/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 
  after : [o * o] --> o 
  cons : [o * o] --> o 
  from : [o] --> o 
  n!6220!6220from : [o] --> o 
  s : [o] --> o 

  from(X) => cons(X, n!6220!6220from(s(X))) 
  after(0, X) => X 
  after(s(X), cons(Y, Z)) => after(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))) 
  after(0, X) >? X 
  after(s(X), cons(Y, Z)) >? after(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 = {after} and Mul = {0, activate, cons, from, n!6220!6220from, s}, and the following precedence: 0 > after > 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 (Star) 
  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] after(0, X) >= X  because [8], by (Star) 
  8] after*(0, X) >= X  because [9], by (Select) 
  9] X >= X  by (Meta) 

  10] after(s(X), cons(Y, Z)) >= after(X, activate(Z))  because [11], by (Star) 
  11] after*(s(X), cons(Y, Z)) >= after(X, activate(Z))  because [12], [15] and [17], by (Stat) 
  12] s(X) > X  because [13], by definition 
  13] s*(X) >= X  because [14], by (Select) 
  14] X >= X  by (Meta) 
  15] after*(s(X), cons(Y, Z)) >= X  because [16], by (Select) 
  16] s(X) >= X  because [13], by (Star) 
  17] after*(s(X), cons(Y, Z)) >= activate(Z)  because after > activate and [18], by (Copy) 
  18] after*(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 [9], by (Select) 

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

  23] activate(n!6220!6220from(X)) > from(X)  because [24], by definition 
  24] activate*(n!6220!6220from(X)) >= from(X)  because activate = from, activate in Mul and [25], by (Stat) 
  25] n!6220!6220from(X) > X  because [26], by definition 
  26] n!6220!6220from*(X) >= X  because [4], by (Select) 

  27] activate(X) >= X  because [28], by (Star) 
  28] activate*(X) >= X  because [4], by (Select) 

We can thus remove the following rules:

  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) >? cons(X, n!6220!6220from(s(X))) 
  after(0, X) >? X 
  after(s(X), cons(Y, Z)) >? after(X, activate(Z)) 
  from(X) >? n!6220!6220from(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 
  after = \y0y1.3 + y0 + y1 
  cons = \y0y1.y0 + 2y1 
  from = \y0.3 + 3y0 
  n!6220!6220from = \y0.y0 
  s = \y0.y0 

Using this interpretation, the requirements translate to:

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

We can thus remove the following rules:

  from(X) => cons(X, n!6220!6220from(s(X))) 
  after(0, X) => X 
  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]):

  after(s(X), cons(Y, Z)) >? after(X, activate(Z)) 
  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 
  after = \y0y1.2y1 + 3y0 
  cons = \y0y1.3 + y0 + 3y1 
  s = \y0.3 + 3y0 

Using this interpretation, the requirements translate to:

  [[after(s(_x0), cons(_x1, _x2))]] = 15 + 2x1 + 6x2 + 9x0 > 2 + 2x2 + 3x0 = [[after(_x0, activate(_x2))]] 
  [[activate(_x0)]] = 1 + x0 > x0 = [[_x0]] 

We can thus remove the following rules:

  after(s(X), cons(Y, Z)) => after(X, activate(Z)) 
  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.