ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)audited 2026-10-02

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.

Four shared values do not force equality

Example

Assume Countable Choice. The distinct nonconstant entire functions $F (z) = e^{z}$ and $G (z) = e^{- z}$ share exactly the four distinct sphere values $0, \infty, 1, - 1$ ignoring multiplicity: both omit $0$ and $\infty$ , and their preimage sets of $1$ and of $- 1$ agree, ${z : e^{z} = 1} = {z : e^{- z} = 1} = 2 π i Z, {z : e^{z} = - 1} = {z : e^{- z} = - 1} = i π (2 Z + 1) .$ No other sphere value is shared. Since $F \neq G$ , the five distinct shared values required by the five-value uniqueness theorem Nevanlinna five-value uniqueness theorem cannot be reduced to four.

Facts & Assumptions

Given: The functions $F (z) = e^{z}$ and $G (z) = e^{- z}$ ; Countable Choice is assumed as in the statement, and the computation below is choice-free.

[F1]

Five-value theorem: if two nonconstant meromorphic functions on $C$ share five distinct sphere values ignoring multiplicity, that is, the preimage sets of each value agree, then they are identically equal (Nevanlinna five-value uniqueness theorem).

[F2]

Fibres of the exponential: $\ker (\exp) = 2 π i Z$ , and $e^{z} = e^{w}$ exactly when $z - w \in 2 π i Z$ ; moreover $e^{i π} = - 1$ ( $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ).

[F3]

The complex exponential is entire and $∣ e^{z} ∣ = e^{Re z} > 0$ (The complex exponential is entire and its complex derivative is itself, $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ).

[F4]

The complex exponential maps $C$ onto $C ∖ {0}$ (The complex exponential maps $C$ onto $C ∖ {0}$ ).

[F5]

Sharing ignoring multiplicity concerns the sets of points only: $\overset{ˉ}{n}$ counts each preimage once and multiplicities are discarded (Truncated value and ramification counts), exactly the convention in the statement of [F1].

Verification

technique · compute the four preimage sets that the two exponentials have in common, verify that no further value is shared, and observe that the functions are distinct

1.1F3F5

(Omitted values) By [F3] both $F$ and $G$ are entire and never vanish; hence the preimage set of $0$ is empty for both, and the preimage set of $\infty$ , that is, the pole set, is empty for both as well. Thus $0$ and $\infty$ are shared values in the sense of [F5] with empty preimage sets.

1.2F2algebra

(The value $1$ ) By [F2], $e^{z} = 1$ if and only if $z \in 2 π i Z$ ; and $e^{- z} = 1$ if and only if $- z \in 2 π i Z$ , which is the same set. Hence the two $1$ -point sets agree and equal $2 π i Z$ .

1.3F2algebra

(The value $- 1$ ) By [F2] and $e^{i π} = - 1$ , $e^{z} = - 1$ if and only if $z - i π \in 2 π i Z$ , that is, $z \in i π (2 Z + 1)$ ; and $e^{- z} = - 1$ if and only if $- z \in i π (2 Z + 1)$ , that is, $z \in i π (2 Z + 1)$ as well, because $- i π (2 k + 1) = i π (2 (- k - 1) + 1)$ . Hence the two $(- 1)$ -point sets agree.

1.4F2algebra

(Distinctness) If $F = G$ then $e = e^{- 1}$ , hence $e^{2} = 1 = e^{0}$ , so by [F2] the number $2$ would lie in $2 π i Z$ ; but $2$ is real and nonzero while every element of $2 π i Z$ is purely imaginary. Thus $F \neq G$ .

1.5F2F4algebra

(No other shared value) Let $a \in \hat{C} ∖ {0, \infty, 1, - 1}$ be a shared value in the sense of [F5]. Since $a \neq 0, \infty$ we may fix $b$ with $e^{b} = a$ by [F4]; the preimage sets of $a$ under $F$ and $G$ are $b + 2 π i Z$ and $- b + 2 π i Z$ by [F2], and agreement forces $b \in - b + 2 π i Z$ , hence $2 b \in 2 π i Z$ ; then $a^{2} = e^{2 b} = 1$ by [F2], so $a = \pm 1$ , contrary to the choice of $a$ . Hence the shared sphere values of $F$ and $G$ are exactly $0, \infty, 1, - 1$ , four in number.

2.1F1step 1.1step 1.2step 1.3step 1.4step 1.5∎

(Sharpness) Steps 1.1-1.5 exhibit two distinct nonconstant meromorphic functions sharing exactly four distinct sphere values ignoring multiplicity, while [F1] guarantees equality as soon as five distinct values are shared. Therefore the number five in the five-value theorem is optimal.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

42 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

I. Laine, Complex Analysis III lecture notes (standard reference, not scraped)