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fix: simp_arith #6608

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Jan 12, 2025
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fix: simp_arith #6608

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30 changes: 18 additions & 12 deletions src/Lean/Meta/Tactic/LinearArith/Nat/Simp.lean
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Original file line number Diff line number Diff line change
Expand Up @@ -50,18 +50,24 @@ def simpCnstr? (e : Expr) : MetaM (Option (Expr × Expr)) := do
if let some arg := e.not? then
let mut eNew? := none
let mut thmName := Name.anonymous
if arg.isAppOfArity ``LE.le 4 then
eNew? := some (← mkLE (← mkAdd (arg.getArg! 3) (mkNatLit 1)) (arg.getArg! 2))
thmName := ``Nat.not_le_eq
else if arg.isAppOfArity ``GE.ge 4 then
eNew? := some (← mkLE (← mkAdd (arg.getArg! 2) (mkNatLit 1)) (arg.getArg! 3))
thmName := ``Nat.not_ge_eq
else if arg.isAppOfArity ``LT.lt 4 then
eNew? := some (← mkLE (arg.getArg! 3) (arg.getArg! 2))
thmName := ``Nat.not_lt_eq
else if arg.isAppOfArity ``GT.gt 4 then
eNew? := some (← mkLE (arg.getArg! 2) (arg.getArg! 3))
thmName := ``Nat.not_gt_eq
match_expr arg with
| LE.le α _ _ _ =>
if α.isConstOf ``Nat then
eNew? := some (← mkLE (← mkAdd (arg.getArg! 3) (mkNatLit 1)) (arg.getArg! 2))
thmName := ``Nat.not_le_eq
| GE.ge α _ _ _ =>
if α.isConstOf ``Nat then
eNew? := some (← mkLE (← mkAdd (arg.getArg! 2) (mkNatLit 1)) (arg.getArg! 3))
thmName := ``Nat.not_ge_eq
| LT.lt α _ _ _ =>
if α.isConstOf ``Nat then
eNew? := some (← mkLE (arg.getArg! 3) (arg.getArg! 2))
thmName := ``Nat.not_lt_eq
| GT.gt α _ _ _ =>
if α.isConstOf ``Nat then
eNew? := some (← mkLE (arg.getArg! 2) (arg.getArg! 3))
thmName := ``Nat.not_gt_eq
| _ => pure ()
if let some eNew := eNew? then
let h₁ := mkApp2 (mkConst thmName) (arg.getArg! 2) (arg.getArg! 3)
if let some (eNew', h₂) ← simpCnstrPos? eNew then
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9 changes: 9 additions & 0 deletions tests/lean/run/grind_t1.lean
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Expand Up @@ -225,3 +225,12 @@ example {P Q : Prop} (q : Q) (w : P = (P = ¬ Q)) : False := by

example (P Q : Prop) : (¬P → ¬Q) ↔ (Q → P) := by
grind

example {α} (a b c : α) [LE α] :
¬(¬a ≤ b ∧ a ≤ c ∨ ¬a ≤ c ∧ a ≤ b) ↔ a ≤ b ∧ a ≤ c ∨ ¬a ≤ c ∧ ¬a ≤ b := by
simp_arith -- should not fail
sorry

example {α} (a b c : α) [LE α] :
¬(¬a ≤ b ∧ a ≤ c ∨ ¬a ≤ c ∧ a ≤ b) ↔ a ≤ b ∧ a ≤ c ∨ ¬a ≤ c ∧ ¬a ≤ b := by
grind
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