Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)
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.

Finite subtraction ambiguity

Statement

Let x=(xN)N≥1 and r=(rN)N≥1 be real sequences, and let L∈R. Here xN−rN→L means that for every ε>0 there is an integer N0 such that ∣xN−rN−L∣<ε for all N≥N0. For every a∈R, the equally valid subtraction rN′=rN−a satisfies

xN−rN′⟶L+a.

Thus existence of a finite subtracted limit alone does not select its additive finite part; no particular divergence expansion or physical renormalization scheme is assumed.

Facts & Assumptions

Given: Real sequences x=(xN)N≥1 and r=(rN)N≥1, a finite real number L, and xN−rN→L in the epsilon definition stated above.

[A1]

For every ε>0, some N0 satisfies ∣xN−rN−L∣<ε for all N≥N0.

Proof

1.1givenA1choose

Fix any real a and any ε>0. By [A1], choose N0 so that ∣xN−rN−L∣<ε for every N≥N0.

2.1givenstep 1.1algebra∎

For every N≥N0, ∣xN−rN′−(L+a)∣=∣xN−(rN−a)−(L+a)∣=∣xN−rN−L∣<ε. Since a and ε were arbitrary, the claimed limit is L+a for every a∈R.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources