/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


--------------------------------------------------------------------------------


YES
We consider the system theBenchmark.

We are asked to determine termination of the following first-order TRS.

  app : [o * o] --> o 
  compose : [] --> o 

  app(app(app(compose, X), Y), Z) => app(X, app(Y, Z)) 

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

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

  app(app(app(compose, X), Y), Z) >? app(X, app(Y, Z)) 

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

The following interpretation satisfies the requirements:

  app = \y0y1.y1 + 2y0 
  compose = 3 

Using this interpretation, the requirements translate to:

  [[app(app(app(compose, _x0), _x1), _x2)]] = 24 + x2 + 2x1 + 4x0 > x2 + 2x0 + 2x1 = [[app(_x0, app(_x1, _x2))]] 

We can thus remove the following rules:

  app(app(app(compose, X), Y), Z) => app(X, app(Y, Z)) 

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.