(i) strange f g = g (f g)
Assume g :: a -> b. Then f :: (a -> b) -> c. But since g :: a -> b, f g :: a, so c = a. Therefore, f :: (a -> b) -> a, and g (f g) :: a. Therefore, strange :: ((a -> b) -> a) -> (a -> b) -> a.
Almost. The return type of strange is the same as the return type of g (the outermost function), namely b. So strange :: ((a -> b) -> a) -> (a -> b) -> b. Dan R J wrote:
Bird 1.6.3 requires deducing type signatures for the functions "strange" and "stranger."
Are my solutions below correct?
(i) strange f g = g (f g)
Assume g :: a -> b. Then f :: (a -> b) -> c. But since g :: a -> b, f g :: a, so c = a. Therefore, f :: (a -> b) -> a, and g (f g) :: a. Therefore, strange :: ((a -> b) -> a) -> (a -> b) -> a.
(ii) stranger f = f f
Assume f :: a -> b. Since "f f" is well-typed, type unification requires a = b. Therefore, f :: a -> a, and stranger :: (a -> a) -> a.
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