YES

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

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