[Haskell-cafe] Function Types

PR Stanley prstanley at ntlworld.com
Mon Oct 22 22:41:57 EDT 2007

f x = x
x :: a
f x :: b
therefore f :: a -> b
x = a and x = b
therefore a = b
therefore f :: a -> a
Simple mappings are easy to work out. It's the more detailed stuff 
I'm not sure about.
f g x y = g x (y x)

Cheers, Paul

At 03:15 23/10/2007, you wrote:
>On 10/22/07, PR Stanley 
><<mailto:prstanley at ntlworld.com>prstanley at ntlworld.com> wrote:
>What are the rules for calculating function types?
>Is there a set procedure ?
>Thanks, Paul
>There must be a set procedure, since otherwise the compiler could 
>not function! =)
>Seriously, though, I'm not exactly sure what you're asking 
>for.  Could you maybe provide a few examples of the kind of thing 
>you're asking about?

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