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

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  +(x,0()) -> x
  +(x,s(y)) -> s(+(x,y))
  +(0(),y) -> y
  +(s(x),y) -> s(+(x,y))
  +(+(x,y),z) -> +(x,+(y,z))
  +(x,y) -> +(y,x)
  s(s(x)) -> s(x)
  s(x) -> s(s(x))
  
   T' = (P union S) with
  
   TRS P:+(+(x,y),z) -> +(x,+(y,z))
         +(x,y) -> +(y,x)
         s(s(x)) -> s(x)
         s(x) -> s(s(x))
  
   TRS S:+(x,0()) -> x
         +(x,s(y)) -> s(+(x,y))
         +(0(),y) -> y
         +(s(x),y) -> s(+(x,y))
  
  S is linear and P is reversible.
  
   CP(S,S) = 
  0() = 0(), s(x) = s(+(x,0())), s(+(0(),x364)) = s(x364), s(+(s(x),x366)) = 
  s(+(x,s(x366))), s(y) = s(+(0(),y)), s(+(x369,0())) = s(x369), s(+(x371,s(y))) = 
  s(+(s(x371),y))
  
   CP(S,P union P^-1) = 
  +(x,y) = +(x,+(y,0())), +(x,z) = +(x,+(0(),z)), x = +(0(),x), +(x,y) = 
  +(+(x,y),0()), y = +(0(),y), s(+(+(x,y),x433)) = +(x,+(y,s(x433))), 
  +(s(+(x,x435)),z) = +(x,+(s(x435),z)), s(+(x,x437)) = +(s(x437),x), 
  +(x,s(+(y,x439))) = +(+(x,y),s(x439)), s(+(y,x441)) = +(s(x441),y), 
  +(y,z) = +(0(),+(y,z)), y = +(y,0()), +(y,z) = +(+(0(),y),z), +(x,z) = 
  +(+(x,0()),z), x = +(x,0()), +(s(+(x447,y)),z) = +(s(x447),+(y,z)), 
  s(+(x449,y)) = +(y,s(x449)), s(+(x451,+(y,z))) = +(+(s(x451),y),z), 
  +(x,s(+(x453,z))) = +(+(x,s(x453)),z), s(+(x455,x)) = +(x,s(x455))
  
   CP(P union P^-1,S) = 
  +(x553,+(x554,0())) = +(x553,x554), +(x556,+(x557,s(y))) = s(+(+(x556,x557),y)), 
  +(0(),x) = x, +(s(y),x) = s(+(x,y)), +(y,0()) = y, +(y,s(x)) = s(+(x,y)), 
  +(x,s(x567)) = s(+(x,s(x567))), +(s(x568),y) = s(+(s(x568),y)), +(x,s(s(y))) = 
  s(+(x,y)), +(s(s(x)),y) = s(+(x,y)), +(+(0(),x572),x573) = +(x572,x573), 
  +(+(s(x),x575),x576) = s(+(x,+(x575,x576)))
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [s](x0) = x0 + 1,
    
    [+](x0, x1) = 2x0 + 2x1 + 5,
    
    [0] = 4
   orientation:
    +(x,0()) = 2x + 13 >= x = x
    
    +(x,s(y)) = 2x + 2y + 7 >= 2x + 2y + 6 = s(+(x,y))
    
    +(0(),y) = 2y + 13 >= y = y
    
    +(s(x),y) = 2x + 2y + 7 >= 2x + 2y + 6 = s(+(x,y))
   problem:
    
   Qed