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

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  +(0(),y) -> y
  +(x,s(y)) -> s(+(y,x))
  +(y,0()) -> y
  +(s(y),x) -> s(+(x,y))
  +(x,y) -> +(y,x)
  
   T' = (P union S) with
  
   TRS P:+(x,y) -> +(y,x)
  
   TRS S:+(0(),y) -> y
         +(x,s(y)) -> s(+(y,x))
         +(y,0()) -> y
         +(s(y),x) -> s(+(x,y))
  
  S is linear and P is reversible.
  
   CP(S,S) = 
  s(y) = s(+(y,0())), 0() = 0(), s(+(x209,0())) = s(x209), s(+(x211,s(y))) = 
  s(+(s(x211),y)), s(y) = s(+(0(),y)), s(+(s(y),x214)) = s(+(y,s(x214))), 
  s(+(0(),x216)) = s(x216)
  
   CP(S,P union P^-1) = 
  y = +(y,0()), x = +(x,0()), s(+(x233,x)) = +(s(x233),x), s(+(x235,y)) = 
  +(s(x235),y), x = +(0(),x), y = +(0(),y), s(+(y,x238)) = +(y,s(x238)), 
  s(+(x,x240)) = +(x,s(x240))
  
   CP(P union P^-1,S) = 
  +(y,0()) = y, +(s(y),x) = s(+(y,x)), +(0(),y) = y, +(x,s(y)) = s(+(x,y))
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [s](x0) = x0,
    
    [+](x0, x1) = 2x0 + 2x1,
    
    [0] = 1
   orientation:
    +(0(),y) = 2y + 2 >= y = y
    
    +(x,s(y)) = 2x + 2y >= 2x + 2y = s(+(y,x))
    
    +(y,0()) = 2y + 2 >= y = y
    
    +(s(y),x) = 2x + 2y >= 2x + 2y = s(+(x,y))
   problem:
    +(x,s(y)) -> s(+(y,x))
    +(s(y),x) -> s(+(x,y))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [s](x0) = x0 + 1,
     
     [+](x0, x1) = 2x0 + 2x1 + 2
    orientation:
     +(x,s(y)) = 2x + 2y + 4 >= 2x + 2y + 3 = s(+(y,x))
     
     +(s(y),x) = 2x + 2y + 4 >= 2x + 2y + 3 = s(+(x,y))
    problem:
     
    Qed