ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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.

A scalar LSTM cell with constant gates

Example

Take $d_{x} = d_{h} = 1$ , set every LSTM weight and bias to zero, allow any scalar input sequence and any initial values $h_{0}, c_{0} \in R$ . Then for every $t \geq 1$ , the input, forget, and output gates equal $1 / 2$ , the candidate update is zero, and

c_{t} = 2^{- t} c_{0}, h_{t} = \frac{1}{2} \tanh (2^{- t} c_{0}) .

Example references

Standard LSTM cell

Goodfellow, Bengio, and Courville, Deep Learning, Chapter 10, §10.10.1

Facts & Assumptions

Given: The scalar LSTM, zero weights and biases, arbitrary scalar inputs, and initial values $h_{0}, c_{0}$ as in the Example.

[F1]

Standard LSTM cell: each gate's argument is its input-weight matrix times $x_{t}$ , plus its hidden-state matrix times $h_{t - 1}$ , plus its bias; the gates use $σ (s) = (1 + e^{- s})^{- 1}$ and coordinatewise $\tanh$ , with $c_{t} = f_{t} ⊙ c_{t - 1} + i_{t} ⊙ g_{t}$ and $h_{t} = o_{t} ⊙ \tanh (c_{t})$ .

Proof

1.1givenF1algebra

Every gate's affine argument is zero because all weights and biases vanish. By [F1], $σ (0) = 1 / (1 + 1) = 1 / 2$ and $\tanh (0) = 0$ , so $i_{t} = f_{t} = o_{t} = 1 / 2$ and $g_{t} = 0$ . Substitution into the cell recurrence gives $c_{t} = c_{t - 1} / 2$ and $h_{t} = \frac{1}{2} \tanh (c_{t})$ .

2.1givenstep 1.1inductionalgebra∎

The recurrence in step 1.1 gives $c_{1} = 2^{- 1} c_{0}$ . If $c_{t - 1} = 2^{- (t - 1)} c_{0}$ , then $c_{t} = c_{t - 1} / 2 = 2^{- t} c_{0}$ ; induction proves this for every $t \geq 1$ . Substituting this expression into the hidden-state identity in step 1.1 gives $h_{t} = \frac{1}{2} \tanh (2^{- t} c_{0})$ .

Depends on

Standard LSTM cell

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

2 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

Goodfellow, Bengio, and Courville, Deep Learning, Chapter 10, §10.10.1 (standard reference, not scraped)