On Dec 18, 2007 1:00 PM, Cristian Baboi <cristian.baboi@gmail.com> wrote:

This is what I "understand" so far ...

Suppose we have these two values:
a) \x->x + x
b) \x->2 * x
Because these to values are equal, all functions definable in Haskell must
preserve this.
This is why I am not allowed to define a function like

h :: (a->b) -> (a->b)
h x = x

Of course you can define h. This is just a more specific (as far as types go) version of 'id', as defined in the Prelude:

   id :: a -> a
   id x = x

where 'a' can be any type, including a function such as (a -> b). You can apply 'id' to either of your functions above, and get back an equivalent function, so each of the following evaluates to 20:

    let f = id (\x -> x+x) in f 10
    let f = id (\x -> 2 * x) in f 10