NO

Problem:
 f(x) -> g(k(x))
 f(x) -> a()
 g(x) -> a()
 k(a()) -> k(k(a()))

Proof:
 Containment Processor: loop length: 1
                        terms:
                         k(a())
                        context: k([])
                        substitution:
                         
  Qed