YES

Problem:
 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

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  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
  
   T' = (P union S) with
  
   TRS P:
  
   TRS S: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
  
  S is left-linear and P is reversible.
  
   CP(S,S) = 
  c(x193) = c(x193), c(B(c(b(x194)))) = B(c(b(c(B(c(b(c(x194)))))))), 
  b(B(x195)) = B(b(x195)), B(B(x196)) = b(x196), B(b(x197)) = b(B(x197)), 
  b(b(x198)) = B(x198), c(B(c(b(c(B(c(b(x199)))))))) = B(c(b(c(x199)))), 
  c(B(c(b(B(c(b(c(B(c(b(x200))))))))))) = B(c(b(c(B(c(b(B(c(b(c(x200))))))))))), 
  b(x201) = B(B(x201)), B(x202) = B(x202), B(x203) = b(b(x203)), b(x204) = 
  b(x204)
  
   CP(S,P union P^-1) = 
  
  
   PCP_in(P union P^-1,S) = 
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [b](x0) = x0,
    
    [c](x0) = 2x0 + 1,
    
    [B](x0) = x0
   orientation:
    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
   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
   String Reversal Processor:
    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) -> 3*
       b0(5) -> 6*
       B1(10) -> 11*
       c0(6) -> 7*
       c0(1) -> 5*
       c0(8) -> 9*
       B0(2) -> 1*
       B0(7) -> 8*
       f30() -> 2*
       4 -> 5,7
       11 -> 6*
       2 -> 1*
       1 -> 3*
       9 -> 10,8
       3 -> 1*
     problem:
      
     Qed