{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-mobius-inversion-finite-divisors-01",
  "mko_id": "mko-mobius-inversion-finite-divisors",
  "title": {
    "zh-Hant": "仿射函數在divisor poset的反演",
    "en": "Divisor-poset inversion of an affine function"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-mobius-inversion-finite-divisors-01",
    "source": {
      "mko_id": "mko-mobius-inversion-finite-divisors",
      "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f",
      "sha256": "903f02acdd799627c36103659ee206c7907a68eedbfb36d6a35a0e9dc6a2ef98",
      "english_sha256": "addf39b281bccdf34cabbe47f1f285b27a3040a3677596baafb9907ab773011d"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "calc"
    ],
    "goal": {
      "zh-Hant": "f(k)=2k+1、n=6時，F(6)=28且反演得到f(6)=13；n=1亦roundtrip。",
      "en": "For f(k)=2k+1 at n=6, F(6)=28 and inversion returns f(6)=13; n=1 roundtrips too."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "f(k)=2k+1，F(m)=Σ_{d|m}f(d)；n=6，使用μ(d)F(6/d)索引。",
          "en": "f(k)=2k+1, F(m)=sum of f(d) over d dividing m; n=6, with indexing mu(d)*F(6/d)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1"
        ],
        "before_tex": "6=2\\cdot3",
        "after_tex": "D(6)=\\{1,2,3,6\\}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "完整prime分解兩指數0或1產生全部四divisors。",
          "en": "The complete factorization and exponents zero or one give all four divisors."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1"
          ],
          "explanation": {
            "zh-Hant": "4、5雖小於6卻不整除6，不能混入。",
            "en": "4 and 5 are below 6 but do not divide it."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "S01"
        ],
        "before_tex": "d=1,2,3,6",
        "after_tex": "f(d)=3,5,7,13\\text{ respectively}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依同一仿射公式逐個精確代入。",
          "en": "Evaluate the same affine formula exactly at every divisor."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01"
          ],
          "explanation": {
            "zh-Hant": "只需這四個f值，不是把1至6全加總。",
            "en": "Only these four f-values are used, not all integers 1 through 6."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S03",
        "uses": [
          "G1",
          "S01",
          "S02"
        ],
        "before_tex": "D(1)=\\{1\\},\\ D(2)=\\{1,2\\},\\ D(3)=\\{1,3\\}",
        "after_tex": "F(1)=3,\\quad F(2)=3+5=8,\\quad F(3)=3+7=10",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "為反演所需每個quotient建立完整forward sums。",
          "en": "Build complete forward sums for each quotient needed by inversion."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01",
            "S02"
          ],
          "explanation": {
            "zh-Hant": "F(3)不含f(2)，因2不整除3。",
            "en": "F(3) excludes f(2) because 2 does not divide 3."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S04",
        "uses": [
          "G1",
          "S01",
          "S02"
        ],
        "before_tex": "D(6)=\\{1,2,3,6\\}",
        "after_tex": "F(6)=3+5+7+13=28",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "完整forward transform包含1與n自身。",
          "en": "The full forward transform includes 1 and n itself."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01",
            "S02"
          ],
          "explanation": {
            "zh-Hant": "只加proper divisors會得15而非28，破壞roundtrip。",
            "en": "Using only proper divisors gives 15, not 28, breaking the roundtrip."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S05",
        "uses": [
          "G1",
          "S01"
        ],
        "before_tex": "d=1,2,3,6",
        "after_tex": "\\mu(d)=1,-1,-1,1;\\quad6/d=6,3,2,1",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "完整UPF給四個μ值，同時列出配對quotients避免索引混合。",
          "en": "Complete factorization gives all four mu values, with paired quotients to prevent index mixing."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01"
          ],
          "explanation": {
            "zh-Hant": "d=6對F(1)，不是對F(6)。",
            "en": "d=6 pairs with F(1), not F(6)."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S06",
        "uses": [
          "G1",
          "S03",
          "S04",
          "S05"
        ],
        "before_tex": "\\sum_{d\\mid6}\\mu(d)F(6/d)",
        "after_tex": "28-10-8+3=13=f(6)",
        "relation": "implies",
        "rule": "mko_definition",
        "conditions": [],
        "explanation": {
          "zh-Hant": "四項精確逆變換重建原值13。",
          "en": "The exact four-term inverse transform reconstructs 13."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S03",
            "S04",
            "S05"
          ],
          "explanation": {
            "zh-Hant": "若改用μ(6/d)F(d)整體換索引亦可，但不能只換半邊。",
            "en": "The fully reindexed mu(6/d)F(d) is also valid; changing only one side is not."
          }
        },
        "leaf_basis": "mko_definition"
      },
      {
        "id": "S07",
        "uses": [
          "G1",
          "S03",
          "S04",
          "S06"
        ],
        "before_tex": "F(6)-F(3)-F(2)+F(1)",
        "after_tex": "f(1)(1-1-1+1)+f(2)(1-1)+f(3)(1-1)+f(6)=f(6)",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "展開所有forward sums，直接看低divisor項係數消去，解釋本例反演機制。",
          "en": "Expand all forward sums to see lower-divisor coefficients cancel, explaining inversion in this instance."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S03",
            "S04",
            "S06"
          ],
          "explanation": {
            "zh-Hant": "每個f只依divisibility出現，不是一般d≤n前綴和。",
            "en": "Terms appear according to divisibility, not an ordinary d≤n prefix sum."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S08",
        "uses": [
          "G1",
          "S03",
          "S05",
          "S07"
        ],
        "before_tex": "n=1,\\quad D(1)=\\{1\\}",
        "after_tex": "F(1)=f(1)=3,\\quad\\mu(1)F(1)=3",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "最小domain點保留μ(1)=1，roundtrip完整。",
          "en": "The smallest domain point retains mu(1)=1 and roundtrips exactly."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S03",
            "S05",
            "S07"
          ],
          "explanation": {
            "zh-Hant": "0、負數或不完整divisors不在此證書域。",
            "en": "Zero, negatives and incomplete divisor lists are outside this certificate."
          }
        },
        "leaf_basis": "calc"
      }
    ],
    "result": {
      "zh-Hant": "f(k)=2k+1、n=6時，F(6)=28且反演得到f(6)=13；n=1亦roundtrip。",
      "en": "For f(k)=2k+1 at n=6, F(6)=28 and inversion returns f(6)=13; n=1 roundtrips too."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "F(6)要加f(4)、f(5)。",
          "en": "F(6) includes f(4),f(5)."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "4與5不是6的divisors。",
          "en": "4 and 5 are not divisors of 6."
        }
      },
      {
        "claim": {
          "zh-Hant": "反演可把每項都配F(d)但保留μ(d)。",
          "en": "Use F(d) with unchanged mu(d) in every inverse term."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "3−8−10+28=13在本例巧合成立，但n=2時3−8=−5≠5；正確需同步換μ為μ(n/d)。",
          "en": "Here 3-8-10+28 happens to be 13, but at n=2,3-8=-5 differs from 5; correct reindexing changes mu to mu(n/d)."
        }
      },
      {
        "claim": {
          "zh-Hant": "只列prime divisors就能反演。",
          "en": "Prime divisors alone suffice for inversion."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "漏1與6會只得−10−8=−18，不是13。",
          "en": "Omitting 1 and 6 leaves -10-8=-18, not 13."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "calc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "假設精確整數算術、帶餘除法、整除與gcd的定義；同餘等式始終綁定明示模數。",
          "en": "Assume exact integer arithmetic, division with remainder, and definitions of divisibility and gcd; every congruence has an explicit modulus."
        }
      },
      {
        "id": "mko_definition",
        "kind": "cited",
        "statement": {
          "zh-Hant": "positive integer n上，若F(n)=sum_{d|n}f(d)，則f(n)=sum_{d|n}mu(d)F(n/d)。等價index形式可用mu(n/d)F(d)，但不得混合。必須枚舉全部positive divisors並由完整UPF取得mu；n=1與mu(1)=1保留。這是divisor-poset convolution，不是d≤n。",
          "en": "For positive integer n, F(n)=sum over d|n of f(d) implies f(n)=sum over d|n of mu(d)*F(n/d). The equivalent indexing mu(n/d)*F(d) is valid but must not be mixed with the first convention. Enumerate every positive divisor and derive mu from complete prime multiplicities. Retain n=1 and mu(1)=1. This is convolution on the divisor poset, not a sum over all d≤n."
        },
        "source": {
          "mko_id": "mko-mobius-inversion-finite-divisors",
          "sha256": "903f02acdd799627c36103659ee206c7907a68eedbfb36d6a35a0e9dc6a2ef98",
          "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f"
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "forward",
        "lhs": "3+5+7+13",
        "rhs": "28",
        "step": "S04"
      },
      {
        "id": "inverse",
        "lhs": "28-10-8+3",
        "rhs": "13",
        "step": "S06"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "正整數6的全部divisors與exact仿射題族；只核這個有限反演證書。",
      "en": "Every divisor of positive integer 6 and an exact affine family; only this finite inversion certificate is claimed."
    },
    "md_symbols": [
      [
        "F",
        "divisor forward transform",
        "divisor forward transform"
      ],
      [
        "f",
        "仿射原函數2k+1",
        "affine function 2k+1"
      ]
    ],
    "md_error_controls": [
      {
        "title": "索引配對 / Index pairing",
        "zh": "同一項是μ(d)F(n/d)，或完整同步改為μ(n/d)F(d)。",
        "en": "Each term is mu(d)F(n/d), or a fully reindexed mu(n/d)F(d)."
      },
      {
        "title": "完整divisors / All divisors",
        "zh": "包括1與n，不以prime或proper列表替代。",
        "en": "Include 1 and n, not just prime or proper divisors."
      }
    ],
    "md_open_items": []
  },
  "revision": {
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    "original_source_sha256": "a3d906c53553c720d08fa4763944293d886bfad9ea166a77f4ae2c1399c65f18",
    "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",
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    "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."
  }
}