NO

Problem:
 f(x,a(g(x))) -> g(f(x,x))
 f(x,g(x)) -> g(f(x,x))
 a(x) -> x
 h(x) -> h(a(h(x)))

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