Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pending
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.

Position-free versus position-encoded token order

Example

Assume the Axiom of Countable Choice. Let X=(x1,x2)∈[0,1]1×2 and define the constant ordered-output map f(X)=(0,1). A position-free unmasked self-attention stack followed by shared tokenwise maps must permute its two output entries when the two input tokens are exchanged, whereas a trainable positional matrix can distinguish the first and second token positions. On the compact square [0,1]2, the position-encoded fixed-length theorem therefore approximates f in integrated Lp error for every 1≤p<∞.

Facts & Assumptions

Given: Countable Choice, the two-token input X, the coordinate-swap permutation P, the constant map f, and position-free or position-encoded Transformer stacks as in the Example.

[F1]

The row-token theorem gives R(PZ)=PR(Z). In the column-token convention used here, define T(X)=R(XT)T; transposition gives T(XP)=T(X)P for every unmasked position-free stack with shared tokenwise maps (Position-free unmasked self-attention is permutation equivariant).

[F2]

Assuming Countable Choice, for fixed n≥2, d≥1, and 1≤p<∞, a two-head, head-size-one Transformer with width-four ReLU feedforward layers, no layer normalization, and a trainable positional matrix approximates every continuous map on a compact domain in integrated Lp error, for 1≤p<∞ (Fixed-width Transformer approximation on compact domains).

Proof

1.1givenF1algebra

Let P swap the two columns. Since f is constant, f(XP)=(0,1), while f(X)P=(1,0). Hence f(XP)≠f(X)P, so f is not permutation equivariant. By [F1], no position-free stack can equal f on both X and XP.

2.1givenF2∎

The map f is continuous on K=[0,1]1×2, which is compact, and n=2,d=1 meet the size hypotheses. Under the assumed Countable Choice, the position-encoded compact-domain branch of [F2] with any prescribed ε>0 gives a position-encoded stack whose integrated Lp error from f is less than ε. Thus positional tags remove the equivariance restriction for this fixed-length approximation problem.

Depends on

Used by

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Dependency tree · two levels

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Sources