I am currently reading about the lambda calculus as well as the Y combinator. I know that for any function $f$, $Yf$ is a fixed-point of $f$, that is $f(Yf) = Yf$. In order to wrap my head around this, I came up with the following example:
$Y := \lambda f. (\lambda x. f(x, x)) (\lambda x. f(x, x))$
$i : = \lambda n. n$
In order to exemplify the equasion $f(Yf) = Yf$, I tried to get from $i(Yi)$ to $Yi$, which only really required one step:
$i(Yi) = \lambda n. n (\lambda f. (\lambda x. f(x, x)) (\lambda x. f(x, x)) \lambda n.n) = \lambda f. (\lambda x. f(x, x)) (\lambda x. f(x, x)) \lambda n.n = Yi$
However, I can't seem to get from $Yi$ to $i(Yi)$. I guess I am not applying $\beta$-reduction correctly. If someone could show me how to get there, I would be so greatful.
Thank you in advance!
Since $i$ is the identity, you have $(Y) i = (\lambda n . n) (Y) i$, which you can rewrite as $(i)(Y)i$, or in your notation, $i (Y i)$.
With lambda calculus, I would bracket the thing to be evaluated. So $Y$ would be expressed as "$\lambda f . (\lambda x . (f) (x) x) \lambda x . (f) (x) x$". Then $\beta$-reduce $(Y)i$ a few times (I'll repeat the LHS for later clarity).
$$ \begin{align} (Y) i &= (\lambda f . (\lambda x . (f) (x) x) \lambda x . (f) (x) x) i \\ (Y) i &= (\lambda x . (i) (x) x) \lambda x . (i) (x) x \\ (Y) i &= (i) (\lambda x . (i) (x) x) \lambda x . (i) (x) x \end{align} $$
Now define $E$ as the expression on the right hand side of the second line. The second line now looks like $(Y)i = E$.
Then the third line looks like $(Y)i = (i)E$, which by definition of $E$ from the second line, expands to $(Y)i = (i)(Y)i$, as required.