YES

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

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