YES

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

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,z)) -> +(+(x,y),z)
  +(+(x,y),z) -> +(x,+(y,z))
  
   T' = (P union S) with
  
   TRS P:+(x,+(y,z)) -> +(+(x,y),z)
         +(+(x,y),z) -> +(x,+(y,z))
  
   TRS S:+(0(),y) -> y
         +(x,0()) -> x
         +(s(x),y) -> s(+(x,y))
         +(x,s(y)) -> s(+(x,y))
  
  S is linear and P is reversible.
  
   CP(S,S) = 
  0() = 0(), s(y) = s(+(0(),y)), s(x) = s(+(x,0())), s(+(x319,0())) = 
  s(x319), s(+(x321,s(y))) = s(+(s(x321),y)), s(+(0(),x324)) = s(x324), 
  s(+(s(x),x326)) = s(+(x,s(x326)))
  
   CP(S,P union P^-1) = 
  +(y,z) = +(+(0(),y),z), +(x,z) = +(+(x,0()),z), +(y,z) = +(0(),+(y,z)), 
  +(x,y) = +(+(x,y),0()), +(x,y) = +(x,+(y,0())), +(x,z) = +(x,+(0(),z)), 
  s(+(x357,+(y,z))) = +(+(s(x357),y),z), +(x,s(+(x359,z))) = +(+(x,s(x359)),z), 
  +(s(+(x361,y)),z) = +(s(x361),+(y,z)), +(x,s(+(y,x364))) = +(+(x,y),s(x364)), 
  s(+(+(x,y),x366)) = +(x,+(y,s(x366))), +(s(+(x,x368)),z) = +(x,+(s(x368),z))
  
   CP(P union P^-1,S) = 
  +(+(0(),x418),x419) = +(x418,x419), +(+(s(x),x421),x422) = s(+(x,+(x421,x422))), 
  +(x423,+(x424,0())) = +(x423,x424), +(x426,+(x427,s(y))) = s(+(+(x426,x427),y))
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [+](x0, x1) = 2x0 + x1 + 2,
    
    [0] = 2,
    
    [s](x0) = x0 + 2
   orientation:
    +(0(),y) = y + 6 >= y = y
    
    +(x,0()) = 2x + 4 >= x = x
    
    +(s(x),y) = 2x + y + 6 >= 2x + y + 4 = s(+(x,y))
    
    +(x,s(y)) = 2x + y + 4 >= 2x + y + 4 = s(+(x,y))
   problem:
    +(x,s(y)) -> s(+(x,y))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [+](x0, x1) = x0 + 4x1,
     
     [s](x0) = x0 + 5
    orientation:
     +(x,s(y)) = x + 4y + 20 >= x + 4y + 5 = s(+(x,y))
    problem:
     
    Qed