Problem:
 +(s(x),y) -> s(+(x,y))
 +(x,s(y)) -> s(+(x,y))
 +(x,y) -> +(y,x)

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  +(s(x),y) -> s(+(x,y))
  +(x,s(y)) -> s(+(x,y))
  +(x,y) -> +(y,x)
  
   T' = (P union S) with
  
   TRS P:+(x,y) -> +(y,x)
  
   TRS S:+(s(x),y) -> s(+(x,y))
         +(x,s(y)) -> s(+(x,y))
  
  S is linear and P is reversible.
  
   CP(S,S) = 
  s(+(x78,s(y))) = s(+(s(x78),y)), s(+(s(x),x81)) = s(+(x,s(x81)))
  
   CP(S,P union P^-1) = 
  s(+(x90,y)) = +(y,s(x90)), s(+(x92,x)) = +(x,s(x92)), s(+(x,x95)) = 
  +(s(x95),x), s(+(y,x97)) = +(s(x97),y)
  
   CP(P union P^-1,S) = 
  +(y,s(x)) = s(+(x,y)), +(s(y),x) = s(+(x,y))
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [+](x0, x1) = 2x0 + 4x1 + 4,
    
    [s](x0) = x0 + 1
   orientation:
    +(s(x),y) = 2x + 4y + 6 >= 2x + 4y + 5 = s(+(x,y))
    
    +(x,s(y)) = 2x + 4y + 8 >= 2x + 4y + 5 = s(+(x,y))
   problem:
    
   Qed