/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.

  0 : [] --> o 
  activate : [o] --> o 
  cons : [o * o] --> o 
  f : [o] --> o 
  g : [o] --> o 
  n!6220!6220f : [o] --> o 
  n!6220!6220g : [o] --> o 
  s : [o] --> o 
  sel : [o * o] --> o 

  f(X) => cons(X, n!6220!6220f(n!6220!6220g(X))) 
  g(0) => s(0) 
  g(s(X)) => s(s(g(X))) 
  sel(0, cons(X, Y)) => X 
  sel(s(X), cons(Y, Z)) => sel(X, activate(Z)) 
  f(X) => n!6220!6220f(X) 
  g(X) => n!6220!6220g(X) 
  activate(n!6220!6220f(X)) => f(activate(X)) 
  activate(n!6220!6220g(X)) => g(activate(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(X) >? cons(X, n!6220!6220f(n!6220!6220g(X))) 
  g(0) >? s(0) 
  g(s(X)) >? s(s(g(X))) 
  sel(0, cons(X, Y)) >? X 
  sel(s(X), cons(Y, Z)) >? sel(X, activate(Z)) 
  f(X) >? n!6220!6220f(X) 
  g(X) >? n!6220!6220g(X) 
  activate(n!6220!6220f(X)) >? f(activate(X)) 
  activate(n!6220!6220g(X)) >? g(activate(X)) 
  activate(X) >? X 

about to try horpo

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[0]] = _|_ 

We choose Lex = {sel} and Mul = {activate, cons, f, g, n!6220!6220f, n!6220!6220g, s}, and the following precedence: sel > activate > f > cons > g > n!6220!6220f > n!6220!6220g > s

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  f(X) > cons(X, n!6220!6220f(n!6220!6220g(X))) 
  g(_|_) >= s(_|_) 
  g(s(X)) >= s(s(g(X))) 
  sel(_|_, cons(X, Y)) >= X 
  sel(s(X), cons(Y, Z)) >= sel(X, activate(Z)) 
  f(X) >= n!6220!6220f(X) 
  g(X) >= n!6220!6220g(X) 
  activate(n!6220!6220f(X)) >= f(activate(X)) 
  activate(n!6220!6220g(X)) > g(activate(X)) 
  activate(X) >= X 

With these choices, we have:

  1] f(X) > cons(X, n!6220!6220f(n!6220!6220g(X)))  because [2], by definition 
  2] f*(X) >= cons(X, n!6220!6220f(n!6220!6220g(X)))  because f > cons, [3] and [5], by (Copy) 
  3] f*(X) >= X  because [4], by (Select) 
  4] X >= X  by (Meta) 
  5] f*(X) >= n!6220!6220f(n!6220!6220g(X))  because f > n!6220!6220f and [6], by (Copy) 
  6] f*(X) >= n!6220!6220g(X)  because f > n!6220!6220g and [3], by (Copy) 

  7] g(_|_) >= s(_|_)  because [8], by (Star) 
  8] g*(_|_) >= s(_|_)  because g > s and [9], by (Copy) 
  9] g*(_|_) >= _|_  by (Bot) 

  10] g(s(X)) >= s(s(g(X)))  because [11], by (Star) 
  11] g*(s(X)) >= s(s(g(X)))  because g > s and [12], by (Copy) 
  12] g*(s(X)) >= s(g(X))  because g > s and [13], by (Copy) 
  13] g*(s(X)) >= g(X)  because g in Mul and [14], by (Stat) 
  14] s(X) > X  because [15], by definition 
  15] s*(X) >= X  because [4], by (Select) 

  16] sel(_|_, cons(X, Y)) >= X  because [17], by (Star) 
  17] sel*(_|_, cons(X, Y)) >= X  because [18], by (Select) 
  18] cons(X, Y) >= X  because [19], by (Star) 
  19] cons*(X, Y) >= X  because [4], by (Select) 

  20] sel(s(X), cons(Y, Z)) >= sel(X, activate(Z))  because [21], by (Star) 
  21] sel*(s(X), cons(Y, Z)) >= sel(X, activate(Z))  because [14], [22] and [24], by (Stat) 
  22] sel*(s(X), cons(Y, Z)) >= X  because [23], by (Select) 
  23] s(X) >= X  because [15], by (Star) 
  24] sel*(s(X), cons(Y, Z)) >= activate(Z)  because sel > activate and [25], by (Copy) 
  25] sel*(s(X), cons(Y, Z)) >= Z  because [26], by (Select) 
  26] cons(Y, Z) >= Z  because [27], by (Star) 
  27] cons*(Y, Z) >= Z  because [28], by (Select) 
  28] Z >= Z  by (Meta) 

  29] f(X) >= n!6220!6220f(X)  because [30], by (Star) 
  30] f*(X) >= n!6220!6220f(X)  because f > n!6220!6220f and [3], by (Copy) 

  31] g(X) >= n!6220!6220g(X)  because [32], by (Star) 
  32] g*(X) >= n!6220!6220g(X)  because g > n!6220!6220g and [33], by (Copy) 
  33] g*(X) >= X  because [4], by (Select) 

  34] activate(n!6220!6220f(X)) >= f(activate(X))  because [35], by (Star) 
  35] activate*(n!6220!6220f(X)) >= f(activate(X))  because activate > f and [36], by (Copy) 
  36] activate*(n!6220!6220f(X)) >= activate(X)  because activate in Mul and [37], by (Stat) 
  37] n!6220!6220f(X) > X  because [38], by definition 
  38] n!6220!6220f*(X) >= X  because [4], by (Select) 

  39] activate(n!6220!6220g(X)) > g(activate(X))  because [40], by definition 
  40] activate*(n!6220!6220g(X)) >= g(activate(X))  because activate > g and [41], by (Copy) 
  41] activate*(n!6220!6220g(X)) >= activate(X)  because activate in Mul and [42], by (Stat) 
  42] n!6220!6220g(X) > X  because [43], by definition 
  43] n!6220!6220g*(X) >= X  because [4], by (Select) 

  44] activate(X) >= X  because [45], by (Star) 
  45] activate*(X) >= X  because [4], by (Select) 

We can thus remove the following rules:

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

  g(0) >? s(0) 
  g(s(X)) >? s(s(g(X))) 
  sel(0, cons(X, Y)) >? X 
  sel(s(X), cons(Y, Z)) >? sel(X, activate(Z)) 
  f(X) >? n!6220!6220f(X) 
  g(X) >? n!6220!6220g(X) 
  activate(n!6220!6220f(X)) >? f(activate(X)) 
  activate(X) >? X 

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

The following interpretation satisfies the requirements:

  0 = 3 
  activate = \y0.y0 
  cons = \y0y1.3 + y0 + 3y1 
  f = \y0.2y0 
  g = \y0.3y0 
  n!6220!6220f = \y0.2y0 
  n!6220!6220g = \y0.y0 
  s = \y0.1 + y0 
  sel = \y0y1.3 + y1 + 2y0 

Using this interpretation, the requirements translate to:

  [[g(0)]] = 9 > 4 = [[s(0)]] 
  [[g(s(_x0))]] = 3 + 3x0 > 2 + 3x0 = [[s(s(g(_x0)))]] 
  [[sel(0, cons(_x0, _x1))]] = 12 + x0 + 3x1 > x0 = [[_x0]] 
  [[sel(s(_x0), cons(_x1, _x2))]] = 8 + x1 + 2x0 + 3x2 > 3 + x2 + 2x0 = [[sel(_x0, activate(_x2))]] 
  [[f(_x0)]] = 2x0 >= 2x0 = [[n!6220!6220f(_x0)]] 
  [[g(_x0)]] = 3x0 >= x0 = [[n!6220!6220g(_x0)]] 
  [[activate(n!6220!6220f(_x0))]] = 2x0 >= 2x0 = [[f(activate(_x0))]] 
  [[activate(_x0)]] = x0 >= x0 = [[_x0]] 

We can thus remove the following rules:

  g(0) => s(0) 
  g(s(X)) => s(s(g(X))) 
  sel(0, cons(X, Y)) => X 
  sel(s(X), cons(Y, Z)) => sel(X, activate(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]):

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

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

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

We can thus remove the following rules:

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

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

The following interpretation satisfies the requirements:

  activate = \y0.2y0 
  f = \y0.2 + 2y0 
  n!6220!6220f = \y0.1 + 2y0 

Using this interpretation, the requirements translate to:

  [[f(_x0)]] = 2 + 2x0 > 1 + 2x0 = [[n!6220!6220f(_x0)]] 
  [[activate(n!6220!6220f(_x0))]] = 2 + 4x0 >= 2 + 4x0 = [[f(activate(_x0))]] 
  [[activate(_x0)]] = 2x0 >= x0 = [[_x0]] 

We can thus remove the following rules:

  f(X) => n!6220!6220f(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]):

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

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

  [[activate(n!6220!6220f(_x0))]] = 9 + 9x0 > 3x0 = [[f(activate(_x0))]] 
  [[activate(_x0)]] = 3x0 >= x0 = [[_x0]] 

We can thus remove the following rules:

  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]):

  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 

Using this interpretation, the requirements translate to:

  [[activate(_x0)]] = 1 + x0 > x0 = [[_x0]] 

We can thus remove the following rules:

  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.