Hey, No, I do not want to show anything. The monoid instance is from my teacher. It is an exercise. Sry for the misunderstanding. LG Alex Am 06.05.2017 um 20:25 schrieb MarLinn:
mappend pb1 pb2 = FunPB $ \n -> (,) n $ msum . (<$>) (uncurry (flip const).((flip runFunPB) n)) $ [pb1,pb2]
This cracks me up. Well played. Well played indeed.
Well all the things you need are right here in this line, right? I mean you sent it because you want to show us what you know, right? There's higher order functions, there's mapping and folding and currying of different kinds and everything. So what's missing?
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