What is a fixed point of the identity combinator I?
+What is a fixed point of the false combinator, KI?
+
It's a theorem of the lambda calculus that every formula has a fixed point. In fact, it will have infinitely many, non-equivalent fixed points. And we don't just know that they exist: for any given formula, we can name many of them.
Yes, even the formula that you're using the define the successor function will have a fixed point. Isn't that weird? Think about how it might be true.
L L L L L L L L L L L L L L L L L L L L L L L L L L
+##Watching Y in action##
+
+For those of you who like to watch ultra slow-mo movies of bullets
+piercing apples, here's a stepwise computation of the application of a
+recursive function. We'll use a function `sink`, which takes one
+argument. If the argument is boolean true (i.e., `\x y.x`), it
+returns itself (a copy of `sink`); if the argument is boolean false
+(`\x y. y`), it returns `I`. That is, we want the following behavior:
+
+ sink false ~~> I
+ sink true false ~~> I
+ sink true true false ~~> I
+ sink true true true false ~~> I
+
+So we make `sink = Y (\f b. b f I)`:
+
+ 1. sink false
+ 2. Y (\fb.bfI) false
+ 3. (\f. (\h. f (h h)) (\h. f (h h))) (\fb.bfI) false
+ 4. (\h. [\fb.bfI] (h h)) (\h. [\fb.bfI] (h h)) false
+ 5. [\fb.bfI] ((\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))) false
+ 6. (\b.b[(\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))]I) false
+ 7. false [(\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))] I
+ --------------------------------------------
+ 8. I
+
+So far so good. The crucial thing to note is that as long as we
+always reduce the outermost redex first, we never have to get around
+to computing the underlined redex: because `false` ignores its first
+argument, we can throw it away unreduced.
+
+Now we try the next most complex example:
+
+ 1. sink true false
+ 2. Y (\fb.bfI) true false
+ 3. (\f. (\h. f (h h)) (\h. f (h h))) (\fb.bfI) true false
+ 4. (\h. [\fb.bfI] (h h)) (\h. [\fb.bfI] (h h)) true false
+ 5. [\fb.bfI] ((\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))) true false
+ 6. (\b.b[(\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))]I) true false
+ 7. true [(\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))] I false
+ 8. [(\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))] false
+
+We've now arrived at line (4) of the first computation, so the result
+is again I.
+
+You should be able to see that `sink` will consume as many `true`s as
+we throw at it, then turn into the identity function after it
+encounters the first `false`.
+
+The key to the recursion is that, thanks to Y, the definition of
+`sink` contains within it the ability to fully regenerate itself as
+many times as is necessary. The key to *ending* the recursion is that
+the behavior of `sink` is sensitive to the nature of the input: if the
+input is the magic function `false`, the self-regeneration machinery
+will be discarded, and the recursion will stop.
+
+That's about as simple as recursion gets.
+
+##Base cases, and their lack##
+
+As any functional programmer quickly learns, writing a recursive
+function divides into two tasks: figuring out how to handle the
+recursive case, and remembering to insert a base case. The
+interesting and enjoyable part is figuring out the recursive pattern,
+but the base case cannot be ignored, since leaving out the base case
+creates a program that runs forever. For instance, consider computing
+a factorial: `n!` is `n * (n-1) * (n-2) * ... * 1`. The recursive
+case says that the factorial of a number `n` is `n` times the
+factorial of `n-1`. But if we leave out the base case, we get
+
+ 3! = 3 * 2! = 3 * 2 * 1! = 3 * 2 * 1 * 0! = 3 * 2 * 1 * 0 * -1! ...
+
+That's why it's crucial to declare that 0! = 1, in which case the
+recursive rule does not apply. In our terms,
+
+ fac = Y (\fac n. iszero n 1 (fac (predecessor n)))
+
+If `n` is 0, `fac` reduces to 1, without computing the recursive case.
+
+There is a well-known problem in philosophy and natural language
+semantics that has the flavor of a recursive function without a base
+case: the truth-teller paradox (and related paradoxes).
+
+(1) This sentence is true.
+
+If we assume that the complex demonstrative "this sentence" can refer
+to (1), then the proposition expressed by (1) will be true just in
+case the thing referred to by *this sentence* is true. Thus (1) will
+be true just in case (1) is true, and (1) is true just in case (1) is
+true, and so on. If (1) is true, then (1) is true; but if (1) is not
+true, then (1) is not true.
+
+Without pretending to give a serious analysis of the paradox, let's
+assume that sentences can have for their meaning boolean functions
+like the ones we have been working with here. Then the sentence *John
+is John* might denote the function `\x y. x`, our `true`.
+
+Then (1) denotes a function from whatever the referent of *this
+sentence* is to a boolean. So (1) denotes `\f. f true false`, where
+the argument `f` is the referent of *this sentence*. Of course, if
+`f` is a boolean, `f true false <~~> f`, so for our purposes, we can
+assume that (1) denotes the identity function `I`.
+
+If we use (1) in a context in which *this sentence* refers to the
+sentence in which the demonstrative occurs, then we must find a
+meaning `m` such that `I m = I`. But since in this context `m` is the
+same as the meaning `I`, so we have `m = I m`. In other words, `m` is
+a fixed point for the denotation of the sentence (when used in the
+appropriate context).
+
+That means that in a context in which *this sentence* refers to the
+sentence in which it occurs, the sentence denotes a fixed point for
+the identity function. Here's a fixed point for the identity
+function:
+
+<pre>
+ Y I
+ (\f. (\h. f (h h)) (\h. f (h h))) I
+ (\h. I (h h)) (\h. I (h h)))
+ (\h. (h h)) (\h. (h h)))
+ ω ω
+ &Omega
+</pre>
+
+Oh. Well! That feels right. The meaning of *This sentence is true*
+in a context in which *this sentence* refers to the sentence in which
+it occurs is Ω, our prototypical infinite loop...
+
+What about the liar paradox?
+
+(2) This sentence is false.
+
+Used in a context in which *this sentence* refers to the utterance of
+(2) in which it occurs, (2) will denote a fixed point for `\f.neg f`,
+or `\f l r. f r l`, which is the `C` combinator. So in such a
+context, (2) might denote
+
+ Y C
+ (\f. (\h. f (h h)) (\h. f (h h))) I
+ (\h. C (h h)) (\h. C (h h)))
+ C ((\h. C (h h)) (\h. C (h h)))
+ C (C ((\h. C (h h))(\h. C (h h))))
+ C (C (C ((\h. C (h h))(\h. C (h h)))))
+ ...
+
+And infinite sequence of `C`s, each one negating the remainder of the
+sequence. Yep, that feels like a reasonable representation of the
+liar paradox.
+
+See Barwise and Etchemendy's 1987 OUP book, [The Liar: an essay on
+truth and circularity](http://tinyurl.com/2db62bk) for an approach
+that is similar, but expressed in terms of non-well-founded sets
+rather than recursive functions.