YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(b(b(x1)))) -> a#(x1)
   a#(b(b(b(x1)))) -> a#(b(a(x1)))
  TRS:
   a(a(x1)) -> x1
   a(b(b(b(x1)))) -> b(b(b(a(b(a(x1))))))
  Arctic Interpretation Processor:
   dimension: 3
   interpretation:
    [a#](x0) = [0 0 0]x0 + [0],
    
              [-& -& 0 ]     [0 ]
    [b](x0) = [1  -& 0 ]x0 + [-&]
              [0  0  -&]     [0 ],
    
              [-& -& 0 ]     [0]
    [a](x0) = [-& -& 0 ]x0 + [0]
              [0  0  1 ]     [1]
   orientation:
    a#(b(b(b(x1)))) = [1 1 1]x1 + [1] >= [0 0 0]x1 + [0] = a#(x1)
    
    a#(b(b(b(x1)))) = [1 1 1]x1 + [1] >= [0 0 1]x1 + [1] = a#(b(a(x1)))
    
               [0 0 1]     [1]           
    a(a(x1)) = [0 0 1]x1 + [1] >= x1 = x1
               [1 1 2]     [2]           
    
                     [0 0 1]     [1]    [0 0 1]     [1]                       
    a(b(b(b(x1)))) = [0 0 1]x1 + [1] >= [0 0 1]x1 + [1] = b(b(b(a(b(a(x1))))))
                     [1 1 2]     [2]    [1 1 2]     [2]                       
   problem:
    DPs:
     a#(b(b(b(x1)))) -> a#(b(a(x1)))
    TRS:
     a(a(x1)) -> x1
     a(b(b(b(x1)))) -> b(b(b(a(b(a(x1))))))
   Arctic Interpretation Processor:
    dimension: 3
    interpretation:
     [a#](x0) = [-& -& 0 ]x0 + [0],
     
               [-& 1  0 ]     [1 ]
     [b](x0) = [-& -& 0 ]x0 + [-&]
               [0  0  -&]     [0 ],
     
               [-& -& 0 ]     [0 ]
     [a](x0) = [-& -& 0 ]x0 + [-&]
               [0  0  1 ]     [0 ]
    orientation:
     a#(b(b(b(x1)))) = [0 0 1]x1 + [1] >= [-& -& 0 ]x1 + [0] = a#(b(a(x1)))
     
                [0 0 1]     [0]           
     a(a(x1)) = [0 0 1]x1 + [0] >= x1 = x1
                [1 1 2]     [1]           
     
                      [0 0 1]     [1]    [0 0 1]     [1]                       
     a(b(b(b(x1)))) = [0 0 1]x1 + [1] >= [0 0 1]x1 + [1] = b(b(b(a(b(a(x1))))))
                      [1 1 2]     [2]    [1 1 2]     [2]                       
    problem:
     DPs:
      
     TRS:
      a(a(x1)) -> x1
      a(b(b(b(x1)))) -> b(b(b(a(b(a(x1))))))
    Qed