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:
 DP Processor:
  DPs:
   1#(q0(1(x1))) -> 1#(q1(x1))
   1#(q0(1(x1))) -> 0#(1(q1(x1)))
   1#(q0(0(x1))) -> 0#(q1(x1))
   1#(q0(0(x1))) -> 0#(0(q1(x1)))
   1#(q1(1(x1))) -> 1#(q1(x1))
   1#(q1(1(x1))) -> 1#(1(q1(x1)))
   1#(q1(0(x1))) -> 0#(q1(x1))
   1#(q1(0(x1))) -> 1#(0(q1(x1)))
   0#(q1(x1)) -> 1#(x1)
   1#(q2(x1)) -> 1#(x1)
   0#(q2(x1)) -> 0#(q0(x1))
  TRS:
   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))
  KBO Processor:
   weight function:
    w0 = 1
    w(1#) = w(q2) = w(0) = w(q1) = w(q0) = w(1) = 1
    w(0#) = 0
   precedence:
    0# > q1 > 1 > 0 > q2 > 1# ~ q0
   problem:
    DPs:
     
    TRS:
     
   Qed