On Tue, May 28, 2013 at 05:21:58PM +0200, Johannes Gerer wrote: > That makes sense. But why does > > instance Monad m => ArrowApply (Kleisli m) > > show that a Monad can do anything an ArrowApply can (and the two are > thus equivalent)?
I've tried to chase around the equivalence between these two before, and I didn't find the algebra simple. I'll give an outline. In non-Haskell notation 1) instance Monad m => ArrowApply (Kleisli m) means that if "m" is a Monad then "_ -> m _" is an ArrowApply. 2) instance ArrowApply a => Monad (a anyType) means that if "_ ~> _" is an ArrowApply then "a ~> _" is a Monad. One direction seems easy: for a Monad m, 1) gives that "_ -> m _" is an ArrowApply. By 2), "() -> m _" is a Monad. It is equivalent to the Monad m we started with. Given an ArrowApply "_ ~> _", 2) shows that "() ~> _" is a Monad. Thus by 1) "_ -> (() ~> _)" is an ArrowApply. I believe this should be the same type as "_ ~> _" but I don't see how to demonstrate the isomorphsim here. Tom _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
