{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-catalan-numbers-01",
  "mko_id": "mko-catalan-numbers",
  "title": {
    "zh-Hant": "Catalan卷積與整除證書",
    "en": "Catalan convolution and a divisibility certificate"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-catalan-numbers-01",
    "source": {
      "mko_id": "mko-catalan-numbers",
      "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f",
      "sha256": "2d847df82bf2c534d59aedb78304a0f59d94a87bf4454e0eee1886df489e21b5",
      "english_sha256": "328d37ca081c47ad3e92abe05ddd5028dfeb23321136b79cc8b3d101c89e814d"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "work"
    ],
    "goal": {
      "zh-Hant": "從C0完整生成C4並核兩個closed forms。",
      "en": "Generate C4 from C0 and check both closed forms."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "C0=1，所有n≥0用C_(n+1)=sum_(i=0)^n C_i C_(n-i)。本例比較n=4的兩個已引用Catalan closed forms。",
          "en": "C0=1 and for every n≥0 use C_(n+1)=sum_(i=0)^n C_i C_(n-i). Compare the two cited Catalan closed forms at n=4."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "work"
        ],
        "before_tex": "C_0=1",
        "after_tex": "C_1=C_0C_0=1",
        "relation": "implies",
        "rule": "work",
        "conditions": [],
        "explanation": {
          "zh-Hant": "先固定從0起算，不混C1 convention。",
          "en": "Fix zero-based indexing first rather than mixing C1 conventions."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "work"
          ],
          "explanation": {
            "zh-Hant": "空物件的C0不是零。",
            "en": "C0 for the empty object is not zero."
          }
        },
        "leaf_basis": "work"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "S01",
          "work"
        ],
        "before_tex": "C_2=C_0C_1+C_1C_0",
        "after_tex": "C_2=2",
        "relation": "implies",
        "rule": "work",
        "conditions": [],
        "explanation": {
          "zh-Hant": "n=1完整兩split。",
          "en": "At n=1 include both splits."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01",
            "work"
          ],
          "explanation": {
            "zh-Hant": "兩項相同仍各有一個index。",
            "en": "Equal terms still have separate indices."
          }
        },
        "leaf_basis": "work"
      },
      {
        "id": "S03",
        "uses": [
          "G1",
          "S02",
          "work"
        ],
        "before_tex": "C_3=C_0C_2+C_1C_1+C_2C_0",
        "after_tex": "C_3=2+1+2=5",
        "relation": "implies",
        "rule": "work",
        "conditions": [],
        "explanation": {
          "zh-Hant": "n=2的三split均列出。",
          "en": "List all three splits at n=2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S02",
            "work"
          ],
          "explanation": {
            "zh-Hant": "中間split不可漏。",
            "en": "Do not omit the middle split."
          }
        },
        "leaf_basis": "work"
      },
      {
        "id": "S04",
        "uses": [
          "G1",
          "S03",
          "work"
        ],
        "before_tex": "C_4=C_0C_3+C_1C_2+C_2C_1+C_3C_0",
        "after_tex": "C_4=5+2+2+5=14",
        "relation": "implies",
        "rule": "work",
        "conditions": [],
        "explanation": {
          "zh-Hant": "n=3的四split完整求得14。",
          "en": "All four splits at n=3 give 14."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S03",
            "work"
          ],
          "explanation": {
            "zh-Hant": "不是只加左右兩端得10。",
            "en": "Adding only endpoints would give the wrong value 10."
          }
        },
        "leaf_basis": "work"
      },
      {
        "id": "S05",
        "uses": [
          "G1",
          "S04",
          "work"
        ],
        "before_tex": "\\binom84=8!/(4!4!)",
        "after_tex": "\\binom84=70,\\quad70\\bmod5=0",
        "relation": "implies",
        "rule": "work",
        "conditions": [],
        "explanation": {
          "zh-Hant": "先算central binomial並核餘數零，再允許整數除法。",
          "en": "Compute the central binomial coefficient and certify remainder zero before integer division."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S04",
            "work"
          ],
          "explanation": {
            "zh-Hant": "不可用截斷商冒充整除證書。",
            "en": "A truncated quotient is not a divisibility certificate."
          }
        },
        "leaf_basis": "work"
      },
      {
        "id": "S06",
        "uses": [
          "G1",
          "S05",
          "work"
        ],
        "before_tex": "\\binom84/(4+1)",
        "after_tex": "70/5=14=C_4",
        "relation": "implies",
        "rule": "work",
        "conditions": [],
        "explanation": {
          "zh-Hant": "有整除資格後核第一closed form在n=4的值。",
          "en": "After divisibility, check the first closed form at n=4."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S05",
            "work"
          ],
          "explanation": {
            "zh-Hant": "此數值吻合不單獨證closed form對全部n。",
            "en": "This numerical agreement alone does not prove the closed form for every n."
          }
        },
        "leaf_basis": "work"
      },
      {
        "id": "S07",
        "uses": [
          "G1",
          "S06",
          "work"
        ],
        "before_tex": "\\binom85=56",
        "after_tex": "\\binom84-\\binom85=70-56=14",
        "relation": "implies",
        "rule": "work",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第二closed form給同一整數並獨立核對。",
          "en": "The second closed form gives the same integer as a cross-check."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S06",
            "work"
          ],
          "explanation": {
            "zh-Hant": "下標n+1=5不能誤用n-1。",
            "en": "The lower index n+1=5 must not be replaced by n-1."
          }
        },
        "leaf_basis": "work"
      },
      {
        "id": "S08",
        "uses": [
          "G1",
          "S07",
          "work"
        ],
        "before_tex": "\\binom{2n}{n+1}=\\binom{2n}{n}\\frac{n}{n+1}\\quad(n\\ge0)",
        "after_tex": "\\binom{2n}{n}-\\binom{2n}{n+1}=\\frac1{n+1}\\binom{2n}{n}\\in\\mathbb Z",
        "relation": "implies",
        "rule": "work",
        "conditions": [],
        "explanation": {
          "zh-Hant": "factorial比值給identity；整數差證一般整除性，n=0用comb(0,1)=0。",
          "en": "The factorial ratio gives the identity; an integer difference proves divisibility, with comb(0,1)=0 at n=0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S07",
            "work"
          ],
          "explanation": {
            "zh-Hant": "這證integrality，不把它誤稱卷積closed form的完整一般證明。",
            "en": "This proves integrality, not the entire general closed-form solution of the convolution."
          }
        },
        "leaf_basis": "work"
      }
    ],
    "result": {
      "zh-Hant": "C4=14，central coefficient70可被5整除，且70-56=14。",
      "en": "C4=14, central coefficient 70 is divisible by 5, and 70-56=14."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "C0=0仍給相同序列。",
          "en": "C0=0 gives the same sequence."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "第一卷積即給C1=0而不是1。",
          "en": "The first convolution would give C1=0 rather than 1."
        }
      },
      {
        "claim": {
          "zh-Hant": "70//5可代替餘數檢查的一般責任。",
          "en": "70//5 replaces the general remainder obligation."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "截斷運算本身不證整除，須另核remainder0。",
          "en": "Truncation itself proves no divisibility; remainder zero must be checked."
        }
      },
      {
        "claim": {
          "zh-Hant": "C4只需C0C3+C3C0=10。",
          "en": "C4 only needs C0C3+C3C0=10."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "中間兩split各給2，正確14。",
          "en": "The two middle splits each add2, giving 14."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "mko_definition",
        "kind": "cited",
        "statement": {
          "zh-Hant": "C_(n+1)=sum_{i=0}^n C_i C_(n-i)；C_n=binomial(2n,n)/(n+1)=binomial(2n,n)-binomial(2n,n+1)。division前必須驗(n+1)整除binomial(2n,n)，不可依integer truncation。domain n≥0且不可混用C_0/C_1 index convention。",
          "en": "C_(n+1)=sum over i=0..n of C_i*C_(n−i), and C_n=binomial(2n,n)/(n+1)=binomial(2n,n)−binomial(2n,n+1). Before division, certify that n+1 divides the central binomial coefficient; integer truncation is not a substitute. The domain is n≥0 and the C_0/C_1 indexing conventions must not be mixed."
        },
        "source": {
          "mko_id": "mko-catalan-numbers",
          "sha256": "2d847df82bf2c534d59aedb78304a0f59d94a87bf4454e0eee1886df489e21b5",
          "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f"
        }
      },
      {
        "id": "work",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限卷積定義；binomial factorial identity與超過上標時comb=0 convention；Catalan兩closed forms以本MKO為已引用事實，非此例證明的一般定理。",
          "en": "Finite convolution definition; binomial factorial identities and comb=0 beyond its upper index; the Catalan closed forms are cited MKO facts, not general theorems proved by this example."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N01",
        "lhs": "5+2+2+5",
        "rhs": "14"
      },
      {
        "id": "N02",
        "lhs": "comb(8,4)%5",
        "rhs": "0"
      },
      {
        "id": "N03",
        "lhs": "comb(8,4)/5",
        "rhs": "14"
      },
      {
        "id": "N04",
        "lhs": "comb(8,4)-comb(8,5)",
        "rhs": "14"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "n=4 exact例與一般整除identity；不證所有Catalan bijections。",
      "en": "An exact n=4 instance and general integrality identity, not every Catalan bijection."
    },
    "md_symbols": [
      [
        "C_n",
        "0-based Catalan序列",
        "zero-based Catalan sequence"
      ]
    ],
    "md_error_controls": [
      {
        "title": "index / Index",
        "zh": "C4使用n=3的卷積與n=4的closed form。",
        "en": "C4 uses n=3 in the convolution and n=4 in the closed form."
      },
      {
        "title": "exact除法 / Exact division",
        "zh": "先70 mod5=0，後70/5=14。",
        "en": "Check70 mod5=0 before 70/5=14."
      }
    ],
    "md_open_items": []
  },
  "revision": {
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    "collection": "coverage-20261021",
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    "original_source_sha256": "ed47d97d69cfdf0cb831c56ec2b237fbcd2f49986cf1490c20647d56623f8552",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "reader_review": "142_of_142_source_bound",
    "wording_review": "original_Lead_decisions_resolved",
    "glossary_reader_paraphrases": [],
    "independent_AI_concurrence": "SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEW",
    "issue_ids": [
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    "parent_manifest_sha256": "d29f4cae8229ebd24f60d496cb1d3cd4ad8c0a4dca9d44daab0c3245c544283a",
    "scope": "Owner-adopted reader revision. Lead and bilingual review concern the recorded text and premises; no new universal or Lean proof, canonical promotion or historical-label rewrite."
  }
}