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:
    [-](x0, x1) = 2x0 + 2x1 + 1,
    
    [d](x0) = 2x0,
    
    [0] = 0,
    
    [s](x0) = x0 + 7
   orientation:
    -(0(),0()) = 1 >= 0 = 0()
    
    -(s(x),0()) = 2x + 15 >= x + 7 = s(x)
    
    d(s(x)) = 2x + 14 >= x = x
    
    -(s(x),s(y)) = 2x + 2y + 29 >= 2x + 2y + 1 = -(x,y)
   problem:
    
   Qed