Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-02 (claude-opus-5)
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≤ is a linear order on N

Statement

The relation ≤ on N, where m≤n  ⟺  ∃k (m+k=n) (Order on the natural numbers), is a linear (total) order: reflexive, antisymmetric, transitive, and total.

Facts & Assumptions

Given: The order m≤n  ⟺  ∃k (m+k=n) (Order on the natural numbers) and addition with m+0=m, m+σ(n)=σ(m+n) (Addition of natural numbers).

[L1]

0+k=k and addition is associative (Left identity for addition, Addition is associative).

[L2]

Cancellation: m+k=n+k⇒m=n (Addition is cancellative).

[L3]

Every nonzero natural is a successor: if j≠0 then j=σ(i) for some i (Every nonzero natural number is a successor).

[L4]

Trichotomy: exactly one of m<n, m=n, m>n holds (Trichotomy of the order on N).

[L5]

Addition is commutative (Addition is commutative).

[L6]

σ(x)≠0 for every x (Peano axiom P1) (The von Neumann naturals form a Peano system).

Proof

technique · direct
1.1

Reflexive: m+0=m, so m≤m.

given
1.2

Transitive: if m≤n and n≤p, say m+k=n and n+j=p, then m+(k+j)=(m+k)+j=n+j=p, so m≤p.

givenL1
1.3

A sum is zero only if both summands are: if k+j=0 with j≠0 then j=σ(i) and k+j=k+σ(i)=σ(k+i)≠0 [L6], a contradiction, so k=j=0.

L1L3L6
1.4

Total: by trichotomy one of m<n, m=n, m>n holds, and each of these gives m≤n or n≤m.

L4
2.1

Antisymmetric: if m≤n and n≤m, say m+k=n and n+j=m, then m+(k+j)=(m+k)+j=n+j=m=m+0; commuting both sides [L5] gives (k+j)+m=0+m, and cancelling the common right summand m [L2] gives k+j=0, hence k=j=0 by step 1.3 and m=n.

givenL1L2L5step 1.3
3.1

Reflexivity, antisymmetry, transitivity, and totality all hold, so ≤ is a linear order on N.

step 1.1step 1.2step 2.1step 1.4∎

Depends on

Used by

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