please elaborate on comment for base:Data.Type.Equality.(==)?
nicolas.frisby at gmail.com
Tue Feb 4 19:32:38 UTC 2014
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" <eir at cis.upenn.edu> 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
> 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!
> On Feb 4, 2014, at 1:08 PM, Nicolas Frisby <nicolas.frisby at gmail.com>
> [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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