f9fb041e04
There was a typo in the proof generation. The weird part is that the proof was valid, but it was very inefficient to check. The proof was valid because ((n:int) ≥ m) reduces to true/false if n and m are integer numerals. Thus, if ((n:int) ≥ m) holds then `trivial` is a valid proof. However, the reduction is extremely inefficient since it relies on computations in unary. In the buggy version, we provided a proof for (a >= 0) where (b >= 0) was expected. However, both types are definitionally equal to true. This is why the proof worked.
16 lines
323 B
Text
16 lines
323 B
Text
open tactic
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example : (1030912003002020:int) ≠ 1021992923928 :=
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by comp_val
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example : -(1030912003002020:int) ≠ 1021992923928 :=
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by comp_val
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example : (1030912003002020:int) ≠ -1021992923928 :=
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by comp_val
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example : -(1030912003002020:int) ≠ 0 :=
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by comp_val
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example : (0:int) ≠ 1021992923928 :=
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by comp_val
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