+
+
+Notes on do_subst stuff
+~~~~~~~~~~~~~~~~~~~~~~~
+Consider
+ data T = forall a. Ord a => T a (a->Int)
+
+ f (T x f) True = ...expr1...
+ f (T y g) False = ...expr2..
+
+When we put in the tyvars etc we get
+
+ f (T a (d::Ord a) (x::a) (f::a->Int)) True = ...expr1...
+ f (T b (e::Ord a) (y::a) (g::a->Int)) True = ...expr2...
+
+After desugaring etc we'll get a single case:
+
+ f = \t::T b::Bool ->
+ case t of
+ T a (d::Ord a) (x::a) (f::a->Int)) ->
+ case b of
+ True -> ...expr1...
+ False -> ...expr2...
+
+*** We have to substitute [a/b, d/e] in expr2! **
+That is what do_subst is doing.
+
+Originally I tried to use
+ (\b -> let e = d in expr2) a
+to do this substitution. While this is "correct" in a way, it fails
+Lint, because e::Ord b but d::Ord a.
+
+So now I simply do the substitution properly using substExpr.
+