[Haskell-cafe] Deducing a type signature

[Dan Weston]

 > (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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