Problem:
 f(x,x) -> plus(plus(x,x),x)
 plus(x,y) -> plus(y,x)

Proof:
 Church Rosser Transformation Processor (to relative problem):
  strict:
   f(x,x) -> plus(plus(x,x),x)
   plus(x,y) -> plus(y,x)
  weak:
   
  original problem:
   f(x,x) -> plus(plus(x,x),x)
   plus(x,y) -> plus(y,x)
  critical peaks: 
   Polynomial Interpretation Processor:
    dimension: 1
    interpretation:
     [plus](x0, x1) = x0 + x1,
     
     [f](x0, x1) = 2x0 + x1 + 2x1x0 + 1
    orientation:
     f(x,x) = 3x + 2x*x + 1 >= 3x = plus(plus(x,x),x)
     
     plus(x,y) = x + y >= x + y = plus(y,x)
    problem:
     strict:
      plus(x,y) -> plus(y,x)
     weak:
      
     original problem:
      f(x,x) -> plus(plus(x,x),x)
      plus(x,y) -> plus(y,x)
    KH confluence processor
     Split input TRS into two TRSs S and T:
     
     TRS S:
      plus(x,y) -> plus(y,x)
     
     TRS T:
      f(x,x) -> plus(plus(x,x),x)
     
     As established above, T/S is terminating.
     T is strongly non-overlapping on S and S is strongly non-overlapping on T
     
     Please install theorem prover 'Prover9' and 'Mace4' for handling more TRSs.
     
      All S-critical pairs are joinable.
     
     We have to check confluence of S.
     
     Church Rosser Transformation Processor (no redundant rules):
      strict:
       plus(x,y) -> plus(y,x)
      weak:
       
      critical peaks: 0
      Closedness Processor (*feeble*):
       
       Qed