Past exam of the computer science course of the University of Oxford / 2015 / Lambda calculus and types / 3 / d

Successive bracket abstractions are processed from the inside out: $[x][y]((\mathbf{K}x)y)$. Retain $I\equiv (\mathbf{SK})\mathbf{K}$ only as a temporary abbreviation for the I combinator. Since $y$ is absent from $\mathbf{K}x$, the first bracket abstraction isFor the outer bracket abstraction, $x$ is absent from $I$ and the S combinator, while it occurs in $\mathbf{K}(\mathbf{K}x)$. HenceSubstituting upward and then expanding $I$ gives the following fully bracketed string, consisting solely of the S combinator, the K combinator and parentheses:For readability, the same result with the usual left association convention and the temporary abbreviation isThe syntactic result comes from the specified bracket abstraction algorithm. Replacing it by the shorter K combinator, despite agreement after two arguments, would not answer the requested computation. As a check using the result of part (b), applying the computed combinatory logic term to $u,v$ gives $\mathbf{K}uv=u$.