NO

Problem:
 H(I(x)) -> K(J(x))
 J(x) -> K(J(x))
 I(x) -> I(J(x))
 J(x) -> J(K(J(x)))

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