+We've already established that the behavior of combinatory terms can be perfectly mimicked by lambda terms: just replace each combinator with its equivalent lambda term, i.e., replace I with `\x.x`, replace K with `\fxy.x`, and replace S with `\fgx.fx(gx)`. How about the other direction? Here is a method for converting an arbitrary lambda term into an equivalent Combinatory Logic term using only S, K, and I. Besides the intrinsic beauty of this mapping, and the importance of what it says about the nature of binding and computation, it is possible to hear an echo of computing with continuations in this conversion strategy (though you wouldn't be able to hear these echos until we've covered a considerable portion of the rest of the course).
+
+Assume that for any lambda term T, [T] is the equivalent combinatory logic term. The we can define the [.] mapping as follows:
+
+ 1. [a] a
+ 2. [(M N)] ([M][N])
+ 3. [\a.a] I
+ 4. [\a.M] KM assumption: a does not occur free in M
+ 5. [\a.(M N)] S[\a.M][\a.N]
+ 6. [\a\b.M] [\a[\b.M]]
+
+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.
+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, `Ky` 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 not have any lambdas in it, we can be sure that we won't end up applying rule 6 again in an infinite loop.)
+
+[Fussy notes: if the original lambda term has free variables in it, so will the combinatory logic translation. Feel free to worry about this, though you should be confident that it makes sense. You should also convince 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). One other detail: this translation algorithm builds expressions that combine lambdas with 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 mixes combinators 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.]
+
+[Various, slightly differing translation schemes from combinatorial logic to the lambda calculus are also possible. These generate different metatheoretical correspondences between the two calculii. Consult Hindley and Seldin for details. Also, note that the combinatorial proof theory needs to be strengthened with axioms beyond anything we've here described in order to make [M] convertible with [N] whenever the original lambda-terms M and N are convertible.]
+
+
+Let's check that the translation of the false boolean behaves as expected by feeding it two arbitrary arguments:
+
+ KIXY ~~> IY ~~> Y
+
+Throws away the first argument, returns the second argument---yep, it works.
+
+Here's a more elaborate example of the translation. The goal is to establish that combinators can reverse order, so we use the **T** combinator, where <code>T ≡ \x\y.yx</code>:
+
+ [\x\y.yx] = [\x[\y.yx]] = [\x.S[\y.y][\y.x]] = [\x.(SI)(Kx)] = S[\x.SI][\x.Kx] = S(K(SI))(S[\x.K][\x.x]) = S(K(SI))(S(KK)I)
+
+We can test this translation by seeing if it behaves like the original lambda term does.
+The orginal lambda term lifts its first argument (think of it as reversing the order of its two arguments):
+
+ S(K(SI))(S(KK)I) X Y ~~>
+ (K(SI))X ((S(KK)I) X) Y ~~>
+ SI ((KK)X (IX)) Y ~~>
+ SI (KX) Y ~~>
+ IY (KXY) ~~>
+ Y X
+
+Voilà: the combinator takes any X and Y as arguments, and returns Y applied to X.