YES

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

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  +(0(),y) -> y
  +(x,0()) -> x
  +(s(x),y) -> s(+(x,y))
  +(x,s(y)) -> s(+(x,y))
  +(x,y) -> +(y,x)
  
   T' = (P union S) with
  
   TRS P:+(x,y) -> +(y,x)
  
   TRS S:+(0(),y) -> y
         +(x,0()) -> x
         +(s(x),y) -> s(+(x,y))
         +(x,s(y)) -> s(+(x,y))
  
  S is left-linear and P is reversible.
  
   CP(S,S) = 
  0() = 0(), s(y) = s(+(0(),y)), s(x) = s(+(x,0())), s(+(x154,0())) = 
  s(x154), s(+(x156,s(y))) = s(+(s(x156),y)), s(+(0(),x159)) = s(x159), 
  s(+(s(x),x161)) = s(+(x,s(x161)))
  
   CP(S,P union P^-1) = 
  y = +(y,0()), x = +(x,0()), x = +(0(),x), y = +(0(),y), s(+(x178,y)) = 
  +(y,s(x178)), s(+(x180,x)) = +(x,s(x180)), s(+(x,x183)) = +(s(x183),x), 
  s(+(y,x185)) = +(s(x185),y)
  
  
   PCP_in(P union P^-1,S) = 
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [+](x0, x1) = 4x0 + 6x1 + 2,
    
    [0] = 0,
    
    [s](x0) = x0 + 3
   orientation:
    +(0(),y) = 6y + 2 >= y = y
    
    +(x,0()) = 4x + 2 >= x = x
    
    +(s(x),y) = 4x + 6y + 14 >= 4x + 6y + 5 = s(+(x,y))
    
    +(x,s(y)) = 4x + 6y + 20 >= 4x + 6y + 5 = s(+(x,y))
   problem:
    
   Qed