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Lemma 16 — the composition of two atomic transformations is atomic

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InfoDerivQT.atomicity_of_composition

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

operational-probabilistic-theoriesquantum-foundationsquantum-information

Let TTT satisfy the six principles and let A,B,CA,B,CA,B,C be systems. If C∈Transf(A,B)\mathcal C\in\mathrm{Transf}(A,B)C∈Transf(A,B) and D∈Transf(B,C)\mathcal D\in\mathrm{Transf}(B,C)D∈Transf(B,C) are atomic, then the sequential composition DC∈Transf(A,C)\mathcal D\mathcal C\in\mathrm{Transf}(A,C)DC∈Transf(A,C) is atomic.

This is Axiom 5′'′ of the paper; it is derived here from pure conditioning and purification and is used repeatedly afterwards.

Preamble
import Mathlib
import Definitions.Def_InfoDerivQT_principles
Formal statement
namespace InfoDerivQT
theorem atomicity_of_composition (T : OPT) (hT : T.SatisfiesPrinciples) (A B C : T.Sys)
    (D₁ : Vec (T.size A) →ₗ[ℝ] Vec (T.size B)) (D₂ : Vec (T.size B) →ₗ[ℝ] Vec (T.size C))
    (h₁ : T.IsAtomicT A B D₁) (h₂ : T.IsAtomicT B C D₂) :
    T.IsAtomicT A C (D₂.comp D₁) := by sorry
end InfoDerivQT
Source
G. Chiribella, G. M. D'Ariano, P. Perinotti, *Informational derivation of quantum theory*, Phys. Rev. A 84, 012311 (2011), https://doi.org/10.1103/PhysRevA.84.012311 (arXiv:1011.6451), p. 012311-12, Sec. IV D, Lemma 16 (Atomicity of composition)
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 statements, with full knowledge of the source paper and of the intended meaning. It is not a blind audit and must not be mistaken for independent testimony; an independent, blind read-back should replace it before the proposal is relied upon.

Fix TTT satisfying all six principles, systems A,B,CA,B,CA,B,C, and linear maps D1:RDA→RDB\mathcal D_1:\mathbb R^{D_A}\to\mathbb R^{D_B}D1​:RDA​→RDB​, D2:RDB→RDC\mathcal D_2:\mathbb R^{D_B}\to\mathbb R^{D_C}D2​:RDB​→RDC​. Assume both are atomic transformations: D1\mathcal D_1D1​ belongs to some test from AAA to BBB, and whenever there are a test {tx}\{t_x\}{tx​} from AAA to BBB and a finite subset X0X_0X0​ with D1=∑x∈X0tx\mathcal D_1=\sum_{x\in X_0}t_xD1​=∑x∈X0​​tx​, every txt_xtx​ with x∈X0x\in X_0x∈X0​ equals p D1p\,\mathcal D_1pD1​ for some p≥0p\ge0p≥0; similarly for D2\mathcal D_2D2​ (tests from BBB to CCC). The conclusion is that the composite map D2∘D1\mathcal D_2\circ\mathcal D_1D2​∘D1​ is an atomic transformation from AAA to CCC in the same sense (it belongs to some test from AAA to CCC, and all its refinements by tests from AAA to CCC are nonnegative multiples of it).

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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