For instance, Szabolcsi argues that reflexive pronouns are argument
duplicators.
+
-
+Notice that the semantic value of *himself* is exactly W.
+The reflexive pronoun in direct object position combines first with the transitive verb (through compositional magic we won't go into here). The result is an intransitive verb phrase that takes a subject argument, duplicates that argument, and feeds the two copies to the transitive verb meaning.
+
+Note that W = S(CI):
+
+ S(CI) =
+ S((\fxy.fyx)(\x.x)) =
+ S(\xy.(\x.x)yx) =
+ (\fgx.fx(gx))(\xy.yx) =
+ \gx.[\xy.yx]x(gx) =
+ \gx.(gx)x =
+ W
+
+Ok, here comes a shift in thinking. Instead of defining combinators as equivalent to certain lambda terms,
+we can define combinators by what they do. If we have the I combinator followed by any expression X,
+I will take that expression as its argument and return that same expression as the result. In pictures,
+
+ IX ~~> X
+
+Thinking of this as a reduction rule, we can perform the following computation
+
+ II(IX) ~~> IIX ~~> IX ~~> X
+
+The reduction rule for K is also straightforward:
+
+ KXY ~~> X
+
+That is, K throws away its second argument. The reduction rule for S can be constructed by examining
+the defining lambda term:
+
+ 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:
+
+ 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.
+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 S's and K's:
+
+ SKKX ~~> KX(KX) ~~> X
+
+So the combinator SKK is equivalent to the combinator I.
+
+Combinatory Logic is what you have when you choose a set of combinators and regulate their behavior with a set of reduction rules. 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 takes to be Turing complete, imagine what a text editor does:
+it transforms 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 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)`. 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. 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 would be able to hear these echos until we've covered a considerable portion of the rest of the course).
+
+Assume that for any lambda term T, [T] is the equivalent combinatory logic term. The we can define the [.] mapping as follows:
+
+ lambda term equivalent SKI term condition
+ ----------- ------------------- ---------
+ 1. [\x.x] I
+ 2. [\x.M] K[M] x does not occur free in M
+ 3. [\x.(M N)] S[\x.M][\x.N]
+ 4. [\x\y.M] [\x[\y.M]]
+ 5. [M N] [M][N]
+
+It's easy to understand these rules based on what S, K and I do. The first rule is obvious.
+The second rule says that if a lambda abstract contains no occurrences of the variable targeted by lambda,
+what the function expressed by that lambda term does it throw away its argument and returns whatever M
+computes: it's the constant function K[M].
+The third and fourth rules say what happens when there are occurrences of the bound variable in the body.
+
+Finally, the fifth rule says what to do for an application (divide and conquer).
These systems 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!