YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(x1)) -> c#(b(x1))
   a#(b(x1)) -> b#(c(b(x1)))
   a#(b(x1)) -> c#(b(c(b(x1))))
   b#(x1) -> a#(x1)
   b#(x1) -> a#(a(x1))
  TRS:
   a(x1) -> x1
   a(b(x1)) -> c(b(c(b(x1))))
   b(x1) -> a(a(x1))
   c(c(x1)) -> x1
  TDG Processor:
   DPs:
    a#(b(x1)) -> c#(b(x1))
    a#(b(x1)) -> b#(c(b(x1)))
    a#(b(x1)) -> c#(b(c(b(x1))))
    b#(x1) -> a#(x1)
    b#(x1) -> a#(a(x1))
   TRS:
    a(x1) -> x1
    a(b(x1)) -> c(b(c(b(x1))))
    b(x1) -> a(a(x1))
    c(c(x1)) -> x1
   graph:
    b#(x1) -> a#(a(x1)) -> a#(b(x1)) -> c#(b(c(b(x1))))
    b#(x1) -> a#(a(x1)) -> a#(b(x1)) -> b#(c(b(x1)))
    b#(x1) -> a#(a(x1)) -> a#(b(x1)) -> c#(b(x1))
    b#(x1) -> a#(x1) -> a#(b(x1)) -> c#(b(c(b(x1))))
    b#(x1) -> a#(x1) -> a#(b(x1)) -> b#(c(b(x1)))
    b#(x1) -> a#(x1) -> a#(b(x1)) -> c#(b(x1))
    a#(b(x1)) -> b#(c(b(x1))) -> b#(x1) -> a#(a(x1))
    a#(b(x1)) -> b#(c(b(x1))) -> b#(x1) -> a#(x1)
   SCC Processor:
    #sccs: 1
    #rules: 3
    #arcs: 8/25
    DPs:
     b#(x1) -> a#(a(x1))
     a#(b(x1)) -> b#(c(b(x1)))
     b#(x1) -> a#(x1)
    TRS:
     a(x1) -> x1
     a(b(x1)) -> c(b(c(b(x1))))
     b(x1) -> a(a(x1))
     c(c(x1)) -> x1
    Arctic Interpretation Processor:
     dimension: 2
     interpretation:
      [b#](x0) = [-3 0 ]x0 + [1],
      
      [a#](x0) = [-4 -1]x0 + [0],
      
                [-& 2 ]     [0]
      [c](x0) = [-2 -1]x0 + [0],
      
                [0  3 ]     [0]
      [b](x0) = [-1 2 ]x0 + [2],
      
                [0  3 ]     [-4]
      [a](x0) = [-3 0 ]x0 + [-&]
     orientation:
      b#(x1) = [-3 0 ]x1 + [1] >= [-4 -1]x1 + [0] = a#(a(x1))
      
      a#(b(x1)) = [-2 1 ]x1 + [1] >= [-2 1 ]x1 + [1] = b#(c(b(x1)))
      
      b#(x1) = [-3 0 ]x1 + [1] >= [-4 -1]x1 + [0] = a#(x1)
      
              [0  3 ]     [-4]           
      a(x1) = [-3 0 ]x1 + [-&] >= x1 = x1
      
                 [2  5 ]     [5]    [2  5 ]     [5]                 
      a(b(x1)) = [-1 2 ]x1 + [2] >= [-1 2 ]x1 + [2] = c(b(c(b(x1))))
      
              [0  3 ]     [0]    [0  3 ]     [-4]           
      b(x1) = [-1 2 ]x1 + [2] >= [-3 0 ]x1 + [-7] = a(a(x1))
      
                 [0  1 ]     [2]           
      c(c(x1)) = [-3 0 ]x1 + [0] >= x1 = x1
     problem:
      DPs:
       a#(b(x1)) -> b#(c(b(x1)))
      TRS:
       a(x1) -> x1
       a(b(x1)) -> c(b(c(b(x1))))
       b(x1) -> a(a(x1))
       c(c(x1)) -> x1
     EDG Processor:
      DPs:
       a#(b(x1)) -> b#(c(b(x1)))
      TRS:
       a(x1) -> x1
       a(b(x1)) -> c(b(c(b(x1))))
       b(x1) -> a(a(x1))
       c(c(x1)) -> x1
      graph:
       
      Qed