YES

Problem:
 +(x,y) -> +(y,x)
 +(x,0()) -> x
 +(x,s(y)) -> s(+(x,y))
 +(0(),y) -> y
 +(s(x),y) -> s(+(x,y))
 +(+(x,y),z) -> +(x,+(y,z))
 s(x) -> s(s(x))
 s(s(x)) -> 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) -> +(y,x)
  +(+(x,y),z) -> +(x,+(y,z))
  s(x) -> s(s(x))
  s(s(x)) -> s(x)
  
   T' = (P union S) with
  
   TRS P:+(x,y) -> +(y,x)
         +(+(x,y),z) -> +(x,+(y,z))
         s(x) -> s(s(x))
         s(s(x)) -> 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 = +(0(),x), +(x,y) = +(x,+(y,0())), +(x,z) = +(x,+(0(),z)), y = +(0(),y), 
  +(x,y) = +(+(x,y),0()), s(+(x,x433)) = +(s(x433),x), s(+(+(x,y),x435)) = 
  +(x,+(y,s(x435))), +(s(+(x,x437)),z) = +(x,+(s(x437),z)), s(+(y,x439)) = 
  +(s(x439),y), +(x,s(+(y,x441))) = +(+(x,y),s(x441)), y = +(y,0()), 
  +(y,z) = +(0(),+(y,z)), x = +(x,0()), +(y,z) = +(+(0(),y),z), +(x,z) = 
  +(+(x,0()),z), s(+(x447,y)) = +(y,s(x447)), +(s(+(x449,y)),z) = +(s(x449),+(y,z)), 
  s(+(x451,x)) = +(x,s(x451)), s(+(x453,+(y,z))) = +(+(s(x453),y),z), 
  +(x,s(+(x455,z))) = +(+(x,s(x455)),z)
  
   CP(P union P^-1,S) = 
  +(0(),x) = x, +(s(y),x) = s(+(x,y)), +(y,0()) = y, +(y,s(x)) = s(+(x,y)), 
  +(x561,+(x562,0())) = +(x561,x562), +(x564,+(x565,s(y))) = s(+(+(x564,x565),y)), 
  +(x,s(s(y))) = s(+(x,y)), +(s(s(x)),y) = s(+(x,y)), +(x,s(x569)) = 
  s(+(x,s(x569))), +(s(x570),y) = s(+(s(x570),y)), +(+(0(),x580),x581) = 
  +(x580,x581), +(+(s(x),x583),x584) = s(+(x,+(x583,x584)))
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [0] = 5,
    
    [+](x0, x1) = 5x0 + 2x1,
    
    [s](x0) = x0 + 6
   orientation:
    +(x,0()) = 5x + 10 >= x = x
    
    +(x,s(y)) = 5x + 2y + 12 >= 5x + 2y + 6 = s(+(x,y))
    
    +(0(),y) = 2y + 25 >= y = y
    
    +(s(x),y) = 5x + 2y + 30 >= 5x + 2y + 6 = s(+(x,y))
   problem:
    
   Qed