Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-01
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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.

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Cauchy's mean-value theorem in quotient form when the denominator derivative is nonzero

Statement

Let a<ba<b. If f,gf,g are continuous on [a,b][a,b], differentiable on (a,b)(a,b), and g(x)0g'(x)\ne0 throughout (a,b)(a,b), then g(b)g(a)g(b)\ne g(a) and there is c(a,b)c\in(a,b) such that f(b)f(a)g(b)g(a)=f(c)g(c).\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(c)}{g'(c)}.

Facts & Assumptions

Proof

technique · direct
1.1

If g(a)=g(b)g(a)=g(b), Rolle gives d(a,b)d\in(a,b) with g(d)=0g'(d)=0, contrary to the hypothesis. Hence g(b)g(a)0g(b)-g(a)\ne0.

L2given
1.2

Cauchy's theorem supplies c(a,b)c\in(a,b) with the cross-product identity in [L1].

L1
2.1

Divide that identity by the two nonzero factors g(b)g(a)g(b)-g(a) and g(c)g'(c) to obtain the quotient formula.

step 1.1step 1.2algebra

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 56 results over 18 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources