YES

Problem:
 q0(a(x1)) -> x(q1(x1))
 q1(a(x1)) -> a(q1(x1))
 q1(y(x1)) -> y(q1(x1))
 a(q1(b(x1))) -> q2(a(y(x1)))
 a(q2(a(x1))) -> q2(a(a(x1)))
 a(q2(y(x1))) -> q2(a(y(x1)))
 y(q1(b(x1))) -> q2(y(y(x1)))
 y(q2(a(x1))) -> q2(y(a(x1)))
 y(q2(y(x1))) -> q2(y(y(x1)))
 q2(x(x1)) -> x(q0(x1))
 q0(y(x1)) -> y(q3(x1))
 q3(y(x1)) -> y(q3(x1))
 q3(bl(x1)) -> bl(q4(x1))

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