dd78012ddd
Although `HEq` was abbreviated as `≍` in #8503, many instances of the form `HEq x y` still remain. Therefore, I searched for occurrences of `HEq x y` using the regular expression `(?<![A-Za-z/@]|``)HEq(?![A-Za-z.])` and replaced as many as possible with the form `x ≍ y`.
114 lines
3.4 KiB
Text
114 lines
3.4 KiB
Text
inductive ListSplit : List α → Type _
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| split l₁ l₂ : ListSplit (l₁ ++ l₂)
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def splitList : (l : List α) → ListSplit l
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| [] => ListSplit.split [] []
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| h :: t => ListSplit.split [h] t
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@[simp] def ListSplit.left {as : List α} : ListSplit as → List α
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| split a b => a
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@[simp] def ListSplit.right {as : List α} : ListSplit as → List α
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| split a b => b
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/-- Helper theorem for justifying termination. -/
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theorem splitList_length (as : List α) (h₁ : as.length > 1) (h₂ : as = bs) : (splitList as).left.length < bs.length ∧ (splitList as).right.length < bs.length := by
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match as with
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| [] => contradiction
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| a :: as => simp +arith [← h₂, splitList]; simp +arith at h₁; assumption
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def len : List α → Nat
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| [] => 0
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| a :: [] => 1
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| l@h₁:(a :: b :: as) =>
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-- Remark: we didn't use `_` because we currently don't have a way for getting a hypothesis stating that the previous two case were not taken here.
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-- h₁ : l = a :: b :: as
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match h₂ : splitList l with
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| ListSplit.split fst snd =>
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-- Remark: `match` refined `h₁`s type to `h₁ : fst ++ snd = a :: b :: as`
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-- h₂ : splitList l ≍ ListSplit.split fst snd
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have := splitList_length (fst ++ snd) (by simp +arith [h₁]) h₁
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-- The following two proofs ase used to justify the recursive applications `len fst` and `len snd`
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have dec₁ : fst.length < as.length + 2 := by subst l; simp +arith [eq_of_heq h₂] at this |- ; simp [this]
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have dec₂ : snd.length < as.length + 2 := by subst l; simp +arith [eq_of_heq h₂] at this |- ; simp [this]
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len fst + len snd
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termination_by xs => xs.length
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-- The equational theorems are
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#check @len.eq_1
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#check @len.eq_2
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#check @len.eq_3
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#check @len.eq_def
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theorem len_nil : len ([] : List α) = 0 := by
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simp [len]
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theorem len_1 (a : α) : len [a] = 1 := by
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simp [len]
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theorem len_2 (a b : α) (bs : List α) : len (a::b::bs) = 1 + len (b::bs) := by
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simp [len, splitList]
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theorem len_cons (a : α) (as : List α) : len (a::as) = 1 + len as := by
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cases as with
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| nil => simp [len_1, len_nil]
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| cons b bs => simp [len_2]
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theorem listlen : ∀ l : List α, l.length = len l := by
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intro l
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induction l with
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| nil => simp [len_nil]
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| cons h t ih =>
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simp [List.length, len_cons, ih]
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rw [Nat.add_comm]
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namespace Ex2
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/--
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`len` example again but with the proofs at `decreasing_by`
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-/
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def len : List α → Nat
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| [] => 0
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| a :: [] => 1
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| l@h₁:(a :: b :: as) =>
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match h₂ : l, h₃ : splitList l with
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| _, ListSplit.split fst snd =>
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len fst + len snd
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termination_by xs => xs.length
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decreasing_by
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all_goals
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have := splitList_length (fst ++ snd) (by simp +arith [h₁]) h₁
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subst h₂
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simp +arith [eq_of_heq h₃] at this |- ; simp [this]
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-- The equational theorems are
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#check @len.eq_1
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#check @len.eq_2
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#check @len.eq_3
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#check @len.eq_def
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theorem len_nil : len ([] : List α) = 0 := by
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simp [len]
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theorem len_1 (a : α) : len [a] = 1 := by
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simp [len]
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theorem len_2 (a b : α) (bs : List α) : len (a::b::bs) = 1 + len (b::bs) := by
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conv => lhs; unfold len
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simp [len, splitList]
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theorem len_cons (a : α) (as : List α) : len (a::as) = 1 + len as := by
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cases as with
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| nil => simp [len_1, len_nil]
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| cons b bs => simp [len_2]
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theorem listlen : ∀ l : List α, l.length = len l := by
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intro l
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induction l with
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| nil => simp [len_nil]
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| cons h t ih =>
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simp [List.length, len_cons, ih]
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rw [Nat.add_comm]
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end Ex2
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