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`.
60 lines
2 KiB
Text
60 lines
2 KiB
Text
/-! # Check basic `calc` error messages and positions -/
|
||
|
||
/-! Basic proof failure -/
|
||
|
||
theorem ex1 (a b c : Nat) (h₁ : a = b) (h₂ : b = c) : a + b = 0 + c + b :=
|
||
calc a + b = b + b := by rw [h₁]
|
||
_ = 0 + c + b := rfl
|
||
|
||
/-! Step term mismatch -/
|
||
|
||
theorem ex2 (a b c : Nat) (h₁ : a = b) (h₂ : b = c) : a + b = 0 + c + b :=
|
||
calc a + b = b + b := by rw [h₁]
|
||
0 + c + b = 0 + c + b := rfl
|
||
|
||
-- fixed
|
||
theorem ex3 (a b c : Nat) (h₁ : a = b) (h₂ : b = c) : a + b = 0 + c + b :=
|
||
calc a + b = b + b := by rw [h₁]
|
||
_ = 0 + b + b := by rw [Nat.zero_add]
|
||
_ = 0 + c + b := by rw [h₂]
|
||
|
||
/-! Initial term mismatch -/
|
||
|
||
theorem ex4 (p : Nat → Prop) (a b : Nat) (h₁ : p a) (h₂ : p b) : p c :=
|
||
calc p a := h₁
|
||
_ := h₂
|
||
|
||
/-! Final term mismatch -/
|
||
|
||
theorem ex5 (p : Nat → Nat → Prop) (a b : Nat) (h₁ : p a b) (h₂ : p b c) : p a c :=
|
||
calc p a b := h₁
|
||
p _ c := h₂
|
||
|
||
/-! Relation with bad signature -/
|
||
|
||
instance {α β γ} : Trans (· ≍ · : α → β → Prop) (· ≍ · : β → γ → Prop) (· ≍ · : α → γ → Prop) where
|
||
trans h₁ h₂ := HEq.trans h₁ h₂
|
||
|
||
theorem ex6 {a : α} {b : β} {c : γ} (h₁ : a ≍ b) (h₂ : b ≍ c) : a ≍ c :=
|
||
calc a ≍ b := h₁
|
||
_ ≍ c := h₂ -- Error because the last two arguments of HEq are not explicit
|
||
|
||
-- fixed
|
||
abbrev HEqRel {α β} (a : α) (b : β) := a ≍ b
|
||
|
||
infix:50 "===" => HEqRel
|
||
|
||
instance {α β γ} : Trans (HEqRel : α → β → Prop) (HEqRel : β → γ → Prop) (HEqRel : α → γ → Prop) where
|
||
trans h₁ h₂ := HEq.trans h₁ h₂
|
||
|
||
theorem ex7 {a : α} {b : β} {c : γ} (h₁ : a ≍ b) (h₂ : b ≍ c) : a === c :=
|
||
calc a === b := h₁
|
||
_ === c := h₂
|
||
|
||
/-!
|
||
By enforcing separating newlines in the `calc` parser, we make sure that the next error reported is at `[`
|
||
instead of it being interpreted as the beginning of an incomplete calc step. -/
|
||
|
||
example : 1 + 1 = 2 := by
|
||
calc 1 + 1 = 1 + 1 := by rfl
|
||
_ = 2 := by rfl [h]
|