YES

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

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