NO

Problem:
 F(D(x),y) -> F(D(x),G(G(y)))
 F(x,E(y)) -> F(G(x),E(y))
 G(x) -> x
 H(I(x)) -> K(J(x))
 J(x) -> K(J(x))
 I(x) -> I(J(x))
 J(x) -> J(K(J(x)))
 S(x,T(x)) -> T(x)

Proof:
 Containment Processor: loop length: 1
                        terms:
                         F(D(x),y)
                        context: []
                        substitution:
                         x -> x
                         y -> G(G(y))
  Qed