equivalent to `(\z. z y) z` is that when a lambda binds a set of
occurrences, it doesn't matter which variable serves to carry out the
binding. Either way, the function does the same thing and means the
-same thing. Look in the standard treatments for discussions of alpha
-equivalence for more detail.]
+same thing.
+Linguistic trivia: some linguistic discussions suppose that alphabetic variance
+has important linguistic consequences (notably Ivan Sag's dissertation).
+Look in the standard treatments for discussions of alpha
+equivalence for more detail. Also, as mentioned below, one of the intriguing
+properties of Combinatory Logic is that alpha equivalence is not an issue.]
This:
[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 `\x.\y.y` is `[\x[\y.y]] = [\x.I] = KI`. In that intermediate stage, we have `\x.I`. It's possible to avoid this, but it takes some careful thought. See, e.g., Barendregt 1984, page 156.]
-Here's an example of the translation:
+Here's a simple example of the translation. We already have a combinator for our true boolean (K is true: it returns its first argument and discards its second argument). What about false?
+
+ [\x\y.y] = [\x[\y.y]] = [\xI] = KI
+
+We can test this translation by feeding it two arbitrary arguments:
+
+ KIXY ~~> IY ~~> Y
+
+Yep, it works.
+
+Here's a more elaborat example of the translation. The goal is to establish that combinators can reverse order, so we use the T combinator, where `T = \x\y.yx`:
[\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)