{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-tree-01",
  "mko_id": "mko-tree",
  "title": {
    "zh-Hant": "樹：判定一個 6 頂點圖是樹，求葉、唯一道路與加邊後的唯一環",
    "en": "Trees: deciding that a 6-vertex graph is a tree, with its leaves, a unique path and the unique cycle after adding an edge"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-tree-01",
    "source": {
      "mko_id": "mko-tree",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "48c7310136a8fd9749e9a9af278b0ade15fdecb8f4cc5657f4f0895eef89aebf",
      "english_sha256": "9b6b7c0d53aa18952a811e36d538baf58041cc6982489dd77fe0f5e4535379f6"
    },
    "mode": "definition_walkthrough",
    "audience_prerequisites": [
      "graph_basics",
      "well_ordering",
      "nat_induction",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "設 G=(V,E)，V={1,2,3,4,5,6}，E={{1,2},{2,3},{2,4},{4,5},{4,6}}。(1) 依 MKO 定義證明 G 是樹（V 非空、連通、無簡單環）；(2) 列出葉；(3) 求從 3 到 6 的道路並證明它唯一；(4) 加入邊 {3,6}，寫出所形成的簡單環並證明它是唯一的簡單環；(5) 對照：三角形加一個孤立點雖有 |E|=|V|−1 卻不是樹；單點無邊圖是樹但沒有葉。",
      "en": "Let G=(V,E) with V={1,2,3,4,5,6} and E={{1,2},{2,3},{2,4},{4,5},{4,6}}. (1) Prove from the MKO's definition that G is a tree (V nonempty, connected, no simple cycle); (2) list its leaves; (3) find the path from 3 to 6 and prove it is unique; (4) add the edge {3,6}, write the simple cycle it creates and prove it is the only simple cycle; (5) contrast: a triangle plus an isolated vertex has |E|=|V|−1 but is not a tree; a single vertex without edges is a tree with no leaf."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "G=(V,E)，V={1,2,3,4,5,6}，E={{1,2},{2,3},{2,4},{4,5},{4,6}}；邊是無向的，每條只列一次。",
          "en": "G=(V,E) with V={1,2,3,4,5,6} and E={{1,2},{2,3},{2,4},{4,5},{4,6}}; edges are undirected and each is listed once."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "MKO 的道路約定：道路不重複頂點；同一頂點到自身的零邊道路合法。葉是度數恰為 1 的頂點。",
          "en": "The MKO's path conventions: a path repeats no vertex; the zero-edge path from a vertex to itself is valid. A leaf is a vertex of degree exactly 1."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "對照圖：H=({1,2,3,4},{{1,2},{2,3},{1,3}})；K=({7},∅)。",
          "en": "Contrast graphs: H=({1,2,3,4},{{1,2},{2,3},{1,3}}) and K=({7},∅)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2",
          "mko_tree_def"
        ],
        "before_tex": "G\\ \\text{is a tree}",
        "after_tex": "V\\neq\\varnothing\\ \\land\\ \\forall u,v\\in V\\ \\exists\\ \\text{path}\\ u\\to v\\ \\land\\ \\neg\\exists\\ \\text{simple cycle}",
        "relation": "definition",
        "rule": "mko_tree_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO 的定義，要證 G 是樹必須證三件事：V 非空、任兩頂點有道路相連、沒有簡單環。只看邊數不在定義之內。",
          "en": "By the MKO's definition, proving that G is a tree requires three things: V is nonempty, any two vertices are joined by a path, and there is no simple cycle. Counting edges is not part of the definition."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "mko_tree_def"
          ],
          "explanation": {
            "zh-Hant": "此步只展開定義，不推出新事實；後面三個條件各自證明。",
            "en": "This step only unfolds the definition and derives nothing new; the three conditions are proved separately later."
          }
        },
        "leaf_basis": "mko_tree_def + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "graph_basics"
        ],
        "before_tex": "E=\\{\\{1,2\\},\\{2,3\\},\\{2,4\\},\\{4,5\\},\\{4,6\\}\\}",
        "after_tex": "G\\ \\text{finite simple},\\ |V|=6,\\ |E|=5,\\ V\\neq\\varnothing",
        "relation": "implies",
        "rule": "graph_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐條檢查：每條邊的兩端都在 V 中且相異，沒有自環；五條邊作為二元素集合兩兩不同，沒有重邊；無向邊不分方向，各列一次（MKO 假設 a1 的正規化）。所以 G 是有限簡單無向圖，|V|=6>0。",
          "en": "Check each edge: both endpoints are in V and distinct, so there are no loops; the five edges are pairwise different as two-element sets, so there are no multiple edges; undirected edges ignore direction and each is listed once (the normalization of MKO assumption a1). So G is a finite simple undirected graph with |V|=6>0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "graph_basics"
          ],
          "explanation": {
            "zh-Hant": "只檢查資料格式，不宣稱任何連通或無環性質。",
            "en": "This only checks the data format; no connectivity or acyclicity is claimed."
          }
        },
        "leaf_basis": "graph_basics + numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "graph_basics"
        ],
        "before_tex": "N(v)\\ \\text{for}\\ v\\in V",
        "after_tex": "\\deg(1),\\deg(2),\\deg(3),\\deg(4),\\deg(5),\\deg(6)=1,3,1,3,1,1",
        "relation": "implies",
        "rule": "graph_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐點列鄰居：N(1)={2}、N(2)={1,3,4}、N(3)={2}、N(4)={2,5,6}、N(5)={4}、N(6)={4}。度數是鄰居個數。握手核對：1+3+1+3+1+1=10=2·5=2|E|（N1）。",
          "en": "List neighbours: N(1)={2}, N(2)={1,3,4}, N(3)={2}, N(4)={2,5,6}, N(5)={4}, N(6)={4}. The degree is the number of neighbours. Handshake check: 1+3+1+3+1+1=10=2·5=2|E| (N1)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "graph_basics",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "每條邊 {u,v} 恰使 v∈N(u) 與 u∈N(v) 各一次，握手和吻合是對鄰居表的獨立核對。",
            "en": "Each edge {u,v} contributes v∈N(u) and u∈N(v) exactly once each; the matching handshake sum is an independent check of the neighbour lists."
          }
        },
        "leaf_basis": "graph_basics + numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "mko_tree_def"
        ],
        "before_tex": "\\{v\\in V:\\ \\deg(v)=1\\}",
        "after_tex": "\\{1,3,5,6\\}",
        "relation": "equivalent",
        "rule": "def_unfold",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依葉的定義（度數恰為 1），對 6 個頂點逐一判斷：1、3、5、6 的度數為 1，2 與 4 的度數為 3。葉集合按遞增列為 {1,3,5,6}。",
          "en": "By the definition of a leaf (degree exactly 1), check all 6 vertices: 1, 3, 5, 6 have degree 1, while 2 and 4 have degree 3. The leaf set, in increasing order, is {1,3,5,6}."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "S03",
            "def_unfold"
          ],
          "explanation": {
            "zh-Hant": "兩向都逐點核對：集合中每個點度數為 1，集合外的 2、4 度數不為 1。",
            "en": "Both directions are checked point by point: each listed vertex has degree 1, and the unlisted 2 and 4 do not."
          }
        },
        "leaf_basis": "def_unfold + set_builder"
      },
      {
        "id": "S05",
        "uses": [
          "S03",
          "G2",
          "walk_path_cycle"
        ],
        "before_tex": "\\text{vertex}\\ 1",
        "after_tex": "(1),\\ (1,2),\\ (1,2,3),\\ (1,2,4),\\ (1,2,4,5),\\ (1,2,4,6)",
        "relation": "implies",
        "rule": "exists_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "從頂點 1 到每個頂點各寫出一條道路。每個序列的相鄰兩點都是 E 中的邊（{1,2}、{2,3}、{2,4}、{4,5}、{4,6}），且沒有重複頂點；(1) 是到自身的零邊道路。",
          "en": "Write one path from vertex 1 to each vertex. In each sequence consecutive vertices form edges of E ({1,2}, {2,3}, {2,4}, {4,5}, {4,6}) and no vertex repeats; (1) is the zero-edge path to itself."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "walk_path_cycle",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "每條道路只需檢查相鄰邊與不重複；這一步不宣稱這些道路唯一。",
            "en": "Each path only needs its edges and the no-repeat condition checked; this step does not claim the paths are unique."
          }
        },
        "leaf_basis": "exists_intro + walk_path_cycle"
      },
      {
        "id": "S06",
        "uses": [
          "S05",
          "L_walk_path"
        ],
        "before_tex": "\\forall u,v\\in V:\\ \\text{walk}\\ u\\to 1\\to v",
        "after_tex": "\\forall u,v\\in V\\ \\exists\\ \\text{path}\\ u\\to v",
        "relation": "implies",
        "rule": "L_walk_path",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對任意 u、v：把 S05 中從 1 到 u 的道路反向（邊無向，反向仍是走道），再接上從 1 到 v 的道路，得到從 u 到 v 的走道。依引理 L_walk_path（S06.1、S06.2 證明），有從 u 到 v 的道路。例如 u=3、v=6：走道 (3,2,1,2,4,6) 重複了 2，引理的刪減給出道路 (3,2,4,6)。所以 G 連通。",
          "en": "For any u, v: reverse the path from 1 to u of S05 (edges are undirected, so the reverse is still a walk), then append the path from 1 to v, giving a walk from u to v. By Lemma L_walk_path (proved in S06.1 and S06.2) there is a path from u to v. For example u=3, v=6: the walk (3,2,1,2,4,6) repeats 2, and the lemma's shortening gives the path (3,2,4,6). So G is connected."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "L_walk_path"
          ],
          "explanation": {
            "zh-Hant": "只需「1 能到每個點」與引理；不必逐一列出 15 對頂點的道路。不宣稱道路唯一。",
            "en": "Only 'vertex 1 reaches every vertex' and the lemma are needed; the 15 vertex pairs need not be listed one by one. Uniqueness of paths is not claimed."
          }
        },
        "leaf_basis": "walk_path_cycle + well_ordering + contradiction"
      },
      {
        "id": "S06.1",
        "uses": [
          "walk_path_cycle",
          "well_ordering"
        ],
        "before_tex": "\\text{the set of lengths of walks}\\ u\\to v\\ \\text{is nonempty}",
        "after_tex": "\\exists\\ \\text{a walk}\\ W\\ \\text{from}\\ u\\ \\text{to}\\ v\\ \\text{of least length}\\ k",
        "relation": "implies",
        "rule": "well_ordering",
        "conditions": [],
        "explanation": {
          "zh-Hant": "假設有從 u 到 v 的走道。所有這種走道的長度構成 ℕ 的非空子集，依良序原理有最小元素 k；取一條長度為 k 的走道 W，其頂點依序為 w_0=u 到 w_k=v。",
          "en": "Assume there is a walk from u to v. The lengths of all such walks form a nonempty subset of ℕ, which by well-ordering has a least element k; take a walk W of length k with vertices w_0=u through w_k=v."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "walk_path_cycle",
            "well_ordering"
          ],
          "explanation": {
            "zh-Hant": "良序原理需要非空，由引理的前提（有走道）保證。",
            "en": "Well-ordering needs nonemptiness, which the lemma's premise (a walk exists) supplies."
          }
        },
        "leaf_basis": "well_ordering"
      },
      {
        "id": "S06.2",
        "uses": [
          "S06.1",
          "contradiction"
        ],
        "before_tex": "w_i=w_j\\ (i<j)\\ \\text{in the least walk}\\ W",
        "after_tex": "\\text{a walk of length}\\ k-(j-i)<k,\\ \\text{contradiction; so}\\ W\\ \\text{is a path}",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [
          "i<j"
        ],
        "explanation": {
          "zh-Hant": "假設 W 重複了頂點，w_i=w_j 且 i<j。刪去 w_{i+1} 到 w_j，剩下 w_0 到 w_i、再接 w_{j+1} 到 w_k；因 w_i=w_j 與 w_{j+1} 相鄰（若 j=k 則走道就停在 w_i=v），這仍是從 u 到 v 的走道，長度 k−(j−i)<k，與 k 最小矛盾。所以 W 不重複頂點，是道路。S06.1 與 S06.2 合起來證得引理 L_walk_path。",
          "en": "Suppose W repeats a vertex, w_i=w_j with i<j. Delete w_{i+1} through w_j, keeping w_0 to w_i followed by w_{j+1} to w_k; since w_i=w_j is adjacent to w_{j+1} (or, if j=k, the walk simply stops at w_i=v), this is still a walk from u to v, of length k−(j−i)<k, contradicting the minimality of k. So W repeats no vertex and is a path. S06.1 and S06.2 together prove Lemma L_walk_path."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06.1",
            "contradiction",
            "walk_path_cycle"
          ],
          "explanation": {
            "zh-Hant": "矛盾只依賴 k 的最小性與刪減後仍是走道；證明中沒有使用引理本身。",
            "en": "The contradiction rests only on the minimality of k and on the shortened sequence still being a walk; the lemma itself is not used."
          }
        },
        "leaf_basis": "contradiction + walk_path_cycle + order_basics"
      },
      {
        "id": "S07",
        "uses": [
          "S03",
          "L_leaf_cycle",
          "delete_vertex"
        ],
        "before_tex": "G\\ \\text{has a simple cycle}",
        "after_tex": "G_1=(\\{2,4\\},\\{\\{2,4\\}\\})\\ \\text{has a simple cycle}",
        "relation": "equivalent",
        "rule": "L_leaf_cycle",
        "conditions": [
          "\\deg(v)\\le 1\\ \\text{at each removal}"
        ],
        "explanation": {
          "zh-Hant": "逐一刪葉，每次都用引理 L_leaf_cycle（S07.1、S07.2 證明）。在 G 中 deg(1)=1，所以 G 有簡單環 ⇔ G−1 有。在 G−1 中，3 的鄰居仍只有 2，度數 1，刪去；接著 5（鄰居只有 4）、6（鄰居只有 4）。每次刪去一條邊，四次之後剩下頂點 {2,4} 與一條邊 {2,4}。等價串接得結論。",
          "en": "Delete leaves one at a time, each time using Lemma L_leaf_cycle (proved in S07.1 and S07.2). In G, deg(1)=1, so G has a simple cycle ⇔ G−1 has one. In G−1, vertex 3 still has 2 as its only neighbour, degree 1, so delete it; then 5 (only neighbour 4) and 6 (only neighbour 4). Each deletion removes one edge, and after four of them the vertices {2,4} and the single edge {2,4} remain. Chaining the equivalences gives the claim."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_leaf_cycle",
            "logic_chain"
          ],
          "explanation": {
            "zh-Hant": "每次套用引理都是等價，等價可串接，所以整步雙向成立；每次刪除前都重新確認該點在當前圖中度數至多 1。",
            "en": "Each application of the lemma is an equivalence and equivalences chain, so the whole step is two-way; before each deletion the degree in the current graph is rechecked to be at most 1."
          }
        },
        "leaf_basis": "logic_chain + delete_vertex + graph_basics"
      },
      {
        "id": "S07.1",
        "uses": [
          "delete_vertex",
          "walk_path_cycle"
        ],
        "before_tex": "C\\ \\text{simple cycle in}\\ G-v",
        "after_tex": "C\\ \\text{simple cycle in}\\ G",
        "relation": "implies",
        "rule": "subset_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "（⇐ 方向）G−v 的頂點與邊都屬於 G，所以同一個頂點序列在 G 中仍滿足簡單環的所有條件。",
          "en": "(The ⇐ direction) The vertices and edges of G−v all belong to G, so the same vertex sequence still meets every condition of a simple cycle in G."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "delete_vertex",
            "walk_path_cycle",
            "subset_def"
          ],
          "explanation": {
            "zh-Hant": "只用子圖的包含關係；這個方向不需要度數條件。",
            "en": "Only the inclusion of the subgraph is used; this direction needs no degree condition."
          }
        },
        "leaf_basis": "subset_def + delete_vertex"
      },
      {
        "id": "S07.2",
        "uses": [
          "walk_path_cycle",
          "graph_basics"
        ],
        "before_tex": "C\\ \\text{simple cycle in}\\ G,\\ \\deg_G(v)\\le 1",
        "after_tex": "v\\notin C,\\ C\\ \\text{simple cycle in}\\ G-v",
        "relation": "implies",
        "rule": "contrapositive",
        "conditions": [
          "k\\ge 3"
        ],
        "explanation": {
          "zh-Hant": "（⇒ 方向）在長度 k≥3 的簡單環中，每個 w_j（0≤j<k）都有兩個環上鄰居 w_{j−1} 與 w_{j+1}（下標以 k 為模），兩者相異，因為 w_0 到 w_{k−1} 兩兩相異且 k≥3；所以 C 上每點在 G 中度數至少 2。逆否：deg_G(v)≤1 的 v 不在 C 上，C 的每條邊都不含 v，所以 C 是 G−v 的簡單環。S07.1 與 S07.2 合起來證得引理 L_leaf_cycle。",
          "en": "(The ⇒ direction) In a simple cycle of length k≥3, each w_j (0≤j<k) has two cycle neighbours w_{j−1} and w_{j+1} (indices mod k), which differ because w_0 to w_{k−1} are pairwise distinct and k≥3; so every vertex of C has degree at least 2 in G. Contrapositively, a vertex v with deg_G(v)≤1 is not on C, no edge of C contains v, and C is a simple cycle of G−v. S07.1 and S07.2 together prove Lemma L_leaf_cycle."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "walk_path_cycle",
            "graph_basics",
            "contrapositive"
          ],
          "explanation": {
            "zh-Hant": "關鍵是 k≥3 保證兩個環上鄰居相異；若允許 k=2 的「環」(u,v,u)，這個論證就失效，所以定義要求 k≥3。證明中沒有使用引理本身。",
            "en": "The key is that k≥3 makes the two cycle neighbours distinct; with a 'cycle' (u,v,u) of length 2 the argument would fail, which is why the definition requires k≥3. The lemma itself is not used."
          }
        },
        "leaf_basis": "contrapositive + graph_basics + walk_path_cycle"
      },
      {
        "id": "S08",
        "uses": [
          "S07",
          "L_leaf_cycle"
        ],
        "before_tex": "G_1=(\\{2,4\\},\\{\\{2,4\\}\\})\\ \\text{has a simple cycle}",
        "after_tex": "(\\{4\\},\\varnothing)\\ \\text{has a simple cycle}",
        "relation": "equivalent",
        "rule": "L_leaf_cycle",
        "conditions": [
          "\\deg_{G_1}(2)=1"
        ],
        "explanation": {
          "zh-Hant": "在 G_1 中 2 只有鄰居 4，度數 1；再用一次引理，刪去 2 得到單點無邊圖 ({4},∅)。",
          "en": "In G_1 vertex 2 has the single neighbour 4, degree 1; one more use of the lemma deletes 2 and leaves the single-vertex graph ({4},∅)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_leaf_cycle"
          ],
          "explanation": {
            "zh-Hant": "引理是等價，雙向成立。",
            "en": "The lemma is an equivalence, so the step is two-way."
          }
        },
        "leaf_basis": "delete_vertex + graph_basics"
      },
      {
        "id": "S09",
        "uses": [
          "S07",
          "S08",
          "walk_path_cycle"
        ],
        "before_tex": "G\\ \\text{has a simple cycle}\\iff(\\{4\\},\\varnothing)\\ \\text{has one}",
        "after_tex": "G\\ \\text{has no simple cycle}",
        "relation": "implies",
        "rule": "logic_chain",
        "conditions": [],
        "explanation": {
          "zh-Hant": "簡單環至少要 3 個相異頂點，({4},∅) 只有一個頂點、沒有邊，所以沒有簡單環。串接 S07、S08 的等價，G 也沒有簡單環。",
          "en": "A simple cycle needs at least 3 distinct vertices; ({4},∅) has one vertex and no edges, so it has no simple cycle. Chaining the equivalences of S07 and S08, G has no simple cycle either."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "S08",
            "logic_chain",
            "walk_path_cycle"
          ],
          "explanation": {
            "zh-Hant": "結論依賴整條等價鏈；若某次刪除的點度數大於 1，鏈就斷了，所以 S07 每次都重新核對度數。",
            "en": "The conclusion relies on the whole chain of equivalences; if some deleted vertex had degree above 1 the chain would break, which is why S07 rechecks degrees each time."
          }
        },
        "leaf_basis": "logic_chain + walk_path_cycle"
      },
      {
        "id": "S10",
        "uses": [
          "S02",
          "S06",
          "S09",
          "mko_tree_def"
        ],
        "before_tex": "V\\neq\\varnothing,\\ \\text{connected},\\ \\text{no simple cycle}",
        "after_tex": "G\\ \\text{is a tree}",
        "relation": "implies",
        "rule": "mko_tree_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S02 給 V 非空，S06 給連通，S09 給無簡單環；三者合起來正是 MKO 的樹定義。",
          "en": "S02 gives V nonempty, S06 gives connectivity and S09 gives no simple cycle; together these are exactly the MKO's definition of a tree."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S02",
            "S06",
            "S09",
            "mko_tree_def",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：G 是樹。足夠條件 H：三個條件同時成立；H⇒G 由定義。H 由 S02、S06、S09 正向證成，沒有用到 |E|=|V|−1，也沒有把 G 當作前提。",
            "en": "Goal G: G is a tree. Sufficient condition H: the three conditions together; H⇒G by the definition. H is proved forwards in S02, S06 and S09, without using |E|=|V|−1 and without using G as a premise."
          }
        },
        "leaf_basis": "mko_tree_def + and_intro_elim"
      },
      {
        "id": "S11",
        "uses": [
          "S03",
          "walk_path_cycle"
        ],
        "before_tex": "3,\\ 6\\in V",
        "after_tex": "(3,2,4,6)\\ \\text{is a path from 3 to 6}",
        "relation": "implies",
        "rule": "exists_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "{3,2}、{2,4}、{4,6} 都在 E 中，四個頂點 3、2、4、6 兩兩相異，所以 (3,2,4,6) 是從 3 到 6 的道路。依 MKO 段落 1，以實際頂點順序輸出，而不是排序後的集合 {2,3,4,6}。",
          "en": "{3,2}, {2,4} and {4,6} are in E and the four vertices 3, 2, 4, 6 are pairwise distinct, so (3,2,4,6) is a path from 3 to 6. Following MKO paragraph 1 it is written in its actual vertex order, not as the sorted set {2,3,4,6}."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "walk_path_cycle",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "只證存在；唯一性在 S12 另證。",
            "en": "Only existence is shown; uniqueness is proved separately in S12."
          }
        },
        "leaf_basis": "exists_intro + walk_path_cycle"
      },
      {
        "id": "S12",
        "uses": [
          "S03",
          "S11",
          "walk_path_cycle"
        ],
        "before_tex": "P\\ \\text{any path from 3 to 6}",
        "after_tex": "P=(3,2,4,6)",
        "relation": "implies",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "設 P 是任一條從 3 到 6 的道路，頂點依序為 p_0=3 到 p_m=6。S12.1 到 S12.3 逐點窮舉每一步所有可能的下一個頂點，證明 P 只能是 (3,2,4,6)。",
          "en": "Let P be any path from 3 to 6 with vertices p_0=3 through p_m=6. S12.1 to S12.3 exhaust every possible next vertex at each stage and show that P can only be (3,2,4,6)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S11",
            "walk_path_cycle",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "窮舉涵蓋每一步的全部鄰居，所以對這一對頂點是完整證明；它不是一般定理的證明。",
            "en": "The exhaustion covers every neighbour at every stage, so it is a complete proof for this pair of vertices; it is not a proof of the general theorem."
          }
        },
        "leaf_basis": "case_analysis + walk_path_cycle + graph_basics"
      },
      {
        "id": "S12.1",
        "uses": [
          "S03",
          "walk_path_cycle"
        ],
        "before_tex": "p_0=3\\neq 6",
        "after_tex": "p_1=2",
        "relation": "implies",
        "rule": "graph_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "3≠6，所以 P 至少有一條邊；p_1 必須是 3 的鄰居，而 N(3)={2}，故 p_1=2。",
          "en": "3≠6, so P has at least one edge; p_1 must be a neighbour of 3, and N(3)={2}, so p_1=2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "graph_basics"
          ],
          "explanation": {
            "zh-Hant": "只用到 3 的鄰居表。",
            "en": "Only the neighbour list of 3 is used."
          }
        },
        "leaf_basis": "graph_basics"
      },
      {
        "id": "S12.2",
        "uses": [
          "S12.1",
          "S03"
        ],
        "before_tex": "p_1=2\\neq 6",
        "after_tex": "p_2=4",
        "relation": "case_split",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "2≠6，P 繼續；p_2∈N(2)∖{3}={1,4}（道路不重複頂點）。情況 p_2=1：1≠6 須再走，但 N(1)∖{3,2}=∅，無路可走，矛盾。所以 p_2=4。",
          "en": "2≠6, so P continues; p_2∈N(2)∖{3}={1,4} (a path repeats no vertex). Case p_2=1: 1≠6, so P must continue, but N(1)∖{3,2}=∅, so there is nowhere to go, a contradiction. Hence p_2=4."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S12.1",
            "S03",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "兩種情況 {1,4} 窮盡了所有可能。",
            "en": "The two cases {1,4} exhaust all possibilities."
          }
        },
        "leaf_basis": "case_analysis + contradiction + graph_basics"
      },
      {
        "id": "S12.3",
        "uses": [
          "S12.2",
          "S03"
        ],
        "before_tex": "p_2=4\\neq 6",
        "after_tex": "p_3=6,\\ m=3,\\ P=(3,2,4,6)",
        "relation": "case_split",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "4≠6，P 繼續；p_3∈N(4)∖{3,2}={5,6}。情況 p_3=5：5≠6 須再走，但 N(5)∖{3,2,4}=∅，矛盾。所以 p_3=6；6 在道路中只出現一次且是終點，故 m=3，P=(3,2,4,6)。",
          "en": "4≠6, so P continues; p_3∈N(4)∖{3,2}={5,6}. Case p_3=5: 5≠6, so P must continue, but N(5)∖{3,2,4}=∅, a contradiction. Hence p_3=6; 6 occurs only once in the path and is its end, so m=3 and P=(3,2,4,6)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S12.2",
            "S03",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "兩種情況 {5,6} 窮盡了所有可能；結論與 S11 的道路相同。",
            "en": "The two cases {5,6} exhaust all possibilities; the result is the path of S11."
          }
        },
        "leaf_basis": "case_analysis + contradiction + graph_basics"
      },
      {
        "id": "S13",
        "uses": [
          "G1",
          "S10",
          "graph_basics"
        ],
        "before_tex": "e=\\{3,6\\}",
        "after_tex": "3,6\\in V,\\ 3\\neq 6,\\ e\\notin E;\\ G+e=(V,\\,E\\cup\\{e\\})",
        "relation": "implies",
        "rule": "graph_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "核對 MKO 假設 a3 的加邊前提：原圖是樹（S10）；3、6 都在 V 且相異；{3,6} 不在 E 的五條邊中。所以 G+e 仍是簡單圖，有 6 條邊。",
          "en": "Check the add-an-edge premise of MKO assumption a3: the original graph is a tree (S10); 3 and 6 are in V and distinct; {3,6} is not among the five edges of E. So G+e is still a simple graph, with 6 edges."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S10",
            "graph_basics"
          ],
          "explanation": {
            "zh-Hant": "若 e 已在 E 中，重列不會產生新邊，前提不成立（見錯誤控制）。",
            "en": "If e were already in E, listing it again would create no new edge and the premise would fail (see the error controls)."
          }
        },
        "leaf_basis": "graph_basics"
      },
      {
        "id": "S14",
        "uses": [
          "S11",
          "S13",
          "walk_path_cycle"
        ],
        "before_tex": "(3,2,4,6)\\ \\text{in}\\ G,\\ e=\\{6,3\\}",
        "after_tex": "C=(3,2,4,6,3)\\ \\text{is a simple cycle in}\\ G+e",
        "relation": "implies",
        "rule": "walk_path_cycle",
        "conditions": [
          "k=4\\ge 3"
        ],
        "explanation": {
          "zh-Hant": "從兩端較小的標籤 3 出發，沿原道路走到 6，再由新邊回到 3：長度 k=4≥3，前四個頂點 3、2、4、6 兩兩相異，{3,2}、{2,4}、{4,6} 在 E 中，{6,3}=e。所以 C 是 G+e 的簡單環，且符合 MKO 段落 1 的輸出約定。",
          "en": "Start at the smaller end label 3, follow the original path to 6, and return to 3 by the new edge: the length is k=4≥3, the first four vertices 3, 2, 4, 6 are pairwise distinct, {3,2}, {2,4}, {4,6} are in E and {6,3}=e. So C is a simple cycle of G+e, written according to the output convention of MKO paragraph 1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11",
            "S13",
            "walk_path_cycle"
          ],
          "explanation": {
            "zh-Hant": "只證存在；唯一性在 S15。",
            "en": "Only existence is shown; uniqueness is in S15."
          }
        },
        "leaf_basis": "walk_path_cycle + exists_intro"
      },
      {
        "id": "S15",
        "uses": [
          "S09",
          "S12",
          "S14"
        ],
        "before_tex": "D\\ \\text{any simple cycle in}\\ G+e",
        "after_tex": "D=C\\ \\text{up to starting point and direction}",
        "relation": "implies",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "若 D 不用 e，它的邊都在 E 中，就是 G 的簡單環，與 S09 矛盾；所以 D 用到 e。因 D 的頂點（除首尾）兩兩相異，3 與 6 各出現一次，e 在 D 中恰用一次。把 D 旋轉、必要時反向，使它從 3 出發、最後一條邊是由 6 回到 3 的 e；去掉這條邊，剩下一條只用 E 中邊、從 3 到 6 的道路。依 S12 它是 (3,2,4,6)，所以 D 就是 C。",
          "en": "If D did not use e, all its edges would be in E and it would be a simple cycle of G, contradicting S09; so D uses e. Since the vertices of D (apart from the repeated endpoint) are pairwise distinct, 3 and 6 each occur once and e is used exactly once. Rotate D, reversing it if needed, so that it starts at 3 and its last edge is e from 6 back to 3; removing that edge leaves a path from 3 to 6 that uses only edges of E. By S12 that path is (3,2,4,6), so D is C."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09",
            "S12",
            "S14",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "唯一性依賴 G 無環（S09）與道路唯一（S12）；環視為相同是指只差起點與方向，這正是 MKO 用固定輸出約定處理的部分。",
            "en": "Uniqueness relies on G having no cycle (S09) and on the unique path (S12); cycles are identified when they differ only in starting point and direction, which the MKO handles by its fixed output convention."
          }
        },
        "leaf_basis": "case_analysis + contradiction + walk_path_cycle"
      },
      {
        "id": "S16",
        "uses": [
          "G3",
          "walk_path_cycle",
          "mko_tree_def"
        ],
        "before_tex": "H=(\\{1,2,3,4\\},\\{\\{1,2\\},\\{2,3\\},\\{1,3\\}\\})",
        "after_tex": "|E_H|=|V_H|-1,\\ \\text{no path}\\ 1\\to 4;\\ H\\ \\text{is not a tree}",
        "relation": "implies",
        "rule": "mko_tree_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "|V_H|=4、|E_H|=3=4−1，邊數與主例同型，但 deg(4)=0。從 1 到 4 的道路至少一條邊（1≠4），最後一條邊含 4，不可能；所以 H 不連通。此外 (1,2,3,1) 是簡單環。H 不是樹。",
          "en": "|V_H|=4 and |E_H|=3=4−1, the same edge count pattern as the main example, but deg(4)=0. A path from 1 to 4 has at least one edge (1≠4) and its last edge contains 4, which is impossible; so H is not connected. Moreover (1,2,3,1) is a simple cycle. H is not a tree."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "walk_path_cycle",
            "mko_tree_def"
          ],
          "explanation": {
            "zh-Hant": "否定樹只需一個條件失敗；這裡兩個條件都失敗。",
            "en": "Showing that a graph is not a tree requires only one failed condition; here two of them fail."
          }
        },
        "leaf_basis": "mko_tree_def + contradiction + graph_basics"
      },
      {
        "id": "S17",
        "uses": [
          "G2",
          "G3",
          "mko_tree_def"
        ],
        "before_tex": "K=(\\{7\\},\\varnothing)",
        "after_tex": "K\\ \\text{is a tree},\\ \\text{leaves}(K)=\\varnothing",
        "relation": "implies",
        "rule": "mko_tree_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "V={7} 非空；唯一的頂點對 (7,7) 由零邊道路 (7) 連接（G2）；簡單環需要至少 3 個相異頂點，所以沒有。K 是樹。但 deg(7)=0≠1，依葉的定義，K 沒有葉（MKO 假設 a2）。",
          "en": "V={7} is nonempty; the only pair (7,7) is joined by the zero-edge path (7) (G2); a simple cycle needs at least 3 distinct vertices, so there is none. K is a tree. But deg(7)=0≠1, so by the definition K has no leaf (MKO assumption a2)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "G3",
            "mko_tree_def"
          ],
          "explanation": {
            "zh-Hant": "依賴「零邊道路合法」與「葉的度數恰為 1」兩個明示約定。",
            "en": "It relies on two stated conventions: the zero-edge path is valid, and a leaf has degree exactly 1."
          }
        },
        "leaf_basis": "mko_tree_def + walk_path_cycle"
      }
    ],
    "result": {
      "zh-Hant": "G 是樹：V 非空，頂點 1 能到每個點故連通，逐次刪葉到單點證明無簡單環。葉為 {1,3,5,6}。從 3 到 6 的唯一道路是 (3,2,4,6)。加入新邊 {3,6} 後恰有一個簡單環 (3,2,4,6,3)。對照：H 有 |E|=|V|−1 但不連通且有環，不是樹；單點圖 K 是樹，葉集合為空。",
      "en": "G is a tree: V is nonempty, vertex 1 reaches every vertex so G is connected, and deleting leaves down to a single vertex shows there is no simple cycle. The leaves are {1,3,5,6}. The unique path from 3 to 6 is (3,2,4,6). After adding the new edge {3,6} there is exactly one simple cycle, (3,2,4,6,3). Contrast: H has |E|=|V|−1 but is disconnected and has a cycle, so it is not a tree; the single-vertex graph K is a tree with an empty leaf set."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "H 有 4 個頂點、3 條邊，所以 H 是樹",
          "en": "H has 4 vertices and 3 edges, so H is a tree"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "頂點 4 孤立，H 不連通；且 (1,2,3,1) 是簡單環。",
          "en": "Vertex 4 is isolated, so H is disconnected; and (1,2,3,1) is a simple cycle."
        }
      },
      {
        "claim": {
          "zh-Hant": "K=({7},∅) 的頂點 7 是葉",
          "en": "The vertex 7 of K=({7},∅) is a leaf"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "deg(7)=0，葉要求度數恰為 1。",
          "en": "deg(7)=0, while a leaf needs degree exactly 1."
        }
      },
      {
        "claim": {
          "zh-Hant": "把 {2,4} 再加入 G 一次，會形成環 (2,4,2)",
          "en": "Adding {2,4} to G once more creates the cycle (2,4,2)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "{2,4} 已在 E 中，邊集不變；(2,4,2) 長度 2<3，不是簡單環。",
          "en": "{2,4} is already in E, so the edge set is unchanged; (2,4,2) has length 2<3 and is not a simple cycle."
        }
      },
      {
        "claim": {
          "zh-Hant": "從 3 到 6 的道路是 (2,3,4,6)",
          "en": "The path from 3 to 6 is (2,3,4,6)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "它不從 3 出發，且 {3,4}∉E；正確的道路是 (3,2,4,6)。",
          "en": "It does not start at 3, and {3,4}∉E; the correct path is (3,2,4,6)."
        }
      },
      {
        "claim": {
          "zh-Hant": "G+{3,6} 有兩個不同的簡單環 (3,2,4,6,3) 與 (6,4,2,3,6)",
          "en": "G+{3,6} has two different simple cycles, (3,2,4,6,3) and (6,4,2,3,6)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "後者只是同一個環換起點並反向，邊集都是 {{3,2},{2,4},{4,6},{6,3}}（S15）。",
          "en": "The second is the same cycle with another starting point and the opposite direction; both have edge set {{3,2},{2,4},{4,6},{6,3}} (S15)."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "graph_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "（簡單無向）圖：G=(V,E)，E 中每條邊是 V 中兩個相異頂點的無序對；頂點度數是與它相連的邊數；握手引理 Σ_v deg(v)=2|E|。",
          "en": "(Simple undirected) graph: G=(V,E), where each edge is an unordered pair of distinct vertices; the degree of a vertex is the number of edges at it; handshake lemma Σ_v deg(v)=2|E|."
        }
      },
      {
        "id": "well_ordering",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "良序原理：ℕ 的每個非空子集都有最小元素。",
          "en": "Well-ordering principle: every nonempty subset of ℕ has a least element."
        }
      },
      {
        "id": "nat_induction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "數學歸納原理：對從 n₀ 起的命題族 P(n)，若 P(n₀) 成立，且對每個 n≥n₀ 都有 P(n)⇒P(n+1)，則所有 n≥n₀ 都有 P(n)。",
          "en": "Principle of induction: for a family P(n) starting at n₀, if P(n₀) holds and P(n)⇒P(n+1) for every n≥n₀, then P(n) holds for every n≥n₀."
        }
      },
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "set_builder",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合建構式：x∈{y∈A ∣ P(y)} 當且僅當 x∈A 且 P(x)。",
          "en": "Set-builder notation: x∈{y∈A ∣ P(y)} if and only if x∈A and P(x)."
        }
      },
      {
        "id": "exists_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃引入：若找到 a∈D 使 P(a) 成立，則 ∃x∈D, P(x)。",
          "en": "∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x)."
        }
      },
      {
        "id": "contradiction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。",
          "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds."
        }
      },
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "logic_chain",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "推論的傳遞：由 P⇒Q 與 Q⇒R 得 P⇒R；⇔ 同理。",
          "en": "Chaining: from P⇒Q and Q⇒R infer P⇒R; likewise for ⇔."
        }
      },
      {
        "id": "subset_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "子集：A⊆B 表示 A 的每個元素都是 B 的元素，即 ∀x (x∈A⇒x∈B)。",
          "en": "Subset: A⊆B means every element of A is an element of B, i.e. ∀x (x∈A⇒x∈B)."
        }
      },
      {
        "id": "contrapositive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "逆否命題：P⇒Q 與 ¬Q⇒¬P 等價。逆命題 Q⇒P 一般不與 P⇒Q 等價。",
          "en": "Contrapositive: P⇒Q is equivalent to ¬Q⇒¬P. The converse Q⇒P is in general not equivalent to P⇒Q."
        }
      },
      {
        "id": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "case_analysis",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分情況論證：若 P∨Q 成立，且 P⇒R、Q⇒R 都已證，則 R 成立。所有情況必須涵蓋全部可能。",
          "en": "Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility."
        }
      },
      {
        "id": "mko_tree_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-tree",
          "sha256": "48c7310136a8fd9749e9a9af278b0ade15fdecb8f4cc5657f4f0895eef89aebf",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在有限簡單無向圖 G=(V,E) 中，樹是 V 非空、任意兩頂點可由道路連接、且沒有簡單環的圖；葉是度數恰為 1 的頂點。",
          "en": "In a finite simple undirected graph G=(V,E), a tree is a graph with V nonempty, every pair of vertices joined by a path, and no simple cycle; a leaf is a vertex of degree exactly 1."
        }
      },
      {
        "id": "walk_path_cycle",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "長度 k 的走道是頂點序列 w_0、w_1 一直到 w_k，其中每對相鄰的 w_{j−1}、w_j 構成一條邊；道路是不重複頂點的走道（k=0 合法）；簡單環是長度 k≥3、w_k=w_0、且 w_0 到 w_{k−1} 兩兩相異的走道。",
          "en": "A walk of length k is a sequence of vertices w_0, w_1 and so on up to w_k in which each consecutive pair w_{j−1}, w_j is an edge; a path is a walk with no repeated vertex (k=0 allowed); a simple cycle is a walk of length k≥3 with w_k=w_0 and w_0 to w_{k−1} pairwise distinct."
        }
      },
      {
        "id": "delete_vertex",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "G−v 的頂點集是 V∖{v}，邊集是 E 中不含 v 的邊；G−v 的走道、道路、環也都是 G 的走道、道路、環。",
          "en": "G−v has vertex set V∖{v} and the edges of E not containing v; every walk, path or cycle of G−v is also one of G."
        }
      },
      {
        "id": "L_walk_path",
        "kind": "lemma",
        "proved_in": [
          "S06.1",
          "S06.2"
        ],
        "statement": {
          "zh-Hant": "若圖中有從 u 到 v 的走道，則有從 u 到 v 的道路。",
          "en": "If a graph has a walk from u to v, it has a path from u to v."
        }
      },
      {
        "id": "L_leaf_cycle",
        "kind": "lemma",
        "proved_in": [
          "S07.1",
          "S07.2"
        ],
        "statement": {
          "zh-Hant": "若 deg_G(v)≤1，則 G 有簡單環 ⇔ G−v 有簡單環。",
          "en": "If deg_G(v)≤1, then G has a simple cycle ⇔ G−v has a simple cycle."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03",
        "lhs": "1+3+1+3+1+1",
        "rhs": "2*5"
      },
      {
        "id": "N2",
        "step": "S02",
        "lhs": "5",
        "rhs": "6-1"
      },
      {
        "id": "N3",
        "step": "S16",
        "lhs": "3",
        "rhs": "4-1"
      },
      {
        "id": "N4",
        "step": "S16",
        "lhs": "2+2+2+0",
        "rhs": "2*3"
      },
      {
        "id": "N5",
        "step": "S06",
        "lhs": "comb(6,2)",
        "rhs": "15"
      },
      {
        "id": "N6",
        "step": "S07",
        "lhs": "5-4",
        "rhs": "1"
      },
      {
        "id": "N7",
        "step": "S14",
        "lhs": "1+3+2+3+1+2",
        "rhs": "2*(5+1)"
      },
      {
        "id": "N8",
        "step": "S06.2",
        "lhs": "5-(3-1)",
        "rhs": "3"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理有限簡單無向圖（MKO 假設 a1），以一個具體的 6 頂點圖為主例。連通性用「走道含道路」引理證明，無環性用「度數至多 1 的頂點不在簡單環上」引理逐步刪葉證明，兩個引理都在本稿證明。從 3 到 6 的道路唯一性只對本例用窮舉證明；一般定理「樹中任兩頂點的道路唯一」與「樹有 |V|−1 條邊」不在本稿證明。不改動 canonical MKO 或英文 edition。",
      "en": "Treats finite simple undirected graphs only (MKO assumption a1), with one concrete 6-vertex graph as the main example. Connectivity is proved with the lemma 'a walk contains a path', and acyclicity by deleting leaves one at a time with the lemma 'a vertex of degree at most 1 lies on no simple cycle'; both lemmas are proved here. Uniqueness of the path from 3 to 6 is proved for this example only, by exhaustion; the general theorems 'any two vertices of a tree have a unique path' and 'a tree has |V|−1 edges' are not proved here. The canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "G=(V,E)",
        "主例的圖",
        "the main example graph"
      ],
      [
        "N(v),\\ \\deg(v)",
        "鄰居集合與度數",
        "neighbour set and degree"
      ],
      [
        "(v_0,v_1,v_2)",
        "以頂點順序寫出的道路或環",
        "a path or cycle written as its vertex sequence"
      ],
      [
        "G-v",
        "刪去頂點 v 及含 v 的邊",
        "delete the vertex v and the edges containing it"
      ],
      [
        "G+e",
        "加入新邊 e",
        "add the new edge e"
      ],
      [
        "H,\\ K",
        "對照用的兩個圖",
        "the two contrast graphs"
      ]
    ],
    "md_error_controls": [
      {
        "title": "只看邊數不能判樹 / Edge count alone does not determine whether a graph is a tree",
        "zh": "MKO 指出的誤解：只看 |E|=|V|−1 就判樹。H（三角形加孤立點 4）有 4 個頂點 3 條邊，卻不連通且有環。本稿判定 G 是樹時完全沒用到邊數，而是直接證明連通與無環（S06、S09）。",
        "en": "The misconception the MKO names: deciding that a graph is a tree from |E|=|V|−1 alone. H (a triangle plus the isolated vertex 4) has 4 vertices and 3 edges, yet is disconnected and has a cycle. This guide decides that G is a tree without using the edge count at all, by proving connectivity and acyclicity directly (S06, S09)."
      },
      {
        "title": "單點樹沒有葉 / A single-vertex tree has no leaf",
        "zh": "K=({7},∅) 是樹，但 deg(7)=0，不是度數 1，所以依本 MKO 的葉約定 K 的葉集合是空集合。把度 0 的點叫作葉是 MKO 指出的另一個誤解。",
        "en": "K=({7},∅) is a tree, but deg(7)=0, not 1, so under this MKO's leaf convention its leaf set is empty. Calling a degree-0 vertex a leaf is another misconception the MKO names."
      },
      {
        "title": "已存在的邊不能再「新增」 / An existing edge cannot be 'added' again",
        "zh": "{2,4} 已在 E 中；把它再列一次，正規化後邊集不變，不會形成環。若把 (2,4,2) 當成環，它長度只有 2，不滿足簡單環 k≥3 的要求。MKO 假設 a3 明定加邊題必須是原本不存在的邊。",
        "en": "{2,4} is already in E; listing it again leaves the normalized edge set unchanged and creates no cycle. Treating (2,4,2) as a cycle fails the requirement k≥3 for a simple cycle. MKO assumption a3 requires the added edge to be absent from the original graph."
      },
      {
        "title": "道路按頂點順序寫 / Write paths in vertex order",
        "zh": "道路是序列，不是集合。(3,2,4,6) 與排序後的 (2,3,4,6) 不同：後者不從 3 出發，而且 {3,4} 不是邊。",
        "en": "A path is a sequence, not a set. (3,2,4,6) differs from the sorted (2,3,4,6): the latter does not start at 3, and {3,4} is not an edge."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "一般定理「樹中任兩頂點的道路唯一」與「有限樹有 |V|−1 條邊」不在本稿證明；S12 的唯一性只對 3 與 6 用窮舉證明。",
        "en": "The general theorems 'any two vertices of a tree have a unique path' and 'a finite tree has |V|−1 edges' are not proved here; the uniqueness in S12 is proved by exhaustion for 3 and 6 only."
      },
      {
        "zh-Hant": "環視為相同的標準是只差起點與方向；MKO 以固定的輸出約定（較小標籤出發、沿原道路）消除這個選擇。",
        "en": "Cycles are identified when they differ only in starting point and direction; the MKO removes this choice by a fixed output convention (start at the smaller label, follow the original path)."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "2705a4400701f09e6a4f469a879e803b4a263ba61ee3436c9788c3e7c6c938dc",
    "original_source_sha256": "2705a4400701f09e6a4f469a879e803b4a263ba61ee3436c9788c3e7c6c938dc",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion.",
    "english_binding_revision": {
      "id": "edition-symbol-bindings-20261101",
      "path": "content/edition-symbol-bindings-20261101/revision.json",
      "sha256": "e9b4c69f2c4732074c37abb624c8ab8a70afc3645c12bf9df2268ef650ad7b6c"
    }
  }
}