-- Fibonacci.lean -- Equivalence of definitions of the Fibonacci function. -- José A. Alonso Jiménez <https://jaalonso.github.io> -- Seville, August 29, 2024 -- --------------------------------------------------------------------- -- --------------------------------------------------------------------- -- In Lean4, the Fibonacci function can be defined as -- def fibonacci : Nat → Nat -- | 0 => 0 -- | 1 => 1 -- | n + 2 => fibonacci n + fibonacci (n+1) -- -- Another more efficient definition is -- def fib (n : Nat) : Nat := -- (loop n).1 -- where -- loop : Nat → Nat × Nat -- | 0 => (0, 1) -- | n + 1 => -- let p := loop n -- (p.2, p.1 + p.2) -- -- Prove that both definitions are equivalent; that is, -- fibonacci = fib -- --------------------------------------------------------------------- -- Proof in natural language -- ========================= -- From the definition of the mirror function, we have the following lemma -- fib_suma : fib (n + 2) = fib n + fib (n + 1) -- -- We need to prove that, for all n ∈ ℕ, -- fibonacci n = fib n -- We will prove this by induction on n -- -- Case 1: Suppose that n = 0. Then, -- fibonacci n = fibonacci 0 -- = 1 -- and -- fib n = fib 0 -- = (loop 0).1 -- = (0, 1).1 -- = 1 -- Therefore, -- fibonacci n = fib n -- -- Case 2: Suppose that n = 1. Then, -- fibonacci n = 1 -- and -- fib 1 = (loop 1).1 -- = (p.2, p.1 + p.2).1 -- donde p = loop 0 -- = ((0, 1).2, (0, 1).1 + (0, 1).2).1 -- = (1, 0 + 1).1 -- = 1 -- Therefore, -- fibonacci n = fib n -- -- Case 3: Suppose that n = m + 2 and that the induction hypotheses hold, -- ih1 : fibonacci n = fib n -- ih2 : fibonacci (n + 1) = fib (n + 1) -- Then, -- fibonacci n -- = fibonacci (m + 2) -- = fibonacci m + fibonacci (m + 1) -- = fib m + fib (m + 1) [por ih1, ih2] -- = fib (m + 2) [por fib_suma] -- = fib n -- Proof with Lean4 -- ================ open Nat set_option pp.fieldNotation false def fibonacci : Nat → Nat | 0 => 0 | 1 => 1 | n + 2 => fibonacci n + fibonacci (n+1) def fib (n : Nat) : Nat := (loop n).1 where loop : Nat → Nat × Nat | 0 => (0, 1) | n + 1 => let p := loop n (p.2, p.1 + p.2) -- Auxiliary lemma -- =============== theorem fib_suma (n : Nat) : fib (n + 2) = fib n + fib (n + 1) := by rfl -- Proof 1 -- ======= example : fibonacci = fib := by ext n -- n : Nat -- ⊢ fibonacci n = fib n induction n using fibonacci.induct with | case1 => -- ⊢ fibonacci 0 = fib 0 rfl | case2 => -- ⊢ fibonacci 1 = fib 1 rfl | case3 n ih1 ih2 => -- n : Nat -- ih1 : fibonacci n = fib n -- ih2 : fibonacci (n + 1) = fib (n + 1) -- ⊢ fibonacci (succ (succ n)) = fib (succ (succ n)) rw [fib_suma] -- ⊢ fibonacci (succ (succ n)) = fib n + fib (n + 1) rw [fibonacci] -- ⊢ fibonacci n + fibonacci (n + 1) = fib n + fib (n + 1) rw [ih1] -- ⊢ fib n + fibonacci (n + 1) = fib n + fib (n + 1) rw [ih2] -- Proof 2 -- ======= example : fibonacci = fib := by ext n -- n : Nat -- ⊢ fibonacci n = fib n induction n using fibonacci.induct with | case1 => -- ⊢ fibonacci 0 = fib 0 rfl | case2 => -- ⊢ fibonacci 1 = fib 1 rfl | case3 n ih1 ih2 => -- n : Nat -- ih1 : fibonacci n = fib n -- ih2 : fibonacci (n + 1) = fib (n + 1) -- ⊢ fibonacci (succ (succ n)) = fib (succ (succ n)) calc fibonacci (succ (succ n)) = fibonacci n + fibonacci (n + 1) := by rw [fibonacci] _ = fib n + fib (n + 1) := by rw [ih1, ih2] _ = fib (succ (succ n)) := by rw [fib_suma] -- Proof 3 -- ======= example : fibonacci = fib := by ext n -- n : Nat -- ⊢ fibonacci n = fib n induction n using fibonacci.induct with | case1 => -- ⊢ fibonacci 0 = fib 0 rfl | case2 => -- ⊢ fibonacci 1 = fib 1 rfl | case3 n ih1 ih2 => -- n : Nat -- ih1 : fibonacci n = fib n -- ih2 : fibonacci (n + 1) = fib (n + 1) -- ⊢ fibonacci (succ (succ n)) = fib (succ (succ n)) simp [ih1, ih2, fibonacci, fib_suma]