枚举类型的COQ相等

Question

I have a finite enumerated type (say T) in COQ and want to check elements for equality. This means, I need a function

bool beq_T(x:T,y:T)  

The only way I managed to define such a function, is with a case by case analysis. This results in lots of match statements and looks very cumbersome. Therefore, I wonder if there is not a simpler way to achieve this.

Answer 1

甚至更简单：...的Scheme Equality for ...生成两个函数，分别返回布尔值和sumbool。

Answer 2

The bad news is that the program that implements beq_T must necessarily consist of a large match statement on both of its arguments. The good news is that you can automatically generate/synthesize this program using Coq's tactic language . For example, given the type:

Inductive T := t0 | t1 | t2 | t3.

You can define beq_T as follows. The first two destruct tactics synthesize the code necessary to match on both x and y . The match tactic inspects the branch of the match that it is in, and depending on whether x = y , the tactic either synthesises the program that returns true or false .

Definition beq_T (x y:T) : bool.
  destruct x eqn:?;
  destruct y eqn:?;
  match goal with 
  | _:x = ?T, _:y = ?T |- _ => exact true
  | _ => exact false
  end.
Defined.

If you want to see the synthesized program, run:

Print beq_T.

Thankfully, Coq already comes with a tactic that does almost what you want. It is called decide equality . It automatically synthesizes a program that decides whether two elements of a type T are equal. But instead of just returning a boolean value, the synthesized program returns a proof of the (in)equality of the two elements.

Definition eqDec_T (x y:T) : {x = y} + {x <> y}.
  decide equality.
Defined.

With that program synthesized, it is easy to implement beq_T .

Definition beq_T' {x y:T} : bool := if eqDec_T x y then true else false.