YES

Problem:
 1(q0(1(x1))) -> 0(1(q1(x1)))
 1(q0(0(x1))) -> 0(0(q1(x1)))
 1(q1(1(x1))) -> 1(1(q1(x1)))
 1(q1(0(x1))) -> 1(0(q1(x1)))
 0(q1(x1)) -> q2(1(x1))
 1(q2(x1)) -> q2(1(x1))
 0(q2(x1)) -> 0(q0(x1))

Proof:
 KBO Processor:
  weight function:
   w0 = 1
   w(q2) = w(0) = w(q1) = w(q0) = w(1) = 1
  precedence:
   q1 > 1 > 0 > q2 > q0
  problem:
   
  Qed