Problem:
 a() -> b()
 f(x,a()) -> f(b(),b())
 f(b(),x) -> f(b(),b())
 f(f(x,y),z) -> f(b(),b())

Proof:
 Nonconfluence Processor:
  terms: f(x,b()) *<- f(x,a()) ->* f(b(),b())
  Qed