YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(x1)) -> c#(x1)
   c#(c(x1)) -> a#(c(x1))
   c#(c(x1)) -> a#(a(c(x1)))
   c#(c(x1)) -> b#(a(a(c(x1))))
   c#(c(x1)) -> b#(b(a(a(c(x1)))))
  TRS:
   a(x1) -> x1
   a(b(x1)) -> c(x1)
   b(x1) -> x1
   c(c(x1)) -> b(b(a(a(c(x1)))))
  TDG Processor:
   DPs:
    a#(b(x1)) -> c#(x1)
    c#(c(x1)) -> a#(c(x1))
    c#(c(x1)) -> a#(a(c(x1)))
    c#(c(x1)) -> b#(a(a(c(x1))))
    c#(c(x1)) -> b#(b(a(a(c(x1)))))
   TRS:
    a(x1) -> x1
    a(b(x1)) -> c(x1)
    b(x1) -> x1
    c(c(x1)) -> b(b(a(a(c(x1)))))
   graph:
    c#(c(x1)) -> a#(c(x1)) -> a#(b(x1)) -> c#(x1)
    c#(c(x1)) -> a#(a(c(x1))) -> a#(b(x1)) -> c#(x1)
    a#(b(x1)) -> c#(x1) -> c#(c(x1)) -> b#(b(a(a(c(x1)))))
    a#(b(x1)) -> c#(x1) -> c#(c(x1)) -> b#(a(a(c(x1))))
    a#(b(x1)) -> c#(x1) -> c#(c(x1)) -> a#(a(c(x1)))
    a#(b(x1)) -> c#(x1) -> c#(c(x1)) -> a#(c(x1))
   SCC Processor:
    #sccs: 1
    #rules: 3
    #arcs: 6/25
    DPs:
     c#(c(x1)) -> a#(c(x1))
     a#(b(x1)) -> c#(x1)
     c#(c(x1)) -> a#(a(c(x1)))
    TRS:
     a(x1) -> x1
     a(b(x1)) -> c(x1)
     b(x1) -> x1
     c(c(x1)) -> b(b(a(a(c(x1)))))
    Arctic Interpretation Processor:
     dimension: 2
     interpretation:
      [c#](x0) = [0  -&]x0 + [1],
      
      [a#](x0) = [-& 2 ]x0 + [0],
      
                [3  0 ]     [0 ]
      [c](x0) = [0  -&]x0 + [-&],
      
                [0 2]     [0]
      [b](x0) = [0 0]x0 + [0],
      
                [0  3 ]     [0 ]
      [a](x0) = [-& 0 ]x0 + [-&]
     orientation:
      c#(c(x1)) = [3 0]x1 + [1] >= [2  -&]x1 + [0] = a#(c(x1))
      
      a#(b(x1)) = [2 2]x1 + [2] >= [0  -&]x1 + [1] = c#(x1)
      
      c#(c(x1)) = [3 0]x1 + [1] >= [2  -&]x1 + [0] = a#(a(c(x1)))
      
              [0  3 ]     [0 ]           
      a(x1) = [-& 0 ]x1 + [-&] >= x1 = x1
      
                 [3 3]     [3]    [3  0 ]     [0 ]        
      a(b(x1)) = [0 0]x1 + [0] >= [0  -&]x1 + [-&] = c(x1)
      
              [0 2]     [0]           
      b(x1) = [0 0]x1 + [0] >= x1 = x1
      
                 [6 3]     [3]    [5 2]     [2]                    
      c(c(x1)) = [3 0]x1 + [0] >= [3 0]x1 + [0] = b(b(a(a(c(x1)))))
     problem:
      DPs:
       
      TRS:
       a(x1) -> x1
       a(b(x1)) -> c(x1)
       b(x1) -> x1
       c(c(x1)) -> b(b(a(a(c(x1)))))
     Qed