YES

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

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  +(0(),y) -> y
  +(s(x),y) -> s(+(x,y))
  +(x,0()) -> x
  +(x,s(x50)) -> s(+(x,x50))
  +(x,y) -> +(y,x)
  +(+(x,y),z) -> +(x,+(y,z))
  
   T' = (P union S) with
  
   TRS P:+(x,y) -> +(y,x)
         +(+(x,y),z) -> +(x,+(y,z))
  
   TRS S:+(0(),y) -> y
         +(s(x),y) -> s(+(x,y))
         +(x,0()) -> x
         +(x,s(x50)) -> s(+(x,x50))
  
  S is left-linear and P is reversible.
  
   CP(S,S) = 
  0() = 0(), s(x50) = s(+(0(),x50)), s(+(x272,0())) = s(x272), s(+(x274,s(x50))) = 
  s(+(s(x274),x50)), s(x) = s(+(x,0())), s(+(0(),x279)) = s(x279), s(+(s(x),x281)) = 
  s(+(x,s(x281)))
  
   CP(S,P union P^-1) = 
  y = +(y,0()), +(y,z) = +(0(),+(y,z)), x = +(x,0()), +(y,z) = +(+(0(),y),z), 
  +(x,z) = +(+(x,0()),z), s(+(x323,y)) = +(y,s(x323)), +(s(+(x325,y)),z) = 
  +(s(x325),+(y,z)), s(+(x327,x)) = +(x,s(x327)), s(+(x329,+(y,z))) = 
  +(+(s(x329),y),z), +(x,s(+(x331,z))) = +(+(x,s(x331)),z), x = +(0(),x), 
  +(x,y) = +(x,+(y,0())), +(x,z) = +(x,+(0(),z)), y = +(0(),y), +(x,y) = 
  +(+(x,y),0()), s(+(x,x339)) = +(s(x339),x), s(+(+(x,y),x341)) = +(x,+(y,s(x341))), 
  +(s(+(x,x343)),z) = +(x,+(s(x343),z)), s(+(y,x345)) = +(s(x345),y), 
  +(x,s(+(y,x347))) = +(+(x,y),s(x347))
  
  
   PCP_in(P union P^-1,S) = 
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [0] = 5,
    
    [+](x0, x1) = x0 + 3x1,
    
    [s](x0) = x0 + 7
   orientation:
    +(0(),y) = 3y + 5 >= y = y
    
    +(s(x),y) = x + 3y + 7 >= x + 3y + 7 = s(+(x,y))
    
    +(x,0()) = x + 15 >= x = x
    
    +(x,s(x50)) = x + 3x50 + 21 >= x + 3x50 + 7 = s(+(x,x50))
   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