784972462a
while trying to help a user who was facing an unhelpful ``` omega did not find a contradiction: [0, 0, 0, 0, 1, -1] ∈ [1, ∞) [0, 0, 0, 0, 0, 1] ∈ [0, ∞) [0, 0, 0, 0, 1] ∈ [0, ∞) [1, -1] ∈ [1, ∞) [0, 0, 0, 1] ∈ [0, ∞) [0, 1] ∈ [0, ∞) [1] ∈ [0, ∞) [0, 0, 0, 1, 1] ∈ [-1, ∞) ``` I couldn’t resist and wrote a pretty-printer for these problem that shows the linear combination as such, and includes the recognized atoms. This is especially useful since oftem `omega` failures stem from failure to recognize atoms as equal. In this case, we now get: ``` omega-failure.lean:19:2-19:7: error: omega could not prove the goal: a possible counterexample may satisfy the constraints d - e ≥ 1 e ≥ 0 d ≥ 0 a - b ≥ 1 c ≥ 0 b ≥ 0 a ≥ 0 c + d ≥ -1 where a := ↑(sizeOf xs) b := ↑(sizeOf x) c := ↑(sizeOf x.fst) d := ↑(sizeOf x.snd) e := ↑(sizeOf xs) ``` and this might help the user make progress (e.g. by using `case x` first, and investingating why `sizeOf xs` shows up twice)
32 lines
911 B
Text
32 lines
911 B
Text
/-!
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Testing omega's failure message
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-/
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example : 0 < 0 := by omega
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example (x : Nat) : x < 0 := by omega
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example (x : Nat) (y : Int) : x < y := by omega
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example (x y : Int) : 5 < x ∧ x < 10 → y > 0 := by omega
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-- this used to fail with y = x, but then omega got better, so now there are unrelated x and y
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-- to make omega fail
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theorem sizeOf_snd_lt_sizeOf_list {α : Type u} {β : Type v} [SizeOf α] [SizeOf β]
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{x y : α × β} {xs : List (α × β)} :
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y ∈ xs → sizeOf x.snd < 1 + sizeOf xs
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:= by
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intro h
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have := List.sizeOf_lt_of_mem h
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have : sizeOf x = 1 + sizeOf x.1 + sizeOf x.2 := rfl
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omega
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example (reallyreallyreallyreally longlonglonglong namenamename : Nat) :
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reallyreallyreallyreally < longlonglonglong + namenamename := by omega
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def a := 1
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example (b c d e f g h i j k l m n o p : Nat) :
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b + c + d + e + f + g + h + i + j + k + l + m + n + o + p < 100 := by omega
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