YES

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

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