NO

Problem:
 0(q0(0(x1))) -> 0(0(q0(x1)))
 0(q0(h(x1))) -> 0(0(q0(h(x1))))
 0(q0(1(x1))) -> 0(1(q0(x1)))
 1(q0(0(x1))) -> 0(0(q1(x1)))
 1(q0(h(x1))) -> 0(0(q1(h(x1))))
 1(q0(1(x1))) -> 0(1(q1(x1)))
 1(q1(0(x1))) -> 1(0(q1(x1)))
 1(q1(h(x1))) -> 1(0(q1(h(x1))))
 1(q1(1(x1))) -> 1(1(q1(x1)))
 0(q1(0(x1))) -> 0(0(q2(x1)))
 0(q1(h(x1))) -> 0(0(q2(h(x1))))
 0(q1(1(x1))) -> 0(1(q2(x1)))
 1(q2(0(x1))) -> 1(0(q2(x1)))
 1(q2(h(x1))) -> 1(0(q2(h(x1))))
 1(q2(1(x1))) -> 1(1(q2(x1)))
 0(q2(x1)) -> q3(1(x1))
 1(q3(x1)) -> q3(1(x1))
 0(q3(x1)) -> q4(0(x1))
 1(q4(x1)) -> q4(1(x1))
 0(q4(0(x1))) -> 1(0(q5(x1)))
 0(q4(h(x1))) -> 1(0(q5(h(x1))))
 0(q4(1(x1))) -> 1(1(q5(x1)))
 1(q5(0(x1))) -> 0(0(q1(x1)))
 1(q5(h(x1))) -> 0(0(q1(h(x1))))
 1(q5(1(x1))) -> 0(1(q1(x1)))
 h(q0(x1)) -> h(0(q0(x1)))
 h(q1(x1)) -> h(0(q1(x1)))
 h(q2(x1)) -> h(0(q2(x1)))
 h(q3(x1)) -> h(0(q3(x1)))
 h(q4(x1)) -> h(0(q4(x1)))
 h(q5(x1)) -> h(0(q5(x1)))

Proof:
 Containment Processor: loop length: 1
                        terms:
                         0(q0(h(x1)))
                        context: 0([])
                        substitution:
                         x1 -> x1
  Qed