LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck pass

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.

A finite straight-line C1 program defines a C1 map on each fixed strict branch

Statement

For a finite straight-line program whose primitives are $C^{1}$ , the recursively defined admissible input domain is open on each fixed strict branch. The scalar output $f$ is $C^{1}$ there, and its total derivative is obtained by the ordinary chain rule through the primitive maps.

Facts & Assumptions

Given: The finite topologically ordered graph and its $C^{1}$ primitive maps.

[F1]

Finite straight-line computational programs and real-arithmetic tapes: each node reads only earlier parent slots, each primitive is $C^{1}$ on an open domain, and the admissible input set requires every parent tuple to lie in that domain.

[F2]

Continuously differentiable maps, local inverses, and local diffeomorphisms: a map is $C^{1}$ when it is totally differentiable at every point and its derivative matrix varies continuously.

[F3]

Total differentiability gives a local $O (∥ h ∥_{2})$ increment bound and therefore continuity: total differentiability at a point implies continuity there.

[F4]

The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ : if $g$ is totally differentiable at $f (a)$ and $f$ is totally differentiable at $a$ , then $g \circ f$ is totally differentiable at $a$ , with derivative $D g (f (a)) D f (a)$ .

Proof

technique · induction on the operation order

1.1F1F3givenconstruct

Begin with the open input domain $R^{n}$ , whose coordinate maps are continuous and $C^{1}$ . Suppose the values through node $j - 1$ are defined and continuous on an open set $Ω_{j - 1}$ . The parent tuple $g_{j} (x) = (z_{p_{j, 1}} (x), \dots, z_{p_{j, r_{j}}} (x))$ is continuous there. Thus $Ω_{j} : = {x \in Ω_{j - 1} : g_{j} (x) \in U_{j}}$ is open in $Ω_{j - 1}$ , hence open in $R^{n}$ , because $U_{j}$ is open. This proves by finite induction that the full admissible domain is open.

2.1F1F2F4step 1.1algebra

For each operation $j$ , assume its parent-node maps are $C^{1}$ on $Ω_{j}$ . Their tuple $g_{j}$ is $C^{1}$ componentwise. Since $z_{j} = ϕ_{j} \circ g_{j}$ , [F4] gives $D z_{j} (x) = D ϕ_{j} (g_{j} (x)) D g_{j} (x)$ . Both matrix factors vary continuously because $ϕ_{j}$ and the parent maps are $C^{1}$ ; hence $D z_{j}$ is continuous and $z_{j}$ is $C^{1}$ by [F2]. Thus every operation value is $C^{1}$ ; the scalar output is $C^{1}$ whether it is one of these operation values or an input coordinate projection.

3.1F1step 2.1∎

On a strict piecewise branch, the branch label is fixed on the open branch region and its primitive formula is $C^{1}$ ; steps 1.1–2.1 therefore apply unchanged on that region. At a boundary the definition in [F1] returns NONDIFFERENTIABLE, so no derivative assertion is made there.

Depends on

Used by

Dependency tree · two levels

10 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

MIT OpenCourseWare 18.S096, Matrix Calculus for Machine Learning and Beyond, complete lecture notes, Chapter 8 §§8.3–8.4.1 (standard reference, not scraped)