/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.

  0 : [] --> o 
  f : [o] --> o 
  g : [o] --> o 
  s : [o] --> o 

  g(s(X)) => f(X) 
  f(0) => s(0) 
  f(s(X)) => s(s(g(X))) 
  g(0) => 0 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  g(s(X)) >? f(X) 
  f(0) >? s(0) 
  f(s(X)) >? s(s(g(X))) 
  g(0) >? 0 

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

The following interpretation satisfies the requirements:

  0 = 0 
  f = \y0.1 + y0 
  g = \y0.1 + y0 
  s = \y0.y0 

Using this interpretation, the requirements translate to:

  [[g(s(_x0))]] = 1 + x0 >= 1 + x0 = [[f(_x0)]] 
  [[f(0)]] = 1 > 0 = [[s(0)]] 
  [[f(s(_x0))]] = 1 + x0 >= 1 + x0 = [[s(s(g(_x0)))]] 
  [[g(0)]] = 1 > 0 = [[0]] 

We can thus remove the following rules:

  f(0) => s(0) 
  g(0) => 0 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  g(s(X)) >? f(X) 
  f(s(X)) >? s(s(g(X))) 

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

The following interpretation satisfies the requirements:

  f = \y0.2y0 
  g = \y0.2y0 
  s = \y0.1 + y0 

Using this interpretation, the requirements translate to:

  [[g(s(_x0))]] = 2 + 2x0 > 2x0 = [[f(_x0)]] 
  [[f(s(_x0))]] = 2 + 2x0 >= 2 + 2x0 = [[s(s(g(_x0)))]] 

We can thus remove the following rules:

  g(s(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]):

  f(s(X)) >? s(s(g(X))) 

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

The following interpretation satisfies the requirements:

  f = \y0.3 + 3y0 
  g = \y0.y0 
  s = \y0.y0 

Using this interpretation, the requirements translate to:

  [[f(s(_x0))]] = 3 + 3x0 > x0 = [[s(s(g(_x0)))]] 

We can thus remove the following rules:

  f(s(X)) => s(s(g(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.