(* Test that adding notations that overlap with the tactic grammar does not
* interfere with Ltac parsing. *)
Module test1.
Notation "x [ y ]" := (fst (id x, id y)) (at level 11).
Goal True \/ (exists x : nat, True /\ True) -> True.
Proof .
intros [|[a [y z]]]; [idtac |idtac ]; try solve [eauto | trivial ; [trivial ]].
Qed .
End test1.
Module test2.
Notation "x [ y ]" := (fst (id x, id y)) (at level 100).
Goal True \/ (exists x : nat, True /\ True) -> True.
Proof .
intros [|[a [y z]]]; [idtac |idtac ]; try solve [eauto | trivial ; [trivial ]].
Qed .
End test2.
Module test3.
Notation "x [ y ]" := (fst (id x, id y)) (at level 1).
Goal True \/ (exists x : nat, True /\ True) -> True.
Proof .
intros [|[a [y z]]]; [idtac |idtac ]; try solve [eauto | trivial ; [trivial ]].
Qed .
End test3.
Module test1'.
Notation "x [ [ y ] ] " := (fst (id x, id y)) (at level 11).
Goal True \/ (exists x : nat, True /\ True) -> True.
Proof .
intros [|[a [y z]]]; [idtac |idtac ]; try solve [eauto | trivial ; [trivial ]].
Qed .
End test1'.
Module test2'.
Notation "x [ [ y ] ]" := (fst (id x, id y)) (at level 100).
Goal True \/ (exists x : nat, True /\ True) -> True.
Proof .
intros [|[a [y z]]]; [idtac |idtac ]; try solve [eauto | trivial ; [trivial ]].
Qed .
End test2'.
Module test3'.
Notation "x [ [ y ] ]" := (fst (id x, id y)) (at level 1).
Goal True \/ (exists x : nat, True /\ True) -> True.
Proof .
intros [|[a [y z]]]; [idtac |idtac ]; try solve [eauto | trivial ; [trivial ]].
Qed .
End test3'.
Messung V0.5 C=95 H=89 G=91
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