CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-25

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 multiplicative identity is positive

In any ordered field $F$ with positive cone $P$ , the multiplicative identity satisfies $0 < 1$ ; that is, $1 \in P$ .

Given: An ordered field $F$ with positive cone $P$ and multiplicative identity $1$ (Field).

[L1]

In any field $1 \neq 0$ (Field).

[L2]

For every $x \neq 0$ the square satisfies $x^{2} > 0$ (Squares of nonzero elements are positive).

[L3]

The identity axiom gives $1 = 1 \cdot 1 = 1^{2}$ (Field).

technique · direct

1.1L1L3

By the field axioms $1 \neq 0$ , and $1 = 1 \cdot 1 = 1^{2}$ .

2.1step 1.1L2

Since $1 \neq 0$ , applying the square lemma with $x = 1$ gives $1^{2} > 0$ .

3.1step 2.1step 1.1L3∎

Because $1 = 1^{2}$ , it follows that $1 > 0$ , i.e. $0 < 1$ .

Depends on