YES

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