[[!toc]]

#Q: How do you know that every term in the untyped lambda calculus has a fixed point?#

A: That's easy: let `T` be an arbitrary term in the lambda calculus.  If
`T` has a fixed point, then there exists some `X` such that `X <~~>
TX` (that's what it means to *have* a fixed point).

<pre><code>let L = \x. T (x x) in
let X = L L in
X &equiv; L L &equiv; (\x. T (x x)) L ~~> T (L L) &equiv; T X
</code></pre>

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 `T`, `Y T` is a fixed point of `T`?#

A: Note that in the proof given in the previous answer, we chose `T`
and then set `X = L L = (\x. T (x x)) (\x. T (x x))`.  If we abstract over
`T`, we get the Y combinator, `\T. (\x. T (x x)) (\x. T (x x))`.  No matter
what argument `T` we feed `Y`, it returns some `X` that is a fixed point
of `T`, by the reasoning in the previous answer.

#Q: So if every term has a fixed point, even `Y` has fixed point.#

A: Right:

<pre><code>let Y = \T. (\x. T (x x)) (\x. T (x x)) in
Y Y &equiv; \T. (\x. T (x x)) (\x. T (x x)) Y
~~> (\x. Y (x x)) (\x. Y (x x))
~~> Y ((\x. Y (x x)) (\x. Y (x x)))
~~> Y (Y ((\x. Y (x x)) (\x. Y (x x))))
~~> Y (Y (Y (...(Y (Y Y))...)))</code></pre>


#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 X such that X = X+1?
That would imply that

    X <~~> succ X <~~> succ (succ X) <~~> succ (succ (succ (X))) <~~> succ (... (succ X)...)

In other words, the fixed point of `succ` is a term that is its own
successor.  Let's just check that `X = succ X`:

<pre><code>let succ = \n s z. s (n s z) in
let X = (\x. succ (x x)) (\x. succ (x x)) in
succ X 
&equiv; succ ( (\x. succ (x x)) (\x. succ (x x)) ) 
~~> succ (succ ( (\x. succ (x x)) (\x. succ (x x))))
&equiv; succ (succ X)
</code></pre>

You should see the close similarity with `Y Y` here.


#Q. So `Y` applied to `succ` returns a number that is not finite!#

A. Yes!  Let's see why it makes sense to think of `Y succ` as a Church
numeral:

	[same definitions]
	succ X
	= (\n s z. s (n s z)) X 
	= \s z. s (X s z)
	= succ (\s z. s (X s z)) ; using fixed-point reasoning
	= \s z. s ([succ (\s z. s (X s z))] s z)
	= \s z. s ([\s z. s ([succ (\s z. s (X s z))] s z)] s z)
	= \s z. s (s (succ (\s z. s (X s z))))

So `succ X` looks like a numeral: 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 `mult`, `minus`, `pow`.  What happens if we try `minus (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 term in question is not
infinite:

	Y succ = (\f. (\x. f (x x)) (\x. f (x x))) (\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 prefac = \f n. iszero n 1 (mult n (f (pred n))) in
	let fac = Y prefac in
	fac 2
	   = [(\f.(\x.f(xx))(\x.f(xx))) prefac] 2
	   = [(\x.prefac(xx))(\x.prefac(xx))] 2
	   = [prefac((\x.prefac(xx))(\x.prefac(xx)))] 2
	   = [prefac(prefac((\x.prefac(xx))(\x.prefac(xx))))] 2
	   = [(\f n. iszero n 1 (mult n (f (pred n))))
		  (prefac((\x.prefac(xx))(\x.prefac(xx))))] 2
	   = [\n. iszero n 1 (mult n ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred n)))] 2
	   = iszero 2 1 (mult 2 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred 2)))
	   = mult 2 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] 1)
	   ...
	   = mult 2 (mult 1 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] 0))
	   = mult 2 (mult 1 (iszero 0 1 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred 0))))
	   = mult 2 (mult 1 1)
	   = mult 2 1
	   = 2

The crucial step is the third from the last.  We have our choice of
either evaluating the test `iszero 0 1 ...`, which evaluates to `1`,
no matter what the ... contains;
or we can evaluate the `Y` pump, `(\x.prefac(xx))(\x.prefac(xx))`, to
produce another copy of `prefac`.  If we postpone evaluting the
`iszero` test, we'll pump out copy after copy of `prefac`, 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 `iszero` predicate, and only produce a fresh copy of
`prefac` if we are forced to.