/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.

  a : [] --> o 
  b : [] --> o 
  f : [o * o] --> o 
  g : [o * o] --> o 

  f(a, X) => g(a, X) 
  g(a, X) => f(b, X) 
  f(a, X) => f(b, X) 

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(a, X) >? g(a, X) 
  g(a, X) >? f(b, X) 
  f(a, X) >? f(b, X) 

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

  [[f(a, _x0)]] = 8 + x0 > 2 + x0 = [[g(a, _x0)]] 
  [[g(a, _x0)]] = 2 + x0 >= 2 + x0 = [[f(b, _x0)]] 
  [[f(a, _x0)]] = 8 + x0 > 2 + x0 = [[f(b, _x0)]] 

We can thus remove the following rules:

  f(a, X) => g(a, X) 
  f(a, X) => f(b, X) 

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(a, X) >? f(b, X) 

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

The following interpretation satisfies the requirements:

  a = 3 
  b = 0 
  f = \y0y1.y0 + y1 
  g = \y0y1.3 + 3y0 + 3y1 

Using this interpretation, the requirements translate to:

  [[g(a, _x0)]] = 12 + 3x0 > x0 = [[f(b, _x0)]] 

We can thus remove the following rules:

  g(a, X) => f(b, X) 

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.