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#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.