ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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.

Solving $\omega + \gamma = \omega\cdot 2$ and dividing $\omega^{2} + \omega + 3$ by $\omega$

Example

Two computations with the two "inverse" operations of this page.

Left subtraction. The equation $\omega + \gamma = \omega \cdot 2$ has exactly one solution, and it is $\gamma = \omega$ . Existence and uniqueness are For $\alpha \le \beta$ there is exactly one ordinal $\gamma$ with $\alpha + \gamma = \beta$ , applicable because $\omega \le \omega \cdot 2$ ; finding the solution is the computation $\omega \cdot 2 = \omega + \omega$ .

Division with remainder. Dividing $\omega^{2} + \omega + 3$ by $\omega$ gives

$\omega^{2} + \omega + 3 \;=\; \omega \cdot (\omega + 1) + 3, \qquad 3 < \omega,$

so the quotient is $\omega + 1$ and the remainder is $3$ , and by the uniqueness in For $\alpha > 0$ every ordinal $\beta$ is $\alpha \cdot \xi + \rho$ with $\rho < \alpha$ , in exactly one way there is no other answer. The step that does the work is $\omega^{2} + \omega = \omega \cdot (\omega + 1)$ , which is left distributivity.

Facts & Assumptions

Given: The ordinals with the operations of Ordinal addition $\alpha + \beta$ , Ordinal multiplication $\alpha \cdot \beta$ and Ordinal exponentiation $\alpha^{\beta}$ , with the conventions $\alpha^{0} = 1$ and $0^{0} = 1$ ; $\omega$ is the least limit ordinal and $3 \in \omega$ ( $\omega$ is the least limit ordinal, The natural numbers $\mathbb{N}$ (von Neumann)).

[L1]

$\mu \cdot \delta^{+} = \mu \cdot \delta + \mu$ and $\mu \cdot 0 = 0$ (Ordinal multiplication $\alpha \cdot \beta$ ); $\mu + 1 = \mu^{+}$ and $\mu + 0 = \mu$ (Ordinal addition $\alpha + \beta$ ); $\mu^{\delta^{+}} = \mu^{\delta} \cdot \mu$ and $\mu^{0} = 1$ (Ordinal exponentiation $\alpha^{\beta}$ , with the conventions $\alpha^{0} = 1$ and $0^{0} = 1$ ).

[L2]

From Monotonicity of ordinal $+$ and $\cdot$ : strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities $0 + \beta = \beta$ and $1 \cdot \beta = \beta$ : $1 \cdot \mu = \mu \cdot 1 = \mu$ (claim (a)); $\nu < \theta$ implies $\mu + \nu < \mu + \theta$ , and $\mu \le \mu + \nu$ (claim (b)).

[L3]

$\mu(\nu + \theta) = \mu\nu + \mu\theta$ (Ordinal multiplication is associative, and $\alpha \cdot (\beta + \gamma) = \alpha\cdot\beta + \alpha\cdot\gamma$ ).

[L4]

$\mu^{1} = \mu$ ( $\alpha^{\beta+\gamma} = \alpha^{\beta}\cdot\alpha^{\gamma}$ and $(\alpha^{\beta})^{\gamma} = \alpha^{\beta\cdot\gamma}$ ; and for $\alpha > 1$ exponentiation is strictly increasing with $\beta \le \alpha^{\beta}$ ).

[L5]

For $\mu \le \nu$ there is exactly one $\gamma$ with $\mu + \gamma = \nu$ (For $\alpha \le \beta$ there is exactly one ordinal $\gamma$ with $\alpha + \gamma = \beta$ ).

[L6]

For $\mu > 0$ every $\nu$ is $\mu \cdot \xi + \rho$ with $\rho < \mu$ , in exactly one way (For $\alpha > 0$ every ordinal $\beta$ is $\alpha \cdot \xi + \rho$ with $\rho < \alpha$ , in exactly one way).

[L7]

$\omega$ is a limit ordinal, so $\omega \ne 0$ ( $\omega$ is the least limit ordinal, Successor and limit ordinals); trichotomy and the elementary ordinal facts (Ordinal (von Neumann), Basic closure properties of ordinals, Trichotomy and well-ordering of the ordinals).

Verification

technique · direct

1.1

$\omega \cdot 2 = \omega \cdot 1^{+} = \omega \cdot 1 + \omega = \omega + \omega$ by [L1] and [L2], and $\omega \le \omega + \omega$ by [L2].

L1L2

1.2

$\omega^{2} = \omega^{1^{+}} = \omega^{1} \cdot \omega = \omega \cdot \omega$ by [L1] and [L4].

L1L4

2.1

Left subtraction: $\omega \le \omega \cdot 2$ by step 1.1, so [L5] gives exactly one $\gamma$ with $\omega + \gamma = \omega \cdot 2$ ; and $\gamma = \omega$ works, since $\omega + \omega = \omega \cdot 2$ by step 1.1, so $\gamma = \omega$ is the solution.

step 1.1L5

2.2

$\omega^{2} + \omega = \omega \cdot \omega + \omega \cdot 1 = \omega \cdot (\omega + 1)$ by step 1.2, [L2] and left distributivity [L3].

step 1.2L2L3

3.1

Division: $\omega > 0$ by [L7], and $\omega^{2} + \omega + 3 = \omega \cdot (\omega + 1) + 3$ by step 2.2, with $3 < \omega$ ; so by the uniqueness in [L6] the quotient of $\omega^{2} + \omega + 3$ by $\omega$ is $\omega + 1$ and the remainder is $3$ .

step 2.2L6L7

4.1

The unique solution of $\omega + \gamma = \omega \cdot 2$ is $\gamma = \omega$ , and dividing $\omega^{2} + \omega + 3$ by $\omega$ gives quotient $\omega + 1$ and remainder $3$ .

step 3.1step 2.1∎

Remarks

Uniqueness is what makes "the answer" meaningful. Both computations exhibit a solution and then quote a uniqueness theorem. Without For $\alpha \le \beta$ there is exactly one ordinal $\gamma$ with $\alpha + \gamma = \beta$ the equation $\omega + \gamma = \omega \cdot 2$ would only be known to have a solution; without For $\alpha > 0$ every ordinal $\beta$ is $\alpha \cdot \xi + \rho$ with $\rho < \alpha$ , in exactly one way the pair $(\omega + 1, 3)$ would be an answer among possibly many. Both theorems are proved from left cancellation, which is the one cancellation law ordinal addition has.

The equation on the other side has no solution at all. There is no $\gamma$ with $\gamma + \omega = \omega + 1$ , because $\gamma + \omega$ is a limit ordinal for every $\gamma$ (claim (g) of Monotonicity of ordinal $+$ and $\cdot$ : strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities $0 + \beta = \beta$ and $1 \cdot \beta = \beta$ ) while $\omega + 1 = \omega^{+}$ is a successor. So ordinal subtraction genuinely exists only on the left, and the same asymmetry is what forces the quotient in For $\alpha > 0$ every ordinal $\beta$ is $\alpha \cdot \xi + \rho$ with $\rho < \alpha$ , in exactly one way to be written on the right of $\alpha$ .

Reading the division off the Cantor normal form. $\omega^{2} + \omega + 3$ has normal form with exponents $2 > 1 > 0$ and coefficients $1, 1, 3$ (Cantor normal form: every nonzero ordinal is $\omega^{\beta_0}\cdot c_0 + \cdots + \omega^{\beta_{k-1}}\cdot c_{k-1}$ with $\beta_0 > \cdots > \beta_{k-1}$ and each $c_i$ a nonzero natural number, in exactly one way). In this instance, dividing by $\omega = \omega^{1}$ has put the term of exponent $0$ into the remainder and lowered each of the two remaining exponents by one: $\omega^{2}$ became $\omega$ and $\omega$ became $1$ , which is exactly the quotient $\omega + 1$ , while $3$ is the constant term. That pattern is what the general division algorithm is doing, but no general statement of it is claimed here.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 57 results over 24 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

Ordinal arithmetic (Wikipedia) (standard reference, not scraped)
T. Jech, Set Theory, 3rd millennium ed., Ch. 2 (Ordinal numbers) (standard reference, not scraped)

Solving ω+γ=ω⋅2\omega + \gamma = \omega\cdot 2ω+γ=ω⋅2 and dividing ω2+ω+3\omega^{2} + \omega + 3ω2+ω+3 by ω\omegaω