ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)

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.
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A Steenrod operation on the universal Thom class

Example

Assume AC, inherited from the cited bundle, cohomology, or operation suppliers. In stable mod-two Thom cohomology, Sq³(U)=w₃U. At rank 3, Sq³(u₃)=w₃(γ₃)u₃; at rank 2 the component is zero because w₃(γ₂)=0 and Sq³(u₂)=0 by instability.

Facts & Assumptions

Given: AC; the stable mod-two Thom cohomology module with its stable class $U$ and component classes $u_{r}$ ; the stable squares $S q^{i} (U) = w_{i} U$ ; and the ranks $2$ and $3$ .

[F1]

The stable-square lemma identifies every component of $S q^{i} (U)$ with $w_{i} (γ_{r}) u_{r}$ and proves compatibility under the inverse-system maps (Stable Steenrod squares on universal Thom cohomology); the top-square formula and instability govern the degree-two class $u_{2}$ , and $w_{3} (γ_{2}) = 0$ for rank reasons.

Verification

1.1givenF1

The stable-square supplier identifies every component of Sq^i(U) with w_i(γ_r)u_r and proves compatibility under the inverse-system maps. The rank bound makes w₃(γ₂)=0; instability kills Sq³ on the degree-2 class u₂.

2.1step 1.1F1∎

At rank 3 the top-square formula agrees with the Thom identity.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

John Milnor and James Stasheff, Characteristic Classes (standard reference, not scraped)