Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28
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.

The chain rule applied to x↦(x2+1)5 and to x↦((3x−1)2+2)3, with the Carathéodory factor written out in closed form in the first case

Example

Numerals denote canonical naturals of R (The canonical natural ι(n)=n⋅1F of a field) and powers are those of Integer powers am.

Claim 1. Let F:R→R be F(x)=(x2+1)5. Then F is differentiable at every c∈R (The derivative f′(c)=lim⁡x→cf(x)−f(c)x−c of f:A→R at a point c∈A that is a limit point of A, and differentiability on a set) and

F′(c)  =  ι(10) c (c2+1)4.

Claim 2. For b∈R the Carathéodory factor (Carathéodory's characterisation: f is differentiable at c if and only if there is φ:A→R, continuous at c, with f(x)−f(c)=φ(x)(x−c) for every x∈A, and then φ is unique and φ(c)=f′(c)) of y↦y5 at b is the polynomial function

φ(y)  =  ∑k<5bk y 4−k,

which satisfies y5−b5=φ(y)(y−b) for every y∈R, is continuous at b, and has φ(b)=ι(5)b4.

Claim 3. Let G:R→R be G(x)=((ι(3)x−1)2+2)3. Then G is differentiable at every c∈R and

G′(c)  =  ι(18) (ι(3)c−1) ((ι(3)c−1)2+2)2.

Claim 2 is included because it makes the mechanism of the chain rule visible: the factor that the proof of The chain rule, in one line from Carathéodory: if g is differentiable at c and f is differentiable at g(c), then f∘g is differentiable at c with (f∘g)′(c)=f′(g(c)) g′(c) takes from Carathéodory's characterisation: f is differentiable at c if and only if there is φ:A→R, continuous at c, with f(x)−f(c)=φ(x)(x−c) for every x∈A, and then φ is unique and φ(c)=f′(c) is, for a power, an explicit polynomial, and no auxiliary case distinction is hidden inside it.

Facts & Assumptions

Given: The functions F, φ and G of the statement, and an arbitrary real c.

[L1]

Chain rule (The chain rule, in one line from Carathéodory: if g is differentiable at c and f is differentiable at g(c), then f∘g is differentiable at c with (f∘g)′(c)=f′(g(c)) g′(c)): with g:A→R, g[A]⊆B, f:B→R, c∈A a limit point of A at which g is differentiable, and b:=g(c)∈B a limit point of B at which f is differentiable, the composite f∘g is differentiable at c with (f∘g)′(c)=f′(g(c))g′(c).

[L4]

Carathéodory's characterisation (Carathéodory's characterisation: f is differentiable at c if and only if there is φ:A→R, continuous at c, with f(x)−f(c)=φ(x)(x−c) for every x∈A, and then φ is unique and φ(c)=f′(c)): f is differentiable at a limit point p of its domain if and only if some η continuous at p satisfies f(y)−f(p)=η(y)(y−p) throughout, and then η(p)=f′(p); the factor is unique.

[L5]

Factorisation of a difference of powers (Factorisation of bn−an, and the resulting Lipschitz estimate): for reals s,t and a natural n≥1, tn−sn=(t−s)∑k<nskt n−1−k (Finite sums and finite products, by recursion).

[L6]

Finite sums (Laws of finite sums and finite products, claim 2): ∑k<nλ=ι(n)λ for a constant λ; and powers combine as bkb n−1−k=b n−1 for k≤n−1 (Laws of integer exponents).

[L8]

Canonical naturals (The canonical natural ι(n)=n⋅1F of a field, Canonical naturals are positive and strictly increasing): ι(mn)=ι(m)ι(n) for naturals m,n≥1, so ι(5)ι(2)=ι(10), ι(2)ι(3)=ι(6) and ι(3)ι(6)=ι(18); and ι(1)=1.

Verification

technique · direct
1.1

Put g(x):=x2+1 and f(y):=y5, both on R. By [L2] and [L3] the function g is differentiable at every real c with g′(c)=ι(2)c, and f is differentiable at every real b with f′(b)=ι(5)b4.

L2L3
1.2

Put u(x):=ι(3)x−1, v(y):=y2+2 and w(z):=z3, all on R. By [L2] and [L3], u′(c)=ι(3), v′(b)=ι(2)b and w′(a)=ι(3)a2 at every real argument.

L2L3
2.1

Claim 2. Fix b∈R and put φ(y):=∑k<5bky 4−k for y∈R. Applying [L5] with n:=5, s:=b and t:=y gives y5−b5=(y−b)∑k<5bky 4−k=φ(y)(y−b) for every real y. As a finite sum of scalar multiples of powers of y, the function φ is a polynomial function and so is continuous at b by [L7]. Finally φ(b)=∑k<5bkb 4−k=∑k<5b4=ι(5)b4 by [L6]. So φ is the factor of [L4] for f at b, and [L4] returns f′(b)=φ(b)=ι(5)b4, in agreement with step 1.1.

L4L5L6L7
2.2

Claim 1. By [L9] every real is a limit point of R, and g maps R into R, so [L1] applies to F=f∘g at any c: F is differentiable at c with F′(c)=f′(g(c)) g′(c)=ι(5)(c2+1)4⋅ι(2)c=ι(10) c (c2+1)4, the last step by [L8].

step 1.1L1L8L9
2.3

Claim 3. By [L1] and [L9], applied first to v∘u and then to w∘(v∘u), the function G=w∘(v∘u) is differentiable at every real c, with (v∘u)′(c)=v′(u(c))u′(c)=ι(2)(ι(3)c−1)⋅ι(3)=ι(6)(ι(3)c−1) and then G′(c)=w′((v∘u)(c)) (v∘u)′(c)=ι(3)((ι(3)c−1)2+2)2⋅ι(6)(ι(3)c−1)=ι(18)(ι(3)c−1)((ι(3)c−1)2+2)2, the collapsing of the numerals by [L8].

step 1.2L1L8L9
3.1

The three claims are verified: claim 1 by step 2.2, claim 2 by step 2.1 and claim 3 by step 2.3.

step 2.1step 2.2step 2.3∎

Remarks

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

54 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