YES

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

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