ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-05 (gpt-5.6-sol-codex-subscription)‡ rests on unproved material (inherited)

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.

‡ Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Cohen's first model: an infinite Dedekind-finite set of reals, ‡ Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and ‡ The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Assuming the Axiom of Choice: $\aleph_0^{\aleph_0} = 2^{\aleph_0}$ and $\lvert \mathbb{R}^{\mathbb{R}} \rvert = 2^{2^{\aleph_0}}$ , computed from the exponent laws and Hessenberg

Example

Assume the Axiom of Choice (The Axiom of Choice). Write $\mathfrak{c} = 2^{\aleph_0}$ , and let $\mathbb{R}^{\mathbb{R}}$ denote the set ${}^{\mathbb{R}}\mathbb{R}$ of all functions $\mathbb{R} \to \mathbb{R}$ , continuity playing no role. Then

$\aleph_0^{\aleph_0} \;=\; 2^{\aleph_0} \;=\; \mathfrak{c}, \qquad \lvert \mathbb{R}^{\mathbb{R}} \rvert \;=\; 2^{2^{\aleph_0}} .$

Both computations are squeezes: an upper and a lower bound that meet, with Hessenberg: $\kappa \otimes \kappa = \kappa$ for every infinite cardinal $\kappa$ , proved in ZF from the canonical well-order of $\kappa \times \kappa$ closing the gap through $\kappa \otimes \kappa = \kappa$ and the second exponent law (Commutativity, associativity, distributivity and monotonicity of $\oplus$ and $\otimes$ , the unit laws, the two exponent laws, and $\kappa \le \lambda$ if and only if $\kappa$ injects into $\lambda$ ) turning a repeated exponent into a product.

The second value is worth reading against the first. There are $\mathfrak{c}$ real numbers and $2^{\mathfrak{c}}$ functions between them, so the set of all real functions is strictly larger than the continuum (Assuming the Axiom of Choice, $2^{\kappa} = \lvert \mathcal{P}(\kappa) \rvert$ , and Cantor's theorem in cardinal form: $\kappa < 2^{\kappa}$ ), by exactly one application of the power operation.

Facts & Assumptions

Given: The Axiom of Choice, so that every exponential below is defined (Cardinal sum $\kappa \oplus \lambda$ , product $\kappa \otimes \lambda$ and exponentiation $\kappa^{\lambda}$ , and why they are written apart from the ordinal operations, The well-ordering theorem).

[L1]

$\kappa \le \lambda$ implies $\kappa^{\mu} \le \lambda^{\mu}$ ; $(\mu^{\nu})^{\rho} = \mu^{\nu \otimes \rho}$ ; and for cardinals $\kappa \le \lambda$ iff $\kappa \preceq \lambda$ (Commutativity, associativity, distributivity and monotonicity of $\oplus$ and $\otimes$ , the unit laws, the two exponent laws, and $\kappa \le \lambda$ if and only if $\kappa$ injects into $\lambda$ ).

[L2]

$\kappa \otimes \kappa = \kappa$ for every infinite cardinal $\kappa$ (Hessenberg: $\kappa \otimes \kappa = \kappa$ for every infinite cardinal $\kappa$ , proved in ZF from the canonical well-order of $\kappa \times \kappa$ ); and $\mu \otimes \nu = \mu$ for an infinite cardinal $\mu$ and a cardinal $\nu$ with $0 \ne \nu \le \mu$ (Absorption: for cardinals $\kappa, \lambda$ with $\kappa$ infinite and $\lambda \le \kappa$ , $\kappa \oplus \lambda = \kappa$ , and $\kappa \otimes \lambda = \kappa$ when $\lambda \ne 0$ ).

[L3]

$\kappa < 2^{\kappa}$ , and $2^{\kappa}$ is a cardinal (Assuming the Axiom of Choice, $2^{\kappa} = \lvert \mathcal{P}(\kappa) \rvert$ , and Cantor's theorem in cardinal form: $\kappa < 2^{\kappa}$ ).

[L4]

$\lvert \mathbb{R}\rvert = 2^{\aleph_0}$ ( $\mathbb{R} \approx \mathcal{P}(\mathbb{N})$ in ZF, by the Cantor set for one injection and by the cuts $\{q \in \mathbb{Q} : q < x\}$ for the other; so $\lvert \mathbb{R} \rvert = 2^{\aleph_0}$ under the Axiom of Choice).

[L5]

$\lvert {}^{B}A \rvert = \lvert A \rvert^{\lvert B \rvert}$ whenever the sets involved have cardinalities, because ${}^{B}A$ respects $\approx$ in both arguments (Cardinal sum $\kappa \oplus \lambda$ , product $\kappa \otimes \lambda$ and exponentiation $\kappa^{\lambda}$ , and why they are written apart from the ordinal operations, Disjoint union, cartesian product, function space and power set respect equinumerosity, and for ordinals $\alpha, \beta$ the sets $\alpha \sqcup \beta$ and $\alpha \times \beta$ carry explicit well-orders, so their cardinalities exist in ZF, A set equinumerous with some ordinal has a least such ordinal, that ordinal is a cardinal, and equinumerous sets get the same one; no choice principle is used, Equinumerous sets, $A \approx B$ and $A \preceq B$ ).

[L6]

$\aleph_0 = \omega$ is an infinite cardinal and $2 \in \omega$ is a cardinal with $2 \le \aleph_0$ (The successor cardinal $\kappa^{+}$ , the alephs $\aleph_\alpha$ , the beths $\beth_\alpha$ , successor and limit cardinals, and the identifications $\aleph_0 = \omega$ and $\aleph_1 = \omega_1$ , Every natural number and $\omega$ are cardinals, every infinite cardinal is a limit ordinal, and on the natural numbers the cardinal operations are the published finite counting operations, with $\lvert A \rvert$ in the finite sense equal to $\lvert A \rvert$ in the cardinal sense, Cardinal (initial ordinal) and cardinality).

[L7]

Ordinals satisfy trichotomy, $\alpha \subseteq \beta$ iff $\alpha \in \beta$ or $\alpha = \beta$ , and $\alpha \subseteq \beta \subseteq \alpha$ forces $\alpha = \beta$ (Trichotomy and well-ordering of the ordinals, Basic closure properties of ordinals).

Verification

technique · direct

1.1

By [L6] and [L3], $2 \le \aleph_0 \le 2^{\aleph_0}$ ; so [L1] gives $2^{\aleph_0} \le \aleph_0^{\aleph_0} \le (2^{\aleph_0})^{\aleph_0}$ , and $(2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0 \otimes \aleph_0} = 2^{\aleph_0}$ by [L1] and [L2]; therefore $\aleph_0^{\aleph_0} = 2^{\aleph_0}$ by [L7].

L1L2L3L6L7

1.2

Writing $\mathfrak{c} = 2^{\aleph_0}$ , [L4] and [L5] give $\lvert \mathbb{R}^{\mathbb{R}} \rvert = \lvert \mathbb{R} \rvert^{\lvert \mathbb{R} \rvert} = \mathfrak{c}^{\mathfrak{c}}$ .

L4L5

2.1

$\mathfrak{c}^{\mathfrak{c}} = (2^{\aleph_0})^{\mathfrak{c}} = 2^{\aleph_0 \otimes \mathfrak{c}} = 2^{\mathfrak{c}}$ : the middle equality is the second exponent law in [L1], and the last is absorption in [L2], applicable because $\mathfrak{c}$ is an infinite cardinal with $0 \ne \aleph_0 \le \mathfrak{c}$ by [L3] and [L6].

step 1.2L1L2L3L6

3.1

So $\aleph_0^{\aleph_0} = 2^{\aleph_0}$ and $\lvert \mathbb{R}^{\mathbb{R}} \rvert = 2^{\mathfrak{c}} = 2^{2^{\aleph_0}}$ .

step 1.1step 2.1∎

Remarks

Why the first computation is a collapse and not a coincidence. Any base between $2$ and $2^{\aleph_0}$ gives the same value when raised to $\aleph_0$ , because the chain of step 1.1 closes on both sides. That is the general phenomenon recorded in FALSE: $\kappa < \lambda$ implies $\kappa^{\mu} < \lambda^{\mu}$ : strict monotonicity in the base is false, and this is the smallest instance.

Where Hessenberg's theorem enters. Twice, both times as $\kappa \otimes \kappa = \kappa$ turning a repeated exponent into a single one: at $\aleph_0$ in step 1.1 and, through absorption, at $\mathfrak{c}$ in step 2.1. Without it neither exponent could be simplified and both computations would stall at an upper bound.

Continuity is irrelevant here, and that is worth saying. $\mathbb{R}^{\mathbb{R}}$ above is the set of all functions, with no regularity assumed. Counting the continuous ones is a different computation, needing tools this page and the pages it rests on do not provide, and no claim about it is made here.

Depends on

Used by

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Sources

Cardinal number — cardinal arithmetic (Wikipedia) (standard reference, not scraped)
Cardinality of the continuum (Wikipedia) (standard reference, not scraped)

Assuming the Axiom of Choice: ℵ0ℵ0=2ℵ0\aleph_0^{\aleph_0} = 2^{\aleph_0}ℵ0ℵ0​​=2ℵ0​ and ∣RR∣=22ℵ0\lvert \mathbb{R}^{\mathbb{R}} \rvert = 2^{2^{\aleph_0}}∣RR∣=22ℵ0​, computed from the exponent laws and Hessenberg