/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.

  a : [o] --> o 
  b : [o] --> o 

  a(b(a(X))) => b(a(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]):

  a(b(a(X))) >? b(a(X)) 

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

The following interpretation satisfies the requirements:

  a = \y0.3y0 
  b = \y0.3 + y0 

Using this interpretation, the requirements translate to:

  [[a(b(a(_x0)))]] = 9 + 9x0 > 3 + 3x0 = [[b(a(_x0))]] 

We can thus remove the following rules:

  a(b(a(X))) => b(a(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.