{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-present-value-discounting-01",
  "mko_id": "mko-present-value-discounting",
  "title": {
    "zh-Hant": "現值與折現：證明折現與複利互逆，並算出 1331 的三年現值",
    "en": "Present value and discounting: proving that discounting inverts compounding, and the three-year present value of 1331"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-present-value-discounting-01",
    "source": {
      "mko_id": "mko-present-value-discounting",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "2e47f9d121504d577795a46f4235f98114b5701015039cc83bef029cead220b0",
      "english_sha256": "209445c349c4cc3d484122872870e437c5723eb7e12b40900d44b653eb99a043"
    },
    "mode": "theorem_proof",
    "audience_prerequisites": [
      "power_laws",
      "frac_rules",
      "mul_inverse_nonzero",
      "order_basics",
      "nat_induction",
      "function_def",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "在固定每期利率 i（1+i>0）與期數 n 下，證明 (1+i)^n>0，因此現值 PV_n(F)=F/(1+i)^n 有定義；證明 PV_n 與複利 A_n(P)=P(1+i)^n 互為反函數。再以 i=1/10（每年）、n=3、F=1331 算出現值 1000，用逐年折現與 roundtrip 交叉核對，並說明把折現寫成乘法或混用年率與月率為何錯誤。",
      "en": "For a fixed rate i per period (1+i>0) and a period count n, prove that (1+i)^n>0, so the present value PV_n(F)=F/(1+i)^n is defined; prove that PV_n and compounding A_n(P)=P(1+i)^n are inverse functions of each other. Then, with i=1/10 per year, n=3 and F=1331, compute the present value 1000, cross-check by year-by-year discounting and a round trip, and explain why writing discounting as a multiplication, or mixing annual and monthly rates, is wrong."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "固定每期利率 i 為有理數且 1+i>0；n 為自然數；期別單位、利率慣例與 n 一致；期間內無額外現金流、費用、稅或四捨五入（MKO 假設 a1）。金額為非負精確有理數（假設 a2）。",
          "en": "The fixed rate i per period is rational with 1+i>0; n is a natural number; period units, rate conventions and n agree; there are no additional cash flows, fees, taxes or rounding during the interval (MKO assumption a1). Amounts are nonnegative exact rationals (assumption a2)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "定義：n 期複利累積 A_n(P)=P(1+i)^n；未來金額 F 的現值 PV_n(F)=F/(1+i)^n。",
          "en": "Definitions: accumulation over n periods A_n(P)=P(1+i)^n; present value of a future amount F, PV_n(F)=F/(1+i)^n."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "例子：每年利率 i=1/10，期數 n=3 年，未來金額 F=1331。",
          "en": "Example: rate i=1/10 per year, n=3 years, future amount F=1331."
        }
      },
      {
        "id": "G4",
        "statement": {
          "zh-Hant": "期別對照：月利率 j=1/100，每月計息。",
          "en": "Period comparison: a monthly rate j=1/100, compounded monthly."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2",
          "mko_pv",
          "mko_compound"
        ],
        "before_tex": "\\text{compounding and discounting}",
        "after_tex": "A_n(P)=P(1+i)^n,\\ PV_n(F)=\\tfrac{F}{(1+i)^n},\\ 1+i>0",
        "relation": "definition",
        "rule": "mko_pv",
        "conditions": [
          "1+i>0"
        ],
        "explanation": {
          "zh-Hant": "依兩個 MKO 固定定義：複利 n 期後的餘額為 P(1+i)^n；未來金額 F 的現值為 F/(1+i)^n。兩者用同一個每期利率 i 與同一個期數 n，期別單位必須一致（G1）。",
          "en": "Fix the definitions from the two MKOs: after n periods of compounding the balance is P(1+i)^n; the present value of a future amount F is F/(1+i)^n. Both use the same rate per period i and the same period count n, and the period units must agree (G1)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "mko_pv",
            "mko_compound"
          ],
          "explanation": {
            "zh-Hant": "此步只陳述定義，不推出新事實；PV_n 是否有定義留待 S03。",
            "en": "This step only states the definitions and derives no new fact; whether PV_n is defined is settled in S03."
          }
        },
        "leaf_basis": "mko_pv + mko_compound + def_intro"
      },
      {
        "id": "S02",
        "uses": [
          "G1"
        ],
        "before_tex": "1+i>0",
        "after_tex": "\\forall n\\in\\mathbb{N}:\\ (1+i)^n>0",
        "relation": "implies",
        "rule": "L_pos_pow",
        "conditions": [
          "1+i>0"
        ],
        "explanation": {
          "zh-Hant": "證明引理 L_pos_pow（a=1+i）：正數的每個自然數次冪都是正數。以兩個子步做歸納。N11 是對 a=11/10、n=0 到 10 的有限核對，證明本身是歸納。",
          "en": "Prove Lemma L_pos_pow (with a=1+i): every natural-number power of a positive number is positive. Two sub-steps form an induction. N11 is a finite check for a=11/10 and n=0 to 10; the proof itself is the induction."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_pos_pow",
            "G1"
          ],
          "explanation": {
            "zh-Hant": "需要 1+i>0；若 1+i<0，奇數次冪為負；若 1+i=0，n≥1 時冪為 0。",
            "en": "It needs 1+i>0; if 1+i<0, odd powers are negative; if 1+i=0, the power is 0 for n≥1."
          }
        },
        "leaf_basis": "nat_induction + order_basics + power_laws"
      },
      {
        "id": "S02.1",
        "uses": [
          "power_laws"
        ],
        "before_tex": "n=0",
        "after_tex": "a^0=1>0",
        "relation": "induction_base",
        "rule": "nat_induction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "基底：a^0=1，而 1>0。",
          "en": "Base: a^0=1, and 1>0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "power_laws",
            "nat_induction"
          ],
          "explanation": {
            "zh-Hant": "只用 a^0=1。",
            "en": "Only a^0=1 is used."
          }
        },
        "leaf_basis": "nat_induction + power_laws"
      },
      {
        "id": "S02.2",
        "uses": [
          "S02.1",
          "power_laws",
          "order_basics"
        ],
        "before_tex": "a^n>0,\\ a>0",
        "after_tex": "a^{n+1}=a^n\\cdot a>0",
        "relation": "induction_step",
        "rule": "order_basics",
        "conditions": [
          "a>0"
        ],
        "explanation": {
          "zh-Hant": "設 a^n>0。由指數律 a^(n+1)=a^n·a。由 0<a^n 兩邊乘正數 a 得 0·a<a^n·a，即 a^(n+1)>0。依歸納原理，對所有 n 成立。S02.1 與 S02.2 證得引理 L_pos_pow。",
          "en": "Assume a^n>0. By the exponent law a^(n+1)=a^n·a. Multiplying 0<a^n by the positive number a gives 0·a<a^n·a, that is a^(n+1)>0. By induction this holds for every n. S02.1 and S02.2 prove Lemma L_pos_pow."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "power_laws",
            "nat_induction"
          ],
          "explanation": {
            "zh-Hant": "乘以正數保序是這一步的關鍵，需要 a>0。",
            "en": "That multiplying by a positive number preserves order is the key, and it needs a>0."
          }
        },
        "leaf_basis": "order_basics + power_laws + mul_zero + nat_induction"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "S01"
        ],
        "before_tex": "(1+i)^n>0",
        "after_tex": "(1+i)^n\\neq 0,\\ \\tfrac{1}{(1+i)^n}\\ \\text{exists}",
        "relation": "implies",
        "rule": "mul_inverse_nonzero",
        "conditions": [],
        "explanation": {
          "zh-Hant": "正數不等於 0，所以 (1+i)^n 有乘法反元素，PV_n(F)=F·(1/(1+i)^n) 對每個 F 都有定義。這就是 MKO 要求「正累積因子 1+i」的作用之一。",
          "en": "A positive number is not 0, so (1+i)^n has a multiplicative inverse and PV_n(F)=F·(1/(1+i)^n) is defined for every F. This is one role of the MKO's requirement of a 'positive accumulation factor 1+i'."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "mul_inverse_nonzero",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "只需要非零；正性另外保證 PV 把非負金額映到非負金額。",
            "en": "Only nonzero is needed here; positivity additionally ensures that PV maps nonnegative amounts to nonnegative amounts."
          }
        },
        "leaf_basis": "mul_inverse_nonzero + order_basics"
      },
      {
        "id": "S04",
        "uses": [
          "S01",
          "S03"
        ],
        "before_tex": "PV_n(A_n(P))",
        "after_tex": "\\tfrac{P(1+i)^n}{(1+i)^n}=P",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "(1+i)^n\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "先複利再以同一因子與期數折現：PV_n(A_n(P))=P(1+i)^n·(1/(1+i)^n)=P·1=P。用到乘法結合律與反元素，條件是 (1+i)^n≠0（S03）。",
          "en": "Compound first, then discount with the same factor and period count: PV_n(A_n(P))=P(1+i)^n·(1/(1+i)^n)=P·1=P. This uses associativity and the multiplicative inverse, under the condition (1+i)^n≠0 (S03)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "mul_inverse_nonzero",
            "mul_assoc"
          ],
          "explanation": {
            "zh-Hant": "每一步都是對所有 P 成立的恆等式，兩邊指同一個數（N12 以 i=1/10、n=3 的一次恆等式核對）。",
            "en": "Every move is an identity for all P, so both sides name the same number (N12 checks the degree-one identity for i=1/10, n=3)."
          }
        },
        "leaf_basis": "frac_rules + mul_assoc + mul_inverse_nonzero + mul_one"
      },
      {
        "id": "S05",
        "uses": [
          "S01",
          "S03"
        ],
        "before_tex": "A_n(PV_n(F))",
        "after_tex": "\\tfrac{F}{(1+i)^n}(1+i)^n=F",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "(1+i)^n\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "反方向：先折現再複利，A_n(PV_n(F))=F·(1/(1+i)^n)·(1+i)^n=F·1=F。",
          "en": "The other order: discount first, then compound, A_n(PV_n(F))=F·(1/(1+i)^n)·(1+i)^n=F·1=F."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "mul_inverse_nonzero",
            "mul_assoc"
          ],
          "explanation": {
            "zh-Hant": "同 S04，對所有 F 成立的恆等式。",
            "en": "As in S04, an identity for all F."
          }
        },
        "leaf_basis": "frac_rules + mul_assoc + mul_inverse_nonzero + mul_one"
      },
      {
        "id": "S06",
        "uses": [
          "S04",
          "S05",
          "S02",
          "function_def"
        ],
        "before_tex": "PV_n\\circ A_n=\\mathrm{id},\\ A_n\\circ PV_n=\\mathrm{id}",
        "after_tex": "PV_n=A_n^{-1}\\ \\text{on}\\ \\mathbb{Q}_{\\ge 0}",
        "relation": "implies",
        "rule": "injective_surjective_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "因 (1+i)^n 是正有理數（S02 與 i 為有理數），A_n 與 PV_n 都把非負有理數映到非負有理數。兩個合成都是恆等映射，所以 A_n 是雙射且 PV_n 是它的反函數：這就是 MKO 所說「此式與固定期率複利互逆」的精確內容。",
          "en": "Since (1+i)^n is a positive rational (S02, with i rational), A_n and PV_n both map nonnegative rationals to nonnegative rationals. Both composites are the identity, so A_n is a bijection and PV_n is its inverse: this is the precise content of the MKO's 'this inverts fixed-period-rate compounding'."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S05",
            "injective_surjective_def",
            "composition_def"
          ],
          "explanation": {
            "zh-Hant": "互逆需要兩個方向的合成都是恆等；只有一個方向只給出單射或滿射。本模型以同一個 i 與 n 保證累積因子相同；代數上只需要兩邊使用相同的正累積因子，不要求參數表示唯一。例如 (1+3)^1=(1+1)^2=4。這不取消期別與利率單位必須一致的建模要求。",
            "en": "Being inverses needs both composites to be the identity; one direction alone gives only injectivity or surjectivity. Using the same i and n ensures a common accumulation factor in this model. Algebraically, the two maps need the same positive factor, not a unique parameterization: for example (1+3)^1=(1+1)^2=4. This does not remove the modeling requirement that period and rate units agree."
          }
        },
        "leaf_basis": "injective_surjective_def + composition_def + function_def"
      },
      {
        "id": "S07",
        "uses": [
          "G3",
          "power_laws"
        ],
        "before_tex": "(1+\\tfrac{1}{10})^3",
        "after_tex": "(\\tfrac{11}{10})^3=\\tfrac{1331}{1000}",
        "relation": "equivalent",
        "rule": "power_laws",
        "conditions": [],
        "explanation": {
          "zh-Hant": "例子的累積因子：1+1/10=11/10>0，(11/10)³=11³/10³=1331/1000（N1）。期別是「年」，i 是每年利率，n=3 是年數，三者一致。",
          "en": "The example's accumulation factor: 1+1/10=11/10>0, and (11/10)³=11³/10³=1331/1000 (N1). The period is the year, i is the rate per year and n=3 counts years, so the three agree."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "power_laws",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "(a/b)^n=a^n/b^n 是指數律，兩邊同值。",
            "en": "(a/b)^n=a^n/b^n is an exponent law; both sides have the same value."
          }
        },
        "leaf_basis": "power_laws + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "S07",
          "S01",
          "S03"
        ],
        "before_tex": "PV_3(1331)=\\tfrac{1331}{1331/1000}",
        "after_tex": "1000",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "\\tfrac{1331}{1000}\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "PV_3(1331)=1331÷(1331/1000)=1331·1000/1331=1000（N2）。三年後的 1331 在每年 10% 的模型下相當於現在的 1000。",
          "en": "PV_3(1331)=1331÷(1331/1000)=1331·1000/1331=1000 (N2). In the model with 10% per year, 1331 three years from now corresponds to 1000 now."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "除以非零分數等於乘以其倒數；精確計算，兩邊同值。",
            "en": "Dividing by a nonzero fraction is multiplying by its reciprocal; an exact computation, so both sides agree."
          }
        },
        "leaf_basis": "frac_rules + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S08",
          "S06",
          "S07"
        ],
        "before_tex": "A_3(1000)",
        "after_tex": "1000\\cdot\\tfrac{1331}{1000}=1331",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "roundtrip 核對：把 1000 以同一因子複利三年，得 1331（N3），回到原本的未來金額，與 S06 的互逆一致。",
          "en": "Round-trip check: compounding 1000 for three years with the same factor gives 1331 (N3), returning the original future amount, as S06 predicts."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確計算；這是 S06 的例子，不是它的證明。",
            "en": "An exact computation; it is an instance of S06, not its proof."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S10",
        "uses": [
          "S07",
          "S08",
          "power_laws"
        ],
        "before_tex": "\\tfrac{1331}{(11/10)^3}",
        "after_tex": "\\tfrac{1331}{11/10}=1210,\\ \\tfrac{1210}{11/10}=1100,\\ \\tfrac{1100}{11/10}=1000",
        "relation": "equivalent",
        "rule": "power_laws",
        "conditions": [
          "\\tfrac{11}{10}\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "逐年折現：由 a³=a·a·a，除以 a³ 等於連續除以 a 三次。1331→1210→1100→1000（N4、N5、N6），與 S08 一致。這對應「每年往回折一期」，說明 n 期折現就是一期折現重複 n 次。",
          "en": "Year-by-year discounting: since a³=a·a·a, dividing by a³ is dividing by a three times in a row. 1331→1210→1100→1000 (N4, N5, N6), agreeing with S08. This corresponds to 'discounting back one period each year' and shows that discounting over n periods is one-period discounting repeated n times."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "power_laws",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "x/(abc)=((x/a)/b)/c 對非零 a、b、c 成立，兩種算法同值。",
            "en": "x/(abc)=((x/a)/b)/c for nonzero a, b, c, so the two computations agree."
          }
        },
        "leaf_basis": "power_laws + frac_rules + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "G1",
          "S03"
        ],
        "before_tex": "i\\ge 0,\\ F\\ge 0",
        "after_tex": "PV_n(F)\\le F",
        "relation": "implies",
        "rule": "L_ge1_pow",
        "conditions": [
          "i\\ge 0",
          "F\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "若 i≥0，則 a=1+i≥1，依引理 L_ge1_pow（兩個子步）a^n≥1。於是 F−PV_n(F)=F(a^n−1)/a^n：分子 F≥0、a^n−1≥0，分母 a^n>0，所以差 ≥0，即現值不超過未來金額。例子中 1000≤1331（N13）。",
          "en": "If i≥0, then a=1+i≥1, and by Lemma L_ge1_pow (two sub-steps) a^n≥1. Then F−PV_n(F)=F(a^n−1)/a^n: the numerator has F≥0 and a^n−1≥0 and the denominator a^n>0, so the difference is ≥0, that is the present value does not exceed the future amount. In the example 1000≤1331 (N13)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_ge1_pow",
            "S02",
            "order_basics",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "i≥0 與 F≥0 是本步的充分前提；若 −1<i<0、n≥1 且 F>0，則 (1+i)^n<1，現值大於未來金額，故不能不加條件推廣 PV≤F。F=0 或 n=0 時仍為等號。",
            "en": "The assumptions i≥0 and F≥0 are sufficient here. If −1<i<0, n≥1 and F>0, then (1+i)^n<1 and the present value exceeds the future amount, so PV≤F cannot be extended without conditions. Equality still holds when F=0 or n=0."
          }
        },
        "leaf_basis": "L_ge1_pow + order_basics + frac_rules"
      },
      {
        "id": "S11.1",
        "uses": [
          "power_laws"
        ],
        "before_tex": "n=0",
        "after_tex": "a^0=1\\ge 1",
        "relation": "induction_base",
        "rule": "nat_induction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "基底：a^0=1≥1。",
          "en": "Base: a^0=1≥1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "power_laws",
            "nat_induction"
          ],
          "explanation": {
            "zh-Hant": "只用 a^0=1。",
            "en": "Only a^0=1 is used."
          }
        },
        "leaf_basis": "nat_induction + power_laws"
      },
      {
        "id": "S11.2",
        "uses": [
          "S11.1",
          "order_basics",
          "power_laws"
        ],
        "before_tex": "a^n\\ge 1,\\ a\\ge 1",
        "after_tex": "a^{n+1}=a^n\\cdot a\\ge a\\ge 1",
        "relation": "induction_step",
        "rule": "order_basics",
        "conditions": [
          "a\\ge 1"
        ],
        "explanation": {
          "zh-Hant": "設 a^n≥1。a≥1>0，把 a^n≥1 兩邊乘正數 a 得 a^n·a≥a，再與 a≥1 傳遞得 a^(n+1)≥1。依歸納原理對所有 n 成立。S11.1 與 S11.2 證得引理 L_ge1_pow。",
          "en": "Assume a^n≥1. Since a≥1>0, multiplying a^n≥1 by the positive a gives a^n·a≥a, and with a≥1 transitivity gives a^(n+1)≥1. By induction this holds for every n. S11.1 and S11.2 prove Lemma L_ge1_pow."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "power_laws",
            "nat_induction"
          ],
          "explanation": {
            "zh-Hant": "需要 a 為正才能保序相乘，並需要 a≥1 才能傳遞到 1。",
            "en": "a must be positive to multiply while preserving order, and a≥1 is needed to pass on to 1."
          }
        },
        "leaf_basis": "order_basics + power_laws + nat_induction"
      },
      {
        "id": "S12",
        "uses": [
          "S07",
          "S08"
        ],
        "before_tex": "1331\\cdot(\\tfrac{11}{10})^3",
        "after_tex": "\\tfrac{1771561}{1000}\\neq 1000=PV_3(1331)",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把折現誤寫成乘上累積因子：1331·1331/1000=1771561/1000（N7），這是 1331 再複利三年的累積值 A_3(1331)，不是現值；它不等於 1000（N8）。由 S06，現值是累積的反運算，所以應除以因子。",
          "en": "Writing discounting wrongly as multiplication by the accumulation factor: 1331·1331/1000=1771561/1000 (N7), which is the accumulation A_3(1331) of 1331 over three more years, not the present value; it differs from 1000 (N8). By S06 the present value is the inverse of accumulation, so one must divide by the factor."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "numeric_evaluation",
            "S08"
          ],
          "explanation": {
            "zh-Hant": "一個數值反例足以說明乘法不是折現；不宣稱兩者在任何情況下都不同（F=0 或 n=0 時相同）。",
            "en": "One numeric counterexample suffices to show multiplication is not discounting; it is not claimed that they always differ (they agree when F=0 or n=0)."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "G1",
          "G4",
          "S07"
        ],
        "before_tex": "\\text{monthly rate}\\ j=\\tfrac{1}{100}",
        "after_tex": "(1+j)^{12}>\\tfrac{112}{100},\\ (1+j)^{36}>(\\tfrac{11}{10})^3",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "期別必須一致：月利率 j=1/100 時，期數要以月計，三年是 n=36，折現時所除的累積因子是 (101/100)^36；它大於年利率 10% 三年的 1331/1000（N10），所以兩者給出不同的現值。另外 (101/100)^12>112/100（N9），所以「月 1%」也不等於「年 12% 每年計息一次」。把年利率配上月數（例如 (11/10)^36），或把月利率乘以 12 當作年有效利率，都違反 G1 的單位一致要求。",
          "en": "Period units must agree: with the monthly rate j=1/100 the periods are counted in months, three years is n=36, and the accumulation factor by which to divide is (101/100)^36; it exceeds 1331/1000 for 10% per year over three years (N10), so the two give different present values. Also (101/100)^12>112/100 (N9), so '1% per month' is not '12% per year compounded once a year'. Pairing an annual rate with a count of months (such as (11/10)^36), or treating twelve times a monthly rate as an effective annual rate, violates the unit agreement required by G1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G4",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "這些是精確的有理數比較；不處理名目與有效利率的正式換算（MKO 的 formal next）。",
            "en": "These are exact rational comparisons; the formal conversion between nominal and effective rates (the MKO's formal next step) is not treated."
          }
        },
        "leaf_basis": "numeric_evaluation + power_laws + order_basics"
      },
      {
        "id": "S14",
        "uses": [
          "S03",
          "S06",
          "S08",
          "S10",
          "S11",
          "S12",
          "S13"
        ],
        "before_tex": "1+i>0,\\ n\\in\\mathbb{N}",
        "after_tex": "PV_n=A_n^{-1};\\ PV_3(1331)=1000\\ \\text{at}\\ i=\\tfrac{1}{10}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "彙整：1+i>0 使 (1+i)^n>0，現值有定義（S02、S03）；折現與複利互為反函數（S06）；每年 10%、三年後的 1331 現值為 1000，逐年折現與 roundtrip 都核對一致（S08、S09、S10）；i≥0 時現值不超過未來金額（S11）；乘以因子或混用期別都會出錯（S12、S13）。",
          "en": "Summary: 1+i>0 makes (1+i)^n>0, so the present value is defined (S02, S03); discounting and compounding are inverse functions (S06); at 10% per year the present value of 1331 due in three years is 1000, confirmed by year-by-year discounting and the round trip (S08, S09, S10); for i≥0 the present value does not exceed the future amount (S11); multiplying by the factor or mixing period units goes wrong (S12, S13)."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S03",
            "S06",
            "S08",
            "S11"
          ],
          "explanation": {
            "zh-Hant": "目標 G：現值有定義、與複利互逆、例子的現值。足夠條件 H：S03、S06、S08（S11 為附帶性質）；H⇒G 由合取。H 由定義、指數律與歸納正向證成，沒有把 G 當作前提；S12、S13 只說明錯誤做法。",
            "en": "Goal G: the present value is defined, inverts compounding, and the example's value. Sufficient condition H: S03, S06 and S08 (with S11 as an additional property); H⇒G by conjunction. H is proved forwards from the definitions, exponent laws and induction; G is never used as a premise, and S12, S13 only illustrate wrong methods."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "當 1+i>0 時，(1+i)^n>0，現值 PV_n(F)=F/(1+i)^n 有定義，且 PV_n 與複利 A_n(P)=P(1+i)^n 互為反函數。每年利率 10%、三年後的 1331，現值為 1331/(11/10)³=1000；逐年折現 1331→1210→1100→1000 與 roundtrip 1000·(11/10)³=1331 都一致。i≥0 時現值不超過未來金額。把折現寫成乘法得到的是累積值 1771561/1000，而期別不一致（年率配月數，或月率 1% 當年率 12%）會給出不同的因子。",
      "en": "When 1+i>0, (1+i)^n>0, the present value PV_n(F)=F/(1+i)^n is defined, and PV_n and compounding A_n(P)=P(1+i)^n are inverse functions. At 10% per year, 1331 due in three years has present value 1331/(11/10)³=1000; year-by-year discounting 1331→1210→1100→1000 and the round trip 1000·(11/10)³=1331 agree. For i≥0 the present value does not exceed the future amount. Writing discounting as multiplication gives the accumulation 1771561/1000 instead, and mismatched periods (an annual rate with a count of months, or 1% monthly taken as 12% annual) give different factors."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "每年 10%，三年後 1331 的現值是 1331×1.1³",
          "en": "At 10% per year, the present value of 1331 due in three years is 1331×1.1³"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "那是累積值 1771561/1000；現值是 1331/1.1³=1000。",
          "en": "That is the accumulation 1771561/1000; the present value is 1331/1.1³=1000."
        }
      },
      {
        "claim": {
          "zh-Hant": "年利率 10%、36 個月後的 F，現值為 F/(1.1)^36",
          "en": "With an annual rate of 10%, the present value of F due in 36 months is F/(1.1)^36"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "年利率要配年數，n=3，不是 36；混用期別違反假設 a1。",
          "en": "An annual rate goes with years, n=3, not 36; mixing periods violates assumption a1."
        }
      },
      {
        "claim": {
          "zh-Hant": "月利率 1% 每月複利，等同年利率 12% 每年複利一次",
          "en": "1% per month compounded monthly is the same as 12% per year compounded once a year"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "(101/100)^12>112/100（N9）。",
          "en": "(101/100)^12>112/100 (N9)."
        }
      },
      {
        "claim": {
          "zh-Hant": "i=−1 時，現值 F/(1+i)^n 為 0",
          "en": "With i=−1, the present value F/(1+i)^n is 0"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "1+i=0 違反 1+i>0；n≥1 時要除以 0，沒有定義。",
          "en": "1+i=0 violates 1+i>0; for n≥1 it would divide by 0, which is undefined."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "power_laws",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整數指數律（底數非零或指數為正整數）：a^m·a^n=a^{m+n}，(a^m)^n=a^{mn}，(ab)^n=a^n b^n，a^0=1。",
          "en": "Integer exponent laws (base nonzero, or exponents positive integers): a^m·a^n=a^{m+n}, (a^m)^n=a^{mn}, (ab)^n=a^n b^n, a^0=1."
        }
      },
      {
        "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": "mul_inverse_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。",
          "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined."
        }
      },
      {
        "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": "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": "function_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "函數：f:A→B 對每個 a∈A 指定恰好一個 f(a)∈B。兩函數相等 ⇔ 定義域、對應域相同且對每個 a 都有相同值。",
          "en": "Function: f:A→B assigns to each a∈A exactly one f(a)∈B. Two functions are equal ⇔ they have the same domain and codomain and agree at every a."
        }
      },
      {
        "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_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "引入定義：可為一個命題、式子或物件取名；之後名稱與被定義者可互相替換。",
          "en": "Introducing a definition: a statement, expression or object may be given a name, after which the name and what it stands for are interchangeable."
        }
      },
      {
        "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": "mul_assoc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法結合律：(uv)w=u(vw)。",
          "en": "Associativity of multiplication: (uv)w=u(vw)."
        }
      },
      {
        "id": "mul_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "id": "injective_surjective_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "單射：f(a)=f(a′) ⇒ a=a′。滿射：每個 b∈B 都有 a∈A 使 f(a)=b。雙射：既單又滿，此時有唯一反函數 f⁻¹ 使 f⁻¹∘f=id_A、f∘f⁻¹=id_B。",
          "en": "Injective: f(a)=f(a′) ⇒ a=a′. Surjective: every b∈B has some a∈A with f(a)=b. Bijective: both; then there is a unique inverse f⁻¹ with f⁻¹∘f=id_A and f∘f⁻¹=id_B."
        }
      },
      {
        "id": "composition_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合成：對 f:A→B 與 g:B→C，g∘f:A→C 定義為 (g∘f)(a)=g(f(a))。",
          "en": "Composition: for f:A→B and g:B→C, g∘f:A→C is defined by (g∘f)(a)=g(f(a))."
        }
      },
      {
        "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": "mko_pv",
        "kind": "cited",
        "source": {
          "mko_id": "mko-present-value-discounting",
          "sha256": "2e47f9d121504d577795a46f4235f98114b5701015039cc83bef029cead220b0",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "相對固定每期利率 i、正累積因子 1+i 與期數 n，未來金額 F 的現值定義為 F/(1+i)^n。此式與固定期率複利互逆；現金流、變動利率與費用需另建模型。",
          "en": "Relative to a fixed rate i per period, positive accumulation factor 1+i and period count n, the present value of a future amount F is defined as F/(1+i)^n. This inverts fixed-period-rate compounding. Cash flows, variable rates and fees require a different model."
        }
      },
      {
        "id": "mko_compound",
        "kind": "cited",
        "source": {
          "mko_id": "mko-compound-interest",
          "sha256": "5e3f823637101eaaf3ba73ed1b448b858ef9d83163c30769859bfcd7bbb22bb8",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在期別、計息時點與每期利率 i 已固定，且期間內無存提款或費用時，複利餘額以 B_(n+1)=(1+i)B_n 更新。若累積因子 1+i>0，n 期後為 B_0(1+i)^n。",
          "en": "Fix periods, interest timing and rate i per period, with no deposits, withdrawals or fees during the modeled interval. Compound balance updates are B_(n+1)=(1+i)B_n. For 1+i>0, the balance after n periods is B_0(1+i)^n."
        }
      },
      {
        "id": "L_pos_pow",
        "kind": "lemma",
        "proved_in": [
          "S02.1",
          "S02.2"
        ],
        "statement": {
          "zh-Hant": "若 a>0，則對每個自然數 n，a^n>0。",
          "en": "If a>0, then a^n>0 for every natural number n."
        }
      },
      {
        "id": "L_ge1_pow",
        "kind": "lemma",
        "proved_in": [
          "S11.1",
          "S11.2"
        ],
        "statement": {
          "zh-Hant": "若 a≥1，則對每個自然數 n，a^n≥1。",
          "en": "If a≥1, then a^n≥1 for every natural number n."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S07",
        "lhs": "(11/10)**3",
        "rhs": "1331/1000"
      },
      {
        "id": "N2",
        "step": "S08",
        "lhs": "1331/(1331/1000)",
        "rhs": "1000"
      },
      {
        "id": "N3",
        "step": "S09",
        "lhs": "1000*(11/10)**3",
        "rhs": "1331"
      },
      {
        "id": "N4",
        "step": "S10",
        "lhs": "1331/(11/10)",
        "rhs": "1210"
      },
      {
        "id": "N5",
        "step": "S10",
        "lhs": "1210/(11/10)",
        "rhs": "1100"
      },
      {
        "id": "N6",
        "step": "S10",
        "lhs": "1100/(11/10)",
        "rhs": "1000"
      },
      {
        "id": "N7",
        "step": "S12",
        "lhs": "1331*(11/10)**3",
        "rhs": "1771561/1000"
      },
      {
        "id": "N8",
        "step": "S12",
        "lhs": "1771561/1000",
        "relation": "!=",
        "rhs": "1000"
      },
      {
        "id": "N9",
        "step": "S13",
        "lhs": "(101/100)**12",
        "relation": ">",
        "rhs": "112/100"
      },
      {
        "id": "N10",
        "step": "S13",
        "lhs": "(101/100)**36",
        "relation": ">",
        "rhs": "(11/10)**3"
      },
      {
        "id": "N11",
        "step": "S02",
        "vars": [
          "n"
        ],
        "range": [
          0,
          10
        ],
        "lhs": "(11/10)**n",
        "relation": ">",
        "rhs": "0"
      },
      {
        "id": "N12",
        "step": "S04",
        "vars": [
          "P"
        ],
        "degree": 1,
        "lhs": "P*(11/10)**3/(11/10)**3",
        "rhs": "P"
      },
      {
        "id": "N13",
        "step": "S11",
        "lhs": "1000",
        "relation": "<=",
        "rhs": "1331"
      },
      {
        "id": "N14",
        "step": "S02.2",
        "lhs": "(11/10)**4",
        "rhs": "(11/10)**3*(11/10)"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 的範圍內工作：期別、利率慣例與 n 一致，1+i>0，期間內無額外現金流、費用、稅或四捨五入（假設 a1）；金額為非負精確有理數，計算不是報價、投資建議或市場預測（假設 a2）。不處理多筆現金流、年金、變動利率或名目與有效利率的換算；不改動 canonical MKO 或英文 edition。",
      "en": "Works inside the MKO's scope: period units, rate conventions and n agree, 1+i>0, and there are no additional cash flows, fees, taxes or rounding during the modeled interval (assumption a1); amounts are nonnegative exact rationals, and the computations are not quotes, investment advice or market forecasts (assumption a2). Multiple cash flows, annuities, variable rates and nominal-to-effective conversion are not treated; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "i",
        "每期固定利率（有理數），1+i>0",
        "the fixed rate per period (rational), 1+i>0"
      ],
      [
        "n",
        "期數，自然數",
        "the number of periods, a natural number"
      ],
      [
        "A_n(P)=P(1+i)^n",
        "n 期複利累積",
        "accumulation by compounding over n periods"
      ],
      [
        "PV_n(F)=\\tfrac{F}{(1+i)^n}",
        "未來金額 F 的現值",
        "the present value of a future amount F"
      ],
      [
        "\\mathbb{Q}_{\\ge 0}",
        "非負有理數",
        "the nonnegative rationals"
      ]
    ],
    "md_error_controls": [
      {
        "title": "折現要除以累積因子，不是乘上它 / Discount by dividing by the accumulation factor, not multiplying by it",
        "zh": "MKO 指出的第一個誤解是把折現乘上累積因子。1331·(11/10)³=1771561/1000 是 1331 再複利三年的結果（S12）；現值是累積的反運算（S06），應為 1331/(11/10)³=1000。檢查方法：把結果再複利同樣期數，應回到原金額（S09）。",
        "en": "The first mistake named by the MKO is multiplying by the accumulation factor when discounting. 1331·(11/10)³=1771561/1000 is 1331 compounded for three more years (S12); the present value is the inverse of accumulation (S06), namely 1331/(11/10)³=1000. The check: compound the result over the same periods and it must return the original amount (S09)."
      },
      {
        "title": "期別與期數必須一致 / Period units and counts must agree",
        "zh": "MKO 指出的第二個誤解是混用年率、月率與不同期數。年利率要配年數，月利率要配月數；(101/100)^12>112/100，所以月 1% 不等於年 12%（S13）。本稿的公式只在 i 與 n 使用同一期別時適用。",
        "en": "The second mistake named by the MKO is mixing annual rates, monthly rates and different period counts. An annual rate goes with a count of years, a monthly rate with a count of months; (101/100)^12>112/100, so 1% per month is not 12% per year (S13). The formula here applies only when i and n use the same period."
      },
      {
        "title": "1+i>0 的作用 / What 1+i>0 does",
        "zh": "1+i>0 保證 (1+i)^n>0（S02），所以可以除（S03），且現值保持非負。i=−1 時 1+i=0，n≥1 的折現要除以 0，沒有定義；這不是「現值為 0」。",
        "en": "1+i>0 guarantees (1+i)^n>0 (S02), so the division is possible (S03) and present values stay nonnegative. With i=−1, 1+i=0 and discounting over n≥1 periods would divide by 0, which is undefined; it is not 'a present value of 0'."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "多筆現金流的時點、年金、名目與有效利率的換算與敏感度分析屬於 MKO 的 formal next，本稿不處理；所有計算只是模型計算，不是報價或投資建議。",
        "en": "Timing of multiple cash flows, annuities, nominal-to-effective rate conversion and sensitivity analysis belong to the MKO's formal next step and are not treated; all computations are model computations, not quotes or investment advice."
      },
      {
        "zh-Hant": "N11 是 n=0 到 10 的有限核對，(1+i)^n>0 對所有 n 的證明是 S02 的歸納；N12 是一次多項式恆等式的網格核對。",
        "en": "N11 is a finite check for n=0 to 10; the proof that (1+i)^n>0 for all n is the induction of S02; N12 is a grid check of a degree-one polynomial identity."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "b712d385bc1ee8edccfd1a652e7db341e2a1b9342c4e958af54cc398fe26a0a5",
    "original_source_sha256": "985912bc73b59962d395f9a20ecb54a42e67fe4bb06bf99ec0459f92ca92d0e0",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M051"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}