NO

Problem:
 f(g(a())) -> f(s(g(b())))
 f(f(x)) -> b()
 g(x) -> f(g(x))

Proof:
 Containment Processor: loop length: 1
                        terms:
                         g(x)
                        context: f([])
                        substitution:
                         x -> x
  Qed