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:
    [b](x0) = x0 + 1,
    
    [a](x0) = x0 + 1,
    
    [s](x0) = 8x0
   orientation:
    s(a(x1)) = 8x1 + 8 >= 8x1 + 1 = a(s(x1))
    
    s(b(a(b(x1)))) = 8x1 + 24 >= 8x1 + 17 = a(s(b(a(x1))))
    
    b(b(a(b(x1)))) = x1 + 4 >= x1 + 4 = b(a(b(a(x1))))
    
    a(a(b(a(x1)))) = x1 + 4 >= x1 + 4 = 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))))
   String Reversal Processor:
    b(a(b(b(x1)))) -> a(b(a(b(x1))))
    a(b(a(a(x1)))) -> b(a(b(a(x1))))
    Matrix Interpretation Processor: dim=4
     
     interpretation:
                [1 0 0 1]     [0]
                [1 0 0 0]     [0]
      [b](x0) = [1 0 0 0]x0 + [1]
                [0 0 1 0]     [0],
      
                [1 0 0 1]     [0]
                [1 0 0 0]     [1]
      [a](x0) = [1 0 0 0]x0 + [0]
                [0 1 0 0]     [0]
     orientation:
                       [3 0 1 2]     [1]    [3 0 1 2]     [0]                 
                       [2 0 1 1]     [1]    [2 0 1 1]     [1]                 
      b(a(b(b(x1)))) = [2 0 1 1]x1 + [2] >= [2 0 1 1]x1 + [0] = a(b(a(b(x1))))
                       [1 0 1 1]     [0]    [1 0 1 1]     [0]                 
      
                       [3 1 0 2]     [1]    [3 1 0 2]     [0]                 
                       [2 1 0 1]     [2]    [2 1 0 1]     [0]                 
      a(b(a(a(x1)))) = [2 1 0 1]x1 + [1] >= [2 1 0 1]x1 + [1] = b(a(b(a(x1))))
                       [1 1 0 1]     [0]    [1 1 0 1]     [0]                 
     problem:
      
     Qed