YES

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

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