Beta reducing $SS(SK)$ using SKI calculus I have an expression to $\beta$-reduce and I managed to brute force it using the $\lambda$-calculus. I was wondering though, if I could make it in less steps, than what I did, using the $SKI$-calculus. For example it is said, that $S$ takes 3 arguments so that $Sxyz = xz (yz)$, but in my case I only have 2 arguments at most, so I can't really apply it.
Is there something I am not aware of, that allows us to use the combinators?
Edit: Am I correct to assume, that $(SK) = (\lambda z. K z)(z)$ which in turn can help us use $S$, so the expression then transforms into $SS(SK) = SS(\lambda z. K z)(z) = Sz((\lambda z.Kz)z)$?
 A: It is true that $S S (SK)$ you cannot apply any rule of the $SKI$-calculus because, as you correctly said, you do not have enough arguments. But you can feed $S S (SK)$ with enough arbitrary arguments (i.e. variables), to see how it behaves, and so you can give a "functional definition" of $S S (SK)$ (in the style of the other rules of the $SKI$-calculus.
Consider $S S (SK)$ applied to two arbitrary arguments, i.e. to two variables. Remember that application is left-associative, i.e. $LMNO = ((LM)N)O$.
Hence, using the rules $Sxyz = xz(yz)$ and $Kxy = x$, you get
\begin{align}
SS(SK)xy &= Sx(SKx)y 
\\
&= xy(SKxy)
\\
&= xy(Ky(xy))
\\
&= xyy
\end{align}
Thus, if we set $W =SS(SK)$, we have the diagonalizing combinator which satisfies the rule
$$Wxy = xyy$$
Note that it is not surprising that to see the behaviour of $SS(SK)$ you have to feed with two arguments, and not one.
Indeed, $S$ requires three arguments, and so the first occurrence of $S$ in $SS(SK)$ is waiting for an argument $x$, but $SK$ is waiting for two arguments! The first argument is provided by $x$ (the third argument that the first occurrence of $S$ in $SS(SK)$ was waiting for), but the second argument is something that you have to add.
Said differently, $SS(SK)x = Sx(SKx)$ but then you get stuck because the first occurrence of $S$ in $Sx(SKx)$ and $SK$ are waiting for a further argument.

Methodologically, it is correct that you can embed the $SKI$-calculus into the $\lambda$-calculus. But you have to do it in the correct way!
It is not true that $SK$ is equivalent to $(\lambda z.Kz)z$, otherwise we would have that $SK$ is equivalent to $Kz$ (since $(\lambda z.Kz)z \to_\beta Kz$) and this is wrong.
In the $\lambda$-calculus, $S = \lambda x \lambda y \lambda z. xz(yz)$ and $K = \lambda x \lambda y. x$. Therefore,
\begin{align}
Kxy &= (\lambda x \lambda y.x)xy 
\\
&\to_\beta (\lambda y.x)y
\\
&\to_\beta x
\end{align}
\begin{align}
SK &= (\lambda x \lambda y \lambda z. xz(yz))K
\\
&\to_\beta \lambda y \lambda z. Kz(yz)
\\
&\to_\beta^* \lambda y \lambda z. z
\end{align}
Thus,
\begin{align}
 SS(SK) &= (\lambda x \lambda y \lambda z. xz(yz)) S (SK)
\\
&\to_\beta (\lambda y \lambda z. Sz(yz))(SK)
\\
&\to_\beta \lambda z. Sz (SKz)
\\
&\to_\beta^* \lambda z. Sz ((\lambda z.\lambda y.y)z)
\\
&\to_\beta \lambda z. Sz (\lambda y.y)
\end{align}
So, $SS(SK) \to_\beta^* \lambda z. Sz (\lambda y.y)$,  and not $SS(SK) \to_\beta^* Sz((λz.Kz)z)$ as you wrote in the OP.
Hence,
\begin{align}
SS(SK) x y &\to_\beta^*  (\lambda z. Sz (\lambda y.y)) x y
\\
&\to_\beta Sx (\lambda y.y) y
\\
&\to_\beta^* xy((\lambda y.y)y)
\\
&\to_\beta xyy
\end{align}
which is coherent with what we found in the $SKI$-calculus.
