60 lines
2.5 KiB
Text
60 lines
2.5 KiB
Text
import data.nat data.prod logic.wf_k
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open nat well_founded decidable prod eq.ops
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-- I use "dependent" if-then-else to be able to communicate the if-then-else condition
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-- to the branches
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definition dite (c : Prop) [H : decidable c] {A : Type} (t : c → A) (e : ¬ c → A) : A :=
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decidable.rec_on H (assume Hc, t Hc) (assume Hnc, e Hnc)
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notation `dif` c `then` t:45 `else` e:45 := dite c t e
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-- Auxiliary lemma used to justify recursive call
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private lemma lt_aux {x y : nat} (H : 0 < y ∧ y ≤ x) : x - y < x :=
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have y0 : 0 < y, from and.elim_left H,
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have x0 : 0 < x, from lt_le_trans y0 (and.elim_right H),
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sub_lt x0 y0
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definition wdiv.F (x : nat) (f : Π x₁, x₁ < x → nat → nat) (y : nat) : nat :=
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dif 0 < y ∧ y ≤ x then (λ Hp, f (x - y) (lt_aux Hp) y + 1) else (λ Hn, zero)
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definition wdiv (x y : nat) :=
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fix wdiv.F x y
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theorem wdiv_def (x y : nat) : wdiv x y = if 0 < y ∧ y ≤ x then wdiv (x - y) y + 1 else 0 :=
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congr_fun (well_founded.fix_eq wdiv.F x) y
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-- There is a little bit of cheating in the definition above.
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-- I avoid the packing/unpacking into tuples.
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-- The actual definitional package would not do that.
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-- It will always pack things.
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definition pair_nat.lt := lex lt lt -- Could also be (lex lt empty_rel)
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definition pair_nat.lt.wf [instance] : well_founded pair_nat.lt :=
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intro_k (prod.lex.wf lt.wf lt.wf) 20 -- allow 20 recursive calls without computing with proofs
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infixl `≺`:50 := pair_nat.lt
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-- Recursive lemma used to justify recursive call
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private lemma plt_aux (p : nat × nat) (H : 0 < pr₂ p ∧ pr₂ p ≤ pr₁ p) : (pr₁ p - pr₂ p, pr₂ p) ≺ p :=
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have aux₁ : pr₁ p - pr₂ p < pr₁ p, from lt_aux H,
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have aux₂ : (pr₁ p - pr₂ p, pr₂ p) ≺ (pr₁ p, pr₂ p), from !lex.left aux₁,
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prod.eta p ▸ aux₂
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definition pdiv.F (p₁ : nat × nat) (f : Π p₂ : nat × nat, p₂ ≺ p₁ → nat) : nat :=
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let x := pr₁ p₁ in
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let y := pr₂ p₁ in
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dif 0 < y ∧ y ≤ x then (λ Hp, f (x - y, y) (plt_aux p₁ Hp) + 1) else (λ Hnp, zero)
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definition pdiv (x y : nat) :=
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fix pdiv.F (x, y)
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theorem pdiv_def (x y : nat) : pdiv x y = if 0 < y ∧ y ≤ x then pdiv (x - y) y + 1 else zero :=
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well_founded.fix_eq pdiv.F (x, y)
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-- Remark: dite and ite are "definitionally equal" when we ignore the proofs.
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-- I used that, when I wrote div_def and pdiv_def.
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theorem dite_ite_eq (c : Prop) [H : decidable c] {A : Type} (t : A) (e : A) :
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dite c (λ Hc, t) (λ Hnc, e) = ite c t e :=
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rfl
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example : pdiv 398 23 = 17 :=
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rfl
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