/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


YES
We consider the system theBenchmark.

We are asked to determine termination of the following first-order TRS.

  0 : [] --> o 
  f : [o * o] --> o 
  g : [o * o] --> o 
  s : [o] --> o 

  g(0, f(X, X)) => X 
  g(X, s(Y)) => g(f(X, Y), 0) 
  g(s(X), Y) => g(f(X, Y), 0) 
  g(f(X, Y), 0) => f(g(X, 0), g(Y, 0)) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  g(0, f(X, X)) >? X 
  g(X, s(Y)) >? g(f(X, Y), 0) 
  g(s(X), Y) >? g(f(X, Y), 0) 
  g(f(X, Y), 0) >? f(g(X, 0), g(Y, 0)) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  0 = 0 
  f = \y0y1.2 + y0 + y1 
  g = \y0y1.3y0 + 3y1 
  s = \y0.3 + 3y0 

Using this interpretation, the requirements translate to:

  [[g(0, f(_x0, _x0))]] = 6 + 6x0 > x0 = [[_x0]] 
  [[g(_x0, s(_x1))]] = 9 + 3x0 + 9x1 > 6 + 3x0 + 3x1 = [[g(f(_x0, _x1), 0)]] 
  [[g(s(_x0), _x1)]] = 9 + 3x1 + 9x0 > 6 + 3x0 + 3x1 = [[g(f(_x0, _x1), 0)]] 
  [[g(f(_x0, _x1), 0)]] = 6 + 3x0 + 3x1 > 2 + 3x0 + 3x1 = [[f(g(_x0, 0), g(_x1, 0))]] 

We can thus remove the following rules:

  g(0, f(X, X)) => X 
  g(X, s(Y)) => g(f(X, Y), 0) 
  g(s(X), Y) => g(f(X, Y), 0) 
  g(f(X, Y), 0) => f(g(X, 0), g(Y, 0)) 

All rules were succesfully removed.  Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.