Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29
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FALSE: (β+γ)⋅α=β⋅α+γ⋅α(\beta + \gamma)\cdot\alpha = \beta\cdot\alpha + \gamma\cdot\alpha for all ordinals

Statement

FALSE. Ordinal multiplication distributes over addition on the right:

(β+γ)⋅α=β⋅α+γ⋅αfor all ordinals α,β,γ.(\beta + \gamma) \cdot \alpha = \beta \cdot \alpha + \gamma \cdot \alpha \qquad \text{for all ordinals } \alpha, \beta, \gamma.

Distributivity on the left is a theorem (Ordinal multiplication is associative, and α⋅(β+γ)=α⋅β+α⋅γ\alpha \cdot (\beta + \gamma) = \alpha\cdot\beta + \alpha\cdot\gamma): α⋅(β+γ)=αβ+αγ\alpha \cdot (\beta + \gamma) = \alpha\beta + \alpha\gamma. The right-hand law is a different statement, and it fails at β=γ=1\beta = \gamma = 1, α=ω\alpha = \omega.

Facts & Assumptions

Given: The ordinals with the operations of Ordinal addition α+β\alpha + \beta and Ordinal multiplication α⋅β\alpha \cdot \beta, and ω\omega the least limit ordinal (ω\omega is the least limit ordinal, Successor and limit ordinals). Here 2=1+2 = 1^{+}, so 1+1=1+=21 + 1 = 1^{+} = 2 by Ordinal addition α+β\alpha + \beta.

[L1]

α⋅0=0\alpha \cdot 0 = 0, α⋅δ+=α⋅δ+α\alpha \cdot \delta^{+} = \alpha \cdot \delta + \alpha, and α⋅λ=⋃{α⋅ξ:ξ∈λ}\alpha \cdot \lambda = \bigcup\{\alpha \cdot \xi : \xi \in \lambda\} for limit λ\lambda (Ordinal multiplication α⋅β\alpha \cdot \beta); α+0=α\alpha + 0 = \alpha and α+1=α+\alpha + 1 = \alpha^{+} (Ordinal addition α+β\alpha + \beta).

[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+β=β0 + \beta = \beta and 1⋅β=β1 \cdot \beta = \beta: 1⋅μ=μ⋅1=μ1 \cdot \mu = \mu \cdot 1 = \mu (claim (a)); ν<θ\nu < \theta implies α+ν<α+θ\alpha + \nu < \alpha + \theta (claim (b)); μ≤ν\mu \le \nu implies μγ≤νγ\mu\gamma \le \nu\gamma (claim (e)).

[L3]

For m,n∈ωm, n \in \omega the ordinal m⋅nm \cdot n lies in ω\omega (claim (a) of On ω\omega the ordinal ++ and ⋅\cdot are the Peano operations: ω\omega is closed under ordinal ++, ⋅\cdot and exponentiation, and for naturals m,nm, n the ordinal m+nm + n and m⋅nm \cdot n are the natural-number sum and product).

[L4]

ω\omega is a limit ordinal, so ⋃ω=ω\bigcup \omega = \omega and 0∈ω0 \in \omega (ω\omega is the least limit ordinal, Successor and limit ordinals); every ordinal is transitive, μ⊆ν\mu \subseteq \nu iff μ∈ν\mu \in \nu or μ=ν\mu = \nu, and μ∉μ\mu \notin \mu (Ordinal (von Neumann), Basic closure properties of ordinals, Trichotomy and well-ordering of the ordinals).

Refutation

technique · direct
1.1

For every n∈ωn \in \omega the ordinal 2⋅n2 \cdot n lies in ω\omega by [L3], hence 2⋅n⊆ω2 \cdot n \subseteq \omega by [L4]; and n=1⋅n≤2⋅nn = 1 \cdot n \le 2 \cdot n by [L2], since 1≤21 \le 2, hence n⊆2⋅nn \subseteq 2 \cdot n.

L2L3L4
1.2

The right-hand side of the claimed law at β=γ=1\beta = \gamma = 1, α=ω\alpha = \omega is 1⋅ω+1⋅ω=ω+ω1 \cdot \omega + 1 \cdot \omega = \omega + \omega by [L2], and ω+ω≠ω\omega + \omega \ne \omega, because 0∈ω0 \in \omega gives ω=ω+0<ω+ω\omega = \omega + 0 < \omega + \omega by [L1] and [L2], while μ∉μ\mu \notin \mu by [L4].

L1L2L4
2.1

The left-hand side is (1+1)⋅ω=2⋅ω=⋃{2⋅n:n∈ω}(1 + 1) \cdot \omega = 2 \cdot \omega = \bigcup\{2 \cdot n : n \in \omega\} by [L1], and that union equals ω\omega: it is contained in ω\omega because each 2⋅n⊆ω2 \cdot n \subseteq \omega by step 1.1, and it contains ω\omega because ω=⋃ω=⋃{n:n∈ω}\omega = \bigcup \omega = \bigcup\{n : n \in \omega\} by [L4] and each n⊆2⋅nn \subseteq 2 \cdot n by step 1.1.

step 1.1L1L4
3.1

Therefore (1+1)⋅ω=ω(1 + 1) \cdot \omega = \omega while 1⋅ω+1⋅ω=ω+ω≠ω1 \cdot \omega + 1 \cdot \omega = \omega + \omega \ne \omega, so the claimed right distributive law fails.

step 2.1step 1.2∎

Remarks

Why the two laws are genuinely different. α⋅(β+γ)\alpha \cdot (\beta + \gamma) is "β+γ\beta + \gamma copies of α\alpha", which is β\beta copies followed by γ\gamma copies, and that is exactly αβ+αγ\alpha\beta + \alpha\gamma; the left law is therefore a statement about concatenating blocks and it is true. (β+γ)⋅α(\beta + \gamma) \cdot \alpha is "α\alpha copies of the block β+γ\beta + \gamma", and interleaving α\alpha copies of a two part block is not the same as α\alpha copies of the first part followed by α\alpha copies of the second. The witness above is the smallest instance of that difference.

The computation is repeated on purpose. The value 2⋅ω=ω2 \cdot \omega = \omega also appears in FALSE: ordinal multiplication is commutative, and it is recomputed here from the limit clause rather than quoted from that item, so that this refutation rests only on definitions and theorems.

The failure is not a failure of associativity. ⋅\cdot is associative (Ordinal multiplication is associative, and α⋅(β+γ)=α⋅β+α⋅γ\alpha \cdot (\beta + \gamma) = \alpha\cdot\beta + \alpha\cdot\gamma); what fails is the interaction of ⋅\cdot with ++ on one particular side. So the ordinals under ++ and ⋅\cdot satisfy every semiring law except commutativity of the two operations and right distributivity, and each of those three failures is refuted separately on this page.

Depends on

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Sources