# 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

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

$W'W'x = xW'W' \neq x(W'W')$
$W'xy=yxx \neq y(xx)\,$
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 $$x$$.