PropositionStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-09-07

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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.

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Chain rule for smooth half-space maps

Statement

If $f : U \to V$ and $g : V \to W$ are smooth maps between relatively open half-space sets, then $g \circ f$ is smooth and $D (g \circ f)_{p} = D g_{f (p)} \circ D f_{p}$ .

Facts & Assumptions

Given: Relatively open half-space sets $U, V, W$ , smooth maps $f : U \to V$ and $g : V \to W$ , and a point $p \in U$ .

[L1]

Half-space smoothness supplies smooth Euclidean extensions near every point (Smooth functions on relatively open half-space sets).

[L2]

The derivatives of two Euclidean extensions agreeing on a relatively open half-space set agree on that set (Half-space extensions agreeing on a relatively open set have the same derivatives there).

[L3]

Total derivatives satisfy the Euclidean chain rule (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

Proof

technique · direct

1.1givenL1choose

By [L1], choose Euclidean extensions near $p$ and $f (p)$ and shrink the first neighbourhood so that its image lies in the domain of the second extension. Their ordinary composite then extends $g \circ f$ near $p$ .

2.1L2L3step 1.1∎

Applying [L3] to the extensions from step 1.1 gives the displayed formula, and [L2] makes the resulting derivatives independent of both extension choices. Hence $g \circ f$ is smooth and the formula is intrinsic.

Chain rule for smooth half-space maps

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources