{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-linear-independence-basis-01",
  "mko_id": "mko-linear-independence-basis",
  "title": {
    "zh-Hant": "線性獨立與基底：以精確消去證明 ℝ² 的一組有序基底，並以兩個反例分開兩條責任",
    "en": "Linear independence and bases: proving an ordered basis of ℝ² by exact elimination, with two counterexamples that separate the two obligations"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-linear-independence-basis-01",
    "source": {
      "mko_id": "mko-linear-independence-basis",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "78c45988a718f6a7ca4acf2aba00ea4df794c8f141cefe822bf72206fa2b2f62",
      "english_sha256": "48c43771b8f6734ece79626851e05a5dca9bb984732eb3f69cffe2c6e364a912"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "vector_space_axioms",
      "cartesian_product_def",
      "eq_congruence",
      "set_extensionality",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "在 ℝ² 中令 v_1=(1,2)、v_2=(3,5)、v_3=(1,1)、w=(4,7)。(1) 以精確消去證明 (v_1,v_2) 線性獨立；(2) 對任意 (p,q) 求出係數，證明 span(v_1,v_2)=ℝ²，因而 (v_1,v_2) 是有序基底；(3) 求 w 的唯一座標；(4) 證明 (v_1) 獨立但不生成 ℝ²，而 (v_1,v_2,v_3) 生成 ℝ² 但不獨立，且 w 在其中的表示不唯一。",
      "en": "In ℝ² let v_1=(1,2), v_2=(3,5), v_3=(1,1), w=(4,7). (1) Prove by exact elimination that (v_1,v_2) is linearly independent; (2) find the coefficients for an arbitrary (p,q), proving span(v_1,v_2)=ℝ², so that (v_1,v_2) is an ordered basis; (3) find the unique coordinates of w; (4) prove that (v_1) is independent but does not span ℝ², while (v_1,v_2,v_3) spans ℝ² but is not independent, and w has more than one representation in it."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "V=ℝ²，向量逐坐標相加與數乘。v_1=(1,2)、v_2=(3,5)、v_3=(1,1)、w=(4,7)（MKO 假設 a1：整數座標有限向量族）。",
          "en": "V=ℝ² with coordinatewise addition and scalar multiplication. v_1=(1,2), v_2=(3,5), v_3=(1,1), w=(4,7) (MKO assumption a1: finite integer-coordinate vector families)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "mko_basis_def"
        ],
        "before_tex": "(u_1,u_2)\\ \\text{a finite family in}\\ V",
        "after_tex": "(u_1,u_2)\\ \\text{independent}\\iff\\big(a u_1+b u_2=0\\Rightarrow a=b=0\\big)",
        "relation": "definition",
        "rule": "mko_basis_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO 展開線性獨立：唯一能使加權和等於零向量的係數是全部為零。所以要證獨立，就是解齊次方程並證明只有平凡解（MKO 段落 0）。",
          "en": "Unfold linear independence by the MKO: the only coefficients making the weighted sum equal to the zero vector are all zero. So proving independence means solving the homogeneous equation and showing that only the trivial solution exists (MKO paragraph 0)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_basis_def"
          ],
          "explanation": {
            "zh-Hant": "只展開定義；平凡解 a=b=0 永遠是解，要證的是沒有別的解。",
            "en": "Only unfolds the definition; the trivial solution a=b=0 always works, and what must be shown is that there is no other."
          }
        },
        "leaf_basis": "mko_basis_def + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "mko_basis_def",
          "p_span_def"
        ],
        "before_tex": "B\\ \\text{basis of}\\ V",
        "after_tex": "B\\ \\text{independent}\\ \\land\\ \\operatorname{span}(B)=V",
        "relation": "definition",
        "rule": "mko_basis_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "基底是兩條責任的合取：線性獨立，且 span 等於整個 V（MKO 段落 1）。之後兩條責任分開證明，並用反例說明任何一條都不能代替另一條。",
          "en": "A basis is the conjunction of two obligations: linear independence, and span equal to all of V (MKO paragraph 1). The two are proved separately below, and counterexamples show that neither can replace the other."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_basis_def",
            "p_span_def"
          ],
          "explanation": {
            "zh-Hant": "合取的兩邊都必須成立；只證一邊不構成基底。",
            "en": "Both sides of the conjunction must hold; proving one side does not make a basis."
          }
        },
        "leaf_basis": "mko_basis_def + and_intro_elim"
      },
      {
        "id": "S03",
        "uses": [
          "G1",
          "S01",
          "cartesian_product_def"
        ],
        "before_tex": "a(1,2)+b(3,5)=(0,0)",
        "after_tex": "a+3b=0\\ \\land\\ 2a+5b=0",
        "relation": "equivalent",
        "rule": "cartesian_product_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐坐標計算 a(1,2)+b(3,5)=(a+3b, 2a+5b)。有序對等於 (0,0) 當且僅當兩個坐標都為 0，所以向量方程等價於兩個實數方程，兩個方程共用同一組 a、b。",
          "en": "Computing coordinatewise, a(1,2)+b(3,5)=(a+3b, 2a+5b). An ordered pair equals (0,0) exactly when both coordinates are 0, so the vector equation is equivalent to two real equations sharing the same a and b."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "cartesian_product_def",
            "vector_space_axioms"
          ],
          "explanation": {
            "zh-Hant": "有序對相等的條件本身是雙向的；逐坐標運算由 G1 規定。",
            "en": "The equality condition for ordered pairs is itself two-way; coordinatewise operations are fixed by G1."
          }
        },
        "leaf_basis": "cartesian_product_def + vector_space_axioms"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "L_row_op"
        ],
        "before_tex": "a+3b=0\\ \\land\\ 2a+5b=0",
        "after_tex": "a+3b=0\\ \\land\\ b=0",
        "relation": "equivalent",
        "rule": "L_row_op",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_row_op（t=2）把第二式換成 (2a+5b)−2(a+3b)=−b=0，方程組等價；S04.2 再把 −b=0 化成 b=0。這是精確有理消去的第一個主元步驟。",
          "en": "By Lemma L_row_op (t=2), replace the second equation by (2a+5b)−2(a+3b)=−b=0, which keeps the system equivalent; S04.2 then turns −b=0 into b=0. This is the first pivot step of exact rational elimination."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_row_op",
            "neg_mul"
          ],
          "explanation": {
            "zh-Hant": "行運算可逆（加回 2 倍第一式），乘以 −1 也可逆，所以解集不變。",
            "en": "The row operation is reversible (add back twice the first equation), and so is multiplying by −1, so the solution set is unchanged."
          }
        },
        "leaf_basis": "L_row_op + neg_mul"
      },
      {
        "id": "S04.1",
        "uses": [
          "eq_congruence"
        ],
        "before_tex": "E_1:\\ \\ell_1=0,\\quad E_2:\\ \\ell_2=0",
        "after_tex": "E_1,\\ E_2\\ \\Longleftrightarrow\\ E_1,\\ \\ell_2-t\\,\\ell_1=0",
        "relation": "equivalent",
        "rule": "eq_congruence",
        "conditions": [],
        "explanation": {
          "zh-Hant": "設某組值滿足 ℓ_1=0 與 ℓ_2=0，則 ℓ_2−tℓ_1=0−t·0=0。反之若滿足 ℓ_1=0 與 ℓ_2−tℓ_1=0，則 ℓ_2=(ℓ_2−tℓ_1)+tℓ_1=0+t·0=0。所以兩個方程組的解集相同，對任意實數 t 都成立。這證得引理 L_row_op。",
          "en": "If some values satisfy ℓ_1=0 and ℓ_2=0, then ℓ_2−tℓ_1=0−t·0=0. Conversely, if they satisfy ℓ_1=0 and ℓ_2−tℓ_1=0, then ℓ_2=(ℓ_2−tℓ_1)+tℓ_1=0+t·0=0. So the two systems have the same solution set, for every real t. This proves Lemma L_row_op."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_congruence",
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "兩個方向都寫出了；關鍵是保留 E_1，否則反向推不回 E_2。",
            "en": "Both directions are written out; the key is that E_1 is kept, otherwise E_2 could not be recovered."
          }
        },
        "leaf_basis": "eq_congruence + mul_zero + add_zero"
      },
      {
        "id": "S04.2",
        "uses": [
          "S04.1",
          "distributive"
        ],
        "before_tex": "(2a+5b)-2(a+3b)=0",
        "after_tex": "-b=0\\ \\Longleftrightarrow\\ b=0",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "展開：2a+5b−2a−6b=−b（N1 為恆等式）。−b=0 兩邊乘 −1 得 b=0，反向再乘 −1。",
          "en": "Expand: 2a+5b−2a−6b=−b (N1 is an identity). Multiplying −b=0 by −1 gives b=0, and multiplying by −1 again reverses it."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "neg_mul"
          ],
          "explanation": {
            "zh-Hant": "展開是恆等式；乘 −1 可逆。",
            "en": "The expansion is an identity; multiplication by −1 is reversible."
          }
        },
        "leaf_basis": "distributive + neg_mul + add_inverse"
      },
      {
        "id": "S05",
        "uses": [
          "S04"
        ],
        "before_tex": "a+3b=0\\ \\land\\ b=0",
        "after_tex": "a=0\\ \\land\\ b=0",
        "relation": "equivalent",
        "rule": "eq_replace",
        "conditions": [],
        "explanation": {
          "zh-Hant": "回代 b=0：a+3·0=a=0。所以齊次方程組只有平凡解。用消去的語言說，兩個方程都有主元，rank 為 2，等於向量個數。",
          "en": "Back-substitute b=0: a+3·0=a=0. So the homogeneous system has only the trivial solution. In the language of elimination, both equations have pivots, so the rank is 2, equal to the number of vectors."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_replace",
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "反向：a=0、b=0 代回 a+3b=0 成立。",
            "en": "Backward: a=0, b=0 satisfy a+3b=0."
          }
        },
        "leaf_basis": "eq_replace + mul_zero + add_zero"
      },
      {
        "id": "S06",
        "uses": [
          "S03",
          "S04",
          "S05",
          "S01"
        ],
        "before_tex": "a v_1+b v_2=0",
        "after_tex": "a=b=0:\\ (v_1,v_2)\\ \\text{independent}",
        "relation": "implies",
        "rule": "def_unfold",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S03 到 S05 是一條等價鏈：a v_1+b v_2=0 ⇔ a=b=0。依 S01 的定義，(v_1,v_2) 線性獨立。第一條基底責任完成。",
          "en": "S03 to S05 form a chain of equivalences: a v_1+b v_2=0 ⇔ a=b=0. By the definition in S01, (v_1,v_2) is linearly independent. The first basis obligation is met."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "S01",
            "logic_chain"
          ],
          "explanation": {
            "zh-Hant": "只用到 ⇒ 方向；獨立性本身不說明 span，span 另在 S08 證明。",
            "en": "Only the ⇒ direction is used; independence by itself says nothing about the span, which is proved separately in S08."
          }
        },
        "leaf_basis": "def_unfold + logic_chain"
      },
      {
        "id": "S07",
        "uses": [
          "G1",
          "cartesian_product_def",
          "L_row_op"
        ],
        "before_tex": "a(1,2)+b(3,5)=(p,q)",
        "after_tex": "a=-5p+3q,\\ \\ b=2p-q",
        "relation": "equivalent",
        "rule": "L_row_op",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對任意目標 (p,q)，向量方程等價於 a+3b=p、2a+5b=q（逐坐標）。同樣的行運算（S07.1）給 b=2p−q，回代（S07.2）給 a=−5p+3q；S07.3 直接驗證這組係數確實給出 (p,q)。",
          "en": "For an arbitrary target (p,q), the vector equation is equivalent to a+3b=p and 2a+5b=q (coordinatewise). The same row operation (S07.1) gives b=2p−q, back-substitution (S07.2) gives a=−5p+3q, and S07.3 verifies directly that these coefficients produce (p,q)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_row_op",
            "eq_congruence"
          ],
          "explanation": {
            "zh-Hant": "每一步都是等價變形，所以這組係數不但是解，而且是唯一解；其後的 S07.3 是額外的直接核對。",
            "en": "Every step is an equivalence, so these coefficients are not only a solution but the only one; S07.3, which follows, is an extra direct check."
          }
        },
        "leaf_basis": "L_row_op + eq_congruence"
      },
      {
        "id": "S07.1",
        "uses": [
          "L_row_op",
          "distributive"
        ],
        "before_tex": "a+3b=p\\ \\land\\ 2a+5b=q",
        "after_tex": "a+3b=p\\ \\land\\ b=2p-q",
        "relation": "equivalent",
        "rule": "L_row_op",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第二式減 2 倍第一式：(2a+5b)−2(a+3b)=q−2p，即 −b=q−2p，兩邊乘 −1 得 b=2p−q。",
          "en": "Subtract twice the first equation from the second: (2a+5b)−2(a+3b)=q−2p, that is −b=q−2p; multiplying by −1 gives b=2p−q."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_row_op",
            "neg_mul"
          ],
          "explanation": {
            "zh-Hant": "與 S04 相同的可逆行運算，只是右邊改為任意實數 p、q（也允許同時為 0）。",
            "en": "The same reversible row operation as in S04, now with an arbitrary real right-hand side (p,q), including (0,0)."
          }
        },
        "leaf_basis": "L_row_op + distributive + neg_mul"
      },
      {
        "id": "S07.2",
        "uses": [
          "S07.1",
          "eq_replace"
        ],
        "before_tex": "a+3b=p,\\ b=2p-q",
        "after_tex": "a=p-3(2p-q)=-5p+3q,\\quad b=2p-q",
        "relation": "equivalent",
        "rule": "eq_replace",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把 b=2p−q 代入第一式並移項：a=p−3(2p−q)=p−6p+3q=−5p+3q。",
          "en": "Substitute b=2p−q into the first equation and move terms: a=p−3(2p−q)=p−6p+3q=−5p+3q."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_replace",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "代入與移項都可逆。",
            "en": "Substitution and moving terms are both reversible."
          }
        },
        "leaf_basis": "eq_replace + distributive + eq_congruence"
      },
      {
        "id": "S07.3",
        "uses": [
          "S07.2"
        ],
        "before_tex": "(-5p+3q)(1,2)+(2p-q)(3,5)",
        "after_tex": "(p,q)",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第一坐標 (−5p+3q)+3(2p−q)=p（N2），第二坐標 2(−5p+3q)+5(2p−q)=q（N3），兩者都是 p、q 的多項式恆等式。所以每個 (p,q) 都是 v_1、v_2 的線性組合。",
          "en": "First coordinate (−5p+3q)+3(2p−q)=p (N2), second coordinate 2(−5p+3q)+5(2p−q)=q (N3), both polynomial identities in p and q. So every (p,q) is a linear combination of v_1 and v_2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "恆等式對所有 p、q 成立；N2、N3 在足夠的格點上驗證一次式恆等。",
            "en": "The identities hold for all p, q; N2 and N3 check these degree-1 identities on a sufficient grid."
          }
        },
        "leaf_basis": "distributive + vector_space_axioms"
      },
      {
        "id": "S08",
        "uses": [
          "S07",
          "p_span_def",
          "set_extensionality",
          "forall_intro"
        ],
        "before_tex": "\\forall (p,q)\\in\\mathbb{R}^2:\\ (p,q)\\in\\operatorname{span}(v_1,v_2)",
        "after_tex": "\\operatorname{span}(v_1,v_2)=\\mathbb{R}^2",
        "relation": "implies",
        "rule": "set_extensionality",
        "conditions": [],
        "explanation": {
          "zh-Hant": "⊆：任何線性組合仍在 ℝ² 中（向量空間對加法與數乘封閉）。⊇：S07 對任意 (p,q) 給出係數，(p,q) 沒有任何限制，依 ∀引入得每個 (p,q) 都在 span 中。兩個包含合起來依外延性得相等。第二條基底責任完成。",
          "en": "⊆: every linear combination stays in ℝ² (a vector space is closed under addition and scalar multiplication). ⊇: S07 gives coefficients for an arbitrary (p,q) with no restriction on it, so ∀-introduction puts every (p,q) in the span. The two inclusions give equality by extensionality. The second basis obligation is met."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "forall_intro",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "生成性是對所有目標的全稱命題；有限個目標的核對不夠，這裡用的是對任意 (p,q) 的公式。",
            "en": "Spanning is a universal statement over all targets; checking finitely many targets would not suffice, and the formula used here is for arbitrary (p,q)."
          }
        },
        "leaf_basis": "set_extensionality + forall_intro + vector_space_axioms"
      },
      {
        "id": "S09",
        "uses": [
          "S02",
          "S06",
          "S08"
        ],
        "before_tex": "(v_1,v_2)\\ \\text{independent},\\ \\operatorname{span}(v_1,v_2)=\\mathbb{R}^2",
        "after_tex": "B=(v_1,v_2)\\ \\text{is an ordered basis of}\\ \\mathbb{R}^2",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S06 與 S08 分別完成 S02 的兩條責任，合取得到 B 是 ℝ² 的有序基底。",
          "en": "S06 and S08 discharge the two obligations of S02, and their conjunction makes B an ordered basis of ℝ²."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "S08",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "兩條都用到了；S12 與 S14 顯示少任何一條都不行。",
            "en": "Both obligations are used; S12 and S14 show that neither can be missing."
          }
        },
        "leaf_basis": "and_intro_elim"
      },
      {
        "id": "S10",
        "uses": [
          "S07",
          "S09",
          "L_unique_coords"
        ],
        "before_tex": "w=(4,7)",
        "after_tex": "w=1\\cdot v_1+1\\cdot v_2,\\ \\ [w]_B=(1,1)\\ \\text{unique}",
        "relation": "implies",
        "rule": "L_unique_coords",
        "conditions": [],
        "explanation": {
          "zh-Hant": "代入 S07 的公式（p=4、q=7）：a=−20+21=1、b=8−7=1（N4、N5）；核對 (1,2)+(3,5)=(4,7)。依引理 L_unique_coords（S10.1），因為 B 獨立，這組座標唯一（MKO 段落 2：有序基底成立後才有唯一座標）。",
          "en": "Substitute into the formula of S07 (p=4, q=7): a=−20+21=1, b=8−7=1 (N4, N5); check (1,2)+(3,5)=(4,7). By Lemma L_unique_coords (S10.1), since B is independent these coordinates are unique (MKO paragraph 2: unique coordinates come only once the ordered basis is established)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "L_unique_coords",
            "S06"
          ],
          "explanation": {
            "zh-Hant": "存在性來自 S07（生成），唯一性來自 S06（獨立）；兩者分工不同。",
            "en": "Existence comes from S07 (spanning) and uniqueness from S06 (independence); the two do different jobs."
          }
        },
        "leaf_basis": "L_unique_coords + numeric_evaluation"
      },
      {
        "id": "S10.1",
        "uses": [
          "S01",
          "vector_space_axioms"
        ],
        "before_tex": "a u_1+b u_2=a' u_1+b' u_2",
        "after_tex": "(a-a')u_1+(b-b')u_2=0\\ \\Rightarrow\\ a=a',\\ b=b'",
        "relation": "implies",
        "rule": "vector_space_axioms",
        "conditions": [
          "(u_1,u_2)\\ \\text{independent}"
        ],
        "explanation": {
          "zh-Hant": "兩邊同減 a' u_1+b' u_2，再用分配律 (a−a')u_1=a u_1−a' u_1 等，得 (a−a')u_1+(b−b')u_2=0。依獨立性的定義（S01），係數 a−a' 與 b−b' 都是 0，所以 a=a'、b=b'。這證得引理 L_unique_coords。",
          "en": "Subtract a' u_1+b' u_2 from both sides and use distributivity, (a−a')u_1=a u_1−a' u_1 and likewise, to get (a−a')u_1+(b−b')u_2=0. By the definition of independence (S01), the coefficients a−a' and b−b' are both 0, so a=a' and b=b'. This proves Lemma L_unique_coords."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "vector_space_axioms"
          ],
          "explanation": {
            "zh-Hant": "唯一性完全依賴獨立性；若族不獨立，S15 顯示表示可以不唯一。",
            "en": "Uniqueness rests entirely on independence; for a dependent family S15 shows that representations need not be unique."
          }
        },
        "leaf_basis": "vector_space_axioms + def_unfold"
      },
      {
        "id": "S11",
        "uses": [
          "S07",
          "G1"
        ],
        "before_tex": "\\det\\begin{pmatrix}1&3\\\\2&5\\end{pmatrix}",
        "after_tex": "1\\cdot5-3\\cdot2=-1\\neq0",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "交叉核對（MKO 假設 a2 允許對有序方形 ℝ² 族使用）：以 v_1、v_2 為列的矩陣行列式是 −1（N6）。S07 的係數可寫成 a=(5p−3q)/(−1)、b=(1·q−2p)/(−1)，分母正是這個行列式；它不為 0，與消去法沒有遇到零主元一致。本稿的結論不依賴行列式判準。",
          "en": "Cross-check (allowed by MKO assumption a2 for ordered square ℝ² families): the matrix with columns v_1, v_2 has determinant −1 (N6). The coefficients of S07 can be written a=(5p−3q)/(−1) and b=(1·q−2p)/(−1), whose denominator is exactly this determinant; it is nonzero, consistent with elimination meeting no zero pivot. The conclusions of this guide do not depend on the determinant criterion."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "numeric_evaluation",
            "S07"
          ],
          "explanation": {
            "zh-Hant": "只是一致性核對；一般的「det≠0 ⇔ 列構成基底」未在本稿證明，也不用於非方形族。",
            "en": "Only a consistency check; the general criterion 'det≠0 ⇔ the columns form a basis' is not proved here and is not used for non-square families."
          }
        },
        "leaf_basis": "numeric_evaluation + frac_rules"
      },
      {
        "id": "S12",
        "uses": [
          "G1",
          "S01",
          "p_span_def"
        ],
        "before_tex": "(v_1)=((1,2))",
        "after_tex": "(v_1)\\ \\text{independent, but}\\ (1,0)\\notin\\operatorname{span}(v_1)",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "獨立：a(1,2)=(0,0) 的第一坐標給 a=0，所以只有平凡解。不生成：若 (1,0)=a(1,2)，第一坐標給 a=1，第二坐標給 2a=0 即 a=0，矛盾。所以 span(v_1)≠ℝ²。獨立而不生成，不是基底。",
          "en": "Independent: the first coordinate of a(1,2)=(0,0) gives a=0, so only the trivial solution exists. Not spanning: if (1,0)=a(1,2), the first coordinate gives a=1 and the second gives 2a=0, that is a=0, a contradiction. So span(v_1)≠ℝ². Independent but not spanning, it is not a basis."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "contradiction",
            "cartesian_product_def"
          ],
          "explanation": {
            "zh-Hant": "否定生成只需一個不在 span 中的目標；兩個坐標方程要求同一個 a，不能各取不同的 a。",
            "en": "Refuting spanning needs one target outside the span; the two coordinate equations require the same a and may not use different values."
          }
        },
        "leaf_basis": "contradiction + cartesian_product_def"
      },
      {
        "id": "S13",
        "uses": [
          "S07",
          "p_span_def"
        ],
        "before_tex": "B'=(v_1,v_2,v_3)",
        "after_tex": "(p,q)=(-5p+3q)v_1+(2p-q)v_2+0\\cdot v_3:\\ \\operatorname{span}(B')=\\mathbb{R}^2",
        "relation": "implies",
        "rule": "p_span_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "B' 包含 v_1、v_2，S07 的係數再加上 v_3 的係數 0，就把任意 (p,q) 寫成 B' 的線性組合。所以 B' 生成 ℝ²。",
          "en": "B' contains v_1 and v_2; the coefficients of S07 together with coefficient 0 on v_3 write any (p,q) as a linear combination of B'. So B' spans ℝ²."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "p_span_def"
          ],
          "explanation": {
            "zh-Hant": "加入向量不會使 span 變小；這裡只宣稱生成，不宣稱獨立。",
            "en": "Adding vectors cannot shrink the span; only spanning is claimed here, not independence."
          }
        },
        "leaf_basis": "p_span_def + add_zero + mul_zero"
      },
      {
        "id": "S14",
        "uses": [
          "S07",
          "S01",
          "G1"
        ],
        "before_tex": "v_3=(1,1)",
        "after_tex": "-2v_1+v_2-v_3=(0,0)\\ \\text{with nonzero coefficients}",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "用 S07 的公式求 v_3 的座標（p=1、q=1）：a=−5+3=−2、b=2−1=1，所以 v_3=−2v_1+v_2（S14.1 核對）。移項得 −2v_1+v_2−v_3=0，係數 (−2,1,−1) 不全為零。依 S01 的否定，B' 不獨立。生成而不獨立，也不是基底。",
          "en": "Use the formula of S07 for the coordinates of v_3 (p=1, q=1): a=−5+3=−2, b=2−1=1, so v_3=−2v_1+v_2 (checked in S14.1). Rearranging gives −2v_1+v_2−v_3=0 with coefficients (−2,1,−1), not all zero. By the negation of S01, B' is not independent. Spanning but not independent, it is not a basis either."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "quantifier_negation",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "否定獨立只需一組非平凡係數；也可說三個向量消去後 rank 為 2，小於向量個數 3。",
            "en": "Refuting independence needs one nontrivial set of coefficients; equivalently, elimination gives rank 2, less than the 3 vectors."
          }
        },
        "leaf_basis": "quantifier_negation + numeric_evaluation"
      },
      {
        "id": "S14.1",
        "uses": [
          "G1"
        ],
        "before_tex": "-2(1,2)+(3,5)-(1,1)",
        "after_tex": "(-2+3-1,\\ -4+5-1)=(0,0)",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐坐標計算：第一坐標 −2+3−1=0（N7），第二坐標 −4+5−1=0（N8）。",
          "en": "Compute coordinatewise: first coordinate −2+3−1=0 (N7), second coordinate −4+5−1=0 (N8)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation",
            "cartesian_product_def"
          ],
          "explanation": {
            "zh-Hant": "精確整數計算，兩邊同一個向量。",
            "en": "Exact integer computation; both sides are the same vector."
          }
        },
        "leaf_basis": "numeric_evaluation + vector_space_axioms"
      },
      {
        "id": "S15",
        "uses": [
          "S10",
          "S14"
        ],
        "before_tex": "w=1\\cdot v_1+1\\cdot v_2+0\\cdot v_3",
        "after_tex": "w=-1\\cdot v_1+2\\cdot v_2-1\\cdot v_3\\ \\ (\\text{a second representation})",
        "relation": "implies",
        "rule": "vector_space_axioms",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把 S14 的零組合加到 S10 的表示上：(1,1,0)+(−2,1,−1)=(−1,2,−1)。核對：−(1,2)+2(3,5)−(1,1)=(−1+6−1, −2+10−1)=(4,7)（N9、N10）。所以 w 在 B' 中有兩種不同表示，座標不唯一；這正是 L_unique_coords 需要獨立性的原因。",
          "en": "Add the zero combination of S14 to the representation of S10: (1,1,0)+(−2,1,−1)=(−1,2,−1). Check: −(1,2)+2(3,5)−(1,1)=(−1+6−1, −2+10−1)=(4,7) (N9, N10). So w has two different representations in B', and coordinates are not unique; this is why L_unique_coords needs independence."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S14",
            "vector_space_axioms"
          ],
          "explanation": {
            "zh-Hant": "加上零向量不改變結果；非平凡的零組合讓每個向量都有無限多種表示（加上它的任意倍數）。",
            "en": "Adding the zero vector changes nothing; a nontrivial zero combination gives every vector infinitely many representations (add any multiple of it)."
          }
        },
        "leaf_basis": "vector_space_axioms + numeric_evaluation"
      },
      {
        "id": "S16",
        "uses": [
          "S09",
          "S10",
          "S12",
          "S14",
          "S15"
        ],
        "before_tex": "\\text{S09, S10, S12, S14, S15}",
        "after_tex": "\\text{basis}=\\text{independent}\\land\\text{spanning};\\ \\text{neither replaces the other}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "總結：(v_1,v_2) 兩條責任都成立，是有序基底，w 的唯一座標為 (1,1)；(v_1) 只有獨立、(v_1,v_2,v_3) 只有生成，兩者都不是基底，後者中 w 的表示不唯一。",
          "en": "Summary: (v_1,v_2) meets both obligations and is an ordered basis, with unique coordinates (1,1) for w; (v_1) is only independent and (v_1,v_2,v_3) is only spanning, so neither is a basis, and in the latter w has more than one representation."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09",
            "S12",
            "S14"
          ],
          "explanation": {
            "zh-Hant": "只合取已證結論。",
            "en": "Only conjoins conclusions already proved."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "(v_1,v_2)=((1,2),(3,5)) 線性獨立（消去只得平凡解），且任意 (p,q)=(−5p+3q)v_1+(2p−q)v_2，故生成 ℝ²，是有序基底；w=(4,7) 的唯一座標是 (1,1)；det=−1 作為交叉核對。(v_1) 獨立但 (1,0) 不在其 span 中；(v_1,v_2,v_3) 生成 ℝ² 但 −2v_1+v_2−v_3=0，w 也可寫成 −v_1+2v_2−v_3。",
      "en": "(v_1,v_2)=((1,2),(3,5)) is linearly independent (elimination gives only the trivial solution), and every (p,q)=(−5p+3q)v_1+(2p−q)v_2, so it spans ℝ² and is an ordered basis; the unique coordinates of w=(4,7) are (1,1); det=−1 serves as a cross-check. (v_1) is independent but (1,0) is not in its span; (v_1,v_2,v_3) spans ℝ² but −2v_1+v_2−v_3=0, and w can also be written −v_1+2v_2−v_3."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "(v_1,v_2,v_3) 生成 ℝ²，所以它是 ℝ² 的基底",
          "en": "(v_1,v_2,v_3) spans ℝ², so it is a basis of ℝ²"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "−2v_1+v_2−v_3=0 是非平凡零組合，族不獨立（S14）。",
          "en": "−2v_1+v_2−v_3=0 is a nontrivial zero combination, so the family is not independent (S14)."
        }
      },
      {
        "claim": {
          "zh-Hant": "(v_1) 線性獨立，所以它是 ℝ² 的基底",
          "en": "(v_1) is linearly independent, so it is a basis of ℝ²"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "(1,0) 不在 span(v_1) 中（S12）。",
          "en": "(1,0) is not in span(v_1) (S12)."
        }
      },
      {
        "claim": {
          "zh-Hant": "w 在 (v_1,v_2,v_3) 中的座標唯一，等於 (1,1,0)",
          "en": "The coordinates of w in (v_1,v_2,v_3) are unique and equal (1,1,0)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "(−1,2,−1) 也給出 w（S15）。",
          "en": "(−1,2,−1) also gives w (S15)."
        }
      },
      {
        "claim": {
          "zh-Hant": "非方形族 (v_1,v_2,v_3) 的「行列式」為零，所以它不生成 ℝ²",
          "en": "The 'determinant' of the non-square family (v_1,v_2,v_3) is zero, so it does not span ℝ²"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "2×3 矩陣沒有行列式（MKO 假設 a2 只對方形 ℝ² 族使用），而且這個族確實生成 ℝ²（S13）。",
          "en": "A 2×3 matrix has no determinant (MKO assumption a2 uses it only for square ℝ² families), and this family does span ℝ² (S13)."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "vector_space_axioms",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "向量空間公理（實數或其他域 F 上）：加法成交換群；純量乘法滿足 a(u+v)=au+av、(a+b)u=au+bu、(ab)u=a(bu)、1u=u。",
          "en": "Vector space axioms (over ℝ or another field F): vectors form a commutative group under addition; scalar multiplication satisfies a(u+v)=au+av, (a+b)u=au+bu, (ab)u=a(bu) and 1u=u."
        }
      },
      {
        "id": "cartesian_product_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "笛卡兒積：A×B 是所有有序對 (a,b)（a∈A、b∈B）的集合；(a,b)=(c,d) ⇔ a=c 且 b=d。",
          "en": "Cartesian product: A×B is the set of ordered pairs (a,b) with a∈A and b∈B; (a,b)=(c,d) ⇔ a=c and b=d."
        }
      },
      {
        "id": "eq_congruence",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等式同用：若 u=v，對兩邊做同一個有定義的運算 f，得 f(u)=f(v)。只對一邊做運算不屬於此規則。",
          "en": "Congruence of equality: if u=v, applying the same defined operation f to both sides gives f(u)=f(v). Operating on one side only is not an instance of this rule."
        }
      },
      {
        "id": "set_extensionality",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合相等：A=B 當且僅當 A⊆B 且 B⊆A，即兩者有相同元素。",
          "en": "Set equality: A=B if and only if A⊆B and B⊆A, i.e. they have the same elements."
        }
      },
      {
        "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": "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": "neg_mul",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "負號與乘法：(−u)v=−(uv)，(−u)(−v)=uv。",
          "en": "Signs and multiplication: (−u)v=−(uv) and (−u)(−v)=uv."
        }
      },
      {
        "id": "mul_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘以零：u·0=0。理由：u·0=u·(0+0)=u·0+u·0，兩邊同減 u·0 得 0=u·0。",
          "en": "Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0."
        }
      },
      {
        "id": "add_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "id": "distributive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。",
          "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu."
        }
      },
      {
        "id": "add_inverse",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。",
          "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)."
        }
      },
      {
        "id": "eq_replace",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。",
          "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true."
        }
      },
      {
        "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": "forall_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。",
          "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain."
        }
      },
      {
        "id": "frac_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。",
          "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0."
        }
      },
      {
        "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": "quantifier_negation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。",
          "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)."
        }
      },
      {
        "id": "mko_basis_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-linear-independence-basis",
          "sha256": "78c45988a718f6a7ca4acf2aba00ea4df794c8f141cefe822bf72206fa2b2f62",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "有限向量族線性獨立，若係數加權和為零只可能所有係數皆零。向量族是空間 V 的基底，必須同時線性獨立且 span 等於 V；兩條責任不可互相代替。",
          "en": "A finite vector family is linearly independent if a weighted sum equal to zero requires every coefficient to be zero. To be a basis of a space V, it must be independent and have span equal to V. Neither obligation replaces the other."
        }
      },
      {
        "id": "p_span_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限族 (u_i) 的 span 是所有線性組合 Σ a_i u_i（a_i∈ℝ）的集合；它是 V 的子集。",
          "en": "The span of a finite family (u_i) is the set of all linear combinations Σ a_i u_i (a_i∈ℝ); it is a subset of V."
        }
      },
      {
        "id": "L_row_op",
        "kind": "lemma",
        "proved_in": [
          "S04.1"
        ],
        "statement": {
          "zh-Hant": "對實數方程組 {E_1, E_2} 與任意實數 t：把 E_2 換成 E_2−tE_1 得到等價的方程組（解集相同）。",
          "en": "For a system of real equations {E_1, E_2} and any real t: replacing E_2 by E_2−tE_1 gives an equivalent system (the same solution set)."
        }
      },
      {
        "id": "L_unique_coords",
        "kind": "lemma",
        "proved_in": [
          "S10.1"
        ],
        "statement": {
          "zh-Hant": "若 (u_1,u_2) 線性獨立且 a u_1+b u_2=a' u_1+b' u_2，則 a=a' 且 b=b'。",
          "en": "If (u_1,u_2) is linearly independent and a u_1+b u_2=a' u_1+b' u_2, then a=a' and b=b'."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S04.2",
        "vars": [
          "a",
          "b"
        ],
        "degree": 1,
        "lhs": "(2*a+5*b)-2*(a+3*b)",
        "rhs": "-b"
      },
      {
        "id": "N2",
        "step": "S07.3",
        "vars": [
          "p",
          "q"
        ],
        "degree": 1,
        "lhs": "(-5*p+3*q)*1+(2*p-q)*3",
        "rhs": "p"
      },
      {
        "id": "N3",
        "step": "S07.3",
        "vars": [
          "p",
          "q"
        ],
        "degree": 1,
        "lhs": "(-5*p+3*q)*2+(2*p-q)*5",
        "rhs": "q"
      },
      {
        "id": "N4",
        "step": "S10",
        "lhs": "-5*4+3*7",
        "rhs": "1"
      },
      {
        "id": "N5",
        "step": "S10",
        "lhs": "2*4-7",
        "rhs": "1"
      },
      {
        "id": "N6",
        "step": "S11",
        "lhs": "1*5-3*2",
        "rhs": "-1"
      },
      {
        "id": "N7",
        "step": "S14.1",
        "lhs": "-2*1+3-1",
        "rhs": "0"
      },
      {
        "id": "N8",
        "step": "S14.1",
        "lhs": "-2*2+5-1",
        "rhs": "0"
      },
      {
        "id": "N9",
        "step": "S15",
        "lhs": "-1*1+2*3-1*1",
        "rhs": "4"
      },
      {
        "id": "N10",
        "step": "S15",
        "lhs": "-1*2+2*5-1*1",
        "rhs": "7"
      },
      {
        "id": "N11",
        "step": "S14",
        "lhs": "-5*1+3*1",
        "rhs": "-2"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理 G1 的整數座標向量族，以精確有理消去計算（MKO 假設 a1）。行列式只對有序方形 ℝ² 族 (v_1,v_2) 作交叉核對，不外推到非方形族（a2），也不證明一般的行列式判準。維度、基底變換與線性映射矩陣不在範圍內。不改動 canonical MKO 或其英文 edition。",
      "en": "Treats only the integer-coordinate vector families of G1, computed by exact rational elimination (MKO assumption a1). The determinant is used only as a cross-check for the ordered square ℝ² family (v_1,v_2), not for non-square families (a2), and the general determinant criterion is not proved. Dimension, change of basis and matrices of linear maps are out of scope. The canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "v_1=(1,2),\\ v_2=(3,5),\\ v_3=(1,1)",
        "給定的整數向量",
        "the given integer vectors"
      ],
      [
        "w=(4,7)",
        "求座標的目標向量",
        "the target vector whose coordinates are sought"
      ],
      [
        "a,b,c",
        "線性組合的係數",
        "coefficients of linear combinations"
      ],
      [
        "(p,q)",
        "ℝ² 中任意目標",
        "an arbitrary target in ℝ²"
      ],
      [
        "\\operatorname{span}(\\cdot)",
        "所有有限線性組合的集合",
        "the set of all finite linear combinations"
      ],
      [
        "B=(v_1,v_2),\\ B'=(v_1,v_2,v_3)",
        "有序基底與過多向量的族",
        "the ordered basis and the family with too many vectors"
      ]
    ],
    "md_error_controls": [
      {
        "title": "只生成或只獨立都不是基底 / Spanning alone or independence alone is not a basis",
        "zh": "MKO 指出的誤解是只因向量生成空間、或只因彼此獨立，就宣稱它們是基底。S12 的 (v_1) 獨立卻不生成；S14 的 (v_1,v_2,v_3) 生成卻不獨立。基底必須兩條責任都通過（S09）。",
        "en": "The misconception named by the MKO is to call vectors a basis just because they span, or just because they are independent. In S12, (v_1) is independent but does not span; in S14, (v_1,v_2,v_3) spans but is not independent. A basis must pass both obligations (S09)."
      },
      {
        "title": "消去時保留原方程 / Keep the original equation during elimination",
        "zh": "行運算等價的前提是保留被用來消去的方程（S04.1）。若把 2a+5b=0 換成 −b=0 後丟掉 a+3b=0，只剩 b=0，就會誤以為 a 可任意，進而誤判為不獨立。",
        "en": "A row operation is an equivalence only if the equation used for elimination is kept (S04.1). If 2a+5b=0 is replaced by −b=0 and a+3b=0 is dropped, only b=0 remains, suggesting that a is free and wrongly indicating dependence."
      },
      {
        "title": "生成要對所有目標 / Spanning must cover every target",
        "zh": "生成是全稱命題：S07 對任意 (p,q) 給出公式，才能推出 span=ℝ²。只核幾個目標（例如 w 與 v_3）都能表示，並不證明生成；(v_1) 就能表示 (2,4) 與 (−1,−2)，卻不能表示 (1,0)。",
        "en": "Spanning is a universal statement: only the formula of S07 for arbitrary (p,q) gives span=ℝ². Checking that a few targets (such as w and v_3) are representable does not prove spanning; (v_1) can represent (2,4) and (−1,−2) but not (1,0)."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "S11 的行列式只作交叉核對；「2×2 矩陣行列式非零 ⇔ 列構成基底」的一般判準未在本稿證明。",
        "en": "The determinant of S11 is only a cross-check; the general criterion '2×2 determinant nonzero ⇔ the columns form a basis' is not proved here."
      },
      {
        "zh-Hant": "維度、基底變換與線性映射矩陣（MKO 的 formal next）不在範圍內。",
        "en": "Dimension, change of basis and matrices of linear maps (the MKO's formal next) are out of scope."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "3daf790affb7a390725ee178abf97c6c54700877e37d65703477e3327f0019fb",
    "original_source_sha256": "e338f1259740f85df598d8def495be6c4b80bc2cbfa19ac852cf13ab5a744260",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M060"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}