/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 
  f : [o] --> o 
  false : [] --> o 
  if : [o * o * o] --> o 
  n!6220!6220f : [o] --> o 
  n!6220!6220true : [] --> o 
  true : [] --> o 

  f(X) => if(X, c, n!6220!6220f(n!6220!6220true)) 
  if(true, X, Y) => X 
  if(false, X, Y) => activate(Y) 
  f(X) => n!6220!6220f(X) 
  true => n!6220!6220true 
  activate(n!6220!6220f(X)) => f(activate(X)) 
  activate(n!6220!6220true) => true 
  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(X) >? if(X, c, n!6220!6220f(n!6220!6220true)) 
  if(true, X, Y) >? X 
  if(false, X, Y) >? activate(Y) 
  f(X) >? n!6220!6220f(X) 
  true >? n!6220!6220true 
  activate(n!6220!6220f(X)) >? f(activate(X)) 
  activate(n!6220!6220true) >? true 
  activate(X) >? X 

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

The following interpretation satisfies the requirements:

  activate = \y0.2y0 
  c = 0 
  f = \y0.y0 
  false = 3 
  if = \y0y1y2.y0 + y1 + 2y2 
  n!6220!6220f = \y0.y0 
  n!6220!6220true = 0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[f(_x0)]] = x0 >= x0 = [[if(_x0, c, n!6220!6220f(n!6220!6220true))]] 
  [[if(true, _x0, _x1)]] = x0 + 2x1 >= x0 = [[_x0]] 
  [[if(false, _x0, _x1)]] = 3 + x0 + 2x1 > 2x1 = [[activate(_x1)]] 
  [[f(_x0)]] = x0 >= x0 = [[n!6220!6220f(_x0)]] 
  [[true]] = 0 >= 0 = [[n!6220!6220true]] 
  [[activate(n!6220!6220f(_x0))]] = 2x0 >= 2x0 = [[f(activate(_x0))]] 
  [[activate(n!6220!6220true)]] = 0 >= 0 = [[true]] 
  [[activate(_x0)]] = 2x0 >= x0 = [[_x0]] 

We can thus remove the following rules:

  if(false, X, Y) => activate(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) >? if(X, c, n!6220!6220f(n!6220!6220true)) 
  if(true, X, Y) >? X 
  f(X) >? n!6220!6220f(X) 
  true >? n!6220!6220true 
  activate(n!6220!6220f(X)) >? f(activate(X)) 
  activate(n!6220!6220true) >? true 
  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 = 0 
  f = \y0.y0 
  if = \y0y1y2.y0 + y1 + y2 
  n!6220!6220f = \y0.y0 
  n!6220!6220true = 0 
  true = 1 

Using this interpretation, the requirements translate to:

  [[f(_x0)]] = x0 >= x0 = [[if(_x0, c, n!6220!6220f(n!6220!6220true))]] 
  [[if(true, _x0, _x1)]] = 1 + x0 + x1 > x0 = [[_x0]] 
  [[f(_x0)]] = x0 >= x0 = [[n!6220!6220f(_x0)]] 
  [[true]] = 1 > 0 = [[n!6220!6220true]] 
  [[activate(n!6220!6220f(_x0))]] = 1 + x0 >= 1 + x0 = [[f(activate(_x0))]] 
  [[activate(n!6220!6220true)]] = 1 >= 1 = [[true]] 
  [[activate(_x0)]] = 1 + x0 > x0 = [[_x0]] 

We can thus remove the following rules:

  if(true, X, Y) => X 
  true => n!6220!6220true 
  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(X) >? if(X, c, n!6220!6220f(n!6220!6220true)) 
  f(X) >? n!6220!6220f(X) 
  activate(n!6220!6220f(X)) >? f(activate(X)) 
  activate(n!6220!6220true) >? true 

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

The following interpretation satisfies the requirements:

  activate = \y0.1 + 3y0 
  c = 0 
  f = \y0.2 + y0 
  if = \y0y1y2.y0 + y1 + y2 
  n!6220!6220f = \y0.1 + y0 
  n!6220!6220true = 0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[f(_x0)]] = 2 + x0 > 1 + x0 = [[if(_x0, c, n!6220!6220f(n!6220!6220true))]] 
  [[f(_x0)]] = 2 + x0 > 1 + x0 = [[n!6220!6220f(_x0)]] 
  [[activate(n!6220!6220f(_x0))]] = 4 + 3x0 > 3 + 3x0 = [[f(activate(_x0))]] 
  [[activate(n!6220!6220true)]] = 1 > 0 = [[true]] 

We can thus remove the following rules:

  f(X) => if(X, c, n!6220!6220f(n!6220!6220true)) 
  f(X) => n!6220!6220f(X) 
  activate(n!6220!6220f(X)) => f(activate(X)) 
  activate(n!6220!6220true) => true 

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.