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


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YES
We consider the system theBenchmark.

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

  f : [o * o] --> o 
  g : [o * o] --> o 
  h : [o] --> o 

  f(X, Y) => g(X, Y) 
  g(h(X), Y) => h(f(X, Y)) 
  g(h(X), Y) => h(g(X, Y)) 

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

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

  f(X, Y) >? g(X, Y) 
  g(h(X), Y) >? h(f(X, Y)) 
  g(h(X), Y) >? h(g(X, Y)) 

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

The following interpretation satisfies the requirements:

  f = \y0y1.1 + y1 + 2y0 
  g = \y0y1.y1 + 2y0 
  h = \y0.1 + y0 

Using this interpretation, the requirements translate to:

  [[f(_x0, _x1)]] = 1 + x1 + 2x0 > x1 + 2x0 = [[g(_x0, _x1)]] 
  [[g(h(_x0), _x1)]] = 2 + x1 + 2x0 >= 2 + x1 + 2x0 = [[h(f(_x0, _x1))]] 
  [[g(h(_x0), _x1)]] = 2 + x1 + 2x0 > 1 + x1 + 2x0 = [[h(g(_x0, _x1))]] 

We can thus remove the following rules:

  f(X, Y) => g(X, Y) 
  g(h(X), Y) => h(g(X, Y)) 

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(h(X), Y) >? h(f(X, Y)) 

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

The following interpretation satisfies the requirements:

  f = \y0y1.y0 + y1 
  g = \y0y1.3 + 3y0 + 3y1 
  h = \y0.y0 

Using this interpretation, the requirements translate to:

  [[g(h(_x0), _x1)]] = 3 + 3x0 + 3x1 > x0 + x1 = [[h(f(_x0, _x1))]] 

We can thus remove the following rules:

  g(h(X), Y) => h(f(X, Y)) 

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.