/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 
  s : [o] --> o 

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

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

  [[f(0)]] = 1 >= 1 = [[s(0)]] 
  [[f(s(0))]] = 2 > 1 = [[s(0)]] 
  [[f(s(s(_x0)))]] = 3 + x0 >= 3 + x0 = [[f(f(s(_x0)))]] 

We can thus remove the following rules:

  f(s(0)) => s(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]):

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

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

The following interpretation satisfies the requirements:

  0 = 0 
  f = \y0.2 + 2y0 
  s = \y0.1 + 3y0 

Using this interpretation, the requirements translate to:

  [[f(0)]] = 2 > 1 = [[s(0)]] 
  [[f(s(s(_x0)))]] = 10 + 18x0 >= 10 + 12x0 = [[f(f(s(_x0)))]] 

We can thus remove the following rules:

  f(0) => s(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]):

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

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

The following interpretation satisfies the requirements:

  f = \y0.2y0 
  s = \y0.3 + 3y0 

Using this interpretation, the requirements translate to:

  [[f(s(s(_x0)))]] = 24 + 18x0 > 12 + 12x0 = [[f(f(s(_x0)))]] 

We can thus remove the following rules:

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