/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/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 
  from : [o] --> o 
  n!6220!6220from : [o] --> o 
  s : [o] --> o 
  sel : [o * o] --> o 

  from(X) => cons(X, n!6220!6220from(s(X))) 
  sel(0, cons(X, Y)) => X 
  sel(s(X), cons(Y, Z)) => sel(X, activate(Z)) 
  from(X) => n!6220!6220from(X) 
  activate(n!6220!6220from(X)) => from(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]):

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

about to try horpo

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

We choose Lex = {sel} and Mul = {0, activate, cons, from, n!6220!6220from, s}, and the following precedence: 0 > sel > activate > from > cons > n!6220!6220from > s

With these choices, we have:

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

  7] sel(0, cons(X, Y)) >= X  because [8], by (Star) 
  8] sel*(0, cons(X, Y)) >= X  because [9], by (Select) 
  9] cons(X, Y) >= X  because [10], by (Star) 
  10] cons*(X, Y) >= X  because [4], by (Select) 

  11] sel(s(X), cons(Y, Z)) >= sel(X, activate(Z))  because [12], by (Star) 
  12] sel*(s(X), cons(Y, Z)) >= sel(X, activate(Z))  because [13], [15] and [17], by (Stat) 
  13] s(X) > X  because [14], by definition 
  14] s*(X) >= X  because [4], by (Select) 
  15] sel*(s(X), cons(Y, Z)) >= X  because [16], by (Select) 
  16] s(X) >= X  because [14], by (Star) 
  17] sel*(s(X), cons(Y, Z)) >= activate(Z)  because sel > activate and [18], by (Copy) 
  18] sel*(s(X), cons(Y, Z)) >= Z  because [19], by (Select) 
  19] cons(Y, Z) >= Z  because [20], by (Star) 
  20] cons*(Y, Z) >= Z  because [21], by (Select) 
  21] Z >= Z  by (Meta) 

  22] from(X) >= n!6220!6220from(X)  because [23], by (Star) 
  23] from*(X) >= n!6220!6220from(X)  because from > n!6220!6220from and [3], by (Copy) 

  24] activate(n!6220!6220from(X)) >= from(X)  because [25], by (Star) 
  25] activate*(n!6220!6220from(X)) >= from(X)  because activate > from and [26], by (Copy) 
  26] activate*(n!6220!6220from(X)) >= X  because [27], by (Select) 
  27] n!6220!6220from(X) >= X  because [28], by (Star) 
  28] n!6220!6220from*(X) >= X  because [4], by (Select) 

  29] activate(X) > X  because [30], by definition 
  30] activate*(X) >= X  because [4], by (Select) 

We can thus remove the following rules:

  from(X) => cons(X, n!6220!6220from(s(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]):

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

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

  [[sel(0, cons(_x0, _x1))]] = 21 + 3x0 + 9x1 > x0 = [[_x0]] 
  [[sel(s(_x0), cons(_x1, _x2))]] = 21 + 3x1 + 9x0 + 9x2 > 3 + 3x0 + 6x2 = [[sel(_x0, activate(_x2))]] 
  [[from(_x0)]] = 1 + x0 >= 1 + x0 = [[n!6220!6220from(_x0)]] 
  [[activate(n!6220!6220from(_x0))]] = 2 + 2x0 > 1 + x0 = [[from(_x0)]] 

We can thus remove the following rules:

  sel(0, cons(X, Y)) => X 
  sel(s(X), cons(Y, Z)) => sel(X, activate(Z)) 
  activate(n!6220!6220from(X)) => from(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]):

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

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

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

We can thus remove the following rules:

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