我现在是Idris的初学者，所以我想寻求帮助。 我有除法的定义：

data Dividesnat : (a : Nat) -> (b : Nat) -> Type where
    Div : (k ** (k * x = y)) -> Dividesnat y x

和质数的定义，基于Dividesnat：

data Prime : (p : Nat) -> Type where
    ConsPrime : LTE 2 p ->
        ((d : Nat) -> Dividesnat p d -> Either (d = 1) (d = p)) ->
        Prime p

现在我想证明2是质数：

prf2IsPrime : Prime (S (S Z))
prf2IsPrime = ConsPrime (LTESucc (LTESucc LTEZero)) prf
    where
        prf : (d : Nat) -> Dividesnat (S (S Z)) d ->
            Either (d = 1) (d = (S (S Z)))
        prf d x = ?prf_rhs

d =（S Z）或d =（S（S Z））的情况非常简单：

prf : (d : Nat) -> Dividesnat (S (S Z)) d ->
            Either (d = 1) (d = (S (S Z)))
        prf Z (Div (x ** pf)) = ?prf_rhs2_3
        prf (S Z) (Div (x ** pf)) = Left Refl
        prf (S (S Z)) (Div (x ** pf)) = Right Refl
        prf (S (S (S _))) (Div (x ** pf)) = ?rr_2

但我不知道如何针对d = Z或d = (S (S (S _)))进行证明。我如何证明这些情况是不可能的？