[Haskell-cafe] Deducing a type signature

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 requiresa = b. Therefore, f :: a -> a, and stranger :: (a -> a) -> a. _________________________________________________________________ Hotmail is redefining busy with tools for the New Busy. Get more from your inbox. http://www.windowslive.com/campaign/thenewbusy?ocid=PID28326::T:WLMTAGL:ON:W...