YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(b(x1))) -> a#(x1)
   a#(b(b(x1))) -> a#(a(x1))
  TRS:
   a(b(a(x1))) -> b(b(b(b(x1))))
   a(b(b(x1))) -> b(b(a(a(x1))))
  Arctic Interpretation Processor:
   dimension: 2
   usable rules:
    a(b(a(x1))) -> b(b(b(b(x1))))
    a(b(b(x1))) -> b(b(a(a(x1))))
   interpretation:
              [-& 0 ]     [1]
    [b](x0) = [1  -&]x0 + [1],
    
              [0  -&]     [-&]
    [a](x0) = [2  0 ]x0 + [2 ],
    
    [a#](x0) = [1 0]x0
   orientation:
    a#(b(b(x1))) = [2 1]x1 + [2] >= [1 0]x1 = a#(x1)
    
    a#(b(b(x1))) = [2 1]x1 + [2] >= [2 0]x1 + [2] = a#(a(x1))
    
                  [2 0]     [2]    [2  -&]     [2]                 
    a(b(a(x1))) = [4 2]x1 + [4] >= [-& 2 ]x1 + [3] = b(b(b(b(x1))))
    
                  [1  -&]     [1]    [1  -&]     [1]                 
    a(b(b(x1))) = [3  1 ]x1 + [3] >= [3  1 ]x1 + [3] = b(b(a(a(x1))))
   problem:
    DPs:
     a#(b(b(x1))) -> a#(a(x1))
    TRS:
     a(b(a(x1))) -> b(b(b(b(x1))))
     a(b(b(x1))) -> b(b(a(a(x1))))
   Restore Modifier:
    DPs:
     a#(b(b(x1))) -> a#(a(x1))
    TRS:
     a(b(a(x1))) -> b(b(b(b(x1))))
     a(b(b(x1))) -> b(b(a(a(x1))))
    Arctic Interpretation Processor:
     dimension: 2
     usable rules:
      a(b(a(x1))) -> b(b(b(b(x1))))
      a(b(b(x1))) -> b(b(a(a(x1))))
     interpretation:
                [-& 1 ]     [0 ]
      [b](x0) = [0  -&]x0 + [-4],
      
                [-1 3 ]     [2 ]
      [a](x0) = [-& 0 ]x0 + [-2],
      
      [a#](x0) = [-3 3 ]x0 + [2]
     orientation:
      a#(b(b(x1))) = [-2 4 ]x1 + [3] >= [-4 3 ]x1 + [2] = a#(a(x1))
      
                    [2  6 ]     [5]    [2  -&]     [1]                 
      a(b(a(x1))) = [-1 3 ]x1 + [2] >= [-& 2 ]x1 + [1] = b(b(b(b(x1))))
      
                    [0  4 ]     [3]    [-1 4 ]     [3]                 
      a(b(b(x1))) = [-& 1 ]x1 + [0] >= [-& 1 ]x1 + [0] = b(b(a(a(x1))))
     problem:
      DPs:
       
      TRS:
       a(b(a(x1))) -> b(b(b(b(x1))))
       a(b(b(x1))) -> b(b(a(a(x1))))
     Qed