slug: cao-zhang-shi-zhao-2026-mabk-optimal-robustness-positivity bibkey: cao2026sizeindependent doi: 10.48550/arXiv.2608.30851 url: https://arxiv.org/abs/2608.30851v1 triage: theorem motivation_gids:

• D5/S3/QuantumBounds/MabkSelfTestingPositivity.result

Cao–Zhang–Shi–Zhao MABK positivity for the two remaining blocks

Problem

Shen Cao, Xingjian Zhang, Fei Shi and Qi Zhao, Size-Independent Robustness in Multipartite Bell Self-Testing, arXiv:2608.30851v1, Supplemental Material, “Efficient Verification of Optimal Lower Bound”, state:

For $n\le 5$ this bound is already established analytically [20, 23]; for $n\ge 6$ the two remaining cases ($h=1,2$) are an open conjecture supported by the numerical evidence above.

The conjecture is that the five-term expression ${\lambda }_{\vec{a}}$ on PDF p. 37 is nonnegative at every point of $[0,\kappa {]}^n$, where $\kappa =1-1/\sqrt{2}$, for every $n\ge 6$ and $h\in \{1,2\}$. The source has complementary index sets $T$ and $J$, of cardinalities $h$ and $d=n-h$. Its convention $h\le d$ follows from these hypotheses. Issue #11561 preregisters the literal expression, full quantifiers, Tier 1 classification and literature check.

The definition lambdaA preserves all five terms, their coefficients, both complementary products in each paired term, and the real square roots. The bijection from Fin n to the source indices is $i\mapsto i+1$. The definitions kappa and claim express the endpoint and complete conjecture. The theorem result : claim proves it.

Motivation

The conjectured parameters are ${\bar{s}}_n=(1+\sqrt{2})/2^{n/2+1}$ and $\bar{\mu }=-1/\sqrt{2}$. The source proves the other block cases analytically, and identifies $h=1,2$ as the remaining scalar obstruction for $n\ge 6$. D5/S3/QuantumBounds/MabkSelfTestingPositivity.result supplies that scalar inequality for every party number in the stated range, without restricting the cube to a grid. The operator reductions and extractability statements are separate, unformalized obligations.

Gap

The preregistered literature check in #11561 records arXiv v1, zero citing works in the Semantic Scholar query, and no entry for this statement in the MathDB queries. Google Scholar and INSPIRE were not reached. These are bounded, source-reported readings; they do not establish a worldwide novelty or priority claim. The source TeX explicitly calls the two cases open. Repository and pinned-library searches find no owner for the same conjecture in the searched scope.

Route

Put $r=\sqrt{2}$, $s=1+r$, $c=s/r$ and, for each coordinate, $x_i=1-v_i$, $y_i=\sqrt{2v_i-v_i^2}$, $a_i=1-c{v}_i$, $b_i=c{v}_i$. For a finite block $E$, use $A_E,B_E,X_E,Y_E$ for their products. The cube gives $a_i\ge x_i^2$, $0\le b_i\le 1/2$, $0\le y_i^2\le 1/2$, $x_i^2+y_i^2=1$ and $b_i\ge (c/2)y_i^2$.

Let $q=X_J$, $B=B_T$, $Y=Y_T$ and $K=(2+r)BY$. Retaining the positive cubic term gives

$${\lambda }_{\vec{a}}\ge 1-s^2{X}_T^2{Y}_J^2+(rB-s^2{Y}^2)q^2+K{q}^3.$$

Finite-set induction proves $Y_J^2\le (1-q^2)2^{1-d}$. The identity $2q^3-3q^2+1=(1-q{)}^2(2q+1)$ gives $q^3\ge (3q^2-1)/2$. Thus

$${\lambda }_{\vec{a}}\ge C(1-q^2)+F{q}^2,$$

where $C=1-K/2-s^2{2}^{1-d}$ and $F=1-s^2{Y}^2+rB+K$. For $h=1$, a denominator-cleared unit-circle factorization proves $F\ge 0$. For $h=2$, $B\ge s^2{Y}^2/8$ and $0\le Y\le 1/2$ give

$$F\ge 1-(5/2+13r/8)Y^2+(5/4+7r/8)Y^3\ge (34-19r)/64>0.$$

The complementary block has $d\ge 5$ for $h=1$ and $d\ge 4$ for $h=2$; the respective bounds on $K$ establish $C\ge 0$. Since $q\in [0,1]$, the final lower bound is nonnegative.

Falsifier

A counterexample to the encoded claim would supply $n\ge 6$, $|T|\in \{1,2\}$ and one point of the whole cube with lambdaA < 0. The kernel-checked theorem excludes such a witness for the stated real expression. A different coefficient, omitted product, altered square root or weakened cube would define a different problem. No such substitution is used in lambdaA.

Evidence

The mathematical source is D5/S3/QuantumBounds/MabkSelfTestingPositivity.lean, with public definitions kappa, lambdaA, claim, and the sole public theorem result. The two private content theorems are product_gap (finite-set induction) and block_bound (one- and two-element scalar estimates). Both are used on the live path of result. The module imports pinned Mathlib only. The proof uses the standard axioms propext, Classical.choice and Quot.sound.

The Scribe source and its emitted Markdown expose the complete expression, quantifiers and cube assumptions. The Proved resolution claim links this dossier to the frozen settling theorem. The Reg mirror Reg/D5/S3/QuantumBounds/MabkSelfTestingPositivity.lean declares the dependent-family realization audit; its residual continuation is open.

Triage

Tier 1: a recent quant-ph conjecture with the full analytic proof for the other block cases and numerical support for the two remaining cases. Resolution: Proved. Admission basis: open-problem-resolution (#11561). The public settling theorem and both private helper theorems have proof_shape: content. Utility is none: the result is an analytic inequality over all $n\ge 6$ and all real cube points, rather than an enumeration, checker, numerical reduction or certified finite instance.

What the settlement shows

• Proved in this module: nonnegativity on the entire cube for both remaining block sizes and every $n\ge 6$. The decisive mechanism is the product gap on $J$, preservation of a positive cubic term, and the lower bound $C(1-q^2)+F{q}^2$. Dependence on the complementary dimension enters through the decreasing factor $2^{1-d}$.
• Proved inside the result’s proof: the one-element unit-circle factorization and the two-element cubic lower estimate for $F$. The generic product-gap helper applies to any nonempty finite set with $u_i+z_i=1$, $u_i\in [0,1]$, $z_i\in [0,1/2]$. These are proof components, not additional public settling theorems.
• Computed: for the source’s boundary configuration $v_T=\kappa$, $v_J=0$, the reduced expression is $0$ when $h=1$ and $(4-\sqrt{2})/8$ when $h=2$. The command python3 -c 'import sympy as s; r=s.sqrt(2); print([s.simplify(1+r*s.Rational(1,2)**h-(1+r)**2/r**(2*h)+(2+r)*s.Rational(1,2)**h/r**h) for h in (1,2)])' returns [0, 1/2 - sqrt(2)/8]. This algebra computation is not a kernel theorem about equality configurations.
• Open: the exact $h=2$ minimum, classification of all $h=1$ equality points, extension beyond the cube, and optimality of the block-size thresholds in this argument. No relaxed-hypothesis theorem is delivered.
• Open as a formalization obligation: the source’s Bell-operator reduction, the other block cases and the extractability conclusions. Under the source’s reductions, the proved scalar conjecture removes its remaining obstruction to optimal MABK robustness for all party numbers; those reductions are literature conclusions outside this module.

ASSUMED-UNVERIFIED

Worldwide priority and exhaustive absence of another answer are not established by the bounded literature searches. The kernel does not authenticate an external paper or its interpretation. The source’s Bell-operator and extractability implications have not been checked in Lean here. An open audit residual does not prove undecidability or inexhaustibility.