I have the following definition for an ordType and infix notations for its boolean comparison operators ==, <b and <=b.
Module Order.
Structure type: Type:= Pack {
sort: Type;
eqb: sort-> sort -> bool;
ltb: sort-> sort -> bool;
eq_P: forall x y, reflect (eq x y)(eqb x y);
ltb_irefl: forall x, ltb x x=false;
ltb_antisym: forall x y,x<>y -> ltb x y =negb (ltb y x);
ltb_trans: forall x y z, ltb x y -> ltb y z -> ltb x z }.
Module Exports.
Coercion sort : type >-> Sortclass.
Notation ordType:= type.
End Exports.
End Order.
Definition eqb := Order.eqb.
Definition ltb := Order.ltb.
Definition leb (T:ordType) := fun (x y:T) => (ltb x y || eqb x y).
Notation "x == y":= (@eqb _ x y)(at level 70, no associativity): bool_scope.
Notation "x <b y":= (@ltb _ x y)(at level 70, no associativity): bool_scope.
Notation " x <=b y" := (@leb _ x y)(at level 70, no associativity): bool_scope.
Now consider the following Ltac definition for show_H which prints the hypothesis whose type is of the form x == y.
Ltac show_H:=
match goal with
| H: ?x == ?y |- _ => idtac H
end.
However, when I use this definition to show the name of hypothesis in the following Lemma it fails with the following message.
Lemma triv (T:ordType)(x y:T): x == y -> y <b x -> 2=3.
Proof. intros. show_H.
(Error: No matching clauses for match)
Why is the parser unable to detect notation == in the hypothesis ?
This is happening because of the coercion is_true -- the H hypothesis is actually H : is_true (x == y).
You can see it if you enable printing the coercions like so:
Set Printing Coercions.