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
     interpretation:
      [c#](x0) = [2 1]x0,
      
                [0 0]     [0]
      [c](x0) = [0 0]x0 + [1],
      
                [0 0]     [0]
      [b](x0) = [2 2]x0 + [3],
      
                [0 0]     [0]
      [a](x0) = [2 0]x0 + [3]
     orientation:
      c#(b(x1)) = [3 3]x1 + [4] >= [2 2]x1 + [2] = c#(c(x1))
      
      c#(b(x1)) = [3 3]x1 + [4] >= [2 1]x1 = c#(x1)
      
              [0 0]     [0]           
      a(x1) = [2 0]x1 + [3] >= x1 = x1
      
                 [2 0]     [3]    [0 0]     [0]        
      a(a(x1)) = [2 2]x1 + [3] >= [2 2]x1 + [3] = b(x1)
      
                 [0 0]     [1]           
      b(c(x1)) = [2 2]x1 + [3] >= x1 = x1
      
                 [2 2]     [3]    [2 2]     [3]                 
      c(b(x1)) = [2 2]x1 + [3] >= [2 2]x1 + [3] = 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