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

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  +(x,0()) -> x
  +(x,s(y)) -> s(+(y,x))
  +(0(),x) -> x
  +(s(y),x) -> s(+(x,y))
  +(x,y) -> +(y,x)
  
   T' = (P union S) with
  
   TRS P:+(x,y) -> +(y,x)
  
   TRS S:+(x,0()) -> x
         +(x,s(y)) -> s(+(y,x))
         +(0(),x) -> x
         +(s(y),x) -> s(+(x,y))
  
  S is linear and P is reversible.
  
   CP(S,S) = 
  0() = 0(), s(y) = s(+(0(),y)), s(+(x209,0())) = s(x209), s(+(x211,s(y))) = 
  s(+(s(x211),y)), s(y) = s(+(y,0())), s(+(0(),x214)) = s(x214), s(+(s(y),x216)) = 
  s(+(y,s(x216)))
  
   CP(S,P union P^-1) = 
  x = +(0(),x), y = +(0(),y), s(+(x233,x)) = +(s(x233),x), s(+(x235,y)) = 
  +(s(x235),y), y = +(y,0()), x = +(x,0()), s(+(y,x238)) = +(y,s(x238)), 
  s(+(x,x240)) = +(x,s(x240))
  
   CP(P union P^-1,S) = 
  +(0(),x) = x, +(s(y),x) = s(+(y,x)), +(x,0()) = x, +(x,s(y)) = s(+(x,y))
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [s](x0) = x0 + 1,
    
    [+](x0, x1) = 2x0 + 2x1 + 3,
    
    [0] = 0
   orientation:
    +(x,0()) = 2x + 3 >= x = x
    
    +(x,s(y)) = 2x + 2y + 5 >= 2x + 2y + 4 = s(+(y,x))
    
    +(0(),x) = 2x + 3 >= x = x
    
    +(s(y),x) = 2x + 2y + 5 >= 2x + 2y + 4 = s(+(x,y))
   problem:
    
   Qed