Problem:
 -(0(),0()) -> 0()
 -(s(x),0()) -> s(x)
 -(x,s(y)) -> -(d(x),y)
 d(s(x)) -> x
 -(s(x),s(y)) -> -(x,y)
 -(d(x),y) -> -(x,s(y))

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  -(0(),0()) -> 0()
  -(s(x),0()) -> s(x)
  d(s(x)) -> x
  -(s(x),s(y)) -> -(x,y)
  -(x,s(y)) -> -(d(x),y)
  -(d(x),y) -> -(x,s(y))
  
   T' = (P union S) with
  
   TRS P:-(x,s(y)) -> -(d(x),y)
         -(d(x),y) -> -(x,s(y))
  
   TRS S:-(0(),0()) -> 0()
         -(s(x),0()) -> s(x)
         d(s(x)) -> x
         -(s(x),s(y)) -> -(x,y)
  
  S is linear and P is reversible.
  
   CP(S,S) = 
  
   CP(S,P union P^-1) = 
  -(x236,y) = -(s(x236),s(y)), -(x237,y) = -(d(s(x237)),y)
  
   CP(P union P^-1,S) = 
  -(d(s(x)),y) = -(x,y)
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [d](x0) = x0,
    
    [s](x0) = 7x0 + 1,
    
    [-](x0, x1) = x0 + 3x1,
    
    [0] = 0
   orientation:
    -(0(),0()) = 0 >= 0 = 0()
    
    -(s(x),0()) = 7x + 1 >= 7x + 1 = s(x)
    
    d(s(x)) = 7x + 1 >= x = x
    
    -(s(x),s(y)) = 7x + 21y + 4 >= x + 3y = -(x,y)
   problem:
    -(0(),0()) -> 0()
    -(s(x),0()) -> s(x)
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [s](x0) = x0 + 4,
     
     [-](x0, x1) = x0 + 4x1 + 1,
     
     [0] = 2
    orientation:
     -(0(),0()) = 11 >= 2 = 0()
     
     -(s(x),0()) = x + 13 >= x + 4 = s(x)
    problem:
     
    Qed