# AMRAL × Lebesgue Universal Covering — Round 23 ## Witness Portfolio Closure, Strict Incidence, and Residual-Frontier Routing **Document ID:** AMRAL-LUC-FC-R23 **Version:** v0.1 **Date:** 2026-09-19 **Research status:** Round 23 / Portfolio closure / Set-cover legality / Horizon routing **Research mode:** Human-Directed + Semi-Autonomous AI Mathematical Research **Research initiation and methodology source:** Neo.K **AI collaborating researcher and primary executor:** Aletheia / ChatGPT, GPT-5.6 Sol **Prerequisite documents:** AMRAL-LUC-FC-R00 v0.2; R01–R22 v0.1 --- # 0. Round Summary Round 22 established the witness proof-cost Pareto portfolio: $$ \mathcal W = \{ B_7, H_{11}, H_{13}, H_{17}, H_{19} \}. $$ Round 23 no longer asks: > Which witness is strongest on average? and instead asks, at the theorem level: > For every cell of the necessity-marked base atlas, has at least one witness's **complete relevant placement root** already been closed by certificate? This round therefore formally distinguishes: $$ \boxed{ \text{Strict Incidence} } $$ from: $$ \boxed{ \text{Quantitative Partial Edge.} } $$ Main results: 1. the strict cell × witness incidence theorem; 2. the witness portfolio closure theorem; 3. the weighted set-cover / facility-location formulation; 4. the Partial Edge Non-Composition Rule; 5. the budget residual frontier; 6. the 12-cell × 5-witness empirically measured portfolio matrix; 7. of the 60 edges at lift depth 16: $$ \boxed{0} $$ are `COMPLETE`; 8. so strict weighted set cover at this budget: $$ \boxed{\text{INFEASIBLE}}; $$ 9. shallow quantitative routing all favors $H_{19}$; 10. but at depth 22, among the first four hardest cells, $B_7$ has already overtaken $H_{19}$ in unresolved placement volume in 3/4 cells; 11. so witness routing must be horizon-dependent; 12. partial witnesses cannot be "pieced together" into a theorem cover; if there is no single-witness `COMPLETE` edge, deeper certification, base refinement, a new witness, or a genuine joint multi-witness certificate is required. This round's determination: $$ \boxed{ \text{STRICT PORTFOLIO INCIDENCE: CLOSED} } $$ $$ \boxed{ \text{SET-COVER LEGALITY: CLOSED} } $$ $$ \boxed{ \text{CURRENT 12-CELL STRICT COVER AT DEPTH 16: INFEASIBLE} } $$ $$ \boxed{ a_{\mathrm{Leb}}\ge0.8350: \text{STILL COMPUTE-DEFERRED} } $$ --- # 1. Necessity cells Let: $$ \mathcal C = \{ C_1,\ldots,C_m \} $$ be the necessity-marked base cells. Each: $$ C_j $$ already carries a Round 19 marker: $$ \exists q_j\in C_j, \qquad A_{\mathcal F_3}(q_j) over the entire base cell $C_j$, witness $K$'s complete relevant placement root has already been proved by a legitimate lift certificate: $$ J_K(H_{\mathcal F_3}(q)) \ge T \qquad \forall q\in C_j. $$ otherwise: $$ I_{jK}=0. $$ --- # 4. What does NOT create an incidence edge None of the following can make: $$ I_{jK}=1. $$ including: - a `PARTIAL` lift tree; - a very small pending fraction; - a very small unresolved placement volume; - a numerical search minimum $>T$; - no counterexample found; - fast tail contraction; - the witness succeeding on some other cell. so: $$ \boxed{ \text{a strict edge is certificate-complete, not confidence-complete}. } $$ --- # 5. Portfolio Closure Theorem ## Theorem 5.1 Suppose every necessity cell: $$ C_j $$ has at least one witness: $$ K(j) $$ such that: $$ I_{j,K(j)}=1. $$ Let: $$ \mathcal B = \{ K(j):j=1,\ldots,m \}. $$ Then for all these necessity cells: $$ \boxed{ \Lambda( \mathcal F_3\cup\mathcal B ) \ge T } $$ holds on that atlas coverage. ### Reason Fix any base configuration: $$ q\in C_j. $$ Because: $$ I_{j,K(j)}=1, $$ so: $$ J_{K(j)}(H(q))\ge T. $$ the full portfolio hull contains that single witness's hull. Hence the full portfolio's placements, whatever they are, also cannot fall below that already-proved lower bound. This is the strict-incidence form, on a finite cell atlas, of the witness-switching argument from Round 07. Q.E.D. --- # 6. Weighted set cover If witness: $$ K $$ has a fixed compiler / verifier cost: $$ f_K, $$ and the cell-witness certificate cost is: $$ c_{jK}, $$ introduce: $$ y_K\in\{0,1\}, $$ $$ x_{jK}\in\{0,1\}. $$ objective: $$ \boxed{ \min \sum_K f_Ky_K + \sum_{j,K}c_{jK}x_{jK}. } $$ subject to: $$ \boxed{ \sum_Kx_{jK}\ge1 \quad \forall j, } $$ and: $$ \boxed{ x_{jK} \le I_{jK}y_K. } $$ This is the witness-portfolio facility-location / weighted set-cover form. --- # 7. Budgeted incidence In practice a certificate has a budget: $$ d, $$ e.g. the maximum lift-tree depth. so write: $$ I_{jK}^{(d)}. $$ If there is no closure within the budget: $$ I_{jK}^{(d)}=0 $$ does not mean the witness can never close it. it only means: $$ \boxed{ \text{not complete at budget }d. } $$ --- # 8. Residual frontier Define: $$ \boxed{ \mathcal R_d = \left\{ j: \max_K I_{jK}^{(d)}=0 \right\}. } $$ Then a necessary condition for strict set cover to be feasible at budget $d$ is: $$ \boxed{ \mathcal R_d=\varnothing. } $$ In fact, on the current finite cell list, it is also sufficient: as long as every cell has at least one strict edge, an assignment can be chosen. --- # 9. Quantitative partial edge For a partial lift, additionally store: $$ \boxed{ Q_{jK}^{(d)} = ( N, P, V_{\rm unres}, t, m_{\rm search}, \rho_{\rm tail} ). } $$ where: - $N$: nodes; - $P$: pending leaves; - $V_{\rm unres}$: unresolved placement volume; - $t$: compute cost; - $m_{\rm search}$: forcing search margin; - $\rho_{\rm tail}$: multi-depth contraction. it is given to the scheduler only. not to the theorem. --- # 10. Partial Edge Non-Composition Rule Two witnesses: $$ K_1,K_2 $$ if both have only a partial lift tree, then one cannot, just because: > $K_1$ is almost entirely closed; > $K_2$ is also almost entirely closed; claim cell closure. The two witnesses have independent placement adversaries. so: $$ \boxed{ \text{PARTIAL}+\text{PARTIAL} \not\Rightarrow \text{STRICT COVER}. } $$ --- # 11. Multi-Witness Synergy Needs a New Certificate The full family: $$ H\cup g_1K_1\cup g_2K_2 $$ may, because of witness-witness interaction, be stronger than any single-witness lower bound. But to exploit this synergy, one must establish: $$ \boxed{ \text{a genuine joint multi-witness certificate} } $$ or some other theorem capable of controlling simultaneous tails. It cannot be smuggled out of two marginal partial trees. --- # 12. Pilot scope A full: $$ 77\times5 $$ same-depth run in the interactive environment exceeds a single round's execution budget. So this round's theorems / grammar are complete, but the empirically measured matrix first takes, from the necessity atlas, the: $$ \boxed{ 12 } $$ cells with the deepest outer marker, with witnesses: $$ 5. $$ for a total of: $$ \boxed{ 60 } $$ edges. The long-run spec has been output, for the local runtime to expand into the complete $77\times5$. --- # 13. Depth-16 strict incidence result lift depth: $$ d=16. $$ Of the 60 edges: $$ \boxed{ 0 } $$ are `COMPLETE`. so: $$ \boxed{ I_{jK}^{(16)}=0 \quad \forall j,K } $$ on this 12-cell pilot. therefore: $$ \boxed{ \mathcal R_{16} = \{1,\ldots,12\}. } $$ strict weighted set cover: $$ \boxed{ \text{INFEASIBLE}. } $$ This is the correct result, not an algorithm failure. --- # 14. Why set cover must not be forced If, at this point, one were to treat: > the witness with the smallest unresolved volume as: $$ I=1, $$ that would pass off a scheduling heuristic as a theorem. Round 23 explicitly forbids this. The correct action is: - deepen the selected witnesses; - further base refinement; - add new witnesses; - or a joint multi-witness certificate. --- # 15. Depth-16 aggregate quantitative edges ## $B_7$ mean nodes: $$ 2401. $$ mean unresolved volume: $$ \boxed{ 9.42\times10^{-4}. } $$ search forcing margin: $$ \approx1.93\times10^{-3}. $$ --- ## $H_{11}$ mean nodes: $$ 3749.5. $$ mean unresolved volume: $$ 4.77\times10^{-4}. $$ forcing margin: $$ 7.47\times10^{-4}. $$ --- ## $H_{13}$ mean nodes: $$ 3644.2. $$ mean unresolved volume: $$ 3.41\times10^{-4}. $$ forcing margin: $$ 6.38\times10^{-4}. $$ --- ## $H_{17}$ mean nodes: $$ 2362.5. $$ mean unresolved volume: $$ 2.66\times10^{-4}. $$ forcing margin: $$ 7.83\times10^{-5}. $$ --- ## $H_{19}$ mean nodes: $$ 2286.3. $$ mean unresolved volume: $$ \boxed{ 2.29\times10^{-4}. } $$ forcing margin: $$ 4.96\times10^{-5}. $$ --- # 16. Shallow proxy routing If only: $$ V_{\rm unres} $$ is used as the short-horizon scheduler metric, for the 12 cells: $$ \boxed{ 12/12 } $$ all select: $$ H_{19}. $$ This is consistent with Round 22's proof-cost Pareto conclusion: > a high-$k$ smooth harmonic is very cheap in shallow root-volume economics. But this is not a strict cover. --- # 17. Horizon crossover Round 20 already showed that $B_7$'s deep-tail contraction is faster. Round 23 pushes the first four hardest cells directly to: $$ d=22. $$ comparing: $$ V_{j,B_7}^{(22)} $$ with: $$ V_{j,H_{19}}^{(22)}. $$ Result: - cell 0: $B_7$ wins; - cell 1: $B_7$ wins; - cell 2: $B_7$ wins; - cell 3: $H_{19}$ wins. so: $$ \boxed{ B_7\text{ has already overtaken in }3/4\text{ of the hardest cells at deep horizon.} } $$ --- # 18. Cell 0 full depth-22 ranking unresolved placement volume: $$ B_7: \boxed{ 5.17\times10^{-5} } $$ $$ H_{19}: 5.43\times10^{-5} $$ $$ H_{17}: 6.27\times10^{-5} $$ $$ H_{13}: 9.24\times10^{-5} $$ $$ H_{11}: 1.09\times10^{-4}. $$ so the depth-16 leader: $$ H_{19} $$ has, by depth 22, already been overtaken by: $$ B_7 $$ --- # 19. Crossover depth For cell $j$ and witnesses $K,L$, define: $$ \boxed{ d^\star_{j;K,L} = \inf \{ d: V_{jK}^{(d)} \le V_{jL}^{(d)} \}. } $$ This is a performance quantity. not a theorem threshold. It tells the scheduler: > how deep do you expect to run? If the horizon is short: choose the cheap witness. If the horizon is long: choose the witness with faster contraction. --- # 20. Portfolio scheduler is horizon-aware so witness routing cannot be ordered by any single value alone, such as: - search forcing; - shallow pending; - orientation period; It should use: $$ \boxed{ \text{estimated total cost-to-COMPLETE}. } $$ including: - current unresolved volume; - tail contraction; - nodes per level; - forcing risk; - negative counterexample; - expected remaining horizon. --- # 21. Strict proof and routing must remain separate ## Proof layer looks only at: $$ I_{jK}\in\{0,1\}. $$ ## Scheduler layer looks at: $$ Q_{jK}. $$ These two layers must not be mixed. Round 23 formally lists this separation as a production invariant. --- # 22. Current residual On the 12-cell pilot, at depth 16: $$ \boxed{ |\mathcal R_{16}|=12. } $$ so Round 23 has no "portfolio closure." what it has obtained is: > the existing portfolio is already enough to form a meaningful routing competition, but there is not yet a single strict complete edge. This directly specifies the next step of work. --- # 23. 77-cell long-run handoff This round is accompanied by: `AMRAL_LUC_FC_Round_23_LONG_RUN_SPEC.json` requiring the local runtime to: 1. run the full 77 necessity cells × 5 witnesses; 2. at common budgets; 3. save the strict incidence and the quantitative edges; 4. deepen only the residual cells; 5. as soon as any `COMPLETE` edges appear, immediately update the facility-location solver; 6. finally hand only the base residual with no strict edge at all to: - deeper B7; - joint multi-witness; - new witness synthesis. --- # 24. Portfolio monotonicity For a fixed cell-witness pair: if the certificate budget only increases without changing the theorem semantics, once: $$ I_{jK}^{(d)}=1, $$ subsequent larger budgets need not turn it back to 0. so the strict incidence graph can grow append-only. except for a negative audit / a change in dependencies. --- # 25. Counterexample effect If witness: $$ K $$ already has a valid negative counterexample on cell $C_j$, then: $$ \boxed{ I_{jK}=0 } $$ is not merely budget-zero, but means this witness cannot be the sole closer for that cell. Such an edge can be directly marked: `DISQUALIFIED` and permanently removed from future single-witness set-cover candidates, until the semantics change. --- # 26. Round 24 ## Portfolio Residual Core and Joint-Witness Necessity The next round should, after the full / local-compute incidence returns: 1. extract: $$ \mathcal R = \{C_j:\forall K,\ I_{jK}=0\}; $$ 2. for the residual cells, compare: - deepening $B_7$; - deepening $H_{19}$ further; - additional portfolio witnesses; 3. if multiple single witnesses all stably have a nonempty tail: - analyze whether their tails can be simultaneously realized by an adversary; 4. only then open: $$ \boxed{ \text{a joint multi-witness placement certificate}. } $$ --- # 27. Shortest Handoff The most important thing about Round 23 is not obtaining a set-cover answer. but cleanly separating: $$ \boxed{ \text{proof cover} } $$ from: $$ \boxed{ \text{portfolio routing} } $$ completely. Currently: $$ \boxed{ \text{strict cover at depth 16 = infeasible}. } $$ but: $$ \boxed{ \text{routing structure = highly informative}. } $$ short horizon: $$ H_{19} $$ is cheap. deep horizon: $$ B_7 $$ begins to overtake. so the production portfolio should be: $$ \boxed{ \text{strict COMPLETE edges for the theorem} + \text{horizon-aware quantitative edges for scheduling}. } $$