YES

Problem:
 a(s(x1)) -> s(a(x1))
 b(a(b(s(x1)))) -> a(b(s(a(x1))))
 b(a(b(b(x1)))) -> a(b(a(b(x1))))
 a(b(a(a(x1)))) -> b(a(b(a(x1))))

Proof:
 String Reversal Processor:
  s(a(x1)) -> a(s(x1))
  s(b(a(b(x1)))) -> a(s(b(a(x1))))
  b(b(a(b(x1)))) -> b(a(b(a(x1))))
  a(a(b(a(x1)))) -> a(b(a(b(x1))))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [a](x0) = 2x0 + 1,
    
    [s](x0) = 3x0 + 1,
    
    [b](x0) = 2x0 + 1
   orientation:
    s(a(x1)) = 6x1 + 4 >= 6x1 + 3 = a(s(x1))
    
    s(b(a(b(x1)))) = 24x1 + 22 >= 24x1 + 21 = a(s(b(a(x1))))
    
    b(b(a(b(x1)))) = 16x1 + 15 >= 16x1 + 15 = b(a(b(a(x1))))
    
    a(a(b(a(x1)))) = 16x1 + 15 >= 16x1 + 15 = a(b(a(b(x1))))
   problem:
    b(b(a(b(x1)))) -> b(a(b(a(x1))))
    a(a(b(a(x1)))) -> a(b(a(b(x1))))
   Bounds Processor:
    bound: 0
    enrichment: match
    automaton:
     final states: {6,1}
     transitions:
      a0(7) -> 8*
      a0(4) -> 5*
      a0(9) -> 6*
      a0(2) -> 3*
      f30() -> 2*
      b0(3) -> 4*
      b0(2) -> 7*
      b0(8) -> 9*
      b0(5) -> 1*
      6 -> 3*
      1 -> 7*
    problem:
     
    Qed