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=3
   
   interpretation:
                   [0]
    [b](x0) = x0 + [0]
                   [1],
    
              [1 0 0]     [0]
    [a](x0) = [0 0 0]x0 + [1]
              [0 1 1]     [0],
    
              [1 0 1]     [0]
    [s](x0) = [0 1 0]x0 + [0]
              [0 0 1]     [1]
   orientation:
               [1 1 1]     [0]    [1 0 1]     [0]           
    s(a(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a(s(x1))
               [0 1 1]     [1]    [0 1 1]     [1]           
    
                     [1 1 1]     [2]    [1 1 1]     [1]                 
    s(b(a(b(x1)))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a(s(b(a(x1))))
                     [0 1 1]     [3]    [0 1 1]     [3]                 
    
                     [1 0 0]     [0]    [1 0 0]     [0]                 
    b(b(a(b(x1)))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = b(a(b(a(x1))))
                     [0 1 1]     [3]    [0 1 1]     [3]                 
    
                     [1 0 0]     [0]    [1 0 0]     [0]                 
    a(a(b(a(x1)))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a(b(a(b(x1))))
                     [0 1 1]     [3]    [0 1 1]     [3]                 
   problem:
    s(a(x1)) -> a(s(x1))
    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: {8,4,1}
     transitions:
      f30() -> 2*
      a0(2) -> 5*
      a0(9) -> 10*
      a0(11) -> 8*
      a0(6) -> 7*
      a0(3) -> 1*
      s0(2) -> 3*
      b0(10) -> 11*
      b0(5) -> 6*
      b0(7) -> 4*
      b0(2) -> 9*
      1 -> 3*
      4 -> 9*
      8 -> 5*
    problem:
     
    Qed