Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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)\sigma(n) is the immediate successor

Statement

For all m,n∈Nm, n \in \mathbb{N}: m<n  ⟺  σ(m)≤nm < n \iff \sigma(m) \le n (Order on the natural numbers). Consequently there is no kk with n<k<σ(n)n < k < \sigma(n): the successor σ(n)\sigma(n) is the immediate successor of nn.

Facts & Assumptions

Given: The order m≤n  ⟺  ∃j (m+j=n)m \le n \iff \exists j\ (m + j = n), with m<nm < n meaning m≤nm \le n and m≠nm \ne n (Order on the natural numbers), and addition with m+0=mm + 0 = m, m+σ(n)=σ(m+n)m + \sigma(n) = \sigma(m + n) (Addition of natural numbers).

[L1]

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

[L2]

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

[L3]

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

[L4]

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

[L5]

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

Proof

technique · direct
1.1

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

givenL1L2
1.2

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

givenL2L3L4L5
2.1

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

step 1.1step 1.2∎

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