2014-11-10 20:46:55 +00:00
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import data.nat
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open nat eq.ops
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definition fib.F (n : nat) : (Π (m : nat), m < n → nat) → nat :=
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nat.cases_on n
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(λ (f : Π (m : nat), m < zero → nat), succ zero)
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(λ (n₁ : nat), nat.cases_on n₁
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(λ (f : Π (m : nat), m < (succ zero) → nat), succ zero)
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(λ (n₂ : nat) (f : Π (m : nat), m < (succ (succ n₂)) → nat),
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2014-11-22 08:15:51 +00:00
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have l₁ : succ n₂ < succ (succ n₂), from lt.base (succ n₂),
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have l₂ : n₂ < succ (succ n₂), from lt.trans (lt.base n₂) l₁,
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2014-11-10 20:46:55 +00:00
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f (succ n₂) l₁ + f n₂ l₂))
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definition fib (n : nat) :=
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well_founded.fix fib.F n
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theorem fib.zero_eq : fib 0 = 1 :=
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well_founded.fix_eq fib.F 0
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theorem fib.one_eq : fib 1 = 1 :=
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well_founded.fix_eq fib.F 1
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theorem fib.succ_succ_eq (n : nat) : fib (succ (succ n)) = fib (succ n) + fib n :=
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well_founded.fix_eq fib.F (succ (succ n))
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2014-11-22 08:15:51 +00:00
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example : fib 5 = 8 :=
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rfl
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example : fib 6 = 13 :=
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rfl
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