi_A,\pi_B,\pi_C,\pi_D$$L_A,L_B,L_C,L_D$Targets
0$\varnothing$$(0,0,112,128)$$(32,32,\frac{376}{5},\frac{424}{5})$$\{1\}$
1$\{3\}$$(120,120,0,0)$$(64,64,\frac{192}{5},\frac{208}{5})$$\{14\}$
2$\{5\}$$(-56,0,56,0)$$(\frac{-16}{5},\frac{96}{5},\frac{1688}{25},\frac{1272}{25})$$\{0\}$
3$\{6\}$$(0,-56,56,0)$$(\frac{96}{5},\frac{-16}{5},\frac{1688}{25},\frac{1272}{25})$$\{0\}$
4$\{7\}$$(\frac{-112}{3},\frac{-112}{3},\frac{224}{3},0)$$(\frac{64}{15},\frac{64}{15},\frac{5624}{75},\frac{1272}{25})$$\{0\}$
5$\{9\}$$(-64,0,0,64)$$(\frac{-32}{5},\frac{96}{5},\frac{1128}{25},\frac{1912}{25})$$\{0\}$
6$\{10\}$$(0,-64,0,64)$$(\frac{96}{5},\frac{-32}{5},\frac{1128}{25},\frac{1912}{25})$$\{0\}$
7$\{11\}$$(\frac{-128}{3},\frac{-128}{3},0,\frac{256}{3})$$(\frac{32}{15},\frac{32}{15},\frac{1128}{25},\frac{6376}{75})$$\{0\}$
8$\{12\}$$(0,0,-8,8)$$(16,16,\frac{168}{5},\frac{232}{5})$$\{0\}$
9$\{13\}$$(-80,0,32,48)$$(\frac{-64}{5},\frac{96}{5},\frac{1448}{25},\frac{1752}{25})$$\{0\}$
10$\{14\}$$(0,-80,32,48)$$(\frac{96}{5},\frac{-64}{5},\frac{1448}{25},\frac{1752}{25})$$\{0\}$
11$\{15\}$$(4,4,116,132)$$(\frac{104}{5},\frac{104}{5},\frac{2288}{25},\frac{2592}{25})$$\{0\}$
12$\{3,7\}$$(40,40,-80,0)$$(\frac{272}{5},\frac{272}{5},\frac{-224}{25},\frac{624}{25})$$\{1\}$
13$\{3,11\}$$(40,40,0,-80)$$(\frac{272}{5},\frac{272}{5},\frac{576}{25},\frac{-176}{25})$$\{1\}$
14$\{3,12\}$$(0,0,120,120)$$(8,8,\frac{384}{5},\frac{416}{5})$$\{8\}$
15$\{3,15\}$$(124,124,4,4)$$(88,88,\frac{616}{25},\frac{664}{25})$$\{1\}$
16$\{5,7\}$$(-56,0,56,0)$$(\frac{-16}{5},\frac{96}{5},\frac{1688}{25},\frac{1272}{25})$$\{0\}$
17$\{5,10\}$$(0,0,0,0)$$(\frac{240}{13},\frac{1152}{65},\frac{13248}{325},\frac{1104}{25})$$\{2,6,41\}$
18$\{5,13\}$$(-56,0,56,0)$$(\frac{-16}{5},\frac{96}{5},\frac{1688}{25},\frac{1272}{25})$$\{0\}$
19$\{5,15\}$$(8,64,120,64)$$(\frac{112}{5},\frac{224}{5},\frac{2328}{25},\frac{1912}{25})$$\{0\}$
20$\{6,7\}$$(0,-56,56,0)$$(\frac{96}{5},\frac{-16}{5},\frac{1688}{25},\frac{1272}{25})$$\{0\}$
21$\{6,9\}$$(0,0,0,0)$$(\frac{1152}{65},\frac{240}{13},\frac{13248}{325},\frac{1104}{25})$$\{3,5,44\}$
22$\{6,14\}$$(0,-56,56,0)$$(\frac{96}{5},\frac{-16}{5},\frac{1688}{25},\frac{1272}{25})$$\{0\}$
23$\{6,15\}$$(64,8,120,64)$$(\frac{224}{5},\frac{112}{5},\frac{2328}{25},\frac{1912}{25})$$\{0\}$
24$\{7,15\}$$(\frac{80}{3},\frac{80}{3},\frac{416}{3},64)$$(\frac{448}{15},\frac{448}{15},\frac{7544}{75},\frac{1912}{25})$$\{0\}$
25$\{9,11\}$$(-64,0,0,64)$$(\frac{-32}{5},\frac{96}{5},\frac{1128}{25},\frac{1912}{25})$$\{0\}$
26$\{9,13\}$$(-64,0,0,64)$$(\frac{-32}{5},\frac{96}{5},\frac{1128}{25},\frac{1912}{25})$$\{0\}$
27$\{9,15\}$$(0,64,64,128)$$(\frac{96}{5},\frac{224}{5},\frac{1768}{25},\frac{2552}{25})$$\{0\}$
28$\{10,11\}$$(0,-64,0,64)$$(\frac{96}{5},\frac{-32}{5},\frac{1128}{25},\frac{1912}{25})$$\{0\}$
29$\{10,14\}$$(0,-64,0,64)$$(\frac{96}{5},\frac{-32}{5},\frac{1128}{25},\frac{1912}{25})$$\{0\}$
30$\{10,15\}$$(64,0,64,128)$$(\frac{224}{5},\frac{96}{5},\frac{1768}{25},\frac{2552}{25})$$\{0\}$
31$\{11,15\}$$(\frac{64}{3},\frac{64}{3},64,\frac{448}{3})$$(\frac{416}{15},\frac{416}{15},\frac{1768}{25},\frac{8296}{75})$$\{0\}$
32$\{12,13\}$$(0,0,-8,8)$$(\frac{96}{5},\frac{96}{5},\frac{1048}{25},\frac{1352}{25})$$\{0\}$
33$\{12,14\}$$(0,0,-8,8)$$(\frac{96}{5},\frac{96}{5},\frac{1048}{25},\frac{1352}{25})$$\{0\}$
34$\{12,15\}$$(64,64,56,72)$$(\frac{224}{5},\frac{224}{5},\frac{1688}{25},\frac{1992}{25})$$\{0\}$
35$\{13,15\}$$(-16,64,96,112)$$(\frac{64}{5},\frac{224}{5},\frac{2088}{25},\frac{2392}{25})$$\{0\}$
36$\{14,15\}$$(64,-16,96,112)$$(\frac{224}{5},\frac{64}{5},\frac{2088}{25},\frac{2392}{25})$$\{0\}$
37$\{3,7,15\}$$(104,104,-16,64)$$(80,80,\frac{416}{25},\frac{1264}{25})$$\{1\}$
38$\{3,11,15\}$$(104,104,64,-16)$$(80,80,\frac{1216}{25},\frac{464}{25})$$\{1\}$
39$\{3,12,15\}$$(4,4,124,124)$$(\frac{224}{23},\frac{224}{23},\frac{8776}{115},\frac{9544}{115})$$\{8,14\}$
40$\{5,7,15\}$$(8,64,120,64)$$(\frac{112}{5},\frac{224}{5},\frac{2328}{25},\frac{1912}{25})$$\{0\}$
41$\{5,10,15\}$$(64,64,64,64)$$(64,64,64,64)$$\varnothing$
42$\{5,13,15\}$$(8,64,120,64)$$(\frac{112}{5},\frac{224}{5},\frac{2328}{25},\frac{1912}{25})$$\{0\}$
43$\{6,7,15\}$$(64,8,120,64)$$(\frac{224}{5},\frac{112}{5},\frac{2328}{25},\frac{1912}{25})$$\{0\}$
44$\{6,9,15\}$$(64,64,64,64)$$(64,64,64,64)$$\varnothing$
45$\{6,14,15\}$$(64,8,120,64)$$(\frac{224}{5},\frac{112}{5},\frac{2328}{25},\frac{1912}{25})$$\{0\}$
46$\{9,11,15\}$$(0,64,64,128)$$(\frac{96}{5},\frac{224}{5},\frac{1768}{25},\frac{2552}{25})$$\{0\}$
47$\{9,13,15\}$$(0,64,64,128)$$(\frac{96}{5},\frac{224}{5},\frac{1768}{25},\frac{2552}{25})$$\{0\}$
48$\{10,11,15\}$$(64,0,64,128)$$(\frac{224}{5},\frac{96}{5},\frac{1768}{25},\frac{2552}{25})$$\{0\}$
49$\{10,14,15\}$$(64,0,64,128)$$(\frac{224}{5},\frac{96}{5},\frac{1768}{25},\frac{2552}{25})$$\{0\}$
50$\{12,13,15\}$$(64,64,56,72)$$(\frac{224}{5},\frac{224}{5},\frac{1688}{25},\frac{1992}{25})$$\{0\}$
51$\{12,14,15\}$$(64,64,56,72)$$(\frac{224}{5},\frac{224}{5},\frac{1688}{25},\frac{1992}{25})$$\{0\}$

Theorem (Fixed-model nonarrival arbitrarily close to patience): The fixed four-player payoff model above satisfies the hypotheses of the attachment. For every \[ \boxed{\quad\frac{3496}{3501}\le\delta<1,\qquad p_\delta=(1-\alpha_\delta)I+\alpha_\delta Q, \quad \alpha_\delta=\frac{3(1-\delta)}{\delta}\quad} \] $p_\delta$ is an equilibrium and $\Pr^{p_\delta}_{\varnothing}(T_{\mathcal G}=\infty)=1$.
Proof: The discount interval is exactly $0<\alpha_\delta\le15/3496$. Proposition Proposition (Finite rational certificate for the displayed model) and Lemma Lemma (A sufficient exact certificate) prove the equilibrium axioms. The set $C_0$ contains no grand state and is closed under $Q$, and therefore under $p_\delta$. Starting from $x_0=\varnothing$, every finite-time state is in $C_0$ with probability one. The countable union of the zero-probability events $\{X_t\notin C_0\}$ has probability zero. Hence the grand coalition is never hit almost surely. The same primitives and the same initial state work for all these discount factors. For example, $\delta_k=1-1/(1000k)$ for $k\ge1$ gives the requested sequence. ∎

VERIFICATION

Independent supplied audit. The supplied verify_coalition.py reconstructs laminar states, recursive flow payoffs and all legal moves, and solves $(4I-3Q)L=\pi$ by exact rational Gauss--Jordan elimination. It does not import the displayed value table or execute repository code. Rerunning it reproduces the supplied coalition_results.json exactly: 52 states, 558 moves including stays, $\eta=24/115$, $B=608/25$, $\alpha_0=15/3496$, and $\delta_{\min}=3496/3501$. It checks supported legal targets, the complete undominated target sets, profitability and favourites in (E1)--(E3) at $\alpha_0$, the four-state closed nongrand class, and the surplus gap 16. These are exact finite checks of this displayed model. Their extension to every $0<\alpha\le\alpha_0$ uses the strict-margin argument in Lemma Lemma (A sufficient exact certificate); it is not an enumeration of an infinite discount interval. The endpoint test and that lemma serve distinct roles.

Adversarial checks. The first attempted time change of a different auxiliary kernel failed (E3) at off-cycle states: additional termination moves became profitable at high patience. That candidate was rejected. The kernel printed above fixes those rows and has been checked independently with both a matrix-based exact calculation and the self-contained verifier below.

The proof uses no uniform mixing bound: positive move probabilities are allowed to converge to zero. The payoff table does not change with $\delta$. The result includes unilateral termination of an inner agreement together with every ancestor. Stays participate in domination even when they are not profitable. The computation retains exact equality in the status-quo tie rule. All off-cycle states are covered; no local-cycle calculation is promoted to a global equilibrium. The finite inverse in (eq:L) is certified by the Bellman identities and stochasticity, not by floating-point inversion. Starting in a grand state gives $T_{\mathcal G}=0$ and is not the initial condition used in the theorem.

Reproducible verification. Copy the following listing to a Python 3 file and run it. Only the standard library is required. It outputs the state and move counts and the exact patience threshold. The interval between discount factors is covered analytically by Lemma Lemma (A sufficient exact certificate), not by numerical interpolation.

\begin{lstlisting}[language=Python]
"""Self-contained certificate for C71-12; Python 3 standard library only. No floating point, numerical search, or external solver is used. """ from fractions import Fraction as F from itertools import combinations from functools import lru_cache N = frozenset('ABCD') AB, CD = frozenset('AB'), frozenset('CD') def subsets(s, minimum=0): s = sorted(s, key=lambda a: (len(a), tuple(sorted(a)))) return [frozenset(t) for r in range(minimum,len(s)+1) for t in combinations(s,r)] coalitions = subsets(N,2) X = [frozenset(t) for r in range(len(coalitions)+1) for t in combinations(coalitions,r) if all(not(a & b) or a <= b or b <= a for a,b in combinations(t,2))] assert len(X) == 52 empty = frozenset() core = [empty, frozenset([AB]), frozenset([AB,CD]), frozenset([CD])] cross = [frozenset([frozenset('AC'),frozenset('BD')]), frozenset([frozenset('BC'),frozenset('AD')])] def top(x): t = [a for a in x if not any(a < b for b in x)] used = frozenset().union(*t) return frozenset(t + [frozenset([i]) for i in N-used]) def block_value(k,p): if k == N: return F(256) if k == AB and p == frozenset([AB,frozenset('C'),frozenset('D')]): return F(240) if k == CD and p == frozenset([AB,CD]): return F(240) if all(len(a)==1 for a in p): if k == frozenset('C'): return F(112) if k == frozenset('D'): return F(128) return F(0) @lru_cache(None) def payoff(x): ans = {}; p = top(x) for k in p: if len(k)==1: ans[next(iter(k))] = block_value(k,p) else: before = payoff(x-frozenset([k])) gain = block_value(k,p)-sum(before[i] for i in k) for i in k: ans[i] = before[i]+gain/len(k) return ans

def kernel(x): if x in core: return {x:F(2,3), core[(core.index(x)+1)%4]:F(1,3)} if x == frozenset([AB,CD,N]): return {x:F(3,8), core[3]:F(3,8), core[2]:F(1,4)} if x in cross: return {x:F(1,4), x|frozenset([N]):F(1,4), **{frozenset([k]):F(1,4) for k in x}} if any(x == y|frozenset([N]) for y in cross): return {x:F(1)} return {x:F(1,2), core[1] if AB in x else empty:F(1,2)}

# A displayed rational Bellman certificate, not a computed approximation. # Each key lists agreements as strings, each value is in A,B,C,D order. DATA = [ ((), ('32', '32', '376/5', '424/5')), (('AB',), ('64', '64', '192/5', '208/5')), (('AC',), ('-16/5', '96/5', '1688/25', '1272/25')), (('BC',), ('96/5', '-16/5', '1688/25', '1272/25')), (('ABC',), ('64/15', '64/15', '5624/75', '1272/25')), (('AD',), ('-32/5', '96/5', '1128/25', '1912/25')), (('BD',), ('96/5', '-32/5', '1128/25', '1912/25')), (('ABD',), ('32/15', '32/15', '1128/25', '6376/75')), (('CD',), ('16', '16', '168/5', '232/5')), (('ACD',), ('-64/5', '96/5', '1448/25', '1752/25')), (('BCD',), ('96/5', '-64/5', '1448/25', '1752/25')), (('ABCD',), ('104/5', '104/5', '2288/25', '2592/25')), (('AB', 'ABC'), ('272/5', '272/5', '-224/25', '624/25')), (('AB', 'ABD'), ('272/5', '272/5', '576/25', '-176/25')), (('AB', 'CD'), ('8', '8', '384/5', '416/5')), (('AB', 'ABCD'), ('88', '88', '616/25', '664/25')), (('AC', 'ABC'), ('-16/5', '96/5', '1688/25', '1272/25')), (('AC', 'BD'), ('240/13', '1152/65', '13248/325', '1104/25')), (('AC', 'ACD'), ('-16/5', '96/5', '1688/25', '1272/25')), (('AC', 'ABCD'), ('112/5', '224/5', '2328/25', '1912/25')), (('BC', 'ABC'), ('96/5', '-16/5', '1688/25', '1272/25')), (('BC', 'AD'), ('1152/65', '240/13', '13248/325', '1104/25')), (('BC', 'BCD'), ('96/5', '-16/5', '1688/25', '1272/25')), (('BC', 'ABCD'), ('224/5', '112/5', '2328/25', '1912/25')), (('ABC', 'ABCD'), ('448/15', '448/15', '7544/75', '1912/25')), (('AD', 'ABD'), ('-32/5', '96/5', '1128/25', '1912/25')), (('AD', 'ACD'), ('-32/5', '96/5', '1128/25', '1912/25')), (('AD', 'ABCD'), ('96/5', '224/5', '1768/25', '2552/25')), (('BD', 'ABD'), ('96/5', '-32/5', '1128/25', '1912/25')), (('BD', 'BCD'), ('96/5', '-32/5', '1128/25', '1912/25')), (('BD', 'ABCD'), ('224/5', '96/5', '1768/25', '2552/25')), (('ABD', 'ABCD'), ('416/15', '416/15', '1768/25', '8296/75')), (('CD', 'ACD'), ('96/5', '96/5', '1048/25', '1352/25')), (('CD', 'BCD'), ('96/5', '96/5', '1048/25', '1352/25')), (('CD', 'ABCD'), ('224/5', '224/5', '1688/25', '1992/25')), (('ACD', 'ABCD'), ('64/5', '224/5', '2088/25', '2392/25')), (('BCD', 'ABCD'), ('224/5', '64/5', '2088/25', '2392/25')), (('AB', 'ABC', 'ABCD'), ('80', '80', '416/25', '1264/25')), (('AB', 'ABD', 'ABCD'), ('80', '80', '1216/25', '464/25')), (('AB', 'CD', 'ABCD'), ('224/23', '224/23', '8776/115', '9544/115')), (('AC', 'ABC', 'ABCD'), ('112/5', '224/5', '2328/25', '1912/25')), (('AC', 'BD', 'ABCD'), ('64', '64', '64', '64')), (('AC', 'ACD', 'ABCD'), ('112/5', '224/5', '2328/25', '1912/25')), (('BC', 'ABC', 'ABCD'), ('224/5', '112/5', '2328/25', '1912/25')), (('BC', 'AD', 'ABCD'), ('64', '64', '64', '64')), (('BC', 'BCD', 'ABCD'), ('224/5', '112/5', '2328/25', '1912/25')), (('AD', 'ABD', 'ABCD'), ('96/5', '224/5', '1768/25', '2552/25')), (('AD', 'ACD', 'ABCD'), ('96/5', '224/5', '1768/25', '2552/25')), (('BD', 'ABD', 'ABCD'), ('224/5', '96/5', '1768/25', '2552/25')), (('BD', 'BCD', 'ABCD'), ('224/5', '96/5', '1768/25', '2552/25')), (('CD', 'ACD', 'ABCD'), ('224/5', '224/5', '1688/25', '1992/25')), (('CD', 'BCD', 'ABCD'), ('224/5', '224/5', '1688/25', '1992/25')), \] L = {frozenset(frozenset(k) for k in fam): dict(zip('ABCD',map(F,values))) for fam,values in DATA} assert set(L)==set(X) Q = {x:kernel(x) for x in X} Pi = {x:payoff(x) for x in X} assert all(sum(q.values())==1 and min(q.values())>0 for q in Q.values()) assert all(sum(Pi[x].values())==256 for x in X if N in x) assert min(256-sum(Pi[x].values()) for x in X if N not in x)==16 for x in X: for k in top(x) & x: b=Pi[x-frozenset([k])] gain=sum(Pi[x][i]-b[i] for i in k) assert all(Pi[x][i]==b[i]+gain/len(k) for i in k) Plus={x:{i:sum(p*L[y][i] for y,p in Q[x].items()) for i in N} for x in X} assert all(L[x][i]==Pi[x][i]/4+F(3,4)*Plus[x][i] for x in X for i in N)

def moves(x): result=[] for a in subsets(top(x),2): k=frozenset().union(*a) result.append((x|frozenset([k]),k)) for k in x: target=frozenset(a for a in x if not k<=a) result.extend((target,frozenset([i])) for i in k) result.extend((x,frozenset([i])) for i in N) return result

def strict(x,y,z,i): return L[y][i]>L[z][i] or (L[y][i]==L[z][i] and y==x and z!=x) eta=None; B=F(0); count=0 for x in X: M=moves(x); count+=len(M) U=[(y,k) for y,k in M if not any( h<=k and all(strict(x,z,y,i) for i in h) for z,h in M)] moving=[(y,k) for y,k in U if y!=x] assert {y for y,k in moving} == set(Q[x])-{x} for y,k in moving: for i in k: gap=L[y][i]-L[x][i] assert gap>0 eta=gap if eta is None else min(eta,gap) B=max(B,max(abs(Plus[x][i]-L[x][i]) for i in N)) assert count==558 and eta==F(24,115) and B==F(608,25) alpha0=eta/(2*B) assert alpha0==F(15,3496) assert F(3)/(3+alpha0)==F(3496,3501) assert all(N not in x and set(Q[x])<=set(core) for x in core) print('PASS: 52 states; 558 moves; all sharing and Bellman identities;') print('all undominated target sets; eta=24/115; B=608/25;') print('equilibrium family certified for 3496/3501 <= delta < 1.') 

\end{lstlisting}

FINAL

LABEL: FULL SOLUTION.

SUBLABEL: FULL PROOF of the existential nonarrival assertion; COUNTEREXAMPLE/DISPROOF of the universal eventual-arrival threshold assertion.

Theorem Theorem (Fixed-model nonarrival arbitrarily close to patience) supplies one fixed four-player model, explicit equilibrium matrices at every $\delta\in[3496/3501,1)$, and probability-one nonarrival from $\varnothing$. The decisive mechanism is endogenous waiting: the equilibrium axioms require positive support, not a lower bound on the frequency with which supported moves occur. This conclusion applies to the literal definitions in the attachment; adding a uniform positive lower bound on transition rates or requiring all positive self-loops to have profitable stay witnesses changes the problem.

Completion Estimate

100%

This is a post hoc, heuristic estimate for the literal fixed-model target. The explicit model and exact certificate prove equilibrium and probability-one nonarrival for the full interval $3496/3501\le\delta<1$, settling the existential assertion and refuting its universal threshold negation. This percentage does not assert a result for stronger transition-rate or stay requirements, or measure novelty or independent community acceptance.

References

References

  • [heitzig] Jobst Heitzig, Does the grand coalition form? Persistence, arrival, and the role of the sharing rule in a dynamic process of nested binding agreements, arXiv:2608.17766v2 (2026). Definitions 2.14--2.15, Remark 6.7 and Proposition 6.8; the old deterministic cycle is not used as an equilibrium certificate here. https://arxiv.org/html/2608.17766v2.
  • [audit] User-supplied Pro audit bundle, Independent checks for the economics "solved" audit, supplied 9 October 2026: README.md, proposed_corrections.md, verify_coalition.py, and coalition_results.json. The stated reviewed snapshot is ccccd798864807c1f3fb8fe36727fe35ee875aeb. Unpublished finite verification and logical annotation; the values reported above were reproduced from the supplied program.

Go to contribution →

Econ-12-A2 · Full solution claimed · LLM-generated · gpt 6 astra pro ·

GPT 6 Astra Pro attempt on Econ-12

Grand-coalition arrival in the patient limit

C71-12: research attempt and adversarial verification

9 October 2026

Tools. Web browsing: YES. Exact rational computation: YES.

FORMAL RESTATEMENT

Work with one fixed finite set $N$, a fixed real payoff table, and fixed positive sharing weights. These primitives may not depend on the discount factor. The state space $\mathcal X$ consists of all laminar families of subsets of $N$ of size at least two; a family is laminar when two members are disjoint or one contains the other. Its top-level partition $P(x)$ consists of maximal agreements and uncovered singleton players. Write $\mathcal G=\{x:N\in x\}$.

Require $V(x)=\sum_i\pi_i(x)$, $V(h)=V^*$ for $h\in\mathcal G$, and $\gamma=\min_{x\notin\mathcal G}(V^*-V(x))>0$. For each top-level agreement $K\in x\cap P(x)$ require \[ \pi_i(x)=\pi_i(x\setminus\{K\})+s_i(x,K) \sum_{j\in K}[\pi_j(x)-\pi_j(x\setminus\{K\})],\quad i\in K, \] where every $s_i(x,K)>0$ and $\sum_{i\in K}s_i(x,K)=1$.

A merge joins at least two blocks of $P(x)$, adds their union $K$, and needs the relevant player set $R=K$. Terminating $K\in x$ by $i\in K$ removes every agreement containing $K$ and has $R=\{i\}$. Player $i$ may also stay, with target $x$ and $R=\{i\}$. These are all moves. Let $M^+(x)$ include stays, and let $M(x)$ exclude them.

For a stochastic matrix $p$ and $0<\delta<1$, set \[ \ell_i=(1-\delta)(I-\delta p)^{-1}\pi_i, \qquad \ell_i^+(x)=\sum_y p_{xy}\ell_i(y). \] The inverse exists: $\sum_{t\ge0}\delta^tp^t$ converges entrywise and in the maximum-row-sum matrix norm. A player ranks targets by $\ell_i$, with the current state strictly preferred to every different target having equal value; all other value ties remain ties. A move $m$ is profitable when $\ell_i(\tau(m))\ge\ell_i^+(x)$ for every $i\in R(m)$. It is dominated when some $m'\in M^+(x)$ has nonempty $R(m')\subseteq R(m)$ and every member of $R(m')$ strictly prefers its target. Let $C(x)$ be the profitable undominated moves. Let $F(x)$ consist of moves in $C(x)$ whose target is weakly best in $C(x)$ for at least one player, whether or not that player is relevant to the move.

An equilibrium satisfies precisely \begin{align*} p_{xy}>0, \ y\ne x&\ \Longrightarrow\ \exists m\in C(x):\tau(m)=y;\tag{E1}\\ C(x)\cap M(x)\ne\varnothing&\ \Longrightarrow\ p_{xx}<1;\tag{E2}\\ m\in F(x)&\ \Longrightarrow\ p_{x,\tau(m)}>0.\tag{E3} \end{align*} No positive lower bound on a nonzero transition probability is imposed. In particular, (E1) does not require a positive diagonal entry to be justified by a profitable stay.

Literal question. Do there exist such fixed primitives, $\delta_k\uparrow1$, equilibrium matrices $p_k$, and initial states $x_k$ such that $\Pr_{x_k}^{p_k}(T_{\mathcal G}=\infty)>0$ for every $k$? Here the hitting time includes time zero. The answer below is affirmative, with $x_k=\varnothing$ for every $k$ and probability equal to one.

Logical correction. The attachment's "equivalently, is it true that every fixed model has a threshold" is the negation of the preceding existential assertion, not the same assertion. The two formulations describe opposite sides of the same decision problem. We prove the existential assertion and disprove the threshold assertion.

QUICK LITERATURE/CONTEXT CHECK

Heitzig's version 2 explicitly withdraws the equilibrium status of the old deterministic cycle under the revised rules: see Remark 6.7 and Proposition 6.8. Some later passages still refer to that superseded construction without the qualification. We use neither those claims nor any equilibrium-existence theorem from that source. We reuse its concrete block-value primitives, then give a different stochastic kernel and verify the stated axioms independently.

Checked primary source. Jobst Heitzig, Does the grand coalition form? Persistence, arrival, and the role of the sharing rule in a dynamic process of nested binding agreements, arXiv:2608.17766v2, especially Definitions 2.14--2.15 and Remark 6.7. https://arxiv.org/html/2608.17766v2. This is a preprint; no novelty or peer-review claim is made for the construction below.

ATTACK PLAN

Proof track. Try to turn the welfare gap into a drift or supermartingale argument and examine closed communicating classes. The obstruction is that (E2)--(E3) require positivity but do not bound transition rates away from zero.

Construction track, selected. Introduce waiting while preserving discounted values by a time-change identity. Specify every off-cycle row, not merely a recurrent cycle. Enumerate all moves using exact fractions, then replace numerical evidence by a rational Bellman certificate and a uniform strict-margin bound.

Tools and their roles. Finite Markov-chain closure proves nonarrival; Bellman equations certify discounted values; diagonal time change separates patience from effective transition frequency; laminar-family enumeration checks the entire domain; subset domination incorporates unilateral deviations and stays; rational gap bounds make the result uniform in $\delta$; independent exact arithmetic detects ties and excludes rounding errors.

Tiny cases. With two players there are only $e=\varnothing$ and $h=\{N\}$. Sharing implies $\pi_i(h)>\pi_i(e)$ for both players. If $a=p_{eh}$ and $b=p_{he}$, subtraction of the two Bellman equations gives \[ \ell_i(h)-\ell_i(e) =\frac{(1-\delta)[\pi_i(h)-\pi_i(e)]} {1-\delta+\delta(a+b)}>0. \] At $h$, every termination is dominated by its player's stay, hence (E1) gives $b=0$. At $e$, merge-all is profitable and undominated; it is best for both players among the candidates. Thus (E3) gives $a>0$, and arrival occurs almost surely. The counterexample uses four players.

WORK

A fixed payoff model

Let $N=\{A,B,C,D\}$. Concatenated letters denote sets, not new players. Define a block-value function $v(K;P)$ to be zero except for \[ \begin{array}{c|r} (K;P)&v(K;P)\\\hline (N;\{N\})&256\\ (AB;AB|C|D)&240\\ (CD;AB|CD)&240\\ (C;A|B|C|D)&112\\ (D;A|B|C|D)&128. \end{array} \] Define $\pi(x)$ recursively on the number of agreements. For a singleton block $\{i\}\in P(x)$ set $\pi_i(x)=v(\{i\};P(x))$. For a nonsingleton block $K\in P(x)$ put $x^-=x\setminus\{K\}$ and set \begin{equation}\label{wsref-0-65-71-3a-72-65-63-75-72-73-69-76-65-2d-70-61-79-6f-66-66} \pi_i(x)=\pi_i(x^-)+\frac{v(K;P(x))-\sum_{j\in K}\pi_j(x^-)}{|K|},\qquad i\in K. \end{equation} Every player belongs to exactly one top-level block, so this defines a single payoff for each player. The recursive calls have fewer agreements, so they terminate. Summing over $K$ shows that its players' total payoff equals $v(K;P(x))$. Therefore $V(x)=\sum_{K\in P(x)}v(K;P(x))$. For grand states this is 256; for nongrand states it is either 240 or 0. Hence $V^*=256$ and $\gamma=16$. Equation (eq:recursive-payoff) is exactly the required sharing identity with $s_i(x,K)=1/|K|>0$. Negative individual payoffs are allowed by the problem.

An explicit auxiliary kernel on every state

Define \[ x_0=\varnothing,\quad x_1=\{AB\},\quad x_2=\{AB,CD\},\quad x_3=\{CD\},\qquad C_0=\{x_0,x_1,x_2,x_3\}. \] Let $Q$ be the following stochastic matrix. All unspecified entries are zero, and the cases are applied in the displayed order.
  1. For $x=x_j\in C_0$, set $Q_{xx}=2/3$ and $Q_{x,x_{j+1\bmod4}}=1/3$.
  2. For $x=\{AB,CD,N\}$, give probabilities $3/8,3/8,1/4$ to $x,x_3,x_2$, respectively.
  3. For $x=\{AC,BD\}$ or $x=\{BC,AD\}$, give probability $1/4$ to each of $x$, $x\cup\{N\}$, and the two states containing just one of the two agreements in $x$.
  4. For either grand state obtained in (iii), set $Q_{xx}=1$.
  5. In every other state give probability $1/2$ to $x$ and $1/2$ to $\{AB\}$ when $AB\in x$, or to $\varnothing$ when $AB\notin x$.
These cases exhaust all states and do not assign the same destination twice within a row. All diagonal entries are positive. Every off-diagonal destination is a legal moving target. To check the last assertion in case (v), a laminar family on four players is a chain unless it has two disjoint pairs. The three possible two-pair families were already handled. Removing the innermost agreement empties a chain; if $AB$ is present, terminating the next larger agreement leaves exactly $\{AB\}$. The remaining special cases are direct merges or chain-rule terminations.

Define the auxiliary value table \begin{equation}\label{wsref-0-65-71-3a-4c} L=\frac14(I-\tfrac34 Q)^{-1}\pi, \qquad H=QL. \end{equation} $Q$ is auxiliary: it is not asserted to be an equilibrium at discount $3/4$.

Time change and the role of undominated moves

For a state $x$, let $U(x)$ be all undominated moves, without imposing profitability, using the target values $L$ and the specified status-quo tie rule.
Lemma (Strict improvement for relevant players): If a moving move $m$ is undominated, then $L_i(\tau(m))>L_i(x)$ for every $i\in R(m)$.
Proof: If $L_i(\tau(m))\le L_i(x)$, staying at $x$ by $i$ is strictly preferred to the different target $\tau(m)$, including when values tie. Its relevant set $\{i\}$ is contained in $R(m)$. This stay therefore dominates $m$. Domination is tested against all moves, not only profitable moves, so no profitability hypothesis on the stay is needed. ∎
Lemma (A sufficient exact certificate): Suppose a finite payoff model and a stochastic matrix $Q$ with positive diagonal have the values in (eq:L). Suppose, at every state, \begin{equation}\label{wsref-0-65-71-3a-73-75-70-70-6f-72-74} \{\tau(m):m\in U(x),\ \tau(m)\ne x\} =\{y\ne x:Q_{xy}>0\}. \end{equation} Let $\eta>0$ be the minimum of $L_i(\tau(m))-L_i(x)$ over all moving $m\in U(x)$ and $i\in R(m)$, and let $B=\max_{x,i}|H_i(x)-L_i(x)|$. When the moving set is nonempty and $B>0$, put $\alpha_0=\min(1,\eta/(2B))$. For every $0<\alpha\le\alpha_0$, the matrix \[ p_\alpha=(1-\alpha)I+\alpha Q \] is an equilibrium at $\delta=3/(3+\alpha)$.
Proof: Lemma Lemma (Strict improvement for relevant players) makes the finite minimum positive. The identity $L=\pi/4+3QL/4$ gives $\pi=4L-3QL$. Substitution shows \[ (1-\delta)\pi+\delta[(1-\alpha)L+\alpha QL]=L \quad\text{when }\alpha=3(1-\delta)/\delta. \] The unique discounted Bellman solution for $p_\alpha$ is consequently $\ell=L$. Its continuation benchmark is $\ell^+(x)=L(x)+\alpha(H(x)-L(x))$. For every relevant player of an undominated moving move, \[ L_i(\tau(m))-\ell_i^+(x) \ge\eta-\alpha B\ge\eta/2>0. \] Thus all undominated moving moves are profitable. Some undominated stays can be unprofitable; this causes no problem because all diagonal entries of $p_\alpha$ are positive.

Every positive off-diagonal transition has an undominated moving witness by (eq:support); that witness is profitable, proving (E1). If $C(x)$ has a moving member, (eq:support) provides a positive off-diagonal entry, proving (E2). Every profitable undominated move has either a supported moving target by (eq:support) or the supported current state. In particular all moves in $F(x)$ have supported targets, proving (E3). This establishes the lemma without an appeal to equilibrium existence. ∎

Proposition (Finite rational certificate for the displayed model): For the payoff model and $Q$ above, (eq:support) holds at all 52 states. The minimum and maximum in Lemma Lemma (A sufficient exact certificate) are \[ \eta=\frac{24}{115},\qquad B=\frac{608}{25},\qquad \alpha_0=\frac{15}{3496}. \]
Proof: The complete rational table below specifies $L$ and the set of undominated moving targets at every state. The displayed values satisfy $L=\pi/4+3QL/4$ row by row. To test domination, enumerate the legal merges, every termination by every eligible player, and each stay, and compare their relevant sets and target values. This finite calculation checks 558 moves. The complete standard-library verifier in Section 5 independently regenerates the laminar families, payoffs and moves and checks every assertion using rational arithmetic. It does not invoke a linear-programming or equilibrium solver. It also checks the two extrema rather than merely testing a sample of discount factors. Together with the table, this is an explicit finite arithmetic certificate. ∎

Complete arithmetic certificate

State identifiers are local to this table. A coalition is encoded by the bit mask $A=1,B=2,C=4,D=8$; for example, $3=AB$ and $15=N$. The last column lists all undominated moving target states, not relevant player sets.

IDAgreements$\pi_A,\pi_B,\pi_C,\pi_D$$L_A,L_B,L_C,L_D$Targets
0$\varnothing$$(0,0,112,128)$$(32,32,\frac{376}{5},\frac{424}{5})$$\{1\}$
1$\{3\}$$(120,120,0,0)$$(64,64,\frac{192}{5},\frac{208}{5})$$\{14\}$
2$\{5\}$$(-56,0,56,0)$$(\frac{-16}{5},\frac{96}{5},\frac{1688}{25},\frac{1272}{25})$$\{0\}$
3$\{6\}$$(0,-56,56,0)$$(\frac{96}{5},\frac{-16}{5},\frac{1688}{25},\frac{1272}{25})$$\{0\}$
4$\{7\}$$(\frac{-112}{3},\frac{-112}{3},\frac{224}{3},0)$$(\frac{64}{15},\frac{64}{15},\frac{5624}{75},\frac{1272}{25})$$\{0\}$
5$\{9\}$$(-64,0,0,64)$$(\frac{-32}{5},\frac{96}{5},\frac{1128}{25},\frac{1912}{25})$$\{0\}$
6$\{10\}$$(0,-64,0,64)$$(\frac{96}{5},\frac{-32}{5},\frac{1128}{25},\frac{1912}{25})$$\{0\}$
7$\{11\}$$(\frac{-128}{3},\frac{-128}{3},0,\frac{256}{3})$$(\frac{32}{15},\frac{32}{15},\frac{1128}{25},\frac{6376}{75})$$\{0\}$
8$\{12\}$$(0,0,-8,8)$$(16,16,\frac{168}{5},\frac{232}{5})$$\{0\}$
9$\{13\}$$(-80,0,32,48)$$(\frac{-64}{5},\frac{96}{5},\frac{1448}{25},\frac{1752}{25})$$\{0\}$
10$\{14\}$$(0,-80,32,48)$$(\frac{96}{5},\frac{-64}{5},\frac{1448}{25},\frac{1752}{25})$$\{0\}$
11$\{15\}$$(4,4,116,132)$$(\frac{104}{5},\frac{104}{5},\frac{2288}{25},\frac{2592}{25})$$\{0\}$
12$\{3,7\}$$(40,40,-80,0)$$(\frac{272}{5},\frac{272}{5},\frac{-224}{25},\frac{624}{25})$$\{1\}$
13$\{3,11\}$$(40,40,0,-80)$$(\frac{272}{5},\frac{272}{5},\frac{576}{25},\frac{-176}{25})$$\{1\}$
14$\{3,12\}$$(0,0,120,120)$$(8,8,\frac{384}{5},\frac{416}{5})$$\{8\}$
15$\{3,15\}$$(124,124,4,4)$$(88,88,\frac{616}{25},\frac{664}{25})$$\{1\}$
16$\{5,7\}$$(-56,0,56,0)$$(\frac{-16}{5},\frac{96}{5},\frac{1688}{25},\frac{1272}{25})$$\{0\}$
17$\{5,10\}$$(0,0,0,0)$$(\frac{240}{13},\frac{1152}{65},\frac{13248}{325},\frac{1104}{25})$$\{2,6,41\}$
18$\{5,13\}$$(-56,0,56,0)$$(\frac{-16}{5},\frac{96}{5},\frac{1688}{25},\frac{1272}{25})$$\{0\}$
19$\{5,15\}$$(8,64,120,64)$$(\frac{112}{5},\frac{224}{5},\frac{2328}{25},\frac{1912}{25})$$\{0\}$
20$\{6,7\}$$(0,-56,56,0)$$(\frac{96}{5},\frac{-16}{5},\frac{1688}{25},\frac{1272}{25})$$\{0\}$
21$\{6,9\}$$(0,0,0,0)$$(\frac{1152}{65},\frac{240}{13},\frac{13248}{325},\frac{1104}{25})$$\{3,5,44\}$
22$\{6,14\}$$(0,-56,56,0)$$(\frac{96}{5},\frac{-16}{5},\frac{1688}{25},\frac{1272}{25})$$\{0\}$
23$\{6,15\}$$(64,8,120,64)$$(\frac{224}{5},\frac{112}{5},\frac{2328}{25},\frac{1912}{25})$$\{0\}$
24$\{7,15\}$$(\frac{80}{3},\frac{80}{3},\frac{416}{3},64)$$(\frac{448}{15},\frac{448}{15},\frac{7544}{75},\frac{1912}{25})$$\{0\}$
25$\{9,11\}$$(-64,0,0,64)$$(\frac{-32}{5},\frac{96}{5},\frac{1128}{25},\frac{1912}{25})$$\{0\}$
26$\{9,13\}$$(-64,0,0,64)$$(\frac{-32}{5},\frac{96}{5},\frac{1128}{25},\frac{1912}{25})$$\{0\}$
27$\{9,15\}$$(0,64,64,128)$$(\frac{96}{5},\frac{224}{5},\frac{1768}{25},\frac{2552}{25})$$\{0\}$
28$\{10,11\}$$(0,-64,0,64)$$(\frac{96}{5},\frac{-32}{5},\frac{1128}{25},\frac{1912}{25})$$\{0\}$
29$\{10,14\}$$(0,-64,0,64)$$(\frac{96}{5},\frac{-32}{5},\frac{1128}{25},\frac{1912}{25})$$\{0\}$
30$\{10,15\}$$(64,0,64,128)$$(\frac{224}{5},\frac{96}{5},\frac{1768}{25},\frac{2552}{25})$$\{0\}$
31$\{11,15\}$$(\frac{64}{3},\frac{64}{3},64,\frac{448}{3})$$(\frac{416}{15},\frac{416}{15},\frac{1768}{25},\frac{8296}{75})$$\{0\}$
32$\{12,13\}$$(0,0,-8,8)$$(\frac{96}{5},\frac{96}{5},\frac{1048}{25},\frac{1352}{25})$$\{0\}$
33$\{12,14\}$$(0,0,-8,8)$$(\frac{96}{5},\frac{96}{5},\frac{1048}{25},\frac{1352}{25})$$\{0\}$
34$\{12,15\}$$(64,64,56,72)$$(\frac{224}{5},\frac{224}{5},\frac{1688}{25},\frac{1992}{25})$$\{0\}$
35$\{13,15\}$$(-16,64,96,112)$$(\frac{64}{5},\frac{224}{5},\frac{2088}{25},\frac{2392}{25})$$\{0\}$
36$\{14,15\}$$(64,-16,96,112)$$(\frac{224}{5},\frac{64}{5},\frac{2088}{25},\frac{2392}{25})$$\{0\}$
37$\{3,7,15\}$$(104,104,-16,64)$$(80,80,\frac{416}{25},\frac{1264}{25})$$\{1\}$
38$\{3,11,15\}$$(104,104,64,-16)$$(80,80,\frac{1216}{25},\frac{464}{25})$$\{1\}$
39$\{3,12,15\}$$(4,4,124,124)$$(\frac{224}{23},\frac{224}{23},\frac{8776}{115},\frac{9544}{115})$$\{8,14\}$
40$\{5,7,15\}$$(8,64,120,64)$$(\frac{112}{5},\frac{224}{5},\frac{2328}{25},\frac{1912}{25})$$\{0\}$
41$\{5,10,15\}$$(64,64,64,64)$$(64,64,64,64)$$\varnothing$
42$\{5,13,15\}$$(8,64,120,64)$$(\frac{112}{5},\frac{224}{5},\frac{2328}{25},\frac{1912}{25})$$\{0\}$
43$\{6,7,15\}$$(64,8,120,64)$$(\frac{224}{5},\frac{112}{5},\frac{2328}{25},\frac{1912}{25})$$\{0\}$
44$\{6,9,15\}$$(64,64,64,64)$$(64,64,64,64)$$\varnothing$
45$\{6,14,15\}$$(64,8,120,64)$$(\frac{224}{5},\frac{112}{5},\frac{2328}{25},\frac{1912}{25})$$\{0\}$
46$\{9,11,15\}$$(0,64,64,128)$$(\frac{96}{5},\frac{224}{5},\frac{1768}{25},\frac{2552}{25})$$\{0\}$
47$\{9,13,15\}$$(0,64,64,128)$$(\frac{96}{5},\frac{224}{5},\frac{1768}{25},\frac{2552}{25})$$\{0\}$
48$\{10,11,15\}$$(64,0,64,128)$$(\frac{224}{5},\frac{96}{5},\frac{1768}{25},\frac{2552}{25})$$\{0\}$
49$\{10,14,15\}$$(64,0,64,128)$$(\frac{224}{5},\frac{96}{5},\frac{1768}{25},\frac{2552}{25})$$\{0\}$
50$\{12,13,15\}$$(64,64,56,72)$$(\frac{224}{5},\frac{224}{5},\frac{1688}{25},\frac{1992}{25})$$\{0\}$
51$\{12,14,15\}$$(64,64,56,72)$$(\frac{224}{5},\frac{224}{5},\frac{1688}{25},\frac{1992}{25})$$\{0\}$

Theorem (Fixed-model nonarrival arbitrarily close to patience): The fixed four-player payoff model above satisfies the hypotheses of the attachment. For every \[ \boxed{\quad\frac{3496}{3501}\le\delta<1,\qquad p_\delta=(1-\alpha_\delta)I+\alpha_\delta Q, \quad \alpha_\delta=\frac{3(1-\delta)}{\delta}\quad} \] $p_\delta$ is an equilibrium and $\Pr^{p_\delta}_{\varnothing}(T_{\mathcal G}=\infty)=1$.
Proof: The discount interval is exactly $0<\alpha_\delta\le15/3496$. Proposition Proposition (Finite rational certificate for the displayed model) and Lemma Lemma (A sufficient exact certificate) prove the equilibrium axioms. The set $C_0$ contains no grand state and is closed under $Q$, and therefore under $p_\delta$. Starting from $x_0=\varnothing$, every finite-time state is in $C_0$ with probability one. The countable union of the zero-probability events $\{X_t\notin C_0\}$ has probability zero. Hence the grand coalition is never hit almost surely. The same primitives and the same initial state work for all these discount factors. For example, $\delta_k=1-1/(1000k)$ for $k\ge1$ gives the requested sequence. ∎

VERIFICATION

Adversarial checks. The first attempted time change of a different auxiliary kernel failed (E3) at off-cycle states: additional termination moves became profitable at high patience. That candidate was rejected. The kernel printed above fixes those rows and has been checked independently with both a matrix-based exact calculation and the self-contained verifier below.

The proof uses no uniform mixing bound: positive move probabilities are allowed to converge to zero. The payoff table does not change with $\delta$. The result includes unilateral termination of an inner agreement together with every ancestor. Stays participate in domination even when they are not profitable. The computation retains exact equality in the status-quo tie rule. All off-cycle states are covered; no local-cycle calculation is promoted to a global equilibrium. The finite inverse in (eq:L) is certified by the Bellman identities and stochasticity, not by floating-point inversion. Starting in a grand state gives $T_{\mathcal G}=0$ and is not the initial condition used in the theorem.

Reproducible verification. Copy the following listing to a Python 3 file and run it. Only the standard library is required. It outputs the state and move counts and the exact patience threshold. The interval between discount factors is covered analytically by Lemma Lemma (A sufficient exact certificate), not by numerical interpolation.

\begin{lstlisting}[language=Python]
"""Self-contained certificate for C71-12; Python 3 standard library only. No floating point, numerical search, or external solver is used. """ from fractions import Fraction as F from itertools import combinations from functools import lru_cache N = frozenset('ABCD') AB, CD = frozenset('AB'), frozenset('CD') def subsets(s, minimum=0): s = sorted(s, key=lambda a: (len(a), tuple(sorted(a)))) return [frozenset(t) for r in range(minimum,len(s)+1) for t in combinations(s,r)] coalitions = subsets(N,2) X = [frozenset(t) for r in range(len(coalitions)+1) for t in combinations(coalitions,r) if all(not(a & b) or a <= b or b <= a for a,b in combinations(t,2))] assert len(X) == 52 empty = frozenset() core = [empty, frozenset([AB]), frozenset([AB,CD]), frozenset([CD])] cross = [frozenset([frozenset('AC'),frozenset('BD')]), frozenset([frozenset('BC'),frozenset('AD')])] def top(x): t = [a for a in x if not any(a < b for b in x)] used = frozenset().union(*t) return frozenset(t + [frozenset([i]) for i in N-used]) def block_value(k,p): if k == N: return F(256) if k == AB and p == frozenset([AB,frozenset('C'),frozenset('D')]): return F(240) if k == CD and p == frozenset([AB,CD]): return F(240) if all(len(a)==1 for a in p): if k == frozenset('C'): return F(112) if k == frozenset('D'): return F(128) return F(0) @lru_cache(None) def payoff(x): ans = {}; p = top(x) for k in p: if len(k)==1: ans[next(iter(k))] = block_value(k,p) else: before = payoff(x-frozenset([k])) gain = block_value(k,p)-sum(before[i] for i in k) for i in k: ans[i] = before[i]+gain/len(k) return ans

def kernel(x): if x in core: return {x:F(2,3), core[(core.index(x)+1)%4]:F(1,3)} if x == frozenset([AB,CD,N]): return {x:F(3,8), core[3]:F(3,8), core[2]:F(1,4)} if x in cross: return {x:F(1,4), x|frozenset([N]):F(1,4), **{frozenset([k]):F(1,4) for k in x}} if any(x == y|frozenset([N]) for y in cross): return {x:F(1)} return {x:F(1,2), core[1] if AB in x else empty:F(1,2)}

# A displayed rational Bellman certificate, not a computed approximation. # Each key lists agreements as strings, each value is in A,B,C,D order. DATA = [ ((), ('32', '32', '376/5', '424/5')), (('AB',), ('64', '64', '192/5', '208/5')), (('AC',), ('-16/5', '96/5', '1688/25', '1272/25')), (('BC',), ('96/5', '-16/5', '1688/25', '1272/25')), (('ABC',), ('64/15', '64/15', '5624/75', '1272/25')), (('AD',), ('-32/5', '96/5', '1128/25', '1912/25')), (('BD',), ('96/5', '-32/5', '1128/25', '1912/25')), (('ABD',), ('32/15', '32/15', '1128/25', '6376/75')), (('CD',), ('16', '16', '168/5', '232/5')), (('ACD',), ('-64/5', '96/5', '1448/25', '1752/25')), (('BCD',), ('96/5', '-64/5', '1448/25', '1752/25')), (('ABCD',), ('104/5', '104/5', '2288/25', '2592/25')), (('AB', 'ABC'), ('272/5', '272/5', '-224/25', '624/25')), (('AB', 'ABD'), ('272/5', '272/5', '576/25', '-176/25')), (('AB', 'CD'), ('8', '8', '384/5', '416/5')), (('AB', 'ABCD'), ('88', '88', '616/25', '664/25')), (('AC', 'ABC'), ('-16/5', '96/5', '1688/25', '1272/25')), (('AC', 'BD'), ('240/13', '1152/65', '13248/325', '1104/25')), (('AC', 'ACD'), ('-16/5', '96/5', '1688/25', '1272/25')), (('AC', 'ABCD'), ('112/5', '224/5', '2328/25', '1912/25')), (('BC', 'ABC'), ('96/5', '-16/5', '1688/25', '1272/25')), (('BC', 'AD'), ('1152/65', '240/13', '13248/325', '1104/25')), (('BC', 'BCD'), ('96/5', '-16/5', '1688/25', '1272/25')), (('BC', 'ABCD'), ('224/5', '112/5', '2328/25', '1912/25')), (('ABC', 'ABCD'), ('448/15', '448/15', '7544/75', '1912/25')), (('AD', 'ABD'), ('-32/5', '96/5', '1128/25', '1912/25')), (('AD', 'ACD'), ('-32/5', '96/5', '1128/25', '1912/25')), (('AD', 'ABCD'), ('96/5', '224/5', '1768/25', '2552/25')), (('BD', 'ABD'), ('96/5', '-32/5', '1128/25', '1912/25')), (('BD', 'BCD'), ('96/5', '-32/5', '1128/25', '1912/25')), (('BD', 'ABCD'), ('224/5', '96/5', '1768/25', '2552/25')), (('ABD', 'ABCD'), ('416/15', '416/15', '1768/25', '8296/75')), (('CD', 'ACD'), ('96/5', '96/5', '1048/25', '1352/25')), (('CD', 'BCD'), ('96/5', '96/5', '1048/25', '1352/25')), (('CD', 'ABCD'), ('224/5', '224/5', '1688/25', '1992/25')), (('ACD', 'ABCD'), ('64/5', '224/5', '2088/25', '2392/25')), (('BCD', 'ABCD'), ('224/5', '64/5', '2088/25', '2392/25')), (('AB', 'ABC', 'ABCD'), ('80', '80', '416/25', '1264/25')), (('AB', 'ABD', 'ABCD'), ('80', '80', '1216/25', '464/25')), (('AB', 'CD', 'ABCD'), ('224/23', '224/23', '8776/115', '9544/115')), (('AC', 'ABC', 'ABCD'), ('112/5', '224/5', '2328/25', '1912/25')), (('AC', 'BD', 'ABCD'), ('64', '64', '64', '64')), (('AC', 'ACD', 'ABCD'), ('112/5', '224/5', '2328/25', '1912/25')), (('BC', 'ABC', 'ABCD'), ('224/5', '112/5', '2328/25', '1912/25')), (('BC', 'AD', 'ABCD'), ('64', '64', '64', '64')), (('BC', 'BCD', 'ABCD'), ('224/5', '112/5', '2328/25', '1912/25')), (('AD', 'ABD', 'ABCD'), ('96/5', '224/5', '1768/25', '2552/25')), (('AD', 'ACD', 'ABCD'), ('96/5', '224/5', '1768/25', '2552/25')), (('BD', 'ABD', 'ABCD'), ('224/5', '96/5', '1768/25', '2552/25')), (('BD', 'BCD', 'ABCD'), ('224/5', '96/5', '1768/25', '2552/25')), (('CD', 'ACD', 'ABCD'), ('224/5', '224/5', '1688/25', '1992/25')), (('CD', 'BCD', 'ABCD'), ('224/5', '224/5', '1688/25', '1992/25')), \] L = {frozenset(frozenset(k) for k in fam): dict(zip('ABCD',map(F,values))) for fam,values in DATA} assert set(L)==set(X) Q = {x:kernel(x) for x in X} Pi = {x:payoff(x) for x in X} assert all(sum(q.values())==1 and min(q.values())>0 for q in Q.values()) assert all(sum(Pi[x].values())==256 for x in X if N in x) assert min(256-sum(Pi[x].values()) for x in X if N not in x)==16 for x in X: for k in top(x) & x: b=Pi[x-frozenset([k])] gain=sum(Pi[x][i]-b[i] for i in k) assert all(Pi[x][i]==b[i]+gain/len(k) for i in k) Plus={x:{i:sum(p*L[y][i] for y,p in Q[x].items()) for i in N} for x in X} assert all(L[x][i]==Pi[x][i]/4+F(3,4)*Plus[x][i] for x in X for i in N)

def moves(x): result=[] for a in subsets(top(x),2): k=frozenset().union(*a) result.append((x|frozenset([k]),k)) for k in x: target=frozenset(a for a in x if not k<=a) result.extend((target,frozenset([i])) for i in k) result.extend((x,frozenset([i])) for i in N) return result

def strict(x,y,z,i): return L[y][i]>L[z][i] or (L[y][i]==L[z][i] and y==x and z!=x) eta=None; B=F(0); count=0 for x in X: M=moves(x); count+=len(M) U=[(y,k) for y,k in M if not any( h<=k and all(strict(x,z,y,i) for i in h) for z,h in M)] moving=[(y,k) for y,k in U if y!=x] assert {y for y,k in moving} == set(Q[x])-{x} for y,k in moving: for i in k: gap=L[y][i]-L[x][i] assert gap>0 eta=gap if eta is None else min(eta,gap) B=max(B,max(abs(Plus[x][i]-L[x][i]) for i in N)) assert count==558 and eta==F(24,115) and B==F(608,25) alpha0=eta/(2*B) assert alpha0==F(15,3496) assert F(3)/(3+alpha0)==F(3496,3501) assert all(N not in x and set(Q[x])<=set(core) for x in core) print('PASS: 52 states; 558 moves; all sharing and Bellman identities;') print('all undominated target sets; eta=24/115; B=608/25;') print('equilibrium family certified for 3496/3501 <= delta < 1.') 

\end{lstlisting}

FINAL

LABEL: FULL SOLUTION.

SUBLABEL: FULL PROOF of the existential nonarrival assertion; COUNTEREXAMPLE/DISPROOF of the universal eventual-arrival threshold assertion.

Theorem Theorem (Fixed-model nonarrival arbitrarily close to patience) supplies one fixed four-player model, explicit equilibrium matrices at every $\delta\in[3496/3501,1)$, and probability-one nonarrival from $\varnothing$. The decisive mechanism is endogenous waiting: the equilibrium axioms require positive support, not a lower bound on the frequency with which supported moves occur. This conclusion applies to the literal definitions in the attachment; adding a uniform positive lower bound on transition rates or requiring all positive self-loops to have profitable stay witnesses changes the problem.

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Econ-12-A1 · Exploration · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Econ-12

Grand-coalition arrival in the patient limit GPT 6 Astra Ultra

Target, source correction, and result

The target asks whether a single fixed nested-coalition payoff model can have nonarriving equilibria for discount factors arbitrarily close to one. The source's revised equilibrium definition includes stays and compares a target with next-period continuation value. Its Remark 6.7 expressly withdraws the earlier deterministic-cycle example under that definition [heitzig]. That example is not used here.

Two results are established: the bad discount factors in any fixed model have only finitely many interval components, and all equilibria in the two-player case arrive at the grand coalition. The first result changes a potentially sparse patient-limit counterexample search into a fixed-support interval search. It does not decide whether such an interval exists for any larger model.

Finite polynomial encoding of the equilibrium conditions

Use the finite state set $X$ of laminar agreement families and the legal move sets from the target. A move $m$ at $x$ has target $\tau(m)$ and relevant set $R(m)\ne\varnothing$. All these sets are finite. Write $G\subseteq X$ for the grand-coalition states and $M(x)$ for the moving moves, with $M^+(x)$ also including all unilateral stays. Fix the real numbers $\pi_i(x)$ once and for all. For a stochastic matrix $p$ and $0<\delta<1$, define numbers $L_i(x),Z_i(x)$ by \begin{equation}\label{wsref-0-62-65-6c-6c} Z_i(x)=\sum_y p_{xy}L_i(y),\qquad L_i(x)=(1-\delta)\pi_i(x)+\delta Z_i(x). \end{equation} These equations have a unique solution: $(I-\delta p)^{-1}=\sum_{t\ge0}\delta^t p^t$ converges in the maximum row-sum norm. Thus $L_i$ is precisely the normalized discounted value and $Z_i$ the continuation value.

For targets $y,z$, define the strict comparison at $x$ by \[ y\succ_i^x z\ \Longleftrightarrow\ [L_i(y)>L_i(z)]\ \lor\ [L_i(y)=L_i(z),\ y=x,\ z\ne x], \] and define weak comparison as the negation of $z\succ_i^x y$. A move belongs to $C(x)$ exactly when \[ \bigwedge_{i\in R(m)}L_i(\tau(m))\ge Z_i(x) \] holds and there is no $m'\in M^+(x)$ with $R(m')\subseteq R(m)$ and $\tau(m')\succ_i^x\tau(m)$ for every $i\in R(m')$. A move in $C(x)$ belongs to $F(x)$ exactly when some player weakly prefers its target to the target of every member of $C(x)$.

Call $E(\delta,p,L,Z)$ the conjunction of stochasticity, (bell), and the following conditions at every state: \begin{align*} p_{xy}>0, y\ne x&\ \Longrightarrow\ \exists m\in C(x):\tau(m)=y,\\ C(x)\cap M(x)\ne\varnothing&\ \Longrightarrow\ p_{xx}<1,\\ m\in F(x)&\ \Longrightarrow\ p_{x,\tau(m)}>0. \end{align*} Every quantifier in the definitions of $C,F$ ranges over an explicitly finite set and can be expanded to a finite conjunction or disjunction. Therefore $E$ is a Boolean combination of polynomial equalities and strict or weak inequalities in $(\delta,p,L,Z)$, with the fixed payoffs as real coefficients. This also keeps the equality tie rule exact.

Lemma (Nonarrival certificate): For a finite stochastic matrix $p$ and a nonempty target set $G$, some initial state has positive probability of never hitting $G$ if and only if there is a nonempty $C\subseteq X\setminus G$ such that \[ p_{xy}=0\quad(x\in C,\ y\notin C). \]
Proof: If $X=G$, neither a nonarrival state nor a nonempty $C$ exists, so the equivalence holds. Otherwise, a chain starting in such a closed set $C$ never leaves it and therefore never hits $G$. Conversely, suppose no such set exists. In the directed support graph, every state outside $G$ has a path to $G$: otherwise the collection of states reachable from it would be a nonempty closed set outside $G$. Remove loops from each chosen path so its length is at most $|X|$. For each initial state outside $G$, the probability of following its path is positive. The minimum $\rho$ of these finitely many positive products is positive. Conditional on avoidance up to the beginning of any block of $|X|$ steps, the probability of hitting $G$ in that block is at least $\rho$. By the Markov property, avoidance for $k|X|$ steps has probability at most $(1-\rho)^k$, which tends to zero. ∎
Theorem (Fixed-model patient-limit dichotomy): Fix any finite payoff model with the moves and equilibrium definition above, in particular any fixed model satisfying the target's surplus and sharing assumptions. Let \[ B=\{\delta\in(0,1):\text{some equilibrium has positive nonarrival probability from some state}\}. \] Then $B$ is a semialgebraic subset of the real line and is a finite union of points and intervals. Exactly one of the following alternatives holds:
  1. There is $\delta_0<1$ with $B\cap(\delta_0,1)=\varnothing$.
  2. There is $\delta_0<1$ with $(\delta_0,1)\subseteq B$. Moreover, one fixed nonempty set $C\subseteq X\setminus G$ and one fixed directed support $S\subseteq X\times X$ can be used for a witnessing equilibrium at every $\delta\in(\delta_0,1)$.
For rational payoff data, it is decidable which alternative holds, with no polynomial-time bound asserted.
Proof: For each pair $(C,S)$, impose $E$, the closed-set equations of the lemma, and $p_{xy}>0$ for $(x,y)\in S$, $p_{xy}=0$ otherwise. Existentially quantify all coordinates of $p,L,Z$ to obtain a set $B_{C,S}$ of discounts.

We use the standard real quantifier-elimination theorem in exactly the following form: a set defined by a first-order formula with real polynomial comparisons is semialgebraic; for rational coefficients an equivalent quantifier-free formula can be computed. This is the Tarski--Seidenberg theorem and its effective form, stated in [basu], Section 2.1. This imported general theorem is not claimed as a new result here. It makes every $B_{C,S}$ semialgebraic, and the preceding lemma gives $B=\bigcup_{C,S}B_{C,S}$, a finite union.

A semialgebraic subset of the real line is a finite union of points and intervals: after quantifier elimination, gather the finitely many nonzero univariate polynomials in its formula. Each has finitely many roots. Between consecutive roots every polynomial has constant sign, so the formula has constant truth value. Zero polynomials do not create additional boundaries.

If $B$ has points arbitrarily close to one, finiteness of the union implies that some $B_{C,S}$ also has points arbitrarily close to one. Its finite interval decomposition must contain an interval ending at one; isolated points cannot accumulate there. This proves alternative 2 with fixed $C,S$. Otherwise alternative 1 holds. They cannot both hold. For rational data, apply effective quantifier elimination to the sentence \[ \exists d\ (0<d<1\ \wedge\ \forall\delta\,[d<\delta<1\Rightarrow\delta\in B]). \] Its truth selects alternative 2; its falsity selects alternative 1. ∎

This is a reduction for each fixed model, not a termination argument for searching over all player counts and all payoffs. Nor does fixed support impose a uniform positive lower bound on transition probabilities.

The two-player model

Proposition (Arrival with two players at every discount): If $N=\{1,2\}$ and the target's strict grand-surplus gap and positive sharing assumptions hold, every equilibrium reaches the grand coalition almost surely from every state, for every $0<\delta<1$.
Proof: There are only two states: $e=\varnothing$ and $g=\{N\}$. The assumptions give $\pi_i(g)-\pi_i(e)=s_i(g,N)\gamma>0$ for both players. Put $a=p_{gg}$ and $b=p_{eg}$. Subtracting the two Bellman equations yields \[ L_i(g)-L_i(e)= \frac{(1-\delta)[\pi_i(g)-\pi_i(e)]}{1-\delta(a-b)}>0, \] because $a-b\in[-1,1]$ and the denominator is at least $1-\delta$. At $g$, any termination to $e$ by player $i$ is dominated by that same player's stay at $g$. Thus no moving target is admissible at $g$, and equilibrium implies $p_{gg}=1$. At $e$, merging to $g$ gives both relevant players at least $Z_i(e)$, since $Z_i(e)$ is a convex combination of $L_i(e)$ and $L_i(g)$. No stay can dominate this merge. It is the sole moving move: each player's stay at $e$ either fails profitability or is dominated only if allowed by the relevant-set rule, but in either case its target is strictly worse than $g$; hence the merge is a favorite whether or not stays remain in $C(e)$. The favorite condition forces $b>0$. Consequently the probability of remaining at $e$ for $t$ steps is $(1-b)^t\to0$. ∎

An adversarial limit check and unresolved gap

The matrices $p_h=\left(\begin{smallmatrix}1-h&h\\0&1\end{smallmatrix}\right)$, $h>0$, all hit the second state almost surely, but their limit as $h\downarrow0$ is the identity and has nonarrival from the first state. This elementary calculation is a warning about a proposed compactness proof: arrival is not preserved when positive edges disappear. These matrices are only a Markov-chain check, not a claimed equilibrium counterexample.

The unresolved task is to find a fixed admissible model whose semialgebraic bad set contains a final interval, or to exclude every such interval for every player count. No rational model was run through full quantifier elimination in this attempt.

References

  • [heitzig] J. Heitzig, Does the grand coalition form? Persistence, arrival, and the role of the sharing rule in a dynamic process of nested binding agreements, arXiv:2608.17766v2 (2026), Definitions 2.14--2.15, Remark 6.7, and Section 8. https://arxiv.org/html/2608.17766v2. These revised definitions and the explicit correction to the old cycle were checked on 9 October 2026.
  • [basu] S. Basu, Algorithms in Real Algebraic Geometry: A Survey, arXiv:1409.1534 (2014), Section 2.1, Theorem 2.1. https://arxiv.org/pdf/1409.1534. Used solely for real quantifier elimination; all equilibrium-specific encoding and finite-chain arguments are proved above.

Completion Estimate

15%

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