YES

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

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