+ 4. @aa = I
+ 5. @aX = KX if a is not in X
+ 6. @a(Xa) = X if a is not in X
+ 7. @a(XY) = S(@aX)(@aY)
+
+Think of `@aX` as a pseudo-lambda abstract. (Hankin and Barendregt write it as <code>λ*a. X</code>; Hindley & Seldin write it as `[a] X`.) It is possible to omit line 6, and some presentations do, but Hindley & Seldin observe that this "enormously increases" the length of "most" translations.
+
+It's easy to understand these rules based on what `S`, `K` and `I` do.
+
+Rule (1) 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.
+
+Rule (2) says that the way to translate an application is to
+first translate the body (i.e., `[X]`), and then prefix a kind of
+temporary psuedo-lambda built from `@` and the original variable.
+
+Rule (3) says that the translation of an application of `X` to `Y` is
+the application of the translation of `X` to the translation of `Y`.
+
+As we'll see, the first three rules sweep through the lambda term,
+changing each lambda to an @.
+
+Rules (4) through (7) tell us how to eliminate all the `@`'s.
+
+In rule (4), if we have `@aa`, we need a CL expression that behaves
+like the lambda term `\aa`. Obviously, `I` is the right choice here.
+
+In rule (5), if we're binding into an expression that doesn't contain
+any variables that need binding, then we need a CL term that behaves
+the same as `\aX` would if `X` didn't contain `a` as a free variable.
+Well, how does `\aX` behave? When `\aX` occurs in the head position
+of a redex, then no matter what argument it occurs with, it throws
+away its argument and returns `X`. In other words, `\aX` is a
+constant function returning `X`, which is exactly the behavior
+we get by prefixing `K`.
+
+Rule (6) should be intuitive; and as we said, we could in principle omit it and just handle such cases under the final rule.
+
+The easiest way to grasp rule (7) is to consider the following claim:
+
+ \a(XY) <~~> S(\aX)(\aY)
+
+To prove it to yourself, just consider what would happen when each term is applied to an argument `a`. Or substitute `\x y a. x a (y a)` in for `S`
+and reduce.
+
+Persuade yourself that if the original lambda term contains no free
+variables --- i.e., is a Lambda Calculus combinator --- then the translation will
+consist only of `S`, `K`, and `I` (plus parentheses).
+
+Various, slightly differing translation schemes from Combinatory Logic to the
+Lambda Calculus are also possible. These generate different meta-theoretical
+correspondences between the two calculi. Consult Hindley & Seldin for
+details.
+
+Also, note that the combinatorial proof theory needs to be