YES

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  b(a(x1)) -> a(c(b(c(b(c(x1))))))
  c(c(x1)) -> a(x1)
  DP Processor:
   DPs:
    b#(a(x1)) -> c#(x1)
    b#(a(x1)) -> b#(c(x1))
    b#(a(x1)) -> c#(b(c(x1)))
    b#(a(x1)) -> b#(c(b(c(x1))))
    b#(a(x1)) -> c#(b(c(b(c(x1)))))
    b#(a(x1)) -> a#(c(b(c(b(c(x1))))))
    c#(c(x1)) -> a#(x1)
   TRS:
    a(x1) -> x1
    b(a(x1)) -> a(c(b(c(b(c(x1))))))
    c(c(x1)) -> a(x1)
   TDG Processor:
    DPs:
     b#(a(x1)) -> c#(x1)
     b#(a(x1)) -> b#(c(x1))
     b#(a(x1)) -> c#(b(c(x1)))
     b#(a(x1)) -> b#(c(b(c(x1))))
     b#(a(x1)) -> c#(b(c(b(c(x1)))))
     b#(a(x1)) -> a#(c(b(c(b(c(x1))))))
     c#(c(x1)) -> a#(x1)
    TRS:
     a(x1) -> x1
     b(a(x1)) -> a(c(b(c(b(c(x1))))))
     c(c(x1)) -> a(x1)
    graph:
     b#(a(x1)) -> c#(b(c(b(c(x1))))) -> c#(c(x1)) -> a#(x1)
     b#(a(x1)) -> c#(b(c(x1))) -> c#(c(x1)) -> a#(x1)
     b#(a(x1)) -> c#(x1) -> c#(c(x1)) -> a#(x1)
     b#(a(x1)) -> b#(c(b(c(x1)))) -> b#(a(x1)) -> a#(c(b(c(b(c(x1))))))
     b#(a(x1)) -> b#(c(b(c(x1)))) -> b#(a(x1)) -> c#(b(c(b(c(x1)))))
     b#(a(x1)) -> b#(c(b(c(x1)))) -> b#(a(x1)) -> b#(c(b(c(x1))))
     b#(a(x1)) -> b#(c(b(c(x1)))) -> b#(a(x1)) -> c#(b(c(x1)))
     b#(a(x1)) -> b#(c(b(c(x1)))) -> b#(a(x1)) -> b#(c(x1))
     b#(a(x1)) -> b#(c(b(c(x1)))) -> b#(a(x1)) -> c#(x1)
     b#(a(x1)) -> b#(c(x1)) -> b#(a(x1)) -> a#(c(b(c(b(c(x1))))))
     b#(a(x1)) -> b#(c(x1)) -> b#(a(x1)) -> c#(b(c(b(c(x1)))))
     b#(a(x1)) -> b#(c(x1)) -> b#(a(x1)) -> b#(c(b(c(x1))))
     b#(a(x1)) -> b#(c(x1)) -> b#(a(x1)) -> c#(b(c(x1)))
     b#(a(x1)) -> b#(c(x1)) -> b#(a(x1)) -> b#(c(x1))
     b#(a(x1)) -> b#(c(x1)) -> b#(a(x1)) -> c#(x1)
    SCC Processor:
     #sccs: 1
     #rules: 2
     #arcs: 15/49
     DPs:
      b#(a(x1)) -> b#(c(b(c(x1))))
      b#(a(x1)) -> b#(c(x1))
     TRS:
      a(x1) -> x1
      b(a(x1)) -> a(c(b(c(b(c(x1))))))
      c(c(x1)) -> a(x1)
     Arctic Interpretation Processor:
      dimension: 2
      usable rules:
       a(x1) -> x1
       b(a(x1)) -> a(c(b(c(b(c(x1))))))
       c(c(x1)) -> a(x1)
      interpretation:
       [b#](x0) = [0 3]x0 + [0],
       
                 [-& 0 ]     [0]
       [b](x0) = [0  1 ]x0 + [0],
       
                 [0  -&]     [-&]
       [a](x0) = [1  0 ]x0 + [2 ],
       
                 [1  0 ]     [2 ]
       [c](x0) = [0  -&]x0 + [-&]
      orientation:
       b#(a(x1)) = [4 3]x1 + [5] >= [3 0]x1 + [3] = b#(c(b(c(x1))))
       
       b#(a(x1)) = [4 3]x1 + [5] >= [3 0]x1 + [2] = b#(c(x1))
       
               [0  -&]     [-&]           
       a(x1) = [1  0 ]x1 + [2 ] >= x1 = x1
       
                  [1 0]     [2]    [1 0]     [2]                       
       b(a(x1)) = [2 1]x1 + [3] >= [2 1]x1 + [3] = a(c(b(c(b(c(x1))))))
       
                  [2 1]     [3]    [0  -&]     [-&]        
       c(c(x1)) = [1 0]x1 + [2] >= [1  0 ]x1 + [2 ] = a(x1)
      problem:
       DPs:
        
       TRS:
        a(x1) -> x1
        b(a(x1)) -> a(c(b(c(b(c(x1))))))
        c(c(x1)) -> a(x1)
      Qed