LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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.

The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$ , $s_{\sigma(k)} = -s_k$ , which satisfies $|s_k| = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv -1$

Statement

Let $\sigma$ be the successor on $\mathbb{N}$ (The natural numbers $\mathbb{N}$ (von Neumann)). There are functions $e, o : \mathbb{N} \to \mathbb{N}$ and a sequence $(s_k)$ of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) with the following properties.

The index maps. $e$ is the unique function with $e_0 = 0$ and $e_{\sigma(j)} = \sigma(\sigma(e_j))$ , and $o$ the unique function with $o_0 = \sigma(0)$ and $o_{\sigma(j)} = \sigma(\sigma(o_j))$ . Both are strictly increasing.
The partition. $\mathbb{N}$ is the disjoint union of the ranges of $e$ and of $o$ : every natural number is $e_i$ for exactly one $i$ or $o_i$ for exactly one $i$ , and never both.
The alternating sequence. $(s_k)$ is the unique sequence of reals with $s_0 = 1, \qquad s_{\sigma(k)} = -s_k \quad (k \in \mathbb{N}).$
Its values. $|s_k| = 1$ for every $k$ , so $(s_k)$ is bounded; and $s_{e_j} = 1, \qquad s_{o_j} = -1 \qquad (j \in \mathbb{N}),$ that is $s \circ e$ is constantly $1$ and $s \circ o$ constantly $-1$ .

This is the sequence usually written $s_k = (-1)^k$ , with $e_j = 2j$ and $o_j = 2j+1$ , presented by the recursions that its proofs actually use. It is collected here once because three separate items on this page and its companion need an alternating or interleaved witness, and rebuilding the recursion inside each of them is what this lemma exists to prevent.

Facts & Assumptions

Given: By the recursion theorem (The recursion theorem) applied to the set $\mathbb{R}$ , the element $1$ and the function $u \mapsto -u$ , the unique sequence $(s_k)$ of reals with $s_0 = 1$ and $s_{\sigma(k)} = -s_k$ ; applied to the set $\mathbb{N}$ , the element $0$ and the function $i \mapsto \sigma(\sigma(i))$ , the unique $e : \mathbb{N} \to \mathbb{N}$ with $e_0 = 0$ and $e_{\sigma(j)} = \sigma(\sigma(e_j))$ ; and applied to $\mathbb{N}$ , the element $\sigma(0)$ and the same function, the unique $o : \mathbb{N} \to \mathbb{N}$ with $o_0 = \sigma(0)$ and $o_{\sigma(j)} = \sigma(\sigma(o_j))$ (The natural numbers $\mathbb{N}$ (von Neumann), Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L1]

Recursion theorem, including its uniqueness clause (The recursion theorem).

[L2]

Induction principle (The principle of mathematical induction).

[L3]

Order on $\mathbb{N}$ : $i < \sigma(i)$ for every $i$ , since $\sigma(i) = i + 1$ gives $i \le \sigma(i)$ and $\sigma(i) \ne i$ ; and the order is transitive and total (Order on the natural numbers, Addition of natural numbers, No natural number equals its own successor, $\le$ is a linear order on $\mathbb{N}$ ).

[L4]

Consecutive comparisons suffice: if $n_i < n_{\sigma(i)}$ for every $i$ then $n$ is strictly increasing (A strictly increasing index map satisfies $n_k \ge k$ ).

[L5]

Absolute value and field arithmetic: $|-u| = |u|$ (Basic properties of the absolute value); $|v| = v$ whenever $v \ge 0$ (Absolute value in an ordered field, Order on the reals); and $-(-u) = u$ (Field).

[L6]

Order in $\mathbb{R}$ : $0 < 1$ (The multiplicative identity is positive), sums of positives are positive and adding a constant preserves the order (Order is preserved by adding a constant and by adding inequalities, Complete ordered field (least-upper-bound property), Ordered field), so $1 - (-1) = 1 + 1 > 0$ and hence $-1 < 1$ ; in particular $1 \ne -1$ .

Proof

technique · induction

1.1

Base case for claim 4: $|s_0| = |1| = 1$ , since $1 > 0$ makes $|1| = 1$ .

givenL5L6base

1.2

Inductive hypothesis: fix $k \in \mathbb{N}$ and assume $|s_k| = 1$ .

1.3

Both index maps satisfy consecutive strict comparisons: $e_j < \sigma(e_j) < \sigma(\sigma(e_j)) = e_{\sigma(j)}$ , and likewise $o_j < o_{\sigma(j)}$ , so $e$ and $o$ are strictly increasing and claim 1 holds, its uniqueness part being the uniqueness clause of the recursion theorem.

givenL1L3L4

1.4

By induction, $s_{e_j} = 1$ for every $j$ : the base case is $s_{e_0} = s_0 = 1$ , and if $s_{e_j} = 1$ then $s_{e_{\sigma(j)}} = s_{\sigma(\sigma(e_j))} = -s_{\sigma(e_j)} = -(-s_{e_j}) = s_{e_j} = 1$ .

givenL1L2L5

1.5

By induction, $s_{o_j} = -1$ for every $j$ : the base case is $s_{o_0} = s_{\sigma(0)} = -s_0 = -1$ , and if $s_{o_j} = -1$ then $s_{o_{\sigma(j)}} = s_{\sigma(\sigma(o_j))} = -(-s_{o_j}) = s_{o_j} = -1$ .

givenL1L2L5

1.6

By induction on $n$ , every natural number satisfies: either $n = e_i$ and $\sigma(n) = o_i$ for some $i$ , or $n = o_i$ and $\sigma(n) = e_{\sigma(i)}$ for some $i$ . The base case is $0 = e_0$ with $\sigma(0) = o_0$ . For the successor step, if $n = e_i$ and $\sigma(n) = o_i$ then $\sigma(n) = o_i$ and $\sigma(\sigma(n)) = \sigma(\sigma(e_i)) = e_{\sigma(i)}$ , which is the second alternative at $\sigma(n)$ ; and if $n = o_i$ and $\sigma(n) = e_{\sigma(i)}$ then $\sigma(n) = e_{\sigma(i)}$ and $\sigma(\sigma(n)) = \sigma(\sigma(o_i)) = o_{\sigma(i)}$ , which is the first alternative at $\sigma(n)$ .

givenL1L2

1.7

The sequence $(s_k)$ is the unique sequence of reals with $s_0 = 1$ and $s_{\sigma(k)} = -s_k$ , by the uniqueness clause of the recursion theorem: this is claim 3.

givenL1

2.1

Successor step for claim 4: $|s_{\sigma(k)}| = |-s_k| = |s_k| = 1$ .

step 1.2L5

2.2

In particular every natural number lies in the range of $e$ or in the range of $o$ , since each alternative of step 1.6 exhibits $n$ as such a value.

step 1.6

2.3

The two ranges are disjoint: if $e_i = o_j$ for some $i, j$ then $1 = s_{e_i} = s_{o_j} = -1$ , contradicting $1 \ne -1$ .

step 1.4step 1.5L6

2.4

Each of $e$ and $o$ is injective, being strictly increasing, so a natural number in the range of $e$ is $e_i$ for exactly one $i$ , and likewise for $o$ .

step 1.3L3

3.1

By the induction principle, $|s_k| = 1$ for every $k \in \mathbb{N}$ ; hence $|s_k| \le 1$ at every index and $(s_k)$ is bounded. Together with steps 1.4 and 1.5 this is claim 4.

step 1.1step 2.1step 1.4step 1.5L2

4.1

Claim 2 follows: by step 2.2 every natural is in one of the two ranges, by step 2.3 not in both, and by step 2.4 the index realising it is unique. Claims 1, 2, 3 and 4 are therefore all established.

step 2.2step 2.3step 3.1step 2.4step 1.3step 1.7discharge-induction∎

Remarks

Why the recursion rather than $(-1)^k$ . Written as a power, every one of the four claims would have to be unwound into the two recursion equations before it could be proved; written as a recursion, each is a two-line induction. The identification with $(-1)^k$ is available (Integer powers $a^m$ ) and is used nowhere.
The parity statement is genuinely proved, not assumed. Claim 2 is where the work is: the covering half is the interleaved induction of step 1.6, which tracks $n$ and $\sigma(n)$ together because neither alone is preserved by the successor, and the disjointness half is settled by the sequence, since $s$ takes the value $1$ on one range and $-1$ on the other and $1 \ne -1$ . Using the sequence to separate the two ranges is shorter than any direct parity argument and needs no arithmetic on $\mathbb{N}$ beyond the successor.
What consumes this lemma. FALSE: a sequence with a convergent subsequence is bounded (the converse of Bolzano-Weierstrass) interleaves a constant sequence with an unbounded one along $e$ and $o$ ; The sequence $(-1)^k(1 + 1/k)$ is bounded with subsequential limit set exactly $\{-1, 1\}$ ↗ multiplies $s$ by a null perturbation to get a sequence with exactly two subsequential limits; and The sequence $1, 1, 2, 1, 3, 1, 4, \dots$ is unbounded and has a convergent subsequence ↗ is the witness for the first of those. The same sequence, built inline, refutes FALSE: every bounded sequence converges on the previous page; that item predates this lemma and is left as it stands.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 58 results over 16 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Parity (mathematics) (Wikipedia) (standard reference, not scraped)
Subsequence (Wikipedia) (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §2.1 and §6.4 (recursive definitions; subsequences) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)