3 ms·

Can you show the derivation steps you take?

• calcnerd256 17y ago
  Sure. Myf = f(Myf) yyf = f(yyf) Suppose yxf = f(xxf) for all x. Now y is simply (lambda x f . f(xxf)) Note that this doesn't get you the applicative-order version we want, but it does satisfy the definition of Y 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)) etc.