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[<michael.soegtrop AT intel.com>]

Dear Coq users,

for records with only assigned elements, the automatically generated induction
principles work, but when I print them it looks like they have an application
of a non function in them. Also if I try to feed the printed term back into
Coq it won't accept it with exectly that error message, which confuses me
quite a bit. Please have a look at the following example:

(* A record with an assigned element. Only one instance of this record exists
*)

Record A:=mkA
{
  a := 1
}.

(* The induction principle for this actually works *)

Lemma TestIndA : forall x:A, a x =1.
Proof.
  induction x.
  reflexivity.
Qed.

(* But when I print the induction principle, I get this *)

Print A_rect.

A_rect =
fun (P : A -> Type) (f : let a := 1 in P mkA) (a : A) =>
match a as a0 return (P a0) with
| {| a := x |} => f x
end
     : forall P : A -> Type, (let a := 1 in P mkA) -> forall a : A, P a

Here the term f in f x is not a function, so I wonder what f x is supposed to
mean. Also if I try to feed this term into Coq like this:

Definition A_rect :=
(fun (P : A -> Type) (f : let a := 1 in P mkA) (a : A) =>
match a as a0 return (P a0) with
| {| a := x |} => f x
end)
     : forall P : A -> Type, (let a := 1 in P mkA) -> forall a : A, P a
.

I get the error message:

Error: Illegal application (Non-functional construction):
The expression "f" of type "let a := 1 in P mkA"
cannot be applied to the term
 "x0" : "nat"

Can someone shed some light on this behaviour? In case this is of interest: I
am using Coq 8.4pl4.

Best regards,

Michael