LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)‡ 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.

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 $\lambda$

Statement

Let $\kappa$ , $\lambda$ , $\mu$ be cardinals (Cardinal (initial ordinal) and cardinality) and let $\oplus$ , $\otimes$ , $\kappa^{\lambda}$ be as in Cardinal sum $\kappa \oplus \lambda$ , product $\kappa \otimes \lambda$ and exponentiation $\kappa^{\lambda}$ , and why they are written apart from the ordinal operations. Clauses (a) to (e) are theorems of ZF; the clauses naming an exponential hold whenever those exponentials are defined, in particular under the Axiom of Choice (The Axiom of Choice).

(a) Comparison. $\kappa \le \lambda$ if and only if $\kappa \preceq \lambda$ , that is, if and only if there is an injection $\kappa \to \lambda$ (Equinumerous sets, $A \approx B$ and $A \preceq B$ ). More generally, if $A$ and $B$ are well-orderable and $A \preceq B$ then $\lvert A \rvert \le \lvert B \rvert$ .

(b) Commutativity and associativity. $\kappa \oplus \lambda = \lambda \oplus \kappa$ , $\kappa \otimes \lambda = \lambda \otimes \kappa$ , $(\kappa \oplus \lambda) \oplus \mu = \kappa \oplus (\lambda \oplus \mu)$ and $(\kappa \otimes \lambda) \otimes \mu = \kappa \otimes (\lambda \otimes \mu)$ .

(c) Distributivity. $\kappa \otimes (\lambda \oplus \mu) = (\kappa \otimes \lambda) \oplus (\kappa \otimes \mu)$ .

(d) Units. $\kappa \oplus 0 = \kappa$ , $\kappa \otimes 0 = 0$ , $\kappa \otimes 1 = \kappa$ , and $\kappa^{0} = 1$ , $\kappa^{1} = \kappa$ , $1^{\kappa} = 1$ , together with $0^{\kappa} = 0$ for $\kappa \ne 0$ . The four exponential unit laws need no choice principle, because the function sets they count are empty, a singleton, or a copy of $\kappa$ .

(e) Monotonicity. If $\kappa \le \lambda$ then $\kappa \oplus \mu \le \lambda \oplus \mu$ and $\kappa \otimes \mu \le \lambda \otimes \mu$ ; and $\kappa^{\mu} \le \lambda^{\mu}$ , and $\mu^{\kappa} \le \mu^{\lambda}$ provided $\mu \ne 0$ .

(f) The two exponent laws. $\kappa^{\lambda \oplus \mu} = \kappa^{\lambda} \otimes \kappa^{\mu}$ and $(\kappa^{\lambda})^{\mu} = \kappa^{\lambda \otimes \mu}$ .

Each clause is an equality or an inequality of cardinals, not merely of sizes: each side is an ordinal, and the claim is that the two ordinals are the same.

Facts & Assumptions

Given: Cardinals $\kappa, \lambda, \mu$ , in ZF; the Axiom of Choice is assumed only where an exponential is written and is not one of the four unit cases.

[L1]

For a well-orderable $X$ , $\lvert X \rvert$ is the least ordinal equinumerous with $X$ , it satisfies $X \approx \lvert X \rvert$ , it is a cardinal, equinumerous sets receive the same one, and $\lvert \alpha \rvert = \alpha$ exactly when $\alpha$ is a cardinal (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).

[L2]

$\sqcup$ , $\times$ and ${}^{B}A$ respect $\approx$ in both arguments (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, claim (a)).

[L3]

$\kappa \oplus \lambda = \lvert \kappa \sqcup \lambda \rvert$ , $\kappa \otimes \lambda = \lvert \kappa \times \lambda \rvert$ , $\kappa^{\lambda} = \lvert {}^{\lambda}\kappa \rvert$ , and these may be computed from any equinumerous representatives (Cardinal sum $\kappa \oplus \lambda$ , product $\kappa \otimes \lambda$ and exponentiation $\kappa^{\lambda}$ , and why they are written apart from the ordinal operations).

[L4]

If $A \preceq B$ and $B \preceq A$ then $A \approx B$ (The Schröder-Bernstein theorem).

[L5]

Ordinals are comparable, exactly one of $\alpha \in \beta$ , $\alpha = \beta$ , $\beta \in \alpha$ holds, $\alpha \subseteq \beta$ if and only if $\alpha \in \beta$ or $\alpha = \beta$ , and $\alpha \notin \alpha$ (Trichotomy and well-ordering of the ordinals, Basic closure properties of ordinals).

[L6]

A composition of bijections is a bijection, a map with a two-sided inverse is a bijection, and a subset inclusion is an injection (Injection, surjection, bijection, Equinumerous sets, $A \approx B$ and $A \preceq B$ ).

[L7]

An ordinal $\kappa$ is a cardinal when no $\alpha \in \kappa$ has $\alpha \approx \kappa$ (Cardinal (initial ordinal) and cardinality).

[L8]

Assuming the Axiom of Choice every set is well-orderable, so every exponential written below is defined (The Axiom of Choice, The well-ordering theorem).

Proof

technique · direct

1.1

First half of (a): if $\kappa \le \lambda$ then $\kappa \subseteq \lambda$ by [L5] and the inclusion is an injection, so $\kappa \preceq \lambda$ ; conversely, if $\kappa \preceq \lambda$ and $\lambda \in \kappa$ then $\lambda \subseteq \kappa$ gives $\lambda \preceq \kappa$ , so $\kappa \approx \lambda$ by [L4], contradicting [L7] since $\lambda \in \kappa$ ; trichotomy then leaves $\kappa \le \lambda$ .

L4L5L6L7

1.2

The maps $(0,\xi) \mapsto (1,\xi)$ , $(1,\eta) \mapsto (0,\eta)$ and $(\xi,\eta) \mapsto (\eta,\xi)$ are their own inverses up to relabelling, hence bijections $\kappa \sqcup \lambda \to \lambda \sqcup \kappa$ and $\kappa \times \lambda \to \lambda \times \kappa$ .

1.3

The maps $(0,(0,\xi)) \mapsto (0,\xi)$ , $(0,(1,\eta)) \mapsto (1,(0,\eta))$ , $(1,\zeta) \mapsto (1,(1,\zeta))$ and $((\xi,\eta),\zeta) \mapsto (\xi,(\eta,\zeta))$ have evident two-sided inverses, hence are bijections $(\kappa \sqcup \lambda) \sqcup \mu \to \kappa \sqcup (\lambda \sqcup \mu)$ and $(\kappa \times \lambda) \times \mu \to \kappa \times (\lambda \times \mu)$ .

1.4

The map $(\xi,(0,\eta)) \mapsto (0,(\xi,\eta))$ , $(\xi,(1,\zeta)) \mapsto (1,(\xi,\zeta))$ is a bijection $\kappa \times (\lambda \sqcup \mu) \to (\kappa \times \lambda) \sqcup (\kappa \times \mu)$ .

1.5

Unit computations: $\kappa \sqcup 0 = \{0\} \times \kappa \approx \kappa$ ; $\kappa \times 0 = \varnothing = 0$ ; $\kappa \times 1 = \kappa \times \{0\} \approx \kappa$ ; ${}^{0}\kappa = \{\varnothing\}$ has the empty function as its only element, so ${}^{0}\kappa \approx 1$ ; ${}^{1}\kappa \approx \kappa$ by $h \mapsto h(0)$ ; ${}^{\kappa}1$ has the constant function $0$ as its only element, so ${}^{\kappa}1 \approx 1$ ; and for $\kappa \ne 0$ there is no function $\kappa \to \varnothing$ , so ${}^{\kappa}0 = \varnothing = 0$ .

1.6

Two bijections of function spaces: $h \mapsto (\xi \mapsto h(0,\xi),\ \eta \mapsto h(1,\eta))$ is a bijection ${}^{\lambda \sqcup \mu}\kappa \to ({}^{\lambda}\kappa) \times ({}^{\mu}\kappa)$ , with inverse gluing a pair back into one function; and $F \mapsto \big((\xi,\eta) \mapsto F(\eta)(\xi)\big)$ is a bijection ${}^{\mu}({}^{\lambda}\kappa) \to {}^{\lambda \times \mu}\kappa$ , with inverse $f \mapsto \big(\eta \mapsto (\xi \mapsto f(\xi,\eta))\big)$ .

1.7

Monotonicity injections, for $\kappa \subseteq \lambda$ : $\kappa \sqcup \mu \subseteq \lambda \sqcup \mu$ and $\kappa \times \mu \subseteq \lambda \times \mu$ and ${}^{\mu}\kappa \subseteq {}^{\mu}\lambda$ are inclusions; and for $\mu \ne 0$ , extending a function by the constant value $0 \in \mu$ on $\lambda \setminus \kappa$ is an injection ${}^{\kappa}\mu \to {}^{\lambda}\mu$ , injective because restricting back to $\kappa$ recovers the original function.

L5L6

2.1

Second half of (a): if $A \preceq B$ with both well-orderable then $\lvert A \rvert \approx A \preceq B \approx \lvert B \rvert$ by [L1], so $\lvert A \rvert \preceq \lvert B \rvert$ by [L6], and step 1.1 applied to these two cardinals gives $\lvert A \rvert \le \lvert B \rvert$ .

step 1.1L1L6

2.2

Claims (b) and (c): by [L1] the sets $\kappa \oplus \lambda$ and $\kappa \sqcup \lambda$ are equinumerous, and likewise for $\otimes$ , so [L2] lets every outer operation be computed on the untagged representatives; steps 1.2, 1.3 and 1.4 then equate the two underlying sets up to $\approx$ , and [L1] gives the same least ordinal on both sides.

step 1.2step 1.3step 1.4L1L2L3

2.3

Claim (d) is step 1.5 read through [L3] and [L1]: each computed set is equinumerous with $\kappa$ , with $1$ , or with $0$ , and its cardinality is the corresponding cardinal by [L1].

step 1.5L1L3

2.4

Claim (f): $\lambda \oplus \mu \approx \lambda \sqcup \mu$ and $\kappa^{\lambda} \approx {}^{\lambda}\kappa$ by [L1], so [L2] and step 1.6 give ${}^{\lambda \oplus \mu}\kappa \approx ({}^{\lambda}\kappa) \times ({}^{\mu}\kappa)$ and ${}^{\mu}(\kappa^{\lambda}) \approx {}^{\lambda \times \mu}\kappa$ ; taking cardinalities through [L1] and [L3] yields $\kappa^{\lambda \oplus \mu} = \kappa^{\lambda} \otimes \kappa^{\mu}$ and $(\kappa^{\lambda})^{\mu} = \kappa^{\lambda \otimes \mu}$ .

step 1.6L1L2L3L8

3.1

Claim (e): each map of step 1.7 is an injection between the underlying sets, so step 2.1 applied to it gives the corresponding inequality of cardinalities, which by [L3] is the stated inequality of cardinals.

step 1.7step 2.1L3L8

4.1

All six claims are established, in ZF except for the exponentials of clauses (e) and (f), which are read under the Axiom of Choice by [L8].

step 2.1step 2.2step 2.3step 2.4step 3.1∎

Remarks

Why clause (a) is the workhorse. Every other clause is proved by writing down a bijection or an injection between two concrete sets; clause (a) is what converts such a map into a statement about the ordinals $\kappa$ and $\lambda$ , and it is the only clause whose proof uses The Schröder-Bernstein theorem. Note the direction of the work there: $\kappa \le \lambda \Rightarrow \kappa \preceq \lambda$ is the trivial half, and the converse is where being a cardinal rather than an arbitrary ordinal is spent.

No cancellation, and no strict monotonicity. Clause (e) gives $\le$ and not $<$ , and that is not a weakness of the proof. The false statements FALSE: $\kappa \oplus \mu = \lambda \oplus \mu$ implies $\kappa = \lambda$ and FALSE: $\kappa < \lambda$ implies $\kappa^{\mu} < \lambda^{\mu}$ , on this page, show that $\kappa \oplus \mu = \lambda \oplus \mu$ does not force $\kappa = \lambda$ and that $\kappa < \lambda$ does not force $\kappa^{\mu} < \lambda^{\mu}$ ; both failures are already visible at $\aleph_0$ .

The exponent laws hold at every value, including the degenerate ones. With $\lambda = \mu = 0$ the first law reads $\kappa^{0} = \kappa^{0} \otimes \kappa^{0}$ , that is $1 = 1 \otimes 1$ , and with $\kappa = 0$ and $\mu = 0$ the second reads $(0^{\lambda})^{0} = 0^{0} = 1$ , both correct from clause (d). Nothing in the proof of clause (f) case-splits on whether an exponent is zero, because the bijections of step 1.6 are between sets of functions and remain bijections when one of the domains is empty.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 73 results over 28 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

K. Kearnes, Cardinal Arithmetic (Fall 2025 course handout) (standard reference, not scraped)
Cardinal number — cardinal arithmetic (Wikipedia) (standard reference, not scraped)
T. Jech, Set Theory, 3rd millennium ed., Ch. 3 (Cardinal numbers) (standard reference, not scraped)

Commutativity, associativity, distributivity and monotonicity of ⊕\oplus⊕ and ⊗\otimes⊗, the unit laws, the two exponent laws, and κ≤λ\kappa \le \lambdaκ≤λ if and only if κ\kappaκ injects into λ\lambdaλ