Yf = f(Yf) for all [combinators] f, defines the desired behavior of Y
Mx = xx for all [combinators] x (note that M then equals SII in SKI combinator calculus)
assume there exists a y such that My = Y
proceeding from there, I can derive Y without having to memorize it
Myf = f(Myf)
yyf = f(yyf) (lambda x f . f(xxf)) My = (lambda x . x x) (lambda x f . f (x x f))
= (lambda x f . f (x x f)) (lambda x f . f (x x f))
= (lambda f . f ((lambda a g . g (a a g)) (lambda a g . g (a a g)) f))