\begin{frame} \frametitle{Towards a Proof of Combinatorial Completeness} \begin{definition}[Abstraction of $x$] \begin{enumerate} \item $[x]t= Kt$, if $t$ is a constant or a variable other than $x$\\ \item $[x]x= I$ \item $[x]tt'= S([x]t)([x]t')$. %, if (i) does not apply \end{enumerate}\pause For $[x_1]([x_2](\ldots ([x_n]t)\ldots ))$ we will write $[x_1x_2\ldots x_n]t$ \end{definition} \pause Hint: $[x]t= Kt$ can also be used if $t$ does not contain $x$. \pause \begin{example} Let $t= [y]yx$ and $t'= [xy]yx$. Then \begin{enumerate} \item $ \begin{array}[t]{l} t \pause= S([y]y)([y]x) \pause= SI(Kx),\pause \end{array} $ \item<6-> $ \begin{array}[t]{ll} t' \pause= [x] t \pause= [x](SI(Kx)) \pause= S([x](SI))([x](Kx))\\ \ \ \pause= S(K(SI))(S([x]K)([x]x)) \pause= S(K(SI))(S(KK)I). \end{array} $ \end{enumerate} \end{example} \end{frame}