Problem:
 H(H(x)) -> K(x)
 H(K(x)) -> K(H(x))

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  H(H(x)) -> K(x)
  H(K(x)) -> K(H(x))
  
   T' = (P union S) with
  
   TRS P:
  
   TRS S:H(H(x)) -> K(x)
         H(K(x)) -> K(H(x))
  
  S is left-linear and P is reversible.
  
   CP(S,S) = 
  H(K(x19)) = K(H(x19)), H(K(H(x20))) = K(K(x20))
  
   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:
    [K](x0) = 4x0 + 6,
    
    [H](x0) = 5x0 + 1
   orientation:
    H(H(x)) = 25x + 6 >= 4x + 6 = K(x)
    
    H(K(x)) = 20x + 31 >= 20x + 10 = K(H(x))
   problem:
    H(H(x)) -> K(x)
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [K](x0) = 4x0,
     
     [H](x0) = 2x0 + 1
    orientation:
     H(H(x)) = 4x + 3 >= 4x = K(x)
    problem:
     
    Qed