YES

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

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  +(0(),y) -> y
  +(s(x),y) -> s(+(x,y))
  +(x,0()) -> x
  +(x,s(x54)) -> s(+(x,x54))
  +(+(x,y),z) -> +(x,+(y,z))
  +(x,y) -> +(y,x)
  
   T' = (P union S) with
  
   TRS P:+(+(x,y),z) -> +(x,+(y,z))
         +(x,y) -> +(y,x)
  
   TRS S:+(0(),y) -> y
         +(s(x),y) -> s(+(x,y))
         +(x,0()) -> x
         +(x,s(x54)) -> s(+(x,x54))
  
  S is left-linear and P is reversible.
  
   CP(S,S) = 
  0() = 0(), s(x54) = s(+(0(),x54)), s(+(x272,0())) = s(x272), s(+(x274,s(x54))) = 
  s(+(s(x274),x54)), 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,z) = +(0(),+(y,z)), y = +(y,0()), +(y,z) = +(+(0(),y),z), +(x,z) = 
  +(+(x,0()),z), x = +(x,0()), +(s(+(x323,y)),z) = +(s(x323),+(y,z)), 
  s(+(x325,y)) = +(y,s(x325)), s(+(x327,+(y,z))) = +(+(s(x327),y),z), 
  +(x,s(+(x329,z))) = +(+(x,s(x329)),z), s(+(x331,x)) = +(x,s(x331)), 
  +(x,y) = +(x,+(y,0())), +(x,z) = +(x,+(0(),z)), x = +(0(),x), +(x,y) = 
  +(+(x,y),0()), y = +(0(),y), s(+(+(x,y),x339)) = +(x,+(y,s(x339))), 
  +(s(+(x,x341)),z) = +(x,+(s(x341),z)), s(+(x,x343)) = +(s(x343),x), 
  +(x,s(+(y,x345))) = +(+(x,y),s(x345)), s(+(y,x347)) = +(s(x347),y)
  
  
   PCP_in(P union P^-1,S) = 
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [+](x0, x1) = x0 + 3x1,
    
    [0] = 5,
    
    [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(x54)) = x + 3x54 + 21 >= x + 3x54 + 7 = s(+(x,x54))
   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