/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.

  f : [o * o] --> o 
  s : [o] --> o 

  f(s(X), Y) => f(X, s(X)) 
  f(X, s(Y)) => f(Y, 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), Y) >? f(X, s(X)) 
  f(X, s(Y)) >? f(Y, X) 

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

The following interpretation satisfies the requirements:

  f = \y0y1.y1 + 2y0 
  s = \y0.1 + 2y0 

Using this interpretation, the requirements translate to:

  [[f(s(_x0), _x1)]] = 2 + x1 + 4x0 > 1 + 4x0 = [[f(_x0, s(_x0))]] 
  [[f(_x0, s(_x1))]] = 1 + 2x0 + 2x1 > x0 + 2x1 = [[f(_x1, _x0)]] 

We can thus remove the following rules:

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