LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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.

Countability of every finite-dimensional integer lattice

Statement

For every positive integer $d$ , the integer lattice $Γ = Z^{d}$ is at most countable. Countability here means finite or in bijection with $N$ , as in Finite, countably infinite, countable, uncountable.

Facts & Assumptions

Given: The usual set $Z$ of signed integers, the von Neumann natural numbers $N$ , and a fixed finite $d \geq 1$ .

[F1]

A set is at most countable when it is finite or equinumerous with $N$ (Finite, countably infinite, countable, uncountable).

[F2]

A nonempty set is at most countable if a surjection from $N$ onto it exists (A nonempty set is at most countable iff it is a surjective image of $N$ ). This criterion uses no choice principle.

[F3]

Every finite Cartesian power of an at most countable set is at most countable (Every finite power of an at most countable set is at most countable).

[F4]

The integers are the quotient of $N \times N$ by $(a, b) \sim (c, d)$ iff $a + d = b + c$ , and the quotient map $π (a, b) = [(a, b)]$ is surjective (The integers as equivalence classes of pairs of naturals).

Proof

technique · direct

Given: $d \geq 1$ and $Γ = Z^{d}$ .

1.1F1F2F3F4construct

The identity map shows that $N$ is at most countable by [F1], so [F3] gives that $N \times N$ is at most countable. This product is nonempty, so [F2] supplies a surjection $s : N \to N \times N$ . By [F4], the quotient map $π : N \times N \to Z$ is surjective; hence $π \circ s : N \to Z$ is a surjection.

2.1step 1.1F1F2

The surjection in step 1.1 makes $Z$ nonempty, and [F2] then shows that $Z$ is at most countable in the sense of [F1].

3.1givenstep 2.1F3∎

Since $d$ is finite and positive, [F3] with $C = Z$ and $m = d$ gives that $Γ = Z^{d}$ is at most countable.

Countability of every finite-dimensional integer lattice

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources