chore: add test
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1 changed files with 24 additions and 29 deletions
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@ -1,35 +1,30 @@
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theorem tst0 (x : Nat) : x + 0 = x + 0 :=
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by {
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generalize x + 0 = y;
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exact (Eq.refl y)
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}
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example (x : Nat) : x + 0 = x + 0 := by
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generalize x + 0 = y
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rfl
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theorem tst1 (x : Nat) : x + 0 = x + 0 :=
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by {
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generalize h : x + 0 = y;
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exact (Eq.refl y)
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}
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example (x : Nat) : x + 0 = x + 0 := by
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generalize h : x + 0 = y
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rfl
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theorem tst2 (x y w : Nat) (h : y = w) : (x + x) + w = (x + x) + y :=
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by {
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generalize h' : x + x = z;
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subst y;
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exact Eq.refl $ z + w
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}
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example (x y w : Nat) (h : y = w) : (x + x) + w = (x + x) + y := by
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generalize h' : x + x = z
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subst y
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rfl
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theorem tst3 (x y w : Nat) (h : x + x = y) : (x + x) + (x+x) = (x + x) + y :=
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by {
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generalize h' : x + x = z;
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subst z;
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subst y;
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exact rfl
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}
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example (x y w : Nat) (h : x + x = y) : (x + x) + (x+x) = (x + x) + y := by
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generalize h' : x + x = z
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subst z
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subst y
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rfl
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theorem tst4 (x y w : Nat) (h : y = w) : (x + x) + w = (x + x) + y :=
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by {
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generalize h' : x + y = z; -- just add equality
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subst h;
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exact rfl
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}
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example (x y w : Nat) (h : y = w) : (x + x) + w = (x + x) + y := by
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generalize h' : x + y = z -- just add equality
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subst h
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rfl
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example (x y w : Nat) (h : y = w) (H : (x + x) + w = (x + x) + y) :
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(x + x) + w = (x + x) + y := by
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generalize h' : x + x = z at H
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exact H
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