Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-02 (claude-opus-5)
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.

Discreteness: σ(n) is the immediate successor

Statement

For all m,n∈N: m<n  ⟺  σ(m)≤n (Order on the natural numbers). Consequently there is no k with n<k<σ(n): the successor σ(n) is the immediate successor of n.

Facts & Assumptions

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

[L1]

Every nonzero natural is a successor: j≠0⇒j=σ(i) (Every nonzero natural number is a successor).

[L2]

σ(m)+i=σ(m+i) (the left successor law, Left successor law for addition) and m+σ(i)=σ(m+i) (the addition recursion).

[L3]

Cancellation: x+k=y+k⇒x=y (Addition is cancellative).

[L4]

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

[L5]

Addition is commutative: x+y=y+x (Addition is commutative).

Proof

technique · direct
1.1

Forward: if m<n then m≤n and m≠n, so m+j=n for some j, with j≠0 since j=0 would give m=n; then j=σ(i) [L1] and n=m+σ(i)=σ(m+i)=σ(m)+i [L2], hence σ(m)≤n.

givenL1L2
1.2

Backward: if σ(m)≤n then σ(m)+i=n, so n=σ(m)+i=σ(m+i)=m+σ(i) [L2], hence m≤n; and m≠n, since m=n would give m+σ(i)=m=m+0, so σ(i)+m=0+m by [L5] and σ(i)=0 by [L3], contradicting [L4]; therefore m<n.

givenL2L3L4L5
2.1

No k satisfies n<k<σ(n): if n<k then σ(n)≤k by the equivalence, so k≥σ(n), contradicting k<σ(n); thus σ(n) is the immediate successor.

step 1.1step 1.2∎

Depends on

Used by

…and 28 more results.

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources