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
    interpretation:
     [a#](x0) = [2 0]x0 + [0],
     
               [0  -&]     [1]
     [b](x0) = [-& -&]x0 + [0],
     
               [0 1]     [0]
     [c](x0) = [0 0]x0 + [1],
     
               [0  -&]     [1]
     [a](x0) = [0  0 ]x0 + [1]
    orientation:
     a#(c(c(x1))) = [3 3]x1 + [4] >= [2 0]x1 + [0] = a#(x1)
     
     a#(c(c(x1))) = [3 3]x1 + [4] >= [2 0]x1 + [3] = a#(a(x1))
     
             [0  -&]     [1]           
     a(x1) = [0  0 ]x1 + [1] >= x1 = x1
     
             [0  -&]     [1]    [0  -&]     [1]              
     a(x1) = [0  0 ]x1 + [1] >= [0  -&]x1 + [1] = c(b(b(x1)))
     
                   [1 1]     [2]    [1 1]     [2]                 
     a(c(c(x1))) = [1 1]x1 + [2] >= [1 1]x1 + [2] = 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