ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pass

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.

Forward and reverse derivatives on a shared-expression graph

Example

Consider the real-arithmetic graph $z_{1} = x + y, z_{2} = z_{1} z_{1}, z_{3} = x y, f = z_{2} + z_{3} .$ At $(x, y) = (1, 2)$ it has output $f = 11$ . Both forward and reverse mode should give gradient $(8, 7)$ ; the repeated parent slot in $z_{2}$ tests that its two edge contributions are both accumulated.

Facts & Assumptions

Given: The graph and input $(x, y) = (1, 2)$ .

[F1]

Finite straight-line computational programs and real-arithmetic tapes: repeated parent slots count as separate edges in the finite real-arithmetic graph.

[F2]

Forward mode computes the exact directional derivative of the real-arithmetic graph: forward mode returns $D f (x, y) (u, v)$ for any direction $(u, v)$ .

[F3]

Reverse mode computes the exact gradient of the real-arithmetic graph: reverse mode returns the exact input gradient of the scalar graph output.

Proof

technique · direct tape evaluation

1.1F1givenalgebra

The primal values are $z_{1} = 1 + 2 = 3$ , $z_{2} = 3 \cdot 3 = 9$ , $z_{3} = 1 \cdot 2 = 2$ , and $f = z_{2} + z_{3} = 11$ .

2.1F1F2step 1.1algebra

For direction $(u, v)$ , forward mode gives ${\dot{z}}_{1} = u + v$ , ${\dot{z}}_{2} = z_{1} {\dot{z}}_{1} + {\dot{z}}_{1} z_{1} = 6 (u + v)$ , ${\dot{z}}_{3} = y u + x v = 2 u + v$ , hence $\dot{f} = 8 u + 7 v$ . By [F2], the gradient is $(8, 7)$ .

2.2F1F3step 1.1algebra

In reverse mode, seed $\overset{ˉ}{f} = 1$ , so ${\overset{ˉ}{z}}_{2} = {\overset{ˉ}{z}}_{3} = 1$ . The product $z_{3} = x y$ adds $2$ to $\overset{ˉ}{x}$ and $1$ to $\overset{ˉ}{y}$ . The repeated slots in $z_{2} = z_{1} z_{1}$ add $z_{1} + z_{1} = 6$ to ${\overset{ˉ}{z}}_{1}$ ; then $z_{1} = x + y$ adds $6$ to each input. The total input cotangents are $(8, 7)$ , consistent with [F3].

3.1step 1.1step 2.1step 2.2∎

The forward value $11$ and both derivative calculations agree with the stated output and gradient.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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

MIT OpenCourseWare 18.S096, Matrix Calculus for Machine Learning and Beyond, complete lecture notes, Chapter 8 §8.3 (standard reference, not scraped)