YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 2
  interpretation:
             [0  1 ]  
   [c](x0) = [-& 0 ]x0,
   
             [2 0]  
   [b](x0) = [3 2]x0
  orientation:
              [4 2]      [0  1 ]                   
   b(b(x1)) = [5 4]x1 >= [-& 0 ]x1 = c(c(c(c(x1))))
   
           [0  1 ]             
   c(x1) = [-& 0 ]x1 >= x1 = x1
   
                 [6 5]      [6 4]                
   b(c(b(x1))) = [7 6]x1 >= [7 6]x1 = b(b(b(x1)))
  problem:
   c(x1) -> x1
   b(c(b(x1))) -> b(b(b(x1)))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [1  -&]  
    [c](x0) = [0  3 ]x0,
    
                
    [b](x0) = x0
   orientation:
            [1  -&]             
    c(x1) = [0  3 ]x1 >= x1 = x1
    
                  [1  -&]                      
    b(c(b(x1))) = [0  3 ]x1 >= x1 = b(b(b(x1)))
   problem:
    
   Qed