YES

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

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  C(x) -> c(x)
  c(c(x)) -> x
  b(b(x)) -> B(x)
  B(B(x)) -> b(x)
  c(B(c(b(c(x))))) -> B(c(b(c(B(c(b(x)))))))
  b(B(x)) -> x
  B(b(x)) -> x
  c(C(x)) -> x
  C(c(x)) -> x
  
   T' = (P union S) with
  
   TRS P:
  
   TRS S:C(x) -> c(x)
         c(c(x)) -> x
         b(b(x)) -> B(x)
         B(B(x)) -> b(x)
         c(B(c(b(c(x))))) -> B(c(b(c(B(c(b(x)))))))
         b(B(x)) -> x
         B(b(x)) -> x
         c(C(x)) -> x
         C(c(x)) -> x
  
  S is linear and P is reversible.
  
   CP(S,S) = 
  c(c(x)) = x, c(x384) = c(x384), c(B(c(b(x385)))) = B(c(b(c(B(c(b(c(x385)))))))), 
  C(x386) = c(x386), b(B(x387)) = B(b(x387)), B(B(x388)) = b(x388), B(b(x389)) = 
  b(B(x389)), b(b(x390)) = B(x390), c(B(c(b(c(B(c(b(x391)))))))) = B(c(b(c(x391)))), 
  c(B(c(b(B(c(b(c(B(c(b(x392))))))))))) = B(c(b(c(B(c(b(B(c(b(c(x392))))))))))), 
  C(B(c(b(c(B(c(b(x393)))))))) = B(c(b(c(x393)))), b(x394) = B(B(x394)), 
  B(x395) = B(x395), B(x396) = b(b(x396)), b(x397) = b(x397), c(x398) = 
  C(x398), c(B(c(b(x399)))) = B(c(b(c(B(c(b(C(x399)))))))), C(x400) = 
  C(x400), x401 = c(c(x401)), c(x402) = c(x402)
  
   CP(S,P union P^-1) = 
  
   CP(P union P^-1,S) = 
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [c](x0) = 2x0 + 1,
    
    [B](x0) = x0,
    
    [C](x0) = 2x0 + 1,
    
    [b](x0) = x0
   orientation:
    C(x) = 2x + 1 >= 2x + 1 = c(x)
    
    c(c(x)) = 4x + 3 >= x = x
    
    b(b(x)) = x >= x = B(x)
    
    B(B(x)) = x >= x = b(x)
    
    c(B(c(b(c(x))))) = 8x + 7 >= 8x + 7 = B(c(b(c(B(c(b(x)))))))
    
    b(B(x)) = x >= x = x
    
    B(b(x)) = x >= x = x
    
    c(C(x)) = 4x + 3 >= x = x
    
    C(c(x)) = 4x + 3 >= x = x
   problem:
    C(x) -> c(x)
    b(b(x)) -> B(x)
    B(B(x)) -> b(x)
    c(B(c(b(c(x))))) -> B(c(b(c(B(c(b(x)))))))
    b(B(x)) -> x
    B(b(x)) -> x
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [c](x0) = x0,
     
     [B](x0) = x0,
     
     [C](x0) = 2x0 + 4,
     
     [b](x0) = x0
    orientation:
     C(x) = 2x + 4 >= x = c(x)
     
     b(b(x)) = x >= x = B(x)
     
     B(B(x)) = x >= x = b(x)
     
     c(B(c(b(c(x))))) = x >= x = B(c(b(c(B(c(b(x)))))))
     
     b(B(x)) = x >= x = x
     
     B(b(x)) = x >= x = x
    problem:
     b(b(x)) -> B(x)
     B(B(x)) -> b(x)
     c(B(c(b(c(x))))) -> B(c(b(c(B(c(b(x)))))))
     b(B(x)) -> x
     B(b(x)) -> x
    Bounds Processor:
     bound: 1
     enrichment: match
     automaton:
      final states: {2,4,3,1}
      transitions:
       B0(9) -> 4*
       B0(2) -> 1*
       B0(5) -> 6*
       c0(6) -> 7*
       c0(3) -> 5*
       c0(8) -> 9*
       b0(2) -> 3*
       b0(7) -> 8*
       f40() -> 2*
       b1(10) -> 11*
       4 -> 5,7
       11 -> 6*
       2 -> 3*
       1 -> 3*
       9 -> 10,8
       3 -> 1*
     problem:
      
     Qed