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diff --git a/exercises/assignment5_answers.mdwn b/exercises/assignment5_answers.mdwn
index 5425ac51..9618cd3c 100644
--- a/exercises/assignment5_answers.mdwn
+++ b/exercises/assignment5_answers.mdwn
@@ -557,36 +557,44 @@ any type `α`, as long as your function is of type `α -> α` and you have a bas
 
     OCAML ANSWERS:
 
-        type ('a) sysf_bool = 'a -> 'a -> 'a
-        let sysf_true : ('a) sysf_bool = fun y n -> y
-        let sysf_false : ('a) sysf_bool = fun y n -> n
+        type 'a sysf_bool = 'a -> 'a -> 'a
+        let sysf_true : 'a sysf_bool = fun y n -> y
+        let sysf_false : 'a sysf_bool = fun y n -> n
 
-        type ('a) sysf_nat = ('a -> 'a) -> 'a -> 'a
-        let sysf_zero : ('a) sysf_nat = fun s z -> z
+        type 'a sysf_nat = ('a -> 'a) -> 'a -> 'a
+        let sysf_zero : 'a sysf_nat = fun s z -> z
 
-        let sysf_iszero (n : 'a sysf_nat) : 'b sysf_bool = n (fun _ -> sysf_false) sysf_true
+        let sysf_iszero (n : ('a sysf_bool) sysf_nat) : 'a sysf_bool = n (fun _ -> sysf_false) sysf_true
         (* Annoyingly, though, if you just say sysf_iszero sysf_zero, you'll get an answer that isn't fully polymorphic.
            This is explained in the comments linked below. The way to get a polymorphic result is to say instead
            `fun next -> sysf_iszero sysf_zero next`. *)
 
         let sysf_succ (n : 'a sysf_nat) : 'a sysf_nat = fun s z -> s (n s z)
-        (* Again, to use this you'll want to call `fun next -> sysf_succ sysf_zero next` *)
-
-        let sysf_pred (n : 'a sysf_nat) : 'a sysf_nat = (* NOT DONE *)
-
-<!--
-        (* Using a System F-style encoding of pairs, rather than native OCaml pairs ... *)
-          let pair a b = fun f -> f a b in
-          let snd x y next = y next in
-          let shift p next = p (fun x y -> pair (sysf_succ x) x) next in (* eta-expanded as in previous definitions *)
-          fun next -> n shift (pair sysf_zero sysf_zero) snd next
-
-    let pair = \a b. \f. f a b in
-    let snd = \x y. y in
-    let shift = \p. p (\x y. pair (succ x) x) in
-    let pred = \n. n shift (pair 0 err) snd in
--->
-
+        (* Again, to get a polymorphic result you'll want to call `fun next -> sysf_succ sysf_zero next` *)
+
+    And here is how to get `sysf_pred`, using a System-F-style encoding of pairs. (For brevity, I'll leave off the `sysf_` prefixes.)
+
+        type 'a natpair = ('a nat -> 'a nat -> 'a nat) -> 'a nat
+        let natpair (x : 'a nat) (y : 'a nat) : 'a natpair = fun f -> f x y
+        let fst x y = x
+        let snd x y = y
+        let shift (p : 'a natpair) : 'a natpair = natpair (succ (p fst)) (p fst)
+        let pred (n : ('a natpair) nat) : 'a nat = n shift (natpair zero zero) snd
+
+        (* As before, to get polymorphic results you need to eta-expand your applications. Witness:
+          # let one = succ zero;;
+          val one : '_a nat = <fun>
+          # let one next = succ zero next;;
+          val one : ('a -> 'a) -> 'a -> 'a = <fun>
+          # let two = succ one;;
+          val two : '_a nat = <fun>
+          # let two next = succ one next;;
+          val two : ('a -> 'a) -> 'a -> 'a = <fun>
+          # pred two;;
+          - : '_a nat = <fun>
+          # fun next -> pred two next;;
+          - : ('a -> 'a) -> 'a -> 'a = <fun>
+        *)
 
 Consider the following list type, specified using OCaml or Haskell datatypes:
 
@@ -778,3 +786,5 @@ and that `bool` is any boolean expression.  Then we can try the following:
         match b with true -> y () | false -> n ()
 
     that would arguably still be relying on the special evaluation order properties of OCaml's native `match`. You'd be assuming that `n ()` wouldn't be evaluated in the computation that ends up selecting the other branch. Your assumption would be correct, but to avoid making that assumption, you should instead first select the `y` or `n` result, _and then afterwards_ force the result. That's what we do in the above answer.
+
+    Question: don't the rhs of all the match clauses in `match b with true -> y | false -> n` have to have the same type? How can they, when one of them is `blackhole` and the other is `blackhole ()`? The answer has two parts. First is that an expression is allowed to have different types when it occurs several times on the same line: consider `let id x = x in (id 5, id true)`, which evaluates just fine. The second is that `blackhole` will get the type `'a -> 'b`, that is, it can have any functional type at all. So the principle types of `y` and `n` end up being `y : unit -> 'a -> 'b` and `n : unit -> 'c`. (The consequent of `n` isn't constrained to use the same type variable as the consequent of `y`.) OCaml can legitimately infer these to be the same type by unifying the types `'c` and `'a -> 'b`; that is, it instantiates `'c` to the functional type had by `y ()`.