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

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  +(0(),x) -> x
  +(x,0()) -> x
  +(+(x,y),z) -> +(x,+(y,z))
  +(x,y) -> +(y,x)
  
   T' = (P union S) with
  
   TRS P:+(+(x,y),z) -> +(x,+(y,z))
         +(x,y) -> +(y,x)
  
   TRS S:+(0(),x) -> x
         +(x,0()) -> x
  
  S is linear and P is reversible.
  
   CP(S,S) = 
  0() = 0()
  
   CP(S,P union P^-1) = 
  +(y,z) = +(0(),+(y,z)), y = +(y,0()), +(y,z) = +(+(0(),y),z), +(x,z) = 
  +(+(x,0()),z), x = +(x,0()), +(x,y) = +(x,+(y,0())), +(x,z) = +(x,+(0(),z)), 
  x = +(0(),x), +(x,y) = +(+(x,y),0()), y = +(0(),y)
  
   CP(P union P^-1,S) = 
  +(x188,+(x189,0())) = +(x188,x189), +(x,0()) = x, +(0(),x) = x, +(+(0(),x196),x197) = 
  +(x196,x197)
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [+](x0, x1) = x0 + x1 + 5,
    
    [0] = 0
   orientation:
    +(0(),x) = x + 5 >= x = x
    
    +(x,0()) = x + 5 >= x = x
   problem:
    
   Qed