YES

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

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  +(0(),y) -> y
  +(s(x),y) -> s(+(x,y))
  +(x,0()) -> x
  +(x,s(y)) -> s(+(x,y))
  
   T' = (P union S) with
  
   TRS P:
  
   TRS S:+(0(),y) -> y
         +(s(x),y) -> s(+(x,y))
         +(x,0()) -> x
         +(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(+(x112,0())) = s(x112), s(+(x114,s(y))) = 
  s(+(s(x114),y)), s(x) = s(+(x,0())), s(+(0(),x119)) = s(x119), s(+(s(x),x121)) = 
  s(+(x,s(x121)))
  
   CP(S,P union P^-1) = 
  
  
   PCP_in(P union P^-1,S) = 
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [+](x0, x1) = 2x0 + 2x1 + 5,
    
    [0] = 1,
    
    [s](x0) = x0
   orientation:
    +(0(),y) = 2y + 7 >= y = y
    
    +(s(x),y) = 2x + 2y + 5 >= 2x + 2y + 5 = s(+(x,y))
    
    +(x,0()) = 2x + 7 >= x = x
    
    +(x,s(y)) = 2x + 2y + 5 >= 2x + 2y + 5 = s(+(x,y))
   problem:
    +(s(x),y) -> s(+(x,y))
    +(x,s(y)) -> s(+(x,y))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [+](x0, x1) = x0 + 2x1 + 7,
     
     [s](x0) = x0 + 6
    orientation:
     +(s(x),y) = x + 2y + 13 >= x + 2y + 13 = s(+(x,y))
     
     +(x,s(y)) = x + 2y + 19 >= x + 2y + 13 = s(+(x,y))
    problem:
     +(s(x),y) -> s(+(x,y))
    Matrix Interpretation Processor: dim=1
     
     interpretation:
      [+](x0, x1) = 6x0 + x1 + 2,
      
      [s](x0) = x0 + 3
     orientation:
      +(s(x),y) = 6x + y + 20 >= 6x + y + 5 = s(+(x,y))
     problem:
      
     Qed