On 9/11/06, Bulat Ziganshin <[EMAIL PROTECTED]> wrote:
Expr Bool = Eq (Expr t) (Expr t) | forall t . Eq t
Still confusing, but less so.
The problem is that it's really backwards. The symbol being defined is "Eq".
Eq (Expr t) (Expr t) | Eq t = Expr Bool
but that doesn't fit well, does it? In my mind, the current mode is a
functional one:
Eq :: Eq t => Expr t -> Expr t -> Expr Bool
And this is perfectly readable. The arrow means that the argument is
an implicit propositional one, but it's still an argument.
--
Taral <[EMAIL PROTECTED]>
"You can't prove anything."
-- Gödel's Incompetence Theorem
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