Great. Thanks. This reminds me of one emphasis of McBride's lectures and keynotes regarding Agda: it's generally a Good Thing for the shape of your type level recursion to match your value level recursion. On Feb 4, 2014 1:20 PM, "Richard Eisenberg" <[email protected]> wrote:
> Say I have > > > data Nat = Zero | Succ Nat > > data SNat :: Nat -> * where > > SZero :: SNat Zero > > SSucc :: SNat n -> SNat (Succ n) > > data SBool :: Bool -> * where > > SFalse :: SBool False > > STrue :: SBool True > > Now, I want > > > eq :: SNat a -> SNat b-> SBool (a == b) > > eq SZero SZero = STrue > > eq (SSucc _) SZero = SFalse > > eq SZero (SSucc _) = SFalse > > eq (SSucc c) (SSucc d) = eq c d > > Does that type check? > > Suppose we have > > > type family EqPoly (a :: k) (b :: k) :: Bool where > > EqPoly a a = True > > EqPoly a b = False > > type instance a == b = EqPoly a b > > (Let's forget that the instance there would overlap with any other > instance.) > > Now, in the last line of `eq`, we have that the type of `eq c d` is `SBool > (e == f)` where (c :: SNat e), (d :: SNat f), (a ~ Succ e), and (b ~ Succ > f). But, does ((e == f) ~ (a == b))? It would need to for the last line of > `eq` to type-check. Alas, there is no way to proof ((e == f) ~ (a == b)), > so we're hosed. > > Now, suppose > > > type family EqNat a b where > > EqNat Zero Zero = True > > EqNat (Succ n) (Succ m) = EqNat n m > > EqNat Zero (Succ n) = False > > EqNat (Succ n) Zero = False > > type instance a == b = EqNat a b > > Here, we know that (a ~ Succ e) and (b ~ Succ f), so we compute that (a == > b) ~ (EqNat (Succ e) (Succ f)) ~ (EqNat e f) ~ (e == f). Huzzah! > > Thus, the second version is better. > > I hope this helps! > Richard > > On Feb 4, 2014, at 1:08 PM, Nicolas Frisby <[email protected]> > wrote: > > [CC'ing Richard, as I'm guessing he's the author of the comment.] > > I have a question regarding the comment on the type > family Data.Type.Equality.(==). > > "A poly-kinded instance [of ==] is *not* provided, as a recursive > definition for algebraic kinds is generally more useful." > > Can someone elaborate on "generally more useful". > > Thank you. > > >
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