CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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.

Two potentials of the same field differ by a constant on each piecewise-C1 path component

Statement

Let $U \subseteq R^{n}$ be open. If $ϕ, ψ : U \to R$ are $C^{1}$ and satisfy $\nabla ϕ = \nabla ψ$ , then $ϕ - ψ$ is constant on every piecewise- $C^{1}$ path component of $U$ .

Facts & Assumptions

Given: The open set and potentials in the Statement.

[L1]

Call $x \sim y$ when some piecewise- $C^{1}$ path in $U$ joins $x$ to $y$ . Constant paths, reversal and concatenation make $\sim$ reflexive, symmetric and transitive, so it is an equivalence relation on $U$ ; its classes are the piecewise- $C^{1}$ path components of $U$ , and a nonempty $U$ is itself piecewise- $C^{1}$ path-connected exactly when it has just one class (Piecewise-C1 path-connected domains, potential functions, conservative fields, and path independence, Reversal, concatenation, closed paths, and oriented piecewise-C1 reparametrizations, Equivalence relation, equivalence class, and the quotient set $A / \sim$ ).

[L2]

The gradient theorem evaluates the line integral of a $C^{1}$ gradient as its endpoint increment (The gradient theorem: the line integral of a gradient is the endpoint increment).

Proof

technique · direct

1.1

Let $x, y$ lie in one piecewise- $C^{1}$ path component, and choose a path $γ$ from $x$ to $y$ as in [L1].

givenL1

2.1

Since $\nabla (ϕ - ψ) = 0$ , [L2] gives $0 = \int_{γ} 0 \cdot d r = (ϕ - ψ) (y) - (ϕ - ψ) (x) .$

givenstep 1.1L2algebra

3.1

Thus $(ϕ - ψ) (y) = (ϕ - ψ) (x)$ . Since $x, y$ were arbitrary within the component, the difference is constant there.

step 2.1

4.1

No equality of the constants on distinct components is asserted, because [L1] supplies no path joining such points.

L1step 3.1∎

Depends on

Used by

Dependency tree · two levels

17 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

J. Lebl, Basic Analysis II, section 9.3 (standard reference, not scraped)