YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 2
  interpretation:
             [0  0 ]  
   [a](x0) = [-& -&]x0,
   
             [0 0]  
   [b](x0) = [0 0]x0,
   
             [0 1]  
   [c](x0) = [0 0]x0
  orientation:
              [1 1]      [0  0 ]                         
   c(c(x1)) = [0 1]x1 >= [-& -&]x1 = a(a(a(b(b(b(x1))))))
   
                    [0 0]      [0 0]                               
   b(b(b(a(x1)))) = [0 0]x1 >= [0 0]x1 = b(b(b(b(b(b(b(b(x1))))))))
   
                    [1 1]      [1 1]                            
   b(b(c(c(x1)))) = [1 1]x1 >= [1 1]x1 = c(c(c(a(a(a(a(x1)))))))
  problem:
   b(b(b(a(x1)))) -> b(b(b(b(b(b(b(b(x1))))))))
   b(b(c(c(x1)))) -> c(c(c(a(a(a(a(x1)))))))
  String Reversal Processor:
   a(b(b(b(x1)))) -> b(b(b(b(b(b(b(b(x1))))))))
   c(c(b(b(x1)))) -> a(a(a(a(c(c(c(x1)))))))
   Bounds Processor:
    bound: 0
    enrichment: match
    automaton:
     final states: {10,1}
     transitions:
      f30() -> 2*
      b0(5) -> 6*
      b0(7) -> 8*
      b0(2) -> 3*
      b0(9) -> 1*
      b0(4) -> 5*
      b0(6) -> 7*
      b0(8) -> 9*
      b0(3) -> 4*
      a0(15) -> 16*
      a0(14) -> 15*
      a0(16) -> 10*
      a0(13) -> 14*
      c0(12) -> 13*
      c0(2) -> 11*
      c0(11) -> 12*
      10 -> 11,12
    problem:
     
    Qed