YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(x1)) -> c#(x1)
   a#(b(x1)) -> a#(c(x1))
   a#(b(x1)) -> b#(a(c(x1)))
   a#(b(x1)) -> b#(b(a(c(x1))))
   c#(c(x1)) -> a#(x1)
  TRS:
   a(x1) -> x1
   a(b(x1)) -> b(b(a(c(x1))))
   b(b(x1)) -> x1
   c(c(x1)) -> a(x1)
  TDG Processor:
   DPs:
    a#(b(x1)) -> c#(x1)
    a#(b(x1)) -> a#(c(x1))
    a#(b(x1)) -> b#(a(c(x1)))
    a#(b(x1)) -> b#(b(a(c(x1))))
    c#(c(x1)) -> a#(x1)
   TRS:
    a(x1) -> x1
    a(b(x1)) -> b(b(a(c(x1))))
    b(b(x1)) -> x1
    c(c(x1)) -> a(x1)
   graph:
    c#(c(x1)) -> a#(x1) -> a#(b(x1)) -> b#(b(a(c(x1))))
    c#(c(x1)) -> a#(x1) -> a#(b(x1)) -> b#(a(c(x1)))
    c#(c(x1)) -> a#(x1) -> a#(b(x1)) -> a#(c(x1))
    c#(c(x1)) -> a#(x1) -> a#(b(x1)) -> c#(x1)
    a#(b(x1)) -> c#(x1) -> c#(c(x1)) -> a#(x1)
    a#(b(x1)) -> a#(c(x1)) -> a#(b(x1)) -> b#(b(a(c(x1))))
    a#(b(x1)) -> a#(c(x1)) -> a#(b(x1)) -> b#(a(c(x1)))
    a#(b(x1)) -> a#(c(x1)) -> a#(b(x1)) -> a#(c(x1))
    a#(b(x1)) -> a#(c(x1)) -> a#(b(x1)) -> c#(x1)
   SCC Processor:
    #sccs: 1
    #rules: 3
    #arcs: 9/25
    DPs:
     c#(c(x1)) -> a#(x1)
     a#(b(x1)) -> c#(x1)
     a#(b(x1)) -> a#(c(x1))
    TRS:
     a(x1) -> x1
     a(b(x1)) -> b(b(a(c(x1))))
     b(b(x1)) -> x1
     c(c(x1)) -> a(x1)
    Arctic Interpretation Processor:
     dimension: 2
     interpretation:
      [c#](x0) = [0 0]x0 + [1],
      
      [a#](x0) = [-& 0 ]x0 + [0],
      
                [0  1 ]     [0]
      [c](x0) = [0  -&]x0 + [0],
      
                [-& 0 ]     [0]
      [b](x0) = [0  0 ]x0 + [2],
      
                [0  1 ]     [-&]
      [a](x0) = [-& 1 ]x0 + [0 ]
     orientation:
      c#(c(x1)) = [0 1]x1 + [1] >= [-& 0 ]x1 + [0] = a#(x1)
      
      a#(b(x1)) = [0 0]x1 + [2] >= [0 0]x1 + [1] = c#(x1)
      
      a#(b(x1)) = [0 0]x1 + [2] >= [0  -&]x1 + [0] = a#(c(x1))
      
              [0  1 ]     [-&]           
      a(x1) = [-& 1 ]x1 + [0 ] >= x1 = x1
      
                 [1 1]     [3]    [1 1]     [2]                 
      a(b(x1)) = [1 1]x1 + [3] >= [1 1]x1 + [2] = b(b(a(c(x1))))
      
                 [0 0]     [2]           
      b(b(x1)) = [0 0]x1 + [2] >= x1 = x1
      
                 [1 1]     [1]    [0  1 ]     [-&]        
      c(c(x1)) = [0 1]x1 + [0] >= [-& 1 ]x1 + [0 ] = a(x1)
     problem:
      DPs:
       a#(b(x1)) -> c#(x1)
       a#(b(x1)) -> a#(c(x1))
      TRS:
       a(x1) -> x1
       a(b(x1)) -> b(b(a(c(x1))))
       b(b(x1)) -> x1
       c(c(x1)) -> a(x1)
     EDG Processor:
      DPs:
       a#(b(x1)) -> c#(x1)
       a#(b(x1)) -> a#(c(x1))
      TRS:
       a(x1) -> x1
       a(b(x1)) -> b(b(a(c(x1))))
       b(b(x1)) -> x1
       c(c(x1)) -> a(x1)
      graph:
       a#(b(x1)) -> a#(c(x1)) -> a#(b(x1)) -> c#(x1)
       a#(b(x1)) -> a#(c(x1)) -> a#(b(x1)) -> a#(c(x1))
      SCC Processor:
       #sccs: 1
       #rules: 1
       #arcs: 2/4
       DPs:
        a#(b(x1)) -> a#(c(x1))
       TRS:
        a(x1) -> x1
        a(b(x1)) -> b(b(a(c(x1))))
        b(b(x1)) -> x1
        c(c(x1)) -> a(x1)
       Arctic Interpretation Processor:
        dimension: 3
        interpretation:
         [a#](x0) = [0  -& -1]x0 + [0],
         
                   [-& 0  -2]     [0 ]
         [c](x0) = [0  1  0 ]x0 + [-&]
                   [-& 1  -1]     [1 ],
         
                   [-1 1  -&]     [1 ]
         [b](x0) = [-1 -& -2]x0 + [-&]
                   [0  -& 0 ]     [-&],
         
                   [0  -2 0 ]  
         [a](x0) = [1  0  0 ]x0
                   [1  -& 1 ]  
        orientation:
         a#(b(x1)) = [-1 1  -1]x1 + [1] >= [-& 0  -2]x1 + [0] = a#(c(x1))
         
                 [0  -2 0 ]             
         a(x1) = [1  0  0 ]x1 >= x1 = x1
                 [1  -& 1 ]             
         
                    [0 1 0]     [1]    [0 1 0]     [1]                 
         a(b(x1)) = [0 2 0]x1 + [2] >= [0 1 0]x1 + [1] = b(b(a(c(x1))))
                    [1 2 1]     [2]    [1 2 1]     [2]                 
         
                    [0  0  -1]     [1]           
         b(b(x1)) = [-2 0  -2]x1 + [0] >= x1 = x1
                    [0  1  0 ]     [1]           
         
                    [0 1 0]     [0]    [0  -2 0 ]          
         c(c(x1)) = [1 2 1]x1 + [1] >= [1  0  0 ]x1 = a(x1)
                    [1 2 1]     [1]    [1  -& 1 ]          
        problem:
         DPs:
          
         TRS:
          a(x1) -> x1
          a(b(x1)) -> b(b(a(c(x1))))
          b(b(x1)) -> x1
          c(c(x1)) -> a(x1)
        Qed