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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.

Recurrent, LSTM, and structured state-space models — Examples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

A scalar LSTM cell with constant gates

Example

Take dx=dh=1, set every LSTM weight and bias to zero, allow any scalar input sequence and any initial values h0,c0∈R. Then for every t≥1, the input, forget, and output gates equal 1/2, the candidate update is zero, and

ct=2−tc0,ht=12tanh⁡(2−tc0).

Facts & Assumptions

Given: The scalar LSTM, zero weights and biases, arbitrary scalar inputs, and initial values h0,c0 as in the Example.

[F1]

Standard LSTM cell: each gate's argument is its input-weight matrix times xt, plus its hidden-state matrix times ht−1, plus its bias; the gates use σ(s)=(1+e−s)−1 and coordinatewise tanh⁡, with ct=ft⊙ct−1+it⊙gt and ht=ot⊙tanh⁡(ct).

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 it=ft=ot=1/2 and gt=0. Substitution into the cell recurrence gives ct=ct−1/2 and ht=12tanh⁡(ct).

2.1givenstep 1.1inductionalgebra∎

The recurrence in step 1.1 gives c1=2−1c0. If ct−1=2−(t−1)c0, then ct=ct−1/2=2−tc0; induction proves this for every t≥1. Substituting this expression into the hidden-state identity in step 1.1 gives ht=12tanh⁡(2−tc0).

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

A scalar state-space layer has matching recurrent and convolutional outputs

Example

Take the scalar state-space model A=0, B=C=1, direct term D=0, step size Δ=1, and initial state x−1=0. For any integer L≥1 and any length-L scalar input sequence (u0,…,uL−1), the bilinear matrices are Aˉ=1 and Bˉ=1, so the recurrence gives xt=xt−1+ut and yt=∑s=0tus for 0≤t<L. The length-L impulse kernel is (1,…,1), and its causal convolution ∑j=0tKjut−j gives the same outputs for 0≤t<L.

Facts & Assumptions

Given: The scalar parameters and zero initial state in the Example, with a length-L input sequence.

[F1]

DPLR structured state-space parameterization: when I−ΔA/2 is invertible, Aˉ=(I−ΔA/2)−1(I+ΔA/2), Bˉ=(I−ΔA/2)−1ΔB, and the zero-state kernel has entries C∗AˉjBˉ.

[F2]

S4 state-space recurrence and convolution procedure: the recurrent mode solves (I−ΔA/2)xt=(I+ΔA/2)xt−1+ΔBut and returns C∗xt.

Proof

1.1givenF1algebra

Here I−ΔA/2=1 and I+ΔA/2=1, so [F1] gives Aˉ=1 and Bˉ=1.

2.1givenF2step 1.1induction

By [F2] and step 1.1, xt=xt−1+ut and yt=xt. The base value is x−1=0; if xt−1=∑s=0t−1us, then xt=xt−1+ut=∑s=0tus. Induction proves yt=∑s=0tus.

3.1F1step 1.1step 2.1algebra∎

By [F1] and step 1.1, every kernel coefficient is C∗AˉjBˉ=1. Therefore its causal convolution at time t is ∑j=0tut−j=∑s=0tus, which equals the recurrent output in step 2.1 after substituting s=t−j.

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

The filter-operation bound is not a training-cost bound

Example

Assume N,r,L≥1 with fixed r, A=Λ−PQ∗ for diagonal Λ∈CN×N and P,Q∈CN×r, B,C∈CN, and Δ>0. Let G=I−ΔA/2 be invertible, let M be the least power of two at least L, set Aˉ=G−1(I+ΔA/2) and Bˉ=G−1ΔB, and supply C~M=(I−AˉM)∗C. Assume the root-grid diagonal resolvents R(z)=(a(1−z)/(1+z)I−Λ)−1 and corrections Ir+Q∗R(z)P are invertible for every Mth root z≠−1, where a=2/Δ; the value at −1, when that root occurs, is handled by its finite formula. In the exact complex arithmetic model, each scalar addition, multiplication, division, and zero test has unit cost, and the filter cost counts evaluation from the supplied C~M, not its formation. A nonzero direct input-to-output term is separate from the kernel.

For numerical evaluation, at each root z≠−1 put β(z)=(1+z)/2 and D(z)=2R(z)/(1+z), and define x=C~M∗DB, u=C~M∗DP, v=Q∗DB, T=Q∗DP, and W=Ir+βT. Matrix infinity norm is maximum absolute row sum, and for a column it is maximum absolute entry. Suppose certified constants b,w,s≥1 bound ∣x∣,∥u∥∞,∥v∥∞,∥T∥∞ by b and ∥W−1∥∞ by w over these roots. For every scalar Cauchy sum used for a scaled block, supply a certified pole/root separation d>0 and a weight bound Awt≥∑j∣aj∣ with Awt/d≤s; also supply the exact index partition λj=−a or λj≠−a so the exceptional constant terms are selected, and form the finite constants and the −1 value separately. Empty Cauchy sums need no separation certificate. Put κ=64rw2(b+1)2s.

For 0<ε≤1/2, assume variable-precision scalar arithmetic with unrestricted exponent range, certified input, root and arc-center approximations, certified enclosures sufficient for the geometric routing decisions, and the finite rational-circuit precision certificate. Explicitly, for the selected finite circuit let T0 bound its arithmetic gates, V0≥1 bound exact intermediate moduli, and 0<D0≤1 bound below every exact divisor modulus; with K0=64(V0+1)2/D02, choose u≤min⁡(1,D0/2,ε/2)/(2K0T0+1). Each supplied input/constant approximation and each scalar arithmetic operation has absolute error at most u; each scalar operation and word access has unit cost. Precision and word lengths may depend on the input separation, correction conditioning, intermediate magnitudes and target accuracy. The operation and word counts exclude bit cost, computing the supplied certificates, and forming C~M.

For the stated DPLR data, the length-L kernel is KL=(C∗AˉjBˉ)j=0L−1. From the supplied C~M, the exact filter procedure evaluates it using at most (r+1)2 roots-of-unity Cauchy products, O((N+L)log⁡2(2+N+L)) exact complex arithmetic operations, and O(N+L) working space. Under the numerical hypotheses above, for 0<ε≤1/2 it returns K^L with ∥K^L−KL∥∞≤ε using O((N+L)log⁡(2+N+L)log⁡(κ/ε)) scalar operations and O(N+L) scalar words; bounded-κ families have the factor log⁡(1/ε). These evaluation counts exclude forming C~M, computing the supplied certificates, and bit cost. The filter procedure does not compute parameter gradients or optimizer updates, so its count is not a bound for a complete training step. Neither the exact arithmetic count nor the conditional numerical guarantee supplies a fixed-word bound or an empirical long-context accuracy guarantee.

Facts & Assumptions

Given: Fixed DPLR parameters, a supplied transformed output vector, and the exact and certified numerical hypotheses in the Example.

[F1]

S4 convolution-filter arithmetic and numerical cost: the exact operation and working-space bounds count evaluation of the filter from supplied C~M; numerical absolute error ε is proved with explicit separation, correction conditioning and variable precision, and a log⁡(κ/ε) factor that reduces to log⁡(1/ε) on bounded-κ families.

[F2]

S4 state-space recurrence and convolution procedure: the filter procedure takes DPLR parameters and C~M and returns the first L kernel coefficients after root evaluation and an inverse transform.

Proof

1.1givenF1F2

The procedure in [F2] receives parameters and returns a kernel. By [F1], the proved exact operation and space bounds count that filter-evaluation procedure with C~M supplied.

2.1F2step 1.1algebra

The procedure's input/output specification in [F2] contains no parameter-gradient computation or optimizer update. Therefore its operation count does not count a larger computation that additionally performs either operation, including a complete training step.

3.1F1step 1.1algebra∎

The exact bound counts arithmetic operations and working space only. The separate numerical guarantee in [F1] specifies its absolute error, conditioning factor and variable-precision scalar model; it requires its own certified hypotheses. Its scalar-operation and word-space counts do not bound bit cost or guarantee a fixed word length, and neither cost theorem asserts empirical long-context accuracy. These additional conclusions therefore do not follow from the filter bounds.

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