[[!toc levels=3]]
## Application to the truth teller/liar paradoxes ##
Curry originally called `Y` the "paradoxical" combinator, and discussed
it in connection with certain well-known paradoxes from the philosophy
literature. The truth-teller paradox has the flavor of a recursive
function without a base case:
(1) This sentence meaning is true.
If we assume that the complex demonstrative "this sentence meaning" can refer
to the very meaning displayed in (1), then that meaning (1) will be true just in
case the thing referred to by *this sentence meaning* 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 values
like the ones we have been working with in the Lambda Calculus. Then the sentence *John
is John* might have as its meaning our `true`, namely `\y n. y`.
Now, the verb phrase in (1) expresses a function from whatever the referent of *this
sentence meaning* is to a boolean. That is, `\m. m true false`, where
the argument `m` is the referent of *this sentence meaning*. Of course, if
`m` is a boolean, `m true false <~~> m`, so for our purposes, we can
assume that the verb phrase of (1) denotes the identity function `I`.
If we use (1) in a context in which *this sentence meaning* refers to the
meaning expressed by the very sentence in which that demonstrative occurs, then we must find a
meaning `m` such that it is equivalent to the application of the verb phrase meaning to itself.
That is, `m <~~> I m`. In other words, `m` is
a fixed point for the meaning of the verb phrase.
That means that in a context in which *this sentence meaning* refers to the
meaning expressed by the sentence in which it occurs, the sentence's meaning is a fixed point for
the identity function. As we observed earlier, *anything* is a fixed point for the identity function.
In particular, each of the boolean values `true` and `false` are fixed points for the identity
function. What fixed point does `Y` give us?
Y I ≡
(\h. (\u. h (u u)) (\u. h (u u))) I ~~>
(\u. I (u u)) (\u. I (u u))) ~~>
(\u. (u u)) (\u. (u u))) ≡
ω ω
Ω
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.
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).
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
Please slow down and make sure that you understand what justified each
of the equalities in the last line.
## Q: How do you know that for any term `N`, `Y N` is a fixed point of `N`? ##
A: Note that in the proof given in the previous answer, we chose `N`
and then set `ξ ≡ H H ≡ (\u. N (u u)) (\u. N (u u))`. If we abstract over
`N`, we get the Y combinator, `\N. (\u. N (u u)) (\u. N (u u))`. No matter
what argument `N` we feed `Y`, it returns some `ξ` that is a fixed point
of `N`, by the reasoning in the previous answer.
## Q: So if every term has a fixed point, even `Y` has fixed point. ##
A: Right:
let Y = \N. (\u. N (u u)) (\u. N (u u)) in
Y Y
≡ \N. (\u. N (u u)) (\u. N (u u)) Y
~~> (\u. Y (u u)) (\u. Y (u u))
~~> Y ((\u. Y (u u)) (\u. Y (u u)))
~~> Y ( Y ((\u. Y (u u)) (\u. Y (u u))))
~~> Y (Y (Y (...(Y (Y Y))...)))
## Q: Ouch! Stop hurting my brain. ##
A: Is that a question?
Let's come at it from the direction of arithmetic. Recall that we
claimed that even `succ`---the function that added one to any
number---had a fixed point. How could there be an `ξ` such that `ξ <~~> succ ξ`?
That would imply that
ξ <~~> succ ξ <~~> succ (succ ξ) <~~> succ (succ (succ ξ)) <~~> succ (...(succ ξ)...)
In other words, the fixed point of `succ` is a term that is its own
successor. Let's just check that `ξ = succ ξ`:
let succ = \n s z. s (n s z) in
let ξ = (\u. succ (u u)) (\u. succ (u u)) in
succ ξ
≡ succ ((\u. succ (u u)) (\u. succ (u u)))
~~> succ (succ ((\u. succ (u u)) (\u. succ (u u))))
≡ succ (succ ξ)
You should see the close similarity with `Y Y` here.
## Q: So `Y` applied to `succ` returns a number that is not finite? ##
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:
; assume same definitions as before
succ ξ
≡ (\n s z. s (n s z)) ξ
~~> \s z. s (ξ s z)
<~~> succ (\s z. s (ξ s z)) ; using fixed-point reasoning
≡ (\n s z. s (n s z)) (\s z. s (ξ s z))
~~> \s z. s ((\s z. s (ξ s z)) s z)
~~> \s z. s (s (ξ s z))
So `succ ξ` looks something like a Church number: it takes two arguments, `s` and `z`,
and returns a sequence of nested applications of `s`...
You should be able to prove that `add 2 (Y succ) <~~> Y succ`,
likewise for `mul`, `sub`, `pow`. What happens if we try `sub (Y
succ) (Y succ)`? What would you expect infinity minus infinity to be?
(Hint: choose your evaluation strategy so that you add two `s`s to the
first number for every `s` that you add to the second number.)
This is amazing, by the way: we're proving things about a term that
represents arithmetic infinity.
It's important to bear in mind the simplest, least-evaluated term we begin with is not
infinitely long:
Y succ = (\h. (\u. h (u u)) (\u. h (u u))) (\n s z. s (n s z))
The way that infinity enters into the picture is that this term has
no normal form: no matter how many times we perform beta reduction,
there will always be an opportunity for more beta reduction. (Lather,
rinse, repeat!)
## Q: That reminds me, what about [[evaluation order]]? ##
A: For a recursive function that has a well-behaved base case, such as
the factorial function, evaluation order is crucial. In the following
computation, we will arrive at a normal form. Watch for the moment at
which we have to make a choice about which beta reduction to perform
next: one choice leads to a normal form, the other choice leads to
endless reduction:
let prefact = \fact n. (zero? n) 1 (mul n (fact (pred n))) in
let fact = Y prefact in
fact 2
≡ [(\h. (\u. h (u u)) (\u. h (u u))) prefact] 2
~~> [(\u. prefact (u u)) (\u. prefact (u u))] 2
~~> [prefact ((\u. prefact (u u)) (\u. prefact (u u)))] 2
~~> [prefact (prefact ((\u. prefact (u u)) (\u. prefact (u u))))] 2
≡ [(\fact n. (zero? n) 1 (mul n (fact (pred n)))) (prefact ((\u. prefact (u u)) (\u. prefact (u u))))] 2
~~> [\n. (zero? n) 1 (mul n ([prefact ((\u. prefact (u u)) (\u. prefact (u u)))] (pred n)))] 2
~~> (zero? 2) 1 (mul 2 ([prefact ((\u. prefact (u u)) (\u. prefact (u u)))] (pred 2)))
~~> mul 2 ([prefact ((\u. prefact (u u)) (\u. prefact (u u)))] 1)
...
~~> mul 2 (mul 1 ([prefact ((\u. prefact (u u)) (\u. prefact (u u)))] 0))
≡ mul 2 (mul 1 ((zero? 0) 1 (mul 1 ([prefact ((\u. prefact (u u)) (\u. prefact (u u)))] (pred 0)))))
~~> mul 2 (mul 1 1)
~~> mul 2 1
~~> 2
The crucial step is the third from the last. We have our choice of
either evaluating the test `(zero? 0) 1 ...`, which evaluates to `1`,
no matter what the ... contains;
or we can evaluate the `Y` pump, `(\u. prefact (u u)) (\u. prefact (u u))`, to
produce another copy of `prefact`. If we postpone evaluating the
`zero?` test, we'll pump out copy after copy of `prefact`, and never
realize that we've bottomed out in the recursion. But if we adopt a
leftmost/call-by-name/normal-order evaluation strategy, we'll always
start with the `zero?` predicate, and only produce a fresh copy of
`prefact` if we are forced to.
## 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. ##
A: OK:
A(m,n) =
| when m == 0 -> n + 1
| else when n == 0 -> A(m-1, 1)
| else -> A(m-1, A(m,n-1))
let A = Y (\A m n. (zero? m) (succ n) ((zero? n) (A (pred m) 1) (A (pred m) (A m (pred n)))))
So for instance:
A 1 2
~~> A 0 (A 1 1)
~~> A 0 (A 0 (A 1 0))
~~> A 0 (A 0 (A 0 1))
~~> A 0 (A 0 2)
~~> A 0 3
~~> 4
`A 1 x` is to `A 0 x` as addition is to the successor function;
`A 2 x` is to `A 1 x` as multiplication is to addition;
`A 3 x` is to `A 2 x` as exponentiation is to multiplication---
so `A 4 x` is to `A 3 x` as hyper-exponentiation is to exponentiation...
## Q: What other questions should I be asking? ##
* What is it about the variant fixed-point combinators that makes
them compatible with a call-by-value evaluation strategy?
* How do you know that the Ackermann function can't be computed
using primitive recursion techniques?
* What *exactly* is primitive recursion?
* I hear that `Y` delivers the/a *least* fixed point. Least
according to what ordering? How do you know it's least?
Is leastness important?
## Q: I still don't fully understand the Y combinator. Can you explain it in a different way?
Sure! Here is another way to derive `Y`. We'll start by choosing a
specific goal, and at each decision point, we'll make a reasonable
guess. The guesses will all turn out to be lucky, and we'll arrive at
a fixed point combinator.
Given an arbitrary term `h`, we want to find a fixed point `X` such that:
X <~~> h X
Our strategy will be to seek an `X` such that `X ~~> h X` (this is just a guess). Because `X` and
`h X` are syntactically different, the only way that `X` can reduce to `h X` is if `X`
contains at least one redex. The simplest way to satisfy this
constraint would be for the fixed point to itself be a redex (again, just a guess):
X ≡ ((\u. M) N) ~~> h X
The result of beta reduction on this redex will be `M` with some
substitutions. We know that after these substitutions, `M` will have
the form `h X`, since that is what the reduction arrow tells us. So we
can refine the picture as follows:
X ≡ ((\u. h (___)) N) ~~> h X
Here, the `___` has to be something that reduces to the fixed point `X`.
It's natural to assume that there will be at least one occurrence of
`u` in the body of the head abstract:
X ≡ ((\u. h (__u__)) N) ~~> h X
After reduction of the redex, we're going to have
X ≡ h (__N__) ~~> h X
Apparently, `__N__` will have to reduce to `X`. Therefore we should
choose a skeleton for `N` that is consistent with what we have decided
so far about the internal structure of `X`. We might like for `N` to
syntactically match the whole of `X`, but this would require `N` to contain itself as
a subpart. So we'll settle for the more modest assumption (or guess) that `N`
matches the head of `X`:
X ≡ ((\u. h (__u__)) (\u. h (__u__))) ~~> h X
At this point, we've derived a skeleton for X on which it contains two
so-far identical halves. We'll guess that the halves will be exactly
identical. Note that at the point at which we perform the first
reduction, `u` will get bound to `N`, which now corresponds to a term
representing one of the halves of `X`. So in order to produce a full `X`,
we simply make a second copy of `u`:
X ≡ ((\u. h (u u)) (\u. h (u u)))
~~> h ((\u. h (u u)) (\u. h (u u)))
≡ h X
Success.
So the function `\h. (\u. h (u u)) (\u. h (u u))` maps an arbitrary term
`h` to a fixed point for `h`.