refactor: Replace I with Identity in Haskell code

The `Identity` is used in other sections. Also, it is defined in base library.

Author: ktgw0316

src/content/3.11/code/haskell/snippet07.hs

@@ -1 +1 @@
-type Exp a b = Lan ((,) a) I b
+type Exp a b = Lan ((,) a) Identity b

src/content/3.11/code/haskell/snippet08.hs

@@ -1,5 +1,5 @@
 toExp :: (a -> b) -> Exp a b
-toExp f = Lan (f . fst) (I ())
+toExp f = Lan (f . fst) (Identity ())
 
 fromExp :: Exp a b -> (a -> b)
-fromExp (Lan f (I x)) = \a -> f (a, x)
+fromExp (Lan f (Identity x)) = \a -> f (a, x)