/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.

  a : [] --> o 
  f : [o * o * o * o * o] --> o 
  g1 : [o * o * o * o] --> o 
  g2 : [o * o * o * o] --> o 
  h : [o * o] --> o 

  f(X, Y, Z, Z, a) => g1(X, X, Y, Z) 
  f(X, Y, Z, a, a) => g1(Y, X, X, Z) 
  f(X, Y, a, a, Z) => g2(X, Y, Y, Z) 
  f(X, Y, a, Z, Z) => g2(Y, Y, X, Z) 
  g1(X, X, Y, a) => h(X, Y) 
  g1(X, Y, Y, a) => h(Y, X) 
  g2(X, Y, Y, a) => h(X, Y) 
  g2(X, X, Y, a) => h(Y, X) 
  h(X, 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, Y, Z, Z, a) >? g1(X, X, Y, Z) 
  f(X, Y, Z, a, a) >? g1(Y, X, X, Z) 
  f(X, Y, a, a, Z) >? g2(X, Y, Y, Z) 
  f(X, Y, a, Z, Z) >? g2(Y, Y, X, Z) 
  g1(X, X, Y, a) >? h(X, Y) 
  g1(X, Y, Y, a) >? h(Y, X) 
  g2(X, Y, Y, a) >? h(X, Y) 
  g2(X, X, Y, a) >? h(Y, X) 
  h(X, X) >? X 

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

The following interpretation satisfies the requirements:

  a = 3 
  f = \y0y1y2y3y4.3 + 3y0 + 3y1 + 3y2 + 3y3 + 3y4 
  g1 = \y0y1y2y3.y0 + y1 + y2 + 3y3 
  g2 = \y0y1y2y3.y0 + y1 + y2 + 3y3 
  h = \y0y1.y0 + y1 

Using this interpretation, the requirements translate to:

  [[f(_x0, _x1, _x2, _x2, a)]] = 12 + 3x0 + 3x1 + 6x2 > x1 + 2x0 + 3x2 = [[g1(_x0, _x0, _x1, _x2)]] 
  [[f(_x0, _x1, _x2, a, a)]] = 21 + 3x0 + 3x1 + 3x2 > x1 + 2x0 + 3x2 = [[g1(_x1, _x0, _x0, _x2)]] 
  [[f(_x0, _x1, a, a, _x2)]] = 21 + 3x0 + 3x1 + 3x2 > x0 + 2x1 + 3x2 = [[g2(_x0, _x1, _x1, _x2)]] 
  [[f(_x0, _x1, a, _x2, _x2)]] = 12 + 3x0 + 3x1 + 6x2 > x0 + 2x1 + 3x2 = [[g2(_x1, _x1, _x0, _x2)]] 
  [[g1(_x0, _x0, _x1, a)]] = 9 + x1 + 2x0 > x0 + x1 = [[h(_x0, _x1)]] 
  [[g1(_x0, _x1, _x1, a)]] = 9 + x0 + 2x1 > x0 + x1 = [[h(_x1, _x0)]] 
  [[g2(_x0, _x1, _x1, a)]] = 9 + x0 + 2x1 > x0 + x1 = [[h(_x0, _x1)]] 
  [[g2(_x0, _x0, _x1, a)]] = 9 + x1 + 2x0 > x0 + x1 = [[h(_x1, _x0)]] 
  [[h(_x0, _x0)]] = 2x0 >= x0 = [[_x0]] 

We can thus remove the following rules:

  f(X, Y, Z, Z, a) => g1(X, X, Y, Z) 
  f(X, Y, Z, a, a) => g1(Y, X, X, Z) 
  f(X, Y, a, a, Z) => g2(X, Y, Y, Z) 
  f(X, Y, a, Z, Z) => g2(Y, Y, X, Z) 
  g1(X, X, Y, a) => h(X, Y) 
  g1(X, Y, Y, a) => h(Y, X) 
  g2(X, Y, Y, a) => h(X, Y) 
  g2(X, X, Y, a) => h(Y, 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]):

  h(X, X) >? X 

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

The following interpretation satisfies the requirements:

  h = \y0y1.1 + y0 + y1 

Using this interpretation, the requirements translate to:

  [[h(_x0, _x0)]] = 1 + 2x0 > x0 = [[_x0]] 

We can thus remove the following rules:

  h(X, 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.