3 #Q: How do you know that every term in the untyped lambda calculus has a fixed point?#
5 A: That's easy: let `T` be an arbitrary term in the lambda calculus. If
6 `T` has a fixed point, then there exists some `X` such that `X <~~>
7 TX` (that's what it means to *have* a fixed point).
12 X = WW = (\x.T(xx))W = T(WW) = TX
15 Please slow down and make sure that you understand what justified each
16 of the equalities in the last line.
18 #Q: How do you know that for any term `T`, `YT` is a fixed point of `T`?#
20 A: Note that in the proof given in the previous answer, we chose `T`
21 and then set `X = WW = (\x.T(xx))(\x.T(xx))`. If we abstract over
22 `T`, we get the Y combinator, `\T.(\x.T(xx))(\x.T(xx))`. No matter
23 what argument `T` we feed Y, it returns some `X` that is a fixed point
24 of `T`, by the reasoning in the previous answer.
26 #Q: So if every term has a fixed point, even `Y` has fixed point.#
30 let Y = \T.(\x.T(xx))(\x.T(xx)) in
31 Y Y = \T.(\x.T(xx))(\x.T(xx)) Y
32 = (\x.Y(xx))(\x.Y(xx))
33 = Y((\x.Y(xx))(\x.Y(xx)))
34 = Y(Y((\x.Y(xx))(\x.Y(xx))))
35 = Y(Y(Y(...(Y(YY))...)))
37 #Q: Ouch! Stop hurting my brain.#
39 A: Let's come at it from the direction of arithmetic. Recall that we
40 claimed that even `succ`---the function that added one to any
41 number---had a fixed point. How could there be an X such that X = X+1?
44 X = succ X = succ (succ X) = succ (succ (succ (X))) = succ (... (succ X)...)
46 In other words, the fixed point of `succ` is a term that is its own
47 successor. Let's just check that `X = succ X`:
49 let succ = \n s z. s (n s z) in
50 let X = (\x.succ(xx))(\x.succ(xx)) in
52 = succ ((\x.succ(xx))(\x.succ(xx)))
53 = succ (succ ((\x.succ(xx))(\x.succ(xx))))
56 You should see the close similarity with YY here.
58 #Q. So `Y` applied to `succ` returns a number that is not finite!#
60 A. Yes! Let's see why it makes sense to think of `Y succ` as a Church
65 = (\n s z. s (n s z)) X
67 = succ (\s z. s (X s z)) ; using fixed-point reasoning
68 = \s z. s ([succ (\s z. s (X s z))] s z)
69 = \s z. s ([\s z. s ([succ (\s z. s (X s z))] s z)] s z)
70 = \s z. s (s (succ (\s z. s (X s z))))
72 So `succ X` looks like a numeral: it takes two arguments, `s` and `z`,
73 and returns a sequence of nested applications of `s`...
75 You should be able to prove that `add 2 (Y succ) <~~> Y succ`,
76 likewise for `mult`, `minus`, `pow`. What happens if we try `minus (Y
77 succ)(Y succ)`? What would you expect infinity minus infinity to be?
78 (Hint: choose your evaluation strategy so that you add two `s`s to the
79 first number for every `s` that you add to the second number.)
81 This is amazing, by the way: we're proving things about a term that
82 represents arithmetic infinity.
84 It's important to bear in mind the simplest term in question is not
87 Y succ = (\f.(\x.f(xx))(\x.f(xx)))(\n s z. s (n s z))
89 The way that infinity enters into the picture is that this term has
90 no normal form: no matter how many times we perform beta reduction,
91 there will always be an opportunity for more beta reduction. (Lather,
94 #Q. That reminds me, what about [[evaluation order]]?#
96 A. For a recursive function that has a well-behaved base case, such as
97 the factorial function, evaluation order is crucial. In the following
98 computation, we will arrive at a normal form. Watch for the moment at
99 which we have to make a choice about which beta reduction to perform
100 next: one choice leads to a normal form, the other choice leads to
103 let prefac = \f n. isZero n 1 (mult n (f (pred n))) in
104 let fac = Y prefac in
106 = [(\f.(\x.f(xx))(\x.f(xx))) prefac] 2
107 = [(\x.prefac(xx))(\x.prefac(xx))] 2
108 = [prefac((\x.prefac(xx))(\x.prefac(xx)))] 2
109 = [prefac(prefac((\x.prefac(xx))(\x.prefac(xx))))] 2
110 = [(\f n. isZero n 1 (mult n (f (pred n))))
111 (prefac((\x.prefac(xx))(\x.prefac(xx))))] 2
112 = [\n. isZero n 1 (mult n ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred n)))] 2
113 = isZero 2 1 (mult 2 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred 2)))
114 = mult 2 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] 1)
116 = mult 2 (mult 1 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] 0))
117 = mult 2 (mult 1 (isZero 0 1 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred 0))))
122 The crucial step is the third from the last. We have our choice of
123 either evaluating the test `isZero 0 1 ...`, which evaluates to `1`,
124 no matter what the ... contains;
125 or we can evaluate the `Y` pump, `(\x.prefac(xx))(\x.prefac(xx))`, to
126 produce another copy of `prefac`. If we postpone evaluting the
127 `isZero` test, we'll pump out copy after copy of `prefac`, and never
128 realize that we've bottomed out in the recursion. But if we adopt a
129 leftmost/call-by-name/normal-order evaluation strategy, we'll always
130 start with the `isZero` predicate, and only produce a fresh copy of
131 `prefac` if we are forced to.
133 #Q. You claimed that the Ackerman 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 Ackerman function using full recursion.#
139 | when m == 0 -> n + 1
140 | else when n == 0 -> A(m-1,1)
141 | else -> A(m-1, A(m,n-1))
143 let A = Y (\A m n. isZero m (succ n) (isZero n (A (pred m) 1) (A (pred m) (A m (pred n))))) in
156 A 1 x is to A 0 x as addition is to the successor function;
157 A 2 x is to A 1 x as multiplication is to addition;
158 A 3 x is to A 2 x as exponentiation is to multiplication---
159 so A 4 x is to A 3 x as hyper-exponentiation is to exponentiation...
161 #Q. What other questions should I be asking?#
163 * What is it about the variant fixed-point combinators that makes
164 them compatible with a call-by-value evaluation strategy?
166 * How do you know that the Ackerman function can't be computed
167 using primitive recursion techniques?
169 * What *exactly* is primitive recursion?
171 * I hear that `Y` delivers the *least* fixed point. Least
172 according to what ordering? How do you know it's least?
173 Is leastness important?
176 ##The simply-typed lambda calculus##
178 The uptyped lambda calculus is pure computation. It is much more
179 common, however, for practical programming languages to be typed.
180 Likewise, systems used to investigate philosophical or linguistic
181 issues are almost always typed. Types will help us reason about our
182 computations. They will also facilitate a connection between logic
185 Soon we will consider polymorphic type systems. First, however, we
186 will consider the simply-typed lambda calculus. There's good news and
187 bad news: the good news is that the simply-type lambda calculus is
188 strongly normalizing: every term has a normal form. We shall see that
189 self-application is outlawed, so Ω can't even be written, let
190 alone undergo reduction. The bad news is that fixed-point combinators
191 are also forbidden, so recursion is neither simple nor direct.
195 We will have at least one ground type, `o`. From a linguistic point
196 of view, think of the ground types as the bar-level 0 categories, that
197 is, the lexical types, such as Noun, Verb, Preposition (glossing over
198 the internal complexity of those categories in modern theories).
200 In addition, there will be a recursively-defined class of complex
201 types `T`, the smallest set such that
203 * ground types, including `o`, are in `T`
205 * for any types σ and τ in `T`, the type σ -->
208 For instance, here are some types in `T`:
214 (o --> o) --> o --> o
220 Given a set of types `T`, we define the set of typed lambda terms <code>Λ_T</code>,
221 which is the smallest set such that
223 * each type `t` has an infinite set of distinct variables, {x^t}_1,
224 {x^t}_2, {x^t}_3, ...
226 * If a term `M` has type σ --> τ, and a term `N` has type
227 σ, then the application `(M N)` has type τ.
229 * If a variable `a` has type σ, and term `M` has type τ,
230 then the abstract <code>λ a M</code> has type σ --> τ.
232 The definitions of types and of typed terms should be highly familiar
233 to semanticists, except that instead of writing σ --> τ,
234 linguists (following Montague, who followed Church) write <σ,
235 τ>. We will use the arrow notation, since it is more iconic.
237 Some examples (assume that `x` has type `o`):
243 Excercise: write down terms that have the following types:
246 (o --> o) --> o --> o
247 (o --> o --> o) --> o
249 #Associativity of types versus terms#
251 As we have seen many times, in the lambda calculus, function
252 application is left associative, so that `f x y z == (((f x) y) z)`.
253 Types, *THEREFORE*, are right associative: if `f`, `x`, `y`, and `z`
254 have types `a`, `b`, `c`, and `d`, respectively, then `f` has type `a
255 --> b --> c --> d == (a --> (b --> (c --> d)))`.
257 It is a serious faux pas to associate to the left for types. You may
258 as well use your salad fork to stir your tea.
260 #The simply-typed lambda calculus is strongly normalizing#
262 If `M` is a term with type τ in Λ_T, then `M` has a
263 normal form. The proof is not particularly complex, but we will not
264 present it here; see Berendregt or Hankin.
266 Since Ω does not have a normal form, it follows that Ω
267 cannot have a type in Λ_T. We can easily see why:
269 Ω = (\x.xx)(\x.xx)
271 Assume Ω has type τ, and `\x.xx` has type σ. Then
272 because `\x.xx` takes an argument of type σ and returns
273 something of type τ, `\x.xx` must also have type σ -->
274 τ. By repeating this reasoning, `\x.xx` must also have type
275 (σ --> τ) --> τ; and so on. Since variables have
276 finite types, there is no way to choose a type for the variable `x`
277 that can satisfy all of the requirements imposed on it.
279 In general, there is no way for a function to have a type that can
280 take itself for an argument. It follows that there is no way to
281 define the identity function in such a way that it can take itself as
282 an argument. Instead, there must be many different identity
283 functions, one for each type.
287 Version 1 type numerals are not a good choice for the simply-typed
288 lambda calculus. The reason is that each different numberal has a
289 different type! For instance, if zero has type σ, then `false`
290 has type τ --> τ --> τ, for some τ. Since one is
291 represented by the function `\x.x false 0`, one must have type (τ
292 --> τ --> τ) --> σ --> σ. But this is a different
293 type than zero! Because each number has a different type, it becomes
294 impossible to write arithmetic operations that can combine zero with
295 one. We would need as many different addition operations as we had
296 pairs of numbers that we wanted to add.
298 Fortunately, the Church numberals are well behaved with respect to
299 types. They can all be given the type (σ --> σ) -->