[[!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).
``````let L = \x. T (x x) in
let X = L L in
X ≡ L L ≡ (\x. T (x x)) L ~~> T (L L) ≡ T X
``````
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:
``````let Y = \T. (\x. T (x x)) (\x. T (x x)) in
Y Y ≡ \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))...)))``````
#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`:
``````let succ = \n s z. s (n s z) in
let X = (\x. succ (x x)) (\x. succ (x x)) in
succ X
≡ succ ( (\x. succ (x x)) (\x. succ (x x)) )
~~> succ (succ ( (\x. succ (x x)) (\x. succ (x x))))
≡ succ (succ X)
``````
You should see the close similarity with `Y Y` here.