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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 $x$ .