LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The product basis is orthonormal and tensor dimensions multiply

Statement

Let $H_{0}, \dots, H_{n - 1}$ be a finite ordered list of finite-dimensional complex inner-product spaces with specified ordered orthonormal bases of sizes $d_{0}, \dots, d_{n - 1}$ , allowing $d_{k} = 0$ . The elementary tensors of the factor bases, ordered lexicographically, form an orthonormal basis of the tensor register, and

$\dim (H_{0} \otimes \dots \otimes H_{n - 1}) = \prod_{k < n} d_{k} .$

For the empty list the product is $1$ and the tensor space is $C$ ; if any $d_{k} = 0$ , the tensor space is zero and the product basis is empty.

Facts & Assumptions

Given: A finite ordered list of finite-dimensional complex inner-product spaces and their specified ordered orthonormal bases.

[F1]

The tensor register's pairing is the product of the factor pairings and its ordered basis is the lexicographic product basis (Ordered tensor registers, their inner product, and wire permutations).

[F2]

For two free modules, elementary tensors of basis vectors form a basis, including an empty basis in a factor (The elementary tensors of two bases form the product basis of the tensor product).

[F3]

The dimension of a tensor product of two finite-dimensional free modules is the product of their dimensions, including dimension zero ( $R^{m} \otimes_{R} R^{n} ≅ R^{m n}$ with the product basis, and $\dim_{F} (V \otimes_{F} W) = \dim_{F} V \dim_{F} W$ ).

[F4]

Finite iterated tensor products are identified across parenthesizations by the canonical maps preserving elementary tensors, and the empty product is the scalar module (Finite iterated tensor products represent multilinear maps independently of parenthesization).

Proof

1.1givenF1algebra

If $n = 0$ , the declared tensor space is $C$ , its product basis is ${1}$ , and its dimension is $1$ , the empty product. If $n = 1$ , the statement is exactly the supplied orthonormal basis of that factor.

1.2givenF1F2F4algebra

For $n \geq 2$ , repeated application of [F2] and the parenthesization identifications in [F4] shows that the elementary tensors of the factor bases form a basis indexed by the Cartesian product of their index sets. By [F1], the inner product of two such basis tensors is the product of their factor inner products; it is $1$ for equal tuples and $0$ for unequal tuples. Therefore this basis is orthonormal.

2.1step 1.1step 1.2F2F3algebra∎

Repeated application of [F3] gives the product of the factor dimensions. If any $d_{k} = 0$ , [F2] makes the product basis empty and both sides are zero; the empty and one-factor cases were handled in step 1.1. Thus the formula holds in every finite case.

Depends on

Used by

Five-local history-state adiabatic path Algorithm
Reversible embedding of a Boolean circuit Algorithm
Binary phase-query convention Definition
Boolean oracle query model for search lower bounds Definition
CNOT and Control-Target Order Definition
Coherent XOR function oracle Definition
CSS code from nested binary codes Definition
Normalized output law for finite quantum query branches Definition
Symmetric and antisymmetric swap-test measurement Definition
Toffoli and Fredkin gates Definition
Toric star and plaquette checks Definition
A reversible AND oracle with clean work Example
A three-stage precision budget Example
Promise, query and cost-model examples Example
QFT circuit order and truncation examples Example
Quantum repetition subspace projector and a distance-one boundary error Example
Qubit dephasing channel in Kraus and Choi forms Example
Register endianness and a two-qubit partial trace Example
A positive Choi matrix yields finite Kraus operators Lemma
A SWAP Is Three CNOTs Lemma
Clean and dirty ancilla identities extend to references and mixed states Lemma
Complete positivity implies Choi positivity Lemma
Computational-basis instruments and conditional states on two registers Lemma
Discarding the hiding register yields a coset-state mixture Lemma
Every finite extension is obtained from a purification by a channel Lemma
Finite ancilla amplification is a well-defined block map Lemma
Fresh-target NOT and OR gadgets Lemma
Operator Norm Is Unchanged by Spectator Registers Lemma
Operator support weight is invariant under wire permutation Lemma
Partial trace contracts trace distance Lemma
Preserving the logical marginal forces a gauge-only channel Lemma
Product operator matrix units form a basis Lemma
Tensor products of operator matrix units form a basis Lemma
Toffoli computes into a clean target and uncomputes it Lemma
Trace distance is unchanged by a fixed tensor factor Lemma
X, CNOT, Toffoli, and Fredkin are involutions Lemma
A finite Kraus sum is completely positive Theorem
A total Boolean XOR oracle is unitary and involutive Theorem
An isometry followed by partial trace is a channel Theorem
Big-endian bit-string indices are bijective Theorem

…and 28 more results.

Dependency tree · two levels

19 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

John Watrous, The Theory of Quantum Information (standard reference, not scraped)
Ronald de Wolf, Quantum Computing: Lecture Notes, Appendix A.6 (standard reference, not scraped)