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.
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.
