YES

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

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