/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


YES
We consider the system theBenchmark.

We are asked to determine termination of the following first-order TRS.

  a!6220!6220f : [o] --> o 
  f : [o] --> o 
  g : [o] --> o 
  h : [o] --> o 
  mark : [o] --> o 

  a!6220!6220f(X) => g(h(f(X))) 
  mark(f(X)) => a!6220!6220f(mark(X)) 
  mark(g(X)) => g(X) 
  mark(h(X)) => h(mark(X)) 
  a!6220!6220f(X) => f(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]):

  a!6220!6220f(X) >? g(h(f(X))) 
  mark(f(X)) >? a!6220!6220f(mark(X)) 
  mark(g(X)) >? g(X) 
  mark(h(X)) >? h(mark(X)) 
  a!6220!6220f(X) >? f(X) 

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

The following interpretation satisfies the requirements:

  a!6220!6220f = \y0.y0 
  f = \y0.y0 
  g = \y0.y0 
  h = \y0.y0 
  mark = \y0.1 + y0 

Using this interpretation, the requirements translate to:

  [[a!6220!6220f(_x0)]] = x0 >= x0 = [[g(h(f(_x0)))]] 
  [[mark(f(_x0))]] = 1 + x0 >= 1 + x0 = [[a!6220!6220f(mark(_x0))]] 
  [[mark(g(_x0))]] = 1 + x0 > x0 = [[g(_x0)]] 
  [[mark(h(_x0))]] = 1 + x0 >= 1 + x0 = [[h(mark(_x0))]] 
  [[a!6220!6220f(_x0)]] = x0 >= x0 = [[f(_x0)]] 

We can thus remove the following rules:

  mark(g(X)) => g(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]):

  a!6220!6220f(X) >? g(h(f(X))) 
  mark(f(X)) >? a!6220!6220f(mark(X)) 
  mark(h(X)) >? h(mark(X)) 
  a!6220!6220f(X) >? f(X) 

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

The following interpretation satisfies the requirements:

  a!6220!6220f = \y0.3 + y0 
  f = \y0.2 + y0 
  g = \y0.y0 
  h = \y0.y0 
  mark = \y0.2 + 3y0 

Using this interpretation, the requirements translate to:

  [[a!6220!6220f(_x0)]] = 3 + x0 > 2 + x0 = [[g(h(f(_x0)))]] 
  [[mark(f(_x0))]] = 8 + 3x0 > 5 + 3x0 = [[a!6220!6220f(mark(_x0))]] 
  [[mark(h(_x0))]] = 2 + 3x0 >= 2 + 3x0 = [[h(mark(_x0))]] 
  [[a!6220!6220f(_x0)]] = 3 + x0 > 2 + x0 = [[f(_x0)]] 

We can thus remove the following rules:

  a!6220!6220f(X) => g(h(f(X))) 
  mark(f(X)) => a!6220!6220f(mark(X)) 
  a!6220!6220f(X) => f(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]):

  mark(h(X)) >? h(mark(X)) 

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

The following interpretation satisfies the requirements:

  h = \y0.1 + y0 
  mark = \y0.3y0 

Using this interpretation, the requirements translate to:

  [[mark(h(_x0))]] = 3 + 3x0 > 1 + 3x0 = [[h(mark(_x0))]] 

We can thus remove the following rules:

  mark(h(X)) => h(mark(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.