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


--------------------------------------------------------------------------------


YES
We consider the system theBenchmark.

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

  a!6220!6220f : [o] --> o 
  a!6220!6220if : [o * o * o] --> o 
  c : [] --> o 
  f : [o] --> o 
  false : [] --> o 
  if : [o * o * o] --> o 
  mark : [o] --> o 
  true : [] --> o 

  a!6220!6220f(X) => a!6220!6220if(mark(X), c, f(true)) 
  a!6220!6220if(true, X, Y) => mark(X) 
  a!6220!6220if(false, X, Y) => mark(Y) 
  mark(f(X)) => a!6220!6220f(mark(X)) 
  mark(if(X, Y, Z)) => a!6220!6220if(mark(X), mark(Y), Z) 
  mark(c) => c 
  mark(true) => true 
  mark(false) => false 
  a!6220!6220f(X) => f(X) 
  a!6220!6220if(X, Y, Z) => if(X, Y, Z) 

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!6220!6220f(X) >? a!6220!6220if(mark(X), c, f(true)) 
  a!6220!6220if(true, X, Y) >? mark(X) 
  a!6220!6220if(false, X, Y) >? mark(Y) 
  mark(f(X)) >? a!6220!6220f(mark(X)) 
  mark(if(X, Y, Z)) >? a!6220!6220if(mark(X), mark(Y), Z) 
  mark(c) >? c 
  mark(true) >? true 
  mark(false) >? false 
  a!6220!6220f(X) >? f(X) 
  a!6220!6220if(X, Y, Z) >? if(X, Y, Z) 

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

The following interpretation satisfies the requirements:

  a!6220!6220f = \y0.2 + 2y0 
  a!6220!6220if = \y0y1y2.y0 + 2y1 + 2y2 
  c = 0 
  f = \y0.1 + 2y0 
  false = 0 
  if = \y0y1y2.y0 + 2y1 + 2y2 
  mark = \y0.2y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[a!6220!6220f(_x0)]] = 2 + 2x0 >= 2 + 2x0 = [[a!6220!6220if(mark(_x0), c, f(true))]] 
  [[a!6220!6220if(true, _x0, _x1)]] = 2x0 + 2x1 >= 2x0 = [[mark(_x0)]] 
  [[a!6220!6220if(false, _x0, _x1)]] = 2x0 + 2x1 >= 2x1 = [[mark(_x1)]] 
  [[mark(f(_x0))]] = 2 + 4x0 >= 2 + 4x0 = [[a!6220!6220f(mark(_x0))]] 
  [[mark(if(_x0, _x1, _x2))]] = 2x0 + 4x1 + 4x2 >= 2x0 + 2x2 + 4x1 = [[a!6220!6220if(mark(_x0), mark(_x1), _x2)]] 
  [[mark(c)]] = 0 >= 0 = [[c]] 
  [[mark(true)]] = 0 >= 0 = [[true]] 
  [[mark(false)]] = 0 >= 0 = [[false]] 
  [[a!6220!6220f(_x0)]] = 2 + 2x0 > 1 + 2x0 = [[f(_x0)]] 
  [[a!6220!6220if(_x0, _x1, _x2)]] = x0 + 2x1 + 2x2 >= x0 + 2x1 + 2x2 = [[if(_x0, _x1, _x2)]] 

We can thus remove the following rules:

  a!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]):

  a!6220!6220f(X) >? a!6220!6220if(mark(X), c, f(true)) 
  a!6220!6220if(true, X, Y) >? mark(X) 
  a!6220!6220if(false, X, Y) >? mark(Y) 
  mark(f(X)) >? a!6220!6220f(mark(X)) 
  mark(if(X, Y, Z)) >? a!6220!6220if(mark(X), mark(Y), Z) 
  mark(c) >? c 
  mark(true) >? true 
  mark(false) >? false 
  a!6220!6220if(X, Y, Z) >? if(X, Y, Z) 

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

The following interpretation satisfies the requirements:

  a!6220!6220f = \y0.2y0 
  a!6220!6220if = \y0y1y2.y0 + 2y1 + 2y2 
  c = 0 
  f = \y0.2y0 
  false = 1 
  if = \y0y1y2.y0 + y2 + 2y1 
  mark = \y0.2y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[a!6220!6220f(_x0)]] = 2x0 >= 2x0 = [[a!6220!6220if(mark(_x0), c, f(true))]] 
  [[a!6220!6220if(true, _x0, _x1)]] = 2x0 + 2x1 >= 2x0 = [[mark(_x0)]] 
  [[a!6220!6220if(false, _x0, _x1)]] = 1 + 2x0 + 2x1 > 2x1 = [[mark(_x1)]] 
  [[mark(f(_x0))]] = 4x0 >= 4x0 = [[a!6220!6220f(mark(_x0))]] 
  [[mark(if(_x0, _x1, _x2))]] = 2x0 + 2x2 + 4x1 >= 2x0 + 2x2 + 4x1 = [[a!6220!6220if(mark(_x0), mark(_x1), _x2)]] 
  [[mark(c)]] = 0 >= 0 = [[c]] 
  [[mark(true)]] = 0 >= 0 = [[true]] 
  [[mark(false)]] = 2 > 1 = [[false]] 
  [[a!6220!6220if(_x0, _x1, _x2)]] = x0 + 2x1 + 2x2 >= x0 + x2 + 2x1 = [[if(_x0, _x1, _x2)]] 

We can thus remove the following rules:

  a!6220!6220if(false, X, Y) => mark(Y) 
  mark(false) => false 

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!6220!6220f(X) >? a!6220!6220if(mark(X), c, f(true)) 
  a!6220!6220if(true, X, Y) >? mark(X) 
  mark(f(X)) >? a!6220!6220f(mark(X)) 
  mark(if(X, Y, Z)) >? a!6220!6220if(mark(X), mark(Y), Z) 
  mark(c) >? c 
  mark(true) >? true 
  a!6220!6220if(X, Y, Z) >? if(X, Y, Z) 

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

The following interpretation satisfies the requirements:

  a!6220!6220f = \y0.1 + 2y0 
  a!6220!6220if = \y0y1y2.y0 + y2 + 2y1 
  c = 0 
  f = \y0.1 + 2y0 
  if = \y0y1y2.y0 + y2 + 2y1 
  mark = \y0.2y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[a!6220!6220f(_x0)]] = 1 + 2x0 >= 1 + 2x0 = [[a!6220!6220if(mark(_x0), c, f(true))]] 
  [[a!6220!6220if(true, _x0, _x1)]] = x1 + 2x0 >= 2x0 = [[mark(_x0)]] 
  [[mark(f(_x0))]] = 2 + 4x0 > 1 + 4x0 = [[a!6220!6220f(mark(_x0))]] 
  [[mark(if(_x0, _x1, _x2))]] = 2x0 + 2x2 + 4x1 >= x2 + 2x0 + 4x1 = [[a!6220!6220if(mark(_x0), mark(_x1), _x2)]] 
  [[mark(c)]] = 0 >= 0 = [[c]] 
  [[mark(true)]] = 0 >= 0 = [[true]] 
  [[a!6220!6220if(_x0, _x1, _x2)]] = x0 + x2 + 2x1 >= x0 + x2 + 2x1 = [[if(_x0, _x1, _x2)]] 

We can thus remove the following rules:

  mark(f(X)) => a!6220!6220f(mark(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!6220!6220f(X) >? a!6220!6220if(mark(X), c, f(true)) 
  a!6220!6220if(true, X, Y) >? mark(X) 
  mark(if(X, Y, Z)) >? a!6220!6220if(mark(X), mark(Y), Z) 
  mark(c) >? c 
  mark(true) >? true 
  a!6220!6220if(X, Y, Z) >? if(X, Y, Z) 

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

The following interpretation satisfies the requirements:

  a!6220!6220f = \y0.3 + 3y0 
  a!6220!6220if = \y0y1y2.y0 + y1 + y2 
  c = 0 
  f = \y0.y0 
  if = \y0y1y2.y0 + y1 + y2 
  mark = \y0.y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[a!6220!6220f(_x0)]] = 3 + 3x0 > x0 = [[a!6220!6220if(mark(_x0), c, f(true))]] 
  [[a!6220!6220if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[mark(_x0)]] 
  [[mark(if(_x0, _x1, _x2))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[a!6220!6220if(mark(_x0), mark(_x1), _x2)]] 
  [[mark(c)]] = 0 >= 0 = [[c]] 
  [[mark(true)]] = 0 >= 0 = [[true]] 
  [[a!6220!6220if(_x0, _x1, _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[if(_x0, _x1, _x2)]] 

We can thus remove the following rules:

  a!6220!6220f(X) => a!6220!6220if(mark(X), c, f(true)) 

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!6220!6220if(true, X, Y) >? mark(X) 
  mark(if(X, Y, Z)) >? a!6220!6220if(mark(X), mark(Y), Z) 
  mark(c) >? c 
  mark(true) >? true 
  a!6220!6220if(X, Y, Z) >? if(X, Y, Z) 

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

The following interpretation satisfies the requirements:

  a!6220!6220if = \y0y1y2.1 + y2 + 2y0 + 2y1 
  c = 0 
  if = \y0y1y2.1 + y2 + 2y0 + 2y1 
  mark = \y0.2y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[a!6220!6220if(true, _x0, _x1)]] = 1 + x1 + 2x0 > 2x0 = [[mark(_x0)]] 
  [[mark(if(_x0, _x1, _x2))]] = 2 + 2x2 + 4x0 + 4x1 > 1 + x2 + 4x0 + 4x1 = [[a!6220!6220if(mark(_x0), mark(_x1), _x2)]] 
  [[mark(c)]] = 0 >= 0 = [[c]] 
  [[mark(true)]] = 0 >= 0 = [[true]] 
  [[a!6220!6220if(_x0, _x1, _x2)]] = 1 + x2 + 2x0 + 2x1 >= 1 + x2 + 2x0 + 2x1 = [[if(_x0, _x1, _x2)]] 

We can thus remove the following rules:

  a!6220!6220if(true, X, Y) => mark(X) 
  mark(if(X, Y, Z)) => a!6220!6220if(mark(X), mark(Y), Z) 

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

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

  mark(c) >? c 
  mark(true) >? true 
  a!6220!6220if(X, Y, Z) >? if(X, Y, Z) 

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

The following interpretation satisfies the requirements:

  a!6220!6220if = \y0y1y2.3 + y0 + y1 + y2 
  c = 0 
  if = \y0y1y2.y0 + y1 + y2 
  mark = \y0.3 + 3y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[mark(c)]] = 3 > 0 = [[c]] 
  [[mark(true)]] = 3 > 0 = [[true]] 
  [[a!6220!6220if(_x0, _x1, _x2)]] = 3 + x0 + x1 + x2 > x0 + x1 + x2 = [[if(_x0, _x1, _x2)]] 

We can thus remove the following rules:

  mark(c) => c 
  mark(true) => true 
  a!6220!6220if(X, Y, Z) => if(X, Y, Z) 

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.