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:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [c](x0) = 2x0,
   
   [B](x0) = x0,
   
   [C](x0) = 4x0 + 8,
   
   [b](x0) = x0
  orientation:
   C(x1) = 4x1 + 8 >= 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))))) = 8x1 >= 8x1 = B(c(b(c(B(c(b(x1)))))))
   
   b(B(x1)) = x1 >= x1 = x1
   
   B(b(x1)) = x1 >= x1 = x1
   
   c(C(x1)) = 8x1 + 16 >= x1 = x1
   
   C(c(x1)) = 8x1 + 8 >= x1 = x1
  problem:
   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
  String Reversal Processor:
   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
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [c](x0) = x0 + 1,
     
     [B](x0) = x0,
     
     [b](x0) = x0
    orientation:
     c(c(x1)) = x1 + 2 >= x1 = x1
     
     b(b(x1)) = x1 >= x1 = B(x1)
     
     B(B(x1)) = x1 >= x1 = b(x1)
     
     c(b(c(B(c(x1))))) = x1 + 3 >= x1 + 3 = 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
    Bounds Processor:
     bound: 1
     enrichment: match
     automaton:
      final states: {2,4,3,1}
      transitions:
       B0(7) -> 8*
       B0(2) -> 1*
       c0(1) -> 5*
       c0(6) -> 7*
       c0(8) -> 9*
       b0(2) -> 3*
       b0(5) -> 6*
       b0(9) -> 4*
       f40() -> 2*
       B1(10) -> 11*
       4 -> 5,7
       11 -> 6*
       2 -> 1*
       1 -> 3*
       9 -> 10,8
       3 -> 1*
     problem:
      
     Qed