Unfortunately I am unable to implement this function I have tried a lot.
f5 :: (Either a b -> c) -> (a -> c, b -> c)
I tried this
f5 k _ (Left x) = k x
f5 _ u (Right y) = u y
Thanks the help in advance .
Let the types be your guide.
Your function should take one argument, a function of type Either a b -> c, and produce a pair of functions, (a -> c, b -> c).
(The function you wrote takes three arguments and doesn't produce a pair.)
That is, you want
f5 f = (a_to_c, b_to_c)
where f :: Either a b -> c, a_to_c :: a -> c and b_to_c :: b -> c.
In order to create a_to_c, assume that you have an a and a function Either a b -> c.
Now, there's (only) one way you can make a function of type a -> c from those two things – create an Either from the a with Left, and then pass it to the "Either function".
Informally,
a_to_c x = f (Left x)
or
a_to_c = f . Left
The same reasoning for b leads to a similar situation with b.
b_to_c x = f (Right x)
or
b_to_c = f . Right
Putting them together
f5 f = (f . Left, f . Right)
or, more verbosely
f5 f = (\x -> f (Left x), \x -> f (Right x))
Either a b -> c) and needs to return a pair of functions so it need to look likef5 f = (\a -> ..., \b -> ...).