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ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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.

Changing both domain and codomain bases of a map F2→F3F^2\to F^3 uses both sides of the formula

Example

Let T:F2→F3T:F^2\to F^3 be T(x,y)=(x+2y,3x−y,x+y)T(x,y)=(x+2y,3x-y,x+y). Use the standard bases B,C\mathcal B,\mathcal C and the new bases

B′=(e0+e1,e1),C′=(f0+f2,f1,f2).\mathcal B'=(e_0+e_1,e_1),\qquad \mathcal C'=(f_0+f_2,f_1,f_2).

Then

[T]B′C′=(322−1−1−1).[T]_{\mathcal B'}^{\mathcal C'}=\begin{pmatrix}3&2\\2&-1\\-1&-1\end{pmatrix}.

Facts & Assumptions

Given: The displayed map and four ordered bases.

[L1]

The two-sided formula is [T]B′C′=PC′←C[T]BCPB←B′[T]_{\mathcal B'}^{\mathcal C'}=P_{\mathcal C'\leftarrow\mathcal C}[T]_{\mathcal B}^{\mathcal C}P_{\mathcal B\leftarrow\mathcal B'} ([T]B′C′=PC′←C[T]BCPB←B′[T]_{\mathcal B'}^{\mathcal C'}=P_{\mathcal C'\leftarrow\mathcal C}[T]_{\mathcal B}^{\mathcal C}P_{\mathcal B\leftarrow\mathcal B'}).

Verification

technique · direct
1.1

Here [T]BC=(123−111)[T]_{\mathcal B}^{\mathcal C}=\begin{pmatrix}1&2\\3&-1\\1&1\end{pmatrix}, PB←B′=(1011)P_{\mathcal B\leftarrow\mathcal B'}=\begin{pmatrix}1&0\\1&1\end{pmatrix}, and PC′←C=(100010−101)P_{\mathcal C'\leftarrow\mathcal C}=\begin{pmatrix}1&0&0\\0&1&0\\-1&0&1\end{pmatrix}. Multiplying in the order of [L1] gives (322−1−1−1)\begin{pmatrix}3&2\\2&-1\\-1&-1\end{pmatrix}.

givenL1
2.1

Independently, T(e0+e1)=(3,2,2)=3(f0+f2)+2f1−f2T(e_0+e_1)=(3,2,2)=3(f_0+f_2)+2f_1-f_2 and T(e1)=(2,−1,1)=2(f0+f2)−f1−f2T(e_1)=(2,-1,1)=2(f_0+f_2)-f_1-f_2, producing the same two coordinate columns.

step 1.1L1∎

Depends on

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