I have the following definition for terms :
Require Import Coq.Arith.Arith.
Require Import Coq.Lists.List.
Require Import Coq.Strings.String.
Import ListNotations.
Definition VarIndex:Type := nat.
Inductive Term : Type :=
|Var : VarIndex -> Term
|Comb: string -> (list Term) -> Term.
(*compare two list *)
Fixpoint beq_list {X:Type} (l l' :list X) (EqX :X -> X -> bool): bool :=
match l with
| [] => match l' with
| [] => true
| a => false
end
| (x::xs) =>
match l' with
|[] => false
|(y::ys) => if (EqX x y) then beq_list xs ys EqX else false
end
end.
Fixpoint length {X : Type} (l : list X) : nat :=
match l with
| nil => 0
| cons _ l' => S (length l')
end.
和一个函数beq_term
来比较两个术语定义如下:
Fixpoint beq_term (t1:Term) (t2:Term) : bool :=
match t1, t2 with
| Var i, Var j => beq_nat i j
| Var _, Comb _ _ => false
|Comb _ _, Var _ => false
|(Comb s1 ts1), Comb s2 ts2 => if(beq_nat (length ts1) (length ts2))
then beq_list ts1 ts2 beq_term
else false
end.
函数beq_term
的定义产生错误消息:
错误:无法猜测
fix
的减少参数。
所以我对如何说服Coq终止感兴趣。
答案 0 :(得分:1)
如果您希望能够在这个简单的示例中使用Coq的语法检查,只需将beq_list
和beq_term
写入单个函数即可。
Fixpoint beq_list (l l' :list Term) : bool :=
match l, l' with
| [], [] => true
| (Var i)::xs, (Var j)::ys => beq_nat i j && beq_list xs ys
| (Comb s1 ts1)::xs, (Comb s2 ts2)::ys => beq_list xs ys
| _,_ => false
end.