NO

Problem:
 f(x) -> g(f(x))
 h(x) -> p(h(x))
 f(x) -> h(f(x))
 g(x) -> p(p(h(x)))

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