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Signatures (1,0,… )(1,0,\dots)(1,0,…) and (1,2,0,… )(1,2,0,\dots)(1,2,0,…) give K=NK=NK=N and K=N2K=N^2K=N2

Proved
HardyFiveAxioms.signature_classical_quantum

by Lucas · Oct 2, 2026 · Mathlib 0df444a (Lean v4.33.1)

combinatoricsmathematical-physicsquantum-foundations

For every N∈NN\in\mathbb NN∈N, with Kx(N)=∑k(Nk)xkK_x(N)=\sum_k\binom Nkx_kKx​(N)=∑k​(kN​)xk​ as in hardy2001_signature:

K(1)(N)=N,K(1,2)(N)=N+2(N2)=N2.K_{(1)}(N)=N,\qquad K_{(1,2)}(N)=N+2\binom N2=N^2 .K(1)​(N)=N,K(1,2)​(N)=N+2(2N​)=N2.

These are the classical signature xClassical=(1,0,0,… )x_{\rm Classical}=(1,0,0,\dots)xClassical​=(1,0,0,…) and the quantum signature xQuantum=(1,2,0,0,… )x_{\rm Quantum}=(1,2,0,0,\dots)xQuantum​=(1,2,0,0,…) of Section 5.

Preamble
import Mathlib
import Definitions.Def_hardy2001_signature
Formal statement
namespace HardyFiveAxioms

/-- Hardy 2001, Section 5 (after Eq. (28)): the classical signature `(1, 0, 0, …)` gives
`K = N` and the quantum signature `(1, 2, 0, 0, …)` gives `K = N²`. -/
theorem signature_classical_quantum (N : ℕ) :
    dofOfSignature [1] N = N ∧ dofOfSignature [1, 2] N = N ^ 2 := by sorry

end HardyFiveAxioms
Source
L. Hardy, *Quantum Theory From Five Reasonable Axioms*, arXiv:quant-ph/0101012v4 (2001), https://arxiv.org/abs/quant-ph/0101012, p. 8, Section 5, Eq. (28) and the following sentence
Read-back

What the Lean code literally says, in plain math · Aristotle (Harmonic)

NON-BLIND READ-BACK — NOT INDEPENDENT TESTIMONY. This read-back was written by the same agent (Aristotle, by Harmonic) that drafted the Lean statement below, with full knowledge of the source paper and of the intended meaning. It was not produced by a blind, independent auditor, so it may share the drafter's blind spots. Reviewers must not treat it as independent confirmation of faithfulness; compare the Lean code against the source directly.

For every natural number NNN: (N1)⋅1=N\binom N1\cdot1=N(1N​)⋅1=N, and (N1)⋅1+(N2)⋅2=N2\binom N1\cdot1+\binom N2\cdot2=N^2(1N​)⋅1+(2N​)⋅2=N2. Equivalently, with the signature function Kx(N)=∑k=1L(Nk)xkK_x(N)=\sum_{k=1}^{L}\binom Nkx_kKx​(N)=∑k=1L​(kN​)xk​ of hardy2001_signature, K[1](N)=NK_{[1]}(N)=NK[1]​(N)=N and K[1,2](N)=N2K_{[1,2]}(N)=N^2K[1,2]​(N)=N2, as natural numbers.

Human review
  • Endorsed by Shuze Chen · Oct 8, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Lucas · Oct 8, 2026

    Confirmed by the mission captain (proposal self-audit).

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