YES

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

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  +(0(),y) -> y
  +(s(0()),y) -> s(+(0(),y))
  s(s(x)) -> x
  +(y,0()) -> y
  +(y,s(0())) -> s(y)
  +(x,y) -> +(y,x)
  
   T' = (P union S) with
  
   TRS P:+(x,y) -> +(y,x)
  
   TRS S:+(0(),y) -> y
         +(s(0()),y) -> s(+(0(),y))
         s(s(x)) -> x
         +(y,0()) -> y
         +(y,s(0())) -> s(y)
  
  S is linear and P is reversible.
  
   CP(S,S) = 
  0() = 0(), s(0()) = s(0()), s(+(0(),0())) = s(0()), s(+(0(),s(0()))) = 
  s(s(0())), s(x253) = s(x253), s(0()) = s(+(0(),0())), s(s(0())) = s(+(0(),s(0())))
  
   CP(S,P union P^-1) = 
  y = +(y,0()), x = +(x,0()), s(+(0(),y)) = +(y,s(0())), s(+(0(),x)) = 
  +(x,s(0())), x = +(0(),x), y = +(0(),y), s(x) = +(s(0()),x), s(y) = 
  +(s(0()),y)
  
   CP(P union P^-1,S) = 
  +(y,0()) = y, +(y,s(0())) = s(+(0(),y)), +(0(),y) = y, +(s(0()),y) = 
  s(y)
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [+](x0, x1) = x0 + x1 + 3,
    
    [0] = 0,
    
    [s](x0) = x0 + 2
   orientation:
    +(0(),y) = y + 3 >= y = y
    
    +(s(0()),y) = y + 5 >= y + 5 = s(+(0(),y))
    
    s(s(x)) = x + 4 >= x = x
    
    +(y,0()) = y + 3 >= y = y
    
    +(y,s(0())) = y + 5 >= y + 2 = s(y)
   problem:
    +(s(0()),y) -> s(+(0(),y))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [+](x0, x1) = 2x0 + 4x1 + 1,
     
     [0] = 0,
     
     [s](x0) = x0 + 1
    orientation:
     +(s(0()),y) = 4y + 3 >= 4y + 2 = s(+(0(),y))
    problem:
     
    Qed