+Well! That feels right. The meaning of *This sentence meaning is true*
+could be `Ω`, our prototypical infinite loop...
+
+### What about the liar? ###
+
+Let's consider:
+
+(2) This sentence meaning is false.
+
+Used in a context in which *this sentence meaning* refers to the meaning expressed by the utterance of
+(2) in which that noun phrase occurs, (2) will denote a fixed point for `\m. neg m`,
+or `\m y n. m n y`, which is the `C` combinator. So in such a
+context, (2) might denote
+
+ Y C
+ (\h. (\u. h (u u)) (\u. h (u u))) C
+ (\u. C (u u)) (\u. C (u u)))
+ C ((\u. C (u u)) (\u. C (u u)))
+ C (C ((\u. C (u u)) (\u. C (u u))))
+ C (C (C ((\u. C (u u)) (\u. C (u u)))))
+ ...
+
+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.
+
+### However... ###
+
+You should be cautious about feeling too comfortable with
+these results. Thinking again of the truth-teller paradox, yes,
+`Ω` is *a* fixed point for `I`, and perhaps it has
+some privileged status among all the fixed points for `I`, being the
+one delivered by `Y` and all (though it is not obvious why `Y` should have
+any special status, versus other fixed point combinators).
+
+But one could observe: look, literally every formula is a fixed point for
+`I`, since
+
+ X <~~> I X
+
+for any choice of `X` whatsoever.
+
+So the `Y` combinator is only guaranteed to give us one fixed point out
+of infinitely many --- and not always the intuitively most useful
+one. (For instance, the squaring function `\x. mul x x` has `0` as a fixed point,
+since `square 0 <~~> 0`, and `1` as a fixed point, since `square 1 <~~> 1`, but `Y
+(\x. mul x x)` doesn't give us `0` or `1`.) So with respect to the
+truth-teller paradox, why in the reasoning we've
+just gone through should we be reaching for just this fixed point at
+just this juncture?
+
+One obstacle to thinking this through is the fact that a sentence
+normally has only two truth values. We might consider instead a noun
+phrase such as
+
+(3) the entity that this noun phrase refers to
+
+The reference of (3) depends on the reference of the embedded noun
+phrase *this noun phrase*. As with (1), it will again need to be some fixed
+point of `I`. It's easy to see that any object is a
+fixed point for this referential function: if this pen cap is the
+referent of the demonstrated noun phrase, then it is the referent of (3), and so on
+for any object.
+
+<!--
+The chameleon nature of (3), by the way (a description that is equally
+good at describing any object), makes it particularly well suited as a
+gloss on pronouns such as *it*. In the system of
+[Jacobson 1999](http://www.springerlink.com/content/j706674r4w217jj5/),
+pronouns denote (you guessed it!) identity functions...
+
+TODO: We haven't made clear how we got from the self-referential (3) to I. We've so far only motivated
+that the meaning of (3) should be *a fixed point* of I, but now you are saying this suggests Jacobson's idea to let it be I itself. Why? Sure that is *a* fixed point for I, but so is everything.
+-->
+
+
+Ultimately, in the context of this course, these paradoxes are more
+useful as a way of gaining leverage on the concepts of fixed points
+and recursion, rather than the other way around.
+
+
+## Q: How do you know that every term in the untyped lambda calculus has a fixed point? ##
+
+A: That's easy: let `N` be an arbitrary term in the lambda calculus. If
+`N` has a fixed point, then there exists some `ξ` such that `ξ <~~>
+N ξ` (that's what it means to *have* a fixed point).
+
+<!-- L is \h u. h (u u); H here is L N -->
+
+ let H = \u. N (u u) in
+ let ξ = H H in
+ ξ ≡ H H ≡ (\u. N (u u)) H ~~> N (H H) ≡ N ξ