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

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  +(x,0()) -> x
  +(s(x),y) -> s(+(y,x))
  +(0(),x) -> x
  +(y,s(x)) -> s(+(x,y))
  +(x,y) -> +(y,x)
  
   T' = (P union S) with
  
   TRS P:+(x,y) -> +(y,x)
  
   TRS S:+(x,0()) -> x
         +(s(x),y) -> s(+(y,x))
         +(0(),x) -> x
         +(y,s(x)) -> s(+(x,y))
  
  S is linear and P is reversible.
  
   CP(S,S) = 
  s(x) = s(+(0(),x)), 0() = 0(), s(+(0(),x208)) = s(x208), s(+(s(x),x210)) = 
  s(+(x,s(x210))), s(x) = s(+(x,0())), s(+(x215,s(x))) = s(+(s(x215),x)), 
  s(+(x217,0())) = s(x217)
  
   CP(S,P union P^-1) = 
  x = +(0(),x), y = +(0(),y), s(+(y,x232)) = +(y,s(x232)), s(+(x,x234)) = 
  +(x,s(x234)), y = +(y,0()), x = +(x,0()), s(+(x239,x)) = +(s(x239),x), 
  s(+(x241,y)) = +(s(x241),y)
  
   CP(P union P^-1,S) = 
  +(0(),x) = x, +(y,s(x)) = s(+(y,x)), +(x,0()) = x, +(s(x),y) = s(+(x,y))
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [+](x0, x1) = 4x0 + 4x1 + 6,
    
    [0] = 0,
    
    [s](x0) = x0 + 4
   orientation:
    +(x,0()) = 4x + 6 >= x = x
    
    +(s(x),y) = 4x + 4y + 22 >= 4x + 4y + 10 = s(+(y,x))
    
    +(0(),x) = 4x + 6 >= x = x
    
    +(y,s(x)) = 4x + 4y + 22 >= 4x + 4y + 10 = s(+(x,y))
   problem:
    
   Qed