[#R865] Fibonacci support four-cycle classification
1Summary
Lean proves that every support square has corner sums q_(t-2), q_t, q_t, q_(t+1) and equal row and column increments q_(t-1).
The proof is kernel-checked in the pinned world and has no local sorry. It first removes the duplicated initial Fibonacci value, proves the required index arithmetic, and then lifts the result to matrix coordinates.
Reported evidence. Replay readiness: partial.
2Verification
Verification material: partial
Part of the replay path is recorded. Check the missing fields before comparing a new run.
- Runtime
- lean-4.33.0-rc1/mathlib4@4608056c77c52468b80773e8dcd585ef821c7c5e+theoremdb@d575c4e2ff28345440c4f8a42bf0178bcb3f6f41b703a45d9d1cbb709036f0dc
Verification source: mathoverflow.net ↗, formal/lean/TheoremDB/Fibonacci/Graph.lean
3Formal statement
lean
lean-4.33.0-rc1/mathlib4@4608056c77c52468b80773e8dcd585ef821c7c5e+theoremdb@d575c4e2ff28345440c4f8a42bf0178bcb3f6f41b703a45d9d1cbb709036f0dctheorem fibonacci_support_four_corner_classification {x X y Y : ℕ} (_hx : x < X) (hy : y < Y) (hmiddle : x + Y ≤ X + y) (hA : IsFibonacci (x + y + 2)) (hB : IsFibonacci (x + Y + 2)) (hC : IsFibonacci (X + y + 2)) (hD : IsFibonacci (X + Y + 2)) : ∃ t : ℕ, 3 ≤ t ∧ x + y + 2 = positiveFib (t - 2) ∧ x + Y + 2 = positiveFib t ∧ X + y + 2 = positiveFib t ∧ X + Y + 2 = positiveFib (t + 1) ∧ X = x + positiveFib (t - 1) ∧ Y = y + positiveFib (t - 1) := by
-- Checked proof in formal/lean/TheoremDB/Fibonacci/Graph.lean4What was measured
- Verification statement
- theorem TheoremDB.Fibonacci.fibonacci_support_four_corner_classification {x X y Y : ℕ} (_hx : x < X) (hy : y < Y) (hmiddle : x + Y ≤ X + y) (hA : IsFibonacci (x + y + 2)) (hB : IsFibonacci (x + Y + 2)) (hC : IsFibonacci (X + y + 2)) (hD : IsFibonacci (X + Y + 2)) : ∃ t : ℕ, 3 ≤ t ∧ x + y + 2 = positiveFib (t - 2) ∧ x + Y + 2 = positiveFib t ∧ X + y + 2 = positiveFib t ∧ X + Y + 2 = positiveFib (t + 1) ∧ X = x + positiveFib (t - 1) ∧ Y = y + positiveFib (t - 1)
5How it connects
Depended on by
- formalization
Depends on
- formalization
6Agent packet
A compact handoff with the evidence boundary, replay manifest, and relation pointers.
View structured packet
json
{
"schema": "theoremdb-agent-record-v1",
"ref": "R865",
"content_hash": null,
"slug": "fib-formalization-four-corner-classification",
"type": "formalization",
"title": "Fibonacci support four-cycle classification",
"summary": "Lean proves that every support square has corner sums q_(t-2), q_t, q_t, q_(t+1) and equal row and column increments q_(t-1).",
"relevance": "For fib problem determinant range; fib problem nonzero support, record fib-formalization-four-corner-classification (“Fibonacci support four-cycle classification”) states a machine-checkable theorem or proof obligation. The record states: Lean proves that every support square has corner sums q_(t-2), q_t, q_t, q_(t+1) and equal row and column increments q_(t-1).",
"relevance_source": "recorded",
"body": "The proof is kernel-checked in the pinned world and has no local sorry. It first removes the duplicated initial Fibonacci value, proves the required index arithmetic, and then lifts the result to matrix coordinates.",
"status": "draft",
"evidence_grade": "unverified_formalization",
"scope": null,
"reproduction": {
"schema": "theoremdb-reproduction-v1",
"readiness": "partial",
"kind": "formalization",
"runtime": "lean-4.33.0-rc1/mathlib4@4608056c77c52468b80773e8dcd585ef821c7c5e+theoremdb@d575c4e2ff28345440c4f8a42bf0178bcb3f6f41b703a45d9d1cbb709036f0dc",
"citation": {
"url": "https://mathoverflow.net/questions/513340/is-the-determinant-of-this-fibonacci-sum-indicator-matrix-always-1-0-or/513372",
"locator": "formal/lean/TheoremDB/Fibonacci/Graph.lean"
},
"missing": [
"source",
"command",
"expected_output"
]
},
"formal_statement": "theorem fibonacci_support_four_corner_classification {x X y Y : ℕ} (_hx : x < X) (hy : y < Y) (hmiddle : x + Y ≤ X + y) (hA : IsFibonacci (x + y + 2)) (hB : IsFibonacci (x + Y + 2)) (hC : IsFibonacci (X + y + 2)) (hD : IsFibonacci (X + Y + 2)) : ∃ t : ℕ, 3 ≤ t ∧ x + y + 2 = positiveFib (t - 2) ∧ x + Y + 2 = positiveFib t ∧ X + y + 2 = positiveFib t ∧ X + Y + 2 = positiveFib (t + 1) ∧ X = x + positiveFib (t - 1) ∧ Y = y + positiveFib (t - 1) := by\n -- Checked proof in formal/lean/TheoremDB/Fibonacci/Graph.lean",
"source": {
"url": "https://mathoverflow.net/questions/513340/is-the-determinant-of-this-fibonacci-sum-indicator-matrix-always-1-0-or/513372",
"locator": "formal/lean/TheoremDB/Fibonacci/Graph.lean"
},
"relations": [
{
"slug": "R863",
"title": "Fibonacci odd square cover",
"object_type": "formalization",
"relation": "depends_on",
"direction": "incoming"
},
{
"slug": "R904",
"title": "Lean definition of the Fibonacci-sum matrix",
"object_type": "formalization",
"relation": "depends_on",
"direction": "outgoing"
}
]
}7Provenance
View source, identifiers, and projection details
- Project
- fibonacci-sum-determinant
- Locator
- formal/lean/TheoremDB/Fibonacci/Graph.lean
- License
- CC-BY-SA-4.0
- Contributors
- Philip Weiss
- Source
- mathoverflow.net ↗
- Public record
- R865
- Stable alias
- fib-formalization-four-corner-classification
- Projection
- Reproduction fields are derived from the immutable record.
A machine-checkable rendering of a statement, with the world it was written against.