YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [B](x0) = x0,
   
   [b](x0) = x0,
   
   [c](x0) = 2x0,
   
   [C](x0) = 2x0
  orientation:
   C(x1) = 2x1 >= 2x1 = c(x1)
   
   c(c(x1)) = 4x1 >= x1 = x1
   
   b(b(x1)) = x1 >= x1 = B(x1)
   
   B(B(x1)) = x1 >= x1 = b(x1)
   
   c(B(c(b(c(x1))))) = 6x1 >= 6x1 = B(c(b(c(B(c(b(x1)))))))
   
   b(B(x1)) = x1 >= x1 = x1
   
   B(b(x1)) = x1 >= x1 = x1
   
   c(C(x1)) = 4x1 >= x1 = x1
   
   C(c(x1)) = 4x1 >= x1 = x1
  problem:
   C(x1) -> c(x1)
   b(b(x1)) -> B(x1)
   B(B(x1)) -> b(x1)
   c(B(c(b(c(x1))))) -> B(c(b(c(B(c(b(x1)))))))
   b(B(x1)) -> x1
   B(b(x1)) -> x1
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [B](x0) = x0,
    
    [b](x0) = x0,
    
    [c](x0) = x0,
    
    [C](x0) = 8x0
   orientation:
    C(x1) = 8x1 >= x1 = c(x1)
    
    b(b(x1)) = x1 >= x1 = B(x1)
    
    B(B(x1)) = x1 >= x1 = b(x1)
    
    c(B(c(b(c(x1))))) = x1 >= x1 = B(c(b(c(B(c(b(x1)))))))
    
    b(B(x1)) = x1 >= x1 = x1
    
    B(b(x1)) = x1 >= x1 = x1
   problem:
    b(b(x1)) -> B(x1)
    B(B(x1)) -> b(x1)
    c(B(c(b(c(x1))))) -> B(c(b(c(B(c(b(x1)))))))
    b(B(x1)) -> x1
    B(b(x1)) -> x1
   String Reversal Processor:
    b(b(x1)) -> B(x1)
    B(B(x1)) -> b(x1)
    c(b(c(B(c(x1))))) -> b(c(B(c(b(c(B(x1)))))))
    B(b(x1)) -> x1
    b(B(x1)) -> x1
    Matrix Interpretation Processor: dim=4
     
     interpretation:
                [1 0 0 0]  
                [0 0 1 0]  
      [B](x0) = [0 0 0 1]x0
                [0 1 0 0]  ,
      
                [1 0 0 0]  
                [0 0 0 1]  
      [b](x0) = [0 1 0 0]x0
                [0 0 1 0]  ,
      
                [1 1 0 0]     [0]
                [0 1 0 0]     [0]
      [c](x0) = [0 0 0 1]x0 + [0]
                [0 0 1 0]     [1]
     orientation:
                 [1 0 0 0]      [1 0 0 0]          
                 [0 0 1 0]      [0 0 1 0]          
      b(b(x1)) = [0 0 0 1]x1 >= [0 0 0 1]x1 = B(x1)
                 [0 1 0 0]      [0 1 0 0]          
      
                 [1 0 0 0]      [1 0 0 0]          
                 [0 0 0 1]      [0 0 0 1]          
      B(B(x1)) = [0 1 0 0]x1 >= [0 1 0 0]x1 = b(x1)
                 [0 0 1 0]      [0 0 1 0]          
      
                          [1 1 1 1]     [2]    [1 1 1 1]     [1]                          
                          [0 0 1 0]     [2]    [0 0 1 0]     [2]                          
      c(b(c(B(c(x1))))) = [0 1 0 0]x1 + [0] >= [0 1 0 0]x1 + [0] = b(c(B(c(b(c(B(x1)))))))
                          [0 0 0 1]     [1]    [0 0 0 1]     [1]                          
      
                              
                              
      B(b(x1)) = x1 >= x1 = x1
                              
      
                              
                              
      b(B(x1)) = x1 >= x1 = x1
                              
     problem:
      b(b(x1)) -> B(x1)
      B(B(x1)) -> b(x1)
      B(b(x1)) -> x1
      b(B(x1)) -> x1
     Arctic Interpretation Processor:
      dimension: 2
      interpretation:
                 [0 2]  
       [B](x0) = [1 3]x0,
       
                 [0 0]  
       [b](x0) = [0 2]x0
      orientation:
                  [0 2]      [0 2]          
       b(b(x1)) = [2 4]x1 >= [1 3]x1 = B(x1)
       
                  [3 5]      [0 0]          
       B(B(x1)) = [4 6]x1 >= [0 2]x1 = b(x1)
       
                  [2 4]             
       B(b(x1)) = [3 5]x1 >= x1 = x1
       
                  [1 3]             
       b(B(x1)) = [3 5]x1 >= x1 = x1
      problem:
       b(b(x1)) -> B(x1)
      Arctic Interpretation Processor:
       dimension: 3
       interpretation:
                  [0  -& 0 ]  
        [B](x0) = [0  -& -&]x0
                  [-& -& -&]  ,
        
                  [0  0  -&]  
        [b](x0) = [3  -& 1 ]x0
                  [-& 0  1 ]  
       orientation:
                   [3 0 1]      [0  -& 0 ]          
        b(b(x1)) = [3 3 2]x1 >= [0  -& -&]x1 = B(x1)
                   [3 1 2]      [-& -& -&]          
       problem:
        
       Qed