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
On May 20, 2010, at 11:03 AM, R J wrote:
stranger f = f f
This doesn't have a type in Haskell.
Suppose f :: a - b
Then if f f made sense, a = (a - b) would be true,
and we'd have an infinite type.
Type the definition into a file, and try loading it
into ghci:
Occurs check: cannot
(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
