Udo Stenzel wrote: Thank you all a lot for helping me, it's amazing how quickly I received these detailed answers!
func2 f g l = filter f (map g l) func2 f g = (filter f) . (map g) -- definition of (.) func2 f g = ((.) (filter f)) (map g) -- desugaring func2 f = ((.) (filter f)) . map -- definition of (.) func2 f = flip (.) map ((.) (filter f)) -- desugaring, def. of flip func2 = flip (.) map . (.) . filter -- def. of (.), twice func2 = (. map) . (.) . filter -- add back some sugar
Aaaah. After learning from Neil's answer and from @pl that (.) is just another infix function, too (well, what else should it be, but it wasn't clear to me) I still wasn't able to come up with that solution without hurting my brain. The desugaring was the bit that was missing. Thanks, I will keep that in mind for other infix functions as well. I tried to work it out on paper again, without looking at your posting while doing it. I did almost the same thing, however, I did not use flip. Instead the last few steps read: = ((.) (filter f)) . map g l = (.)((.) . filter f)(map) g l -- desugaring = (.map)((.) . filter f) g l -- sweeten up = (.map) . (.) . filter g l -- definition of (.) I guess that's possible as well?
The general process is called "lambda elimination" and can be done mechanically. Ask Goole for "Unlambda", the not-quite-serious programming language; since it's missing the lambda, its manual explains lambda elimination in some detail. I think, all that's needed is flip, (.) and liftM2.
Will do, thank you! Cheers, Julien