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Lean.Meta.Tactic.Simp.Rewrite

Wrapper for invoking discharge? method. It checks for maximum discharge depth, create trace nodes, and ensure the generated proof was successfully assigned to x.

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            Remark: the parameter tag is used for creating trace messages. It is irrelevant otherwise.

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              For (← getConfig).index := true, use discrimination tree structure when collecting simp theorem candidates.

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                For (← getConfig).index := false, Lean 3 style simp theorem retrieval. Only the root symbol is taken into account. Most of the structure of the discrimination tree is ignored.

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                      @[inline]
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                          Given a match-application e with MatcherInfo info, return some result if at least of one of the discriminants has been simplified.

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                                        Discharge procedure for the ground/symbolic evaluator.

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                                          Try to unfold ground term in the ground/symbolic evaluator.

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                                                  Invoke ground/symbolic evaluator from simp. It uses the seval theorems and simprocs.

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                                                    Try to unfold ground term in the ground/symbolic evaluator.

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                                                          Return true if e is of the form (x : α) → ... → s = t → ... → False

                                                          Recall that this kind of proposition is generated by Lean when creating equations for functions and match-expressions with overlapping cases. Example: the following match-expression has overlapping cases.

                                                          def f (x y : Nat) :=
                                                            match x, y with
                                                            | Nat.succ n, Nat.succ m => ...
                                                            | _, _ => 0
                                                          

                                                          The second equation is of the form

                                                          (x y : Nat) → ((n m : Nat) → x = Nat.succ n → y = Nat.succ m → False) → f x y = 0
                                                          

                                                          The hypothesis (n m : Nat) → x = Nat.succ n → y = Nat.succ m → False is essentially saying the first case is not applicable.

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                                                            Tries to solve e using unifyEq?. It assumes that isEqnThmHypothesis e is true.

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