/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.

  active : [o] --> o 
  f : [o] --> o 
  g : [o] --> o 
  h : [o] --> o 
  mark : [o] --> o 
  ok : [o] --> o 
  proper : [o] --> o 
  top : [o] --> o 

  active(f(X)) => mark(g(h(f(X)))) 
  active(f(X)) => f(active(X)) 
  active(h(X)) => h(active(X)) 
  f(mark(X)) => mark(f(X)) 
  h(mark(X)) => mark(h(X)) 
  proper(f(X)) => f(proper(X)) 
  proper(g(X)) => g(proper(X)) 
  proper(h(X)) => h(proper(X)) 
  f(ok(X)) => ok(f(X)) 
  g(ok(X)) => ok(g(X)) 
  h(ok(X)) => ok(h(X)) 
  top(mark(X)) => top(proper(X)) 
  top(ok(X)) => top(active(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]):

  active(f(X)) >? mark(g(h(f(X)))) 
  active(f(X)) >? f(active(X)) 
  active(h(X)) >? h(active(X)) 
  f(mark(X)) >? mark(f(X)) 
  h(mark(X)) >? mark(h(X)) 
  proper(f(X)) >? f(proper(X)) 
  proper(g(X)) >? g(proper(X)) 
  proper(h(X)) >? h(proper(X)) 
  f(ok(X)) >? ok(f(X)) 
  g(ok(X)) >? ok(g(X)) 
  h(ok(X)) >? ok(h(X)) 
  top(mark(X)) >? top(proper(X)) 
  top(ok(X)) >? top(active(X)) 

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

The following interpretation satisfies the requirements:

  active = \y0.2 + 3y0 
  f = \y0.2 + 3y0 
  g = \y0.y0 
  h = \y0.y0 
  mark = \y0.1 + 2y0 
  ok = \y0.2 + 3y0 
  proper = \y0.y0 
  top = \y0.y0 

Using this interpretation, the requirements translate to:

  [[active(f(_x0))]] = 8 + 9x0 > 5 + 6x0 = [[mark(g(h(f(_x0))))]] 
  [[active(f(_x0))]] = 8 + 9x0 >= 8 + 9x0 = [[f(active(_x0))]] 
  [[active(h(_x0))]] = 2 + 3x0 >= 2 + 3x0 = [[h(active(_x0))]] 
  [[f(mark(_x0))]] = 5 + 6x0 >= 5 + 6x0 = [[mark(f(_x0))]] 
  [[h(mark(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[mark(h(_x0))]] 
  [[proper(f(_x0))]] = 2 + 3x0 >= 2 + 3x0 = [[f(proper(_x0))]] 
  [[proper(g(_x0))]] = x0 >= x0 = [[g(proper(_x0))]] 
  [[proper(h(_x0))]] = x0 >= x0 = [[h(proper(_x0))]] 
  [[f(ok(_x0))]] = 8 + 9x0 >= 8 + 9x0 = [[ok(f(_x0))]] 
  [[g(ok(_x0))]] = 2 + 3x0 >= 2 + 3x0 = [[ok(g(_x0))]] 
  [[h(ok(_x0))]] = 2 + 3x0 >= 2 + 3x0 = [[ok(h(_x0))]] 
  [[top(mark(_x0))]] = 1 + 2x0 > x0 = [[top(proper(_x0))]] 
  [[top(ok(_x0))]] = 2 + 3x0 >= 2 + 3x0 = [[top(active(_x0))]] 

We can thus remove the following rules:

  active(f(X)) => mark(g(h(f(X)))) 
  top(mark(X)) => top(proper(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]):

  active(f(X)) >? f(active(X)) 
  active(h(X)) >? h(active(X)) 
  f(mark(X)) >? mark(f(X)) 
  h(mark(X)) >? mark(h(X)) 
  proper(f(X)) >? f(proper(X)) 
  proper(g(X)) >? g(proper(X)) 
  proper(h(X)) >? h(proper(X)) 
  f(ok(X)) >? ok(f(X)) 
  g(ok(X)) >? ok(g(X)) 
  h(ok(X)) >? ok(h(X)) 
  top(ok(X)) >? top(active(X)) 

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

The following interpretation satisfies the requirements:

  active = \y0.y0 
  f = \y0.3y0 
  g = \y0.1 + 2y0 
  h = \y0.3y0 
  mark = \y0.1 + y0 
  ok = \y0.3 + 3y0 
  proper = \y0.3y0 
  top = \y0.3y0 

Using this interpretation, the requirements translate to:

  [[active(f(_x0))]] = 3x0 >= 3x0 = [[f(active(_x0))]] 
  [[active(h(_x0))]] = 3x0 >= 3x0 = [[h(active(_x0))]] 
  [[f(mark(_x0))]] = 3 + 3x0 > 1 + 3x0 = [[mark(f(_x0))]] 
  [[h(mark(_x0))]] = 3 + 3x0 > 1 + 3x0 = [[mark(h(_x0))]] 
  [[proper(f(_x0))]] = 9x0 >= 9x0 = [[f(proper(_x0))]] 
  [[proper(g(_x0))]] = 3 + 6x0 > 1 + 6x0 = [[g(proper(_x0))]] 
  [[proper(h(_x0))]] = 9x0 >= 9x0 = [[h(proper(_x0))]] 
  [[f(ok(_x0))]] = 9 + 9x0 > 3 + 9x0 = [[ok(f(_x0))]] 
  [[g(ok(_x0))]] = 7 + 6x0 > 6 + 6x0 = [[ok(g(_x0))]] 
  [[h(ok(_x0))]] = 9 + 9x0 > 3 + 9x0 = [[ok(h(_x0))]] 
  [[top(ok(_x0))]] = 9 + 9x0 > 3x0 = [[top(active(_x0))]] 

We can thus remove the following rules:

  f(mark(X)) => mark(f(X)) 
  h(mark(X)) => mark(h(X)) 
  proper(g(X)) => g(proper(X)) 
  f(ok(X)) => ok(f(X)) 
  g(ok(X)) => ok(g(X)) 
  h(ok(X)) => ok(h(X)) 
  top(ok(X)) => top(active(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]):

  active(f(X)) >? f(active(X)) 
  active(h(X)) >? h(active(X)) 
  proper(f(X)) >? f(proper(X)) 
  proper(h(X)) >? h(proper(X)) 

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

The following interpretation satisfies the requirements:

  active = \y0.3y0 
  f = \y0.2 + y0 
  h = \y0.y0 
  proper = \y0.3y0 

Using this interpretation, the requirements translate to:

  [[active(f(_x0))]] = 6 + 3x0 > 2 + 3x0 = [[f(active(_x0))]] 
  [[active(h(_x0))]] = 3x0 >= 3x0 = [[h(active(_x0))]] 
  [[proper(f(_x0))]] = 6 + 3x0 > 2 + 3x0 = [[f(proper(_x0))]] 
  [[proper(h(_x0))]] = 3x0 >= 3x0 = [[h(proper(_x0))]] 

We can thus remove the following rules:

  active(f(X)) => f(active(X)) 
  proper(f(X)) => f(proper(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]):

  active(h(X)) >? h(active(X)) 
  proper(h(X)) >? h(proper(X)) 

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

The following interpretation satisfies the requirements:

  active = \y0.3y0 
  h = \y0.2 + y0 
  proper = \y0.3y0 

Using this interpretation, the requirements translate to:

  [[active(h(_x0))]] = 6 + 3x0 > 2 + 3x0 = [[h(active(_x0))]] 
  [[proper(h(_x0))]] = 6 + 3x0 > 2 + 3x0 = [[h(proper(_x0))]] 

We can thus remove the following rules:

  active(h(X)) => h(active(X)) 
  proper(h(X)) => h(proper(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.