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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29
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For α≤β there is exactly one ordinal γ with α+γ=β

Statement

Let α and β be ordinals (Ordinal (von Neumann)) with α≤β. Then there is exactly one ordinal γ with

α+γ=β,

namely the order type of the set β∖α of ordinals lying in β but not in α, taken with the membership order (Every well-order has a unique order type).

This is subtraction on the left: the unknown sits on the right of the + sign, which is the side on which ordinal addition is strictly increasing and cancellative (Monotonicity of ordinal + and ⋅: strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities 0+β=β and 1⋅β=β). Subtraction on the other side does not exist in general: there is no ordinal γ at all with γ+ω=ω+1, since γ+ω is a limit ordinal for every γ while ω+1 is a successor.

No choice principle is used.

Facts & Assumptions

Given: Ordinals α≤β, that is α⊆β. Every subset of a well-order carries the inherited order, again a well-order (Well-order and well-ordered set).

[L1]

ot(W)=ot(I)+ot(W∖I) for every well-order W and every initial segment I of it (claim (b) of α+β is the order type of α followed by β).

[L2]

Every well-order is order isomorphic to exactly one ordinal, its order type (Every well-order has a unique order type).

[L3]

An initial segment is a downward closed subset (Initial segment of a well-order); an ordinal is a transitive set strictly well ordered by ∈, so it is a well-order (Ordinal (von Neumann), Well-order and well-ordered set).

[L4]

α⊆β if and only if α∈β or α=β (Basic closure properties of ordinals, claim (f)); exactly one of μ∈ν, μ=ν, ν∈μ holds (Trichotomy and well-ordering of the ordinals).

[L5]

Left cancellation: α+γ=α+γ′ implies γ=γ′; and α+λ is a limit ordinal whenever λ is (claims (b) and (g) of Monotonicity of ordinal + and ⋅: strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities 0+β=β and 1⋅β=β, with + as in Ordinal addition α+β).

Proof

technique · direct
1.1

α is an initial segment of the well-order (β,∈): it is a subset of β because α≤β, and it is downward closed in β because x∈y∈α gives x∈α by transitivity of α.

L3L4
1.2

For an ordinal μ the identity is an order isomorphism of μ onto μ, so ot(μ)=μ by the uniqueness in [L2].

L2L3
2.1

Put γ=ot(β∖α), which exists by [L2] since β∖α is a subset of the well-order β; then [L1] applied to W=β and I=α gives β=ot(β)=ot(α)+ot(β∖α)=α+γ.

step 1.1step 1.2L1L2L3
3.1

If also α+γ′=β then α+γ′=α+γ, so γ′=γ by [L5]; hence exactly one such γ exists, and it is ot(β∖α).

step 2.1L5∎

Remarks

The proof is a picture. β is a copy of α followed by whatever is left, and "whatever is left" is β∖α. Clause (b) of α+β is the order type of α followed by β says exactly that the order type of a well-order split at an initial segment is the sum of the two order types, so no recursion is needed at all.

Why the hypothesis α≤β cannot be dropped. α≤α+γ always holds (claim (b) of Monotonicity of ordinal + and ⋅: strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities 0+β=β and 1⋅β=β), so α+γ=β forces α≤β. The theorem is therefore sharp: the equation is solvable exactly when the hypothesis holds.

The other-sided equation. The claim in the Statement that no γ satisfies γ+ω=ω+1 uses only that γ+ω is a limit ordinal, which is claim (g) of Monotonicity of ordinal + and ⋅: strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities 0+β=β and 1⋅β=β, and that ω+1=ω+ is a successor. Right subtraction, when it exists, is also not unique: 0+ω=1+ω=ω, so the equation γ+ω=ω has at least two solutions (FALSE: β<γ implies β+α<γ+α).

Where it is used. Existence of the remainder in For α>0 every ordinal β is α⋅ξ+ρ with ρ<α, in exactly one way is a direct application, and that theorem in turn is what extracts the coefficients of a Cantor normal form (Cantor normal form: every nonzero ordinal is ωβ0⋅c0+⋯+ωβk−1⋅ck−1 with β0>⋯>βk−1 and each ci a nonzero natural number, in exactly one way).

Depends on

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Sources