YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 3
  interpretation:
             [0  0  0 ]  
   [c](x0) = [-& -& 0 ]x0
             [-& 0  -&]  ,
   
             [1  -& 0 ]  
   [b](x0) = [0  -& -&]x0
             [1  -& 0 ]  ,
   
             [0  0  1 ]  
   [a](x0) = [-& 0  1 ]x0
             [-& -& 0 ]  
  orientation:
           [0  0  1 ]             
   a(x1) = [-& 0  1 ]x1 >= x1 = x1
           [-& -& 0 ]             
   
              [2  -& 1 ]      [1  -& 0 ]             
   a(b(x1)) = [2  -& 1 ]x1 >= [1  -& 0 ]x1 = c(b(x1))
              [1  -& 0 ]      [0  -& -&]             
   
                 [0  0  1 ]      [0  0  1 ]                   
   a(c(c(x1))) = [-& 0  1 ]x1 >= [-& 0  1 ]x1 = c(c(a(a(x1))))
                 [-& -& 0 ]      [-& -& 0 ]                   
  problem:
   a(x1) -> x1
   a(c(c(x1))) -> c(c(a(a(x1))))
  KBO Processor:
   weight function:
    w0 = 1
    w(c) = 1
    w(a) = 0
   precedence:
    a > c
   problem:
    
   Qed