I spent far too long on this.
We have A x y = x (y y) aka A x = x . M and M x = x x. How do we
construct an expression for a combinator that, given x produces M (x . M)? This is M . A, but how do we express that in terms of M
and A?
Of course, we don't. We're not showing a construction of this bird,
only showing that it exists. C1 means that, given that we have M and
A, we have M . A.
Duh.
Of course, this "sage bird" is a fixed-point combinator, which should be fun for future chapters...