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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.