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

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  +(x,0()) -> x
  +(s(x),y) -> s(+(x,y))
  +(0(),x) -> x
  +(y,s(x)) -> s(+(y,x))
  +(x,y) -> +(y,x)
  
   T' = (P union S) with
  
   TRS P:+(x,y) -> +(y,x)
  
   TRS S:+(x,0()) -> x
         +(s(x),y) -> s(+(x,y))
         +(0(),x) -> x
         +(y,s(x)) -> s(+(y,x))
  
  S is linear and P is reversible.
  
   CP(S,S) = 
  s(x) = s(+(x,0())), 0() = 0(), s(+(x208,0())) = s(x208), s(+(x210,s(x))) = 
  s(+(s(x210),x)), s(x) = s(+(0(),x)), s(+(s(x),x215)) = s(+(x,s(x215))), 
  s(+(0(),x217)) = s(x217)
  
   CP(S,P union P^-1) = 
  x = +(0(),x), y = +(0(),y), s(+(x232,y)) = +(y,s(x232)), s(+(x234,x)) = 
  +(x,s(x234)), y = +(y,0()), x = +(x,0()), s(+(x,x239)) = +(s(x239),x), 
  s(+(y,x241)) = +(s(x241),y)
  
   CP(P union P^-1,S) = 
  +(0(),x) = x, +(y,s(x)) = s(+(x,y)), +(x,0()) = x, +(s(x),y) = s(+(y,x))
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [s](x0) = x0 + 6,
    
    [+](x0, x1) = 4x0 + x1 + 1,
    
    [0] = 4
   orientation:
    +(x,0()) = 4x + 5 >= x = x
    
    +(s(x),y) = 4x + y + 25 >= 4x + y + 7 = s(+(x,y))
    
    +(0(),x) = x + 17 >= x = x
    
    +(y,s(x)) = x + 4y + 7 >= x + 4y + 7 = s(+(y,x))
   problem:
    +(y,s(x)) -> s(+(y,x))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [s](x0) = x0 + 1,
     
     [+](x0, x1) = x0 + 2x1 + 5
    orientation:
     +(y,s(x)) = 2x + y + 7 >= 2x + y + 6 = s(+(y,x))
    problem:
     
    Qed