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`.
95 lines
3.1 KiB
Text
95 lines
3.1 KiB
Text
set_option warningAsError false
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set_option grind.debug true
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set_option grind.debug.proofs true
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example (a b : Nat) (f : Nat → Nat) : (h₁ : a = b) → f a = f b := by
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grind
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example (a b : Nat) (f : Nat → Nat) : (h₁ : a = b) → (h₂ : f a ≠ f b) → False := by
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grind
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example (a b : Nat) (f : Nat → Nat) : a = b → f (f a) ≠ f (f b) → False := by
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grind
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example (a b c : Nat) (f : Nat → Nat) : a = b → c = b → f (f a) ≠ f (f c) → False := by
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grind
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example (a b c : Nat) (f : Nat → Nat → Nat) : a = b → c = b → f (f a b) a ≠ f (f c c) c → False := by
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grind
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example (a b c : Nat) (f : Nat → Nat → Nat) : a = b → c = b → f (f a b) a = f (f c c) c := by
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grind
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example (a b c d : Nat) : a = b → b = c → c = d → a = d := by
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grind
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example (a b c d : Nat) : a ≍ b → b = c → c ≍ d → a ≍ d := by
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grind
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example (a b c d : Nat) : a = b → b = c → c ≍ d → a ≍ d := by
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grind
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opaque f {α : Type} : α → α → α := fun a _ => a
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opaque g : Nat → Nat
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example (a b c : Nat) : a = b → g a ≍ g b := by
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grind
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example (a b c : Nat) : a = b → c = b → f (f a b) (g c) = f (f c a) (g b) := by
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grind
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example (a b c d e x y : Nat) : a = b → a = x → b = y → c = d → c = e → c = b → a = e := by
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grind
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namespace Ex1
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opaque f (a b : Nat) : a > b → Nat
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opaque g : Nat → Nat
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example (a₁ a₂ b₁ b₂ c d : Nat)
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(H₁ : a₁ > b₁)
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(H₂ : a₂ > b₂) :
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a₁ = c → a₂ = c →
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b₁ = d → d = b₂ →
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g (g (f a₁ b₁ H₁)) = g (g (f a₂ b₂ H₂)) := by
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grind
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end Ex1
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namespace Ex2
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def f (α : Type) (a : α) : α := a
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example (a : α) (b : β) : (h₁ : α = β) → (h₂ : a ≍ b) → f α a ≍ f β b := by
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grind
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end Ex2
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example (f g : (α : Type) → α → α) (a : α) (b : β) : (h₁ : α = β) → (h₂ : a ≍ b) → (h₃ : f = g) → f α a ≍ g β b := by
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grind
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set_option trace.grind.debug.proof true in
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example (f : {α : Type} → α → Nat → Bool → Nat) (a b : Nat) : f a 0 true = v₁ → f b 0 true = v₂ → a = b → v₁ = v₂ := by
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grind
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set_option trace.grind.debug.proof true in
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example (f : {α : Type} → α → Nat → Bool → Nat) (a b : Nat) : f a b x = v₁ → f a b y = v₂ → x = y → v₁ = v₂ := by
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grind
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set_option trace.grind.debug.proof true in
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theorem ex1 (f : {α : Type} → α → Nat → Bool → Nat) (a b c : Nat) : f a b x = v₁ → f c b y = v₂ → a = c → x = y → v₁ = v₂ := by
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grind
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#print ex1
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example (n1 n2 n3 : Nat) (v1 w1 : Vector Nat n1) (w1' : Vector Nat n3) (v2 w2 : Vector Nat n2) :
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n1 = n3 → v1 = w1 → w1 ≍ w1' → v2 = w2 → v1 ++ v2 ≍ w1' ++ w2 := by
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grind
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example (n1 n2 n3 : Nat) (v1 w1 : Vector Nat n1) (w1' : Vector Nat n3) (v2 w2 : Vector Nat n2) :
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n1 ≍ n3 → v1 = w1 → w1 ≍ w1' → v2 ≍ w2 → v1 ++ v2 ≍ w1' ++ w2 := by
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grind
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theorem ex2 (n1 n2 n3 : Nat) (v1 w1 v : Vector Nat n1) (w1' : Vector Nat n3) (v2 w2 w : Vector Nat n2) :
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n1 ≍ n3 → v1 = w1 → w1 ≍ w1' → v2 ≍ w2 → w1' ++ w2 ≍ v ++ w → v1 ++ v2 ≍ v ++ w := by
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grind
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#print ex2
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