# Page 197 Location: Chapter 21, solution to Problem 2, second paragraph

## It is (solution, second method)

Using the second method, you should get the solution $W'W'$, where $W'$ is the converse warbler -- $W'xy = yxx$. If you get $CW(CW)$ you are also right, since $CW$ is a converse warbler. You can easily check that $W'W'x=x(W'W')$.

## It should become

Using the second method, you should get the solution $CL(CL)$, equivalent to the longer $S,K,I$-expression $A_2A_2$ where $A_2=S(K(SI))(S(KK)(SII))$.

## Short explanation

The second method requires first to find a bird satisfying $A_2yx=x(yy)$; either by the algorithm illustrated in Chapter 18 or by direct derivation, one finds the above result

A_2 = CL = S(K(SI))(S(KK)(SII))\;,

leading to the right solution.

The expression given in the text does not fulfill the requirement: indeed,

W'W'x = xW'W' \neq x(W'W')

because

W'xy=yxx \neq y(xx)\,

where the parentheses are required if one is to honour the original expression given in the Problem of finding a bird $A$ such that $Ax=xA$ for any $x$.