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L1-6-formal: φ-表示系统建立的形式化证明

机器验证元数据​

type: lemma
verification: machine_ready
dependencies: ["L1-5-formal.md", "L1-4-formal.md", "D1-8-formal.md"]
verification_points:
  - existence_proof
  - uniqueness_proof
  - bijection_establishment
  - encoding_efficiency

核心引理​

引理 L1-6（φ-表示系统的建立）​

PhiRepresentationSystem : Prop ≡
  ∀n ∈ ℤ⁺ . ∃!I ⊂ ℕ . 
    n = ∑_{i∈I} F_i ∧ NonConsecutive(I)

where
  F_i = Modified Fibonacci (F₁=1, F₂=2, F₃=3, F₄=5, ...)
  NonConsecutive(I) ≡ ∀i,j ∈ I . i ≠ j → |i-j| ≥ 2

辅助定义​

修正的Fibonacci数列​

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

// Note: F(n) = Standard_Fib(n+1)

φ-表示​

PhiRepresentation : Type ≡ 
  { I : Set ℕ | NonConsecutive(I) }

encode : ℤ⁺ → PhiRepresentation
decode : PhiRepresentation → ℤ⁺

decode(I) ≡ ∑_{i∈I} F(i)

存在性证明​

引理 L1-6.1（贪心算法）​

GreedyAlgorithm : Prop ≡
  ∀n ∈ ℤ⁺ . GreedyDecomposition(n) is valid φ-representation

where
  GreedyDecomposition(n) ≡
    if n = 0 then ∅
    else let k = max{i : F(i) ≤ n} in
         {k} ∪ GreedyDecomposition(n - F(k))

证明​

Proof of existence:
  By strong induction on n.
  
  Base: n = 1, use I = {1}, since F(1) = 1
  
  Inductive step: Assume true for all m < n.
  Let k = max{i : F(i) ≤ n}
  
  Case 1: n = F(k)
    Then I = {k} works
    
  Case 2: n > F(k)
    Let r = n - F(k) < n
    By IH, r has φ-representation J
    
    Claim: k ∉ J and k-1 ∉ J
    Proof: If k-1 ∈ J, then F(k-1) ≤ r = n - F(k)
           So F(k) + F(k-1) ≤ n
           But F(k) + F(k-1) = F(k+1) ≤ n
           Contradicts maximality of k
           
    Therefore I = {k} ∪ J is valid ∎

唯一性证明​

引理 L1-6.2（唯一性）​

UniquenessTheorem : Prop ≡
  ∀n ∈ ℤ⁺ . ∀I,J ⊂ ℕ .
    (decode(I) = n ∧ NonConsecutive(I)) ∧
    (decode(J) = n ∧ NonConsecutive(J)) →
    I = J

证明​

Proof by contradiction:
  Assume I ≠ J with decode(I) = decode(J) = n
  
  Let k = max(I △ J) (symmetric difference)
  WLOG assume k ∈ I, k ∉ J
  
  Key observation: ∀i > k . i ∈ I ↔ i ∈ J
  
  Consider partial sums:
  S_I = ∑_{i∈I, i≤k} F(i) = F(k) + ∑_{i∈I, i<k} F(i)
  S_J = ∑_{j∈J, j≤k} F(j)
  
  Since total sums equal: S_I = S_J
  
  Claim: This is impossible
  Proof: 
    - If k-1 ∈ J: violates non-consecutive with some j > k
    - If k-1 ∉ J: Then S_J < F(k) ≤ S_I
    
  Contradiction in both cases ∎

双射建立​

引理 L1-6.3（编码双射）​

EncodingBijection : Prop ≡
  ∃φ : BinaryStringsNo11 ↔ ℤ⁺ .
    Bijective(φ)

where
  BinaryStringsNo11 = {b ∈ {0,1}* : "11" ∉ substrings(b)}
  
  φ(b₁b₂...bₙ) = ∑_{i=1}^n bᵢ·F(i)

证明​

Proof of bijection:
  1. Well-defined: Clear from definition
  
  2. Injective: 
     If φ(b) = φ(b'), then same φ-representation
     By uniqueness, same index sets
     Therefore b = b'
     
  3. Surjective:
     By existence, every n has φ-representation I
     Define b where bᵢ = 1 iff i ∈ I
     Then φ(b) = n ∎

编码效率​

引理 L1-6.4（长度界限）​

EncodingLength : Prop ≡
  ∀n ∈ ℤ⁺ . 
    |φ-repr(n)| = ⌊log_φ(n)⌋ + O(1)

where
  |φ-repr(n)| = max{i : i ∈ I, decode(I) = n}
  φ = (1+√5)/2

证明​

Proof of length bound:
  Let k = max{i : i ∈ φ-repr(n)}
  
  By greedy property: F(k) ≤ n < F(k+1)
  
  Using Binet formula:
    F(k) ≈ φᵏ/√5
    
  Therefore:
    φᵏ/√5 ≲ n ≲ φᵏ⁺¹/√5
    
  Taking logarithms:
    k ≈ log_φ(n√5) = log_φ(n) + O(1) ∎

算术运算​

引理 L1-6.5（加法算法）​

PhiAddition : Prop ≡
  ∃add : PhiRepr × PhiRepr → PhiRepr .
    ∀I,J . decode(add(I,J)) = decode(I) + decode(J)

Algorithm:
  1. Merge index sets: K = I ∪ J
  2. Handle duplicates: if i ∈ I ∩ J, use F(i)+F(i) = F(i+1)+F(i-2)
  3. Recursively eliminate violations of non-consecutive constraint

系统性质​

定理：完备性​

CompletenessTheorem : Prop ≡
  PhiRepresentationSystem forms complete number system with:
  1. Unique representation (Zeckendorf)
  2. Efficient encoding/decoding
  3. Arithmetic operations
  4. Order preservation

机器验证检查点​

检查点1：存在性验证​

def verify_existence(max_n):
    for n in range(1, max_n):
        repr = greedy_decomposition(n)
        
        # 验证和
        if sum(fib[i] for i in repr) != n:
            return False, n, "sum mismatch"
            
        # 验证非连续性
        sorted_repr = sorted(repr)
        for i in range(len(sorted_repr)-1):
            if sorted_repr[i+1] - sorted_repr[i] < 2:
                return False, n, "consecutive indices"
                
    return True, None, None

检查点2：唯一性验证​

def verify_uniqueness(n):
    # 尝试所有可能的表示
    all_reprs = []
    
    def find_all_representations(target, max_idx, current):
        if target == 0:
            all_reprs.append(current.copy())
            return
            
        for i in range(max_idx, 0, -1):
            if fib[i] <= target:
                # 检查非连续约束
                if not current or min(current) - i >= 2:
                    current.add(i)
                    find_all_representations(target - fib[i], i-2, current)
                    current.remove(i)
    
    find_all_representations(n, find_max_fib_index(n), set())
    return len(all_reprs) == 1

检查点3：双射验证​

def verify_bijection(max_bits):
    # 生成所有no-11串
    no11_strings = generate_no11_strings(max_bits)
    
    # 映射到整数
    mapped_ints = set()
    for s in no11_strings:
        n = phi_encode(s)
        if n in mapped_ints:
            return False, "not injective"
        mapped_ints.add(n)
    
    # 检查连续性
    expected = set(range(1, len(no11_strings) + 1))
    return mapped_ints == expected

检查点4：编码长度验证​

def verify_encoding_length(test_values):
    phi = (1 + math.sqrt(5)) / 2
    
    for n in test_values:
        repr = greedy_decomposition(n)
        actual_length = max(repr) if repr else 0
        
        expected_length = math.floor(math.log(n, phi)) + 1
        
        # 允许O(1)偏差
        if abs(actual_length - expected_length) > 2:
            return False, n, actual_length, expected_length
            
    return True, None, None, None

实用函数​

def greedy_decomposition(n):
    """贪心算法计算φ-表示"""
    result = []
    fib = generate_fibonacci(n)
    
    i = len(fib) - 1
    while n > 0 and i >= 1:
        if fib[i] <= n:
            result.append(i)
            n -= fib[i]
            i -= 2  # 跳过相邻
        else:
            i -= 1
            
    return result

def phi_encode(binary_string):
    """从二进制串计算对应整数"""
    total = 0
    for i, bit in enumerate(binary_string, 1):
        if bit == '1':
            total += fib[i]
    return total

形式化验证状态​

  • 引理语法正确
  • 存在性证明完整
  • 唯一性证明严格
  • 双射性质明确
  • 最小完备