There is no qubit [[11,3,d]] stabilizer code with d>=3 admitting eight independent commuting Pauli generators of weight at most four. This includes degenerate and non-CSS codes. Consequently, together with the published weight-five construction, W_opt(11,3,3)=5.

This is a root-assembled complete proof candidate. Parts I–III restate the supplied prior necessary-condition proof and two native analytic lemmas in full; Part IV uses the root enumeration as reconstructed and reproduced in the native template certificate. The assembly, especially the reduction from degenerate codes, has not yet passed an independent full-proof audit. No independent-discovery or novelty claim is made.

PART I. Elimination of degeneracy and reduction to three incidence profiles.

Work throughout with binary Pauli spaces modulo scalar phases. Let S have n=11, rank m=8, k=3 and distance at least three, and suppose it has a basis of weight at most four. The proof specializes and refines the incidence method of Wei et al., arXiv:2601.19848v2, Appendix F.1–F.2. The star-space and isolated-edge bounds, and the final local-parity obstruction, are attributed to that source. The refinement used here is the strict star-space inequality after excluding small stabilizer-state factors.

1. There is no weight-one stabilizer and no rank-two stabilizer-state factor on two qubits. Indeed, deleting such a factor preserves k=3 and distance and does not increase generator weights: restrictions of the original generators span the residual space, and an independent subset is a basis. The supplied generator-weight bound ceil((11n+k+1)/(4(n-k))) is five at both (n,k)=(10,3) and (9,3), a contradiction. For the two-qubit assertion, two commuting stabilizers with independent local labels on each of the same two qubits span a Lagrangian Pauli space there. The restriction of every element of S lies in this space; hence it is a tensor factor and deletion applies. Signs do not affect these binary arguments.

2. Every qubit has full two-dimensional local Pauli span. Otherwise a nonidentity one-qubit Pauli commutes with S, so distance at least three forces it into S, contrary to step 1. Thus each column in any generator basis has degree at least two, and the labels on a degree-two column are distinct. Let L be the span of weight-two stabilizers and ell=dim L. By ordinary basis extension, choose a basis with ell weight-two rows spanning L and the remaining rows of weight at most four.

3. Form the degree-two multigraph on the eight generator rows, with each degree-two column an edge. Write e for its edge count, v for its number of nonisolated vertices, b=8-v, and s=2e-v. At an edge q=(i,j), the syndrome map from the one-qubit Pauli space to coordinates i,j is an isomorphism. Write E_(i,q) for the single-qubit error with syndrome the unit vector at i. For each row i, products E_(i,q) E_(i,q0) over incident q other than a fixed q0 are independent weight-two stabilizers, because their syndrome is zero and distance is at least three. Their space K_i has dimension deg(i)-1. The K_i are in direct sum: on an edge (i,j), the possible labels from the two endpoint spaces are independent. Consequently K=sum_i K_i is a subspace of L of dimension s, supported on degree-two columns.

There are no parallel edges. Two edges q,r on the same endpoints would give the stabilizers E_(i,q)E_(i,r) and E_(j,q)E_(j,r), a forbidden rank-two factor as in step 1. No basis row can be supported entirely on degree-two columns: at any edge (i,j) incident on such a row i, rows i,j anticommute, and cannot meet elsewhere in the support of i because that would be a parallel edge. This contradicts global commutation. If ell>0 and s=ell, then K=L, and the weight-two rows of our adapted basis would be supported entirely on degree-two columns. Hence

    c := ell-s >= 1 if ell>0; and c>=0 always.

4. We record explicitly the published incidence inequality used with this strict refinement. Put t=sum_(q:r_q>=4)(r_q-3), I=sum_q r_q=33-e+t, and A=32-2ell-I>=0. Let M be the number of isolated edges of the degree-two graph. Its nonisolated components have at most twice their total contribution to s edges, so M>=e-2s. (For a connected nonisolated component with h edges and u vertices, h>=u-1, h>=2, and h<=2(2h-u).)

An isolated edge has two rows with distinct local labels. Commutation forces another column on which these rows also have distinct labels. If a degree-three partner exists, choose it: a Pauli on that partner commuting with its third row has a nonzero syndrome on the two endpoints, which can be matched by a Pauli on the isolated edge. Their product is a weight-two stabilizer. These vectors, one per such edge, are independent of one another and of K, since each has nonzero support on its own isolated edge and K vanishes there. Thus their number G3 is at most ell-s=c.

Assign each remaining isolated edge to a partner of degree r>=4. Such a column receives at most floor(r/2)<=2(r-3) assignments because the endpoints of isolated edges are disjoint. Hence M<=c+2t. Combining this with M>=e-2s and 2e=8-b+s gives 8<=2ell+s+b+4t<=3ell+b+4t. Define H=3ell+b+4t-8>=0. A direct substitution of the definitions gives the specialized version of Eq. (80) in the source:

    4 = (4*4+1)*8-12*11 = H+3ell+b+2c+4A.

If ell>0, the right side is at least 3ell+2c>=5. Therefore ell=0: there are no weight-two stabilizers.

5. Now take ANY basis of weight at most four; it is automatically L-adapted. The same construction gives s=0, so the degree-two graph is a matching and e<=4. With ell=0, I=33-e+t<=32 gives A=e-t-1>=0. No matching edge has a degree-three partner, since its resulting weight-two stabilizer would contradict ell=0. The assignment bound thus gives e=M<=2t. Enumerating these three integer inequalities yields exactly

    (e,t,A)=(2,1,0),(3,2,0),(4,2,1),(4,3,0).

6. The case (e,t,A)=(4,2,1) is impossible by the local-parity obstruction used in the source, which we spell out. The four matching edges cover all eight rows. Equality e=2t forces every column of degree at least four to have degree exactly four and to receive exactly two assignments: floor(r/2)<2(r-3) for r>=5, and all capacity must be used. There are exactly two such columns. Their supports are unions of two disjoint matching pairs, and together cover all eight rows once. Each row therefore belongs to exactly one degree-two column and exactly one degree-four column.

Every degree-three column has labels X,Y,Z in some order. Full local rank excludes three identical labels. If they were P,P,Q with P!=Q, the one-qubit error P on that column would have unit syndrome at the Q row. A Pauli on that row's degree-two matching column has the same syndrome, yielding a forbidden weight-two stabilizer.

For any row, summing its pairwise commutation equations over all other rows says its total local anticommutation degree is even. Its degree-two column contributes one, and each degree-three column contributes two. Its unique degree-four column must therefore contribute an odd number. In a degree-four column containing two assigned matching pairs, the labels within each pair are distinct. The label multiplicities are thus either 2+2 or 2+1+1. A label of multiplicity mu anticommutes with 4-mu other labels, so in each pattern at least two rows have even local anticommutation degree. This is a contradiction.

7. The surviving profiles are (e,t,A)=(2,1,0),(3,2,0),(4,3,0). In all cases A=0, so I=32: all eight rows have weight exactly four in ANY weight-at-most-four basis. If S contained a weight-three vector, append it to a chosen weight-at-most-four basis and extract a basis containing that vector; this is possible by linear-algebra basis exchange and all selected vectors still have weight at most four. That would contradict I=32. Combined with steps 1 and 4, S therefore has no nonidentity stabilizer of weight one, two, or three.

This is a necessary-condition theorem. It neither constructs such a code nor excludes the three remaining profiles, and does not settle the original existence question. The source already implies ell<=1 directly from Eq. (80); that weaker statement is not presented as a new general bound here.


PART II. Exclusion of the perfect-matching profile (e,t)=(4,3).

For a Pauli error P, define its syndrome s(P) in F_2^m by s(P)_i=<P,g_i>. The error hypothesis implies that every nonidentity single-qubit error has nonzero syndrome. Moreover, nonidentity single-qubit errors on distinct qubits have distinct syndromes: otherwise their product is a weight-two nonidentity Pauli commuting with every generator.

No column has degree zero or one, since such a column admits a nonidentity single-qubit error with zero syndrome. More generally, no column is monochromatic among its nonidentity entries: an error with that common Pauli label would have zero syndrome.

Let e count the degree-two columns. Their perfect-matching property gives m=2e. The two nonidentity labels of each such column must be distinct. If its incident rows are i and j, its three nonidentity single-qubit errors therefore have syndromes u_i, u_j, and u_i+u_j, where u_i denotes the ith standard basis vector of F_2^m. Consequently every unit syndrome is realized on a degree-two column.

A column of degree d>=3 cannot have label multiplicities (d-1,1,0), in any order. An error with its majority label would have unit syndrome at the minority row, duplicating the syndrome of an error on a distinct, degree-two column. Thus every degree-three column has one occurrence of each of X,Y,Z.

For a column c and an incident row i, let a_ic count the other incident rows whose local label differs from that of i. These are exactly the local anticommutations involving row i. If c has degree d and the label at i has multiplicity mu, then a_ic=d-mu. For a nonincident row set a_ic=0. Write o(c) for the number of rows for which a_ic is odd.

A degree-two column contributes an odd a_ic at both of its incident rows. A degree-three column contributes a_ic=2 at each incident row, so o(c)=0. For degree four, the only permitted multiplicity partitions are 2+2 and 2+1+1, giving o(c)=0 and o(c)=2, respectively. Thus o(c)<=2. For degree five, o(c) is even by the handshaking identity applied to the graph whose vertices are incident rows and whose edges join unequal labels. Hence o(c)<=4. For d>=6, o(c)<=d<=2(d-3). In all cases with d>=4 we therefore have o(c)<=2(d-3).

For each row i, pairwise commutation implies that sum_c a_ic is even: it equals the sum over j!=i of the number of columns at which rows i and j anticommute, and each of those numbers is even. Row i receives exactly one odd contribution from its matching column, and only even contributions from degree-three columns. It must therefore receive an odd number, and in particular at least one, of odd contributions from columns of degree at least four. Summing over rows gives m<=sum_{c:deg(c)>=4} o(c)<=2t. Because m=2e, this proves t>=e.

All columns have degree at least two, so D=sum_c deg(c)=3n-e+t>=3n. In the eleven-qubit, eight-generator application, the assumed weight cap gives D<=8*4=32, whereas the proved bound gives D>=3*11=33. The contradiction is the exact inequality 32<33.

For application to a stabilizer code of distance at least three, the error hypothesis holds whenever the stabilizer contains no nonzero vector of weight one or two: any such low-weight centralizer vector would otherwise be either a low-weight stabilizer or a logical operator contradicting the distance. Under this hypothesis, four degree-two columns forming a matching on eight rows are a perfect matching, so the (4,3) profile is excluded. No assumption about the validity of the supplied broader profile reduction is needed for this implication.

PART III. Exclusion of (2,1), and reduction of (3,2) to two templates.

Write the nonidentity labels in each column as X,Y,Z. The syndrome of a single-qubit Pauli P at column c is the subset of rows whose labels at c are nonidentity and different from P. The hypothesis on weight-one and weight-two centralizer vectors implies that all three single-qubit syndromes at every column are nonempty and distinct, and that syndromes from distinct columns never coincide. Indeed, a zero syndrome gives a weight-one centralizer vector, and equal syndromes at distinct columns give a weight-two centralizer vector. In particular, every column uses at least two distinct nonidentity labels.

A degree-two column on rows a,b consequently has different labels and supplies the three syndromes {a}, {b}, {a,b}. Call {a,b} a matching edge. No degree-three column can contain both endpoints of a matching edge. To prove this, if all three labels differ, {a,b} is one of its syndromes. If exactly two labels agree, its syndromes include both the repeated-label pair and the singleton carrying the other label. When the repeated-label pair is {a,b}, its pair syndrome is forbidden; otherwise its singleton is one of a,b and is forbidden. These exhaust the possibilities.

Here is the closed-quartet obstruction used below. Suppose U consists of the four endpoints of two matching edges. Suppose each row r in U occurs in its matching column, in one common degree-four column H with support exactly U, and in exactly two degree-three columns, and in no other columns. Then these assumptions are impossible.

To prove this obstruction, a degree-three column meeting U has at most two rows in U: any three rows of U contain a complete matching edge. For a row r in U and a degree-three column T containing r, define A_r(T) to be the set of rows outside U whose labels at T differ from the label at r. These are precisely the outside rows with which r anticommutes at T. Its matching column and H contribute no anticommutation between r and outside rows. Commutation of the generator rows therefore implies A_r(T)=A_r(T') for the two degree-three columns T,T' containing r.

This common set is nonempty. If T has just one row in U, an empty external anticommutation set would make all three labels equal. If T has two rows in U, an empty set means the outside row has the same label as r; either all labels agree or the other U row is the unique different label, producing a forbidden singleton syndrome in U. All these possibilities are excluded.

The common set cannot have size two. Such a column has r and two outside rows, each with label different from r. If those outside labels agree, {r} is a forbidden syndrome. Otherwise all three labels differ. Thus both T and T' would have exactly the same three-row support and three different labels, duplicating their pair syndromes. Consequently A_r(T)=A_r(T')={u} for some outside row u.

A degree-three column containing r with external anticommutation set {u} has one of two forms. Form A has three different labels and support {r,s,u}, with s in U. Form B has its unique different label at u, while the other two rows, one being r, carry the repeated label. To check completeness, a three-distinct-label column can have only one outside row because its external anticommutation set is a singleton. In a two-label column, the uniquely labeled row cannot be in U, because its singleton syndrome would be forbidden. It must therefore be the sole external antineighbor u.

The two columns through r cannot both have Form A, because each would have syndrome {r,u}. They cannot both have Form B, because each would have syndrome {u}. Hence each r belongs to exactly one column of each form. Now take a Form A column on {r,s,u}. For both r and s its external anticommutation set is {u}. Their respective Form B columns must therefore both supply singleton syndrome {u}. Syndrome uniqueness forces these to be the same column. That Form B column has support {r,s,u}, with r and s carrying its repeated label, so it supplies syndrome {r,s}. The original Form A column also supplies {r,s}. This contradiction proves the closed-quartet obstruction.

For each matching edge {a,b}, its own degree-two column contributes one anticommutation between a and b. Other matching columns contribute none, and degree-three columns cannot contain both endpoints. Commutation thus requires an odd number of higher-degree columns at which both endpoints occur with different labels. Say that such a column serves the edge.

In profile (i), the sole degree-four column must serve both matching edges. Its support is therefore exactly their four endpoints U. Each row of U has weight four, with one incidence in its matching column and one in this degree-four column, leaving exactly two degree-three incidences. The closed-quartet obstruction excludes this profile.

In profile (ii), the sole degree-five column would have to serve all three disjoint matching edges. That requires at least six rows in its support, contradicting its degree five.

Finally consider profile (iii). Each matching edge must be served by exactly one of the two degree-four columns: the required number is odd and there are only two candidates. A degree-four column can serve at most two disjoint matching edges. Therefore, after naming the columns H_1,H_2 appropriately, H_1 serves two edges and H_2 serves the third. The support U of H_1 is exactly the four endpoints of its two edges. H_2 contains the two endpoints of the third edge, which are outside U, so its intersection with U has size at most two. If that intersection were empty, each row of U would again have precisely two degree-three incidences, and the closed-quartet obstruction would apply. The intersection therefore has size one or two.

It remains to show that a two-row intersection cannot be a matching edge. More generally, a degree-four column whose support consists of two complete matching edges must have different labels on each edge. If an edge had equal label P on both endpoints, the syndrome of P at that column would be a nonempty subset of the other matching edge: nonempty because the column uses at least two labels. Each such subset is already a syndrome of that other edge's degree-two column, a contradiction. Thus, if H_2 intersected U in a complete matching edge, it would serve that edge as well as the third edge. The former edge would then be served by both H_1 and H_2, contradicting the odd-parity requirement. This proves every stated conclusion.

PART IV. Exact exclusion of the final two templates.

We give an exact finite-enumeration certificate, together with the justification of its search space and every pruning rule. The supplied root computation reports exclusion of these templates. The self-contained checker below reconstructs that calculation; its successful execution validates the computational step. It was reproduced twice by the native template verifier; the present full-proof audit must rerun it independently.

For a column with row support U, partition U into the sets G_0,G_1,G_2 carrying its three possible nonidentity Pauli labels. Its three single-qubit error syndromes are U\G_0, U\G_1, U\G_2, regarded as eight-bit masks. The anticommutation contribution of that column consists of precisely the unordered row pairs with different nonidentity labels. Overall row commutation means that the symmetric difference of all these contributions is empty.

The absence of centralizer vectors of weight one or two is equivalent to the 33 single-qubit error syndromes being nonzero and pairwise distinct. Indeed, a zero syndrome gives a commuting weight-one error, and equal syndromes at different qubits give a commuting weight-two error. Equal syndromes at the same qubit also give a commuting nonidentity weight-one error. Conversely any weight-two error is the sum of two single-qubit errors on distinct qubits.

Every column therefore has at least two label classes. Independent permutations of the three Pauli labels at each qubit preserve both its syndrome set and its anticommutation contribution. Restricted-growth label strings enumerate every partition into two or three classes exactly once. In particular each degree-two column can be fixed to labels 0,1. These three columns contribute syndromes {0},{1},{0,1}, and similarly for the other two matching pairs, and contribute odd commutation parity on the three matching edges.

No admissible degree-three column can contain both endpoints of a matching edge. For three distinct labels, the edge itself is a syndrome. For two labels, if the endpoints have equal labels their pair is a syndrome, and otherwise one endpoint singleton is a syndrome. Each repeats a fixed syndrome. A degree-three column with three different labels is otherwise admissible relative to the fixed columns: there are C(8,3)-3·6=38 such supports. A degree-three column with two labels has a singleton syndrome at its uniquely labelled row. That row must be 6 or 7. For either choice, its other two rows can be any pair of the remaining seven except the three matching edges, giving 2·(C(7,2)-3)=36 possibilities. Thus exactly 74 canonical degree-three columns survive the fixed-syndrome test. The checker also verifies this equality directly.

For each template, the checker enumerates every canonical labelling of both degree-four columns and rejects repeated syndromes. It also rejects any residual odd parity on a matching edge, because no remaining degree-three column contains such an edge. It computes the remaining incidence demand of row i as 4 minus its incidences in the five fixed columns. These demands sum to 18, so precisely six degree-three columns remain to be chosen.

The recursive search maintains these row demands, the accumulated syndrome set, and the outstanding commutation parity. A candidate column is eligible only if all its incident rows still have positive demand and none of its syndromes has already occurred. A row requiring more columns than the number of eligible columns touching it cannot be completed: a column type cannot be repeated because that would repeat syndromes. Likewise an outstanding parity edge absent from the union of eligible anticommutation contributions cannot be corrected. These are necessary-condition prunings only.

When demands remain, the search chooses a row of positive demand and branches over every eligible column incident to it. Every completion must contain at least one such column; after removing that column, its other columns give a completion of the corresponding recursive state. Thus induction on total remaining demand proves exhaustive coverage. Memoization merges only identical states and introduces no pruning assumptions. Zero demands are accepted exactly when parity is zero.

The certificate asserts that each template has 36 admissible higher-column labellings and that none has a completion. The counts and the decisive zero-completion equalities are printed explicitly after their assertions. These finite checks, with the exhaustive-coverage argument above, exclude precisely the two templates in the claim. No earlier structural or degeneracy reduction is used.

PART V. Assembly back to the original quantifiers.

Suppose eight independent commuting rows of weight at most four define an eleven-qubit distance-at-least-three stabilizer code with k=3; degeneracy is allowed at this point. Part I proves that its stabilizer has no nonzero vector of weight one or two. Distance therefore implies that its entire centralizer has no such vector, exactly the hypothesis in Parts II–IV. Part I also proves all rows have weight four in the chosen basis, the degree-two graph is a matching, and the only profiles are (2,1),(3,2),(4,3). Part II excludes (4,3) because 32<33. Part III excludes (2,1).

For (3,2), t=2 partitions the excess higher-column degree as either one degree-five column or two degree-four columns. The degree-five option is excluded in Part III. With two degree-four columns there are three degree-two and six degree-three columns. Name the matching edges01,23,45 so the first higher column serves01 and23 and has support0123. The second contains45 and intersects0123 in one or two rows. With a one-row intersection, swap endpoints and the first two edges to name the common row0; its fourth row must be one of the two unmatched rows6,7, so name it6. Its support is0456. With a two-row intersection, Part III shows the two rows cannot be a matching edge; relabel endpoints to name them0,2. Its support is0245. These are precisely the two templates exhausted in Part IV. Thus no allowable case survives, contradicting existence.

Only the published generator-weight lower bound in Theorem4 of Wei et al., arXiv:2601.19848v2, is used externally in Part I to exclude deleting a one- or two-qubit stabilizer-state factor: at(n,k)=(10,3) it gives ceil(114/28)=5 and at(9,3) ceil(103/24)=5. All other incidence and parity arguments are stated here. The exact checker establishes the final finite combinatorial step; it does not by itself certify the preceding analytic reduction. The already-published weight-five construction recorded in the QIQC statement then gives W_opt(11,3,3)=5, though no new construction is claimed.

