lists-monad tweaks

Signed-off-by: Jim Pryor <profjim@jimpryor.net>

diff --git a/list_monad_as_continuation_monad.mdwn b/list_monad_as_continuation_monad.mdwn
index 0edec25..200b4b0 100644 (file)
--- a/list_monad_as_continuation_monad.mdwn
+++ b/list_monad_as_continuation_monad.mdwn
@@ -88,7 +88,7 @@ need to posit a store `s` that we can apply `u` to.  Once we do so,
 however, we won't have an `'a`; we'll have a pair whose first element
 is an `'a`.  So we have to unpack the pair:
 
-       ... let (a, s') = u s in ... (f a) ...
+       ... let (a, s') = u s in ... f a ...
 
 Abstracting over the `s` and adjusting the types gives the result:
 
@@ -222,7 +222,7 @@ fold that function over its type `'a` members, and that's where we can get at th
 
        ... u (fun (a : 'a) (b : 'b) -> ... f a ... ) ...
 
-In order for `u` to have the kind of argument it needs, the `fun a b -> ... f a ...` has to have type `'a -> 'b -> 'b`; so the `... f a ...` has to evaluate to a result of type `'b`. The easiest way to do this is to collapse (or "unify") the types `'b` and `'d`, with the result that `f a` will have type `('c -> 'b -> 'b) -> 'b -> 'b`. Let's postulate an argument `k` of type `('c -> 'b -> 'b)` and supply it to `(f a)`:
+In order for `u` to have the kind of argument it needs, the `fun a b -> ... f a ...` has to have type `'a -> 'b -> 'b`; so the `... f a ...` has to evaluate to a result of type `'b`. The easiest way to do this is to collapse (or "unify") the types `'b` and `'d`, with the result that `f a` will have type `('c -> 'b -> 'b) -> 'b -> 'b`. Let's postulate an argument `k` of type `('c -> 'b -> 'b)` and supply it to `f a`:
 
        ... u (fun (a : 'a) (b : 'b) -> ... f a k ... ) ...