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