Problem:
 b(w(x)) -> w(w(w(b(x))))
 w(b(x)) -> b(x)
 b(b(x)) -> w(w(w(w(x))))
 w(w(x)) -> w(x)

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  b(w(x)) -> w(w(w(b(x))))
  w(b(x)) -> b(x)
  b(b(x)) -> w(w(w(w(x))))
  w(w(x)) -> w(x)
  
   T' = (P union S) with
  
   TRS P:
  
   TRS S:b(w(x)) -> w(w(w(b(x))))
         w(b(x)) -> b(x)
         b(b(x)) -> w(w(w(w(x))))
         w(w(x)) -> w(x)
  
  S is linear and P is reversible.
  
   CP(S,S) = 
  w(w(w(w(b(x73))))) = b(w(x73)), b(w(w(w(b(x74))))) = w(w(w(w(w(x74))))), 
  b(b(x75)) = w(w(w(b(b(x75))))), w(b(x76)) = w(b(x76)), w(w(w(w(w(x77))))) = 
  b(b(x77)), b(w(w(w(w(x78))))) = w(w(w(w(b(x78))))), b(w(x79)) = w(w(w(b(w(x79))))), 
  w(w(x80)) = w(w(x80))
  
   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:
    [b](x0) = x0 + 1,
    
    [w](x0) = x0
   orientation:
    b(w(x)) = x + 1 >= x + 1 = w(w(w(b(x))))
    
    w(b(x)) = x + 1 >= x + 1 = b(x)
    
    b(b(x)) = x + 2 >= x = w(w(w(w(x))))
    
    w(w(x)) = x >= x = w(x)
   problem:
    b(w(x)) -> w(w(w(b(x))))
    w(b(x)) -> b(x)
    w(w(x)) -> w(x)
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [b](x0) = 4x0 + 6,
     
     [w](x0) = x0 + 4
    orientation:
     b(w(x)) = 4x + 22 >= 4x + 18 = w(w(w(b(x))))
     
     w(b(x)) = 4x + 10 >= 4x + 6 = b(x)
     
     w(w(x)) = x + 8 >= x + 4 = w(x)
    problem:
     
    Qed