YES

Problem:
 +(x,y) -> +(y,x)
 +(0(),y) -> y
 +(x,0()) -> x
 +(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
  +(x,0()) -> x
  +(s(x),y) -> s(+(x,y))
  +(y,s(x)) -> s(+(y,x))
  +(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
         +(x,0()) -> x
         +(s(x),y) -> s(+(x,y))
         +(y,s(x)) -> s(+(y,x))
  
  S is linear and P is reversible.
  
   CP(S,S) = 
  0() = 0(), s(x) = s(+(0(),x)), s(x) = s(+(x,0())), s(+(x370,0())) = 
  s(x370), s(+(x372,s(x))) = s(+(s(x372),x)), s(+(0(),x375)) = s(x375), 
  s(+(s(x),x377)) = s(+(x,s(x377)))
  
   CP(S,P union P^-1) = 
  y = +(y,0()), +(y,z) = +(+(0(),y),z), +(x,z) = +(+(x,0()),z), x = +(x,0()), 
  +(y,z) = +(0(),+(y,z)), x = +(0(),x), +(x,y) = +(+(x,y),0()), y = +(0(),y), 
  +(x,y) = +(x,+(y,0())), +(x,z) = +(x,+(0(),z)), s(+(x424,y)) = +(y,s(x424)), 
  s(+(x426,+(y,z))) = +(+(s(x426),y),z), +(x,s(+(x428,z))) = +(+(x,s(x428)),z), 
  s(+(x430,x)) = +(x,s(x430)), +(s(+(x432,y)),z) = +(s(x432),+(y,z)), 
  s(+(x,x435)) = +(s(x435),x), +(x,s(+(y,x437))) = +(+(x,y),s(x437)), 
  s(+(y,x439)) = +(s(x439),y), s(+(+(x,y),x441)) = +(x,+(y,s(x441))), 
  +(s(+(x,x443)),z) = +(x,+(s(x443),z))
  
   CP(P union P^-1,S) = 
  +(y,0()) = y, +(0(),x) = x, +(y,s(x)) = s(+(x,y)), +(s(x),y) = s(+(y,x)), 
  +(+(0(),x533),x534) = +(x533,x534), +(+(s(x),x536),x537) = s(+(x,+(x536,x537))), 
  +(x546,+(x547,0())) = +(x546,x547), +(x549,+(x550,s(x))) = s(+(+(x549,x550),x))
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [0] = 2,
    
    [+](x0, x1) = 2x0 + x1 + 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))
    
    +(y,s(x)) = x + 2y + 4 >= x + 2y + 4 = s(+(y,x))
   problem:
    +(y,s(x)) -> s(+(y,x))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [+](x0, x1) = x0 + 4x1 + 7,
     
     [s](x0) = x0 + 1
    orientation:
     +(y,s(x)) = 4x + y + 11 >= 4x + y + 8 = s(+(y,x))
    problem:
     
    Qed