当我的函数输出nat类型时，为什么Coq期望一个类型ID？

Question

我正在浏览Imp教程，发现我什至无法编译提供的代码软件基础：

Coercion AId : string >-> aexp.
Coercion ANum : nat >-> aexp.
Definition bool_to_bexp (b: bool) : bexp :=
  if b then BTrue else BFalse.
Coercion bool_to_bexp : bool >-> bexp.

Bind Scope aexp_scope with aexp.
Infix "+" := APlus : aexp_scope.
Infix "-" := AMinus : aexp_scope.
Infix "*" := AMult : aexp_scope.
Bind Scope bexp_scope with bexp.
Infix "<=" := BLe : bexp_scope.
Infix "=" := BEq : bexp_scope.
Infix "&&" := BAnd : bexp_scope.
Notation "'!' b" := (BNot b) (at level 60) : bexp_scope.

(** We can now write [3 + (X * 2)] instead  of [APlus 3 (AMult X 2)],
    and [true && !(X <= 4)] instead of [BAnd true (BNot (BLe X 4))]. *)

(* ================================================================= *)
(** ** Evaluation *)

(** The arith and boolean evaluators are extended to handle
    variables in the obvious way, taking a state as an extra
    argument: *)

Fixpoint aeval (st : state) (a : aexp) : nat :=
  match a with
  | ANum n => n
  | AId x => st x                                (* <----- NEW *)
  | APlus a1 a2 => (aeval st a1) + (aeval st a2)
  | AMinus a1 a2  => (aeval st a1) - (aeval st a2)
  | AMult a1 a2 => (aeval st a1) * (aeval st a2)
  end.

我得到的错误：

In environment
aeval : state -> aexp -> nat
st : state
a : aexp
x : string
The term "x" has type "string" while it is expected to have type "id".

In environment
aeval : state -> aexp -> nat
st : state
a : aexp
x : string
The term "x" has type "string" while it is expected to have type "id".

如果输出类型是nat，为什么Coq期望类型id不应该期望类型nat？