False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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FALSE: $(β + γ) \cdot α = β \cdot α + γ \cdot α$ for all ordinals

Statement

FALSE. Ordinal multiplication distributes over addition on the right:

$(β + γ) \cdot α = β \cdot α + γ \cdot α for all ordinals α, β, γ .$

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

Facts & Assumptions

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

[L1]

$α \cdot 0 = 0$ , $α \cdot δ^{+} = α \cdot δ + α$ , and $α \cdot λ = ⋃ {α \cdot ξ : ξ \in λ}$ for limit $λ$ (Ordinal multiplication $α \cdot β$ ); $α + 0 = α$ and $α + 1 = α^{+}$ (Ordinal addition $α + β$ ).

[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 $0 + β = β$ and $1 \cdot β = β$ : $1 \cdot μ = μ \cdot 1 = μ$ (claim (a)); $ν < θ$ implies $α + ν < α + θ$ (claim (b)); $μ \leq ν$ implies $μ γ \leq ν γ$ (claim (e)).

[L3]

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

[L4]

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

Refutation

technique · direct

1.1L2L3L4

For every $n \in ω$ the ordinal $2 \cdot n$ lies in $ω$ by [L3], hence $2 \cdot n \subseteq ω$ by [L4]; and $n = 1 \cdot n \leq 2 \cdot n$ by [L2], since $1 \leq 2$ , hence $n \subseteq 2 \cdot n$ .

1.2L1L2L4

The right-hand side of the claimed law at $β = γ = 1$ , $α = ω$ is $1 \cdot ω + 1 \cdot ω = ω + ω$ by [L2], and $ω + ω \neq ω$ , because $0 \in ω$ gives $ω = ω + 0 < ω + ω$ by [L1] and [L2], while $μ \notin μ$ by [L4].

2.1step 1.1L1L4

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

3.1step 2.1step 1.2∎

Therefore $(1 + 1) \cdot ω = ω$ while $1 \cdot ω + 1 \cdot ω = ω + ω \neq ω$ , so the claimed right distributive law fails.

Remarks

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

The computation is repeated on purpose. The value $2 \cdot ω = ω$ 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 $α \cdot (β + γ) = α \cdot β + α \cdot γ$ ); 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.

FALSE: $(β + γ) \cdot α = β \cdot α + γ \cdot α$ for all ordinals

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources

FALSE: (β+γ)⋅α=β⋅α+γ⋅α(β+γ)⋅α=β⋅α+γ⋅α for all ordinals

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources

FALSE: $(β + γ) \cdot α = β \cdot α + γ \cdot α$ for all ordinals