{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-at-most-countable-set-01",
  "mko_id": "mko-at-most-countable-set",
  "title": {
    "zh-Hant": "三條自然數列的全稱編碼",
    "en": "A universal code for three natural-number tracks"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-at-most-countable-set-01",
    "source": {
      "mko_id": "mko-at-most-countable-set",
      "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f",
      "sha256": "9c7748686b7974341920a59d07628f20017ef6d65476d751b2feb7424f7b3b57",
      "english_sha256": "a73faf7f1c04553efad774692eab18a2de59165f0fedaac3ba03ee75c7276def"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "calc"
    ],
    "goal": {
      "zh-Hant": "N₀×{0,1,2}以c(n,j)=3n+j單射到N₀，故至多可數。",
      "en": "N0 times {0,1,2} injects into N0 by c(n,j)=3n+j and is at most countable."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "N₀={n∈Z:n≥0}；X=N₀×{0,1,2}；c(n,j)=3n+j。",
          "en": "N0={n in Z:n≥0}; X=N0 times {0,1,2}; c(n,j)=3n+j."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1"
        ],
        "before_tex": "(n,j)\\in X",
        "after_tex": "n\\ge0,\\quad j\\in\\{0,1,2\\}\\Rightarrow3n+j\\in N_0",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "編碼是整數且非負，所以codomain宣告正確。",
          "en": "The code is an integer and nonnegative, so its codomain is valid."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1"
          ],
          "explanation": {
            "zh-Hant": "若允許n負數便不一定進入N₀；參數域不可省。",
            "en": "Allowing negative n could leave N0, so the parameter domain matters."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "S01"
        ],
        "before_tex": "c(n,j)=c(m,k)",
        "after_tex": "3(n-m)=k-j",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對任意兩輸入假設像相同，移項得3的倍數等於小差。",
          "en": "For arbitrary inputs with equal codes, rearranging gives a multiple of 3 equal to a small difference."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01"
          ],
          "explanation": {
            "zh-Hant": "這裡沒有抽樣n或m，是全稱碰撞分析。",
            "en": "No n or m is sampled; this is a universal collision analysis."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S03",
        "uses": [
          "G1",
          "S02"
        ],
        "before_tex": "j,k\\in\\{0,1,2\\}",
        "after_tex": "-2\\le k-j\\le2,\\quad3\\mid(k-j)\\Rightarrow k-j=0",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "範圍內唯一3的倍數是0，因此j=k。",
          "en": "Zero is the only multiple of 3 in that range, forcing j=k."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S02"
          ],
          "explanation": {
            "zh-Hant": "兩端界−2與2都必須納入；沒有±3所以結論成立。",
            "en": "Both endpoints -2 and 2 are included; neither 3 nor -3 occurs."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S04",
        "uses": [
          "G1",
          "S02",
          "S03"
        ],
        "before_tex": "3(n-m)=0,\\quad j=k",
        "after_tex": "n=m,\\quad(n,j)=(m,k)",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "整數中3非零，消去3得到n=m，故c全稱單射。",
          "en": "Since 3 is nonzero in the integers, cancellation gives n=m, so c is universally injective."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S02",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "相等像已推出輸入兩座標都相等，不只第一座標。",
            "en": "Equal images force equality of both coordinates, not just the first."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S05",
        "uses": [
          "G1",
          "S01",
          "S04"
        ],
        "before_tex": "c:X\\hookrightarrow N_0",
        "after_tex": "X\\text{ is at most countable}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "這個全稱單射直接滿足至多可數定義，不依有限測試。",
          "en": "The universal injection directly meets the definition of at most countable, without finite testing."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01",
            "S04"
          ],
          "explanation": {
            "zh-Hant": "不需要先證滿射；至多可數也容許有限集合。",
            "en": "Surjectivity is not needed for this conclusion; finite sets are at most countable too."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S06",
        "uses": [
          "G1",
          "S04"
        ],
        "before_tex": "r\\in N_0",
        "after_tex": "r=3q+j,\\quad q\\in N_0,\\quad j\\in\\{0,1,2\\}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "非負整數帶餘除法亦給每個r的前像，因此此特定c其實是雙射。",
          "en": "Nonnegative division with remainder also gives a preimage for every r, so this particular c is bijective."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S04"
          ],
          "explanation": {
            "zh-Hant": "額外滿射證書才允許稱X可數無限；前一步單射本身只證至多可數。",
            "en": "This extra onto certificate permits countable infinitude here; the injection alone established only at most countable."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S07",
        "uses": [
          "G1"
        ],
        "before_tex": "Y=\\{a,b,c\\},\\quad e(a)=0,e(b)=1,e(c)=2",
        "after_tex": "e:Y\\hookrightarrow N_0,\\quad |Y|=3",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "完整三元素映射的像互異，給有限集合的至多可數證書。",
          "en": "The full three-element map has distinct images, certifying a finite set as at most countable."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1"
          ],
          "explanation": {
            "zh-Hant": "Y至多可數但不是可數無限，因此兩個術語不能混同。",
            "en": "Y is at most countable but not countably infinite, distinguishing the terms."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S08",
        "uses": [
          "G1",
          "S05"
        ],
        "before_tex": "b(n,j)=3n",
        "after_tex": "b(0,0)=b(0,1)=0,\\quad c(0,0)=0\\ne1=c(0,1)",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "另一編碼b碰撞，只否證b這個witness；已證的c仍認證同一X。",
          "en": "Another code b collides, refuting only b; the proven c still certifies the same X."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S05"
          ],
          "explanation": {
            "zh-Hant": "一個壞編碼不能推出X不可數，因已有完整反證性良好編碼。",
            "en": "One bad code cannot imply uncountability when a valid full code already exists."
          }
        },
        "leaf_basis": "calc"
      }
    ],
    "result": {
      "zh-Hant": "N₀×{0,1,2}以c(n,j)=3n+j單射到N₀，故至多可數。",
      "en": "N0 times {0,1,2} injects into N0 by c(n,j)=3n+j and is at most countable."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "b(n,j)=3n是單射。",
          "en": "b(n,j)=3n is injective."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "(0,0)與(0,1)都送到0。",
          "en": "(0,0) and (0,1) both map to 0."
        }
      },
      {
        "claim": {
          "zh-Hant": "b失敗表示X不可數。",
          "en": "Failure of b means X is uncountable."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "c(n,j)=3n+j已全稱證單射。",
          "en": "c(n,j)=3n+j is already universally injective."
        }
      },
      {
        "claim": {
          "zh-Hant": "有限Y至多可數，因此Y可數無限。",
          "en": "Finite Y is at most countable, hence countably infinite."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "Y只有三個元素，不與無限N₀雙射。",
          "en": "Y has three elements and cannot biject with infinite N0."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "calc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "假設整數帶餘除法、非負整數與單射定義；至多可數以存在到N₀的單射定義。",
          "en": "Assume integer quotient/remainder arithmetic, nonnegative integers and injectivity; at most countable means admitting an injection into N0."
        }
      },
      {
        "id": "mko_definition",
        "kind": "cited",
        "statement": {
          "zh-Hant": "本輪以完整有限集合證書或整數仿射編碼的全稱單射證書處理。某個編碼失敗只否證該witness，不證集合不可數。",
          "en": "This slice uses complete finite-set certificates or universal injectivity certificates from integer affine encodings. A failed encoding refutes that particular witness, not countability of the set."
        },
        "source": {
          "mko_id": "mko-at-most-countable-set",
          "sha256": "9c7748686b7974341920a59d07628f20017ef6d65476d751b2feb7424f7b3b57",
          "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f"
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "small_difference",
        "lhs": "3",
        "rhs": "2",
        "step": "S03",
        "relation": ">"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "有限集合證書加上全稱整數仿射編碼；不提出不可數性結論。",
      "en": "Finite-set certificates and a universal integer affine code; no uncountability conclusion."
    },
    "md_symbols": [
      [
        "N_0",
        "含0的自然數",
        "natural numbers including zero"
      ],
      [
        "c",
        "仿射整數編碼",
        "integer affine code"
      ],
      [
        "j",
        "三條列索引",
        "three-track index"
      ]
    ],
    "md_error_controls": [
      {
        "title": "參數域 / Parameter domain",
        "zh": "c的證明固定n≥0及0≤j≤2；允許j=3會和下一條列碰撞。",
        "en": "The proof fixes n≥0 and 0≤j≤2; allowing j=3 collides with the next track."
      },
      {
        "title": "見證與集合 / Witness versus set",
        "zh": "編碼失敗只否證該函數；不可直接標集合不可數。",
        "en": "Failure refutes that function, not countability of the set."
      }
    ],
    "md_open_items": []
  },
  "revision": {
    "id": "ocme-new142-reader-adoption-20261021",
    "collection": "coverage-20261021",
    "source_commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f",
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    "original_source_sha256": "bcd07b59b5e83e1216ba1fb8c780353d2a97cf3c8fd58ec701dee7656e7f25d5",
    "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": [],
    "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."
  }
}