# Page 186

Location: Chapter 19, final Exercise

## Exercise (after the Appendix)

In terms of $B$, $C$, $S$, and $I$, find a combinator $A$ satisfying the condition $Axyz=xz(zy)$. The problem should be divided into three parts: ...

A = C(BBS)(CI)

## Short explanation

As suggested by the text, and according to the algorithm given in the Appendix, the solution proceeds in three steps:

First, a distinguished $z$-eliminate $A_1$ of the expression $X=xz(zy)$ must be found. By Rule 3a, this in turn requires to find:

• a distinguished $z$-eliminate of $xz$, which is $x$ (Rule 2);
• a distinguished $z$-eliminate of $zy$, which by Rule 3c is $CIy$.

Hence it is $A_1=Sx(CIy)$.

The second step consists of finding a distinguished $y$-eliminate $A_2$ of $A_1$: by Rule 3b (and Rule 2), one has immediately

A_2 = B(Sx)(CI)\;.

Finally, one needs a distinguished $x$-eliminate $A$ of $A_2$: again according to Rule 3c, to find it one must have a distinguished $x$-eliminate of $B(Sx)$. This latter term is $BBS$ (applying Rule 3b).

Application of Rule 3c then yields:

A = C(BBS)(CI)\;,

as can be verified by feeding the equation $Axyz=xz(zy)$ to the combinator-finder program (with the BCSI command-line option) as well as by explicit computation:

\begin{align} Axyz =~& C(BBS)(CI)xyz=BBSx(CI)yz=B(Sx)(CI)yz=\\ =~& Sx(CIy)z=xz(CIyz)=xz(Izy)=xz(zy)\;. \end{align}