3 ## Application to the truth teller/liar paradoxes ##
5 Curry originally called `Y` the "paradoxical" combinator, and discussed
6 it in connection with certain well-known paradoxes from the philosophy
7 literature. The truth-teller paradox has the flavor of a recursive
8 function without a base case:
10 (1) This sentence meaning is true.
12 If we assume that the complex demonstrative "this sentence meaning" can refer
13 to the very meaning displayed in (1), then that meaning (1) will be true just in
14 case the thing referred to by *this sentence meaning* is true. Thus, (1) will
15 be true just in case (1) is true, and (1) is true just in case (1) is
16 true, and so on. If (1) is true, then (1) is true; but if (1) is not
17 true, then (1) is not true.
19 Without pretending to give a serious analysis of the paradox, let's
20 assume that sentences can have for their meaning boolean values
21 like the ones we have been working with in the Lambda Calculus. Then the sentence *John
22 is John* might have as its meaning our `true`, namely `\y n. y`.
24 Now, the verb phrase in (1) expresses a function from whatever the referent of *this
25 sentence meaning* is to a boolean. That is, `\m. m true false`, where
26 the argument `m` is the referent of *this sentence meaning*. Of course, if
27 `m` is a boolean, `m true false <~~> m`, so for our purposes, we can
28 assume that the verb phrase of (1) denotes the identity function `I`.
30 If we use (1) in a context in which *this sentence meaning* refers to the
31 meaning expressed by the very sentence in which that demonstrative occurs, then we must find a
32 meaning `m` such that it is equivalent to the application of the verb phrase meaning to itself.
33 That is, `m <~~> I m`. In other words, `m` is
34 a fixed point for the meaning of the verb phrase.
36 That means that in a context in which *this sentence meaning* refers to the
37 meaning expressed by the sentence in which it occurs, the sentence's meaning is a fixed point for
38 the identity function. As we observed earlier, *anything* is a fixed point for the identity function.
39 In particular, each of the boolean values `true` and `false` are fixed points for the identity
40 function. What fixed point does `Y` give us?
43 (\h. (\u. h (u u)) (\u. h (u u))) I ~~>
44 (\u. I (u u)) (\u. I (u u))) ~~>
45 (\u. (u u)) (\u. (u u))) ≡
49 Well! That feels right. The meaning of *This sentence meaning is true*
50 could be `Ω`, our prototypical infinite loop...
52 ### What about the liar? ###
56 (2) This sentence meaning is false.
58 Used in a context in which *this sentence meaning* refers to the meaning expressed by the utterance of
59 (2) in which that noun phrase occurs, (2) will denote a fixed point for `\m. neg m`,
60 or `\m y n. m n y`, which is the `C` combinator. So in such a
61 context, (2) might denote
64 (\h. (\u. h (u u)) (\u. h (u u))) C
65 (\u. C (u u)) (\u. C (u u)))
66 C ((\u. C (u u)) (\u. C (u u)))
67 C (C ((\u. C (u u)) (\u. C (u u))))
68 C (C (C ((\u. C (u u)) (\u. C (u u)))))
71 And infinite sequence of `C`s, each one negating the remainder of the
72 sequence. Yep, that feels like a reasonable representation of the
75 See Barwise and Etchemendy's 1987 OUP book, [The Liar: an essay on
76 truth and circularity](http://tinyurl.com/2db62bk) for an approach
77 that is similar, but expressed in terms of non-well-founded sets
78 rather than recursive functions.
82 You should be cautious about feeling too comfortable with
83 these results. Thinking again of the truth-teller paradox, yes,
84 `Ω` is *a* fixed point for `I`, and perhaps it has
85 some privileged status among all the fixed points for `I`, being the
86 one delivered by `Y` and all (though it is not obvious why `Y` should have
87 any special status, versus other fixed point combinators).
89 But one could observe: look, literally every formula is a fixed point for
94 for any choice of `X` whatsoever.
96 So the `Y` combinator is only guaranteed to give us one fixed point out
97 of infinitely many --- and not always the intuitively most useful
98 one. (For instance, the squaring function `\x. mul x x` has `0` as a fixed point,
99 since `square 0 <~~> 0`, and `1` as a fixed point, since `square 1 <~~> 1`, but `Y
100 (\x. mul x x)` doesn't give us `0` or `1`.) So with respect to the
101 truth-teller paradox, why in the reasoning we've
102 just gone through should we be reaching for just this fixed point at
105 One obstacle to thinking this through is the fact that a sentence
106 normally has only two truth values. We might consider instead a noun
109 (3) the entity that this noun phrase refers to
111 The reference of (3) depends on the reference of the embedded noun
112 phrase *this noun phrase*. As with (1), it will again need to be some fixed
113 point of `I`. It's easy to see that any object is a
114 fixed point for this referential function: if this pen cap is the
115 referent of the demonstrated noun phrase, then it is the referent of (3), and so on
119 The chameleon nature of (3), by the way (a description that is equally
120 good at describing any object), makes it particularly well suited as a
121 gloss on pronouns such as *it*. In the system of
122 [Jacobson 1999](http://www.springerlink.com/content/j706674r4w217jj5/),
123 pronouns denote (you guessed it!) identity functions...
125 TODO: We haven't made clear how we got from the self-referential (3) to I. We've so far only motivated
126 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.
130 Ultimately, in the context of this course, these paradoxes are more
131 useful as a way of gaining leverage on the concepts of fixed points
132 and recursion, rather than the other way around.
135 ## Q: How do you know that every term in the untyped lambda calculus has a fixed point? ##
137 A: That's easy: let `N` be an arbitrary term in the lambda calculus. If
138 `N` has a fixed point, then there exists some `ξ` such that `ξ <~~>
139 N ξ` (that's what it means to *have* a fixed point).
141 <!-- L is \h u. h (u u); H here is L N -->
143 let H = \u. N (u u) in
145 ξ ≡ H H ≡ (\u. N (u u)) H ~~> N (H H) ≡ N ξ
148 Please slow down and make sure that you understand what justified each
149 of the equalities in the last line.
151 ## Q: How do you know that for any term `N`, `Y N` is a fixed point of `N`? ##
153 A: Note that in the proof given in the previous answer, we chose `N`
154 and then set `ξ ≡ H H ≡ (\u. N (u u)) (\u. N (u u))`. If we abstract over
155 `N`, we get the Y combinator, `\N. (\u. N (u u)) (\u. N (u u))`. No matter
156 what argument `N` we feed `Y`, it returns some `ξ` that is a fixed point
157 of `N`, by the reasoning in the previous answer.
160 ## Q: So if every term has a fixed point, even `Y` has fixed point. ##
164 let Y = \N. (\u. N (u u)) (\u. N (u u)) in
166 ≡ \N. (\u. N (u u)) (\u. N (u u)) Y
167 ~~> (\u. Y (u u)) (\u. Y (u u))
168 ~~> Y ((\u. Y (u u)) (\u. Y (u u)))
169 ~~> Y ( Y ((\u. Y (u u)) (\u. Y (u u))))
170 ~~> Y (Y (Y (...(Y (Y Y))...)))
174 ## Q: Ouch! Stop hurting my brain. ##
176 A: Is that a question?
178 Let's come at it from the direction of arithmetic. Recall that we
179 claimed that even `succ`---the function that added one to any
180 number---had a fixed point. How could there be an `ξ` such that `ξ <~~> succ ξ`?
181 That would imply that
183 ξ <~~> succ ξ <~~> succ (succ ξ) <~~> succ (succ (succ ξ)) <~~> succ (...(succ ξ)...)
185 In other words, the fixed point of `succ` is a term that is its own
186 successor. Let's just check that `ξ = succ ξ`:
188 let succ = \n s z. s (n s z) in
189 let ξ = (\u. succ (u u)) (\u. succ (u u)) in
191 ≡ succ ((\u. succ (u u)) (\u. succ (u u)))
192 ~~> succ (succ ((\u. succ (u u)) (\u. succ (u u))))
195 You should see the close similarity with `Y Y` here.
198 ## Q. So `Y` applied to `succ` returns a number that is not finite? ##
200 A. Well, if it makes sense to think of it as a number at all. It doesn't have the same structure as our encodings of finite Church numbers. But let's see if it behaves like they do:
202 ; assume same definitions as before
204 ≡ (\n s z. s (n s z)) ξ
206 <~~> succ (\s z. s (ξ s z)) ; using fixed-point reasoning
207 ≡ (\n s z. s (n s z)) (\s z. s (ξ s z))
208 ~~> \s z. s ((\s z. s (ξ s z)) s z)
209 ~~> \s z. s (s (ξ s z))
211 So `succ ξ` looks something like a Church number: it takes two arguments, `s` and `z`,
212 and returns a sequence of nested applications of `s`...
214 You should be able to prove that `add 2 (Y succ) <~~> Y succ`,
215 likewise for `mul`, `sub`, `pow`. What happens if we try `sub (Y
216 succ) (Y succ)`? What would you expect infinity minus infinity to be?
217 (Hint: choose your evaluation strategy so that you add two `s`s to the
218 first number for every `s` that you add to the second number.)
220 This is amazing, by the way: we're proving things about a term that
221 represents arithmetic infinity.
223 It's important to bear in mind the simplest, least-evaluated term we begin with is not
226 Y succ = (\h. (\u. h (u u)) (\u. h (u u))) (\n s z. s (n s z))
228 The way that infinity enters into the picture is that this term has
229 no normal form: no matter how many times we perform beta reduction,
230 there will always be an opportunity for more beta reduction. (Lather,
234 ## Q. That reminds me, what about [[evaluation order]]? ##
236 A. For a recursive function that has a well-behaved base case, such as
237 the factorial function, evaluation order is crucial. In the following
238 computation, we will arrive at a normal form. Watch for the moment at
239 which we have to make a choice about which beta reduction to perform
240 next: one choice leads to a normal form, the other choice leads to
243 let prefact = \fact n. (zero? n) 1 (mul n (fact (pred n))) in
244 let fact = Y prefact in
246 ≡ [(\h. (\u. h (u u)) (\u. h (u u))) prefact] 2
247 ~~> [(\u. prefact (u u)) (\u. prefact (u u))] 2
248 ~~> [prefact ((\u. prefact (u u)) (\u. prefact (u u)))] 2
249 ~~> [prefact (prefact ((\u. prefact (u u)) (\u. prefact (u u))))] 2
250 ≡ [(\fact n. (zero? n) 1 (mul n (fact (pred n)))) (prefact ((\u. prefact (u u)) (\u. prefact (u u))))] 2
251 ~~> [\n. (zero? n) 1 (mul n ([prefact ((\u. prefact (u u)) (\u. prefact (u u)))] (pred n)))] 2
252 ~~> (zero? 2) 1 (mul 2 ([prefact ((\u. prefact (u u)) (\u. prefact (u u)))] (pred 2)))
253 ~~> mul 2 ([prefact ((\u. prefact (u u)) (\u. prefact (u u)))] 1)
255 ~~> mul 2 (mul 1 ([prefact ((\u. prefact (u u)) (\u. prefact (u u)))] 0))
256 ≡ mul 2 (mul 1 ((zero? 0) 1 (mul 1 ([prefact ((\u. prefact (u u)) (\u. prefact (u u)))] (pred 0)))))
261 The crucial step is the third from the last. We have our choice of
262 either evaluating the test `(zero? 0) 1 ...`, which evaluates to `1`,
263 no matter what the ... contains;
264 or we can evaluate the `Y` pump, `(\u. prefact (u u)) (\u. prefact (u u))`, to
265 produce another copy of `prefact`. If we postpone evaluating the
266 `zero?` test, we'll pump out copy after copy of `prefact`, and never
267 realize that we've bottomed out in the recursion. But if we adopt a
268 leftmost/call-by-name/normal-order evaluation strategy, we'll always
269 start with the `zero?` predicate, and only produce a fresh copy of
270 `prefact` if we are forced to.
273 ## Q. You claimed that the Ackermann function couldn't be implemented using our primitive recursion techniques (such as the techniques that allow us to define addition and multiplication). But you haven't shown that it is possible to define the Ackermann function using full recursion. ##
279 | when m == 0 -> n + 1
280 | else when n == 0 -> A(m-1, 1)
281 | else -> A(m-1, A(m,n-1))
283 let A = Y (\A m n. (zero? m) (succ n) ((zero? n) (A (pred m) 1) (A (pred m) (A m (pred n)))))
289 ~~> A 0 (A 0 (A 1 0))
290 ~~> A 0 (A 0 (A 0 1))
295 `A 1 x` is to `A 0 x` as addition is to the successor function;
296 `A 2 x` is to `A 1 x` as multiplication is to addition;
297 `A 3 x` is to `A 2 x` as exponentiation is to multiplication---
298 so `A 4 x` is to `A 3 x` as hyper-exponentiation is to exponentiation...
300 ## Q. What other questions should I be asking? ##
302 * What is it about the variant fixed-point combinators that makes
303 them compatible with a call-by-value evaluation strategy?
305 * How do you know that the Ackermann function can't be computed
306 using primitive recursion techniques?
308 * What *exactly* is primitive recursion?
310 * I hear that `Y` delivers the/a *least* fixed point. Least
311 according to what ordering? How do you know it's least?
312 Is leastness important?