/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/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 
  activate : [o] --> o 
  f : [o] --> o 
  g : [o] --> o 
  n!6220!6220a : [] --> o 
  n!6220!6220f : [o] --> o 

  f(f(a)) => f(g(n!6220!6220f(n!6220!6220a))) 
  f(X) => n!6220!6220f(X) 
  a => n!6220!6220a 
  activate(n!6220!6220f(X)) => f(activate(X)) 
  activate(n!6220!6220a) => a 
  activate(X) => 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(f(a)) >? f(g(n!6220!6220f(n!6220!6220a))) 
  f(X) >? n!6220!6220f(X) 
  a >? n!6220!6220a 
  activate(n!6220!6220f(X)) >? f(activate(X)) 
  activate(n!6220!6220a) >? a 
  activate(X) >? X 

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

The following interpretation satisfies the requirements:

  a = 0 
  activate = \y0.2y0 
  f = \y0.3 + y0 
  g = \y0.y0 
  n!6220!6220a = 0 
  n!6220!6220f = \y0.2 + y0 

Using this interpretation, the requirements translate to:

  [[f(f(a))]] = 6 > 5 = [[f(g(n!6220!6220f(n!6220!6220a)))]] 
  [[f(_x0)]] = 3 + x0 > 2 + x0 = [[n!6220!6220f(_x0)]] 
  [[a]] = 0 >= 0 = [[n!6220!6220a]] 
  [[activate(n!6220!6220f(_x0))]] = 4 + 2x0 > 3 + 2x0 = [[f(activate(_x0))]] 
  [[activate(n!6220!6220a)]] = 0 >= 0 = [[a]] 
  [[activate(_x0)]] = 2x0 >= x0 = [[_x0]] 

We can thus remove the following rules:

  f(f(a)) => f(g(n!6220!6220f(n!6220!6220a))) 
  f(X) => n!6220!6220f(X) 
  activate(n!6220!6220f(X)) => f(activate(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]):

  a >? n!6220!6220a 
  activate(n!6220!6220a) >? a 
  activate(X) >? X 

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

The following interpretation satisfies the requirements:

  a = 3 
  activate = \y0.3 + 3y0 
  n!6220!6220a = 1 

Using this interpretation, the requirements translate to:

  [[a]] = 3 > 1 = [[n!6220!6220a]] 
  [[activate(n!6220!6220a)]] = 6 > 3 = [[a]] 
  [[activate(_x0)]] = 3 + 3x0 > x0 = [[_x0]] 

We can thus remove the following rules:

  a => n!6220!6220a 
  activate(n!6220!6220a) => a 
  activate(X) => 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.