/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.

  activate : [o] --> o 
  c : [o] --> o 
  d : [o] --> o 
  f : [o] --> o 
  g : [o] --> o 
  h : [o] --> o 
  n!6220!6220d : [o] --> o 
  n!6220!6220f : [o] --> o 

  f(f(X)) => c(n!6220!6220f(g(n!6220!6220f(X)))) 
  c(X) => d(activate(X)) 
  h(X) => c(n!6220!6220d(X)) 
  f(X) => n!6220!6220f(X) 
  d(X) => n!6220!6220d(X) 
  activate(n!6220!6220f(X)) => f(X) 
  activate(n!6220!6220d(X)) => d(X) 
  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(X)) >? c(n!6220!6220f(g(n!6220!6220f(X)))) 
  c(X) >? d(activate(X)) 
  h(X) >? c(n!6220!6220d(X)) 
  f(X) >? n!6220!6220f(X) 
  d(X) >? n!6220!6220d(X) 
  activate(n!6220!6220f(X)) >? f(X) 
  activate(n!6220!6220d(X)) >? d(X) 
  activate(X) >? X 

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

The following interpretation satisfies the requirements:

  activate = \y0.1 + y0 
  c = \y0.1 + y0 
  d = \y0.y0 
  f = \y0.1 + y0 
  g = \y0.y0 
  h = \y0.3 + 3y0 
  n!6220!6220d = \y0.y0 
  n!6220!6220f = \y0.y0 

Using this interpretation, the requirements translate to:

  [[f(f(_x0))]] = 2 + x0 > 1 + x0 = [[c(n!6220!6220f(g(n!6220!6220f(_x0))))]] 
  [[c(_x0)]] = 1 + x0 >= 1 + x0 = [[d(activate(_x0))]] 
  [[h(_x0)]] = 3 + 3x0 > 1 + x0 = [[c(n!6220!6220d(_x0))]] 
  [[f(_x0)]] = 1 + x0 > x0 = [[n!6220!6220f(_x0)]] 
  [[d(_x0)]] = x0 >= x0 = [[n!6220!6220d(_x0)]] 
  [[activate(n!6220!6220f(_x0))]] = 1 + x0 >= 1 + x0 = [[f(_x0)]] 
  [[activate(n!6220!6220d(_x0))]] = 1 + x0 > x0 = [[d(_x0)]] 
  [[activate(_x0)]] = 1 + x0 > x0 = [[_x0]] 

We can thus remove the following rules:

  f(f(X)) => c(n!6220!6220f(g(n!6220!6220f(X)))) 
  h(X) => c(n!6220!6220d(X)) 
  f(X) => n!6220!6220f(X) 
  activate(n!6220!6220d(X)) => d(X) 
  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]):

  c(X) >? d(activate(X)) 
  d(X) >? n!6220!6220d(X) 
  activate(n!6220!6220f(X)) >? f(X) 

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

The following interpretation satisfies the requirements:

  activate = \y0.y0 
  c = \y0.3 + 3y0 
  d = \y0.y0 
  f = \y0.y0 
  n!6220!6220d = \y0.y0 
  n!6220!6220f = \y0.3 + y0 

Using this interpretation, the requirements translate to:

  [[c(_x0)]] = 3 + 3x0 > x0 = [[d(activate(_x0))]] 
  [[d(_x0)]] = x0 >= x0 = [[n!6220!6220d(_x0)]] 
  [[activate(n!6220!6220f(_x0))]] = 3 + x0 > x0 = [[f(_x0)]] 

We can thus remove the following rules:

  c(X) => d(activate(X)) 
  activate(n!6220!6220f(X)) => f(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]):

  d(X) >? n!6220!6220d(X) 

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

The following interpretation satisfies the requirements:

  d = \y0.3 + 3y0 
  n!6220!6220d = \y0.y0 

Using this interpretation, the requirements translate to:

  [[d(_x0)]] = 3 + 3x0 > x0 = [[n!6220!6220d(_x0)]] 

We can thus remove the following rules:

  d(X) => n!6220!6220d(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.