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haskellequational-reasoning

Function evaluation result

Im trying to evaluate manually fc [f1, f2] (\x -> 2) 3 but I don't realize how to match the three parameters: [f1, f2], (\x -> 2) and 3 with the function definition, any help?

The function definition:

fc xss = \f -> let ope x y = x . f . y in foldr1 ope xss

f1, f2 :: Int -> Bool
f1 x = (x `mod` 3) == 0
f2 x = (x `mod` 5) == 0

Thanks,
Sebastián.

Solution

  • Well, first do some η-expansion

    fc :: [a->b] -> (b->a) -> a->b
fc xss f = let ope x y = x . f . y in \q -> foldr1 ope xss q
 fc xss f q = let ope x y = x . f . y in foldr1 ope xss q

    Then perhaps inline the let

     fc xss f q = foldr1 (\x y -> x . f . y) xss q

    You can write [f1, f2] as (==0).(`mod`3) : (==0).(`mod`5) : [].

    Now it should be easy enough, remembering that a right-fold simply replaces all : with the fold function.