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

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  +(0(),y) -> y
  +(s(x),y) -> s(+(x,y))
  +(y,0()) -> y
  +(y,s(x)) -> s(+(y,x))
  +(x,y) -> +(y,x)
  
   T' = (P union S) with
  
   TRS P:+(x,y) -> +(y,x)
  
   TRS S:+(0(),y) -> y
         +(s(x),y) -> s(+(x,y))
         +(y,0()) -> y
         +(y,s(x)) -> s(+(y,x))
  
  S is linear and P is reversible.
  
   CP(S,S) = 
  0() = 0(), s(x) = s(+(0(),x)), s(+(x208,0())) = s(x208), s(+(x210,s(x))) = 
  s(+(s(x210),x)), s(x) = s(+(x,0())), s(+(0(),x215)) = s(x215), s(+(s(x),x217)) = 
  s(+(x,s(x217)))
  
   CP(S,P union P^-1) = 
  y = +(y,0()), x = +(x,0()), s(+(x232,y)) = +(y,s(x232)), s(+(x234,x)) = 
  +(x,s(x234)), x = +(0(),x), y = +(0(),y), s(+(x,x239)) = +(s(x239),x), 
  s(+(y,x241)) = +(s(x241),y)
  
   CP(P union P^-1,S) = 
  +(y,0()) = y, +(y,s(x)) = s(+(x,y)), +(0(),y) = y, +(s(x),y) = s(+(y,x))
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [+](x0, x1) = x0 + 3x1 + 7,
    
    [0] = 3,
    
    [s](x0) = x0 + 7
   orientation:
    +(0(),y) = 3y + 10 >= y = y
    
    +(s(x),y) = x + 3y + 14 >= x + 3y + 14 = s(+(x,y))
    
    +(y,0()) = y + 16 >= y = y
    
    +(y,s(x)) = 3x + y + 28 >= 3x + y + 14 = s(+(y,x))
   problem:
    +(s(x),y) -> s(+(x,y))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [+](x0, x1) = 2x0 + 4x1 + 7,
     
     [s](x0) = x0 + 1
    orientation:
     +(s(x),y) = 2x + 4y + 9 >= 2x + 4y + 8 = s(+(x,y))
    problem:
     
    Qed