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:
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. 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.)
+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 `\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.]
+[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.]
-Here's an example of the translation:
+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 `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)