YES(?,O(n^1))

Problem:
 q0(0(x1)) -> 0'(q1(x1))
 q1(0(x1)) -> 0(q1(x1))
 q1(1'(x1)) -> 1'(q1(x1))
 0(q1(1(x1))) -> q2(0(1'(x1)))
 0'(q1(1(x1))) -> q2(0'(1'(x1)))
 1'(q1(1(x1))) -> q2(1'(1'(x1)))
 0(q2(0(x1))) -> q2(0(0(x1)))
 0'(q2(0(x1))) -> q2(0'(0(x1)))
 1'(q2(0(x1))) -> q2(1'(0(x1)))
 0(q2(1'(x1))) -> q2(0(1'(x1)))
 0'(q2(1'(x1))) -> q2(0'(1'(x1)))
 1'(q2(1'(x1))) -> q2(1'(1'(x1)))
 q2(0'(x1)) -> 0'(q0(x1))
 q0(1'(x1)) -> 1'(q3(x1))
 q3(1'(x1)) -> 1'(q3(x1))
 q3(b(x1)) -> b(q4(x1))

Proof:
 Bounds Processor:
  bound: 1
  enrichment: match
  automaton:
   final states: {10,9,8,7,6,5,4}
   transitions:
    b1(12) -> 13*
    q41(19) -> 20*
    q41(21) -> 22*
    q41(11) -> 12*
    q00(2) -> 4*
    q00(1) -> 4*
    q00(3) -> 4*
    00(2) -> 6*
    00(1) -> 6*
    00(3) -> 6*
    0'0(2) -> 7*
    0'0(1) -> 7*
    0'0(3) -> 7*
    q10(2) -> 5*
    q10(1) -> 5*
    q10(3) -> 5*
    1'0(2) -> 8*
    1'0(1) -> 8*
    1'0(3) -> 8*
    10(2) -> 1*
    10(1) -> 1*
    10(3) -> 1*
    q20(2) -> 9*
    q20(1) -> 9*
    q20(3) -> 9*
    q30(2) -> 10*
    q30(1) -> 10*
    q30(3) -> 10*
    b0(2) -> 2*
    b0(1) -> 2*
    b0(3) -> 2*
    q40(2) -> 3*
    q40(1) -> 3*
    q40(3) -> 3*
    1 -> 19*
    2 -> 11*
    3 -> 21*
    13 -> 10*
    20 -> 12*
    22 -> 12*
  problem:
   
  Qed