/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/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 
  a!6220!6220f : [o] --> o 
  c : [o] --> o 
  f : [o] --> o 
  g : [o] --> o 
  mark : [o] --> o 

  a!6220!6220f(f(a)) => c(f(g(f(a)))) 
  mark(f(X)) => a!6220!6220f(mark(X)) 
  mark(a) => a 
  mark(c(X)) => c(X) 
  mark(g(X)) => g(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(f(a)) >? c(f(g(f(a)))) 
  mark(f(X)) >? a!6220!6220f(mark(X)) 
  mark(a) >? a 
  mark(c(X)) >? c(X) 
  mark(g(X)) >? g(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 = 0 
  a!6220!6220f = \y0.1 + y0 
  c = \y0.y0 
  f = \y0.1 + y0 
  g = \y0.y0 
  mark = \y0.2 + 2y0 

Using this interpretation, the requirements translate to:

  [[a!6220!6220f(f(a))]] = 2 >= 2 = [[c(f(g(f(a))))]] 
  [[mark(f(_x0))]] = 4 + 2x0 > 3 + 2x0 = [[a!6220!6220f(mark(_x0))]] 
  [[mark(a)]] = 2 > 0 = [[a]] 
  [[mark(c(_x0))]] = 2 + 2x0 > x0 = [[c(_x0)]] 
  [[mark(g(_x0))]] = 2 + 2x0 >= 2 + 2x0 = [[g(mark(_x0))]] 
  [[a!6220!6220f(_x0)]] = 1 + x0 >= 1 + x0 = [[f(_x0)]] 

We can thus remove the following rules:

  mark(f(X)) => a!6220!6220f(mark(X)) 
  mark(a) => a 
  mark(c(X)) => c(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(f(a)) >? c(f(g(f(a)))) 
  mark(g(X)) >? g(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 = 3 
  a!6220!6220f = \y0.3 + 3y0 
  c = \y0.y0 
  f = \y0.1 + 3y0 
  g = \y0.y0 
  mark = \y0.3y0 

Using this interpretation, the requirements translate to:

  [[a!6220!6220f(f(a))]] = 33 > 31 = [[c(f(g(f(a))))]] 
  [[mark(g(_x0))]] = 3x0 >= 3x0 = [[g(mark(_x0))]] 
  [[a!6220!6220f(_x0)]] = 3 + 3x0 > 1 + 3x0 = [[f(_x0)]] 

We can thus remove the following rules:

  a!6220!6220f(f(a)) => c(f(g(f(a)))) 
  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(g(X)) >? g(mark(X)) 

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

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

We can thus remove the following rules:

  mark(g(X)) => g(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.