(\x. \z. z x) y
+(\x. \y. y x) y
+(\z. (\z. z z) y
+
+ λ ... ___ ...
+ ^ 
+ ______
+
+
+Here there are no bound variables, but there are *bound positions*. We can regard formula like (a) and (b) as just helpfully readable ways to designate these abstract structures.
+
+A version of this last approach is known as **de Bruijn notation** for the lambda calculus.
+
+It doesn't matter which of these approaches one takes; the logical properties of the systems are exactly the same. It just affects the particulars of how one states the rules for substitution, and so on. And whether one talks about expressions being literally "syntactically identical," or whether one instead counts them as "equivalent modulu alphaconversion."
+
+(In a bit, we'll discuss other systems that lack variables. Those systems will not just lack variables in the sense that de Bruijn notation does; they will furthermore lack any notion of a bound position.)
+
+
Syntactic equality, reduction, convertibility
=============================================
@@ 13,8 +63,12 @@ Define T to be `(\x. x y) z`. Then T and `(\x. x y) z` are syntactically equal,
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:
@@ 117,7 +171,7 @@ Combinatory Logic is what you have when you choose a set of combinators and regu
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).
+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 wouldn't 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:
@@ 134,11 +188,17 @@ The second rule says that the way to translate an application is to translate th
first element and the second element separately.
The third rule should be obvious.
The fourth rule should also be fairly selfevident: 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 variablesi.e., is a combinatorthen 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 variablesi.e., is a combinatorthen 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 argumentyep, 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)
@@ 154,6 +214,12 @@ The orginal lambda term lifts its first argument (think of it as reversing the o
Viola: the combinator takes any X and Y as arguments, and returns Y applied to X.
+One very nice property of combinatory logic is that there is no need to worry about alphabetic variance, or
+variable collisionsince there are no (bound) variables, there is no possibility of accidental variable capture,
+and so reduction can be performed without any fear of variable collision. We haven't mentioned the intricacies of
+alpha equivalence or safe variable substitution, but they are in fact quite intricate. (The best way to gain
+an appreciation of that intricacy is to write a program that performs lambda reduction.)
+
Back to linguistic applications: one consequence of the equivalence between the lambda calculus and combinatory
logic is that anything that can be done by binding variables can just as well be done with combinators.
This has given rise to a style of semantic analysis called Variable Free Semantics (in addition to
@@ 255,16 +321,29 @@ This question highlights that there are different choices to make about how eval
With regard to Q3, it should be intuitively clear that `\x. M x` and `M` will behave the same with respect to any arguments they are given. It can also be proven that no other functions can behave differently with respect to them. However, the logical system you get when etareduction is added to the proof theory is importantly different from the one where only betareduction is permitted.
MORE on extensionality
+If we answer Q2 by permitting reduction inside abstracts, and we also permit etareduction, then where none of y_{1}, ..., y_{n} occur free in M, this:
If we answer Q2 by permitting reduction inside abstracts, and we also permit etareduction, then where neither `y` nor `z` occur in M, this:
+\x y_{1}... y_{n}. M y_{1}... y_{n}
 \x y z. M y z

will etareduce by two steps to:
+will etareduce by n steps to:
\x. M
+The logical system you get when etareduction is added to the proof system has the following property:
+
+> if `M`, `N` are normal forms with no free variables, then M ≡ N
iff `M` and `N` behave the same with respect to every possible sequence of arguments.
+
+That is, when `M` and `N` are (closed normal forms that are) syntactically distinct, there will always be some sequences of arguments L_{1}, ..., L_{n}
such that:
+
+M L_{1} ... L_{n} x y ~~> x
+N L_{1} ... L_{n} x y ~~> y
+
+
+That is, closed normal forms that are not just betareduced but also fully etareduced, will be syntactically different iff they yield different values for some arguments. That is, iff their extensions differ.
+
+So the proof theory with etareduction added is called "extensional," because its notion of normal form makes syntactic identity of closed normal forms coincide with extensional equivalence.
+
+
The evaluation strategy which answers Q1 by saying "reduce arguments first" is known as **callbyvalue**. The evaluation strategy which answers Q1 by saying "substitute arguments in unreduced" is known as **callbyname** or **callbyneed** (the difference between these has to do with efficiency, not semantics).
When one has a callbyvalue strategy that also permits reduction to continue inside unapplied abstracts, that's known as "applicative order" reduction. When one has a callbyname strategy that permits reduction inside abstracts, that's known as "normal order" reduction. Consider an expression of the form: