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I am working the Exercism Haskell Track and one of the questions asks for a custom set implementation. I studied some community solutions to see what other folks are doing, and came across something that I don't quite understand.
You can test out things like that at https://pointfree.io. It takes a lambda with all the arguments present, like \ a b c -> f (g a b c), and it produces ((f .) .) . g.
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