YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 3
  interpretation:
             [0 0 0]  
   [c](x0) = [0 0 0]x0
             [0 0 0]  ,
   
             [0  -& -&]  
   [b](x0) = [0  -& 2 ]x0
             [-& -& -&]  ,
   
             [1 0 0]  
   [a](x0) = [1 0 3]x0
             [0 0 0]  
  orientation:
           [1 0 0]             
   a(x1) = [1 0 3]x1 >= x1 = x1
           [0 0 0]             
   
           [1 0 0]      [0  -& -&]          
   a(x1) = [1 0 3]x1 >= [0  -& 2 ]x1 = b(x1)
           [0 0 0]      [-& -& -&]          
   
                 [2 2 2]      [2 2 2]                      
   a(b(c(x1))) = [2 2 2]x1 >= [2 2 2]x1 = c(c(a(b(a(x1)))))
                 [2 2 2]      [2 2 2]                      
   
           [0 0 0]             
   c(x1) = [0 0 0]x1 >= x1 = x1
           [0 0 0]             
  problem:
   a(x1) -> x1
   a(b(c(x1))) -> c(c(a(b(a(x1)))))
   c(x1) -> x1
  String Reversal Processor:
   a(x1) -> x1
   c(b(a(x1))) -> a(b(a(c(c(x1)))))
   c(x1) -> x1
   Matrix Interpretation Processor: dim=2
    
    interpretation:
               [1 2]  
     [c](x0) = [0 1]x0,
     
               [1 1]     [0]
     [b](x0) = [0 2]x0 + [2],
     
                    [1]
     [a](x0) = x0 + [0]
    orientation:
                  [1]           
     a(x1) = x1 + [0] >= x1 = x1
     
                   [1 5]     [5]    [1 5]     [2]                    
     c(b(a(x1))) = [0 2]x1 + [2] >= [0 2]x1 + [2] = a(b(a(c(c(x1)))))
     
             [1 2]             
     c(x1) = [0 1]x1 >= x1 = x1
    problem:
     c(x1) -> x1
    Arctic Interpretation Processor:
     dimension: 3
     interpretation:
                [2 0 0]  
      [c](x0) = [0 2 0]x0
                [0 0 1]  
     orientation:
              [2 0 0]             
      c(x1) = [0 2 0]x1 >= x1 = x1
              [0 0 1]             
     problem:
      
     Qed