-[WARNING: the mapping from the lambda calculus to Combinatory Logic
-has been changed since the class in which it was presented. It now
-matches the presentation in Barendregt. The revised version is
-cleaner, and more elegant. If you spent a lot of time working to
-understand the original version, there's good news and bad news. The
-bad news is that things have changed. The good news is that the new
-version described the same mapping as before, but does it in a cleaner
-way. That is, the CL term that a given lambda term maps onto hasn't
-changed, only the details of how that CL term gets computed. Sorry if
-the changeup causes any distress!]
+<!--
+Assume that for any lambda term T, [T] is the equivalent Combinatory Logic term. Then we can define the [.] mapping as follows:
+
+ 1. [a] a
+ 2. [(M N)] ([M][N])
+ 3. [\a.a] I
+ 4. [\a.M] K[M] when a does not occur free in M
+ 5. [\a.(M N)] S[\a.M][\a.N]
+ 6. [\a\b.M] [\a[\b.M]]
+
+If the recursive unpacking of these rules ever direct you to "translate" an `S` or a `K` or an `I`, introduced at an earlier stage of translation, those symbols translate themselves.
+
+It's easy to understand these rules based on what `S`, `K` and `I` do.
+
+The first rule says that variables are mapped to themselves. If the original lambda expression had no free variables in it, then any such translations will only be temporary. The variable will later get eliminated by the application of other rules. (If the original lambda term *does* have free variables in it, so too will the final Combinatory Logic translation. Feel free to worry about this, though you should be confident that it makes sense.)
+
+The second rule says that the way to translate an application is to translate the first element and the second element separately.
+
+The third rule should be obvious.
+
+The fourth rule should also be fairly self-evident: since what a lambda term such as `\x. y` does it throw away its first argument and return `y`, that's exactly what the Combinatory Logic translation should do. And indeed, `K y` is a function that throws away its argument and returns `y`.
+
+The fifth rule deals with an abstract whose body is an application: the `S` combinator takes its next argument (which will fill the role of the original variable a) and copies it, feeding one copy to the translation of `\a. M`, and the other copy to the translation of `\a. N`. This ensures that any free occurrences of a inside `M` or `N` will end up taking on the appropriate value.
+
+Finally, the last rule says that if the body of an abstract is itself an abstract, translate the inner abstract first, and then do the outermost. (Since the translation of `[\b. M]` will have eliminated any inner lambdas, we can be sure that we won't end up applying rule 6 again in an infinite loop.)
+
+Persuade yourself that if the original lambda term contains no free variables --- i.e., is a combinator --- then the translation will consist only of `S`, `K`, and `I` (plus parentheses).
+
+(Fussy note: this translation algorithm builds intermediate expressions that combine lambdas with primitive combinators. For instance, the translation of our boolean `false` (`\x y. y`) is `[\x [\y. y]] = [\x. I] = KI`. In the intermediate stage, we have `\x. I`, which has a combinator in the body of a lambda abstract. It's possible to avoid this if you want to, but it takes some careful thought. See, e.g., Barendregt 1984, page 156.)
+
+...
+Here's a more elaborate example of the translation. Let's say we want to establish that combinators can reverse order, so we use the **T** combinator (`\x y. y x`):
+
+ [\x y. y x] =
+ [\x [\y. y x]] =
+ [\x. S [\y. y] [\y. x]] =
+ [\x. (SI) (K x)] =
+ S [\x. SI] [\x. K x] =
+ S (K(SI)) (S [\x. K] [\x. x]) =
+ S (K(SI)) (S(KK)I)
+-->
+
+(*Warning* This is a different mapping from the Lambda Calculus to Combinatory Logic than we presented in class (and was posted here earlier). It now matches the presentation in Barendregt 1984, and in Hankin Chapter 4 (esp. pp. 61, 65) and in Hindley & Seldin Chapter 2 (esp. p. 26). In some ways this translation is cleaner and more elegant, which is why we're presenting it.)