NO

Problem:
 a() -> b()
 f(a(),b()) -> f(a(),a())
 f(b(),a()) -> f(a(),a())
 f(a(),a()) -> c()
 g(x) -> f(x,x)

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