YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(x1)) -> c#(x1)
   a#(b(x1)) -> a#(c(x1))
   a#(b(x1)) -> a#(a(c(x1)))
   a#(b(x1)) -> b#(a(a(c(x1))))
   a#(b(x1)) -> b#(b(a(a(c(x1)))))
  TRS:
   a(x1) -> x1
   a(b(x1)) -> b(b(a(a(c(x1)))))
   b(b(x1)) -> x1
   c(c(x1)) -> x1
  TDG Processor:
   DPs:
    a#(b(x1)) -> c#(x1)
    a#(b(x1)) -> a#(c(x1))
    a#(b(x1)) -> a#(a(c(x1)))
    a#(b(x1)) -> b#(a(a(c(x1))))
    a#(b(x1)) -> b#(b(a(a(c(x1)))))
   TRS:
    a(x1) -> x1
    a(b(x1)) -> b(b(a(a(c(x1)))))
    b(b(x1)) -> x1
    c(c(x1)) -> x1
   graph:
    a#(b(x1)) -> a#(c(x1)) -> a#(b(x1)) -> b#(b(a(a(c(x1)))))
    a#(b(x1)) -> a#(c(x1)) -> a#(b(x1)) -> b#(a(a(c(x1))))
    a#(b(x1)) -> a#(c(x1)) -> a#(b(x1)) -> a#(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)
    a#(b(x1)) -> a#(a(c(x1))) -> a#(b(x1)) -> b#(b(a(a(c(x1)))))
    a#(b(x1)) -> a#(a(c(x1))) -> a#(b(x1)) -> b#(a(a(c(x1))))
    a#(b(x1)) -> a#(a(c(x1))) -> a#(b(x1)) -> a#(a(c(x1)))
    a#(b(x1)) -> a#(a(c(x1))) -> a#(b(x1)) -> a#(c(x1))
    a#(b(x1)) -> a#(a(c(x1))) -> a#(b(x1)) -> c#(x1)
   SCC Processor:
    #sccs: 1
    #rules: 2
    #arcs: 10/25
    DPs:
     a#(b(x1)) -> a#(c(x1))
     a#(b(x1)) -> a#(a(c(x1)))
    TRS:
     a(x1) -> x1
     a(b(x1)) -> b(b(a(a(c(x1)))))
     b(b(x1)) -> x1
     c(c(x1)) -> x1
    Arctic Interpretation Processor:
     dimension: 2
     interpretation:
      [a#](x0) = [-& 0 ]x0 + [0],
      
                [0  0 ]  
      [c](x0) = [0  -&]x0,
      
                [0  -&]     [0]
      [b](x0) = [2  0 ]x0 + [3],
      
                [0  0 ]     [0 ]
      [a](x0) = [-& 2 ]x0 + [-&]
     orientation:
      a#(b(x1)) = [2 0]x1 + [3] >= [0  -&]x1 + [0] = a#(c(x1))
      
      a#(b(x1)) = [2 0]x1 + [3] >= [2  -&]x1 + [0] = a#(a(c(x1)))
      
              [0  0 ]     [0 ]           
      a(x1) = [-& 2 ]x1 + [-&] >= x1 = x1
      
                 [2 0]     [3]    [2 0]     [0]                    
      a(b(x1)) = [4 2]x1 + [5] >= [4 2]x1 + [3] = b(b(a(a(c(x1)))))
      
                 [0  -&]     [0]           
      b(b(x1)) = [2  0 ]x1 + [3] >= x1 = x1
      
                 [0 0]             
      c(c(x1)) = [0 0]x1 >= x1 = x1
     problem:
      DPs:
       a#(b(x1)) -> a#(a(c(x1)))
      TRS:
       a(x1) -> x1
       a(b(x1)) -> b(b(a(a(c(x1)))))
       b(b(x1)) -> x1
       c(c(x1)) -> x1
     EDG Processor:
      DPs:
       a#(b(x1)) -> a#(a(c(x1)))
      TRS:
       a(x1) -> x1
       a(b(x1)) -> b(b(a(a(c(x1)))))
       b(b(x1)) -> x1
       c(c(x1)) -> x1
      graph:
       a#(b(x1)) -> a#(a(c(x1))) -> a#(b(x1)) -> a#(a(c(x1)))
      Arctic Interpretation Processor:
       dimension: 4
       interpretation:
        [a#](x0) = [0  -2 1  0 ]x0 + [0],
        
                  [-& 0  -& -&]     [0 ]
                  [0  -1 1  1 ]     [-&]
        [c](x0) = [-& -1 -& -&]x0 + [-&]
                  [-& -1 -& -&]     [-2],
        
                  [-& 0  -& -&]     [1 ]
                  [0  -& 0  -&]     [0 ]
        [b](x0) = [-& 0  -& -&]x0 + [0 ]
                  [-1 0  0  0 ]     [-&],
        
                  [0  -& 1  -&]     [0 ]
                  [0  0  0  1 ]     [0 ]
        [a](x0) = [-2 -& 0  0 ]x0 + [-1]
                  [0  -& 0  1 ]     [0 ]
       orientation:
        a#(b(x1)) = [-1 1  0  0 ]x1 + [1] >= [-2 0  -1 -1]x1 + [0] = a#(a(c(x1)))
        
                [0  -& 1  -&]     [0 ]           
                [0  0  0  1 ]     [0 ]           
        a(x1) = [-2 -& 0  0 ]x1 + [-1] >= x1 = x1
                [0  -& 0  1 ]     [0 ]           
        
                   [-& 1  -& -&]     [1]    [-& 0  -& -&]     [1]                    
                   [0  1  1  1 ]     [1]    [0  1  1  1 ]     [1]                    
        a(b(x1)) = [-1 0  0  0 ]x1 + [0] >= [-& 0  -& -&]x1 + [0] = b(b(a(a(c(x1)))))
                   [0  1  1  1 ]     [1]    [0  1  1  1 ]     [1]                    
        
                   [0  -& 0  -&]     [1]           
                   [-& 0  -& -&]     [1]           
        b(b(x1)) = [0  -& 0  -&]x1 + [0] >= x1 = x1
                   [0  0  0  0 ]     [0]           
        
                   [0  -1 1  1 ]     [0 ]           
                   [-1 0  0  0 ]     [0 ]           
        c(c(x1)) = [-1 -2 0  0 ]x1 + [-&] >= x1 = x1
                   [-1 -2 0  0 ]     [-2]           
       problem:
        DPs:
         
        TRS:
         a(x1) -> x1
         a(b(x1)) -> b(b(a(a(c(x1)))))
         b(b(x1)) -> x1
         c(c(x1)) -> x1
       Qed