Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-25
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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.

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Squares of nonzero elements are positive

Statement

Let F be an ordered field (Ordered field) with positive cone P. For every a∈F with a≠0 we have a2>0, where a2=a⋅a. In particular, every nonzero square is positive.

Facts & Assumptions

Given: An ordered field F with positive cone P, and an element a∈F with a≠0.

[L1]

x>0  ⟺  x∈P, and by trichotomy for a≠0 exactly one of a∈P, −a∈P holds (Ordered field).

[L2]

P is closed under multiplication: if u,v∈P then uv∈P (axiom O2 of Ordered field).

Proof

technique · cases
1.1

Since a≠0, trichotomy leaves exactly two possibilities: a∈P or −a∈P.

L1
2.1

Case a>0: then a∈P, so closure gives a⋅a∈P, i.e. a2>0.

assume-case posstep 1.1L2L1
2.2

Case a<0: then −a∈P, so closure gives (−a)(−a)∈P; since (−a)(−a)=a⋅a=a2 by L3, we get a2∈P, i.e. a2>0.

assume-case negstep 1.1L2L3L1
3.1

The cases a>0 and a<0 exhaust all of a≠0 by trichotomy, and both yield a2>0.

cases-exhaustivestep 2.1step 2.2∎

Depends on

Used by

Dependency tree · two levels

4 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