3 Q: How do you know that every term in the untyped lambda calculus has
6 A: That's easy: let `T` be an arbitrary term in the lambda calculus. If
7 `T` has a fixed point, then there exists some `X` such that `X <~~>
8 TX` (that's what it means to *have* a fixed point).
12 X = WW = (\x.T(xx))W = T(WW) = TX
14 Q: How do you know that for any term T, YT is a fixed point of T?
16 A: Note that in the proof given in the previous answer, we chose `T`
17 and then set `X = WW = (\x.T(xx))(\x.T(xx))`. If we abstract over
18 `T`, we get the Y combinator, `\T.(\x.T(xx))(\x.T(xx))`. No matter
19 what argument `T` we feed Y, it returns some `X` that is a fixed point
20 of `T`, by the reasoning in the previous answer.
22 Q: So if every term has a fixed point, even Y has fixed point.
26 let Y = \T.(\x.T(xx))(\x.T(xx)) in
27 Y Y = \T.(\x.T(xx))(\x.T(xx)) Y
28 = (\x.Y(xx))(\x.Y(xx))
29 = Y((\x.Y(xx))(\x.Y(xx)))
30 = Y(Y((\x.Y(xx))(\x.Y(xx))))
31 = Y(Y(Y(...(Y(YY))...)))
33 Q: Ouch! Stop hurting my brain.
35 A: Let's come at it from the direction of arithmetic. Recall that we
36 claimed that even `succ`---the function that added one to any
37 number---had a fixed point. How could there be an X such that X = X+1?
40 X = succ X = succ (succ X) = succ (succ (succ (X))) = succ (... (succ X)...)
42 In other words, the fixed point of `succ` is a term that is its own
43 successor. Let's just check that `X = succ X`:
45 let succ = \n s z. s (n s z) in
46 let X = (\x.succ(xx))(\x.succ(xx)) in
48 = succ ((\x.succ(xx))(\x.succ(xx)))
49 = succ (succ ((\x.succ(xx))(\x.succ(xx))))
52 You should see the close similarity with YY here.
54 Q. So `Y` applied to `succ` returns infinity!
56 A. Yes! Let's see why it makes sense to think of `Y succ` as a Church
61 = (\n s z. s (n s z)) X
63 = succ (\s z. s (X s z)) ; using fixed-point reasoning
64 = \s z. s ([succ (\s z. s (X s z))] s z)
65 = \s z. s ([\s z. s ([succ (\s z. s (X s z))] s z)] s z)
66 = \s z. s (s (succ (\s z. s (X s z))))
68 So `succ X` looks like a numeral: it takes two arguments, `s` and `z`,
69 and returns a sequence of nested applications of `s`...
71 You should be able to prove that `add 2 (Y succ) <~~> Y succ`,
72 likewise for `mult`, `minus`, `pow`. What happens if we try `minus (Y
73 succ)(Y succ)`? What would you expect infinity minus infinity to be?
74 (Hint: choose your evaluation strategy so that you add two `s`s to the
75 first number for every `s` that you add to the second number.)
77 This is amazing, by the way: we're proving things about a term that
78 represents arithmetic infinity. It's important to bear in mind the
79 simplest term in question is not infinite:
81 Y succ = (\f.(\x.f(xx))(\x.f(xx)))(\n s z. s (n s z))
83 The way that infinity enters into the picture is that this term has
84 no normal form: no matter how many times we perform beta reduction,
85 there will always be an opportunity for more beta reduction. (Lather,
88 Q. That reminds me, what about [[evaluation order]]?
90 A. For a recursive function that has a well-behaved base case, such as
91 the factorial function, evaluation order is crucial. In the following
92 computation, we will arrive at a normal form. Watch for the moment at
93 which we have to make a choice about which beta reduction to perform
94 next: one choice leads to a normal form, the other choice leads to
97 let prefac = \f n. isZero n 1 (mult n (f (pred n))) in
100 = [(\f.(\x.f(xx))(\x.f(xx))) prefac] 2
101 = [(\x.prefac(xx))(\x.prefac(xx))] 2
102 = [prefac((\x.prefac(xx))(\x.prefac(xx)))] 2
103 = [prefac(prefac((\x.prefac(xx))(\x.prefac(xx))))] 2
104 = [(\f n. isZero n 1 (mult n (f (pred n))))
105 (prefac((\x.prefac(xx))(\x.prefac(xx))))] 2
106 = [\n. isZero n 1 (mult n ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred n)))] 2
107 = isZero 2 1 (mult 2 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred 2)))
108 = mult 2 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] 1)
110 = mult 2 (mult 1 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] 0))
111 = mult 2 (mult 1 (isZero 0 1 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred 0))))
116 The crucial step is the third from the last. We have our choice of
117 either evaluating the test `isZero 0 1 ...`, which evaluates to `1`,
118 or we can evaluate the `Y` pump, `(\x.prefac(xx))(\x.prefac(xx))`, to
119 produce another copy of `prefac`. If we postpone evaluting the
120 `isZero` test, we'll pump out copy after copy of `prefac`, and never
121 realize that we've bottomed out in the recursion. But if we adopt a
122 leftmost/call by name/normal order evaluation strategy, we'll always
123 start with the isZero predicate, and only produce a fresh copy of
124 `prefac` if we are forced to.
126 Q. You claimed that the Ackerman function couldn't be implemented
127 using our primitive recursion techniques (such as the techniques that
128 allow us to define addition and multiplication). But you haven't
129 shown that it is possible to define the Ackerman function using full
136 | when m == 0 -> n + 1
137 | else when n == 0 -> A(m-1,1)
138 | else -> A(m-1, A(m,n-1))
140 let A = Y (\A m n. isZero m (succ n) (isZero n (A (pred m) 1) (A (pred m) (A m (pred n))))) in
153 A 1 x is to A 0 x as addition is to the successor function;
154 A 2 x is to A 1 x as multiplication is to addition;
155 A 3 x is to A 2 x as exponentiation is to multiplication---
156 so A 4 x is to A 3 x as super-exponentiation is to exponentiation...