/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.

  0 : [] --> o 
  half : [o] --> o 
  log : [o] --> o 
  s : [o] --> o 

  half(0) => 0 
  half(s(s(X))) => s(half(X)) 
  log(s(0)) => 0 
  log(s(s(X))) => s(log(s(half(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]):

  half(0) >? 0 
  half(s(s(X))) >? s(half(X)) 
  log(s(0)) >? 0 
  log(s(s(X))) >? s(log(s(half(X)))) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  half = \y0.y0 
  log = \y0.3y0 
  s = \y0.3 + 2y0 

Using this interpretation, the requirements translate to:

  [[half(0)]] = 0 >= 0 = [[0]] 
  [[half(s(s(_x0)))]] = 9 + 4x0 > 3 + 2x0 = [[s(half(_x0))]] 
  [[log(s(0))]] = 9 > 0 = [[0]] 
  [[log(s(s(_x0)))]] = 27 + 12x0 > 21 + 12x0 = [[s(log(s(half(_x0))))]] 

We can thus remove the following rules:

  half(s(s(X))) => s(half(X)) 
  log(s(0)) => 0 
  log(s(s(X))) => s(log(s(half(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]):

  half(0) >? 0 

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

The following interpretation satisfies the requirements:

  0 = 0 
  half = \y0.3 + 3y0 

Using this interpretation, the requirements translate to:

  [[half(0)]] = 3 > 0 = [[0]] 

We can thus remove the following rules:

  half(0) => 0 

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.