YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 3
  interpretation:
             [0  0  -&]  
   [a](x0) = [-& -& 0 ]x0
             [-& 0  -&]  ,
   
             [0  0  1 ]  
   [b](x0) = [0  0  1 ]x0
             [-& -& 0 ]  
  orientation:
                 [0  0  1 ]      [0  0  1 ]                   
   a(a(b(x1))) = [0  0  1 ]x1 >= [0  0  1 ]x1 = b(b(a(a(x1))))
                 [-& -& 0 ]      [-& -& 0 ]                   
   
                 [1 1 2]      [0  0  0 ]                   
   b(a(b(x1))) = [1 1 2]x1 >= [-& 0  -&]x1 = a(a(a(a(x1))))
                 [0 0 1]      [-& -& 0 ]                   
  problem:
   a(a(b(x1))) -> b(b(a(a(x1))))
  String Reversal Processor:
   b(a(a(x1))) -> a(a(b(b(x1))))
   Matrix Interpretation Processor: dim=3
    
    interpretation:
               [1 0 0]     [0]
     [a](x0) = [0 0 1]x0 + [0]
               [0 1 1]     [1],
     
               [1 0 1]     [1]
     [b](x0) = [0 1 0]x0 + [0]
               [0 0 1]     [0]
    orientation:
                   [1 1 2]     [3]    [1 0 2]     [2]                 
     b(a(a(x1))) = [0 1 1]x1 + [1] >= [0 1 1]x1 + [1] = a(a(b(b(x1))))
                   [0 1 2]     [2]    [0 1 2]     [2]                 
    problem:
     
    Qed