YES

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

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