YES

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