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

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  +(x,0()) -> x
  +(x,s(y)) -> s(+(x,y))
  +(0(),y) -> y
  +(s(x21),y) -> s(+(x21,y))
  +(x,y) -> +(y,x)
  
   T' = (P union S) with
  
   TRS P:+(x,y) -> +(y,x)
  
   TRS S:+(x,0()) -> x
         +(x,s(y)) -> s(+(x,y))
         +(0(),y) -> y
         +(s(x21),y) -> s(+(x21,y))
  
  S is left-linear and P is reversible.
  
   CP(S,S) = 
  0() = 0(), s(x21) = s(+(x21,0())), s(+(0(),x173)) = s(x173), s(+(s(x21),x175)) = 
  s(+(x21,s(x175))), s(y) = s(+(0(),y)), s(+(x178,0())) = s(x178), s(+(x180,s(y))) = 
  s(+(s(x180),y))
  
   CP(S,P union P^-1) = 
  x = +(0(),x), y = +(0(),y), s(+(x,x197)) = +(s(x197),x), s(+(y,x199)) = 
  +(s(x199),y), y = +(y,0()), x = +(x,0()), s(+(x202,y)) = +(y,s(x202)), 
  s(+(x204,x)) = +(x,s(x204))
  
  
   PCP_in(P union P^-1,S) = 
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [s](x0) = x0 + 1,
    
    [+](x0, x1) = 2x0 + 2x1 + 5,
    
    [0] = 4
   orientation:
    +(x,0()) = 2x + 13 >= x = x
    
    +(x,s(y)) = 2x + 2y + 7 >= 2x + 2y + 6 = s(+(x,y))
    
    +(0(),y) = 2y + 13 >= y = y
    
    +(s(x21),y) = 2x21 + 2y + 7 >= 2x21 + 2y + 6 = s(+(x21,y))
   problem:
    
   Qed