On Tue, Jul 27, 2010 at 8:32 AM, Henning Thielemann <schlepp...@henning-thielemann.de> wrote: > John Lato schrieb: >> Hello, >> >> I was wondering today, is this generally true? >> >> instance (Monad m, Monoid a) => Monoid (m a) where >> mempty = return mempty >> mappend = liftM2 mappend >> >> I know it isn't a good idea to use this instance, but assuming that >> the instance head does what I mean, is it valid? Or more generally is >> it true for applicative functors as well? I think it works for a few >> tricky monads, but that's not any sort of proof. I don't even know >> how to express what would need to be proven here. > > I translate 'valid' and 'true' to "Is 'm a' a Monoid, given that 'm' is > a Monad and 'a' is a Monoid?" If this is the question then we have to > show the Monoid laws for (m a), namely
Thanks very much, this is what I was unable to express properly (hence the informal description). > left identity: forall x. mappend mempty x = x > right identity: forall x. mappend x mempty = x > associativity: > forall x y z. > (x `mappend` y) `mappend` z = x `mappend` (y `mappend` z) So the task now is to prove that these laws hold, given the instance I defined above, the monoid laws for 'a', and the monad laws for 'm'. Or alternatively using applicative laws for 'm' for the more general case. I'll see how I get on from here then. Thanks, John _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
