Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge 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.

Complementarity, momentum balance, and energy loss at a scalar impact

Statement

For any resolved output of First-impact step for a one-dimensional unilateral contact, let q∗ be the reported gap (zero at an impact, the free-flight endpoint gap otherwise), let J be the reported impulse, and let v∗ be the reported endpoint velocity. Then q∗≥0, J≥0, and q∗J=0. On a free-flight branch, J=0 and mv∗=mv+mah. On an impact branch at time s, with incoming velocity v−=v+as, the post-impact velocity satisfies mv+=mv−+J and 12m(v+)2−12m(v−)2=−12m(1−e2)(v−)2≤0. When a=0, the full resolved step has nonincreasing kinetic energy. For a≠0, the displayed energy statement concerns only the instantaneous impact jump; work during free flight is not covered by it.

Facts & Assumptions

Given: The inputs and resolved output of the algorithm in the Statement.

[F1]

First-impact step for a one-dimensional unilateral contact: an impact output has q∗=0, v−<0, J=−m(1+e)v−, and v+=−ev−; a free-flight output has J=0, q∗=q+hv+12ah2≥0, and v∗=v+ah.

Proof

1.1F1givenalgebra

On an impact branch, [F1] gives q∗=0, J=−m(1+e)v−>0, and v+=−ev−≥0, since m>0, e∈[0,1], and v−<0. Hence q∗J=0. On a free-flight branch, [F1] gives q∗≥0 and J=0, so complementarity also holds.

1.2F1givenalgebra

On a free-flight branch, mv∗=m(v+ah)=mv+mah. On an impact branch, [F1] gives mv−=m(v+as) and J=−m(1+e)v−. Therefore mv−+J=mv−−m(1+e)v−=−mev−=mv+, using v+=−ev−.

2.1F1step 1.1givenalgebra∎

On an impact branch, [F1] gives (v+)2=e2(v−)2, so the kinetic-energy difference is 12m(e2−1)(v−)2=−12m(1−e2)(v−)2≤0. If a=0 and the branch is free flight, v∗=v and kinetic energy is unchanged; impact branches have the nonincrease just established. For nonzero a, no claim is made about total free-flight work.

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