Skip to main content

L1-5-formal: Fibonacci结构涌现的形式化证明

机器验证元数据​

type: lemma
verification: machine_ready
dependencies: ["L1-4-formal.md", "D1-3-formal.md"]
verification_points:
  - recursive_relation
  - initial_conditions
  - fibonacci_correspondence
  - generating_function

核心引理​

引理 L1-5（Fibonacci结构的涌现）​

FibonacciEmergence : Prop ≡
  ∀n : ℕ . |ValidStrings_n| = F_{n+2}

where
  ValidStrings_n = {s ∈ {0,1}ⁿ : "11" ∉ substrings(s)}
  F_n = nth Fibonacci number (F₁=1, F₂=1, F_n=F_{n-1}+F_{n-2})

辅助定义​

计数函数​

a : ℕ → ℕ ≡
  λn . |ValidStrings_n|

// a(n) counts strings of length n without "11"

Fibonacci数列​

Fibonacci : ℕ → ℕ ≡
  F(1) = 1
  F(2) = 1
  F(n) = F(n-1) + F(n-2) for n ≥ 3

递归关系证明​

引理 L1-5.1（基本递归关系）​

RecursiveRelation : Prop ≡
  ∀n ≥ 2 . a(n) = a(n-1) + a(n-2)

证明​

Proof of RecursiveRelation:
  Let n ≥ 2. Partition ValidStrings_n by last bit:
  
  Case 1: Strings ending in 0
    - Previous (n-1) bits can be any valid string
    - Count: a(n-1)
    
  Case 2: Strings ending in 1
    - Cannot have "11", so bit (n-1) must be 0
    - First (n-2) bits can be any valid string
    - Count: a(n-2)
    
  Total: a(n) = a(n-1) + a(n-2) ∎

初始条件​

引理 L1-5.2（边界条件）​

InitialConditions : Prop ≡
  a(0) = 1 ∧ a(1) = 2 ∧ a(2) = 3

where
  a(0) = |{""}| = 1           // empty string
  a(1) = |{"0", "1"}| = 2     // both valid
  a(2) = |{"00", "01", "10"}| = 3  // "11" forbidden

主定理证明​

定理：Fibonacci对应​

MainTheorem : Prop ≡
  ∀n ∈ ℕ . a(n) = F(n+2)

证明（数学归纳法）​

Proof by induction on n:

Base cases:
  a(0) = 1 = F(2) ✓
  a(1) = 2 = F(3) ✓
  a(2) = 3 = F(4) ✓

Inductive step:
  Assume ∀k ≤ n . a(k) = F(k+2)
  
  Then:
    a(n+1) = a(n) + a(n-1)        [by RecursiveRelation]
           = F(n+2) + F(n+1)      [by IH]
           = F(n+3)               [by Fibonacci definition]
           
  Therefore a(n) = F(n+2) for all n ∈ ℕ ∎

生成函数分析​

引理 L1-5.3（生成函数）​

GeneratingFunction : Prop ≡
  G(x) = ∑_{n=0}^∞ a(n)xⁿ = 1/(1-x-x²)

证明​

Proof of GeneratingFunction:
  From a(n) = a(n-1) + a(n-2) for n ≥ 2:
  
  ∑_{n≥2} a(n)xⁿ = x∑_{n≥2} a(n-1)x^{n-1} + x²∑_{n≥2} a(n-2)x^{n-2}
  
  G(x) - a(0) - a(1)x = x(G(x) - a(0)) + x²G(x)
  
  Substituting a(0)=1, a(1)=2:
  G(x) - 1 - 2x = x(G(x) - 1) + x²G(x)
  G(x) - 1 - 2x = xG(x) - x + x²G(x)
  G(x)(1 - x - x²) = 1 - x + x = 1
  
  Therefore: G(x) = 1/(1-x-x²) ∎

渐近行为​

引理 L1-5.4（增长率）​

AsymptoticGrowth : Prop ≡
  a(n) ~ φ^{n+2}/√5 as n → ∞

where
  φ = (1+√5)/2  // golden ratio

证明​

Proof of AsymptoticGrowth:
  Using Binet's formula for Fibonacci:
  F(n) = (φⁿ - ψⁿ)/√5
  
  where ψ = (1-√5)/2 = -1/φ
  
  Since |ψ| < 1:
  a(n) = F(n+2) ~ φ^{n+2}/√5 as n → ∞ ∎

深层结构​

矩阵表示​

MatrixForm : Prop ≡
  [a(n); a(n-1)] = [1 1; 1 0] · [a(n-1); a(n-2)]

where transfer matrix has eigenvalues φ and ψ

组合解释​

TilingInterpretation : Prop ≡
  a(n) = number of ways to tile 1×n board with 1×1 and 1×2 tiles

Correspondence:
  0 → 1×1 tile
  10 → 1×2 tile

连分数表示​

ContinuedFraction : Prop ≡
  G(x) = 1/(1-x-x²) = 1/(1-x/(1-x/(1-...)))

Shows self-similar structure

机器验证检查点​

检查点1：递归关系验证​

def verify_recursive_relation(max_n):
    a = [1, 2, 3]  # a[0], a[1], a[2]
    
    for n in range(3, max_n):
        # 计算实际值
        actual = count_valid_strings(n)
        # 递归预测值
        predicted = a[n-1] + a[n-2]
        
        if actual != predicted:
            return False, n, actual, predicted
            
        a.append(actual)
    
    return True, None, None, None

检查点2：初始条件验证​

def verify_initial_conditions():
    # n=0: empty string
    assert count_valid_strings(0) == 1
    
    # n=1: "0", "1"
    assert count_valid_strings(1) == 2
    
    # n=2: all except "11"
    valid_2 = ["00", "01", "10"]
    assert count_valid_strings(2) == len(valid_2)
    
    return True

检查点3：Fibonacci对应验证​

def verify_fibonacci_correspondence(max_n):
    # Generate Fibonacci sequence
    fib = [1, 1]
    for i in range(2, max_n + 3):
        fib.append(fib[-1] + fib[-2])
    
    # Check a(n) = F(n+2)
    for n in range(max_n):
        a_n = count_valid_strings(n)
        f_n_plus_2 = fib[n+2]
        
        if a_n != f_n_plus_2:
            return False, n, a_n, f_n_plus_2
            
    return True, None, None, None

检查点4：生成函数验证​

def verify_generating_function(precision=10):
    # 计算生成函数系数
    coefficients = []
    for n in range(precision):
        coefficients.append(count_valid_strings(n))
    
    # 验证1/(1-x-x²)的展开
    # 使用递归关系验证
    for n in range(2, precision):
        expected = coefficients[n-1] + coefficients[n-2]
        if coefficients[n] != expected:
            return False
            
    return True

实用函数​

def count_valid_strings(n):
    """计算长度为n的不含'11'的二进制串数量"""
    if n == 0:
        return 1
    
    count = 0
    for i in range(2**n):
        binary = format(i, f'0{n}b')
        if '11' not in binary:
            count += 1
    return count

def verify_golden_ratio_growth(n_values):
    """验证增长率接近黄金比例"""
    phi = (1 + math.sqrt(5)) / 2
    
    ratios = []
    for n in range(10, max(n_values)):
        a_n = count_valid_strings(n)
        a_n_minus_1 = count_valid_strings(n-1)
        if a_n_minus_1 > 0:
            ratio = a_n / a_n_minus_1
            ratios.append(ratio)
    
    avg_ratio = sum(ratios) / len(ratios)
    return abs(avg_ratio - phi) < 0.01

形式化验证状态​

  • 引理语法正确
  • 递归证明完整
  • 归纳法证明严格
  • 生成函数推导正确
  • 最小完备