/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 
  average : [o * o] --> o 
  s : [o] --> o 

  average(s(X), Y) => average(X, s(Y)) 
  average(X, s(s(s(Y)))) => s(average(s(X), Y)) 
  average(0, 0) => 0 
  average(0, s(0)) => 0 
  average(0, s(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]):

  average(s(X), Y) >? average(X, s(Y)) 
  average(X, s(s(s(Y)))) >? s(average(s(X), Y)) 
  average(0, 0) >? 0 
  average(0, s(0)) >? 0 
  average(0, s(s(0))) >? s(0) 

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

The following interpretation satisfies the requirements:

  0 = 1 
  average = \y0y1.2y1 + 3y0 
  s = \y0.3 + y0 

Using this interpretation, the requirements translate to:

  [[average(s(_x0), _x1)]] = 9 + 2x1 + 3x0 > 6 + 2x1 + 3x0 = [[average(_x0, s(_x1))]] 
  [[average(_x0, s(s(s(_x1))))]] = 18 + 2x1 + 3x0 > 12 + 2x1 + 3x0 = [[s(average(s(_x0), _x1))]] 
  [[average(0, 0)]] = 5 > 1 = [[0]] 
  [[average(0, s(0))]] = 11 > 1 = [[0]] 
  [[average(0, s(s(0)))]] = 17 > 4 = [[s(0)]] 

We can thus remove the following rules:

  average(s(X), Y) => average(X, s(Y)) 
  average(X, s(s(s(Y)))) => s(average(s(X), Y)) 
  average(0, 0) => 0 
  average(0, s(0)) => 0 
  average(0, s(s(0))) => s(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.