-Combinatory Logic like manipulations.
-
-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] K[M] 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 combinatory 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
+Combinatory Logic-like manipulations.
+
+[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!]
+
+In order to establish the correspondence, we need to get a bit more
+official about what counts as an expression in CL. We'll endow CL
+with an infinite stock of variable symbols, just like the lambda
+calculus, including `x`, `y`, and `z`. In addition, `S`, `K`, and `I`
+are expressions in CL. Finally, `(XY)` is in CL for any CL
+expressions `X` and `Y`. So examples of CL expressions include
+`x`, `(xy)`, `Sx`, `SK`, `(x(SK))`, `(K(IS))`, and so on. When we
+omit parentheses, the assumption will be left associativity, so that
+`XYZ == ((XY)Z)`.
+
+It may seem wierd to allow variables in CL. The reason that is
+necessary is because we're trying to show that every lambda term can
+be translated into an equivalent CL term. Since some lambda terms
+contain free variables, we need to provide a translation for free
+variables. As you might expect, it will turn out that whenever the
+lambda term in question contains no free variables (i.e., is a
+combinator), its translation in CL will also contain no variables.
+
+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. [\aX] @a[X]
+ 3. [(XY)] ([X][Y])
+
+ 4. @aa I
+ 5. @aX KX if a is not in X
+ 6. @a(XY) S(@aX)(@aY)
+
+Think of `@aX` as a psuedo-lambda abstract.
+
+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 transtlation 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 (6) 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`.
+
+The easiest way to grasp rule (6) is to consider the following claim:
+
+ \a(XY) <~~> S(\aX)(\aY)
+
+To prove it to yourself, just substitute `(\xyz.xz(yz))` in for `S`
+and reduce.
+
+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).
+
+Various, slightly differing translation schemes from Combinatory Logic to the
+Lambda Calculus are also possible. These generate different metatheoretical
+correspondences between the two calculi. Consult Hindley and Seldin for
+details.
+
+Also, note that the combinatorial proof theory needs to be