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diff --git a/exercises/_assignment7.mdwn b/exercises/_assignment7.mdwn
index d351b1b5..793a15e4 100644
--- a/exercises/_assignment7.mdwn
+++ b/exercises/_assignment7.mdwn
@@ -1,50 +1,18 @@
-# Assignment 6 (week 7)
+## Baby monads
 
-## Evaluation order in Combinatory Logic
+(Depends on lecture notes for safe division by zero.)
 
-1. Give a term that the lazy evaluators (either the Haskell evaluator,
-or the lazy version of the OCaml evaluator) do not evaluate all the
-way to a normal form, i.e., that contains a redex somewhere inside of
-it after it has been reduced.
+Write a function `lift'` that generalized the correspondence between +
+and `add'`: that is, `lift'` takes any two-place operation on integers
+and returns a version that takes arguments of type `int option`
+instead, returning a result of `int option`.  In other words, `lift'`
+will have type:
 
-<!-- reduce3 (FA (K, FA (I, I))) -->
+	(int -> int -> int) -> (int option) -> (int option) -> (int option)
 
+so that `lift' (+) (Some 3) (Some 4)` will evalute to `Some 7`.
+Don't worry about why you need to put `+` inside of parentheses.
+You should make use of `bind'` in your definition of `lift'`:
 
-2. One of the 
-[[criteria we established for classifying reduction strategies|
-topics/week3_evaluation_order]]
-strategies is whether they reduce subexpressions hidden under lambdas.
-That is, for a term like `(\x y. x z) (\x. x)`, do we reduce to
-`\y.(\x.x) z` and stop, or do we reduce further to `\y.z`?  Explain
-what the corresponding question would be for CL. Using either the
-OCaml CL evaluator or the Haskell evaluator developed in the wiki
-notes, prove that the evaluator does reduce expressions inside of
-"functional" CL expressions.  Then provide a modified evaluator that
-does not perform reductions in those positions.
-
-<!-- just add early no-op cases for Ka and Sab -->
-
-3. In the previous homework, one of the techniques for controlling
-evaluation order was wrapping expressions in a `let`: `let x = blah in
-foo`, you could be sure that `blah` would be evaluated by the time the
-interpreter considered `foo` (unless you did some fancy footwork with
-thunks).  That suggests the following way to try to arrive at eager
-evaluation in our Haskell evaluator for CL:
-
-    reduce4 t = case t of
-      I -> I
-      K -> K
-      S -> S
-      FA a b -> 
-        let b' = reduce4 b in
-        let a' = reduce4 a in
-        let t' = FA a' b' in
-          if (is_redex t') then reduce4 (reduce_one_step t')
-                           else t'                                
-
-Will this work?  That is, will `reduce4 (FA (FA K I) skomega)` go into
-an infinite loop?  Run the code to find out, if you must, but write
-down your guess (and your rationale) first.
-
-<!-- Doesn't work: infinite loop. -->
-
+	let bind' (u: int option) (f: int -> (int option)) =
+		match u with None -> None | Some x -> f x;;