YES

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:
 String Reversal Processor:
  0(q0(x1)) -> q1(0'(x1))
  0(q1(x1)) -> q1(0(x1))
  1'(q1(x1)) -> q1(1'(x1))
  1(q1(0(x1))) -> 1'(0(q2(x1)))
  1(q1(0'(x1))) -> 1'(0'(q2(x1)))
  1(q1(1'(x1))) -> 1'(1'(q2(x1)))
  0(q2(0(x1))) -> 0(0(q2(x1)))
  0(q2(0'(x1))) -> 0(0'(q2(x1)))
  0(q2(1'(x1))) -> 0(1'(q2(x1)))
  1'(q2(0(x1))) -> 1'(0(q2(x1)))
  1'(q2(0'(x1))) -> 1'(0'(q2(x1)))
  1'(q2(1'(x1))) -> 1'(1'(q2(x1)))
  0'(q2(x1)) -> q0(0'(x1))
  1'(q0(x1)) -> q3(1'(x1))
  1'(q3(x1)) -> q3(1'(x1))
  b(q3(x1)) -> q4(b(x1))
  KBO Processor:
   weight function:
    w0 = 1
    w(q4) = w(b) = w(q3) = w(q2) = w(1) = w(1') = w(0') = w(q1) = w(q0) = w(0) = 1
   precedence:
    b > q2 > q4 > 0' > q0 > 0 > 1 > 1' > q1 > q3
   problem:
    
   Qed