`S ≡ \fgx.fx(gx)`

-S takes three arguments, duplicates the third argument, and feeds one copy to the first argument and the second copy to the second argument. So:
+`S` takes three arguments, duplicates the third argument, and feeds one copy to the first argument and the second copy to the second argument. So:
SFGX ~~> FX(GX)
If the meaning of a function is nothing more than how it behaves with respect to its arguments,
-these reduction rules capture the behavior of the combinators S, K, and I completely.
+these reduction rules capture the behavior of the combinators `S`, `K`, and `I` completely.
We can use these rules to compute without resorting to beta reduction.
-For instance, we can show how the I combinator is equivalent to a
-certain crafty combination of Ss and Ks:
+For instance, we can show how the `I` combinator's behavior is simulated by a
+certain crafty combination of `S`s and `K`s:
SKKX ~~> KX(KX) ~~> X
-So the combinator `SKK` is equivalent to the combinator I.
+So the combinator `SKK` is equivalent to the combinator `I`. (Really, it could be `SKy` for any `y`.)
These reduction rule have the same status with respect to Combinatory
Logic as beta reduction and eta reduction, etc., have with respect to
@@ -138,30 +138,30 @@ Logic are considerably more simple than, say, beta reduction. Also, since
there are no variables in Combiantory Logic, there is no need to worry
about variable collision.
-Combinatory Logic is what you have when you choose a set of combinators and regulate their behavior with a set of reduction rules. As we said, the most common system uses S, K, and I as defined here.
+Combinatory Logic is what you have when you choose a set of combinators and regulate their behavior with a set of reduction rules. As we said, the most common system uses `S`, `K`, and `I` as defined here.
###The equivalence of the untyped lambda calculus and combinatory logic###
We've claimed that Combinatory Logic is equivalent to the lambda calculus. If
-that's so, then S, K, and I must be enough to accomplish any computational task
-imaginable. Actually, S and K must suffice, since we've just seen that we can
-simulate I using only S and K. In order to get an intuition about what it
+that's so, then `S`, `K`, and `I` must be enough to accomplish any computational task
+imaginable. Actually, `S` and `K` must suffice, since we've just seen that we can
+simulate `I` using only `S` and `K`. In order to get an intuition about what it
takes to be Turing complete, recall our discussion of the lambda calculus in
terms of a text editor. A text editor has the power to transform any arbitrary
text into any other arbitrary text.
The way it does this is by deleting, copying, and reordering characters. We've
-already seen that K deletes its second argument, so we have deletion covered.
-S duplicates and reorders, so we have some reason to hope that S and K are
+already seen that `K` deletes its second argument, so we have deletion covered.
+`S` duplicates and reorders, so we have some reason to hope that `S` and `K` are
enough to define arbitrary functions.
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)`. So the behavior of any combination of
+equivalent lambda term, i.e., replace `I` with `\x.x`, replace `K` with `\fxy.x`,
+and replace `S` with `\fgx.fx(gx)`. So the behavior of any combination of
combinators in Combinatory Logic can be exactly reproduced by a lambda term.
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.
+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
@@ -181,15 +181,15 @@ Assume that for any lambda term T, [T] is the equivalent combinatory logic term.
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
+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.)
+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.]
+[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
@@ -264,13 +264,13 @@ in two books in the 1990's.
A final linguistic application: Steedman's Combinatory Categorial Grammar, where the "Combinatory" is
from combinatory logic (see especially his 2012 book, Taking Scope). Steedman attempts to build
a syntax/semantics interface using a small number of combinators, including T ≡ `\xy.yx`, B ≡ `\fxy.f(xy)`,
-and our friend S. Steedman used Smullyan's fanciful bird
+and our friend `S`. Steedman used Smullyan's fanciful bird
names for the combinators, Thrush, Bluebird, and Starling.
Many of these combinatory logics, in particular, the SKI system,
are Turing complete. In other words: every computation we know how to describe can be represented in a logical system consisting of only a single primitive operation!
-The combinators K and S correspond to two well-known axioms of sentential logic:
+The combinators `K` and `S` correspond to two well-known axioms of sentential logic:
###A connection between Combinatory Logic and Sentential Logic###
@@ -285,7 +285,7 @@ are complete for the implicational fragment of intuitionistic logic.
(To get a complete proof theory for *classical* sentential logic, you
need only add one more axiom, constraining the behavior of a new connective "not".)
The way we'll favor for viewing the relationship between these axioms
-and the S and K combinators is that the axioms correspond to type
+and the `S` and `K` combinators is that the axioms correspond to type
schemas for the combinators. This will become more clear once we have
a theory of types in view.