/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.

  2nd : [o] --> o 
  activate : [o] --> o 
  cons : [o * o] --> o 
  from : [o] --> o 
  n!6220!6220cons : [o * o] --> o 
  n!6220!6220from : [o] --> o 
  n!6220!6220s : [o] --> o 
  s : [o] --> o 

  2nd(cons(X, n!6220!6220cons(Y, Z))) => activate(Y) 
  from(X) => cons(X, n!6220!6220from(n!6220!6220s(X))) 
  cons(X, Y) => n!6220!6220cons(X, Y) 
  from(X) => n!6220!6220from(X) 
  s(X) => n!6220!6220s(X) 
  activate(n!6220!6220cons(X, Y)) => cons(activate(X), Y) 
  activate(n!6220!6220from(X)) => from(activate(X)) 
  activate(n!6220!6220s(X)) => s(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]):

  2nd(cons(X, n!6220!6220cons(Y, Z))) >? activate(Y) 
  from(X) >? cons(X, n!6220!6220from(n!6220!6220s(X))) 
  cons(X, Y) >? n!6220!6220cons(X, Y) 
  from(X) >? n!6220!6220from(X) 
  s(X) >? n!6220!6220s(X) 
  activate(n!6220!6220cons(X, Y)) >? cons(activate(X), Y) 
  activate(n!6220!6220from(X)) >? from(activate(X)) 
  activate(n!6220!6220s(X)) >? s(activate(X)) 
  activate(X) >? X 

about to try horpo

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

We choose Lex = {} and Mul = {2nd, activate, cons, from, n!6220!6220cons, n!6220!6220from, n!6220!6220s, s}, and the following precedence: 2nd > activate > from > cons = n!6220!6220cons > n!6220!6220from > s > n!6220!6220s

With these choices, we have:

  1] 2nd(cons(X, n!6220!6220cons(Y, Z))) >= activate(Y)  because [2], by (Star) 
  2] 2nd*(cons(X, n!6220!6220cons(Y, Z))) >= activate(Y)  because 2nd > activate and [3], by (Copy) 
  3] 2nd*(cons(X, n!6220!6220cons(Y, Z))) >= Y  because [4], by (Select) 
  4] cons(X, n!6220!6220cons(Y, Z)) >= Y  because [5], by (Star) 
  5] cons*(X, n!6220!6220cons(Y, Z)) >= Y  because [6], by (Select) 
  6] n!6220!6220cons(Y, Z) >= Y  because [7], by (Star) 
  7] n!6220!6220cons*(Y, Z) >= Y  because [8], by (Select) 
  8] Y >= Y  by (Meta) 

  9] from(X) > cons(X, n!6220!6220from(n!6220!6220s(X)))  because [10], by definition 
  10] from*(X) >= cons(X, n!6220!6220from(n!6220!6220s(X)))  because from > cons, [11] and [13], by (Copy) 
  11] from*(X) >= X  because [12], by (Select) 
  12] X >= X  by (Meta) 
  13] from*(X) >= n!6220!6220from(n!6220!6220s(X))  because from > n!6220!6220from and [14], by (Copy) 
  14] from*(X) >= n!6220!6220s(X)  because from > n!6220!6220s and [11], by (Copy) 

  15] cons(X, Y) >= n!6220!6220cons(X, Y)  because cons = n!6220!6220cons, cons in Mul, [16] and [17], by (Fun) 
  16] X >= X  by (Meta) 
  17] Y >= Y  by (Meta) 

  18] from(X) >= n!6220!6220from(X)  because [19], by (Star) 
  19] from*(X) >= n!6220!6220from(X)  because from > n!6220!6220from and [11], by (Copy) 

  20] s(X) > n!6220!6220s(X)  because [21], by definition 
  21] s*(X) >= n!6220!6220s(X)  because s > n!6220!6220s and [22], by (Copy) 
  22] s*(X) >= X  because [12], by (Select) 

  23] activate(n!6220!6220cons(X, Y)) >= cons(activate(X), Y)  because [24], by (Star) 
  24] activate*(n!6220!6220cons(X, Y)) >= cons(activate(X), Y)  because activate > cons, [25] and [28], by (Copy) 
  25] activate*(n!6220!6220cons(X, Y)) >= activate(X)  because activate in Mul and [26], by (Stat) 
  26] n!6220!6220cons(X, Y) > X  because [27], by definition 
  27] n!6220!6220cons*(X, Y) >= X  because [16], by (Select) 
  28] activate*(n!6220!6220cons(X, Y)) >= Y  because [29], by (Select) 
  29] n!6220!6220cons(X, Y) >= Y  because [30], by (Star) 
  30] n!6220!6220cons*(X, Y) >= Y  because [17], by (Select) 

  31] activate(n!6220!6220from(X)) > from(activate(X))  because [32], by definition 
  32] activate*(n!6220!6220from(X)) >= from(activate(X))  because activate > from and [33], by (Copy) 
  33] activate*(n!6220!6220from(X)) >= activate(X)  because activate in Mul and [34], by (Stat) 
  34] n!6220!6220from(X) > X  because [35], by definition 
  35] n!6220!6220from*(X) >= X  because [12], by (Select) 

  36] activate(n!6220!6220s(X)) > s(activate(X))  because [37], by definition 
  37] activate*(n!6220!6220s(X)) >= s(activate(X))  because activate > s and [38], by (Copy) 
  38] activate*(n!6220!6220s(X)) >= activate(X)  because activate in Mul and [39], by (Stat) 
  39] n!6220!6220s(X) > X  because [40], by definition 
  40] n!6220!6220s*(X) >= X  because [12], by (Select) 

  41] activate(X) >= X  because [42], by (Star) 
  42] activate*(X) >= X  because [12], by (Select) 

We can thus remove the following rules:

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

  2nd(cons(X, n!6220!6220cons(Y, Z))) >? activate(Y) 
  cons(X, Y) >? n!6220!6220cons(X, Y) 
  from(X) >? n!6220!6220from(X) 
  activate(n!6220!6220cons(X, Y)) >? cons(activate(X), Y) 
  activate(X) >? X 

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

The following interpretation satisfies the requirements:

  2nd = \y0.3 + 3y0 
  activate = \y0.y0 
  cons = \y0y1.y0 + y1 
  from = \y0.3 + 3y0 
  n!6220!6220cons = \y0y1.y0 + y1 
  n!6220!6220from = \y0.y0 

Using this interpretation, the requirements translate to:

  [[2nd(cons(_x0, n!6220!6220cons(_x1, _x2)))]] = 3 + 3x0 + 3x1 + 3x2 > x1 = [[activate(_x1)]] 
  [[cons(_x0, _x1)]] = x0 + x1 >= x0 + x1 = [[n!6220!6220cons(_x0, _x1)]] 
  [[from(_x0)]] = 3 + 3x0 > x0 = [[n!6220!6220from(_x0)]] 
  [[activate(n!6220!6220cons(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[cons(activate(_x0), _x1)]] 
  [[activate(_x0)]] = x0 >= x0 = [[_x0]] 

We can thus remove the following rules:

  2nd(cons(X, n!6220!6220cons(Y, Z))) => activate(Y) 
  from(X) => n!6220!6220from(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]):

  cons(X, Y) >? n!6220!6220cons(X, Y) 
  activate(n!6220!6220cons(X, Y)) >? cons(activate(X), Y) 
  activate(X) >? X 

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

The following interpretation satisfies the requirements:

  activate = \y0.2y0 
  cons = \y0y1.2 + 2y0 + 2y1 
  n!6220!6220cons = \y0y1.1 + y1 + 2y0 

Using this interpretation, the requirements translate to:

  [[cons(_x0, _x1)]] = 2 + 2x0 + 2x1 > 1 + x1 + 2x0 = [[n!6220!6220cons(_x0, _x1)]] 
  [[activate(n!6220!6220cons(_x0, _x1))]] = 2 + 2x1 + 4x0 >= 2 + 2x1 + 4x0 = [[cons(activate(_x0), _x1)]] 
  [[activate(_x0)]] = 2x0 >= x0 = [[_x0]] 

We can thus remove the following rules:

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

  activate(n!6220!6220cons(X, Y)) >? cons(activate(X), Y) 
  activate(X) >? X 

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

The following interpretation satisfies the requirements:

  activate = \y0.3y0 
  cons = \y0y1.y0 + y1 
  n!6220!6220cons = \y0y1.3 + 3y0 + 3y1 

Using this interpretation, the requirements translate to:

  [[activate(n!6220!6220cons(_x0, _x1))]] = 9 + 9x0 + 9x1 > x1 + 3x0 = [[cons(activate(_x0), _x1)]] 
  [[activate(_x0)]] = 3x0 >= x0 = [[_x0]] 

We can thus remove the following rules:

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

  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.